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YANPEI LIU 


INTRODUCTORY MAP THEORY 


Kapa & Omega, Glendale, AZ 
USA 


2010 





Yanpei LIU 


Institute of Mathematics 
Beijing Jiaotong University 
Beijing 100044, P.R.China 
Email: ypliu@bjtu.edu.cn 


Introductory Map Theory 


Kapa & Omega, Glendale, AZ 
USA 


2010 


This book can be ordered in a paper bound reprint from: 


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Peer Reviewers: 


L.F.Mao, Chinese Academy of Mathematics and System Science, P.R.China. 
J.L.Cai, Beijing Normal University, P.R.China. 

H.Ren, East China Normal University, P.R.China. 

R.X.Hao, Beijing Jiaotong University, P.R.China. 


Copyright 2010 by Kapa & Omega, Glendale, AZ and Yanpei Liu 


Many books can be downloaded from the following Digital Library of Science: 


http: //www.gallup.unm.edu/~smarandache/eBooks-otherformats.htm 


ISBN: 978-1-59973-134-6 


Printed in America 


Preface 


Maps as a mathematical topic arose probably from the four color 
problem|Bir1, Orel] and the more general map coloring problem|HiC1, 
Rinl, Liu11] in the mid of nineteenth century although maps as poly- 
hedra which go back to the Platonic age. I could not list references 
in detail on them because it is well known for a large range of readers 
and beyond the scope of this book. Here, I only intend to present 
a comprehensive theory of maps as a rigorous mathematical concept 
which has been developed mostly in the last half a century. 

However, as described in the book[Liu15| maps can be seen as 
graphs in development from partition to permutation and as a basis 
extended to Smarandache geometry shown in [Mao3-4]. This is why 
maps are much concerned with abstraction in the present stage. 

In the beginning, maps as polyhedra were as a topological, or ge- 
ometric object even with geographical consideration|Kem1]. The first 
formal definition of a map was done by Heffter from [Hef1] in the 19th 
century. However, it was not paid an attention by mathematicians 
until 1960 when Edmonds published a note in the AMS Notices with 
the dual form of Heffter’s in |Edm1,Liu3]. 

Although this concept was widely used in literature as [Liul—2, 
Liu4—6, Rin1-3, Stal-2, et all, its disadvantage for the nonorientable 
case involved does not bring with some convenience for clarifying some 
related mathematical thinking. 

Since Tutte described the nonorientability in a new way [Tut1- 
3], a number of authors begin to develop it in combinatorization of 
continuous objects as in [Lit1, Liu7-10, Vin1-2, et al]. 

The above representations are all with complication in construct- 
ing an embedding, or all distinct embeddings of a graph on a surface. 


lv Preface 


However, the joint tree model of an embedding completed in recent 
years and initiated from the early articles at the end of seventies in 
the last century by the present author as shown in [Liul—2] enables us 
to make the complication much simpler. 

Because of the generality that an asymmetric object can always 
be seen with some local symmetry in certain extent, the concepts of 
graphs and maps are just put in such a rule. In fact, the former is 
corresponding to that a group of two elements sticks on an edge and 
the later is that a group of four elements sticks on an edge such that 
a graph without symmetry at all is in company with local symmetry. 
This treatment will bring more advantages for observing the structure 
of a graph. Of course, the later is with restriction of the former because 
of the later as a permutation and the former as a partition. 

The joint tree representation of an embedding of a graph on 
two dimensional manifolds, particularly surfaces(compact 2-manifolds 
without boundary in our case), is described in Chapter I for simplifying 
a number of results old and new. 

This book contains the following chapters in company with re- 
lated subjects. 

In Chapter I, the embedding of a graph on surfaces are much 
concerned because they are motivated to building up the theory of 
abstract maps related with Smarandache geometry. 

The second chapter is for the formal definition of abstract maps. 
One can see that this matter is a natural generalization of graph em- 
bedding on surfaces. 

The third chapter is on the duality not only for maps themselves 
but also for operations on maps from one surface to another. One 
can see how the duality is naturally deduced from the abstract maps 
described in the second chapter. 

The fourth chapter is on the orientability. One can see how the 
orientability is formally designed as a combinatorial invariant. The 
fifth chapter concentrates on the classification of orientable maps. The 
sixth chapter is for the classification of nonorientable maps. 

From the two chapters: Chapter V and Chapter VI, one can see 


Preface v 


how the procedure is simplified for these classifications. 

The seventh chapter is on the isomorphisms of maps and pro- 
vides an efficient algorithm for the justification and recognition of the 
isomorphism of two maps, which has been shown to be useful for de- 
termining the automorphism group of a map in the eighth chapter. 
Moreover, it enables us to access an automorphism of a graph. 

The ninth and the tenth chapters observe the number of distinct 
asymmetric maps with the size as a parameter. In the former, only 
one vertex maps are counted by favorite formulas and in the latter, 
general maps are counted from differential equations. More progresses 
about this kind of counting are referred to read the recent book|Liu7] 
and many further articles|«Bax1, BeG1, CaL1-2, ReL1-3, etc]. 

The next chapter, Chapter XI, only presents some ideas for ac- 
cessing the symmetric census of maps and further, of graphs. This 
topic is being developed in some other directions|KwL1-2] and left as 
a subject written in the near future. 

From Chapter XII through Chapter XV, extensions from basic 
theory are much concerned with further applications. 

Chapter XII discusses in brief on genus polynomial of a graph 
and all its super maps rooted and unrooted on the basis of the joint 
tree model. Recent progresses on this aspect are referred to read the 
articles [Liu13-15, LiP1, WaL1-2, ZhL1-2, ZuL1, etc]. 

Chapter XIII is on the census of maps with vertex or face par- 
titions. Although such census involves with much complication and 
difficulty, because of the recent progress on a basic topic about trees 
via an elementary method firstly used by the author himself we are 
able to do a number of types of such census in very simple way. This 
chapter reflects on such aspects around. 

Chapter XIV is on graphs that their super maps are particularly 
considered for asymmetrical and symmetrical census via their semi- 
automorphism and automorphism groups or via embeddings of graphs 
given [Liu19, MaL1, MaT1, MaW1, etc]. 

Chapter XV, is on functional equations discovered in the census 
of a variety of maps on sphere and general surfaces. Although their 


vi Preface 


well definedness has been done, almost all of them have not yet been 
solved up to now. 


Three appendices are compliment to the context. One provides 
the clarification of the concepts of polyhedra, surfaces, embeddings, 
and maps and their relationship. The other two are for exhaustively 
calculating numerical results and listing all rooted and unrooted maps 
for small graphs with more calculating results compared with those 
appearing in [Liu14], |Liu17] and |Liu19]. 

From a large amount of materials, more than hundred observa- 
tions for beginners probably senior undergraduates, more than hun- 
dred exercises for mainly graduates of master degree and more than 
hundred research problems for mainly graduates of doctoral degree are 
carefully designed at the end of each chapter in adapting the needs of 
such a wide range of readers for mastering, extending and investigat- 
ing a number of ways to get further development on the basic theory 
of abstract maps. 


Although I have been trying to design this book self contained as 
much as possible, some books such as [DiM1], [Mss1] and [GaJ1] might 
be helpful to those not familiar with basic knowledge of permutation 
groups, topology and computing complexity as background. 


Since early nineties of the last century, a number of my former 
and present graduates were or are engaged with topics related to this 
book. Among them, I have to mention Dr. Ying Liu[LpL1], Dr. Yuan- 
qiu Huang[|HuL1], Dr. Junliang Cai[CaL1-2], Dr. Deming Li|LiL1], 
Dr. Han Ren[ReL1-3], Dr. Rongxia Hao[HaC1, HaL1], Dr. Zhaox- 
iang Li|LiQ1-2], Dr. Linfan Mao|MaL1, MaT1, MaW1], Dr. Er- 
ling Wei[WiL1-2], Dr. Weili He[HeL1], Dr. Liangxia Wan[WaL1-2], 
Dr. Yichao Chen|CnL1, CnR1], Dr. Yan Xu|XuL1-2], Dr. Wen- 
zhong Liu[LwL1-2]|, Dr. Zeling Shao[ShL1], Dr. Yan Yang[YaL1-2], 
Dr. Guanghua Dong{DoL1], Ms. Ximei Zhao[ZhL1-2], Mr. Lifeng 
Li[LiP1], Ms. Huiyan Wang[WgL1], Ms. Zhao Chai[CiL1], Mr. Zi- 
long Zhu[ZuL1], et al for their successful work related to this book. 


On this occasion, I should express my heartiest appreciation of 
the financial support by KOSEF of Korea from the Com?MaC (Com- 


Preface vu 


binatorial and Computational Mathematics Research Center) of the 
Pohang University of Science and Technology in the summer of 2001. 
In that period, the intention of this book was established. Moreover, 
I should be also appreciated to the Natural Science Foundation of 
China for the research development reflected in this book under its 
Grants(60373030, 10571013, 10871021). 


Y.P. Liu 
Beijing, China 
Jan., 2010 


Contents 


POAC. ca scc eames ENVIO o aM UpR NEUEN EDU a RS iii 
Chapter I Abstract Embeddings ............................ 1 
[I.1 Graphs and networks use a a ag ace xr aC cb doin atrae dea ce a os 1 

L2 AG MERMTTT-—-——————— 9 
iin Cds 25 s68 eer ien t duri dod bb 16 

1.4 Abstract representation ............ 0.00 ccc cece e nee 22 

L5 Smarandache 2-manifolds with map geometry ......... 27 
Activities on Chapter I .............................sess 32 

D5 Observations used Cao ater ido c aac eae aca edd ce c a 32 

Er D e E E E E E 34 

IS Reséearchês:srirssecrsisriisirrid Cada bed OR pound 37 
Chapter II Abstract Maps ..................... 00 cee eee eee Al 
[ll Ground S016. eras e rdeda ive ane an iaiia euet dx 41 

I2 Basi permutations essesi qaa Y asit ede d abe daa dara 43 

[I.3 Conjugate BIO. oo a9 RO au Que ROPA ag RR 46 

I4 TEGBSILTVEOBXIOITIS 5 uec dk Ier RC er Ip E E E 40 

IL5 Included angles.. n... qu schemi ca ide ro ata? enrich 55 
Activities on Chapter II ................................ 57 

[I.6 Observations. tet atias.ose eae eni at dar datio eed ad eee taree 57 

WD ee hee atu et ee went TTE E I TOI TT TT 58 

ILS ResearchESeereeisreredenionit ernaia na eta oy Eia 59 
Chapter III Duality .s.cucidese eases lotti hr o XCR e CR eoe d 65 


DELI Dual DS senesini anene tbi ru ade arsit ad padre RA 65 


X Contents 


II.2 Deletion of an edge us uomexexa xu erae ed d a x eb a redes 72 

TILES Addition of an CACC. 16.20. deGeeuktewdetetideauatadadd 85 

HIA Basie transformati oss siniesigivastacegaesatxnadeeaw’ 96 
Activities on Chapter III ............................se. 98 
[II.5 Observations. uc daa s tin prt n a qitod Sr da Oa ora 98 

ILU EXErCigES s sodatn Exe E RM vd sees Red Padi vd 99 

[HT TRGSOREC DO e erscinas quac tanya dee dp ac opui a vay det 101 
Chapter IV — Orientability .......................Lsssessss 103 
IN d Ornéntation ua sica aca don at Fo a deb S fec rae t har 103 

1352 Basic GuivaleliCeus s quem aua t ri o Redon he 30 9 ae cay 107 

IS 9 Euler characteristic ssa eene Sa a epe RX eadavassdea cs 113 

IV.4 Pattern examples ssa inane aan eave SO OR RR) 116 
Activities on Chapter IV ...........................s. 119 
Es QC NEST T ET OE OT 119 

T5. Exercises ote ries dui ice diesen d n o oni depu m bo aa dic 120 

IV rc Researches: eo cem ekle EO RC RC E CP ec e db oes 122 
Chapter V Orientable Maps ....................... suse. 124 
V.1 Butterflies usus tase dard i b aee RR LR eg oar rat 124 

352 Simplified DubbertliQg. viua dasiooben ER ERA I eee ao does 126 

Veo Reduced rules iuoaaueus d e? Vie waa nir ara q apto edad 130 

V.4 Ornentable principled) order hore EC PpCOO e a 134 

Veo Qrentable PONS: ziehe pda por dex Edu eoe 137 
Activities on Chapter V isse ek xa rarae das 139 

V.6 Observations Lia auis cte dot eoo herd eet tul 139 

Quid 169 1: er araa raS a N OO OUT LOTO Dll S 140 

A S Researches 5.4500 pupa ibt doce dcl opui dc ape Rice EXE LUCR 142 
Chapter VI Nonorientable Maps ........................ 145 


BAN IZ IU OTETTOTTEEOQTE TORTE 145 


Contents xi 


VI.2 Simplified DAIIOS. :scwsaeyvsdsn esaecue i aue ades xad 149 
VI.3 Nonorientable rules iua ua qe iratiag ea eda ow oer as 151 
VLA Nonorientable principles «usos cese mr Rs 156 
VI.5 Nonorientable perius, iussi eerte e 157 
Activities on Chapter VI ........................ssess 159 
VI:S Observations ks aa od Race Pda ade e d rada qr ibas d 159 
VLO ExerciSES PT 160 
DANN. 01: Mm 162 
Chapter VII  Isomorphisms of Maps .................... 164 
MIL EACOmiti AUIID uoces aw bolo DE Ea d acl A ODE A oto 164 
VIIL.2 Isomorphism OHODPOEL, sa dai o S rea ed ea dura 168 
VILS Hee ni MTM" 172 
MESS Justification ecd param nha a dci ru dd RE RE ELE ew ak 177 
VIL Pattern examples... ci reeacea nde rk OR P S e nian 180 
Activities on Chapter VII ..........................se. 185 
VIO ODSOPVOUOPIB o uada ex tod ERO bar acatecta Pars qnos 185 
VIDT Ss cece usb piedad mb dez vua tite Ru acr prp ard; 186 
VILS: Researches ceo zd vxo arceat XR Re xo Pici c aao 188 
Chapter VIII Asymmetrization ......................... 190 
VLE AG OHIOTISDESHEIS eo ee eta pha etam irt bereit tnb 190 
VIIL2 Upper bound of group order............LsLssuuueuue. 193 
VIIL3 Determination of the group...................000- 196 
VULA ROODE S mesire minaaa usec natn ee etui dud subi 201 
Activities on Chapter VIII .........................s. 206 
MILL QOS VA HONS s us a in iq e Perna c epa re opa eg s 206 
MILIA BISOBCISEBS s cued t dp pe VC RRPER C rc Pe Re Pee E Rs 207 


ATI. F6ear hes yak esee S Roe ee acd Ren eem Eo s 209 


xli Contents 


Chapter IX Rooted Petal Bundles ....................... 212 
IX.1 Orientable petal bundles... n.n.. Lor re 212 

[IX.2 Planar pedal bundles «544» 9223 e re heard 2IT 

IX.3 Nonorientable pedal bundles....................sse. 220 

IX.4 The number of pedal bundles....................... 226 
Activities on Chapter IX .............................. 230 
[X.5 Observations. ........ ununun nananana nananana 230 

[X.6 Exercises 410g Da mesi vases be dat ESE KESER ESNA NEE ERDRE Ea 231 

IX. 0 Resear Chics. cessie era PX VOCE HOER VICES E SER RUP ex ra Ro 202 
Chapter X  Asymmetrized Maps ......................... 235 
X.1 Orientable equation........... 0... cece eee nes 235 

X.2 Planar rooted I DES e io pst duuybur dehet dri rrara 243 

A.9 Nonorientable equation :2. 2 voor ter RE RR RR RA RR 250 

X.4 Gross GuabtlOfb. v au Sog serasa saunu abes don debo OU 255 

Ao The number of rooted mapa-ssese sat ex bns 258 
Activities on Chapter X ............................... 261 

X.6 Observations TEE 261 

bm cut RRRREPR 262 

X8 Researches s conc agar eta esr acl Papi Sept edic abd RR i 265 
Chapter XI Maps with Symmetry ....................... 268 
AL] Symmetric relation |. eovescs ones hoa E hee oO hk 268 

XI An annHlcablOleessensequad rest Ibsiea ora RE E d. 270 

XI.3 Symmetric principle >... cu antowy aire, dane aca acne DR 212 

XI.4 General examples........... 0... cc ccc eee ee ee 274 
Activities on Chapter XI .........................Lsss. 278 
XI.5 CDS GB aeuo ert leger nis Ee enr Pb Ehe a d 278 
RAD  PRMNNCE T PITT 279 


AXI T ReséarchéS. Lese wie chee deck arate a weet e Aten Po POR weld 280 


Contents xiii 


Chapter XII Genus Polynomials ......................... 282 
AIL1 Associate surfaces 1i. cessere ert eres 282 

AIL Layer division of a surface Lec o pet E v ERR 285 

AL Handle polymomials 22a pide ded debo er eode 289 

XII.4 Crosscap polynomials ua seien eame coun Fee ec es n 290 
Activities on Chapter XII ..........................ss 292 
XIL.5 Observations. bore nd ducentas pb ROS Roe RU Id dod 292 

PU Gores PRINS actu anes OTI IT LITT 293 

X.T CSC 001. ss sec crnctndcteenaakiensdk tated earners 294 
Chapter XIII Census with Partitions .................... 297 
PTL Planted trees aue dedere dace asm dida Soit cine: 297 

Al 2 Hamiltonian cubic Mapes discs ee bo Le rh RR ARORAA 305 

ATTE S Halin maps aereo es esie qe REA Era tog adie ele POR 307 

XIIL4 Biboundary inner rooted maps ................sue 310 

II General maps ce eectesce qu tad biciaces VAL CARA E aces 315 

XIHI.6 Pane ean: acus acr ai ended he raton Rc p Iu weg 917 
Activities on Chapter XIII ............................ 323 
AIL. E TODSEPUAUIONSS osa arra ta da wires Strap e CR pro pos 323 

RAULS BXGPCISOS sad queas P Eh POS na ian dator mda E E 324 

AULD Researches Lu doses paca is aa xe x ORC IE OR re ecl 325 
Chapter XIV Super Maps of a Graph ................... 326 
XIV.1 Semi-automorphisms on a graph .................. 326 

XIV.2 Automorphisms on a graph scs ovesoerriecet Ene dn 329 

XIV.3 Relationships oio qd eic doe P Desc RR Ra m ORAL 332 

XIV.4 Nonisomorphic super maps................00000ee- 334 

XIV.5 Via rooted super MANS .2iiaacdccesnerganeascae dares 336 
Activities on Chapter XIV .........................s 341 


XIV .G Observallofis.l.2 26r red e Ee x esent 341 


xlv Contents 


PAN AC ECGs S s casses Ped y ebd e me dd Edd SE Se ds 342 
XIV:8 Researches Pm 334 
Chapter XV Equations with Partitions ................. 345 
XV.1 The meson functional xxv esee xe ees 345 
XV.2 General maps on the sphere.............Luuuuusuuus 350 
XV.3 Nonseparable maps on the sphere.................. 303 
XV.4 Maps without cut-edge on furfaces................. 307 
XV.5 Eulerian maps on the sphere.................00000- 361 
XV.6 Eulerian maps on Surfaces... acceso etre 365 
Activities on Chapter XV ........................ssss. 370 
XV.7 CbservatiOfigesscie ee d va e Deren Od Re ia e 370 
DoW oy BOT o MMC orl 
beg X1 040 state ened cian ener ai he ahaoe 372 


Appendix I Concepts of Polyhedra, Surfaces, 


Embeddings and Maps 52 eme 374 
DT Pol Waite care eae bol eae paren aE 374 
Pa ees aver ono Oe owe oo eee ee 377 
Ax.I.3 To DOOOLHE Sos w id dade aar Quiet go M e co ool eed 381 
CY Ri PH" 384 


Appendix II Table of Genus Polynomials for 


Embeddings and Maps of Small Size............ 389 
Ax.IL1 Triconnected cubic graphs...................00005 389 
A IL2- Dot Sie xoc iocos Suc sex dea tances hoes nd eu es 398 
AILES WHOIS Ss wdc cre veces pear kese EREE E RES oe 401 
oe TEE Link Dultidlóges cs exercere tan caet bce ER prati e 403 
Ax.IL5 Complete bipartite graphs....................0005 405 


Ax.ILO Complete graphs. «ce ehe h ge teens ntn 407 


Contents XV 


Appendix III Atlas of Rooted and Unrooted Maps 


for Small €Papligosad esses x AR aad rA ARRA 409 

Ax.IIL1 Bouquets Bm of size 4> M> 1................. 409 
Ax.IIL2 Link bundles Lm of size 7 > m > 3....... ees 415 
Ax.IIL3 Complete bipartite graphs Km,n, 4 > m,n > 3...432 

Ax. ILA Wheels W, of order 6 > n > 4... severe 437 

Ax 111.5 Complete graphs Kn, 5 > m > 4... esee 447 
Ax.III.6 Triconnected cubic graphs of size in [6,15]....... 452 
Biblogřaphy ssuresansde s heeatenoa sis du adbn Sa du EDSR sa ad uar dtd 472 


Terminology cr kiant boa OnE vp" 480 


Chapter I 
Abstract Embeddings 


e A graph is considered as a partition on the union of sets obtained 
from each element of a given set the binary group B = {0,1} sticks 
on. 


e A surface, i.e., a compact 2-manifold without boundary in topol- 
Ogy, ls seen as a polygon of even edges pairwise identified. 


e An embedding of a graph on a surface is represented by a joint 
tree of the graph. A joint of a graph consists of a plane extended 
tree with labelled cotree semi-edges. Two semi-edges of a cotree 
edge has the same label as the cotree edge with a binary index. 
An extended tree is compounded of a spanning tree with cotree 
semi-edges. 


e Combinatorial properties of an embedding in abstraction are par- 
ticularly discussed for the formal definition of a map. 


L1 Graphs and networks 


Let X bea finite set. For any x € X, the binary group B = {0,1} 
sticks on x to obtain Bx = {x(0), z(1)). x(0) and z(1) are called the 
ends of x, or Bx. If Bx is seen as an ordered set (x(0), z(1)), then 


2 Chapter I Abstract Embeddings 


x(0) and z(1) are, respectively, initial and terminal ends of x. Let 


Ade vo Hn (1.1) 


rcx 


i.e., the disjoint union of all Bx, x € X. X is called the ground set. 

A (directed) pregraph is a partition Par= (Pi, Po, ---} of the 
ground set AX ,1.e., 

A= ye, (1.2) 
i>1 

Bx (or (x(0), x(1))), or simply denoted by z itself, x € X, is called an 
(arc) edge and P;, i > 1, a node or vertex. 

A (directed) pregraph is written as G — (V, E) where V —Par 
and 

E = B(X) ={Bal|x € X] 


(= {(x(0), x(1))] € Xj). 


If X is a finite set, the (directed) pregraph is called finite; otherwise, 
infinite. In this book, (directed) pregraphs are all finite. 

If X = Ø, then the (directed) pregraph is said to be empty as 
well. 

An edge (arc) is considered to have two semiedges each of them 
is incident with only one end (semiarcs with directions of one from 
the end and the other to the end). An edge (arc) is with two ends 
identified is called a selfloop (di-selfloop); otherwise, a link (di-link). If 
t edges (arcs) have same ends (same direction) are called a multiedge 
(multiarc), or t-edge (t-arc). 


Example 1.1 There are two directed pregraphs on X = (x], 
1 B. 


Par; = ((z(0)), {x(1) }}; 
Par; = {{x(0), x(1)}}. 


They are all distinct pregraphs as well as shown in Fig.1.1. 


L1 Graphs and networks 3 


Xx 
T 
Parı Pars 
Fig.1.1 Directed pregraphs of 1 edge 


Further, pregraphs of size 2 are observed. 


Example 1.2 On X = {21,22}, the 15 directed pregraphs are 
as follows: 


Par, = ((1(0)). 00100). {2(0)}, {z2(1)}}; 
Par, = {{21(0), x1(1)}, {x2(0)}, 122(1) 1: 
Pars = {{21(0), x2(0)}, {z1(1)}, 102(1) 1: 
Par, = {{21(0), xo(1)}, {z1(1)}, 122(0) 1; 
Pars = {{21(0)}, {21(1), £2(0)}, {za(1) 1: 
Pars = {{21(0)}, {21(1), 29(1) }, 122(0) 1; 
Par; = {{21(0)}, {21(1)}, {z2(1), z2(0)}}; 
Pars = {{1(0), x1(1), z2(0)}, 122(1) } }: 
Parg = {{1(0), 21(1), z2(1)}, 122(0) fF: 
Pario = {{21(0), 22(0); z2(1)}, {£1(1)}}; 
Pari = {{21(0)}, {21(1), %2(0), z2(1)}}; 
Paria = {{x1(0), 21(1), va(0), vo(1) }}; 
Paris = {{21(0), x1(1)}, {22(0), vo(1) 3: 
Paria = {{21(0), x2(0)}, {21(1), a(1) fF: 
Paris = {{21(0), x2(1)}, {1(1), v2(0) fF. 


Among the 15 directed pregraphs, Pars, Par4, Pars; and Pare 
are 1 pregraph; Pars and Parg are 1 pregraph; Parjo and Par; are 1 
pregraph; Par;4 and Parı5 are 1 pregraph; and others are 1 pregraph 
each. Thus, there are 9 pregraphs in all(as shown in Fig.1.2). 


4 Chapter I Abstract Embeddings 


T T 
y y { ] x 1 2 


Par, Paro Para 
Tı T2 "o we I1 T2 
Par4 Pars Parę 
X1 V V 
Parz Parg Parg 
x 
L] M? I 
Pario Pari Parj2 
x 

zl 7 2i qi 
Paria Pari4 Paris 


Fig.1.2 Directed pregraphs of 2 edges 


Now, Par= (P4, Po,---} and B are, respectively, seen as a map- 
ping z => P, z € P, i > 1 anda mapping z +> Z, Z Æ z, {z,z} € 
B(X). The composition of two mappings a and (3 on a set Z is defined 


L1 Graphs and networks 5 


to be the mapping 
(aß)z = U ay, z € Z. (1.3) 


yeBz 


Let Vypargy be the semigroup generated by Par=Par(X) and B = 
B(X). Since the mappings a =Par and B have the property that 
y € az & z € ay, it can be checked that for any z, y € B(X), what is 
determined by 
dy € Vipargy, 2 € vy 

is an equivalence. If B( X) itself is a equivalent class, then the semi- 
group V (pa; g} is called transitive on X' = B(X). A (directed)pregraph 
with V pag transitive on X is called a (directed ) graph . 


A (directed)pregraph G = (V, E) that for any two vertices u,v € 
V, there exists a sequence of edges e1, €2,---,e, for the two ends of e;, 
i = 2,3,---,s—1, are in common with those of respective e;_; and e;44 
where u and v are, respectively, the other ends of e; and eg, is called 
connected . Such a sequence of edges is called a trail between u and 
v. A trail without edge repetition is a walk. A walk without vertex 
repetition is a path. A trail, walk, or path with u = v is, respectively, 
a travel, tour, or circuit. 


Theorem 1.1 A (directed)pregraph is a (directed)graph if, and 
only if, it is connected. 


Proof Necessity. Since Par’ = Par, k > 1, and B* = B, k > 1, 
by the transitivity, for any two elements y, z € X, there exists y such 
that z € yy and there exists an integer n > 0 such that 


y = (BPar)" B = (BPar) --- (BPar) B, (1.4) 


TL 


where BPar appears for n times. Therefore, the (directed)pregraph is 
connected. 

Sufficiency. If a (directed)pregraph is connected, i.e., for any 
two elements x,y € X, their incident vertices u,v € V, have edges 
e1, €2,°°*,@s, Such that e;, 2 = 2,83,---,s5— 1, is in common with e;. 4 


6 Chapter I Abstract Embeddings 


and e;4,1. Of course, u and v are, respectively, the ends of e1 and es. 
Thus, y € yz wherey — (ParB)?B. This implies that the semigroup 
V,parg; is transitive on A. Therefore, the (directed)pregraph is a 
(directed)graph. O 


It is seen from the theorem that (directed) graphs here are, in 
fact, connected (directed) graphs in most textbooks. Because discon- 
nectedness is rarely necessary to consider, for convenience all graphs, 
embeddings and then maps in what follows are defined within con- 
nectedness in this book. 

A network N is such a graph G = (V, E) with a real function 
w(e) € R,e € E on E, and hence write N = (G;w). Usually, a 
network N is denoted by the graph G itself if no confusion occurs. 


Finite recursion principle On a finite set A, choose ag € A 
as the initial element at the Oth step. Assume a; is chosen at the ith, 
i > 0, step with a given rule. If not all elements available from a; are 
not yet chosen, choose one of them as a;,4, at the i+ 1st step by the 
rule, then a chosen element will be encountered in finite steps unless 
all elements of A are chosen. 


Finite restrict recursion principle Ona finite set A, choose 
ag € A as the initial element at the Oth step. Assume a; is chosen at 
the ith, 2 > 0, step with a given rule. If a; is not available from aj, 
choose one of elements available from a; as a;,4 at the i + Ist step by 
the rule, then ap will be encountered in finite steps unless all elements 
of A are chosen. 


The two principles above are very useful in finite sets, graphs and 
networks, even in a wide range of combinatorial optimizations. 

A G = (V, E) with V = Vi + V2 of both V; and Vz independent, 
i.e., its vertex set is partitioned into two parts with each part having 
no pair of vertices adjacent, is called bipartite. 


Theorem 1.2 A graph G = (V, E) is bipartite if, and only if, 
G has no circuit with odd number of edges. 


Proof Necessity. Since G is bipartite, start from vy € V ini- 


L1 Graphs and networks 7 


tially chosen and then by the rule from the vertex just chosen to one 
of its adjacent vertices via an edge unused and then marked by used, 
according to the finite recursion principle, an even circuit (from bipar- 
tite), or no circuit at all, can be found. From the arbitrariness of vo 
and the way going on, no circuit of G is with odd number of edges. 
Sufficiency. Since all circuits are even, start from marking an 
arbitrary vertex by 0 and then by the rule from a vertex marked 
by b € B = {0,1} to mark all its adjacent vertices by b = 1 — b, 
according to the finite recursion principle the vertex set is partitioned 
into Vo = (v € V| marked by 0} and V; = (v € V| marked by 1}. By 
the rule, Vy and Vj are both independent and hence G is bipartite. L 


From this theorem, a graph without circuit is bipartite. In fact, 
from the transitivity, any graph without circuit is a tree. 

On a pregraph, the number of elements incident to a vertex is 
called the degree of the vertex. A pregraph of all vertices with even 
degree is said to be even . If an even pregraph is a graph, then it is 
called a Euler graph. 


Theorem 1.3 A pregraph G = (V, E) is even if, and only if, 
there exist circuits C1, C5, --- , C4, on G such that 


E-Ci tC» C, (1.5) 





where n is a nonnegative integer. 


Proof Necessity. Since all the degrees of vertices on G are even, 
any pregraph obtained by deleting the edges of a circuit from G is still 
even. From the finite recursion principle, there exist a nonnegative 
integer n and circuits Ci, C5, --- , Cn, on G such that (1.5) is satisfied. 

Sufficiency. Because a circuit contributes 2 to the degree of each 
of its incident vertices, (1.5) guarantees each of vertices on G has even 
degree. Hence, G is even. [] 


The set of circuits (C;|1l € i € n} of G in (1.5) is called a 
circuit partition, or written as Cir=Cir(G). Two direct conclusions of 
Theorem 1.3 are very useful . One is the case that G is a graph. The 


8 Chapter I Abstract Embeddings 


other is for G is a directed pregraph. Their forms and proofs are left 
for the reader. 

Let N — (G;w) be a network where G — (V, E) and w(e) — 
—w(e) € Zn = (0,1,---,n — 1}, ie., mod n, n > 1, integer group. 
For examples, Zı = {0}, Zə = B = {0,1} etc. Suppose x, = —z, € 
Zn, v € V, are variables. Let us discuss the system of equations 


Eu + £y = w(e) (mod n), e = (uv) € E (1.6) 


On Zn. 


Theorem 1.4 System of equations(1.6) has a solution on Z, 
if, and only if, there is no circuit C such that 


X wle) ) £0 (mod n) (1.7) 


ecC 
on N. 


Proof Necessity. Assume C is a circuit satisfying (1.7) on N. 
Because the restricted part of (1.6) on C has no solution, the whole 
system of equations (1.6) has to be no solution either. Therefore, N 
has no such circuit. This is a contradiction to the assumption 

Sufficiency. Let £o = a € Zn, start from vg € V reached. Assume 
vi € V reached and x; = a; at step i. Choose one of e; = (vi, vi41) € E 
without used(otherwise, backward 1 step as the step i). Choose v;+1 
reached and e; used with a;,, = a; + w(e;) at step i + 1. If a circuit 
as {e€9,€1,---, e, ej = (vj, Uj41),0 € j € lvi = vo, occurs within a 
permutation of indices, then from (1.7) 


Qj41 = aj + w(ei) 
= a1 + wle) + wle) 


I.2 Surfaces 9 


Because the system of equations obtained by deleting all the equations 
for all the edges on the circuit from (1.6) is equivalent to the original 
system of equations (1.6), in virtue of the finite recursion principle a 
solution of (1.6) can always be extracted . O 


When n = 2, this theorem has a variety of applications. In [Liu5], 
some applications can be seen. Further, its extension on a nonAbelian 
group can also be done while the system of equations are not yet linear 
but quadratic. 


1.2 Surfaces 


In topology, a surface is a compact 2-dimensional manifold with- 
out boundary. In fact, it can be seen as what is obtained by identifying 
each pair of edges on a polygon of even edges pairwise. 


For example, in Fig.1.3, two ends of each dot line are the same 
point. The left is a sphere, or the plane when the infinity is seen as 
a point. The right is the projective plane. From the symmetry of the 
sphere, a surface can also seen as a sphere cutting off a polygon with 
pairwise edges identified. 


The two surfaces in Fig.1.3 are formed by a polygon of two edges 
pairwise as a. 





Sphere(Plane) Projective plane 


Fig.1.3 Sphere and projective plane 


10 Chapter I Abstract Embeddings 


Surface closed curve axiom A closed curve on a surface has 
one of the two possibilities: one side and two sides. 


A curve with two sides is called a double side curve ; otherwise, a 
single side curve . As shown in Fig.1.3, any closed curve on a sphere is 
a double side curve(In fact, this is the Jordan curve axiom). However, 
it is different from the sphere for the projective plane. there are both 
a single(shown by a dot line) and a double side curve. 

How do we justify whether a closed curve on a surface is of single 
side, or not? 

In order to answer this question, the concept of contractibility 
of a curve has to be clarified. If a closed curve on a surface can be 
continuously contracted along one side into a point, then it is said to 
be contractible, or homotopic to 0. 





Torus Klein bottle 


Fig.1.4 Torus and Klein bottle 


It is seen that a single side curve is never homotopic to 0 and 
a double side curve is not always homotopic to 0. For example, in 
Fig.1.4, the left, i.e., the torus, each of the dot lines is of double side 
but not contractible. The right, i.e., the Klein bottle, all the dot lines 
are of single side , and hence, none of them is contractible. 

A surface with all closed curves of double side is called orientable; 
otherwise, nonorientable . 


For example, in Fig.1.3, the sphere is orientable and the projec- 


I.2 Surfaces 11 


tive plane is nonorientable. In Fig.1.4, the torus is orientable and the 
Klein bottle is nonorientable. 

The maximum number of closed curves cutting along without 
destroying the continuity on a surface is called the pregenus of the 
surface. 

In view of Jordan curve axiom, there is no such closed curve on 
the sphere. Thus, the pregenus of sphere is 0. On the projective plane, 
only one such curve is available (each of dot lines is such a closed curve 
in Fig.1.3) and hence the pregenus of projective plane is 1. 

Similarly, the pregenera of torus and Klein bottle are both 2 as 
shown in Fig.1.4. 


Theorem 1.5 The pregenus of an orientable surface is a non- 
negative even number. 


A formal proof can not be done until Chapter 5. Based on this 
theorem, the genus of an orientable surface can be defined to be half 
its pregenus, called the orientable genus. The genus of a nonorientable 
surface, called nonorientable genus, is its pregenus itself. 

The sphere is written as aa^! where a^! is with the opposite 
direction of a on the boundary of the polygon. Thus, the projective 
plane, torus and Klein bottle are, respectively, aa, aba~!b~! and aabb. 
In general, 


p 
Op = ajbja; b7 | 
l T 
= abia bi lasboa; b3 - . aybya;, b; ! 
and 
q 
Qe = li Aili = 41010202 * ` AgAg (1.9) 


i=l 
denote, respectively, a surface of orientable genus p and a surface of 
nonorientable genus q. Of course, Oo, Q1, O1 and Qə are, respectively, 
the sphere, projective plane, torus and Klein bottle. 

It is easily checked that whenever an even polygon is with a pair 
of its edges in the same direction, the polygon represents a nonori- 


12 Chapter I Abstract Embeddings 


entable surface. Thus, all Op, p > 0, orientable and all Q,, q = 1, are 
nonorientable. 

Forms (1.8) and (1.9) are said to be standard. 

If the form of a surface is defined by its orientability and its genus, 
then the operations 1-3 and their inverses shown as in Fig.1.5-7, do 
not change the surface form. 





Fig.1.5 Operation 1: Aaa | < A 





Fig.1.6 Operation 2: AabBab < AcBc 


Fig.1.7 Operation 3: AB 4 ( (Aa)( a 1B) 


In fact, what is determined under these operations is just a topo- 
logical equivalence, denoted by ~top. 


Notice that A and B are all linear order of letters and permitted 


I.2 Surfaces 13 


to be empty as degenerate case in these operations. 
'The parentheses stand for cyclic order when more than one cyclic 
orders occur for distinguishing from one to another. 


Relation 0 On a surface (A, B), if A is a surface itself then 
(A, B) = ((A)z(B)z 7) = ((A)(B)). 


Relation 1 (AxByCaz !Dy )) ~top ((ADCB)(zxyr |y !)). 

Relation 2 (AzBz) ^y ((AB™')(xz)). 

Relation 3 (Axzyzy !z |) ~top ((A)(zx)(yy)(zz)). 

In the four relations, A, B, C, and D are permitted to be empty. 
B^ = b7! . -b3 'b7'b7' is also called the inverse of B = bibob3 - - bs, 


s > 1. Parentheses are always omitted when unnecessary to distin- 
guish cyclic or linear order. 





On a surface S, the operation of cutting off a quadrangle aba! 57! 
and then identifying each pair of edges with the same letter is called 
a handle as shown in the left of Fig.1.8. 

If the quadrangle aba~!b~! is replaced by aa, then such an oper- 
ation is called a crosscap as shown in the right of Fig.1.8. 





Handle Crosscap 


Fig.1.8 Handle and crosscap 


The following theorem shows the result of doing a handle on an 
orientable surface. 


Theorem 1.6 What is obtained by doing a handle on an ori- 
entable surface is still orientable with its genus 1 added. 


14 Chapter I Abstract Embeddings 


Proof Suppose S is the surface obtained, then 


p 
B ovs | [ a0; 5; cap ibat! ( Relation 0) 
i=1 
p 
pad. si aed ; 
~top ] [aio b; xx 0ga55185,115,5, (Relation 1) 
i=1 
p 
eis EET 
^" top ] [ aita; b; Qp410p 4105 10544 (Operation 1) 
i1 
p+1 


= li ajbja; ^b; |. 
i=l 


This is the theorem. [] 


In the above proof, x and x! are a line connecting the two 


boundaries to represent the surface as a polygon shown in Fig.1.8. 
'This procedure can be seen as the degenerate case of operation 3. 


In what follows, observe the result by doing a crosscap on an 
orientable surface. 


Theorem 1.7 Onan orientable surface of genus p, p > 0, what 
is obtained by doing a crosscap is nonorientable with its genus 2p + 1. 


Proof Suppose N is the surface obtained, then 


p 
N Step li a;bia; |b; xaax | (Relation 0) 
i=1 
p 
~top li ajbja; b; xx cc (Relation 2) 
i=1 
p 
^"top li ajbja; 1b; xx" cc, cc36363 (Relation 3) 
i=2 
2p+1 
~top II cici. (Relation 3 by p — 1 times). 
i=1 


I.2 Surfaces 15 


This is the theorem. O 


By doing a handle on a nonorientable surface, 2 more genus 
should be added with the same nonorientability. 


Theorem 1.8 On a nonorientable surface, what is obtained by 
doing a handle is nonorientable with its genus 2 added. 


Proof Suppose N is the obtained surface, then 


q 
N op li aja;zaba tb tx! (Relation 0) 
i-l 
q 
™top II ajajzz laba |b | (Relation 1) 
i=1 
q 
~top li aja;aba !b ! (Operation 1) 
i=l 
q+2 
~top li cici. (Relation 3). 


i—1 


This is the theorem. O 


By doing a crosscap on a nonorientable surface, 1 more genus 
produced with the same nonorientability 


Theorem 1.9 On a nonorientable surface, what is obtained by 
doing a crosscap is nonorientable with its genus 1 added. 


16 Chapter I Abstract Embeddings 


Proof Suppose N is the obtained surface, then 


q 
N “top II aja;zaarx | (Relation 0) 
i=l 
q 
^" top li ajajzx ‘aa (Relation 2) 
i=l 
q 
™top II a,;ajaa (Operation 1) 
i=1 
qi 
^" top li aiai. (Relation 3). 


i=l 


This is the theorem. E 


I.3 Embedding 


Let G = (V, E, V = (vy vo, --- v4], be a graph. A point in the 
3-dimensional space is represented by a real number t as the parame- 
ter, e.g., (x,y, z) = (t, t2, t3). Write the vertices as 


Ui = (£i, Yi, zi) = CER 
such that t; Æ tj, i # j, 1 <i,j <n, and an edge as 
(u,v) =ut Av, 1€ A € I, 


i.e., the straight line segment between u and v. Because for any four 
vertices vj, vj, vj and vy, 


Li Xj Li — XQ Li — Tk 
det | yi—Vj Wi—W Yi Yk 


£j Zj £i — 2| £j — Zk 


tj—tj tj—t ti— tr 
=det | 2-0 -P P-E 


j 
fa PP 9-5 


L3 Embedding 17 


m (ti E DOR = ti) (ti — tr) (tk = ti) (tk = bb = tı) * 0, 1.8: the four 
points are not coplanar, any two edges in G has no intersection inner 
point. 

A representation of a graph on a space with vertices as points 
and edges as curves pairwise no intersection inner point is called an 
embedding of the graph in the space. If all edges are straight line 
segments in an embedding, then it is called a straight line embedding. 
Thus, any graph has a straight line embedding in the 3-dimensional 
space. Similarly, A surface embedding of graph G is a continuous 
injection 4G of an embedding of G on the 3-dimensional space to a 
surface S such that each connected component of S — uG is homotopic 
to 0. The connected component is called a face of the embedding. In 
early books, a surface embedding is also called a cellular embedding. 
Because only a surface embedding is concerned with in what follows, 
an embedding is always meant a surface embedding if not necessary 
to specify. 

A graph without circuit is called a tree. A spanning tree of a 
graph is such a subgraph that is a tree with the same order as the 
graph. Usually, a spanning tree of a graph is in short called a tree on 
the graph. For a tree on a graph, the numbers of edges on the tree 
and not on the tree are only dependent on the order of the graph. 
They are, respectively, called the rank and the corank of the graph. 
The corank is also called the Betti number, or cyclic number by some 
authors. 


The following procedure can be used for finding an embedding 
on a surface. 


First, given a cyclic order of all semiedges at each vertex of G, 
called a rotation. Find a tree(spanning, of course) T' on G and distin- 
guish all the edges not on T by letters. Then, replace each edge not 
of T' by two articulate edges with the same letter. 

From this procedure, G is transformed into G without changing 
the rotation at each vertex except for new vertices that are all ar- 
ticulate. Because G is a tree, according to the rotation, all lettered 
articulate edges of G form a polygon with pairs of edges, and hence 


18 Chapter I Abstract Embeddings 


a surface in correspondence with a choice of indices on each pair of 
the same letter. For convenience, G with a choice of indices of pair in 
the same letter is called a joint tree of G. 


Theorem 1.10 A graph G = (V, E) can always embedded into 
a surface of orientable genus at most |/9/2]|, or of nonorientable genus 
at most 0, where 8 is the Betti number of G. 

Proof It is seen that any joint tree of G is an embedding of G 
on the surface determined by its associate polygon. From (1.8) for 
the orientable case, the surface has its genus at most |20/4| = |8/2 |]. 
From (1.9) for the nonorientable case, the surface has its genus at most 


28/2 = B. o 


In Fig.1.9, graph G and one of its joint tree are shown. Here, the 
spanning tree T is represented by edges without letter. a, b and c are 
edges not on T. Because the polygon is 


abcacb ~top c lb tcbaa (Relation 2) 


~top aabbcc (Theorem 1.7), 


the joint tree is, in fact, an embedding of G on a nonorientable surface 
of genus 3. 





b c 


G G 


Fig.1.9 Graph and its joint tree 


Because any graph with given rotation can always immersed in 
the plane in agreement with the rotation, each edge has two sides. As 
known, embeddings of a graph on surfaces are distinguished by the 
rotation of semiedges at each vertex and the choice of indices of the 
two semiedges on each edge of the graph whenever edges are labelled 


L3 Embedding 19 


by letters. Different indices of the two semiedges of an edge stand 
for from one side of the edge to the other on a face boundary in an 
embedding. 


Theorem 1.11 A tree can only be embedded on the sphere. 
Any graph G except tree can be embedded on a nonorientable surface. 
Any graph G can always be embedded on an orientable surface. Let 
no(G) be the number of distinct embeddings on orientable surfaces, 
then the number of embeddings on all surfaces is 


2" 9no(G), no(G) = T=)", (1.10) 


i22 


where O(G) is the Betti number and n; is the number of vertices of 
degree 7 in G. 


Proof On a surface of genus not 0, only a graph with at least 
a circuit is possible to have an embedding. Because a tree has no 
circuit, it can only embedded on the sphere. Because a graph not 
a tree has at least one circuit, from Theorem 1.10 the second and 
the third statements are true. Since distinct planar embeddings of a 
joint tree of G with the indices of each letter different correspond to 
distinct embeddings of G on orientable surfaces and the number of 
distinct planar embeddings of joint trees is 


no(G) = [TC - 0» 
i22 
Further, since the indices of letters on the B(G) edges has 2°) of 
choices for a given orientable embedding and among them only one 
choice corresponds to an orientable embedding, the fourth statement 
is true. [] 


For an embedding u(G) of G on a surface, let v(uG), e(uG) and 
@(wG) are, respectively, its vertex number, or order, edge number, or 
size and face number, or coorder. 


Theorem 1.12 Fora surface S, all embeddings 4(G) of a graph 
G have Eul(uG) = v(uG) — e(uG) + d(uG) the same, only dependent 


20 Chapter I Abstract Embeddings 


on S and independent of G. Further, 


2— 2p, p 2 0, 
when © has orientable genus p; 
Eul(uG) — jc wd (1.11) 


when S has nonorientable genus q. 


Proof For an embedding u(G) on S, if it has at least 2 faces, 
then by connectedness it has 2 faces with a common edge. From the 
finite recursion principle, by the inverse of Operation 3 an embedding 
u(G,) of Gi on S with only 1 face on S is found. It is easy to check 
that Eul(uG) —Eul(uG). Similarly, by the inverse of Operation 2 an 
embedding u(Go) of Go on S with only 1 vertex is found. It is also 
easy to check that Eul(wG) =Eul(wG’). Further, by Operation 1 and 
Relations 1-3, it is seen that Eul(uGo) =Eul(O,), p > 0; or Eul(Q,), 
q > 1 according as S is an orientable surface in (1.8); or not in (1.9). 
From the arbitrariness of G, the first statement is proved. 

By calculating the order, size and coorder of Op, p > 0; or Qq, 
q > 1, (1.11) is soon obtained. So, the second statement is proved. O 


According to this theorem, for an embedding u(G) of graph G, 
Eul(uG) is called its Euler characteristic, or of the surface it is on. 
Further, g(wG) is the genus of the surface u(G) is on. 

If a graph G is with the minimum length of circuits c, then 
from Theorem 1.12 the genus y(G) of an orientable surface G can be 
embedded on satisfies the inequality 


v(G) - «1 2) etie (1.12) 


ce 
2 2 — 2 


and the genus 4(G) of a nonorientable surface G can be embedded on 
satisfies the inequality 


2— (V(G) - « - 5) €3(0) <8. (1.13) 


If a graph has an embedding with its genus attaining the lower(upper) 
bound in (1.12) and (1.13), then it is called down(up)-embeddable. In 


L3 Embedding 21 


fact, a graph is up-embeddable on nonorientable, or orientable sur- 
faces according as it has an embedding with only 1 face, or at most 2 
faces. 


Theorem 1.13 All graphs but trees are up-embeddable on 
nonorientable surfaces. 

Further, if a graph has an embedding of nonorientable genus / 
and an embedding of nonorientable genus k, | < k, then for any 2, 
| « i « k, it has an embedding of nonorientable genus t. 


Proof For an arbitrary embedding of a graph G on a nonori- 
entable surface, let T' be its corresponding joint tree. From the nonori- 
entability, the associate 20 (G)-gon P has at least 1 letter with different 
indices(or same power of its two occurrences!). If P = Q4, q = B(G), 
then the embedding is an up-embedding in its own right. Otherwise, 
by Relation 2, or Relation 3 if necessary, whenever s^!s or stst oc- 
curs, it is, respectively, replaced by ss or sts- t. In virtue of no letter 
missed in the procedure, from the finite recursion principle, P' = Qg, 
q = B(G), is obtained. This is the first statement. 

From the arbitrariness of starting embedding in the procedure 
of proving the first statement by only using Relation 2 instead of 
Relation 3 (AststB eo, Ass ! Btt by Relation 2), because the genus 
of the surface is increased 1 by 1, the the second statement is true. O 


The second statement of this theorem is also called the inter- 
polation theorem. The orientable form of interpolation theorem is 
firstly given by Duke[Duk1]. The maximum(minimum) of the genus 
of surfaces (orientable or nonorientable) a graph can be embedded 
on is call the maximum genus(minimum genus) of the graph. The- 
orem 1.13 shows that graphs but trees are all have their maximum 
genus on nonorientable surfaces the Betti number with the interpola- 
tion theorem. The proof would be the simplest one. However, for the 
orientable case, it is far from simple. many results have been obtained 
since 1978(see |Liu1-2], [LiuL1], [HuanL1] and [LidL1]) in this aspect. 
On the determination of minimum genus of a graph, only a few of 
graphs with certain symmetry are done(see Chapter 12 in |Liu5-6]). 


22 Chapter I Abstract Embeddings 


L4 Abstract representation 
Let G = (V, X), V = (vi, v3, , Un}, 
X = (z,,25,:-, Lm} CV{X}V = {{u, v} Vu, v € V), 


be a graph. For an embedding u(G) of G on a surface, each edge has 
not only two ends as in G but also two sides. Let o be the operation 
from one side to the other and 8 be the operation from one end to 
the other. From the symmetry between the two ends and between the 
two sides, 

a^ ee um] (1.14) 


where 1 is the identity. By considering that the result from one side 
to the other and then to the other end and the result from one end to 
the other and then to the other side are the same, t.e., 


Bo = a. (1.15) 


Further, it can be seen that K = (1,0, 8, y}, y = a, is a group, 
called the Klein group where 


(ab)? = (a8)(a8) = (aBa) 6 


— (aab) = (aa) (8) =, (1.16) 


Thus, an edge x € X of G in an embedding pu(G) of G becomes 
Ka = {x,ax, Bx, yz}, as shown in Fig.1.10. 


x Ba 


—— -  ——— 


y m 
ax apa 


Fig.1.10 An edge sticking on K 
In fact, let 


X=) Kx; (1.17) 
i=1 


L4 Abstract representation 23 


where summation stands for the disjoint union, then a and (@ can both 
be seen as a permutation on 4, i.e., 


m 


a =| [aa a 8 = aa a 


i=1 iA 
The vertex x is on deals with the rotation as 

{(x, Px, Dp es -), (ax, aP Ix, oP is Jh (1.18) 
as shown in Fig.1.11(when its degree is 4). 


pes 


BPx 














Px | aPx 
ax ox E Px BP x 
Ba <= ij) ePi ABP 2a 
aP3x Pea 
o] BP3x 





Fig.1.11 The rotation at a vertex 


It is seen that P is also a permutation on X. The set of elements 
in each cycle of this permutation is called an orbit of an element in 
the cycle. For example, the orbit of element x under permutation P 
is denoted by (x)p. From (1.18), 


(x)p ( \(ax)p =, LEX. (1.19) 
The two cycles at a vertex in an embedding have a relation as 
(az, Pax, Paz,- -) 
=(axz,aP !z, o/P ?z, ..-) (1.20) 
—a(z,P lx, P ?z,...). 
For convenience, one of the two cycles is chosen to represent the vertex, 


l; Cs 


(x, Px, P?y... 27 


24 Chapter I Abstract Embeddings 


Or 
(ox, P tr, P 7, - --). 


Theorem 1.14 aP = Pta. 


Proof By multiplying the two sides of (1.20) by o from the left 
and then comparing the second terms on the two sides, 


aPa =P. 


By multiplying its two sides by a from the right, the theorem is soon 
obtained. [] 


Since a and f are both permutations on ¥, y = aß and P* = Py 
are permutations on A as well. Let 


(e P aP ige) 


be the cycle of P* involving x. From the symmetry between x and 
x, the cycle of P* involving (x is 


(B, P* x, P** Bx, 3e -) 


which has the same number of elements as that involving x does. 
Because P*(Gx) = PaG(Gx) = Pax and from Theorem 1.14 


Paz = aP 1z = ay(yP x 


= ayP* Iz 
= BP's, 
we have 
P* (G2) = BP* "a. (1.21) 


Furthermore, because P**(Gx) = P*(P*(Gx)) and from (1.21) 
P*(P*(Ba)) = P'(8P* 2), 
by (1.21) for P*~'x instead of x, we have 


P* (Bx) = (P (P*^2)) = B(P* 72). 


L4 Abstract representation 25 


On the basis of the finite restrict recursion principle, a cycle is found. 
Therefore, 


(Bx, P* Bv, P Bx, -.-) = B(a,P* x, P* "z,---). (1.22) 
This implies 
c)» (fo = 0, (1.23) 
for z € X. 
Theorem 1.15 (P* = P* !g. 
Proof A direct result of (1.22). O 
Based on (1.22), it is seen that the face involving x of the em- 
bedding represented by P is 
((z, P*z, P*^z, .--), (Gx,P* "Bx, P* -- ]. (1.24) 
Similarly to vertices, based on (1.22) and (1.23), the face can be 
represented by one of the two cycles in (1.24). 


Example 1.3 Let G = K4, i.e. , the complete set of order 4. 
Given its rotation 


Ua, y, z), (Bz, l, yw), (4l, u, By), (Bx, Ww, au) y, 
as shown in Fig 1.12. Its two faces are (x, Gu, Gl, yz) and 


(y, Qu, AW, al, VY, e; Bus, "yas. 


N 


Fig.1.12 A rotation of Ky 


26 Chapter I Abstract Embeddings 


Thus, it is an embedding of K4 on the torus 
O, = (ABA !B ^J) 


as shown in Fig.1.13. 


ys 


V1 


of Xx 


UA v — 


Á 
EN 


v3 


|B 
Fig.1.13 Embedding determined by rotation 


Further, another rotation of K4 is chosen for getting another 
embedding of K4. 


Example 1.4(Continuous to Example 1.3) Another embedding 
of K4 is shown as in Fig.1.14. Its rotation is 
{(x, y, 2), (8z, L, yw), (u, vy, 71), (Bz, w, yu) }. 


Its two faces are 
(x, Bu, pl yz) 
and 
(az, w, Bz, ay, au, aw, al, By). 


This is an embedding of Ky on the Klein bottle 


Nz = (ABA B) ~top= (AABB) 


L5 Smarandache 2-manifolds with map geometry 27 


as shown in Fig.1.14. 


ys 


V1 


of NX 


UA v2 —— 


A A 


K 
EN 


U3 


B |) ey 


Fig.1.14 Embedding distinguished by rotation 


Such an idea is preferable to deal with combinatorial maps via 
algebraic but neither geometric nor topological approaches. 


I5 Smarandache 2-manifolds with map geometry 


Smarandache system The embedding of a graph on surface 
enables one to construct finitely Smarandache 2-manifolds, i.e., map 
geometries on surfaces. 

A rule in a mathematical system (X; R) is said to be Smaran- 
dachely denied if it behaves in at least two different ways within the 
same set X, i.e., validated and invalided, or only invalided but in mul- 
tiple distinct ways. 

A Smarandache system (X; R) is a mathematical system which 
has at least one Smarandachely denied rule in R (see [Mao4] for de- 
tails). Particularly, if (3; ) is nothing but a metric space (M; p), 
then such a Smarandache system is called a Smarandache geometry, 
seeing references [Maol|-[Mao4] and [Smal]—|Sma2). 


Example 1.5(Smarandache geometry) Let R? be a Euclidean 
plane, points A,B € R? and l a straight line, where each straight 


28 Chapter I Abstract Embeddings 


line passes through A will turn 30° degree to the upper and passes 
through 5 will turn 30? degree to the down such as those shown in 
Fig.1.15. Then each line passing through A in Fı will intersect with 
l, lines passing through B in F5 will not intersect with l and there is 
only one line passing through other points does not intersect with /. 





Fi 4 4996-77 











Fig.1.15 


Then such a geometry space R? with queer points A and B is 
a Smarandache geometry since the axiom given a line and a point 
exterior this line, there is one line parallel to this line is now replaced 
by none line, one line and infinite lines. 

A more general way for constructing Smarandache geometries 
is by Smarandache multi-spaces ({Mao3]). For an integer m > 2, let 
(Xy; R4), (Eo; R3), +++; (Em; Rm) be m mathematical systems different 
two by two. A Smarandache multi-space is a pair (35 R) with 


i=1 


i=l 
Such a multi-space naturally induce a graph structure with 
V(G) = iX de, ey Xl 


Example 1.16(|Mao5]) Let n bean integer, Z; = ({0,1,2,---,n— 
1, +) an additive group (modn) and P = (0,1,2,---,n — 1) a permu- 
tation. For any integer 7,0 < i < n — 1, define 








Zin = P(Z) 
such that P'(k) +; P'(I) = P'(m) in Zj41 if k+l =m in Zi, where +; 


L5 Smarandache 2-manifolds with map geometry 29 


denotes the binary operation +; : (P'(Kk), P'(I)) —^ P'(m). Then we 


with O = {+;, 0 € i € n-1] isa Smarandache multi-space underlying 
a graph Kn, where Z; = Z4 for integers 1 <i € m. 


Map geometry A nice model on Smarandache geometries, 
namely s-manifolds on the plane was found by Iseri in [Ise1l] defined 
as follows, which is in fact a case of map geometry. 


An s-manifold is any collection C(T', n) of these equilateral trian- 
gular disks T;,1 < i < n satisfying the following conditions: 

(i) each edge e is the identification of at most two edges ej, e; in 
two distinct triangular disks T;, Tj, 1 < i,j <n andi Æj; 

(ii) each vertex v is the identification of one vertex in each of 
five, six or seven distinct triangular disks. 


These vertices are classified by the number of the disks around 
them. A vertex around five, six or seven triangular disks is called an 
elliptic vertex, an Euclidean vertex or a hyperbolic vertex, respectively. 











Fig.1.16 


In a plane, an elliptic vertex O, a Euclidean vertex P and a 
hyperbolic vertex Q and an s-line L1, Lə or L3 passes through points 
O, P or Q are shown in Fig.1.16(a), (b), (c), respectively. 

The map geometry is gotten by endowing an angular function 
u: V(M) — |0, 47) on an embedding M for generalizing Iseri’s model 


30 Chapter I Abstract Embeddings 


on surfaces following, which was first introduced in |Mao2]. 


Map geometry without boundary Let M be a combinato- 
rial map on a surface S with each vertex valency> 3 and u : V(M) > 
[0, 4), i.e., endow each vertex u, u € V(M) with a real number u(u), 0 < 
ulu) < DOR The pair (M, u) is called a map geometry without bound- 
ary, ulu) an angle factor on u and orientable or non-orientable if M 
is orientable or not. 


Map geometry with boundary Let (M,u) be a map geom- 
etry without boundary, faces fi, fo,--:, f; € F(M), 1 < L< e(M) — 
1. If S(M) \ (fu fo, fi) is connected, then (M, u)™ = (S(M) \ 
{ fi, foi fit, p) is called a map geometry with boundary fi, fo, fi, 
and orientable or not if (M, u) is orientable or not, where S(M) de- 
notes the underlying surface of M. 





pu (u)u(u) = 2n pu (u)u(u) > 2n 
Fig.1.17 


Certainly, a vertex u € V(M) with py(u)u(u) < 27, = 27 or 
> 2r can be also realized in a Euclidean space R?, such as those 
shown in Fig.1.17. 

A point u in a map geometry (M, u) is said to be elliptic, Eu- 
clidean or hyperbolic if py (u)u(u) < 27, py (u)u(u) = 2x or py(u)u(u) > 
2r. If u(u) = 60°, we find these elliptic, Euclidean or hyperbolic ver- 
tices are just the same in Iseri’s model, which means that these s- 
manifolds are a special map geometry. If a line passes through a point 
£u E W with the entering ray and equal to 
180° only when u is Euclidean. For convenience, we always assume 
that a line passing through an elliptic point turn to the left and a hy- 
perbolic point to the right on the plane. Then we know the following 





u, it must has an angle 


L5 Smarandache 2-manifolds with map geometry 31 


results. 


Theorem 1.16 Let M be an embedding on a locally orientable 
surface with |M| > 3 and py(u) > 3 for Vu € V(M). Then there exists 
an angle factor u : V(M) — [0,47) such that (M, u) is a Smarandache 
geometry by denial the axiom (E5) with axioms (E5) for Euclidean, 
(L5) for hyperbolic and (R5) for elliptic. 


Theorem 1.17 Let M be an embedding on a locally orientable 
surface with order> 3, vertex valency> 3 and a face f € F(M). Then 
there is an angle factor u : V(M) — [0, 47) such that (M, u)! is a 
Smarandache geometry by denial the axiom (E5) with axioms (E5) 
for Euclidean, (L5) for hyperbolic and (R5) for elliptic. 


A complete proof of Theorems 1.16-1.17 can be found in refer- 
ences in [Mao2-4]. It should be noted that the map geometry with 
boundary is in fact a generalization of Klein model for hyperbolic 
geometry, which uses a boundary circle and lines are straight line seg- 
ment in this circle, such as those shown in Fig.1.18. 


uu 
s 
A 
p 
di. 
Ea. 


Fig.1.18 


Activities on Chapter I 


1.6 Observations 


O1.1 Let X be a finite set and B be a binary set. Is (Bx|v € 
X} a pregraph or a graph? If unnecessary, what condition does a 
pregraph, or a graph satisfy? 
O1.2 Let X be a finite set, B be a binary set and 
Me Ba 
LEX 


For a permutation w on 4 such that V? = 1, is ((z,yz]|x € X} a 
graph? If unnecessary, when is it a graph? 


O1.3 How many orientable, or nonorientable surfaces can a 
hexagon represent? List all of them. 


O1.4 How many orientable, or nonorientable surfaces can a 2k- 
gon represent? How to List them all. 


O1.5 In [Liul], an embedding of a graph G on the nonori- 
entable surface of genus 3(G) is constructed in a way from a specific 
tree on G. Now, how to get such an embedding from any tree on G. 


O1.6 Suppose G has an embedding on an orientable surface 
of genus k. If k is not the maximum genus of G, how to find an 
embedding of G on an orientable surface of genus k + 1. 

O1.7 Any embedding of a graph G — (V, X) is a permutation 
on {1,a, 8, o 8X = X as shown by (1.17). However, a, 8 and y = a 
are each a permutation on ¥, but not an embedding of G in general. 
When does each of them determine an embedding of G. 


I.6 Observations 33 


O1.8 If permutation P on X is an embedding of a graph, to 
show that for any x € X, there does not exist an integer m such that 
ax = pra. 


O1.9 If permutation P on X is an embedding of a graph, to 
show 


PaP =a. 


O1.10 Observe that permutation 8 as P satisfies both O1.8 
and 01.9. However, permutation a as P satisfies 01.9 but not O1.8. 


O1.11 Let S be a set of permutations on ¥ and Wg be the 
group generated by S. If for any x,y € X, there exists a Y € Vg such 
that x = wy, then the group Ys is said to be transitive on X. Observe 
that if permutation P on X is an embedding of a graph, then group 
Vr,I-—1P,o,0), is transitive on X. 


An embedding of a graph G = (V, E) can be combinatorially 
represented by a rotation system o(V) on the vertex set V of G and 
a function on the edge set as A: E — B, B = {0,1}, denoted by 
GLA). 

In order to determine the faces of an embedding, an immersion 
of the graph and a rule should be established. 

An immersion of a graph is such a representation of the graph in 
the plane that vertices injects into the plane on an imaged circle and 
edges are straight line segments between their two ends. 


Travel and traverse rule From a point on one side of an edge, 
travel as long as on the same side until at the middle of a edge with 
A = 1, then traverse to the other side. 


O1.12 By the T' T-rule(i.e., the travel and traverse rule) on 
an immersion, the initial side the starting point is on can always be 
encountered to get a travel as a set of edges met on the way. 


O1.13 By the TT-rule on an immersion of a graph, a set of 
travels can always be found for any edge occurs exactly twice. 


34 Activities on Chapter I 


On a graph G — (V, E), a subset of edges C C E that V has a 
2-partition V = Vi + V2 with the property: 


C = {(u,v) € Flue Vive Vo} (1.25) 
is called a cocycle of G. 


O1.14 For an immersion G,(A) of graph G = (V, E), the em- 
bedding of G determined by G, (A) is orientable if, and only if, the set 
telVe € E, € (e) = 1} is a cocycle of G. 


1.7 Exercises 


E1.1 For a graph G, prove that G has no odd circuit(a circuit 
with odd number of edges) if, and only if, for a tree on G, G has no 
odd fundamental circuit. 

E1.2 Prove that a graph G — (V, E) has no odd fundamental 
circuit if, and only if, E itself is a cocycle. 

Let [ be a nonAbelian group. The identity is denoted by 1. 
Write To = (£|£? = 1,€ € T}, ie., the set of all elements of order 2. 
For a pregraph G = (V, E), let z, € I'o,v € V, be variables on the 
vertex set V and w(e) € To, e € E be a weight function on the edge 
set E. On the network N = (G; w), its incidence equation is 


Tuly = w(e) ec E. (1.26) 


E1.3 Prove that if the incidence equation has a solution, then 
it has at least |l'o|, i.e. , the number of elements in To, solutions. 


E1.4 Prove that the incidence equation has a solution if, and 
only if, G has no circuit C such that 


w(C) = [[ «o #1. 


e€C 


E1.5 Let o(G) be the number of connected components on pre- 
graph G. Prove that if the incidence equation has a solution, then it 


I7 Exercises 35 


has 
Pol" 


solutions. 


E1.6 For an orientable surface, provide a procedure for deter- 
mining its orientable genus and then estimate an upper bound of the 
operations necessary. 


E1.7 For a nonorientable surface, provide a procedure for de- 
termining its nonorientable genus and then estimate an upper bound 
of the number of operations necessarily used. 


E1.8 Let G bea graph of order 2 with all its edges selfloops but 
only one. Prove that G is not up-embeddable on orientable surfaces 
if, and only if, each vertex is incident with odd number of selfloops. 


E1.9 According to the orientable and nonorientable genera, list 
all embeddings of K4, the complete graph of order 4. 


A graph is called i-separable if it has i, i > 1, vertices such that 
it is not connected anymore when the 7 vertices with their incident 
edges are deleted. A set of i vertices separable without a proper subset 
separable for a graph is called an i-cut of the graph. A graph which 
has an i-cut without (i — 1)-cut is said to be i- connected. 


E1.10 Prove that if a 3-connected graph G is planar (embed- 
dable on the sphere), then it has exact 2 embeddings on the sphere. 


For a planar graph G — (V, E), and u,v € V, if G has the form 


Q= Mo. X (u, v) E E; (1.27) 
Gil JG» ~~ {(u,v)}, M (u, v) g B 
and 
Gi( 6G» = (u,v) (1.28) 


such that G and Gə are both with at least 2 edges, then {u,v} is 
called a splitting pair of G, and Gi and Gg, its splitting block. If a 
splitting block at a splitting pair has no proper subgraph is still a 
splitting block, then it is said to be standard|Macl]. 


36 Activities on Chapter I 


E1.11 Prove the following three statements. 

(i) Let A and B be two standard splitting blocks of a splitting 
pair, then they are no edge in common; 

(ii) For a splitting pair of a 2-connected planar graph G — (V, E), 
the standard splitting block decomposition of the edge set E, i.e., 


E-ELJEU--Uz 


such that E; N E; = 0, 1 < i # j € s, and the induced subgraphs of 
Ej;, lxi € s, are all standard splitting blocks of the splitting pair, is 
unique. 

(iii) Let b; be the number of standard splitting blocks of the 
ith splitting pair, 2 = 1,2,---,m, m be the number of splitting pairs, 
then the number of embeddings of a 2-connected planar graph on the 
sphere is 


obi o [[@ u D. 
j=l 


For a graph G, let II(G) be the set of rotation systems of inner 
vertices on a joint tree and w,(i), m € II, be the boundary polygon 
with its 9(G) pairs of letters whose indices are determined by a binary 
number i with G(G) digits by the rule: the two indices of the lth letter 
are same or different according as the lth digit of is 0 or 1. Define 


EM 
£u,()) = M. aa (1.29) 
k=-6(G) 


where xz is an undeterminate and 


l; if T ) Nto O , —k; 
a een Un er 84 (1.30) 
0, otherwise. 
Now, let 
Qr = {w,(i)| 0 € i < 299 — 1) (1.31) 
and 


£(0,) = 2. E(w,(7)). (1.32) 


I.8 Researches 37 


E1.12 Prove that the coefficient of z^ in 


&G)— M; &(0;) 


n€llI(G) 
is the number of embeddings of G on the surface of relative genus k, 
-6(G) < k < |). 
A graph of order 2 without selfloop is called a link bundle. 
E1.13 Let Lm be the link bundle of size m > 1. Determine 
(Lm). 
A graph of order 1 is also called a bouquet, or a loop bundle. 


E1.14 Let Bm be a bouquet of size m, m > 1. Determine 
E(Bm). 

A graph of order 2 is called a bipole. Of course, a link bundle is 
a bipole which has no selfloop. 


E1.15 Let P, be a bipole of size m, m > 1. Determine (Pm). 


1.8 Researches 


The set(repetition at most twice of elements permitted) of edges 
appearing on a travel can be shown to have a partition of each subset 
forming still a subtravel except probably the travel itself. Such a 
partition is called a decomposition of a travel into subtravels. However, 
it is not yet known if any travel can be decomposed into tours except 
only the case that its induced graph has a cut-edge. 


R1.1 Prove, or disprove, the conjecture that a travel with at 
most twice occurrences of an edge in a graph has a decomposition 
into tours if, and only if, the induced subgraph of the travel is without 
cut-edge. 


Because a circuit is restricted from a tour by no repetition of a 
vertex, the following conjecture would look stronger the last one. 


R1.2 Prove, or disprove, the conjecture that a travel with at 


38 Activities on Chapter I 


most twice occurrences of an edge in a graph has a decomposition into 
circuits if, and only if, the induced subgraph of the travel is without 
cut-edge. 


However, it can be shown from Theorem 1.3 that any tour has 
a decomposition into circuits. The above two conjectures are, in fact, 
equivalent. Because a cut-edge is never on a circuit, the necessity 
is always true. A travel with three occurrences of an edge permitted 
does not have a decomposition into circuits in general. For example, on 
the graph determined by Par= {{x(0), y(0)}, (z(1), y(1)}}, the travel 
zx ry | where x = (x(0), z(1)) and y = (y(0), y(1)) has no circuit 
decomposition. 

Furthermore, the two conjectures are apparently right when the 
graph is planar because each face boundary of its planar embedding 
is generally a tour whenever without cut-edge. 


R1.3 For a given graph G and an integer p, p > 0, find the 
number n,(G) of embeddings of G on the orientable surface of genus 
p. 

The aim is at the genus distribution of embeddings of G on ori- 
entable surfaces, i.e., the polynomial 


[c/2] 
Po(G) = 5 | m(G)2”, 


p=0 
where o is the Betti number of G. 


For p = 0, no(G) can be done based on |Liu6]. If G is planar, 
O1.11 provides the result for 2-connected case. Others can also be 
derived. As to justify if a graph is planar, a theory can be seen from 
Chapters 3,5 and 7 in [Liu5]. 

Generally speaking, not easy to get the complete answer in a 
short period of time. However, the following approach would be avail- 
able to access this problem. Choose a special type of graphs, for 
instance, a wheel(a circuit Cn all of whose vertices are adjacent to an 
extra vertex), a generalized Halin graph(a circuit with a disjoint tree 


I.8 Researches 39 


except for all articulate vertices forming the vertex set of the circuit) 
and so forth. 

Of course, the technique and theoretical results in 1.3 can be 
employed to calculate the number of distinct embeddings of a graph 
by hand and by computer. 


R1.4  Orientable single peak conjecture. The coefficients of the 
polynomial in R1.3 are of single peak , i.e., they are from increase to 
decrease as p runs from 0 to |a/2]|(> 3), n, (G). 


'The purpose here is to prove, or disprove the conjecture not nec- 
essary to get all n,(G), 0 € p € [c /2](2 3). 


R1.5 Determine the number of distinct embeddings, which have 
one, or two faces, of a graph on orientable surfaces. 


R1.6 Fora given graph G and an integer q, q > 1, find the num- 
ber (G) of distinct embeddings on nonorientable surfaces of genus 
q. 

The aim is at the genus polynomial of embeddings of G on nonori- 
entable surfaces: 


Py(G) = 5 (G), 


where o is the Betti number of G. 


Some pre-investigations for G is that a wheel, or a generalized 
Halin graph can firstly be done. 


R1.7 Fora graph G, justify if it is embeddable on the projective 
plane, and then determine 744(G) according to the connectivity of G. 


R1.8 For a graph embeddable on the projective plane, determine 
how many sets of circuits such that for each, all of its circuits are 
essential if, and only if, one of them is essential in an embedding of G 
on the projective plane. 


R1.9  Nonorientable single peak conjecture. The coefficients of 
the polynomial in R1.6 are of single peak in the interval [o, 0] where 
c is the Betti number of G. 


40 Activities on Chapter I 


R1.10 For a given type of graphs G and an integer p, find the 
number of distinct embeddings of graphs in G on the orientable surface 
of genus p. Further, determine the polynomial 


lo(G)/2| 
Po(G) = $, mp (G)a” 
p=0 
where c(G) = max{a(G)|G € G}. 


R1.11 For a given type of graphs G and an integer q, q > 1, 
find the number of embeddings of graphs in G on the nonorientable 
surface of genus g. Further, determine the polynomial 


o(G) 


Py(G) = 5 fig(G)x4 


where o(G) = max{a(G)|G € G}. 


R1.12 For a set of graphs with some fixed invariants, extract 
sharp bounds(lower or upper) of the orientable minimum genus and 
sharp bounds(lower or upper) of orientable maximum genus. 


Here, invariants are chosen from the order (vertex number), size(edge 
number), chromatic number (the minimum number of colors by which 
vertices of a graph can be colored such that adjacent vertices have 
distinct colors), crossing number(the minimum number of crossing in- 
ner points among all planar immersions of a graph), thickness (the 
minimum number of subsets among all partitions of the edge set such 
that each of the subsets induces a planar graph), and so forth. 


R1.13 For a set of graphs and a set of invariants fixed, pro- 
vide sharp bounds(lower or upper) of minimum nonorientable genus 
of embeddings of graphs in the set. 


Chapter II 
Abstract Maps 


e A ground set is formed by the Klein group K = {1, a, 8, a3} stick- 
ing on a finite set X. 


e A basic permutation is such a permutation on the ground set that 
no element x is in the same cycle with az. 


e The conjugate axiom on a map is determined by each vertex con- 
sisting of two conjugate cycles for o, as well as by each face con- 
sisting of two conjugate cycles for 9. 


e The transitive axiom on a map is from the connectedness of its 
underlying graph. 


e An included angle is determined by either a vertex with one of its 
incident faces, or a face with one of its incident vertices. 


II.1 Ground sets 


Given a finite set X = (z1,25,:--, £m}, called the basic set, its 
elements are distinct. Two operations œ and @ on X are defined 
as for any z € X, az # Bx, ax, Gx d X and o?z = afar) = z, 
B?x = B(8x) = x. Further, define a3 = fa = y such that for any 
x € X, yx ~#ax, px and yx d X. 


42 Chapter II Abstract Maps 


Let aX = [ox|Vvzr € X}, 8X = iüx|Vx € X} and yX = 
{ya|\Vx € X}, then o, B and y determine a bijection between X and, 
respectively, aX, GX and yX. By a bijection is meant a one to one 
correspondence between two sets of the same cardinality. In other 
words, 

X()ex-x(|ex-x[(]|yx-8 
aX ()BX = BX()\yX 24X[(]ox - (2.1) 
JaX| = [BX| = |yX| = |X]. 
The set X UaX UGX UyX, or briefly A = V(X), is called the ground 
set. 

Now, observe set K = 4{1,a,8,y}. Its elements are seen as 
permutations on the ground set X. Here, 1 is the identity. From 
o? =p =1 and af = Ba, 34? = (aB)(aB) = a(88)a = o? = 1, and 
hence 


a= (Bx, Vay); 


=: 
=: 


Il 
H 
= 
Il 
ra 


(£i, azi) 


(zi, Bai) | | (ovi, voi); (2.2) 


ll 
=: 
= 


II 
H 
m 


jai 


Si 


(zi, Yi) lio. aTi). 

i=1 
It is easily seen that K is a group, called Klein group because it is 
isomorphic to the group of four elements discovered by Klein in geom- 


sc 


=: 


Il 
H 


etry. 
For any z € X, let Kx = {x, ax, Bx, yx}, called a quadricell. 


Theorem 2.1 For any basic set X, its ground set is 
Ty in (2.3) 
LEX 
where the summation represents the disjoint union of sets. 


Proof From (2.1), for any z,y € X and z Z y, 
Ke( Ky = (x, arf, 32) ( Ku, ay, Bv, yy} = 9. 


IL2 Basic permutations 43 


From (2.1) again, 





|) Kc2 X c oX - 8X e 4X =X. 
rcx 


Therefore, (2.3) is true. O 


Furthermore, since for any x € X, 
Ring eA eR ow) = he, (2.4) 


we have 


Xy Hye Koc Ki 


ycoa X zEßX teyX 


This implies that the four elements in a quadricell are with symmetry. 


II.2 Basic permutations 


Let Per(X), or briefly Per, be a permutation on the ground set ¥. 
Because of bijection, according to the finite strict recursion principle, 
for any x € X, there exists a minimum positive integer k(x) such that 


Per (95 = Ds 
i.e., Per contains the cyclic permutation , or in short cycle, 
C= Per?z, +--+ , Per* (9715). 


Write {x£ }per as the set of all elements in the cycle (z)pa. Such a set is 
called the orbit of x under permutation Per. The integer k(x) is called 
the order of x under permutation Per. 

If for any x € X, 


az É {x}Per, (2.5) 


then the permutation Per is said to be basic to a. 


Example 2.1 From (2.2), for permutations a and D on the 
ground set, o is not basic, but 8 is basic. 


44 Chapter II Abstract Maps 


Let Par= (Xi, Xo,---,X,} be a partition on the ground set X, 
then 


Per = | [(x) (2.6) 
i=1 

determines a permutation on 4, called induced from the partition Par. 
Here, (X;), 1 € i € s, stands for a cyclic order arranged on X;. This 
shows that a partition (.X;|1 <i € s} has 

Ixi- 212 Ie - 0^ (2.7) 

s-l i23 
induced permutations. In (2.7), n;, à > 3, is the number of subsets in 
Par with 7 elements. 


Theorem 2.2 Let Par= (X;|]1 € i € s) be a partition on the 
ground set V(X). If Par has an induced permutation basic, then all 
of its induced permutations are basic. Further, a partition Par has its 
induced permutation basic if, and only if, for x € X, there does not 
exist Y € Par such that 


{x ax} or {8x, yz} C YAN Kr: (2.8) 


Proof Because the basicness of a permutation is independent of 
the order on cycles, the first statement is proved. 

Assume an induced permutation Per of a partition Par is basic. 
From (2.5), for any x € X, in virtue of 


Ke ={z,axr}+ 18r; yr}, 


no Y €Par exists such that (2.8) is satisfied. This is the necessity of 
the second statement. 

Conversely, because for any x € X, no Y € Par exists such that 
(2.8) is satisfied, it is only possible that x and az are in distinct subsets 
of partition Par. Therefore, ax Z (x]pa. Based on (2.5), this is the 
sufficiency of the second statement. [] 


On the basis of this theorem, induced basic permutations can be 
easily extracted from a partition of the ground set. 


IL2 Basic permutations 45 


Example 2.2 Let ¥ = {z,az,(@x,yx} = Kx. There are 15 
partitions on X as 


Par, = (iz); {az}, {8r}, Uy]; 

Pato =4 {e097}, {Orhi 

Pars = {{x, Bx}, {az}, {yr}}; 

Par, = (iz, yx}, {ax}, (82) 5 

Pars = (1682, ax}, {x}, {yz}}: 

Parę = ((x (ox, yx}, {Gx}}; 

Par; = {{z}, {ax}, (82, yx}}; 

Pare ed wr ore) 

Parg = (iz, Bx, yx}, (6x) ); 

Pari = {{z, ax, yx}, {ax}}; 

Part = liz. 102,09 0mFE 

Pario = {{x£, ax, Bx, yz E 

Pars = {1r arh (eae) | 

Parya = {{z, Bx}, fax, yz] ; 

Paris = {{z, 7x}, fax, Bx}}. 
From Theorem 2.2, induced basic permutations can only be extracted 
from Par,, Pars, Pary, Pars, Parę, Parj, and Par;; among them. Since 
each of these partitions has no subset with at least 3 elements, from 


(2.7) it only induces 1 basic permutation. Hence, 7 basic permutations 
are induced in all. 


Based on Theorem 2.2, a partition that induces a basic permu- 
tation is called basic as well. 

For a partition Par on X, if every pair of x and oz, x € X, deals 
with the element x in Par, then this partition determines a pregraph 
if any. For example, in Example 2, there are only 


Pon = Pars = Pary = Pate ={{7},4 0211: 
Partie = Paria = Paris = {{x, 8£}} 


form 2 premaps of size 1 and others meaningless among the 15 par- 


46 Chapter II Abstract Maps 


titions. Further, each of the 2 premaps is a graph. The result is the 
same as in Example 1.1. 


II.3 Conjugate axiom 


Let Per; and Per» be two permutations on the ground set A. If 
for z € X, 


(Peroz)per, = Per(z)pg-: = Pero(®) pop, (2.9) 
then the two orbits (x)per, and (Perox)pa, of Per, are said to be con- 
jugate. 

For a permutation 7 on the ground set X, if 


aP =P ‘a, (2.10) 


then (P, a)(or for the sake of brevity, P) is called satisfying the con- 
jugate axiom. 


Theorem 2.3 For a basic permutation P on the ground set X, 
the two orbits (x)p and (ax)p for any x € X are conjugate if, and 
only if, (P, a) satisfies the conjugate axiom. 


Proof Necessity. Because of orbits (x)p and (ax)p conjugate, 
from (2.9), (ax)p = a(x)5'. Hence, Pax = aP 12, i.e., Pa = «P1. 
This implies (2.10). 

Sufficiency. Since P satisfies (2.10), 

P(oz) = P(o'P)P-!z = P(P-la)P-!z 
= (PP aP lax = aP™tr. 
By induction, assume that P!(ax) = aP~'x, 1 > 1, then we have 
Plax) = PP'(ox) = PPr 
= P(aPP')P s ( then by (2.10)) 
(PP ap Vz 


sap ity, 


II.3 Conjugate axiom 4T 


Hence, (ax)p = a(x)5'. From (2.9), orbits (z)p and (ax)p are conju- 
gate. This is the sufficiency. [] 


Unlike Theorem 2.2, for a partition on the ground set X, from 
one of its induced permutations satisfying the conjugate axiom, it can 
not be deduced to others. 

A premap, denoted by (A5, P), is such a basic permutation P 
on the ground set X that the conjugate axiom is satisfied for (P, a). 


Example 2.3 By no means any basic partition is in companion 
with a basic permutation. Among the 7 basic partitions as shown in 
Example 2.2, only the induced permutations of Par,, Parj4 and Paris 
are premaps. 


Because (P, 3) is not necessary to satisfy the conjugate axiom 
on a premap (4X, P), a is called the first operation and (3, the second 
. Thus, X should be precisely written as Vg if necessary. 

Based on the basicness and Theorem 2.3, any premap P has the 
form as 

II o}. (2.11) 
LEX p 
where Xp is the set of distinct representatives for the conjugate pairs 
{{x}p, {x}5} of cycles in P. And further, 


x= Y {xhp{a}p. (2.12) 


For convenience, one of two conjugate orbits in {{a}p, {x }p} is chosen 
to stand for the pair itself as a vertex of the premap. 


Example 2.4 Let X = {21,22}, then 


X = (x1am 21,21, 25, AX2, BL, YL}. 
Choose 
Pi = (21, 8x1) (o3, 21) (22) (o2) (822) (2) 
and 
Pa = (z1, 821, T2) (o1, AX2, yz1) (x2) (^yza). 


48 Chapter II Abstract Maps 


The former has 3 vertices (x1, 971), (v2) and (8x3). The latter has 2 
vertices (2, 02, £2) and (zs). 


Lemma 2.1 If permutation P on Xag is a premap, then 
P'8—g8p* (2.13) 
where P* = Py, y = af. 


Proof Because P*3 = Pa = Pa, from the conjugate axiom, 


p*g —op 
= BBoP^' (Al 6? = 1) 
= B((PaB) ^) 
= gp *, 
Therefore, the lemma holds. C 


This lemma tells us that although 8 does not satisfy the conju- 
gate axiom for permutation P in general, 9 does satisfy the conjugate 
axiom for permutation 7*. 


Lemma 2.2 If permutation P on A, is a premap, then per- 
mutation P* = Py is basic for 6. 


Proof Because the 4 elements in a quadricell are distinct, r Æ 
Bats 

Case 1 P*x # Gx. Otherwise, from P*r = Gx, Pa(Gxr) = 
P*x = Bx. A contradiction to that P is basic for a. 

Case 2 (P*)*x Z Bx. Otherwise, P*z = P* !8z. From Lemma 
2.1, P*x = BP*x. A contradiction to that P*x and GP*z are in the 
same quadricell. 

In general, assume by induction that Case l: (P*)'y Z Bx, 1 < 
| € k, k 2 2, is proved. To prove 

Case k+1 (P*)'*!z 4 Bx. Otherwise, P*"z = P* 18x. From 
Lemma 2.1, P*^ !(P*z) = 8(P*z). A contradiction to the induction 
hypothesis. 


II.4 Transitive axiom 40 


In all, the lemma is proved. [] 


Theorem 2.4 Permutation P on X is a premap (Xap, P) if, 
and only if, permutation P* = Py is a premap (A54, P*). 


Proof Necessity. Since permutation P is a premap (4,5, P), 
P is basic for a and satisfies the conjugate axiom. From Lemma 2.2 
and Lemma 2.1, P* is basic for @ and satisfies the conjugate axiom. 
Hence, P* is a premap (A4, P"). 

Sufficiency. Because P** = P*y = P, the sufficiency is right. O 


On the basis of Theorem 2.4, the vertices of premap P* are de- 
fined to be the faces of premap P. The former is called the dual of 
the latter. Since P** = P, the latter is also the dual of the former. 


Example 2.5 For the two premaps Pı and P> as shown in 
Example 2.4, we have 
Pt = (z1,021)(821, yz1) (22, Yr) (822, 022) 
and 
P5 = (71, 023,95, 943) (Da, (35, Dto, ^21). 


Because T has 2 vertices (r1, axı) and (x2, yx2) and P3 has 1 vertex 
(24, a4, £2, yX2), we seen that premaps Pı and P» have, respectively, 
2 faces and 1 face. 


II.4 ‘Transitive axiom 


For a set of permutations T = (r;|1 € i € k}, k > 1 on X, let 


s k 
Wr = {yv = III. im) EZ meEn,s >21}, (214) 


l=1 j=1 


where Z is the set of integers and II is the set of all permutations on 
{1,2,---, k}, k>1. 


50 Chapter II Abstract Maps 


Since all elements in V7 are permutations on X, they are closed 
for composition(or in other word, multiplication) with the associative 
law but without the commutative law. 

Further, it is seen that a permutation in V7 if, and only if, its 
inverse is in V. The identity is the element in V7 when all i;(7) = 0, 
7 € ILin (2.14). Therefore, V7 is a group in its own right, called the 
generated group of T. 

Let Y be a permutation group on X. If for any z,y € X, there 
exists an element Y € V such that x = wy, then the group V is said 
to be transitive , or in other words, the group W satisfies the transitive 
axiom. 

Now, consider a binary relation on X, denoted by ~y, that for 
any z,y € X, 

zr yyy eJ eY. xax. (2.15) 


Because the relation ~y determined by (2.15) for a permutation 
group V on & is a equivalence, & is classified into classes as V/~yw. 


Theorem 2.5 A permutation group V on X is transitive if, and 
only if, for the equivalence ~y determined by (2.15), |¥/ ~y | = 1. 


Proof Necessity. From the transitivity, for any x,y € X, there 
exists Y € V such that x = wy. In view of (2.15), for any z, y € X, 


x ~y y. Hence, for ^, |X/ ey | = 1. 

Sufficiency. Because |X/ ~y | = 1, for any z,y € X, exists 
i» € V such that x = wy. Therefore, the permutation group V on X 
is transitive. L 


For a premap, the pregraph with the same vertices and edges as 
the premap is is called its under pregraph. Conversely, the premap is 
a super premap of its under pregraph. 

Let (Xag, P) be a premap. If permutation group V;, J = 
(o, B, P), on the ground set X, g is transitive, i.e., with the transitive 
axiom, then the premap is called a map . 


Lemma 2.3 Let M = (Xag, P) be a premap. For any x,y € 


II.4 Transitive axiom 51 


Xap, exists Y € Uz, J = (o, 0, P), such that x = wy if, and only if, 
there is a path from the vertex v, x is with to the vertex v, y is with 
in the under pregraph of M. 


Proof Necessity. In view of (2.14) with the conjugate axiom, 

write 
V = a Pra Ba Pa” GB MY Ba Pha, 

where 6;,0; € {0,1}, 1 < i < s, andl; € Z, 1 € i € s. Because 
as Phas and a? Pha?! are, respectively, acting on vertices vy and vy, 
V determines a trail from v, to v, of s — 1 edges. Since there is a trail 
between two vertices if, and only if, there is a path between them, the 
necessity is done. 

Sufficiency. let (us, Us_1, +++, V1),Us = Vy, V1 = Vz, be a path from 
vy to v; in the under pregraph of M. Then, there exist ô; o; € {0,1}, 
1<i<s, andl; E€ Z, 1€ 4 € s, such that 


y — a Pig. -" Bad P's a 
and z = wy. From (2.14), the sufficiency is done . [] 


Theorem 2.6 A premap is a map if, and only if, its under 
pregraph is a graph. 


Proof From the transitive axiom and Lemma 2.3, its under pre- 
graph is a graph. This is the necessity. 

Conversely, from the connectedness and Lemma 2,3 , the premap 
satisfies the transitive axiom and hence its under pregraph is a graph. 
This is the sufficiency. [] 


Example 2.6 In Fig.1.2, the pregraph determined by Par; has 
2 super premaps: 


P, = (z1)(oa1) (y21) (821) (2, x2) (0/2, 822) 


and 
P, = (z1) (azı) (y21) (821) (x2, 822) (o2; *y2) 


as shown in Fig.2.1. 


52 Chapter II Abstract Maps 


(22, 22) (z2, Bx) 


Pi P» 


Fig.2.1 Two super premaps 


From Theorem 2.6, none of the two super premaps is a map in 
Fig.2.1. 


However, the pregraph determined by Parı2 is a graph in Fig.1.2. 
From Theorem 2.6, each of its super premaps is a map as shown in 
the following example. 


Example 2.7 In Fig.2.2, there are 273! = 24 distinct embed- 
dings of the graph determined by Par» in Fig.1.2. On the associate(or 
boundary) polygon of the joint tree, the pair of a letter is defined to be 
of distinct powers when x and yx appear; the same power otherwise. 


yi yr T2 
vU vU vU 
f NE Ya NE fe 
YVL2Q T2 YyT2 


(a) Oo (b) Oo (c) O1 


II.4 Transitive axiom 


x A 
Tj 
U 
f£ o NOn 
T2 


(e) O1 


Ba 
TIN 7 
vU 
$ Nyx2 
X2 


(h) Ni 


x2 
my Z 
vU 
7 b 
T2 


(k) No 


yTı 
TIN 7 
vU 
25 b 
T2 


(n) Ni 


wy 
TIN 
vU 
f Nu 
"yw 


(f) Oo 


T2 
U 
N X1 
x2 


(i) No 


A . Nro 
Via X 


54 


ON OZ ANZ 
N a2 4 Noam 
Jor AN 


(p) Ni (q) No 


hd M 
r1\ Z7 aN Z7 
v U 
4 Nu» f£ No 
Bx» T2 


(s) No (t) No 


We ANZ 
ANS AX 
(v 


) No (w) Ni 


Fig.2.2 All embeddings of a graph 


Chapter II Abstract Maps 


NY 


(r) Mi 


WE d 
gn Ly 
Joan © 


(u) Ni 


Bx 
EX J 
A . Nro 
Bx2 


(x) No 


In Fig.2.2, the graph determined by Parjo(Fig.1.2) has 6 ori- 
entable embeddings (a—f). Here, (a), (b), (d) and (f) are the same 
map on the sphere Op ^o (%12_). And, (c) and (e) are the same map 


on the torus O4 ~top (tirar. 


are, in fact, 2 maps. 


151). Hence, such 6 distinct embeddings 


Among the 18 nonorientable embeddings, 10 are on the projective 
plane and 8 are on the Klein bottle. On the projective plane, (g), (h), 
(i), (D, (m), (n), (p) and (r) are the same map (Ny ep (x12, /xo23)). 


II.5 Included angles 55 


And, (u) and (w) are another map (Nj = (21%2%12%2)). On the Klein 
bottle, (i), (k), (o) and (q) are the same map (N2 = (xizsz1 'a)). 
And, (s), (t), (v) and (x) are another map (No ~top (%1%1%2%2)). 
Therefore, there are only 4 maps among the 18 embeddings. 


II.5 Included angles 


Let M = (Xa B(X), P) be a premap. Write k = |X5,5(.X)/ ov, |, 
J = 1oa,B,P],and 
k k 


Xa p(X) = 3 Aa), X= ^. xi 


i=l i=1 
where A4,5,(X) € Xa8/ ~w,, o; and fj; are, respectively, a and 8 
restricted on Æw; s, (Xi), à = 1,2,---, k. Further, 
k 
M — 3 Mi, Mi = (Xs (X), Pi), (2.16) 
i=l 
where M; is a map and P; is P restricted on X4, a (Xi), i — 1,2,---, k. 
This enables us only to discuss maps without loss of generality. 
Lemma 2.4 Any map (X, P) has that (Pa)? = 1. 
Proof From the conjugate axiom, 
(Pa)? = (Pa)(Pa) = P(aP)a 
= (PP (ao) = 1. 


This is the conclusion of the lemma. O 


Lemma 2.5 Any map (¥,P) has that (P*8)? = 1. 


Proof Because P*3 = Py3 = Pa( 66) = Po, from Lemma 2.4, 
the conclusion is obtained. O 


On the basis of the above two lemmas, on a map M = (Xag, P), 
any z € Xa g has that 


(r, Pos = (F Drm) (2.17) 


56 Chapter II Abstract Maps 


Thus, (x, Paz), or (P*Gx,x), is called an included angle of the map. 
For an edge Ka = {x, ax, Bx, yx} of M, (x, ax] and {8x, yx} are its 
two ends , or semiedges. And, (x, 8x} and {az, yx} are its two sides, 
or cosemiedges. 


Theorem 2.7 Fora (4(X),P), let V = (v|v = (x) pU (ox jp, 


Vr € X} and F = (flf = {x}p-U {Bx}p-, Vr € X). If z, and v; 
are, respectively, in v and f as representatives, then 


A= p» aP Ix, 4 {Pry,aty}+--- 
veV 


+ (Plz, aP~*2,}) 
= (zn BP ay} + {Pray Bas} +. 


feF 


+ {Pt lay, BP* rg). 


(2.18) 


Proof From X = „eyv and the conjugate axiom, 


v = (zy, Pazo} + {Px,,PaPx,}+--- 
HOP zu PaP zu) 
= (z,, P^ Iz.) + {Ptn aty} +>: 
HIP aP tuk 


This is the first equality. 
The second equality can similarly be derived from X = } jep f 
and Lemma 2.1(the conjugate axiom for P* with 8). O 


It is seen from the theorem that the numbers of included angles, 
semiedges and co-semiedges are, each, equal to the sum of degrees of 
vertices. Since every edge has exactly 2 semiedges, this number is 2 
times the size of the map. 


Activities on Chapter II 


II.6 Observations 


O2.1 For a set of 4k elements, observe how many ground sets 
can be produced. 


O2.2 By a set of 12 elements as an example, observe how many 
basic permutations satisfy the conjugate axiom. 


O2.3 In Example 2, observe which nonbasic partition has a 
permutation with the conjugate axiom. 


O2.4 Provide two embeddings with the same under graph. 


O2.5 For an embedding, observe if its mirror image is the same 
as itself. How about for a map? 


O2.6 For a map (A,,5, P), observe the orbits of permutations 
Pa and aP. Whether, or not, they are a map. 


O2.7 For a map (A5, P), observe the orbits of permutations 
PG and BP. Whether, or not, they are a map. 


O2.8 On a map (4,5, P), whether, or not, the permutation 
Pa is a map on the same ground set. 


A map with each of its face a quadrangle is called a quadrangu- 
lation. A map with only triangular faces is a triangulation. 


In general, a map with all vertices of the same degree is said to 
be vertex regular, or primal regular. If all faces are of the same degree, 
the the map is said to be face regular , or dual regular. A primal 
regular map with its vertex degree i, i > 1, is called an i-map. A 


58 Activities on Chapter II 


dual regular map with its face degree 7, 7 > 1, is called a 7*-map. A 
triangulation and a quadrangulation are, respectively, a 3*-map and a 
4*-map in their own right. 


O2.9 Whether, or not, the under graph of a 4*-map is always 
bipartite. Furthermore, whether, or not, the under graph of a (2k)*- 
map, k > 3, is always bipartite. 


O2.10 Observe that for any integer i, ? > 1, whether, or not, 
there always exists an 7-map and an i*-map. 


If the degree of any vertex(or face) of a map is only an integer 
i, or j, j Zz i, i,j > 1, then the map is called a (i, j)-map (or (i*, 7*)- 
map). Similarly, the meanings of a (i, 7*)-map and a (i*,7)-map are 
known. 


O2.11 For any i,j, i Æ 
(i, j)-map, a (i*, j*)-map, or a (i, 


j = 1, whether, or not, there is a 
j')map. 


II.7 Exercises 


E2.1 For two permutations Per; and Per» on a set of 4 elements 
with Per,;? = Per? = 1, list all the generated groups V (Per, Pero} 
Show if each of them is isomorphic to the Klein group. 


E2.2 Let A and u be permutations on the set A and M = u? = 
1. Prove that if the generated group V5, is isomorphic to the Klein 
group, then A is a ground set if, and only if, A = Ay + A» +--+ Ag, 
k > 1, such that Aly, and uļ|4, are both have two orbits, and A|4, Æ 
Bx L9. 








—1 


E2.3 For a map (45,5, P), prove that a(x),p = (ax)p., 


E2.4 For a map (Xag, P), prove that (z),p and (Gx),p are 
conjugate. 


E2.5 To prove that all planar embeddings of K4, i.e., the com- 
plete graph of order 4, are the same map. 


A map is said to be nonseparable if its under graph is nonsepa- 


II.8 Researches 59 


rable, i.e. , no cut-vertez(its deletion with incident edges destroys the 
connectedness in a graph). 


E2.6 List all nonseparable maps of size 4. 


A map with all vertices of even degree is called a Euler map. If 
the face set of a map can be partitioned into two parts each of which 
no two faces have an edge in common, then the face partition is called 
a edge independent 2-partition . 


E2.7 Provide and prove a condition for the face set of a Euler 
map having an edge independent 2-partition. 


E2.8 Prove the following statements: 
(i) The permutation 8 on the ground set ¥ is a map if, and only 
it = Kr: 
(ii) The permutation a on the ground set A is a map if, and 
only if, 
X = Ke = {x, pz, az, yz}, 


i.e., B is the first operation. 


E2.9 Fora map M = (a6, P), 

(i) Provide a map M and an integer i, i > 2, such that P’ is not 
a map; 

(ii) For i, i > 2, provide the condition such that P’ is still a 
map. 

E2.10 Let C and D be, respectively, the sets of 3-maps and 


3*-maps. For the size given, provide a 1—to-1 correspondence between 
them. 


IL.8 Researches 


From Theorem 1.10 in Chapter I, any graph has an embedding on 
a surface (orientable or nonorientable). However, if an embedding is 
restricted to a particular property, then the existence is still necessary 
to investigate. If a map has each of its faces partitionable into circuits, 


60 Activities on Chapter II 


then it is called a favorable map. If a graph has an embedding which is 
a favorable map, then the embedding is also said to be favorable|Liu12]. 


R2.1 Conjecture. Any graph without cut-edge has a favorable 
embedding. 


It is easily checked and proved that a graph with a cut-edge does 
not have a favorable embedding. However, no graph without cut- 
edge is exploded to have no favorable embedding yet. Some types of 
graphs have be shown to satisfy this conjecture such as Kn, n > 3; 
Kmn, m,n > 2, Qn, n 7 2, planar graphs without cut-edge etc. 

A map which has no face itself with a common edge is said to 
be preproper . It can be shown that all preproper maps are favorable. 
However, the converse case is unnecessary to be true. 


R2.2 Conjecture. Any graph without cut-edge has a preproper 
embedding. 


Similarly, it is also known that any graph with a cut-edge does 
not have a preproper embedding. And, Kn, n > 3; Kmn, m,n > 2, 
Qn, n > 2, planar graphs without cut-edge etc are shown to satisfy 
the conjecture as well. 


Furthermore, if a map has each of its faces a circuit itself, then it 
is called a proper map, or strong map. Likewise, proper embedding, or 
strong embedding . It can be shown that all proper maps are preproper. 
However, the converse case is unnecessary to be true. 


R2.3 Conjecture. Any graph without cut-edge has a proper 
embedding. 


For proper embeddings as well, it is known that any graph with 
a cut-edge does not have a proper embedding. And, Kn, n > 3; Kmn, 
m,n > 2, Qn, n 2 2, planar graphs without cut-edge etc are all shown 
to satisfy this conjecture. 

Although conjectures R2.1—R2.3 are stronger to stronger, be- 
cause R2.3 has not yet shown to be true, or not, the the two formers 
are still meaningful. 


If a favorable(proper) embedding of a graph of order n has at 


II.8 Researches 61 


most n — 1 faces, then it said to be of small face . 

Now, it is known that triangulations on the sphere have a small 
face proper embedding. Because triangulations of order n have exactly 
3n — 6 edges and 2n — 4 faces, all the small face embeddings are not 
yet on the sphere for n > 4. 


R2.4 Conjecture. Any graph of order at least six without cut- 
edge has a small face proper embedding. 


Because it is proved that K; has a proper embedding only on the 
surfaces of orientable genus 1(torus) and nonorientable genus at most 
2(Klein bottle) |WeL1], they have at least 5 faces and hence are not of 
small face. 


R2.5 Conjecture. Any nonseparable graph of order n has a 
proper embedding with at most n faces. 


If a map only has i-faces and i + 1-faces, 3 < i < n — 1, then it 
is said to be semi-regular. 


R2.6 Conjecture. Any nonseparable graph of order n, n > 7, 
has a semi-regular proper embedding. 


In fact, if a graph without cut-edge has a cut-vertex, then it can 
be decomposed into nonseparable blocks none of which is a link itself. 
If this conjecture is proved, then it is also right for a graph without 
cut-edge. Some relationships among these conjectures and more with 
new developments can be seen in [Liu12]. 


R2.7 For an integer i > 3, provide a necessary and sufficient 
condition for a graph having an 7-embedding, or 2*-embedding. Par- 
ticularly, when 7 = 3,4 and 5. 


First, start from 2 = 3 with a given type of graphs. For instance, 
choose G — K,, the complete graph of order n. For 3*-embedding, on 
the basis of Theorem 1.12( called Euler formula), a necessary condition 
for Ky, n > 3, having an 3*-embedding is 

—1 
n- 20 D 5 =2— 2p and 36 — n(n - 1), 
where $ and p are, respectively, the face number of an 3*-embedding 


62 Activities on Chapter II 


and the genus of the orientable surface the embedding is on. It is 
known that the condition is still sufficient. 
If nonorientable surfaces are considered, the necessary condition 


n- 20 D 5g = 2~q and 36 = n(n - 1), 
u (n — 3)(n — 4) 
n —3)(n—4 
a — — 3 


where q is the nonorientable genus of the surface an 3*-embedding 
is on is not sufficient anymore for g > 1. because when n = 7, K; 
would have an 3*-embedding on the surface of nonorientable genus 
q = 2. However, it is shown that Ky is not embeddable on the Klein 
bottle([Lemma 4.1 in |Liu11]). It has been proved that except only 
for this case, the necessary condition is also sufficient. 

More other types of graphs, such as the complete bipartite graph 
Kmmn, n-cube Qn and so forth can also be seen in [Liu11]. 


R2.8 For 3 < i,j < 6, recognize if a graph has an (i, j)- 
embedding ( or (i*, j*)-embedding). 
More generally, investigate the upper or/and the lower bounds 


of i and j such that for a given type of graphs having an (i*, j*)- 
embedding. 


R2.9 Given two integers 7,7 not less than 7, justify if a graph 
has an (i, j)-embedding(or (7*, j*)-embedding). 

If a proper map has any pair of its faces with at most 1 edge in 
common, then it is called a polygonal map . 


R2.10 Conjecture. Any 3-connected graph has an embedding 
which is a polygonal map. 


From (a) and (b) in Fig.2.3, this conjecture is not valid for non- 
separable graphs. The two graphs are nonseparable. The graph in (a) 
has a multi-edge, but that in (b) does not. It can be checked that, 
none of them has a polygonal embedding. 


II.8 Researches 63 


(a) (b) 


Fig.2.3 Two graphs without polygonal embedding 


If an embedding of a graph has its genus(orientable or nonori- 
entable) minimum among all the embeddings of the graph, then it 
is called a minimum (orientable or nonorientable) genus embedding. 
Based on the Euler formula, a minimum genus embedding has its face 
number maximum. So, a minimum (orientable or nonorientable) genus 
embedding is also called a maximum (orientable or nonorientable) face 
number embedding. Maximum face number implies that the average 
length of faces is smaller, and hence the possibility of faces being 
circuits is greater. This once caused to guess that minimum genus 
embeddings were all proper. However, a nonproper minimum genus 
embedding of a specific graph can be constructed by making 1 face 
as greater as possible with all other faces as less as possible. In fact, 
for torus and projective plane, all maximum face number embeddings 
are shown to be proper. For surfaces of big genus, a specific type of 
graphs were provided for all of their maximum face number embed- 
dings nonproper|Zhal]. 


R2.11 Conjecture. Any nonseparable regular graph has a max- 
imum face number embedding which is a proper map. 


A further suggestion is to find an embedding the lengths of all 
faces are nearly equal. The difference between the maximum length 


64 Activities on Chapter II 


and the minimum length of faces in a map is call the equilibrium of 
the map. An embedding of a graph with its equilibrium minimum is 
called an equilibrious embedding of the graph. 


R2.12 Conjecture. Any 3-connected graph has an equilibrious 
embedding which is proper. 


An approach to access the conjecture is still for some types of 
graphs, e.g., planar graphs, Halin graphs, Hamiltonian graphs, further 
graphs embeddable to a surface with given genus etc. 


Chapter III 


Duality 


e The dual of a map (A45, P) is the map (A4, Pap) and vice versa. 


e The deletion of an edge in a map is the contraction of the corre- 
sponding edge in the dual map and vice versa. 


e The addition of an edge to, the inverse of deleting an edge in, a map 
is splitting off its corresponding edge on, the inverse of contracting 
an edge in, the dual map and vice versa. 


e The deletion of an edge with its inverse, the addition, and the 
dual of deletion, the contraction of an edge with its inverse, split- 
ting off an edge are restricted on the same surface to form basic 
transformations. 


II.1 Dual maps 


On the basis of IL2, for a basic permutation 7 on the ground 
set Xag, (Xag, P) is a premap if, and only if, (X44, P*) is a premap 
where P* = Py, y = af( Theorem 2.4). The latter is called the dual 
of the former. Since 


P” = P* Ba = (PagB)Ba = P(aBBa) = P, 


the former is also the dual of the latter. 


66 Chapter III Duality 


Because the transitivity of two elements in the ground set X45 
on a premap (Xap, P) under the group Yy, J = {P, a, 8), determine 
an equivalence, denoted by ~y,, the restriction of P on a class 


Xa pl Yyy 


is called a transitive block. 


Theorem 3.1 Premap M» = (X, Per;) is the dual of premap 
Mı = (4, Peri) if, and only if, 


K TNI — x / Wy (3.1) 


where 
Jo = (Pers, a, 8), J = (Peri, a, 8), 


and 
Per? = Peryy = Peri“, y = af. 


Proof Necessity. Since M» is the dual of Mı, Pers = Perjo/f. 
From Per, = PerjaG € V;, V; = Yz. Hence, (3.1) holds. This is 
the necessity. 

Sufficiency. Since Mj is a premap and Pers = Peria, M» is also 
a premap, and then the dual of M; by considering Theorem 2.4. This 
is the sufficiency. [] 


From this theorem, the duality between Mı and M» induces a 
1-to-1l correspondence between their transitive blocks in dual pair. 
Because each transitive block is a map, it leads what the dual map of 
a map is. The representation of a premap by its transitive blocks is 
called its transitive decomposition. 


Example 3.1 Map 
Li = ({x, ax, zx, YTY, DESDE y= ap, 


and its dual . 
Lp = ({x, 0x, ax, yz}, (x, ax)). 


Or, in the form as 


II.1 Dual maps 67 


Lı = (e,v) and L1 = (e, f), 
where v = (x, Gx), f = (x, ox), 
e = (z,oz, Bx, yz), y= af, 


and 
e* = [z,0r,ozx,yr), 


as shown in Fig.3.1. 


In the following figure, two As and two Bs are, respectively, iden- 
tified on the surface. 





Fig.3.1 Map and its dual 


This figure shows what a dual pair of maps looks like. It is a 
generalization of a dual pair of maps on the plane. 


A map with its under pregraph a selfloop is called a loop map. 
It is seen that L4 and its dual Li in Fig.3.1 are both loop maps. 

In a premap M = (X,P), if a vertex v = {(x)p, (ax)p} is 
transformed into two vertices 


v, = ((z, Px, ---, Pix), (az, aPiz, ---,oPz)) 
= ((x)p, (ax)p} 


68 Chapter III Duality 


and 
U9 = Pity, pity. LN xD), (aPi*ty, aP zx, pues oJ *?3)) 
= ((P^*!z)p, (o/P?*1g)p), j = 0, 





with other vertices unchanged for permutation 7 becoming permuta- 
tion P’. It is seen that permutation 7' is basic and with the conjugate 
axiom as well. Hence, M’ = (AX, P") is also a premap. Such an op- 
eration is called cutting a vertex. If elements at vı are not transitive 
with elements at v in M’, then elements at v; and elements at və are 
said to be cuttable in M. ‘The vertex v is called a cutting vertex in M; 
otherwise, noncuttable . 

If there are two elements cuttable in a map, the the map is said 
to be cuttable in its own right. 

In virtue of Theorem 3.1, cuttability and noncuttability are con- 
cerned with only maps without loss of the generality of premaps. 


Lemma 3.1 A map M is cuttable if, and only if, its dual M* 
is cuttable. 


Proof Necessity. From M = (X, P) cuttable, vertex 


is assumed to cut into two vertices as 
v = (x, Px, ---, Pix) 
and . . 
vg = (Pi **n, Pia, ... Pa) 
for obtaining premap M' = (X, P") = Mi + M» where vj is on M; = 
(Xi, Pi), A = X + A5, Pj is the restriction of P’ on Xj, = 1,2. It can 
be checked that Mı and M» are both maps. Thus, on M* = (X,P*), 


vertex 
v* = (z, A, yP!z, PH, B, yP7 1a) 


can be cut into two vertices 


vt = (z, A, YP!) 


IIL.1 Dual maps 69 


and 


us (Pitta, B, yP lr) 


where A and B are, respectively, linear orders of elements in X; and 
AX». This attains the premap M*' = (X,P") = Mi + M3 where 
A = X 4X, P*; is the restriction of P*' on %,i = 1,2. The necessity 
is obtained. 

Sufficiency. From the duality, it is deduced from the necessity. 


If two elements in the ground set of a map M are not transitive 
in M' obtained by cutting a vertex on M, they are said to be cuttable; 
otherwise, noncuttable. It can be checked that the noncuttability de- 
termines an equivalence, denoted by ~ne on the ground set of M. The 
restriction of M on each 


Xap “ne 


is called a noncuttable block. 

If all noncuttable blocks and all cutting vertices of a map deal 
with vertices such that two vertices are adjacent if, and only if, one is 
a noncuttable block and the other is a cutting vertex incident to the 
block, then the graph obtained in this way is called a cutting graph of 
the map. It is easily shown that the cutting graph of a map is always 
a tree. 

For a face f = (x)p, of a premap M = (4’,P), if there has, and 
only has, an integer l > 0 for transforming f into 


f= (z, Trta (P4)'x) 


and 

f- (PA E rg (Py) x) 
such that x and (P)'*!z are not transitive at all, then f is called a 
cutting face of M. From the procedure in the proof of Lemma 3.1, For 
a cutting vertex of a premap, there has, and only has, a corresponding 
cutting face in the dual of the premap. 


Theorem 3.2 Two maps M and N are mutually dual if, and 
only if, their cutting graph are the same and the corresponding non- 


70 Chapter III Duality 


cuttable blocks are mutually dual such that a cutting vertex of one 
corresponds to a cutting face of the other. 


Proof Necessity. Because maps M and N are mutually dual, 
from the procedure in the proof of Lemma 3.1, there is a 1-to-1 corre- 
spondence between their noncuttable blocks such that two correspond- 
ing blocks are mutually dual. There is also a 1-to-1 correspondence 
between their cutting vertices such that the cyclic orders of their blocks 
at two corresponding cutting vertices are in correspondence. There- 
fore, their cutting graphs are the same. This is the necessity. 

Sufficiency. Because maps M and N have the same cutting 
graph, a tree of course, in virtue of the correspondence between cut- 
ting vertices and cutting faces, the sufficiency is deduced from Lemma 
3.1. [] 


Note 3.1 Tow trees are said to be the same in the theorem 
when trees as maps(planar of course) are the same but not the iso- 
morphism of trees as graphs(the latter can be deduced from the former 
but unnecessary to be true from the latter to the former). 


On a premap M = (4,P), if an edge Kx is incident with two 
faces, i.e., 
yx € (x )p4 U (Ox) ps, 


then it is said to be single ; otherwise, i.e., 


yv € ijo, U (O1) p, 


(with only one face), double. An edge with distinct ends is called a 
link; otherwise, a loop . Clearly, single link, single loop, double link 
and double loop. Further, a double link is called a harmonic link, or 
singular link according as yx € (x)p,, or not. Similarly, a single loop 
is called a harmonic loop, or singular loop according as yx € (x)py, or 
not. 


Theorem 3.3 For an edge e, = {z, oz, Gx,yx} of premap 
M = (Xa B, P) and its corresponding edge ež = (x, Bx, ax, yx} of the 
dual M* = (Koa P*), P* = P4, 


IIL1 Dual maps 71 
(i) e; is a single link if, and only if, e* is a single link; 
(ii) e is a harmonic link if, and only if, e* is a harmonic loop; 
(iii) e, is a singular link if, and only if, e* is a singular loop; 


(iv) e; is a double loop if, and only if, e* is a double loop. 


Proof Necessity. (i) Because e, is a link, (x)p and (yx)p belong 
to distinct vertices. And because e; is a single edge, (x)p. and (yx)p, 
belong to distinct face. By the duality, ež is a single link as well. 


(ii) Because e, is a double link, in spite of (x)p and (yx)p belong- 
ing to distinct vertices, yx, or ax € (z)p.. And because of harmonic 


k 


link, the only opportunity is yz € (r)p. From the duality, ež is a 


harmonic loop. 


(iii) Because e, is a singular link, in spite of (x)p and (yx)p 
belonging to distinct vertices, ax € (x)p«. In virtue of a as the second 
operation of M*, e% is a singular loop. 


(iv) Because e, is a double loop, Gx € (x)p and ax € (x)p.. 
From the symmetry between o and 0, M and M*, ež is a double loop 
as well. 


Sufficiency. From the symmetry in duality, i.e., M = (M*)*, it 
is obtained from the necessity. [] 


On the basis of this theorem, the classification and the dual re- 
lationship among edges are shown in Table 3.1. 


12 Chapter II Duality 




































































L Single à Single L 
i i 
D D 
n o Harmonic Harmonic o n 
u u 
b b 
k l Singular Singular il k 
e e 
S S 
i Harmonic Harmonic i 
n n 
L g g L 
ő 7 Singular Singular e. ô 
o o 
Double ———— Double 
P P 
" * 
M Dual relation M 


Table 3.1 Duality between edges 
In the table above, harmonic links will be classified into segmen- 
tation edges and terminal links and harmonic loops into shearing loops 
and terminal loops in III.2 to have additional two dual pairs of edges: 


segmentation edges and shearing loops, terminal links and terminal 
loops. 


II.2 Deletion of an edge 

Let M = (Xa p(X), P)be a premap and e; = Kx = (x, az, pr, 
yx}, € X, an edge. 

What is obtained by deleting the edge e, from M is denoted by 


M-a (290) Er) (3.2) 


where P_, is the permutation restricted from P on Xa g(X) — Ka. 


Lemma 3.2 Permutation P_, is determined in the following 


II.2 Deletion of an edge 73 


way as when e, is not a selfloop, 
Px(and aP~'z), 
if y = P !r(and aPx); 
Psy = 4 Pyz(fl aP yz), (3.3) 
if y = P yz(and aPyz); 
Py, otherwise, 


and when e, is a selfloop with yx € (x)p, 
Px(and aP1z), 
if y =P !z(and aPzx); 
P uy = < Pyz(and o/P- az), (3.4) 
if y =P !rz(and aPyz); 
Py, otherwise, 
otherwise, i.e., yx € (x)p, yx is replaced by (x in (3.4). 


Proof When e, is not a selfloop. Because only vertices (r)p and 
(yx)p are, respectively, changed in M — e, from M as 


(P'2)p_, = (Poe, Pa, Pa) 
and 
(Pyr), m (P^ yz, Pyz, a) Pz) 


(Fig.3.2(a)=(b)). This implies (3.3). 
When e; is a selfloop with yx € (r)p. Because only vertex (x)p 
is changed in M — e, from M as 


(P^!z)p | = (P Iz, Px,- , P yz, Pya,---,P 2) 
(Fig.3.2(c)(d)), or 
(P !z)p | — (P Ig, Pte P 18x, Pha, +++, Px) 


(Fig.3.2(c)—(d) in parentheses) according as yx € (x)p, or not. 
The former is (3.4). The latter is what is obtained from (3.4) 
with yx is replaced by Oz. [] 


74 


Chapter II Duality 


In Fig.3.2, the left two figures are parts of the original map and 
the right two figures, the results by deleting the edge Kz. 


Further, Fig.3.3-9 are all like this without specification. 


Pes 
Pyx NS LZ 


m d 
poi 


x 








xcd 


(P Z182) 
DK CU. 








(c) 


Plyg 
Pys "S LZ 


=. 





p | 
Ply 


(P 18m) 
TP wr 
Px Y Z 


Tur 
ET lo 
Pig 





(d) 


Fig.3.2 Deletion of an edge 


Lemma 3.3 For a premap M = (4,P), 


M — e, = (8 — 


II.2 Deletion of an edge T5 


Kz,P ,)is also a premap. And, the number of transitive blocks 
in M — e, is not less than that in M. 


Proof Because P is basic for a, from Lemma 3.2 P. , is also 
basic for a. Because P satisfies the conjugate axiom for a, from 
Lemma 3.2 and Theorem 2.3 P_, is also satisfies the conjugate axiom 
for a. The first statement is done. 

Because any nontransitive pair of elements in M is never transi- 
tive in M — e,, the second statement is done. E 


If e, is an edge of a premap M such that M — e, has more 
transitive blocks than M does, then e; is called a segmentation edge. If 
an edge has its one end formed by only one semiedge of the edge itself, 
then it is called an terminal link. From the symmetry of elements in a 
quadricell, (x) can be assumed as the 1-vertex incident with a terminal 
link without loss of generality. Since (Py)yz = Px = x, yx € (x)p4. 
Hence, a terminal link is always a harmonic link. However, a harmonic 
link is unnecessary to be a terminal link. This point can be seen in 
the following theorem. 


Theorem 3.4 Fora map M = (X,P), M—e, = (X—Kz,P.,) 
is a map if, and only if, e, is not a harmonic link of M except for 
terminal link. 


Proof When (z,*x) € (x)p4, i.e., e; is a terminal link, Because 
no isolated vertex in any premap, from Lemma 3.3, M — e, = (X — 
Kz,P.,)isa map. In what follows, this case is not considered again. 

Necessity. Suppose M — e, is a map, but e, is a harmonic link of 
M. Because P4^!z and Py Iz + x((x, yx) € (x)py) for group Vy, 
J' = (P..,,0, 3}, are not transitive on the set ¥ — Kx, M — e, is not 
a map. This is a contradiction to the assumption. 

Sufficiency. Because M is a map, P_, is basic. From Theorem 
2.3, P_, satisfies the conjugate axiom. Then, based on Table 3.1, two 
cases should be discussed for the transitivity. 

(i) When e; is a single edge(including single loops!) or singular 
link of M. Because e, is not a cut-edge of its under graph G(M), 


16 Chapter III Duality 


G(M — ez) is connected. From Theorem 2.6, P_, for group V is 
transitive on the set A — Ka. Thus, M — e, is a map. 
(ii) When e, is a double loop of M. Because 


(Pa) !a)p , = ((Py)~*z, Tad, PES 
BPBz, P*yz, ---, (Py) az), 
we have Px = B(Py)x with (Py) tax and P8x = B(8Pxz) are 


transitive in M — e,. Hence, M — e, is a map as well. [] 


From the proof of the theorem, a much fundamental conclusion 
is soon deduced. 


Corollary 3.1 In à map M, an edge e, is a segmentation edge 
if, and only if, e, is a harmonic link except for terminal links. And, 
e, is a harmonic link if, and only if, it is a cut-edge of graph G(M ). 


Proof To prove the first statement. 

Necessity. Because e, is a segmentation edge, G(M — e,) is not 
connected. e, is only a link. On the basis of Table 3.1, e, is also a 
double edge. And, because e, is not singular, e, is only harmonic. 
Clearly, a terminal link is not a segmentation edge in its own right. 

Sufficiency. Because e, is not a terminal link, Px is distinct form 
x and yx is distinct from Pyx. And, because e, is a harmonic link, 
Px and P^z are not transitive in M — ez. Thus, e, is a segmentation 
edge. 

To prove the second statement. Because it can be shown that ex 
is a terminal link of M if, and only if, e, is an articulate edge of graph 
G(M), a cut-edge as well. This statement is deduced from the first 
statement. L] 


Let M = (Xa p(X), P) be a pregraph and e; = Kx = (x, az, Gx, 
yx}, € X, be an edge. The contraction of ey from M, denoted by 
M ee, = (Xa p(X) = Ka, Pex); 
is defined to be the dual of M* — ež where ež = (x, Bx, ax, yx}, the 


corresponding edge of e, in the dual M* of M. In other words, Per = 
FI 


III.2 Deletion of an edge 77 


Lemma 3.4 Per is determined by the following (i-iii): 
(i) When e, is a link. For y € X,9(X) — Ka, 


Pyz(and P^ !z), if y= P^!z(and aPyz); 
Percy = < Pz(and oP-wz), if y= P^ yz(and oPz); (3.5) 
Py, otherwise, 
shown as in Fig.3.3(a)—9(b). 
(ii) When e; is a harmonic loop. For y € %,3(X) — Ka, 
Pyx(and o/P^!z), if y= P^ !z(and aP yz); 
P..y = 4 Pr(and aP az), if y= P^ yz(and aPz); (3.6) 
Py, otherwise, 
shown as in Fig.3.3(c)=—> (d). 
(iii) When e, is a singular , or double loop. For y € Xa p(X) — 
A. 
o/P- 8x(and o/P^!z), 
if y = P Iz(and P^ !6z); 
Pag =< Prud. (3.7) 
if y = aPx(and aP pz); 
Py, otherwise, 
shown as in Fig.3.3(e)=—>(f). 


Proof (i) When e, is a link. In the dual M* of M, from the 
duality, 


(a) p» = (a, Pyx, (PyY^, +++, (Py) 2) 
and 
(yx) p+ = (yx, Pa, (PyY a, - --, (Py) ya), 
Or 
(n)pees Un Pau oes UP) ab qm Pu PP 


and hence P*,, is only different from P* = Py at vertices 


(Pyx)p:, = (Pyr, (PyYz, ---, (Py) 12) 


78 Chapter III Duality 


and 
(Pz)p., = (Px, (Py)*y2,--+, (Py) ya) 


or at vertex 
(Px) px, = (PIE 32999 (Pa) ya, Pa, as (Py) 1a) 


with their conjugations according as e, is single, or double. By con- 
sidering Per = P* y. 


Perly) = Pal P yx) P* Y(P ya) 
= P'(Py yz) = Px 


for y = P^ yz and 


Perly) = TP Ix) = P* A4(P Ix) 
=P (Psy) ix = Pyx 


for y = P ‘x. From the conjugate axiom, the cases for y = aP ‘x 
and y = a(P yz) in the parentheses of (3.5) are also obtained. Then, 
for other y, 


Pay) = P'.yy = (Py)yy = Py 


in the both cases. Therefore, (3.5) is true. 

(ii) When e; is a harmonic loop. In a similar way to (3.5) for e; 
single, (3.6) is also obtained. 

ii) When e, is a singular, or double loop. In the dual M* of M, 


(x) p» = (x, Pyz, (PyYz, +++, (Py) 1x) 


and 
(ax) p: = (ax, PIAL, (Py az, my (Py) lox), 
Or 
(x) p» = Ge Pyz, ttt3 (Py) lax, QT, Pad; E (Py) x), 
and hence 


(z)p*, = (Pyz, (Pyz, ---, (Py) 12) 


II.2 Deletion of an edge 79 


and 
(ax)p», = (Pyaz, (Py)’az, +++, (Py) lox) 
or 
(x)p», = (P»yz,---, (Py) lox, Pyaz,---,(Py)7'z) 
with their conjugations according as e, is singular, or double. By 
considering Pes = P*.7, 
Pat) =P el a Pr) =P nmtapa 
= P* (Pax) = P* ((Py) lax) 
=Pron = Pox 
for y = aPx and 
Posl y) = Pa = P* (P's) 
= PŁ (P3) te) = Paz 
= Paix = oP 18x 
for y = P-!x. From the conjugate axiom, the cases for y = P x and 
a@P Gx are also obtained. Then, for all other y, 


Palu) = P' yy = (Py)yy = Py. 
This is (3.7). O 


NZ 


4 a 


pd 


Pte 





nee 


80 


R 
Ptr 
Ppa 
L NI 
m c P a 
== 
IN 5 
Px 
PlBr 
S 


(e) 


Chapter II Duality 


Ptr 
| Pyx 
Px | 
Playa 


Ply 
| Tz 
| Px 
* . NY 
TP Bx 


Sg-1 


(f) 


Fig.3.3 Contraction of an edge 


IIL2 Deletion of an edge 81 


From Lemma 3.4, it is seen that in the constriction of edge e, on 
a premap only if e, is not a selfloop, two vertices (z)p and (yx)p are 
composed of one vertex 


(P7lz)p,, = (Px, P»yz, ---, P ys, Pax,---,P 72) 


(Fig.3.3(a)— (b)); if e; is a harmonic loop, vertex (x)p is divided into 
two vertices 
(P^ ix)p,. = (P Iz, Pya P72) 
and 
(P^ yz)p,, = (P ya, Pa, «P ^na) 


(Fig.3.2(c)=(d)); and if e, is a singular, or double loop, vertex (x)p 
becomes vertex 


(Px)p,, = (Pzp P 18x, P t -- , o/P Bx) 
(Fig.3.3(e)>(f)). 


Lemma 3.5 For a premap M, M ee, is always a premap. And, 
the number of transitive blocks in M e e, is not less than that in M. 


Proof From Lemma 3.3 and the duality, the first statement is 
true. Because any nontransitive pair of elements in M is never tran- 
sitive in M ee, from Lemma 3.4, the second statement is true. L] 


If a harmonic loop e; has (x) py = (x), or (yx)py = (yx), then it 
is called a terminal loop. If the two elements of a co-semiedge appear 
in a vertex in succession, then the edge is called a twist loop. 


Lemma 3.6 For an edge e, of a map M, ex is a terminal loop 


if, and only if, ež is an terminal link in M*. And, e, is a twist loop if, 
and only if, ež is a twist loop. 


Proof A direct result deduced from the duality. [] 
Theorem 3.5 For an edge e, of a map M = (X4 X), P), 


M ee, is a map if, and only if, e; is not a harmonic loop but terminal 
loop. 


82 Chapter II Duality 


Proof Because for a terminal loop ez, M ee, is always a map. 
In what follows, this case is excluded. 

Necessity. Suppose M ee, is a map but e, is a harmonic loop. 
Since ež is a harmonic link in M*(Table 3.1), from Theorem 3.4 and 
Lemma 3.1, P~'x and Pz are, respectively, belonging to two distinct 
transitive blocks of M. From Lemma 3.4(ii), M ee; has two transitive 
blocks. This contradicts to that M ee, is a map. 

Sufficiency. Since e, is not a harmonic loop, only two cases should 
be considered as e; is not a loop or e; is a singular loop. For the former, 
in spite of a single or double edge, from Lemma 3.4(i), M ee, is a map. 
For the latter, from Lemma 3.4(iii), M ee, = M — e, is also a map. 

Therefore, the theorem is done. [] 


If a loop e, has that P^ !v and Px are in distinct noncuttable 
blocks, then it is called a shearing loop . From Theorem 3.5, all shear- 
ing loops are harmonic. However, the converse case is unnecessarily 
true. 


Corollary 3.2 In a map M, an edge e; is a shearing loop if, 
and only if, ež is a harmonic, but not terminal loop in M*. 


Proof A direct result of Theorem 3.5. E 


Theorem 3.6 The dual of premap M —e, is the premap M*ee*, 
where M* is the dual of M and ež in M* is the corresponding edge of 
e, in M. 

Proof(1) Because M* ee? is the dual of (M*)* —e'2—-M — ex, 
by the symmetry of the duality the theorem holds. [] 


However, if the contraction of e; on M is defined by (3.5-7), then 
the theorem can also be proved. 


Proof(2) Based on Table 3.1, four cases should be discussed. 
(i) In M, e, is a single link, and hence ež is a single link in M*. 
Since 


(Pyx)p 4, = (P72) pz, (Pr)p = (PX) pz, 


II.2 Deletion of an edge 83 


and (Pyx)p_, = (P*z)p: y, 
(M — e,)* = M* © æ. 


(ii) In M, e, is a harmonic link, and hence ež is a harmonic loop 
in M*(Dually, in M, ex is a harmonic loop, and hence e; is a harmonic 
link in M*). Now, (z)p, = (x)p». According as e; is a terminal link 
or not, a transitive block of M becomes one or two transitive blocks 
in M — e,. Meanwhile, According as e? is a terminal loop or not, a 
transitive block of M* becomes one or two transitive blocks of M* e ež. 
By considering the changes in vertices and faces, (M — er)“ = M* e ež 
is found. 


(iii) In M, e, is a singular link, and hence ež is a singular loop 
in M*(Dually, In M, e, is a singular loop, and hence ež is a singular 
link in M*). Since 

(P~'x)p_, = (P^'z)p;, 


and 
(P^ yz)p , = (P^ yx)p; ,, 


in view of (Pyx)p_,y = (Pyx)p:, we have (M —e;)' = M*eej. 


(iv) In M, e, is a double loop, and hence e? is a double loop in 
M*. Since 
(Px)p_, = (Pz)p;, 
and 
(Papa = (Px)p:., 
we have (M — ez)“ = M*ee;. O 
Corollary 3.3 In a map, an edge is a harmonic link if, and 
only if, the corresponding edge in its dual is a harmonic loop. And, 


an edge is a segmentation edge if, and only if, the corresponding edge 
in its dual is a shearing loop. 


Proof A direct result of Theorem 3.6. [] 


84 Chapter III Duality 


Example 3.2 Map M = (Kz + Ky + Kz,P) where 


P = (a, By, yz) (y, 2,72) 


and its dual M* = (K*z + K*y + K*z, P*) where 


D* = Py = CA AT, AZ, VY, z); 


are, respectively, shown in Fig.3.4(a) and (b). Here, K = {1, a, 86, y} 
and K* = {1, 3,a,7} are used to distinguish o and f. 


Map M — e; = (Ky + Kz, P.) where P_z = (By,yz)(y, 2) and 
its dual (M — e,)* = (K*y + K*z,(P_z)*), where (P_z)* = Pry = 
(y,Gz,ay, yz), are, respectively, shown in Fig.3.4(c) and (d). It is 
easily seen that (M — e,)* = M* è ež. 











II.3 Addition of an edge 85 

















Fig.3.4 Duality between deletion and contraction 


IIIL.3 Addition of an edge 


Let M = (Xap, P) be a premap, e; = Kx = {z, az, Bx, yx}, and 
x € Xag. Write as 


M + ez = (Xag Kx, P), 


where P,, is determined from P in the following manner. For any 
y € X and two given angles (l, Pal) and (t, Pat), if l and t are 
not at the same vertex, or at the same vertex and e, as a harmonic 
loop(assume t € (l)p without loss of generality), then 


t(and ax), if y = x(and at); 
Pat(and x), if y = ax(and aPat); 
Pizy = 4 l(and Bx), if y = yx(and al); (3.8) 
Pal(and yx), if y = Gx(and aPal); 
Py, otherwise, 


86 Chapter III Duality 





/UÀ Xx 


> 
\ JAN 
\~ f AN 


Pal 





(a) For (3.8) (b) For (3.9) 


Fig.3.5 Appending an edge 


shown in Fig.3.5(a), otherwise, i.e., e, is a double, or singular loop 
(assume t € (l)p without loss of generality), 


t(and ax), if y = x(and at); 
Pat(and x), if y = ox(ando'Pat); 
Pizy = $ Wanda), if y = yz(and al); (3.9) 
Pal(and Gx), if y = yz(and aPal); 
Py, otherwise, 


as shown in Fig.3.5(b). 


Such a transformation from M into M + e, is called appending 
an edge eyr. 


Lemma 3.7 For à premap M = (4,P), M+e, = (¥ + 
Kz,P,,)is also a premap. And, the number of transitive blocks 
in M 4- e, is not greater than that in M. 


Proof From (3.8) and (3.9), P,, is basic. In virtue of Theorem 
2.3, it suffices to show that the orbits of ,, are partitioned into 
conjugate pairs for the conjugate axiom. In fact, if | and t are at 
distinct vertices, then 7?,, is obtained from P in replacing two vertices 


II.3 Addition of an edge 87 


(t)p and (l)p by respective 
(t)p,, = (t, Pt,---, Pat, x) 


and 
(U)p,, = (, Pl,---,aPal, yx), 


or Oz is substituted for yx. If | and t are at the same vertex, then 
P, is obtained from P in replacing the vertex (l)p by 


(lp,, — (L Pl,---,aPat, x,t, Pt,---,aPal, yz) 


or 


(p, = (U, Pl, ---,aPat, r,t, Pt,---,aPal, Bx) 


according as e, is a harmonic loop or not. This shows that the orbits 
of P,, are partitioned into conjugate pairs for a. E 


Note 3.2 On the degenerate case t = l, if e, is a harmonic 
loop, then 
(Dg, = (L, PL oPal yx, £); 


otherwise, i.e., e; is a twist loop, 


OG, =U, F eo Pal. bre): 


Theorem 3.7 For a premap, not a map, M = (X5, P), the 
number of transitive blocks of M + e, is less than that of M if, and 
only if, e; ia a segmentation edge. If M is a map, then M + e, is also 
a map. 


Proof Since the number of components of graph G(M + e;) is 
less than that of G(M) if, and only if, e, is a cut-edge which is not 
articulate, the first statement is deduced from Corollary 3.1. 

Because the transitivity between two elements in the ground set 
of M under appending an edge is unchanged, the second statement is 
valid. [] 


Note 3.3 Let M' = (X + Kx, P) = M + e; for M = (€,P). 
Because P' ,, = P, M is obtained by the deletion of the edge e, from 


88 Chapter III Duality 


M', i.e., M = M'—e,. Therefore, the operation of appending an edge 
on a premap is the inverse of the corresponding edge deletion. 


Now, another operation for increasing by an edge on a premap 
is considered. This is the splitting an edge seen as the inverse of edge 
contraction. 

Let M = (Xa B, P) be a premap. Suppose (l, Pal) and (t, Pat) 
are two angles. For x £ Xag, let M oe, = (Xag + Kx, Por), where 
Pox is determined by P in the following manner. The transformation 
from M into M oe, is called splitting an edge e, and ez, the splitting 
edge of M. 


Lemma 3.8 Let!c {t}pU {at}p. Ifl ¢ (t)p, U (Bt)p., then 


Pat(or ax), if y = x(or aPat); 
l(or x), if y = az(or al), 
Pory = 4 Pol(or Gx), if y = yx(or aPal); (3.10) 
t(or yx), if y = Gx(or at); 
Py, otherwise. 


The edge e, is a single link as shown in Fig.3.6; Otherwise, t.e., | € 
(t) py U (Gt) p,, then 


Pat(or ax), if y = x(or aPat); 
l(or x), if y= az(or al); 
Pory = $ t(or Bx), if y = Oz (or at); (3.11) 
Pol(or Bx), if y = yz(or aPal); 
Py, otherwise, 


or yx replaced by Gz to attain, respectively, e, as a singular or har- 
monic link shown in Fig.3.7. 


III.3 Addition of an edge 89 


y 
X 


ALA 
7N: 
Da 
zu 


G 
E 


(b) Moe; 


Fig.3.6 1 Z (t)py U (Ot)p, 


uv 


d 
EN 
^ 
N 














Ey 
ü 
x 
Q 

4 


\ 
x 
^N 
A 


3 
E 


(a) M (b) Moe; 


Fig3.7 | € (t)p, U (Bt)p-, 


90 Chapter III Duality 


Proof Since | € {t}p U {at}p, l and t are at the same vertex. 
Thus, e; is a link. If | ¢ (t)p, U (Bt)p4, ie., | is in a face different 
from that t is in, or in other words, e, is single, then by the reason as 
(Pox)ex is different from only the vertex 


(Pont) Paje = (Pat, ol, Pal,---, at) 


where Pozz = Pat shown in Fig.3.6, from Lemma 3.4(i), (Por)exr = P. 
Otherwise, according as e, is singular or harmonic, (Pox)ex is different 
from only the vertex 


(Pozx)tp,.)., = (Pat, ol, Pal, ---, at) 


or 


(Pat,---,al,t,---,aPal) 
shown in Fig.3.7. From Lemma 3.4(i) again, (Por)ex = P. O 


Lemma 3.9 Let! ¢ (t)p U {at}p. If t and l are not transitive 
on M, then 
Pol(or ax), if y = x(or o/Pal); 
t(or x), if y= az(or at); 
Pat =< Pul(or pr); iu yrtor Pot; (3.12) 
l(or yx), if y = Gz(or al); 
Py, otherwise 
as shown in Fig.3.8 where e, is a harmonic loop. 
Otherwise, i.e., t and / are transitive on M, then 
Pol(or ax), if y = x(or aPal); 
Pat(or x), if y= ax(or aPat); 
Pay = 4 Wor Bx), if y = yz(or al); (3.13) 
t(or yx), if y = Bor at); 
Py, otherwise, 


as shown in Fig.3.9 where e; is a singular, or harmonic loop according 
as it is incident with two faces, or one face. 


II.3 Addition of an edge 91 


mal DNV Aa 
ee a s 


(a) M (b) M o ex 


Fig.3.8 t and l nontransitive 


RA dox, D | Les 
a 
oe | Mo oe 





(a) M (b) Moe; 
Fig.3.9 t and l transitive 
Proof Since | € {t}p U {at}p, l and t are at the same vertex. 


Thus, e; is a link. If | ¢ (t)p, U (Gt)p,, i.e., lis in a face different 
from that t is in, or in other words, e, is single, then by the reason as 


92 Chapter II Duality 


(Poxr)ex is different from only the vertex 
(Pork) (Poser = (Pat,-++, al, Pad, ---, at) 


where Pozz = Pat shown in Fig.3.6, from Lemma 3.4(i), (Por)ex = P. 
Otherwise, according as e, is singular or harmonic, (Pox)ex is different 
from only the vertex 


(Port Pda = (Pat,:-+,al, Pal, ---, at) 


or 
(Pat,---,al,t,---,aPal) 
shown in Fig.3.7. From Lemma 3.4(i) again, (Por)ex = P. O 


Lemma 3.10 If M — (X,7P) is a premap, then for any x € X, 
Moe, = (X + Ka, Pox) is also a premap. And, Moe, has the number 
of its transitive blocks not greater than M does. 


Proof From Lemmas 3.8-9, permutation P., is basic and parti- 
tioned into conjugate pairs for a. Then by Theorem 2.3, M oz is also 
a premap. This is the first statement. Because splitting an edge does 
not changing the transitivity of any pair of elements in X, the second 
statement holds. L 


Lemma 3.11 Edge e, is a harmonic loop on Moe; if, and only 
if, Port and P,yx are not transitive on M. 


Proof Necessity. Because the splitting edge e, is a harmonic 
loop on M, it is in the case (3.12). As shown in Fig. 3.8, P,,x(— L) 
and P.,ax(= t) are not transitive on M. Hence, by the symmetry 
among elements in Kx, Porz and Porya are not transitive on M either. 

Sufficiency. Because P,,0r(— l) and P,,ax(= t) are not tran- 
sitive on M, only l ¢ {t}p U {at}p is possible. This is the case for 
(3.12). Thus, e; is a harmonic loop. O 


Theorem 3.8 For a premap M = (Xap, P) not a map, the 
number of transitive blocks in M o e, is less than that in M if, and 
only if, e, is a harmonic loop. If M is a map, then M oe, is also a 
map. 


IIL3 Addition of an edge 93 


Proof From Lemma 3.11, the number of transitive blocks in 
M o e, is less than that in M if, and only if, e, is a segmentation 
edge in M -—- e,. This is the first statement. Because splitting an edge 
in a map does not changing the transitivity, the second statement is 
obtained. [] 


Lemma 3.12 Edgee, = {z, az, Gx, yx} is appended in premap 
M if, and only if, edge ež = {x, Gx, ax, yx} is split to in premap M*. 

Proof Necessity. (1) If e, is a single edge of M + ez, i.e., yx ¢ 
(z)p,,4 U (0x) p,,7, then (x)p;, and (yz)p;, are two vertices on (M + 
€r)“. Hence, from (3.10-11), ež is the splitting edge from the vertex 


(Pi) Pp = US NR UIS Pu OMS (Pin) ya) 


on M*. 
(2) Otherwise, i.e., e; is a double edge on M + ex. Thus, yx € 
(z)p, 4 U (Bz)p,,,. Two cases should be considered. 


(i) If yz € (x)p,,4, then e, is a segmentation edge on M + 
€x. From Corollary 3.1, P?,v and P!,yz are not transitive on M. 
Furthermore, from (3.12), ež is a splitting edge(a harmonic loop) on 
M*. 


* 
x 


(ii) If yx € (Gx)p,,,, then from (3.13), e? is a splitting edge on 
M*. 


Sufficiency. (1) If ež is not a loop on M*oe?, then from (3.10-11), 
M has a face 


(Pottery = Phat (P5) m Phys em OP) YE): 
From (3.6), e, is an appending edge between 
angle (aP, ‘a, P*,yx) and angle (aP*, "yz, P?) 
in a face of M, as shown in (3.8-9). 


(2) Otherwise, i.e., ež is a loop. From (3.12-13), there are two 


faces on M with an angle each. Such two angles determine the ap- 
pending edge e, on M. [] 


94 Chapter III Duality 


Theorem 3.9 The dual of a premap M + e, is the premap 
M* oe? where M* is the dual of M and ej is the dual edge in M* oe; 
corresponding to e; in M + e;. 


Proof A directed result of Lemma 3.4. [] 


From what has been discussed above, both the following diagrams 


(Tab: P) c (Tap tas Pa 
q q (3.14) 
Con ar We dq XP 


and 
(Mog P) => (Mus Ka, Pe) 


ü ü (3.15) 
EP eo. ASR Peer) 
are commutative. 


Example 3.3 Map M = (Kx--Ky,P), P = (x,y)(yy, yx) and 

its dual M* = (K*z + K*y,P*), 
P* = Py = (x,y, az, yy) (y, yz), 

are, respectively, shown in (a) and (b) of Fig.3.10. Notice that œ and 
B have distinguished roles in K = {1,a, 8, y} and K* = {1, 8, a, 7}. 

Map M +e; = (Kz + Ky + Kz, P42), 

Pig = (a y, a), Ba, yy) 

and its dual (M + e,)* = (K*z + K*y + K*z, (P42)®), 


(Pi = Pirs (2, Bz, vy, yz) yo. By), 


are, respectively, shown in (c) and (d) of Fig.3.10. According to (3.13), 
e? is the splitting edge at the pair of angles (x, ay) and (yy, 8x) in 


M*. Therefore, 
M* oe, = (K*x + K*y + K*z, (z, zx, az,vyy)(y,yz)). 


II.3 Addition of an edge 


If By is seen as y, then M* oet = 





K 
AUS 


(c) M+e, 


95 


(M +e)". 











(d) (M + ez)* 


Fig.3.10 Duality between appending and splitting an edge 


Based on Theorem 3.9, in a premap M = (X, P), splitting an 
edge e, attains M o e; which is just M* + e? obtained by appending 
the edge e; in its dual M* = (X*, P"). 


96 Chapter III Duality 
IIL4 Basic transformation 


In a premap M = (X,5,P), the deletion of a single edge e; 
is called basic deleting an edge and its result is denoted by M — 
e,. The contraction of a link e, is called basic contracting an edge, 
and its result is denoted by M ey ez. The two operations are, in all, 
called basic subtracting an edge. Similarly, appending a single edge 
is called basic appending an edge, and splitting a link is called basic 
splitting an edge. Such two operations are, in all, called basic adding 
an edge. Apparently, M +, ez and M oy e, are the results of basic 
adding an edge e, on M in their own right. Basic subtracting and 
basic adding an edge are in all called basic transformation. From what 
we have known above, A premap becomes another premap under basic 
transformation. 


Theorem 3.10 Suppose M' is a premap obtained by basic 
transformation from premap M, then M’ is a map if, and only if, 
M is à map. 


Proof Because a single edge is never a harmonic link, from The- 
orem 3.4 the theorem holds for basic deleting an edge. Because a link 
is never a harmonic loop, from Theorem 3.5, the theorem holds for 
basic contracting an edge. Then, from Theorem 3.7-8, the theorem 
holds for basic adding an edge. [] 


Furthermore, for basic transformation, the following conclusion 
can also be done. 


Theorem 3.11 Let M = (X,P) bea map and M* = (X*,7P*), 
its dual. Then, for any single edge e; in M, (M —» e,)* = M*ey e? 
and for a single edge e, not in M, (M +, e,)* = M* op ež. Conversely, 
for any link e; in M, M ey e; = M* —, ež and for a link e, not in M, 
M op ey = M" Hye. 


Proof Based on the duality between edges as shown in Table 
3.1, the statements are meaningful. From Theorem 3.6 and Theorem 
3.9, the fist statement is true. In virtue of the duality, the second 


IIL.4 Basic transformation 


statement is true. 


97 


[] 


From this theorem, the following two diagrams are seen to be 


commutative: 
(Tap: P) 
EP] 
and 
(Tap P) 
GP 


(Aap 8, Pu) 
* 
us (X* Iu e ex) 


———— Mag- Kr, Pres) 


+pex 


* 


* 9567. 


@ber 


(3.16) 


(3.17) 


On the basis of basic transformation, an equivalence can be es- 
tablished for classifying maps in agreement with the classification of 


surfaces. 


Activities on Chapter III 


III.5 Observations 


O3.1 Observe the condition for two permutations Per; and Per» 
satisfying (3.1) on the same ground set. 


O3.2 Given a tree(e.g., the star of 5 edges), observe how many 
maps are there for their dual maps all having the tree as under graph. 


O3.3 Write the dual of map M = (Kz, (x, Bx)). If a map has 
its dual with the same under graph, then it is said to be self-dual for 
the graph. Observe if M is self-dual for its under graph. 


O3.4 Provide a map which is cuttable and its under graph with 
8, cut-vertex. 


03.5 How to distinguish the cuttability of a map and the sep- 
arability of its under graph? 


O3.6 Is the under graph of the dual of a Eulerian map bipar- 
tite? If yes, explain the why; otherwise, by an example. 


O3.7 Is the dual of a preproper map(no double edge) always a 
preproper map? If yes, explain the why; otherwise, by an example. 


O3.8 Is the dual of a proper map(each face formed by a circuit 
in its under graph) always a proper map? If yes, explain the why; 
otherwise, by an example. 


03.9 Is the dual of a polygonal map(two face with at most one 
edge in common) always a polygonal map? If yes, explain the why; 


III.6 Exercises 99 


otherwise, by an example. 


03.10 Isthe under graph of the dual of a map with 3-connected 
under graph still 3-connected? If yes, explain the why; otherwise, by 
an example. 


III.6 Exercises 


E3.1 Prove that the under graph of a map M = (X,P) isa 
tree if, and only if, its dual M* has the following three properties: 

(i) M* has only one vertex; 

(ii) For any x € X, x and yz are in the same orbit of P*; 

(iii) For any y € (x)p., there is no subsequence z, y, yx, yy or 
x, Yy, yx, y in its cycle. 


For a map M = (X (X), P) and $ C X, let M| Ks] = (Ks, P[Ks]) 
where [Ks] is the restriction of P on Ks = KS, and M[Ks] is said 
to be induced on Ks from M. Generally speaking, M[Ks] is not a 
map, but always a premap. A cocircuit of a graph with all of its edges 
incident to the same vertex is called a proper cocircuit . 


E3.2 Let M — Eg = M|X — Ks], Es = le;|Vxz € S). Prove 
that M — Es, S C X, is a map if, and only if, there is no proper 
cocircuit of G|M] on graph G(M |Ks]). 

A proper circuit of a map M is such a set C of edges that C* = 
{e*|\Ve € C) is a proper cocircuit of G|M*]. 


E3.3 Let H = G/M] and My be the set of all maps whose 
under graphs are H. Prove that if C is a proper circuit of a map M, 
then it is a proper circuit of all N € My. 


E3.4 Let M e Es = (M* — E%)* where M* is the dual of M 
and Ez = [ei|Vr € S). Prove that M e Es is a map if, and only if, 
Eg is a proper circuit of M. 


E3.5 Prove that a map M is on the sphere if, and only if, each 
face of its dual M* corresponds to a proper cocircuit of M. 


100 Activities on Chapter III 


If a map has its dual Eulerian, then it is called a dual Eulerian 
map. 


E3.6 Prove that a map is a dual Eulerian map if, and only if, 
each of its faces is incident with even number of edges. 


If a preproper dual Eulerian map has each of its faces partition- 
able into circuits, then it is called a even assigned map . A map is 
called bipartite when its under graph is bipartite. 


E3.7 Prove that a dual Eulerian map is bipartite if, and only 
if, it is even assigned. 


E3.8 Prove that a quadrangulation is bipartite if, and only if, 
it is without loop. 


For a loopless quadrangulation Q = (44,5, Q), from E3.8, its 
vertex set V can be partitioned into two subsets Vi and V5 such that 
the two ends of each edge are never in the same subset. Such a subset 
of vertices is called an independent set. X it and d stand for 


the sets of elements in Xa g incident to, respectively, Vj and V;|Dehl, 
Gaul, MuS1]. 


E3.9 Let o = 8Qw, y — aß. Prove that (cr, Qox), x € Xap, 
is an angle. 


Angles (oz,0 Qox) and (x, Qox) as shown in E3.9 are called an 
independent pair . Thus, each face in a quadrangulation has exactly 
two independent pairs of angles. 


E3.10 Let Kı = {1, a1, 61,71}, 1 = o11. And, a; = Q and 
Bı = GQy7(= c as shown in E3.9). Prove 
(i) Kı is the Klein group of four elements; 
Gi) Xi, = 95 55 Kw 
v; €Vi ye(vj o 
(ii) Qi = (s Qi) is a map, where Q; is the restriction of 
O on xo 


cu 
Similarly to E3.10, from V5, another map Q» can be deduced. Q1 
and Q» are called the incident pair of the quadrangulation. 


III.7 Researches 101 


E3.11 Prove that the two maps in the incident pair of a quad- 
rangulation are mutually dual. 


E3.12 Prove that any planar quadrangulation is loopless. 


E3.13 Let A and B be the sets of, respective all planar quad- 
rangulations and all dual pairs of planar maps. Establish a 1-to-1 
correspondence between A and B(i.e., bijection). 


IIL7 Researches 


For a map, if the basic deletion of an edge can not be done 
anymore, then the map is said to be basic deleting edge irreducible. 
similarly, if the basic contraction of an edge can not be done on a 
map anymore, then the map is said to be basic contracting irreducible. 


R3.1 Given the size, determine the number of self-dual maps 
as an integral function of the size, or provide a way to list all the self- 
dual maps of the same size and deduce a relation among the numbers 
of different sizes. 


R3.2 Given the size, determine the number of maps all basic 
deleting irreducible as an integral function of the size, or provide a 
way to list all such maps with the same size and deduce a relation 
among the numbers of different sizes. 


R3.3 Given the size, determine the number of maps all basic 
contracting irreducible as an integral function of the size, or provide 
a way to list all such maps with the same size and deduce a relation 
among the numbers of different sizes. 


R3.4 For any given graph, determine the number of maps all 
basic deleting irreducible with the same under graph, or provide a way 
to list all such maps with the same size and deduce a relation among 
the numbers of different sizes. 


R3.5 For any given map, determine the number of all basic 
deleting irreducible maps obtained from the map by basic deletion, or 


102 Activities on Chapter III 


provide a way to list all such maps with the same size and deduce a 
relation among the numbers of different sizes. 


R3.6 Fora given graph, determine the number of maps all basic 
contracting irreducible with the same under graph, or provide a way to 
list such maps and deduce a relation among the numbers of different 
sizes. 


R3.7 For a given map, determine the number of all basic con- 
tracting irreducible maps obtained from the map by basic contraction, 
or provide a way to list such maps and deduce a relation among the 
numbers of different sizes. 


If a map is basic both deleting and contracting irreducible, then 
it is said to be basic subtracting irreducible. 


R3.8 Given the size, determine the number of basic subtracting 
irreducible maps as an integral function of the size, or provide a way 
to list all such maps with the same size and deduce a relation among 
the numbers of different sizes. 


R3.9 Fora given graph, determine the number of maps all basic 
subtracting irreducible with the same under graph, or provide a way 
to list such maps and deduce a relation among the numbers of different 
sizes. 


R3.10 Find a relation between triangulations and quadrangu- 
lations. 


If a map has each of its faces pentagon, then it is called a quin- 
quangulation. Similarly, the meaning of a hexagonalization. 


R3.11 Justify whether or not a triangulation has a spanning 
quinquangulation or hexagonalization. If do, determine its number. 


R3.12 Even assigned conjecture: A bipartite graph without 
cut-edge has a super map even assigned. 


Chapter IV 
Orientability 


e The orientability is determined by the orientation for each edge 
with two sides; otherwise, nonorientability. 


e The basic equivalence is defined via basic transformations to show 
that the orientability is an invariant in an equivalent class. This 
equivalence is, in fact, the elementary equivalence on surfaces. 


e The Euler characteristic is also shown to be an invariant in an 
equivalent class. 


e Two examples show that none of the orientability(nonorientability 
as well) and the Euler characteristic can determine the equivalent 
class. 


IV.1 Orientation 


Let M = (X44, P) be a map(from Theorem 2.6, without loss of 
generality for a premap), and Y7, I = (y, P}, y = aß = Ga, be the 
group generated by the set of permutations /. Now, it is known that 
the number of orbits of P on X,,5 is double the number of vertices on 
M and the number of orbits of P^ on X4, is double the number of 
faces on M. Because P, Py € Vr, the number of orbits of the group 
Vr; on Xag is not greater than any of their both. 


104 Chapter IV Orientability 


Lemma 4.1 The number of orbits of the group V; on Xap is 
not greater than 2. 


Proof Because Py € Wy, for any x € Xa, {r}p, € {r)en 
Here, {x}p, and {x}, are the orbits of, respectively, the permutation 
Py and the group V; on Xap. For any chosen element z € Xag, from 
P € V, for any y € {x}p,, (y) e4 € (x]w,, and from y € v, 


Uv», € ovde, € {rhor 


In view of Theorem 2.6, at least half of elements at each vertex belong 
to (x)w,. Therefore, (x), contains at least half of elements in X, 5. 
Similarly, {ax}, contains at least half of elements in 4X, 5. 

In consequence, based on the basicness of P for o, V; has at 
most 2 orbits on A a. O 


According to this lemma, a map M = (ag, P) has only two 
possibilities: group Vr is with two, or one, orbits on Xag. The former 
is called orientable, and the later, nonorientable . 

From the proof of the lemma, an efficient algorithm can be es- 
tablished for determining all the orbits of group V; on the ground 
set. 

Actually, in an orientable map, because V; has two orbits for o, 
the ground set is partitioned into two parts of equal size. It is seen 
from Lemma 4.1 that each quadricell (7.e., edge) is distinguished by 
two elements in each of the two orbits. And, the two elements of an 
edge in the same orbit have to be with different ends of the edge. 
Thus, each of the two orbits determines the under graph of the map. 


Example 4.1 Consider map M = (X,P) where 





X = Kr + Ky+ Kz+Ku+ Kv+ Kw 


and 
P = (a, y, z)(yz, u,v) (Yu, yy, w)(yw, yu, yz) 


as shown in Fig.4.1(a). Its two faces are 


(2, ^99, yv, yz) 


IV.1 Orientation 105 


and 
(yx, Y, W, YU, V, VYY, Z, u). 


In fact, for this map, group V; has two orbits. One is 


(x, yw, yv, yz, YL, y, w, yu, v, yy, z, u]. 


The other is what is obtained from it by multiplying o to each of all 
its elements. Thus, M is orientable. Fig.4.1(b) shows that M is an 
embedding of the complete graph of order 4 on the torus (yuy lu). 









































Fig.4.1 An embedding of Ky 


Corollary 4.1 If V;, ] = (P, af], has two orbits on A5, then 
they are conjugate for both o and f. 


Proof It is known from Lemma 4.1 that the two obits have the 
same number of elements, i.e., half of Xa g. Because y € {x}w, if, and 
only if, ay € (ax]«, and for any Kx, ox and Bx are always in the 
same orbit of V, this implies that {ar}y, = (Ox)w, different from 
[x y, and hence the conclusion. O 


Example 4.2 Consider map N = (X, Q) where 





X = Kr + Ky+ Kz+ Ku+ Kv+ Kw 


and 


Q = (x,y, z)(yz, u, v) (yu, By, w) (vw, yu, ya) 


106 Chapter IV Orientability 


as shown in Fig.4.2(a). 
That is obtained from the map M in Fig.4.1(a) in the replacement 
of cycle (^v, yy, w) by cycle (zv, Gy, w). Here, N has also two faces 


(z, yw, yv, yz) 
and 
(yz, y, Bv, Qu, Bu, YY, Z, u). 


Because By € {y}ay € (v) w,, ,,; from the corollary group V, 9} has 
only one orbit, t.e., 


hpa = &. 


Therefore, N is nonorientable. It is seen from Fig.4.2(b) that N is an 
embedding of the complete graph of order 4 on the surface 


(yuyu') ep (yyuu), 


i.e., Klein bottle . 









































Fig.4.2 An embedding of Ky on the Klein bottle 


Theorem 4.1 A map M = (X, P) is nonorientable if, and only 
if, there exists an element x € X such that Gx € {x}w,, or ax € {x}u, 
where I = (y, P}. 


Proof Necessity. Suppose ax ¢ {x}w,, then V; has at least two 
orbits. However from Lemma 4.1, it has exactly two orbits . Thus, M 
is never nonorientable. The necessity holds 


IV.2 Basic equivalence 107 


Sufficiency. Because Gx € [x]w,, from Corollary 4.1 it is only 
possible to have {x} = X, i.e., V; has only one orbit. Hence, M is 
nonorientable. This is the sufficiency. [] 


This theorem enables us to justify the nonorientability and hence 
the orientability of a map much simple. If there exists a face (x)p,, 
denoted by S,, such that ox € S,, or there exists a vertex (r)p, 
denoted by S,, such that Gx € S, on M, then M is nonorientable(as 
shown in Example 2). Otherwise, From y € $, via acting P, or y, for 
getting z Z Sz, Sx is extended into 


S,U{z}p,U {z}p 


which is seen as a new S to see if y,ay € S, or y, Gy € S. If it does, 
then M is nonorientable; otherwise, do the extension until |S| = |X|/2, 
or S= X. 


Theorem 4.2 A map M = (X,7) is orientable if, and only if, 
its dual M* = (X*, P*) is orientable. 


Proof Because P* = Py € Woo (y = ap = Ba), Vip = 
Yi py. So, for any z € A = X*, (zjwe,,,, = (]w,, ,.,. This is the 
conclusion of the theorem. [] 


IV.2 Basic equivalence 


First, observe the effect for the orientability of a map via basic 
transformation. 

For a map M = (X,7P) and its edge ez, let M —p e, and M eye, 
be, respectively, obtained by basic deleting and basic contracting the 
edge e, on M. From Theorem 3.10, M —» e; and M ej e, are both a 
map . 


108 Chapter IV Orientability 


SS : Pe — 


ls 
8] 


(a) Single link e; (b) Single loop e; 

















Fig.4.3 Basic deleting an edge 


Lemma 4.2 If M' is the map obtained by basic subtracting an 
edge from M, then M' is orientable if, and only if, M is orientable. 


Proof First, to prove the theorem for M' = M —, eg. 

Necessity. From M' = M — e, = (X', P) orientable, group 
V' = Wy, py has two orbits on X' = X—Kz, i.e., {Px}w and {Pax}w. 
Because e; is single, Pyx € {Px}w and PGx € (PoxWw. So, group 
V = V.» has two orbits 


[je = {Pxr}w U (x, yr} 


and 
(axis = {Par}w U (oz, Br} 


on ¥, i.e., M is orientable. 

Sufficiency. Because e, is a single link (Fig.4.3(a)), or single loop 
(Fig.4.3(b)), in virtue of that group V has two orbits {x} and (ox]w 
on X, group WV’ has two orbits 


{Pr}w = (z]e — (x, yr} 


and 
[Por jw = {ar} — lov, Bx} 


on X’, i.e., M' is orientable. 

Then, to prove the theorem for M' = M e, e,. On the basis 
of Theorem 3.11, the result is directly deduced from that for M' = 
M —p Er. E 


109 


IV.2 Basic equivalence 
Whenever that two new angles occur in the deletion of an edge 
with 4 angles lost is noticed, the edge appending as the inverse of 
deletion is done between the two angles. And then the same case 
comes for basic deleting and basic appending an edge. In this sense, 
Lemma 4.3 in what follows is seen as a direct result of Lemma 4.2. 
However, it is still proved in an independent way. 


For basic appending an edge, since the edge is only permitted 
to be a single link or a single loop, this operation is, in fact, done by 


putting the edge in the same face. 
Let map M = (X, P) have a face 
(y)Py = (Yo Vi; * +, Ys) 


= M +p ez 


where yo = y, yı = (Py)y, +++, Ys = (Py) !y. Denote 


M +; ey 


when appending the edge e, in between angles (y, Pay) and (yi, Pay;), 
0i xs. From (3.10), M +i e; = M ty e, 0 < i < s, are all 





maps(Fig.4.4). 
Eme QE Eae 
A Vi-i SS Yl-1 
m z= 
ZEN P= 
(b) M +i ex 


110 


Chapter IV Orientability 











-— = 
; A. ] A i 
(d) M +141 ex 


(c) M +i ex 
Fig.4.4 Basic appending an edge 


Lemma 4.3 Maps M+ ;e, = M+ per, 0 <i € s, are orientable 


if, and only if, M is orientable. 
sr Peg 


Proof necessity. Since M' = M +; ex 
are all orientable, group V' = Wy. py has two orbits (zw and (ow 
on Xt’ = X + Kx. Because e, is a single link (Fig.4.4(a) and (c)), or 


single loop (Fig.4.4(b) and (d)), 
P'r € (x)w and P'az € (ox. 


Hence, group V = Y; p} has two orbits 
{P't}y = {x}w — ix yx} 


and 
{P'ax}y = far}w — (oz, Bx} 
on X. This implies that M is orientable. 
Sufficiency. Since e, is a single link (Fig.4.4(a) and (c)), or single 


loop (Fig.4.4(b) and (d)), the two orbits 
{r}w = {y}v + {2,72} 


and 
lox jw = {ay}y + (oc, Gx} 
of group V' on X" are deduced from the two orbits {y}w and {ay}w 
[] 


of group V on X. Therefore, M' is orientable. 


IV.2 Basic equivalence 111 


As for basic splitting an edge, whenever that two new angles 
occur in the contraction of an edge with 4 angles lost is noticed, the 
edge splitting seen as the inverse of contraction is done between the 
two angles. 

Next, consider how to list all possibilities for basic splitting from 
a given angle. 

For a map M = (#,P), let 


(y)p = (Yo, Yi; tts Yl-1,Ul,°°°; Üs) 


yo = y, s > 0, be a vertex. Denote by M o;e, the result obtained from 
M by basic splitting an edge between angles (y, Pay) and (yi, ay;_1) 
where y = yo and Pay = ays. From Theorem 3.10, M o;e, = M+ pez, 
0 € i € s, are all maps (Fig.4.5). 














AA ANN á AAN 









































(c) M o e (d) M os e 
Fig.4.5 Basic splitting an edge 
Lemma 4.4 Fora map M = (%,P) and x ¢ X, map M oiez = 
M +p ez, 0 € i € s, is orientable if, and only if, M is orientable. 


Proof Necessity. Because M' = M o; ez = (X',P),0x i s, 
is orientable, group Y = V, py has two obits {x}w and {ar} on 


112 Chapter IV Orientability 


AX" = X + Kr. Since e, is a single link (Fig.4.5(b), (c) and (d)), or a 
double link (Fig.4.5(a) and (c)), 
P'r € {x}w and P'az € {ar}w. 
Therefore, group V = V, p; has two orbits 
UP'zY, = (xw — {x, yx} and (P'ox)y = {ar}w — (oc, Bx} 
on X. This implies that M is orientable. 

Sufficiency. Because e, is a single link (Fig.4.5(b), (c) and (d)), 
or a double link (Fig.4.5(a) and (c)), the two orbits (xw = {y}w+ 
{x, yx} and {ar}w = {ay}wt+{az, Bx} of group V' on X' are deduced 
from the two orbits {y}w and {ay}w of group V on X. Therefore, M' 
is orientable . O 


Corollary 4.2 If M' is the map obtained by basic adding an 
edge from map M, then M' is orientable if, and only if, M is orientable. 


Proof A direct result of Lemma 4.3 and Lemma 4.4. E 


The operation of basic appending an edge between two successive 
angles of a face in a map is also called increasing duplition (Fig.4.4(b) 
and (d)), and its inverse operation, decreasing duplition. And du- 
ally, the operation of basic splitting an edge is also called increasing 
subdivision(Fig.4.5(b) and (d)), and its inverse operation, decreasing 
subdivision. 


Corollary 4.3 A premap M' obtained by increasing duplition, 
increasing subdivision, decreasing duplition, or decreasing subdivision 
from a map M is still a map with the same orientability of M. 


Proof The results for decreasing duplition and decreasing sub- 
division are derived from Lemma 4.2. Those for increasing duplition 
and increasing subdivision are from Corollary 4.2. [] 


If map Mı can be obtained from map M^» via a series of basic 
adding and/or basic subtracting an edge, then they are called mutually 
basic equivalence, denoted by Mı ~pe Mo. 


IV.3 Euler characteristic 113 


Theorem 4.3 If maps Mı ~pe M», then Mj is orientable if, 
and only if, M» is orientable. 


Proof A direct result of Lemma 4.2 and Corollary 4.2. [] 
Since ~pe is an equivalent relation, maps are partitioned into 
classes of basic equivalence, in short equivalent class. Theorem 4.3 


shows that the orientability of maps is an invariant in the same equiv- 
alent class. 


IV.3 Euler characteristic 


For a map M = (X,), let v = v(M), € = e(M) and d= d(M) 
are, respectively, the order(vertex number), size(edge number) and 
co-order(face number) of M, then 


x(M) =v—e+¢ (4.1) 
is called the Euler characteristic of M. 
Theorem 4.4 Let M* be the dual of a map M, then 
x(M7) = x(M). (4.2) 


Proof Because v(M*) = $(M), e(M*) = e(M) and ¢(M*) = 
v(M), (4.2) is obtained from (4.1). O 


Lemma 4.5 For a map M = (X,P) and an edge ez, £ € X, 
let M — e, and M ee, be, respectively, obtained from M by deleting 
and contracting the edge e,, then 


X(M — ez), if e, is single; 
«M) =4 (4.3) 


X(M e e;), if e, is a link. 


Proof From Theorem 3.11 and Theorem 4.4, only necessary to 
consider for one of M — e, and M e e}. Here, the former is chosen. To 
prove x(M — ez) = x(M) for e, single. 


114 Chapter IV Orientability 


Because e; is single, v(M — ez) = v(M), (M — ez) = e( M) — 1 
and ó(M — er) = ¢(M) — 1. From (4.1), 
X(M — ex) = v(M) — (e(M) — 1) + (6(M) — 1) 
= v(M) — (M) + &(M) 
- x(M). 


This is just what is wanted to get. [] 


Corollary 4.4 For any map M, x(M) < 2. 


Proof By induction on the co-order ¢(M). If M has only one 
face, i.e., (M) = 1, then 


In view of the connectedness, 
e(M) > v(M) — 1. 
In consequence, 
x(M) € v(M) — (v(M)—1) 1 — 2. 


Thus, the conclusion is true for 9(M) = 1. 

In general, i.e., 96(M) > 2. Because of the transitivity on a map, 
there exists a single edge e, on M. From Lemma 4.5, M' = M — e, has 
X( M^) = x(M). Since 6(M") = 6(M) —1, by the induction hypothesis 
x(M") € 2. That is x(M) < 2, the conclusion. O 


For an indifferent reception, because the order, size and co-order 
of a map can be much greater as the map is much enlarged. The con- 
clusion would be unimaginable. In fact, since the deletion od a single 
edge does not change the connectivity with the Euler characteristic 
unchange and the size of a connected graph is never less than its order 
minus one, this conclusion becomes reasonable. 


Corollary 4.5 For basic subtracting an edge e; on a map M, 
x(M —» ex) = x(M) and x(M e, ex) = x(M). 


IV.3 Euler characteristic 115 


Proof A direct result of Lemma 4.5. [] 


Lemma 4.6 Fora map M = (4,P) and an edge ez, x ¢ X, let 
M + e and M o e, be obtained from M via, respectively, appending 
and splitting the edge e,, then 


«un - 1 


X(M + e), if e, is single; 


4.4 
X(M o ez), if e, is a link. us 


Proof From Theorem 3.11 and Theorem 4.4, only necessary to 
consider for one of M +e, and M oez. The former is chosen. To prove 
x(M + ex) = x(M). 

Because e, is single, then v(M + e;) = v(M), e(M + er) = 
e(M) +1 and ó(M + e,) = ¢(M) +1. From (4.1), 

x(M + ex) ae (e(M) + 1) + (6(M) +1) 
moe ) - (M) + (M) 
x(M). 


Therefore, the lemma is true. [] 


Corollary 4.6 For basic adding an edge e; on a map M, x( M+» 
er) = x(M) and x(M op e;) = x(M). 


Proof A direct result of Lemma 4.6. [] 


For a map M = (4,7) and an edge ez, x € X, let Mj.) and 
Mp4; be obtained from M by, respectively, increasing subdivision and 
increasing duplition for edge ez, and Miesz] and Mj 4, by, respectively, 
decreasing subdivision and decreasing duplition fox edge e,. From 
Corollary 4.3, they are all maps. 


Corollary 4.7 For increasing subdivision and increasing dupli- 
tion, 
X(Mj) = x(M); x(Mpaj) = x(M) (4.5) 
and for decreasing subdivision and decreasing duplition, 


X(Miex}) = x(M); x(Mpia) = x(M). (4.6) 


116 Chapter IV Orientability 


Proof Because increasing subdivision and increasing duplition 
are a special type of basic adding an edge, from Corollary 4.5, (4.5) 
holds. Because decreasing subdivision and decreasing duplition are 
a special type of basic subtracting an edge, from Corollary 4.6, (4.6) 
holds. The corollary is obtained. [] 


The following theorem shows that the Euler characteristic is an 
invariant in the basic equivalent classes of maps. 


Theorem 4.5 If maps M; ~pe Mo, then 


x(M1) = x(M3). (4.7) 


Proof Because the basic transformation consists of basic sub- 
tracting and basic adding an edge, from Corollary 4,5 and Corollary 
4.6, (4.7) is obtained. [] 


IV.4 Pattern examples 


Pattern 4.1 Consider the map M = (X, P) where XY = Ka + 
Ky 4 Kz 4 Ku 4 Kw + Kl and 





P = (x,y, z)(al, yz, Qw)(Bl, yy, au)(w, yu, 8x), 


shown in Fig.1.13. 
By deleting the single edge e, on M, let M; = (X1, 1) = M—be;, 
then Xj = Ky + Kz + Ku+ Kw + Kl and 





P, = (y, z)(al, yz, Bw) (BL, yy, oru) (w, yu). 


By contracting the double link e; on Mi, let Mo = (Xz, P2) = 
M; e, ez, then Xo = Ky + Ku + Kw + Kl and 





Po = (y, Bw, al)(8l, yy, oru) (w, yu). 


By contracting the double link e; on Mə, let M3 = (A3, P3) = 
Mp €y ej, then 43 = Ky + Ku + Kw and P; = (y, Bw, yy, ou) (w, yu). 


IV.4 Pattern examples 117 


By contracting the double link e, on M3, let M4 = (A4, P4) = 
Ms € eu, then X4, = Ky + Kw and Py = (y, Bw, yy, aw). 

Now, M, has only one vertex and only one face and hence any 
basic transformation for subtracting an edge can not be done. It is a 
map on the torus (Fig.4.6). 


Fig.4.6 A map basic equivalent to M 


Pattern 4.2 Again, consider the map N — (X, Q) where X — 
Ket+Ky+ Kz- Ku+ Kw+ KI and 





Q = (x,y, z)(al, yz, Bw) (Gl, By, au)(w, yu, Bx), 


as shown in Fig.1.14. 
By deleting the single edge e; on M, let Ny = (41, Q1) = N—pez, 
then 4% = Ky + Kz + Ku + Kw + Kl and 


Q; = (y, z) (al, yz, Bw) (BL, By, au) (w, yu). 


By contracting the double link e; on Nj, let No = (A5, Q2) = 
N, @, e,, then X = Ky + Ku + Kw + KI and 


Qə = (y, Bw, al) (Bl, By, au) (w, yu). 


By contracting the double link e; on No, let N3 = (A35, Q3) = 
No €y ej, then A3 = Ky + Ku + Kw and 








Qa = (y, pw, By, ou) (w, yu). 


118 Chapter IV Orientability 


Finally, By contracting the double link e, on Ns, let N4 
(A4, Q4) = N3 ey ey, then X4, = Ky + Kw and Q4- (y, fw, By, aw). 

Now, the basic transformation can not be done anymore on N4. 
N4 is a map on the Klein bottle, as shown in Fig.4.7. 


B 
1 
w 
XL Vv 
A A 
By 
B 


Fig.4.7 A map basic equivalent to N 


From the two patterns, it is seen that M4 ^^y. N4, and hence 
M ^y. N. Although their Euler characteristic are the same, i.e., 


x(M) = x(M4) =1 -2+1 = x(N1) = x(N), 


their orientability are different. 


Activities on Chapter IV 


IV.5 Observations 


04.1 Fora map M = (X,,4, P), observe how many orbits does 
the group Yia p} have on the ground set Yag? What condition is it 
for its transitivity. 


O4.2 Fora map M = (A,,4, P), observe how many orbits does 
the group Vygp.;, ^ = ap, have on the ground set Xag? What 
condition is it for its transitivity. 


04.3 Fora map M = (X,,4, P), observe how many orbits does 
the group V5», have on the ground set Xag? What condition is it 
for its transitivity. 


O4.4 Fora map M = (A,,4, P), observe how many orbits does 
the group Vyapy, y = o, have on the ground set Xag? What 
condition is it for its transitivity. 

04.5 If map M = (4,P) is orientable, is M — e, always ori- 
entable for any ez, r € X? If yes, explain the reason; otherwise, by 
an example. 

04.6 If map M = (X,7P) is nonorientable, is M — e, always 
nonorientable for any ez, x € X? If yes, explain the reason; otherwise, 
by an example. 


04.7 If map M = (X,7) is orientable, is M e e, always ori- 
entable for any ez, x € X? If yes, explain the reason; otherwise, by 
an example. 


04.8 If map M = (X,P) is nonorientable, is M e e, always 


120 Activities on Chapter IV 


nonorientable for any ez, x € X? If yes, explain the reason; otherwise, 
by an example. 

04.9 If map M = (4,P) is orientable, is M + e, always ori- 
entable for any ez, x ¢ X? If yes, explain the reason; otherwise, by 
an example. 


04.10 If map M = (4,P) is nonorientable, is M + ey always 
nonorientable for any ez, x ¢ X? If yes, explain the reason; otherwise, 
by an example. 

04.11 If map M = (X,7P) is orientable, is M o e, always ori- 
entable for any ez, x ¢ X? If yes, explain the reason; otherwise, by 
an example. 


04.12 If map M = (X, P) is nonorientable, is M o e, always 
nonorientable for any ez, x ¢ X? If yes, explain the reason; otherwise, 
by an example. 

04.13 Show by example that a face of an orientable map M 
does not correspond to a cocycle on its dualM*. 


IV.6 Exercises 


E4.1 ‘Try to prove Lemma4.1 in three different manners. 


E4.2 For a map M, prove that there exists a nonnegative inte- 


ger n and basic transformations 71, 7,::-,75, such that 
M* = | [rM (4.8) 
i=1 


where M* is the dual of M. 


E4.3 Foramap M = (Xap, P), the group Yip}, y = aß, with 
two orbits on X45 is known. Prove that 


x(M) = 0 (mod 2). (4.9) 


E4.4 Prove that a mapM = (AX,,5, P) is nonorientable if, and 


IV.6 Exercises 121 


only if, there exist x,y € X such that 
Ky N TE iual > 2. 


E4.5 Try to prove Corollary 4.4 in two different manners. 


E4.6 Ifa map M = (X45,P) has only one face, prove that 
M is nonorientable if, and only if, there exists an x € X such that 
ax € [z)p, where y = af. 

For a map M = (Xag, P), let A be the set of all orbits of P on 
Xag. Graph Gu = (V, E) where V = A and 


E = ((A, B)3x € YX,x € A and yz € B). 
And, Gy is called the subsidiary graph of M. 


E4.7 Fora map M = (Xa p, P), prove that the group V(45pj 
has two orbits on Xag if, and only if, the subsidiary graph Gm of M 
has two connected components. 


For a map M = (Aap P), y = a, let fi = ((zi)p4 (Cri) Py}, 
i = 1,2,---,¢, be all the faces of M. If a set S = {s;|1 € i € $) of 
orbits of permutation Py on X», satisfies |S N f;| = 1,i = 1,2,---,¢, 
then S ia called a face representative of M. 

Let graph Gs = (V, E) be with V = S as a face representative 
of M and e — (s,t) € E as a pair of faces s and t with an edge in 
common. On E, define a weight function 


0, there exists x € s such that yz € t; 

w(e) — | 
1, otherwise 

for e = (s,t) € E. And, (Gs, w) is called an associate net of M. 

On an associate net (Gs, w), Gs = (V, E), of a map M, if there 
exists a label (v) € {0,1} for vertex v € V such that for any e = 
(u,v) € E, 

l(u) + l(v) = w(e) (mod 2), 
then the associate net (Gs, w) is said to be balanced. 


E4.8 For a map M, prove that if one of its associate nets is 
balanced, then all of its associate net are balanced. 


122 Activities on Chapter IV 


E4.9 Prove that a map is orientable if, and only if, the map 
has an associate net balanced. 


For a graph G = (V, E), A C E, if V has a 2-partition, i.e., 
V «WU V3 and Vj NV = Q, such that 


A= ((u,v) € E|u € Vi, v € Va}, 


then A is called a cocycle of G. 


E4.10 Prove that a map M is orientable if, and only if, for a 
face representative S of M, its associate net (Gs, w) has (e € E|w(e) = 
1} as a cocycle. 


EA4.11 Prove that a map M has its Euler characteristic 2 4 
if, and only if, each of its faces corresponds to a cocycle of the under 
graph of its dual M*. 


IV.7 Researches 


R4.1 Characterize that the under graph of a map has an super 
map with its Euler characteristic 1. 


R4.2 Characterize that the under graph of a map has an super 
map with its Euler characteristic 0. 


RA4.3 For any orientable map, characterize that the under graph 
of the map has an super orientable map with its Euler characteristic 
0. 


R4.4 For a vertex regular map and a given integer g < 1, char- 
acterize that the under graph of the map has a super map with its 
Euler characteristic g. 


R4.5 For a vertex regular orientable map and a given integer 
g < 0, characterize that the under graph of the map has a super map 
with its Euler characteristic 2g. 


A graph which has a spanning circuit ia called a Hamiltonian 
graph. Such a spanning circuit is called a Hamiltonian circuit of the 


IV.7 Researches 123 


graph. If a map has its under graph Hamiltonian, then it is called a 
Hamiltonian map. 


R4.6 For a Hamiltonian map and a given integer g < 1, char- 
acterize that the under graph of the map has a super nonorientable 
map with its Euler characteristic g. 


R4.7 For a Hamiltonian map and a given integer g < 0, char- 
acterize that the under graph of the map has a super orientable map 
with its Euler characteristic 2g. 


For a vertex 3-map(or cubic map), if it has only i-face and j-face, 
iz j, i,j > 3, then it is called an (i, j) ;-map. 
R4.8 For a given integer g < 1, determine the number of 


(3, 4) j-map of order n(n > 1) with Euler characteristic g. 


R4.9 For a given integer g € 1, determine the number of 
(4, 5) j-map of order n(n > 1) with Euler characteristic g. 


R4.10 For a given integer g < 1, determine the number of 
(5,6) j-map of order n(n > 1) with Euler characteristic g. 


R4.11 Given a graph G of order n(n > 4), determine the con- 
dition for G have a super (n — 1, n) s-map. 


On a (n — 1, n) j-map of order n(n > 4), let 61 be the number of 
(n — 1)-faces. If its Euler characteristic is g € 1, then n and $; should 
satisfy the following condition: 


(n — 1)l(n(n — g) + $1), (4.10) 
i.e., n— 1 is a facer of n(n — g) + 1. 


R4.12 Given an integer g < 1, for any positive numbers n and 
ġı satisfying (4.10), determine if there exists a (n — 1, n);-map with 
its Euler characteristic g. 


Chapter V 
Orientable Maps 


e Any irreducible orientable map under basic subtracting edges is 
defined to be a butterfly. However, an equivalent class may have 
more than 1 butterflies. 


e The simplified butterflies are for the standard orientable maps to 
show that each equivalent class has at most 1 simplified butterfly. 


Reduced rules are for transforming a map(unnecessary to be ori- 
entable) into another butterfly, if orientable, in the same equivalent 
class. A basic rule is extracted for deriving all other rules. 


e Principles only for orientable maps are clarified to transform any 
map to a simplified butterfly in the same equivalent class. Hence, 
each equivalent class has at least 1 simplified butterfly. 


e Orientable genus instead of the Euler characteristic is an invariant 
in an equivalent class to show that orientable genus itself determine 
the equivalent class. 


V.1 Butterflies 


On the basis of Chapter IV, this chapter discusses orientable 
maps with a standard form in each of basic equivalent classes. If an 


V.1 Butterflies 125 


orientable map has only one vertex and only one face, then it is called 
a butterfly. 


Lemma 5.1 In each of basic equivalent classes, there exists a 
map with only one vertex. 


Proof For a map M = (A55, P), if M has at least two vertices, 
from the transitive axiom, there exists an x € Xa g such that (x)p and 
(yx)p, y = af, determine two distinct vertices. Because e, is a link, 
by basic contracting ez, M' = M ey e, ~pe M. Then, M' has one 
vertex less than M does. In view of Theorem 3.10, M' is also a map. 
If M' does not have only one vertex, the procedure is permitted to go 
on with M" instead of M. By the finite recursion principle a map M" 
with only one vertex can be found such that M’ ~pe M. This is the 
lemma. L 


A map with only one vertex is also called a single vertex map , 
or in brief, a petal bundle . 


Lemma 5.2 In a basic equivalent class of maps, there exists a 
map with only one face. 


Proof For a map M = (Xap, P), if M has at least two faces, 
from the transitive axiom, there exists an x € Xa 8 such that (x). and 
(yx)py, y = af, determine two distinct faces. Because e, is single, 
by basic deleting ez, M' = M —p e, ^y. M. Now, M’ has one face 
less than M does. From Theorem 3.10, M' is still à map . Thus, if 
M' is not with only one face, this procedure is allowed to go on with 
M' instead of M. By the finite recursion principle, a map M" with 
only one face can be finally found such that M’ ~pe M. The lemma 
is proved. [] 


In fact, on the basis of Theorem 3.6, Lemma 5.1 and Lemma 
5.2 are mutually dual. Furthermore, what should be noticed is the 
independence of the orientability for the two lemmas. 


126 Chapter V Orientable Maps 


Theorem 5.1 For any orientable map M, there exists a but- 
terfly H such that H ~pe M. 


Proof If M has at least two vertices, from Lemma 5.1, there 
exists a single vertex map L ~pe M. In virtue of Theorem 4.3, L is 
still orientable. if L has at least two faces, from Lemma 5.2, there 
exists a single face map H ~pe L. In virtue of Theorem 4.3, H is still 
orientable. Since H has, finally, both one vertex and one face, H is a 
butterfly. Therefore, H ~pe L ~pe M. This is the theorem. [] 


This theorem enables us only to discuss butterflies for the Basic 
equivalence classes of maps without loss of generality. 


V.2 Simplified butterflies 
Let O; = (Xk, Tr), k > 0, where 


4a 


u k 
iiec 3 C Ky), 35 k 1 61) 


i=l 


and 


(0), = k = 0; 


— k 
E (I [Go wo vos vu), 24 & > 1. p 


i=l 


It is east to check that all Oi, k > 0, are maps. And, they are called 
O-standard maps. When k = 1 and 2, O; and O» are, respectively 
given in (a) and (b) of Fig.5.1. 


V.2 Simplified butterflies 127 


T1 
yi 


yy2 | 
Vi ^w yai 
Ur E 

— < 
vy. yv2 DW 
Á | i 
71 y2 
r2 


(a) Oi (b) O2 


Fig.5.1 Two O-standard maps 


Note 5.1 When k = 0, Oo = (00,0) is seen as the degenerate 
case of a map with no edge. For example, what is obtained by basic 
deleting an edge on £y = (Ka, (x, yx)) is just Oo. Usually, it is seen 
as the map with only one vertex without edge, or called the trivial 
map. 


Lemma 5.3 For any k > 0, O-standard map QO, is orientable. 
Proof When k = 0, from Oo = (Kz, (x, yx)) —p ex, 
Oo ™be (Kr; (2, yx). 

Because (zjw(,.,,,, = (v, yr} and [o], = lom, Bx) are two 
orbits, (Kx, (x, yx)) is orientable. In view of Theorem 4.3, Oy is ori- 
entable. 

For k > 1, assume, by induction, that Oy 4 = (A4, Jk-1) is 
orientable. From (5.1) and (5.2), group V5, ,.,j has two orbits as 

Unibw, ug = {x yrl <i <k-1} 
and 
[ozije;, 4} = {02i Brill <i E k — 1}. 
For Oy = (Ax, Jr), from (5.2), Jy = (Ji); tis Yk, YLk, YYk). Group 
Vry} has only two orbits as 
{Ci} v5.4} = {T1} es, 1 U Uis Yk Ys Yu) 
= {rn yal <i < kj 


128 Chapter V Orientable Maps 


and 
{axri}u,, 4} = lomijeu, ,,) U loti; our, Bxx, Fyn} 
= {ax;, bxi < i < k — 1} losa Dx. DUI 
= for, Gali < k}. 
Therefore, Og, k > 1, are all orientable. [] 


Lemma 5.4 For any k > 0, O-standard map Ox has only one 
face. 


Proof When k = 0, since Op = (Ka, (x,7yx)) —y e;, Oo should 
have one face which is composed from the two faces (x) and (ax) of 
(Kz,(x,vyx)). Therefore, Op has only one face(seen as a degenerate 


case because of no edge). 
For any Os = C sz 1; trom (5.1) and (5:2), 


(Xy = ED oYUi 315 Uis ooo Tk, Y Uk Y ks Yk) 


is a face of O}. However, since 


T 
{rial = gl; 


O; has only this face. [] 


From (5.2), each O-standard map has only one vertex (Oy is the 
degenerate case of no incident edge). Based on the above two lemmas, 
any O-standard map is a butterfly. Because of the simplicity in form 
for them, they are called simplified butterflies . Since for any k > 0, 
simplified butterfly O; has 2k edges, one vertex and one face, its Euler 


characteristic is 
(Ox) = 2 — 2k. (5.3) 


Lemma 5.5 For any two simplified butterflies O; and O;, 7, j > 
0, Oi ~pe O; if, and only if, i = j. 


Proof Because the sufficiency, t.e., the former O; ~a O; is de- 
rived from the latter i = j, is natural, only necessary to prove the 
necessity. 


V.2 Simplified butterflies 129 


By contradiction. Suppose i Æ j, but O; ~pe O;. Because of the 
basic equivalence, from Theorem 4.5, 


x(Oi) = x(O;). 
However, from (5.3) and the condition i Z j, 


x(Oi) = 2 — 2i 2 2 — 2j = x(O;). 
This is a contradiction. O 


Theorem 5.2 In each of the basic equivalent classes of ori- 
entable maps, there is at most one map which is a simplified butterfly. 


Proof By contradiction. Suppose simplified butterflies O; and 
O;, i Æ j, i,j = 0, are in the same class. However, this is a contradic- 
tion to Lemma 5.5. [] 


In the next two sections of this chapter, it will be seen that in 
each basic equivalent class of orientable maps, there is at least one 
map which is a simplified butterfly. 

On the basis of Theorem 4.5, two butterflies of the same size 
have the same Euler characteristic. Do they all simplified butterflies? 
However, the answer is negative! 


Example 5.1 Observe map M = (4, J) where 
X = Kzı + Kyi + Krz + Kye 


and 
J= (zr yi; La, Yo, YL1, VY1, YL2, "Yy2)- 


Because the face 


(11) 74 = Uis "YU, 92, ^YU2, t1, Y1, "2, y») 


has 8 elements, half the elements of ground set, M has only one face. 
Hence, M is a butterfly, but not a simplified butterfly. Actually, the 
simplified butterfly with the sane Euler characteristic of M is O». 


130 Chapter V Orientable Maps 


V.3 Reduced rules 


Although butterflies are necessary to find a representative for 
each basic equivalent class of orientable maps, single vertex maps are 
restricted in this section for such a classification based on Lemma 5.1. 

For convenience, the basic equivalence between two maps are not 
distinguished from that between their basic permutations. In other 
words, (A, P1) ~pe (A5, P5) stands for P, ~pe P5. 


Lemma 5.6 For a single vertex map M = (X, J), if 
I = (R,x, yz, 8S) 
where R and S are two linear orders on X, then 
J pe CF); (5.4) 


as shown from (a) to (b) in Fig.5.2. 


Proof Because J = (R,2,72,8), (Jyye = Je = ym, i.e., 
(yx) 7, = (ya) is a face. Because e; is a single edge, by basic deleting 
e; on M, M'—M —ye, — (X — Krt, J'), J’ =(R,S). From M ex 
M UP tige d RB O 


This lemma enables us to transform a single vertex map into 
another single vertex map with one face less in a basic equivalent 
class. 





Fig.5.2 Reduced rule (5.4) 


V.3 Reduced rules 






















































































yri 
— — n E WN 
YY | -——— Gi —— 
C E 
D yz 




















Fig.5.3 Reduced rule (5.5) 


Lemma 5.7 BHor(O0,5,2),1 — Ux, B, C, 7% D. yy, E) 
where A, B, C, D and E are all linear orders on X, then 


d Nbe [A D CD. B, asus qe n, 
as shown from (a) to (b) in Fig.5.3. 


(5.5) 
Proof Four steps are considered for each step as a claim. 


Claim 1 J ~re (E, A,2,2, D, C, yz, yz, B). 


Proof For the angle pair (az, Jx) and (Gz, Jyx) of 
J = (A, m, B,y, C, yz, D, yy, E), 
by basic splitting e;(a link), get 
I ewe (D, yy, E, A, x, 2) Uye B, y, ©, yx). 


Then, since e, is a link, by basic contracting e, on J, get 


di “be Jo = KON E 2D Come. 


This is the conclusion of Claim 1. Here, Jı and Jz are, respectively, 
shown in (a) and (b) of Fig.5.4. 


Claim 2 Jorn (y, A, x, B, E, yy, D,C,y2). 


Proof For the pair of angle (az, J2z) and angle between E and 
A on Rh 


(E, A, x, z, D,C, yx, yz, B), by basic splitting e,(a link), 
get Jo c» B = (A, z,z,y)(vy, D,C, yz, yz, B, E). Then, since e; is 


a link, by basic contracting e; on J3, get 


J3 ^v bc JA E (y; A m, B, E, qu D, C, vx). 


131 


132 Chapter V Orientable Maps 


This is the conclusion of Claim 2. Here, 73 and J4 are, respectively, 
shown in (a) and (b) of Fig.5.5. 


Claim 3 Ja ~ve (B, E, z, A, y, yz, yy, D,C) 


























































































































Fig.5.6 For Claim 3 


Proof For the angle pair (ay, Jay) and (o ‘yy, yy) on Ja 


V.3 Reduced rules 133 


(y, A,x, B, E,yy, D,C, yx), by basic splitting e,(a link), get 
JA epe J5 = (A, x, B, E, z\(yz, yy, D, C, y£, y). 
Then, since ez is a link, by basic contracting e, on J3, get 
Js ~ve Jo = (B, E, z, A, y, yz, yy, D, C). 


This is the conclusion of Claim 3. Here, Js and Js are, respectively, 
shown in (a) and (b) of Fig.5.6. 


Claim 4 Js; ~pe (A, D, C, B, E, z, 2, yz, yz). 
Proof For the angle pair (az, Jez) and (ay, yz) of 
Js = (B, E, z, A, y, yz, vy, D, C), 
by basic splitting e, (a link), get 
Ie ~ve Iz = (A, y, x) (ym, yz, yy, D, C, B, E, z). 
Then, since e, is a link, by basic contracting e, on J7, get 
Ur ~ve Jg = (A, DC, B, E, s £, yz, ye). 


This is the conclusion of Claim 4. Here, Jz and Jg are, respectively, 
shown in (a) and (b) of Fig.5.7. 















































Fig.5.7 Claim 4 
On the basis of the four claims above, 


T ~pe So ~pe Ja ~be Jo ~be Jg 
= (A, D,C, B, E, x,y, yz, yy). 


134 Chapter V Orientable Maps 
This is (5.5). O 


An attention should be paid to that Lemma 5.6 and Lemma 5.7 
are both valid for orientable and nonorientable maps. They are called 
reduced rules for maps. More precisely, They are explained as in the 
following. 


Reduced rule 1 A map with its basic permutation J is basic 
equivalent to what is obtained by leaving off such a successive elements 
(xz, ysr), rE X, on J. 


Reduced rule 2 A map with its basic permutation J in the 
form as (A, z.B, y, C, yx, D, yy, E) is basic equivalent to what is ob- 
tained by interchanging the linear order B between x and y and the 
linear order D between yx and yy, and then leaving off x,y, yx and 
yy and putting (z, y, yz, yy) behind E on J. 


V.4 Orientable principles 


'This section is centralized on discussing the basic equivalent 
classes of orientable maps. The main purpose is to extract that there 
is at least one simplified butterfly in each class. From the first sec- 
tion of this chapter, it is known that each class is considered for only 
butterflies without loss of generality. 


Theorem 5.3 For a butterfly M = (Xag, ), y = af, if no 
x,y € X exist such that J = (A, x, B,y, C, yz, D, yy, E) where A, B, 
C, D and E are all linear orders on Xag, then 


M ~re Op, (5.6) 
i.e., the trivial map(the degenerate simplified butterfly without edge). 


Proof For convenience, in a cyclic permutation J on Xag, if two 
elements x,y € Xa g arein the formas J = (A, x, B,y,C, yx, D, yy, E), 
then they are said to be interlaced; otherwise, parallel. 


V.4 Orientable principles 135 


Claim If any two elements are parallel on J, then there is an 
element z € Xag such that (z, yx) C J, ie., (z,"yz) is a segment of 
J itself. 


Proof By contradiction. If no such an elements exists on AX, 
then for any zı € X, there is a nonempty linear order Bı on X such 
that J = (Aj, z, Bi, yz, C1) where A; and C are some linear orders 
on Xa g. Because Bı Z 0, for any x2 € By, on the basis of orientability 
and r9 and 2, parallel, the only possibility is yr. € Bı. From the 
known condition, there is also a linear order B» Z Ø on Xag such that 
Bı = (Ao, £2, Bo, 422, C3) where A> and C3 are segments on Bj, i.e., 
some linear orders on X. Such a procedure can only go on to the 
infinity. This is a contradiction to the finiteness of Xag. Hence, the 
claim is true. 


If J Æ Ú, then from the claim, there exists an element x in J 
such that J = (A,z,yz, B). However, because (yz), = (yz) is a 
face in its own right, J has to be with at least two faces. This is 
a contradiction to that M = (4, J) is a butterfly. Hence, the only 
possibility is J = 0, i.e., (5.6) holds. O 


Actually, this theorem including the claim in its proof is valid for 
any orientable single vertex map. Therefore, it can be seen that the 
reduced rule(Lemma 5.6) and the following corollary are valid for any 
map (orientable or nonorientable) as well. 


Corollary 5.1 Let S = (A, x, yx, B) be a segment on a vertex 
of a map M. And let M’ be obtained from M by substituting (A, B) 
for S and afterward deleting Kx from the ground set. Then, M' is a 
map. And, M’ e. M. 


Proof Because it is easy to check that M’ = M —y M, from 
Theorem 3.10, M' is a map. This is the first statement. In view of 
basic deletion of an edge as a basic transformation, M’ ~pe M. [] 


Corollary 5.2 Let S be a segment at a vertex of a map M. 
If for each element x in S, yx is also in S and any two elements in 


136 Chapter V Orientable Maps 


S are not interlaced, then there exists an element y in S such that 
S = (A, y, yy, B). 


Proof In the same way of proving Theorem 5.3, the conclusion 
is soon obtained. [] 


Theorem 5.4 Ina butterfly M = (2,5, J), if there are x,y € 

Xa g such that J = (A, x, B, y, C, yz, D, yy, E), then there is an inte- 
ger k > 1 such that 

Mac Or: (5.7) 


i.e., the simplified butterfly with 2k edges. 
Proof Based on Reduced rule 2(Lemma 5.7), 


J Nig (ADO BE m uT y). 


Let H = (A, D,C, B, E). From Corollary 5.1, assume H is not in the 
form as S without loss of generality. From Corollary 5.2 and Theorem 
5.3, H has two possibilities: H = Ø, or there exist two elements zi 
and y; interlaced in A. 

If the former, then J ~pe (2, y, yz, yy), ie. , M ~a Oi. Oth- 
erwise, i.e., the latter, then KA = (A, X1; By, Yis C, ^Y 31; IA, YY; Fi). 
An attention should be paid to that E, = (Fi, x,y, yx, yy). In this 
case, from Lemma 5.7, 


vi ~be (Ai, Dy C^ Bis E, X1; Y1, Yv1, yyı) 
= (A1, IA, C, Bı, fu LLY, YT, VY, Ti, Y1, "1, yi). 


Let Hı = (Aj, Di, Ci, By, F3), then for Hı instead of H, go on 
the procedure. According to the principle of finite recursion, it is only 
possible to exists an integer k > 1 such that (5.7) holds. E 


This theorem shows that each basic equivalent class of orientable 
maps has at least one map which is a simplified butterfly. 

By considering Theorem 5.2, each basic equivalent class of ori- 
entable maps has, and only has, an integer k > 0 such that the sim- 
plified butterfly of size k is in the class 


V.5 Orientable genus 137 


V.5 Orientable genus 


Although Euler characteristic of a map is an invariant for basic 
transformation, a basic equivalent class of maps can not be determined 
by itself. This is shown from the map M in Example 4.1 of section 
IV.4 and the map N in Example 4.2 of section IV.4. They both 
have the same Euler characteristic. However, they are not in the 
same basic equivalent class of maps because M is orientable and N is 
nonorientable. 

Now, a further invariant should be considered for a class of ori- 
entable maps under the basic equivalence. 


Theorem 5.5 For any orientable map M = (4, P), there has, 
and only has, an integer k > 0, such that its Euler characteristic is 


x(M) = 2 — 2k. (5.8) 


Proof From Theorem 5.1 and Theorem 4.5, only necessary to 
discuss butterflies. From Theorem 5.2 and Theorem 5.4, M has, and 
only has, an integer k > 0, such that M ~pe Og. Therefore, from 
(5.3), 

X(M) = 2 — 2k. 


This is (5.8). O 


Corollary 5.3 The Euler characteristic of an orientable map 
M = (X, P) is always an even number, i.e., 


X(M) — 0 (mod 2). (5.9) 
Proof A direct result of Theorem 5.5. [] 


Since Euler characteristic is an invariant of a basic equivalent 
class, the integer k in (5.8) is an invariant as well. From Theorem 5.5, 
k determines a basic equivalent class for orientable maps. Since each 
orientable map in the basic equivalent class determined by k can be 
seen as an embedding of its under graph on the orientable surface of 


138 Chapter V Orientable Maps 


genus k, k is also called the genus, or more precisely, orientable genus 
of the map. Of course, only an orientable map has the orientable 
genus. 

From what has been discussed above, it is seen that although 
Euler characteristic can not determine the basic equivalent class for 
all maps, the Euler characteristic can certainly determine the basic 
equivalent class for orientable maps. 


Activities on Chapter V 


V.6 Observations 


O5.1 Think, is there a butterfly which has 3 edges? Further, 
is there a butterfly with some odd number of edges? If yes, provide 
an example. Otherwise, explain the reason. 


O5.2 Observe that any butterfly with 2 edges is a simplified 
butterfly. Explain the reason. 


O5.3 Provide a butterfly of at most 5 edges , which is not a 
simplified butterfly. 


O5.4 Observe that is there a superfluous operation among the 
four operations: basic deleting, basic appending, basic contracting 
and basic splitting an edge in the basic transformation for the basic 
equivalence? If no, indicate the role of each of them. If yes, indicate 
the superfluous operation with the why. 


O5.5 Think, do some three of the four operations: basic delet- 
ing, basic appending, basic contracting and basic splitting an edge in 
the basic transformation determine an equivalence? If do, provide an 
example. Otherwise, explain the reason. 


O5.6 Among the eight operations of the above four with in- 
creasing duplition, decreasing duplition, increasing subdivision and 
decreasing subdivision, how many groups of these operations are there 
for determining the basic equivalence. 


O5.7 Can an equivalence be determined by basic deleting and 
basic appending an edge? If yes, observe some of invariants under the 


140 Activities on Chapter V 


equivalence. Otherwise, explain the reason. 


O5.8 Can an equivalence be determined by basic contracting 
and basic splitting an edge? If yes, observe some of invariants under 
the equivalence. Otherwise, explain the reason. 


O5.9 Can an equivalence be determined by basic increasing 
duplition and decreasing duplition of an edge? If yes, observe some of 
invariants under the equivalence. Otherwise, explain the reason. 


O5.10 Can an equivalence be determined by basic increasing 
subdivision and decreasing subdivision of an edge? If yes, observe 
some of invariants under the equivalence. Otherwise, explain the rea- 
son. 


O5.11 Can the procedure of proving Lemma 5.7 by four steps 


be improved to that by three steps? If yes, provide a proof of three 
steps. Otherwise, explain the why. 


V.7 Exercises 


E5.1 By basic deleting and basic appending an edge only, prove 
(A, z, B, y, C, yz, D, yy, E) ^w (A, D, C, B, E, z, y, yz, yy). 


E5.2 Prove that a map has its orientable genus 1 if, and only 
if, 
M ~he (x, Y, 4, yy) (Bz, bt, bz, at). 
E5.3 Provide two maps of order 3 with orientable genus 1. And, 
explain they are distinct. 


For a set of operations A and a set of maps B not necessary to 
be closed under A, if for a map M € B there is no such a map N € B 
of size less than the size of M that N can be obtained from M by 
operations in A, then M is called irreducible under A. 


E5.4 For an integer k > 0, determine all the irreducible sin- 
gle vertex maps of orientable genus k under basic deleting and basic 
appending an edge. 


V.7 Exercises 141 


E5.5 For an integer n > 1, determine all the irreducible ori- 
entable single vertex maps of size k under basic deleting and basic 
appending an edge. 


E5.6 For an integer n > 1, determine all the irreducible ori- 
entable single vertex maps of size k under basic contracting and basic 
splitting an edge. 


E5.7 For an integer k > 0, determine all the irreducible single 
vertex maps of orientable genus k under basic contracting and basic 
splitting an edge. 


E5.8 Prove that a complete graph K, of order n > 3 has an 
orientable single face embedding if and only if, ("5') is even. 


E5.9 Prove that a complete graph K, of order n > 3 has an 
orientable two face embedding if and only if, a is odd. 


E5.10 Prove that a complete bipartite graph Km,n of order n + 
m > 3 has an orientable single face embedding if, and only if, (m — 
1)(n — 1) is even. 


E5.11 Prove that a complete bipartite graph Km,n of order n + 
m > 3 has an orientable two face embedding if, and only if, (m — 
1)(n — 1) is odd. 


An n-cubeis the graph formed by the skeleton of the n-dimensional 
cuboid. The order of an n-cube is 2” and the size, n2"-. 


E5.12 Prove that any n-cube, n > 2, has an two face embed- 
ding. 


A map of orientable genus 0 is also said to be planar. 


E5.13 Prove that a map M is planar if, and only if, each face 
of M corresponds to a cocycle(See 1.6) in the under graph G(M*) of 
its dual M*. 


E5.14 Prove that a map M = (Xap, P) is orientable if, and 


142 Activities on Chapter V 


only if, P can be transformed into cyclic permutation J such that 


k k 
I = (TL [Gov [Loz vm) (5.10) 
i=l i—1 


V.8 Researches 


If the travel formed by a face in a map can be partitioned into 
tours (travel without edge repetition), then the face is said to be pan- 
tour. À map with all of its faces pan-tour is called a pan-tour map . 
Because any tour can be partitioned into circuits, a pan-tour map is, 
in fact, a favorable map as mentioned in 2.8. A pre-proper embedding 
corresponds to what is called a tour map because each face forms a 
tour in its under graph. 


R5.1 Characterize and recognize that a graph has a super map 
which is an orientable pan-tour map. 


R5.2 Characterize and recognize that a graph has a super map 
which is an orientable tour map. 


R5.3  Orientable pan-tour conjecture. Prove, or improve, that 
any nonseparable graph has a super map which is an orientable pan- 
tour map. 


R5.4  Orientable tour conjecture. Prove, or improve, that any 
nonseparable graph has a super map which is an orientable tour map. 


R5.5 Characterize and recognize that a graph has a super map 
which is an orientable pre-proper map. 


R5.6 Orientable proper map conjecture. Prove, or improve, 
that any nonseparable graph has a super map which is an orientable 
proper map. 


The orientable minimum pan-tour genus , usually called ori- 
entable pan-tour genus, of a graph is the minimum among all orientable 
genera of its super pan-tour maps. Similarly, the orientable pan-tour 


V.8 Researches 143 


marimum genus of a graph is the maximum among all orientable gen- 
era of its super pan-tour maps. 


R5.7 Determine the orientable maximum pan-tour genus of a 
graph. 


R5.8 Determine the orientable maximum tour genus of a graph. 
R5.9 Determine the orientable pan-tour genus of a graph. 
R5.10 Determine the orientable tour genus of a graph. 


Although many progresses has been made on determining the 
orientable maximum genus of a graph, the study on determining ori- 
entable (maximum) pan-tour genus, or orientable (maximum) tour 
genus of a graph does not lead to any notable result yet. This sug- 
gests to investigate their bounds(upper or lower) for some class of 
graphs. 


R5.11  Characterize the class of graphs in which each graph 
has its orientable maximum pan-tour genus equal to its orientable 
maximum genus. Find the least upper bound of the absolute difference 
between the orientable maximum pan-tour genus and the orientable 
maximum genus for a class of graphs with the two genera not equal. 


R5.12  Characterize the class of graphs in which each graph has 
its orientable maximum tour genus equal to its orientable maximum 
genus. Find the least upper bound of the absolute difference between 
the orientable maximum pan-tour genus and the orientable maximum 
genus for a class of graphs with the two genera not equal. 


R5.13 Characterize the class of graphs in which each graph has 
its orientable pan-tour genus equal to its orientable genus. Find the 
least upper bound of the absolute difference between the orientable 
pan-tour genus and the orientable genus for a class of graphs with the 
two genera not equal. 


R5.14 Characterize the class of graphs in which each graph 
has its orientable tour genus equal to its orientable genus. Find the 
least upper bound of the absolute difference between the orientable 


144 Activities on Chapter V 


tour genus and the orientable genus for a class of graphs with the two 
genera not equal. 


Chapter VI 
Nonorientable Maps 


e Any irreducible nonorientable map under basic subtraction of an 
edge is defined to be a barfly. However, an equivalent class may 
have more than 1 barflies. 


e The simplified barflies are for standard nonorientable maps to show 
that each equivalent class has at most 1 simplified barfly. 


e Nonorientable rules are for transforming a barfly into another barfly 
in the same equivalent class. A basic rule is extracted for deriving 
from one to all others. 


e Principles only for nonorientable maps are clarified to transform 
any nonorientable map to a simplified barfly in the same equivalent 
class. Hence, each equivalent class has at least 1 simplified barfly. 


e Nonorientable genus instead of the Euler characteristic is an invari- 
ant in an equivalent class to show that nonorientable genus itself 
determines the equivalent class. 


VI.1 Barflies 


This chapter concentrate on discussing the basic equivalent classes 
of nonorientable maps by extracting a representative for each class. 
On the basis of Lemma 5.1 and Lemma 5.2, Only maps with a single 


146 Chapter VI Nonorientable Maps 


vertex and a single face are considered for this purpose without loss 
of generality. A nonorientable map with both a single vertex and a 
single face is called a barfly. The barfly with only one edge is the map 
consisted of a single twist loop, i.e., NU = (Ka, (x, Bxr)). 


Fig.6.1 Barflies with two edges 


Example 6.1 ‘Two barflies of size two. Let Nn” = (Kr + 
Ky, Z4) and NP = (Kx + Ky, T2)(shown, respectively, in (a) and (b) 
of Fig.6.1) where 


Ti = (£, Y, Y£, BY), To = G5 Da ug. Du). 


Because of 
(tiny = (x, By, ax, ay) 
and 


(2) ty = Gs OQ, y, ay), 
each of Nn? and NP has exactly one face. And, since 
(tomm = (z)v,,, = Kat Ky, 
they are both nonorientable. 


As mentioned in the last section, for convenience, the scope of 
maps considered here for the specific purpose should be enlarged to 
all nonorientable one vertex maps from barflies. 


VI.1 Barflies 147 


Lemma 6.1 For a single vertex map M = (Xa»e, P), M is 
nonorientable if, and only if, there exists an x € Xa p such that Bx € 
{x}p. 

Proof Necessity. By contradiction. Assume that for any x € 
Xap, there always has yx € {x}p, y = af, then az ¢ {r}u,p,,- From 
4.1, M is not nonorientable. 

Sufficiency. Since x € Xag and Bx € {x}p, from Corollary 4.1 
and only one vertex, Yip} has only one orbit on X. Hence, M is 
nonorientable. L 


This lemma can easily be employed for checking the nonori- 
entability of a one vertex map. 


Example 6.2 Six barflies of three edges. Let N89) —UNgq 
Ky + Kz,Z;), i = 1,2,---,6(shown, respectively, in (a,b,---,f) of 
Fig.6.2) where 
x, Bx, y, By, 2, Bz), 
7,y, 2, By, Bz, Bz), 
2, y, 2, Bz, yz, By), 
T, Bx, Y, 2, Bz, By), 
T, Y, Bx, 2, By, Bz), 

£, Y, Z, Px, yy, yz). 

Because Gx € {x}z7, € Ew d ^, — ag, i= 1,2,---,6, from Lemma 
6.1, they are all nonorientable. Since 


= 
dy Si 
T= ( 
ga mST 
T= 
dg eT 


Dt opo ag. ao); 
L,Y, 2, "Y, y, oz), 

z, By, ax, yz, Bz, ay), 
3, 0x Y, YZ, 02, da. 
£, ay, BZ,, YX, y, az), 


225 = e7 OZ, By, QTL, YY; rar 


=| 
= 
= 
= 


the maps (Kx + Ky + Kz, Ti), i = 1,2,---,6, are all with only one 
face. Therefore, they are all barflies. 


148 Chapter VI Nonorientable Maps 


Theorem 6.1 For any nonorientable map M, there exists a 
barfly N such that 
M ^y N. (6.1) 


(e) NO) (f) NO 


Fig.6.2 Barflies of three edges 


Proof From Lemma 5.1, by basic transformation M can be 
transformed into a single vertex map. From Theorem 4.3, in view of 


VI.2 Simplified barflies 149 


the nonorientability of M, the single vertex map is also nonorientable. 
Then from Lemma 5.2, by basic transformation the single vertex can 
be transformed into a single face map. From Theorem 4.3, this map 
is also nonorientable and hence a barfly N which satisfies(6.1). O 


This theorem enables us to restrict ourselves only to transform a 
barfly into another barfly under the basic equivalence. 


VI.2 Simplified barflies 


Let Qi = (X, Zi), l > 1, where 


l 
Ape Kay, (6.2) 
i=1 


and 
l 


Tı = (J [ (xi, Be), (6.3) 
i=] 
they are called N-standard map . When k = 1, 2, 3 and 4, Q; = N™, 
(s = NO. Qs = NO and Q4 are, respectively, shown in (a), (b), (c) 
and (d) of Fig.6.3. 


Lemma 6.2 For any | > 1, N-standard maps Q; are all nonori- 
entable. 


Proof Because all Qj, | > 1, are single vertex map and G2, € 
{x1}, L> 1, from Lemma 6.1, they are all nonorientable. [] 


Lemma 6.3 For any | > 1, N-standard maps Q are all with 
only one face. 


Proof Because Qj = NO, Q4 = NO? and Q3 = N98. from 
the two examples above, they are all with only one face. Their faces 
are (21), = (21, 021), (€1) try = ((11)1,, 22, 022) = (1, 021, 12, 02) 
and (215, = ( (51) T9523, 095) = (91,021, Ta, 025, 23; QU). 


150 Chapter VI Nonorientable Maps 


Assume, by induction, that 
[ilie quies (kūri ted Oba, seres AE qoe 
for | > 4. Since Z; = ((z1)3, ,, 21, xi), 
(z1)z4 = (Gina (Ziy)azi-1,-+:). 


And since (Zry)oxiLi = bxi = zi, (iy) = Zjo(fxi) = axı and 
(Lyon = 16x = xy, 


(Xr — Cyan TI, ox) 


— (245 OX1,°°°, 2-1, (11-1, Tl, aa). 


Therefore, all N-standard maps are with only one face. [] 
| | Bay 


(a) Ni (b) No 


T1 
T1 


Buy 


ENS "N l 
EDS 7A Ne 


Bag Bug 


(c) Ns (d) Na 


Fig.6.3 Simplified barfly 


VI.3 Nonorientable rules 151 


Because each N-standard map has only one face, from the two 
lemmas above, it is known that for any N-standard map is a barfly. 
And because each of such barflies has a simpler form, it is called a 
simplified barfly. 

Since for any | > 1, the simplified barfly Q; is with l edges, 1 
vertex and 1 face, its Euler characteristic is 


x(Q) 22-1. (6.4) 


Theorem 6.2 For any basic equivalent class of nonorientable 
maps, there exists at most one map which is a simplified barfly. 


Proof By contradiction. Assume that there are two simplified 
barflies Q; and Q;, i # j, i,j > 1, in a basic equivalent class of 
nonorientable maps. From Theorem 4.5 and (6.4), 


x(Qi) =2-i=2- j = x(Q;). 
This implies i = 7. A contradiction to the assumption. El 
In the following sections, it will be shown that there exists at 


least one map which is a simplified barfly in each basic equivalent 
class of nonorientable maps. 


VI.3 Nonorientable rules 


As mentioned above, this section is for establishing two basic 
rules of transforming a nonorientable single vertex map into another 
nonorientable single vertex map within basic equivalence. 


Lemma 6.4 Fora nonorientable single vertex map M = (Xap, T), 
if Z = (A, x, B, Bx, C) where A, B and C are segments of linear order 
in the cycle Z on Xag, then 


T ^. (A,aB", C, 2, Br). (6.5) 


Note 6.1 For a segment B = (Zz,T?z,- --, Z?z), Bx = Tir, 


152 Chapter VI Nonorientable Maps 


of linear order in the cycle of Z on Z, 5, from Theorem 2.3, 


(aX "p. Toga s T (ad^ 
= (o?z, aD? 1x, -- , o x) (6.6) 
sab 


Proof Two steps expressed by claims are considered for trans- 
forming a nonorientable single vertex map into another nonorientable 
single vertex map under the basic equivalence. 


Claim 1 (A,z, B, 8z,C) ~pe (A, 9B 1, By, aC" 1, y). 

Proof By basic splitting an edge e, between the two angles 
(az, Zx) and (C, A)(i.e., the angle between C and A) on Z = (A,x, B, 
Bx, C), 

L ™be (A, T, y) (YY; B, pz, C) 
Uum ac yack) 
= (y, A, x) (yz, o B1, By, aC"), 
as shown in (a) and (b) of Fig.6.4. 

Because e, is a link in Z} = (y, A, z)( yz, o.B !, 8y, aC 1), by 
basic contracting e,, 

Tı ^vbc (y, A, oB^l, By, aC!) 
= (A, oB^!, By, aC", y), 


as shown in (c) of Fig.6.4. 
Claim 2 7» ~pe (A, a B-t, C, Bx, x) where 
T, = (A,aB !, By, aC 1,3). 
Proof By basic splitting e, between the two angles (yy, T28y) 
and (ay, Izy) on (A,aB™, By, aC", y), 


To XYbe (A, aB, By, v) (ya, aC, y) 
= (A, aBn S, By, g) (ay, C, Bx) 
= Gs A, aB}, By) (ay, C, Bur); 


VL3 Nonorientable rules 153 


as shown in (d) and (e) of Fig.6.4. 
Because e, is a link in Z3 = (x, A, aB^!, By) (ay, C, Bx), by basic 
contracting €y, 
T3 ^w (2, A,B, C, Bx) 
= (A,aB",C, Bx, x), 


as shown in (f) of Fig.6.4. 


From Claim 1 and Claim 2, 


1 ^vbc Ti XY be To ~Y be Ts 
^^c (A, B^ 1, C, Bax, x). 


This is (6.5). O 


On the basis of the procedure in the proof of the lemma, the two 
claims show the following rules as basic equivalent transformations. 


Nonorientable rule 1 On a nonorientable map M = (A,,5, P) 
unnecessary to have a single vertex, if Gx € (x)p, then the map M' 
obtained by translating r and (x in a direction via, respectively, seg- 
ments C and D, and then by substituting aC~! and aD! for, respec- 
tively, C(without Gx) and D(without x) on (z)p is basic equivalent to 
M, ie., M! ^ M. 


























aB! 






























































154 Chapter VI Nonorientable Maps 
=A > BE 




















Fig.6.4 Claim 1 and Claim 2 


This is, in fact, the Claim 1 above. However, from the proof of 
Claim 2 a much simpler rule can be extracted. 


Nonorientable rule 2 On a nonorientable map M = (X,7) 
unnecessary to have a single vertex, if Gx € (x)p, then the map M’ 
obtained by translating x(or Gx) via a segment C, and then by sub- 
stituting aC"! for C without 6x(or x) on (x)p is basic equivalent to 
M, ie., M' ~pe M. 


It is seen that Nonorientable rule 1 can be done by employing 
Nonorientable rule 2 twice. Therefore, Nonorientable rule 2 is funda- 
mental. From this point of view, the proof of Lemma 6.4 can be done 
only by Nonorientable rule 2. 


Lemma 6.5 For a nonorientable single vertex map (4,5, Z), 


VL3 Nonorientable rules 155 


y = a6, it 
L= (A, z, B£,Y, zZ, YY, yz) 


where A is a segment of linear order on Xag, then 
Zoos (Am, Dau. Dy 2, De. (6.7) 


Proof By basic splitting e; between the two angles (ax, 8x) and 
(az, yy) on (A, c, Bx, y, 2, "yy. yz), 


T = (A, £, Br, y, 2, YY, 72) 

~be (YY, 72, A, m, t) (t, B, y, 2) 

= (yy, yz, A, x, t) (yv, Bt, az, ay) 

= (t, yy, yz, A, c) ys, Bt, az, oy). 
Because e, is a link in 7, = (t, yy, yz, A, x) yz, Gt, az, ay), by con- 
tracting ezr, 

T, ~be (t, "yy, yz, A, Bt, az, ay) 

= (A, Bt, az, ay, t, yy, yz). 

By substituting (A, 8t), (ay, t, yy) and (Ø) for , respectively, A, B and 
C in (6.5), 

T4 ~pe (A, Bt, By, at, y, z, Bz). 
Further, by substituting (A, Gt), (at) and (z, 8z) for, respectively, A, 
B and C in (6.5), 


d ^vbc (A, t, Bt, y, Bus 2,02) 
= (At Dora Dues Oa. 


This is (6.7). O 
This lemma shows that in a map M = (a8, P), Y = aß, if 


(x)p = (x, Bx, Y, Z, VY, VZ, A), 


then the map obtained by substituting (y, Gy, z, Bz) for (y, z, yy, yz) 
on (xz)p is basic equivalent to M, i.e., M' ~be M. This is usually 
called nonorientable rule 3. 


156 Chapter VI Nonorientable Maps 


Actually, nonorientable rule 3 can be also deduced from Nonori- 
entable rule 2. Although Nonorientable rule 2 is fundamental, Nonori- 
entable rule 1 and nonorientable rule 3 are more convenient for recur- 
sion. 


VI.4 Nonorientable principles 


In this section, barflies are only considered for this classification 
because it has been known that there is no loss of generality for general 
nonorientable maps. 


Lemma 6.6 Ina barfly N = (Xag, T), there exists an element 
x € X, such that 
T = (A, x, B, Bx, C), (6.8) 


where A, B and C are segments of Z on X, 5. 


Proof By contradiction. Since A, B and C are permitted to 
be empty, if no x € X such that Z satisfies (6.8), then from only one 
vertex, for any x € X, it is only possible that yx € (x)z and Bx £ (x)z. 
Therefore, (z)y,,., and (Gx)w,,., are the two orbits of Vz 4 on X. 
'Thus, M is orientable. This is a contradiction to the nonorientability 
of a barfly. o 


Theorem 6.3 For any barfly N = (Xa g,T), there exists an 
integer | > 1 such that 
T ~pe Qı- (6.9) 


Proof From Lemma 6.6 and Lemma 6.4, it can assumed that 


L ~be (A, | [tv;, 62;)), 
j=l 
where 2 is as great as possible in this form. Naturally, i > 1. 
From the maximality of 7 and only one vertex, x € A if, and only 
if, yx € A. 


VI.5 Nonorientable genus 157 


Two cases have to be discussed. 

Case 1 If no element in A is interlaced, then from Corollary 
5.2 and Corollary 5.1, (6.9) holds. Here, l = i. 

Case 2 Otherwise, by Lemma 5.7(the reduced rule). it can be 
assumed that 


i t 

Z ^» (B, | | (23, 625 | [( 25 vus v3); 
j=1 j=1 

where t is as great as possible in this form. Naturally, t > 1. From 

the maximality of t, no element in B is interlaced. By Corollary 5.2 

and Corollary 5.1, 


i t 

T ~pe ( qK Eu li Ux Ae 
j=l j=l 

By nonorientable rule 3, 


2t--i 
I ~pe (| | (27, 82) =D 
jal 
From (6.2) and (6.3), this is (6.9) where l = 2t + i. O 


On the basis of Theorem 6.1 and Theorem 6.3, it is know that 
there is at least one simplified barfly in each of basic equivalent classes 
for nonorientable maps. 


VI.5 Nonorientable genus 


Now, let us go back to general nonorientable maps for the in- 
variants of determining the basic equivalent classes for nonorientable 
maps. 


Theorem 6.4 For any nonorientable map N = (X,P) ina 
basic equivalent class, there has, and only has, an integer l > 1 such 
that the Euler characteristic 


id AD mde d (6.10) 


158 Chapter VI Nonorientable Maps 


Proof From Theorem 6.3, there is a simplified barfly in a basic 
equivalent class of barflies. From Theorem 6.1, in each basic equivalent 
class of nonorientable maps, there has an integer l > 1 such that Q; is 
in this class. On the other hand, from Theorem 6.2, only Q; is in this 
class. Therefore, from (6.4) and Theorem 4.5, (6.10) is obtained. O 


This integer | = 2 — y(N) > 1 is called the nonorientable genus 
of the class N is in, or of N. 

Now, it is seen from Chapters IV, V and VI that if the orientabil- 
ity of a map is defined to be 1, when the map is orientable; —1, when 
the map is nonorientable, then the relative genus of a map is the prod- 
uct of its orientability and its absolute genus (orientable genus, if the 
map is orientable; nonorientable genus, if the map is nonorientable). 
Thus, a basic equivalent class of maps(orientable and nonorientable) 
is determined by only its relative genus. 


Activities on Chapter VI 


V.6 Observations 


O6.1 Think, is the map obtained by deleting an edge on a 
nonorientable map still nonorientable? If yes, explain the reason. Oth- 
erwise, provide an example. 


O6.2 Think, the absolute genus of a map obtained via deleting 
an edge on a map(orientable and nonorientable) is at most 1 less than 
that of the original map. Explain the why. 


O6.3 Think, is the map obtained by contracting an edge on 
a nonorientable map still nonorientable? If yes, explain the reason. 
Otherwise, provide an example. 


O6.4 Think, the absolute genus of the map obtained via con- 
tracting an edge on a map(orientable and nonorientable) is at most 1 
less than that of the original map. Explain the why. 


O6.5 Observe if a nonorientable map can always be trans- 
formed into a barfly via only basic deleting and basic appending an 
edge. If it can, explain the reason. Otherwise, discuss what type of 
nonorientable maps can be done. 


O6.6 Observe if a nonorientable map can always be trans- 
formed into a barfly via only basic contracting and basic splitting 
an edge. If it can, explain the reason. Otherwise, discuss what type 
of nonorientable maps can be done. 


O6.7 Observe if there are other standard maps than simpli- 
fied barflies for the classification of nonorientable maps under basic 


160 Activities on Chapter VI 


equivalence. 


O6.8 Consider how to derive the nonorientable rule 3 only from 
the nonorientable rule 2. 


O6.9 For a map M = (X45, P), if the linear order 
(A, T; B; Gz, C) c (x)p 


is replaced by (A, aB^ 1, C, x, Bx) to transform the permutation P on 
Xa g into permutation P’ on Xag, then M’ = (X, P") is also a map. 
Is M' basic equivalent to M? If yes, explain the reason. Otherwise, 
provide an example. 


O6.10 Observe that all the nonorientable rules 1-3 are valid 
for any nonorientable map not necessary to be of single vertex. 


VILT Exercises 


E6.1 By basic deleting and basic appending edge, prove that 
(A, z, B, oz, C) ^ (A, x, B, C, o). 


= (Xab, P) and an 
element x € Xap, the linear order (x,y, yY, yy} € (x)p is replaced 
by (x, Bx, y, By) C (x)p for obtaining M' = (X, P’). Prove that 
M' ~pe M. 


E6.3 By basic deleting and basic appending an edge, prove 


E6.2 For a given nonorientable map M 


(A, £, Y, YT, VY, 2, az) ™be (A, T, AL, Y, QY, Z, az). 


E6.4 List all barflies of three edges rather than those in Exam- 
ple 2. 


E6.5 Prove that for any two barflies, one can always be trans- 
formed into another only by the nonorientable rule 2. 


The irreducibility appearing in what follows is in agreement with 
that in section V.7. 


VI.7 Exercises 161 


E6.6 Determine all irreducible barflies under the basic deleting 
and basic appending an edge. 


E6.7 For a given integer / > 1, determine all the irreducible 
maps of absolute genus / under the basic deleting and basic appending 
an edge. 


E6.8 Determine all irreducible barflies under the basic con- 
tracting and basic splitting an edge. 


E6.9 For a given integer / > 1, determine all the irreducible 
maps of absolute genus / under the basic contracting and basic split- 
ting an edge. 


E6.10 Prove that for any nonorientable map M = (4, P), 
there has, and only has, an integer l > 1 such that 


S 


(I [(ai, Yi, bti, "YYyi), X s--15 Bts31); 


i=1 

P be M 1=2s+1,s>0; (6.11) 
CI [Gs va 8m 9). W l = 25,5 > 1. 
i=1 


Here, when s = 0, 


E 


i=l 


E6.11 Prove that for any nonorientable map M = (4, P), 
there has, and only has, an integer / > 1 such that 


P ™be (24; 22,777, Xk, Dus dE ga, £21) 


k 1 
= qI Ti, li Ba) 
i=1 i=k 


by the nonorientable rules instead of Lemma 6.4. 


E6.12 Prove that for any graph G, but a tree, G has its maxi- 
mum nonorientable genus 


Iu(G) = (G) — v(G) +1 (6.12) 


162 Activities on Chapter VI 


where e(G) and v(G) are, respectively, the size(edge number) and the 
order(vertex number) of G. 


VI.8 Researches 


Similarly to Chapter V, among all nonorientable embeddings of 
a graph, the one with minimum (maximum) of absolute genus is called 
a minimum (maximum) genus embedding. 

The genus of a minimum(maximum) genus embedding on nonori- 
entable surfaces for a graph is called the minimum (maximum) nonori- 
entable genus of the graph. 

The minimum nonorientable genus of a graph is also called the 
nonorientable genus of the graph. If the minimum genus embedding is 
a nonorientable pan-tour(favorable) map, the the genus is called the 
nonorientable pan-tour(favorable) genus. 

And the likes, nonorientable pan-tour maximum genus, nonori- 
entable tour genus (or nonorientable preproper genus), nonorientable 
tour maximum genus, etc. 


R6.1 Justify and recognize if a graph has a nonorientable em- 
bedding which is a pan-tour map. 


R6.2 Justify and recognize if a graph has a nonorientable em- 
bedding which is a tour map. 


R6.3 Determine the least upper bound and the greatest lower 
bound of the nonorientable pan-tour genus(or genera) for a graph(or 
a set of graphs). 


R6.4 Determine the least upper bound and the greatest lower 
bound of the nonorientable tour genus(or genera) for a graph(or a set 
of graphs). 


R6.5 Determine the least upper bound and the greatest lower 
bound of the nonorientable proper genus(or genera) for a graph(or a 
set of graphs). 


Because it looks no much possibility to get a result simple as 


VLS Researches 163 


shown in (6.12) for determining the nonorientable pan-tour maximum 
genus, nonorientable tour maximum genus and nonorientable proper 
maximum genus of a graph in general, only some types of graphs are 
available to be considered for such kind of result. 


R6.6 Determine the least upper bound and the greatest lower 
bound of the nonorientable pan-tour maximum genus(or genera) for a 
graph(or a set of graphs). 


R6.7 Determine the least upper bound and the greatest lower 
bound of the nonorientable tour maximum genus(or genera) for a 
graph(or a set of graphs). 


R6.8 Determine the least upper bound and the greatest lower 
bound of the nonorientable proper maximum genus(or genera) for a 
graph(or a set of graphs). 


R6.9  Nonorientable pan-tour conjecture (prove, or disprove). 
Any nonseparable graph has a nonorientable embedding which is a 
pan-tour map. 


R6.10 Nonorientable tour map conjecture (prove, or disprove). 
Any nonseparable graph has a nonorientable embedding which is a 
tour map. 


R6.11  Nonorientable proper map conjecture (prove,or disprove). 
Any nonseparable graph has a nonorientable embedding which is a 
proper map. 


R6.12  Nonorientable Small face proper map conjecture(prove, 
or disprove). A nonseparable graph of order n has a nonorientable 
embedding which is a proper map with n — 1 faces. 


Chapter VII 


Isomorphisms of Maps 


e An isomorphism is defined for the classification of maps. A map is 
dealt with an isomorphic class of embeddings of the under graph 
of the map. 


e Two maps are isomorphic if, and only if, their dual maps are iso- 
morphic with the same isomorphism. 


e T'wo types of efficient algorithms are designed for recognizing if 
two maps are isomorphic. 


e Primal trail codes, or dual trail codes are used for justifying the 
isomorphism of two maps. 


e Two pattern examples show how to to recognize and justify if two 
maps are isomorphic. 


VIL1 Commutativity 


In view of topology, the basic equivalent classes of maps are, in 
fact, a type of topological equivalent classes of 2-dimensional closed 
compact manifolds without boundary, or in brief surfaces. 

Two embeddings of a graph explained in Chapter I are distinct 
if they are treated as 1-dimensional complexes to be non-equivalent 
under a topological equivalence. 


VIL1 Commutativity 165 


If a map is dealt with an embedding of a graph on a surface, then 
two distinct maps are, of course, distinct embeddings of their under 
graph. However, the conversed case is not necessary to be true. 

This Chapter is intended to introduce a type of combinatorial 
equivalence which is still seen as a type of topological equivalence but 
different from that for embeddings of a graph. 

In general, the equivalence between two maps can be deduced 
from that between two embeddings of their under graph. However, 
the coversed case is not necessary to be true. 

For two maps Mi = (A4,4(X1), Pi) and Mz = (As,5( X2), P2), if 
there exists a 1—to-1 correspondence (i.e., bijection) 


T : Xap Xi) — Xa p(X) 
between Xa 5(.X1) and Xa (X2) such that for any z € Xag X1), 
Tog) =ar z); (Bx) era c Pim) = Prin. (7.1) 
then 7 is called an isomorphism from M to M». 
Lemma 7.1 If is an isomorphism from Mı to M», then its 
inverse T~! exists, and 7^! is an isomorphism from M» to Mj. 


Proof Sincer is a bijection, 7^! exists. And, 7^! is also a 1-to-1 
correspondence from M» to Mi. For any y € Xa p(X2), let x = 7 ly € 
Xq,3(X1). Because y = Tx and 7 is an isomorphism for M; to M», 
from (T.1), 


Thar) = Qy, T(Bx) — By, T(Pi2) E Poy. 


Further, because 7^! exists, then 


T (By) = Bx = B(r y), 
T (Py) —Tiz-—T(rl). 


is an isomorphism from M» to Mj. = 


r (ay)-—arca(r y), 
1 


This implies 7^! 


1 


Based on this lemma, 7, or 7T ^ can be called an isomorphism 


between M, and Mo. 


166 Chapter VII Isomorphisms of Maps 


Examle 7.1 Let Mı = (X1, P1) where 
(= Kzit Kyt Ka+ Ku 





and 
Pi = (tutii) (U1, ya) (Cuy: 
and M» = (A5, P2) where 


Xz = Kt + Ky + Kzı + Kuz 





and 
P» = (yo, 22, £2) (YU2, Bx2) (auz, Bz2)(By2) 
as shown in Fig.7.1. 
First, let r(z1) = x9. from the first two relations in (7.1) and the 
property of Klein group, if 7 is an isomorphism between M; and M», 
then 


T(oui)e arm) = 5, 
T(Bz1) = B(ra1) = B», 
T(yzi) = v(21) = 712, 


i.e., TUR 44) = Kus. 
Then, from the third relation of (7.1), 
T(y1) = r(Piz1) = Pyr(z1) = Pore = ys. 
Thus, T(A yi) = Ky». Similarly, from 
r(z1) = T(Piy1) = Pot (yi) = Pays = 22, 
T(K 2) = Kz, and from 
T(ui) = T(Pryzi) = Pot (y21) = Pyyzo = us, 
T(Kuj) = Kus. 
Finally, check that if the 1—0-1 correspondence 7 from 4X, to 
Xə satisfies TP; = P3. In fact, from the conjugate axiom, it is only 
necessary to have 
TPy = (TH TY1, TA) Ts TVA) (TO, ryz) TI) 
= (Xa, Y2, 22) (u2, 22) (Gua, yz) (Vy2) 
= (Yo, 22, 3) (YUz, Bx2)(AU2, 322) (Bye) 
= P3: 


VIL1 Commutativity 167 


Therefore, 7 is an isomorphism between M; and Mj. 

















Bur | T" 
—_— Í 
j ü i 
(a) Mi (b) M5 


Fig.7.1 Two isomorphic maps 


Note 7.1 If the two maps Mı and M» in Example 1 are, re- 
spectively, seen as embeddings of their under graphs G, and Go, then 
they are distinct. If Ka is represented by x = (ætt, x7!) where 
at! = {x azr} and x^! = {8x, yx}, then the vertices 1G, have their 
rotation as 





E n e, a uj), (zi E ul D, (yi P, 


and hence uG; is on the projective plane (u1, u1). And the vertices of 
[G5 have their rotation as 





(235, ES Uy (z1 E uj p’ (zi E ul p? (yi 2 


and hence G2 is on the projective plane (u1, u1) as well. 
However, the induced 1-to-1 correspondence r|,(r|,(s1) = s», 
s = r,y, z, u) from u(Gi) to u(G2) has 


T]. rh yt at) = (ae ye, 2a") v (ad, a ya). 
This implies that uG and 1G» are distinct. 


Theorem 7.1 Let M; = (A&,,5(.X1), P1) and M» = (Xa ol X2), P2) 
be two maps. For a bijection 7 : Vo,g(X1) —9 | A,5(.X2), T is an iso- 


168 Chapter VII Automorphisms of Maps 


morphism if, and only if, the diagrams 


Xapi) —— Kap(Xa) 
| "| (7.3) 
Xq,a(X1) ———— Xa,(X2) 


for q = 9» = a, 11 = N = D, and for n = Pı and no = Po, are all 
commutative , i.e., all paths with the same initial object and the same 
terminal object have the same effect. 


Proof Necessity. From the first relation in (7.1), for any x € 
Xa pl Xı), T(ax) = a(rz). That is to say the result of composing the 
mappings on the direct path 


Xa p(X) ^ Xa p(X) ^ Xa p(X) 
is the same as the result of composing the mappings on the direct path 
Xa pl X1) 4 a,a(X2) 5 o,8 (X2). 


Therefore, (7.2) is commutative for 7; = 7» = a. 

Similarly, from the second and the third relations in (7.1), the 
commutativity for rj = no = D, and for n = Pı and y» = P; are 
obtained. 

Sufficiency. On the basis of (7.2), the three relations in (7.1) can 
be induced from the commutativity for m = r» = a, for m = m = f, 
and for m = Pı and nz = P». This is the sufficiency E 


VIL2 Isomorphism theorem 


Because the isomorphism between two maps determines an equiv- 
alent relation, what has to be considered for the equivalence is the 
equivalent classes, called isomorphic classes of maps. Two maps are 
said to be different if they are in different isomorphic classes. In order 
to clarify the isomorphic classes of maps, invariants should be inves- 
tigated. In this and the next sections, a sequence of elements with 


VIL2 Isomorphism theorem 169 


its length half the cardinality of the ground set. In fact, this implies 
that the isomorphic class can be determined by a polynomial of degree 
as a linear function of half the cardinality of the ground set for both 
orientable and nonorientable maps. 


Lemma 7.2 If two maps Mı and M» are isomorphic, then Mj 
is orientable if, and only if, M» is orientable. 


Proof Let M; = (Xi, Pi), i = 1,2. Assume 7 is an isomorphism 
from Mı to Mə. From (7.2), ra = ar,TB = Br and TP, = Por, ie., 
rar =a, 707  0drPg = Ps. 

Necessity. Since Mj is orientable, from Theorem 4.1, permutation 
group V; = Yip, a} has two orbits (v1), and (o1)y,, 21 € X, on A. 
And, since rat~! = a and TOTH = fj, 

Tyr =la) = r(or Tor 
= (rat!) (7677) = a 
By considering 7P,7~! = Po, for any v, € V, 
mr = We € Wo. 


Therefore, V» also has two orbits on %2, i.e., (v2)w, and (@x2)y,, where 

£2 = TX, € A». This implies that Mə is orientable as well. 
Sufficiency. Because of the symmetry of 7 between Mı and Mo, 

the sufficiency is deduced from the necessity. [] 


For a map M = (X,7) where v(M), e(M) and ¢(M) stand for, 
respectively, the order(vertex number), the size(edge number) and the 
coorder(face number) of M. 


Lemma 7.3 If two maps Mı and M» are isomorphic, then 


v(Mi) = v(Ms), €(Mi) = (M3), (Mi) = 6(M3). — (7.3) 


Proof Let M; = (Xi, Pi), i = 1,2. Assume 7 is an isomorphism 
from Mj to M». From the commutativity for r4 = Pı and nz = P» in 


170 Chapter VII Automorphisms of Maps 


(7.2), TP47-! = P». Then, for any integer n > 1 by induction, 


FUP ur =r 


Therefore, for any zı € A4, TX] = 2», 


T(z1)p, = (TT1)rpir- = (£2) pp. 


Because a 1-to-1 correspondence on vertices between M, and M» is 
induced from this, v( M) = v(M). 
Similarly, from rr ! = * and (Pi r = (P 


T(Piy)r t = Poy. 
Further, for any integer n > 1, r(P1y)"r! = (Poy)". This provides 


T(X1) Pry = (TT1) Piy- = (22)0 


as a 1-to-1 correspondence on faces between M; and M». Therefore, 


(Mı) = (Mp). 


Finally, from ror ^! 


=a and rfr! = 8 and hence ryr ^! = 4, 


for any z; € Aj, x» = Tx, implies TKx, = Kay. This provides 
a 1-to-1 correspondence on edges between M; and Mə. therefore, 
e(Mi) = e(M3). o 


For a map M = (X, P), the Euler characteristic given by (4.1) 
is X(M) = v(M) — «(M) + 6(M) where v(M), e(M) and ¢(M) are, 
respectively, the order, the size and the co-order of M. 


Corollary 7.1 If two maps Mı and Mə are isomorphic, then 


x(Mi) = x(M3). (7.4) 


VIL2 Isomorphism theorem EE 


Proof A direct result of Lemma 7.3. E 


For a map M = (X5, P), let M* = (X7 5, P*) be the dual of M. 
It is, from Chapter III, known that M* = (Xa, PY). 


Theorem 7.2 Maps M; and M» are isomorphic if, and only if, 
their duals Mf and M5 are isomorphic. 


Proof Let M; = A i = 1,2, then M7 = (CP 
i = 1,2, where AU = X and Pt = Pq, i = 1,2. 

Necessity. Suppose 7 is an isomorphic between M; and Mo, then 
from Theorem 7.1, 


tat =o; TET "=p, Par | = P». 


On the basis of this, for any xı € a = a and %2 = T71 € AD. = 


Xa 
TK x, = T{£1, 821, 021, y$i] = rx, TB2 4, Tax, TYL} 
= (22, Üz2, 05, 13) = K* T9, 
and 


TP -—T(PWaui-rTYyrOj 
= (rPir--) (rr) = Pay 
= Pp. 
This implies that the diagram 


x0 T y a 


m (7.5) 


B 
(1)* T (2)* 
2^ ————————À5 A a 


7] 


are all commutative for 71 = 1j = 0, for m = 1j = a, and for m = TX 
and 72 = P3. therefore, from Theorem 7.1, 7 is an isomorphism 
between Mf and M5 in its own right. 

Sufficiency. from the symmetry of duality, the sufficiency is de- 
duced form the necessity. [] 


ES. Chapter VII Automorphisms of Maps 


— (x 
Let M; = (4555 


and P = Piy, i = LR 


, Pi), and M? = uu F where a = A 


l 


Corollary 7.2 A bijection 7 : x ) — x9 is an isomorphism 
between maps Mı and Mg if, and only iE is an isomorphism between 
maps M; and M5. 


Proof A direct result in the proof of Theorem 7.2. [] 


VIL3 Recognition 


Although some invariants are provided, they are still far from 
determining an isomorphism between two maps in the last section. 

In fact, it will be shown in this section that an isomorphism 
between two maps can be determined by the number of invariants 
dependent on their size, i.e., a sequence of invariants in a number as 
a function of their size. 

In order to do this, algorithms are established for justifying and 
recognizing if two maps are isomorphic. In other words, an isomor- 
phism can be found between two maps if any; or no isomorphism exits 
at all otherwise. 

Generally speaking, since the ground set of a map is finite, i.e., 
its cardinality is 4e, € is the size of the map, in a theoretical point of 
view, there exists a permutation which corresponds to an isomorphism 
among all the (4c)! permutations if any, or no isomorphism at all 
between two maps otherwise. However, this is a impractical way even 
on a modern computer. 

Our purpose is to establish an algorithm directly with the amount 
of computation as small as possible without counting all the permu- 
tations. 

Here, two types of algorithms are presented. One is called vertez- 
algorithm based on (7.2). Another is called face-algorithm based on 
(7.5). 

Their clue is as follows. For two maps M; = (43, P1) and Mz = 


VIL3 Recognition 173 


(A45, P2), from Lemma 7.3, only necessary to consider |À3| = |%| 
because the cardinality is an invariant under an isomorphism. 

First, choose x, € AX, and y; € Xə (a trick should be noticed 
here!). 

Then, start, respectively, from zı and yı on Mı and M» by a 
certain rule (algorithms are distinguished by rules ). Arrange the 
orbits Milta and IA ia as cycles. If 


721) Vera, = wo (7.6) 


can be induced from y; = 7(z1), then stop. Otherwise, choose another 
yıla trick!). Go no the procedure on M» until every possible y; has be 
chosen. 

Finally, if stops at the latter, then it is shown that M; and M» are 
not isomorphic, and denoted by M; Z Mə; otherwise, an isomorphism 
between M; and M^» is done from (7.6), denoted by M, = M». 


Algorithm 7.1 Based on vertices, determine if two maps are 
isomorphic. 

Given two maps P = (X,7) and Q = (Y, Q), and their order, 
size and co-order are all equal(otherwise, not isomorphic!). In conve- 
nience, for any x € X, let |x| = |[x)p|, ie. , the valency of vertex 
(x)p. 

Initiation Given x € X, choose y € y. Let r(x) = y and 
TKzr = Ky. Label both x and y by 1. Naturally, Kz = Ky = K1 = 
(1, o1, 81, 41) (Here, the number 1 deals with a symbol!). Label (x)p 
by 0, then x = 1 is the first element coming to vertex 0. By (v, ty) 
denote that t, is the first element coming to vertex v. 

Let S be a sequence of symbols storing numbers and symbols and 
l, the maximum of labels on all the edges with a label. Here, S = Ó, 
| — 1 and the minimum of labels among all labelled but not passed 
vertices n = 0. If vertex (y1)p = (1)p, the maximum vertex label 
m = 0; otherwise, label vertex (y1)p by 1, m = 1. 


Proceeding When all vertices are labelled as used, then go to 
Halt (1). 


174 Chapter VII Automorphisms of Maps 


For n, let sp and sg be, respectively, the number of edges without 
label on (ytn)p and (4t,)o. 

If sp Z: Sg, when no y can be chosen, then goto Halt (2); other- 
wise, choose another y and then goto Initiation. 

In the direction starting from ytn, label those edges by /+1,---,/+ 
S,8 = Sp = Sg È 0 in order. Thus, two linear orders of elements with 
numbers labelled 


(ts, PYtn, ttg P-MWt,) 
and 


Fir Qytn, ar) Q wo 


are obtained. 

If the two are not equal, when no y is available to choose, then 
goto Halt (2); otherwise, choose another y and then goto Initiation. 

Put this linear order into S as last part and then substitute the 
extended sequence for S. In the meantime, label K(l-4- 1), K(l 4- 2), 
s K(l4- s) on P and Q. Substitute l + s for l. Mark vertex n as 
used. Substitute n 4- 1 for n. Let r be the number of vertices without 
label in 


(o (E L))ps s (E 489, 


and label them as m + 1, - -,m +r in order. Substitute m + r for m. 
Go on the Proceeding. 


Halt (1) Output S. (2) P and Q are not isomorphic. 


About Algorithm 7.1, from the way of choosing y, each element 
in the ground set is passed through at most once. So there exists 
a constant c such that the amount of computation is at most c|AX |. 
Since the worst case is for y chooses all over the ground set ,the 
total amount of computation is at most c|.X|?. Because of |X| = 4e 
where e is the size of the map, this amount is with its order as the size 
squared, i.e., O(€?). 

As described above, if checking all possibilities of |V|!, by Stirling 


VIL3 Recognition 175 


formula, 
I| ~ Vre P y|- 
>> O(e!) >> O(e) 
soe) 


when |V| = |X| = 4e is large enough. Thus, this algorithm is much 
efficient. 


Algorithm 7.2 Based on faces, determine if two maps are iso- 
morphic. 

Given two maps P = (Xap, P) and Q = (Vag, Q), and their 
order, size and co-order are all equal(otherwise, not isomorphic!). For 
convenience, let ¥ = Xag, Y = Y, and for any x € X, let |x| = 
l{z}p,|, i.e., the valency of face (x)p, where y = af. 


Initiation Given z € X, choose y € Y. Let r(x) = y and 
TKr = Ky. Label both x and y by 1. Naturally, Kx = Ky = 
K1 = (1,01, 81, 41] (Here, the number 1 deals with a symbol!). Label 
(z)piga by 0, then x = 1 is the first element coming to face 0. By (t) 
denote that t£; is the first element coming to face f. 

Let T' be a sequence of symbols storing numbers and symbols 
and /, the maximum of labels over all the edges with a label. Here, 
T = 0, l = 1 and the minimum of labels among all labelled but not 
passed faces n = 0. If face (y1)p, = (1)p,, the maximum face label 
m = 0; otherwise, label face (y1)p, by 1, m — 1. 


Proceeding When all faces are labelled as used, then go to Halt 
(1). 

For n, let sp and sg be, respectively, the number of edges without 
label on (7tn)py and (415) o4. 

If sp # Sg, when no y can be chosen, then goto Halt (2); other- 
wise, choose another y and then goto Initiation. 

In the direction starting from ytn, label those edges by /+1,---,/+ 
S,8 = Sp = Sg È 0 in order. Thus, two linear orders of elements with 
numbers labelled 


has Pyryts, UU Py ytn) 


176 Chapter VII Automorphisms of Maps 


and 
(yin, Dyin, ++, Qu Ita) 

are obtained. 

If the two are not equal, when no y is available to choose, then 
goto Halt (2); otherwise, choose another y and then goto Initiation. 

Put this linear order into S as last part and then substitute the 
extended sequence for S. In the meantime, label A(/ +1), K(l 4 1), 
---, K(l+ s) on P and Q. Substitute l + s for l. Mark n as used. 
Substitute n + 1 for n. Let r be the number of vertices without label 
in 


(y(L-4- 1))p,---, (y+ 8), 
and label them as m + 1, - -,m +r in order. Substitute m + r for m. 
Go on the Proceeding. 


Put this linear order into T' as last part and then substitute the 
extended sequence for T. In the meantime, label K(I +1), K(l 4- 2), 
5s K(L+ s) on P and Q. Substitute | + s for l. Mark face n as used. 
Substitute n 4- 1 for n. Let r be the number of faces without label in 


(*(L4- L)p, (E 8)», 


and label them as m + 1,-- m +r in order. Substitute m + r for m. 
Go on the Proceeding. 


Halt (1) Output T. (2) P and Q are not isomorphic. 


About Algorithm 7.2, it can be seen as the dual of Algorithm 
7.1. The amount of its computation is also estimated as O(c?). 


Note 7.2 This two algorithms suggest us that whenever a cyclic 
order of edges at each vertex is given, an efficient algorithm for justi- 
fying and recognizing if two graphs are isomorphic within the cyclic 
order at each vertex can be established. By saying an algorithm effi- 
cient, it is meant that there exists an constant c such that the amount 
of its computation is about O(c^), € is the size of the graphs. 

If without considering the limitation of a cyclic order at each ver- 
tex, no efficient algorithm for an isomorphism of two graphs has been 


VILA Justification 177 


found yet up to now. However, a new approach is, from what has been 
discussed here, provided for further investigation of an isomorphism 
between two graphs. 


VILA Justification 


In this section, it is shown that the two algorithms described in 
the last section can be used for justifying and recognizing whether, or 
not, two maps are isomorphic. 


Lemma 7.4 Let S and T are, respectively, the outputs of Al- 
gorithm 7.1 and Algorithm 7.2 at Halt (1), then 

(i) Elements in S and T are all in the same orbit of group Yip} 
on X. 

(ii) S forms an orbit of group Usp, on X if, and only if, T 
forms an orbit of group Yip} on «; 

(ii) S forms an orbit of group V». on X if, and only if, for 
any z E S, yz ES. 


Proof (i) From the proceedings of the two algorithms, it is seen 
that from an element only passes through y and P( Algorithm 7.1), or 
y and P^(Algorithm 7.2) for getting an element in S, or T. Because 
y, P, Py € VU rp and ~? = 1 , elements in S and T are all in the same 
orbit of group Vp. on «X. 

(ii) Necessity. Because S forms an orbit of group V». on X, 
and from Algorithm 7.1, S contains half the elements of X, by Lemma 
4.1, group Yip, has two orbits on A. This implies in the orientable 
case. Thus, from (i), T forms an orbit of group ipy on X as well. 

Sufficiency. On the basis of duality, it is deduced from the neces- 
sity. 

(ii) Necessity. Since S forms an orbit of group Yip} on ¥ and 
5 contains only half the elements of X, by Lemma 4.1, group Vp. 
has two orbits on X. From the orientability, for any z € S, yx € S. 

Sufficiency. Since for any x € 5, yx € S, and S only contains 


178 Chapter VII Automorphisms of Maps 


half the elements of X, by Corollary 4.1, it is only possible that S 
itself forms an orbit of group V». on X. [] 


For nonorientable maps, such two algorithms have their outputs 
S and T also containing half the elements of ¥ but not forming an 
orbit of group V (p... 


Lemma 7.5 Let S and T are, respectively, the outputs of Al- 
gorithm 7.1 and Algorithm 7.2 at Halt (1). And , let Gs and Gr 
be, respectively, the graphs induced by elements in S and T, then 
Gs = Gr = G(P). 


Proof From Lemma 7.4(i), by the procedures of the two algo- 
rithms, because the intersection of each of S and T with any quadricell 
consists of two elements incident the two ends of the edge, S, T' as 
well, is incident to all edges with two ends of each edge in map P. 

Therefore, Gs = Gr = G(P). O 


Theorem 7.3 The output S of Algorithm 7.1 at Halt (1) in- 
duces an isomorphism between maps P and Q. Halt (2) shows that 
maps P and Q are not isomorphic. 


Proof Let r be a mapping from X to Y such that the image 
and the co-image are with the same label. From the transitivity of a 
map, 7 is a bijection. Because 7T Kr = Krz, x € X, then rar | =a 
and rÜr ! = B. And in the Proceeding, for labelling a vertex (x)p, 
T(r)p = (Trx)g. From Lemma 7.5, this implies that 7PT ! = Q. 
Based on Theorem 7.1, 7 is an isomorphism between P and Q. This 
is the first statement. 

By contradiction to prove the second statement. Assume that 
there is an isomorphism 7 between P and Q. If r(x) = y, then by 
Algorithm 7.1 the procedure should terminate at Halt (1). However, 
a termination at Halt (2) shows that for any x € 4X, there is no 
elements in Y corresponding to x in an isomorphism between maps P 
and Q, and hence it is impossible to terminate at Halt (1). This is a 
contradiction. 


VILA Justification 179 


Therefore, the theorem is true. [] 


Although the theorem below has its proof with a similar reason- 
ing, in order to understand the precise differences the proof is still in 
a detailed explanation. 


Theorem 7.4 The output T of Algorithm 7.2 at Halt (1) in- 
duces an isomorphism between maps P and Q. Halt (2) shows that 
maps P and Q are not isomorphic. 


Proof Let r be a mapping from X to Y such that the image 
and the co-image are with the same label. From the transitivity of a 
map, T is a bijection. Because T Kx = Krz, x € X, then rar ! = o 
and rÜT ! = 8. And in the Proceeding, for labelling a face (x)p,, 
T(x)p, = (rz)g4. From Lemma 7.5, this implies that TP»yr ! = Qy. 
Based on Theorem 7.2, 7 is an isomorphism between P and Q. This 
is the first statement. 

By contradiction to prove the second statement. Assume that 
there is an isomorphism 7 between P and Q. If r(x) = y, then by 
Algorithm 7.2 the procedure should terminate at Halt (1). However, 
a termination at Halt (2) shows that for any x € Æ, there is no 
elements in Y corresponding to x in an isomorphism between maps P 
and Q, and hence it is impossible to terminate at Halt (1). This is a 
contradiction. 

Therefore, the theorem is true. H 


If missing what is related to y in Algorithm 7.1 and Algorithm 
7.2, then for any map M = (X, P), the procedures will always termi- 
nate at Halt (1). Thus, their outputs S and T are, respectively, called 
a primal trail code and a dual trail code of M. When an element x and 
a map P should be indicated, they are denoted by respective S,(P) 
and T}(P). 


Theorem 7.5 Let P = (X,P) and Q = (V, Q) be two given 
maps. Then, they are isomorphic if, and only if, for any x € X 
chosen, there exists an element y € Y such that S (P) = S,(Q), or 


180 Chapter VII Automorphisms of Maps 


T,(P) = T,(Q). 


Proof Necessity. Suppose 7 is an isomorphism between maps 
P = (#,P) and Q = (¥,Q). For the given element r € X, let 
y = iud From Theorem 7.3, or Theorem 7.4, $,(P) = S,(Q), or 
T,(P) = T,(Q). 

Sufficiency. From Theorem 7.3, or Theorem 7.4, it is known 
that by Algorithm 7.1, or Algorithm 7.2, their outputs induces an 
isomorphism between P = (X, P) and Q = (Y, Q). O 


Note 7.3 In justifying whether, or not, two maps are isomor- 
phic, the initial element xr can be chosen arbitrarily in one of the 
two maps to see if there is an element y in the other such that 
Se P) = 540) or TIL P) = TAQ): This enables us to do for some 
convenience. 


In addition, based on Theorem 7.5, all isomorphisms between 
two maps can be found if any. 


VIL5 Pattern examples 


Here, two pattern examples are provided for further understand- 
ing the procedures of the two algorithms described in the last section. 


Pattern 7.1 Justify whether, or not, two maps M; = (A3, P1) 
and M» = (A5, P5) are isomorphic where 


X= Krit Ky, Pi = (zy yy By)(yoi) 
and 
X = Kra + Kyo, P» = (yo, £2, By2) (yv2). 


First, for Mi, choose x = z,. By Algorithm 7.1, find S,(Mj). 
Let 


P, = (zy, yi, By) (y21) = wv. 


VIL5 Pattern examples 181 


Initiation 
xı = 1, Kay = {1,a1, 61,41), u=0,v= 1, 
Seg lL=0, m=1. 
Proceeding 
Step 1 Pı = (1, y1, 8y1) (71). 
yı = 2, Ky = 12 a2 02h u = 0, v = L 
S02 802), l=2, se mL 
Step2 P, = (1,2,82)(31). 
ues og SL, 
S = (1,2, 82, y1), [52,2 1, w= L 
Halt (1) Output: S,(Mi) = S = (1,2, 82, y1). 
Then, for Mə, because a link should correspond to a link and 
a vertex should correspond to a vertex with the same valency, y has 


only two possibilities for choice, i.e., zo and ox». Choose y = x». By 
Algorithm 7.1, find S,(M»). Let 


P» = (ya, £2, By2)(yx2) = uv. 
Initiation 


z2 = 1, Kz = {1,a1, 61, y1}, u=0,v= 1, 
S=0, l=0, m=k 


Proceeding 
Step 1 Pa= (yi, 1, 8y2) (1). 


By» = 2, K By. = 12,02,02,521. u = 0, U= L 
S = (1,2, 62), 122, n2 1, m— 1. 


Step 2 'P = (2,1, 82)(71). 


u=0, v=1, 
S = (1,2, 82, y1}, L=2, n=1, mL 


182 Chapter VII Automorphisms of Maps 

Halt (1) Output: 5,0M5) 9 5 = 11,2,02, 1X. 

Since S,(M1) = S,(M»3) and y = x2, an isomorphism from Mi to 
M^» is found as 7}: 

Ti Kai = Kx, T Ky = K yp. 
Then, choose y = ox». By Algorithm 7.1, find S,(M»). Let 
P5 = (axo, AY2, YY2) (Üx2) = uv. 
Initiation 


az = 1, Kaz = {1,a1, 01,1}, u= 0,v = 1, 
S20. l=0, m=1. 


Proceeding 
Step 1 Py = (1, ayz, yyz) (31). 


ayz = 2, Kay: = 12,02,82, 21, u= 0, v= 1, 
91,960), ET m e, 


Step 2 Pi = (1,2, 82)(71). 
u=0, v= 1, 
S = (1,2, 82, y1}, L=2, n=1, w= 1, 
Halt (1) Output: S,(M2) = S = (1,3,82,71): 
Since S,(M1) = S,(M») and y = oz», an isomorphism from Mı 
to Mo is found as 73: 
ToK a4 = Karte, ToKyy = Kaye. 


In consequence, there are two isomorphisms between M; and Mo 
above in all. Since 2 € S,(Mi) but y2 ¢ S,(M;), by Lemma 7.4(iii), 
Mı, Mə as well, is nonorientable. 


Pattern 7.2 Justify whether, or not, Mı = (43, P1) and M5 = 
(Xə, P5) are isomorphic where 


A= hay t Kyn Tp ovo) mn Va) 


VIL5 Pattern examples 183 


and 
A» = Kus Kyo, Po = (yo, 22, YY2) (722). 
First, for Mi, choose x = gı. By Algorithm 7.2, find T;(Mj). 
Let 
Pry = (21,741, ) (V1) = f9. 
Initiation 
Xj = L, Kx, = {1,a1, 81, y1}, f = 0, g = 1, 
Tel1e0 m=0. 
Proceeding 
Step 1 Piy = (1,71, y1) (791). 
y= 2; Ky = {2, a2, 82,72), f = 0, g= 1, 
Tei(Lo1,2$, (22, 21, mel. 
Step 2 Piy = (1,71, 2)(72). 
509,921 
PSA i527 q els m= 1. 

Halt (1) Output: TaM) & T = 1,41, 2,42). 

Then, for M», because a link should be corresponding to a link 
and a vertex should be corresponding to a vertex with the same va- 
lency, y only has two possibilities for choosing, i.e. , r9 and az». 
Choose y = x». By Algorithm 7.2, find T,(M»). Let 

Pay = (x2, 7X2, YY2) (yo) = fg. 

Initiation 

LQ = 1, Kx = {1,a1, DL y1}, f = 0, g = 1, 
T=), l=0, m=0. 
Proceeding 
Step 1. Poy = (1,71, yy) (a). 


yy = 2, Kyy = 12,02, 82,42), f 20, g— 1, 
T=(1,7L2) l=2 n=l; m=] 


184 Chapter VII Automorphisms of Maps 


Step 2 Poy = (1,71, 2)(72). 


f=0; g= 1, 
Pai). 82 n=1, m= 1. 


Halt (1) Output: T,(M2) = T = (1,71, 2, 72). 


Since T;(M1) = T,(M2) and y = x, an isomorphism from M; to 
M^» is found as 7}: 


Ty Kg, Kay, MK yy = Ky. 
Then, choose y = ax. By Algorithm 7.2, find T,(M2). Let 
Pay = (a, Bx», aye) (Gy2) = f9. 
Initiation 


ax = 1, Kax = {1,a1, 61,71}, f=0,g=1, 
Te 120, 20. 


Proceeding 
Step 1 Poy = (1,71, aye) (Byz). 


ays = 2, Kyy = {2, a2, 82,72}, f=0, g=1, 
T'ecLet2Lgje wel quel 


Step 2 Poy = (1,71, 2)(72). 


f=0, 9=1, 
T = Lo d A doma, n=1, mem 


Halt (1) Output: T,(M5) = T = (1,31, 2,72). 


Since 7,(Mi) = T,(M») and y = ox», an isomorphism from Mı 
to Me» is found as m: 


To)Kx| = Kart, T» Ky1 = Kaye. 


In consequence, there are two isomorphisms between M; and Mo 
in all. By Lemma 7.4(iii), Mi, M» as well, is orientable. 


Activities on Chapter VII 


VII.6 Observations 


O7.1 Observe that whether, or not, an isomorphism between 
two maps is always mapping a link to a link and a loop to a loop. If 
it is, describe the reason. Otherwise, by an example. 


O7.2 Observe that whether, or not, an isomorphism between 
two maps is always mapping an element incident with a vertex of 
valency 7 to an element incident with a vertex of valency i. If it is, 
describe the reason. Otherwise, by an example. 


O7.3 If missing rar ! = a or TT! = 8 in (7.2), whether, or 
not, T is still an isomorphism. If it is, describe the reason. Otherwise, 
by an example. 


O7.4 Provide two distinct embeddings of a graph which are 
two isomorphic maps. 


O7.5 Observe that how many non-isomorphic maps among all 
embeddings of the complete graph of order 4. 


O7.6 List all non-isomorphic maps of size 3 and find the dis- 
tribution by relative genus. 


O7.7 Explain the differences between non-isomorphic maps with 
the same under graph and distinct embeddings of the graph by exam- 
ples. 


O7.8 Explain the differences between non-isomorphic graphs 
and distinct embeddings of these graphs by examples. 


186 Activities on Chapter VII 


O7.9 Let 7 and 7» are two isomorphisms between two maps. 
Observe whether, or not, their composition 7175 l7 is also an isomor- 
phism. If it is, describe the reason. Otherwise, by an example. 


O7.10 Observe some algebraic properties on the set of all iso- 
morphisms between two maps. 


O7.11 Observe that for two maps Mı = (43,1) and Mə = 
(A, P2), is the composition of two permutations Pı and P» a map? If 
it is, describe the reason. Otherwise, by an example. 


O7.12 Observe some algebraic properties of the sets of all maps 
and premaps on the same ground set under the composition of per- 
mutations. 


VII.7 Exercises 


If an edge is with its two ends of valencies ? and j, then it is called 
a (i, j)-edge, 0 < i, j € 2e. If its two incident faces are of valencies | 
and s, then it is called a (l, s)'-edge, 0 € s,t < 2e. Here, e is the size 
of a map. 


E7.1 Let m;;(M) and n;;(M) are, respectively, the numbers of 
(i, j)-edge and (i, 7)*-edge in map M. Prove that if maps M; and M» 
are isomorphic, then for any i and j, 0 € i, j € 2e, mi(Mi) = mjj(M3) 
and ni; (Mi) = nij( Mə). 


E7.2 Prove that a bijection between the basic sets of two maps 
is an isomorphism of the two maps if, and only if, it induces both the 
correspondences between their vertices and between their faces. 


E7.3 Design an algorithm which is different from Algorithm 7.1 
and Algorithm 7.2 for justifying an isomorphism between two maps 
such that its computation amount is in the same order as their’s. 


E7.4 Prove that Algorithm 7.1 and Algorithm 7.2 are with the 
computation order O(e) in justifying an isomorphism of two maps 
which have a triangular face, or a vertex of valency 3 where e€ is the 


VIL.7 Exercises 187 


size of the maps. 


E7.5 Prove that Algorithm 7.1 and Algorithm 7.2 are with the 
computation order O(e) in justifying an isomorphism of two maps 
which have only one —articulate vertea(a vertex of valency 1), or only 
one loop where e is the size of the maps. 


E7.6 Determine the number of non-isomorphic butterflies of 
size m > 1, or establish a method to list them. 


E7.7 Determine the number of non-isomorphic barflies of size 
m > 1, or establish a method to list them. 


E7.8 On the basis of Algorithm 7.1, design an algorithm for 
justifying an isomorphism between two planar graphs(not maps!), and 
estimate its computation order. 


E7.9 On the basis of Algorithm 7.2, design an algorithm for 
justifying an isomorphism between two planar graphs(not maps!), and 
estimate its computation order. 


For any map M, let T be a spanning tree of its under graph 
G(M). Each co-tree edge is partitioned into two semi-edges seen as 
edges. Because what is obtained is just a tree when each of such 
semi-edges is seen with a new articulate vertex, it is a joint tree cor- 
responding to an embedding of its under graph G(M), also called an 
joint tree of M. 

If x and (Gz are in the same direction for an edge X = Kx 
partitioned along the joint tree, then it is said to be with the same 
sign, denoted by X and X; otherwise, different signs, denoted by X 
and X-!, or X^! and X. The cyclic order of letters with signs of such 
semi-edges partitioned into is called an joint sequence of the map. 


E7.10 Prove that a graph G is planar if, and only if, there 
exists an joint sequence of maps whose under graph is G such that 
each letter is with different signs and no two letters are interlaced. 


E7.11 Prove that for any complete graph Kn, n > 1, there 


is no joint tree for all maps whose under graph are K, such that it 
corresponds to a simplified butterfly. 


188 Activities on Chapter VII 


VILS Researches 


R7.1 Discuss whether, or not, there exits a number, indepen- 
dent on the size of a map considered, of invariants within isomorphism 
of maps for justifying and recognizing an isomorphism between two 
maps. 


R7.2 For a given graph G and an integer g, determine the 
number of distinct embeddings of G on the surface of relative genus 
g, and the number of non-isomorphic maps among them. 


R7.3 For a given type of graphs G and an integer g, find the 
number of distinct embeddings of graphs in G on the surface of relative 
genus g, and the number of non-isomorphic maps among them. 


R7.4 Determine the number of non-isomorphic triangulations 
of size m > 3. 


R7.5 Determine the number of non-isomorphic quadrangula- 
tions of size m > 4. 


R7.6 For an integral vector (no, n4, --- , Noi- -), find the num- 
ber of non-isomorphic Euler planar maps each of which has n»; vertices 
of valency 2i, à > 1. 


Because it can be shown that two graphs G4 and Gs are isomor- 
phic if, and only if, for a surface they can be embedded into, there 
exist embeddings p1(G 1) and u3(G3) isomorphic, this enables us to 
investigate the isomorphism between two graphs. The aim is at an 
efficient algorithm if any. 


R7.7 Suppose map Mj is an embedding of Gi on an orientable 
surface of genus g, justify whether, or not, there is an embedding M» 
of graph Gə such that Mə and M; are isomorphic. 


R7.8 Suppose map Mj is an embedding of G, on a non-orientable 
surface of genus g, justify whether, or not, there is an embedding M» 
of graph Gə such that Mə and M; are isomorphic. 


R7.9 According to [Liul], any graph with at least a circuit has 
a non-orientable embedding with only one face. Justify whether, or 


VII.8 Researches 189 


not, two graphs G4 and Gs have two respective single face embeddings 
which are isomorphic. 


R7.10 Justify whether, or not, a graph has two distinct single 
face embeddings which are isomorphic maps. 


A graph is called up-embeddable if it has an orientable embedding 
of genus which is the integral part of half the Betti number of the 
graph. Because of the result in [Liul], unnecessary to consider the 
up-embeddability for non-orientable case. 


R7.11 Determine the up-embeddability and the maximum ori- 
entable genus of a graph via its joint sequences. 


R7.12 For a given graph G and an integer g, justify whether, 
or not, the graph G has an embedding of relative genus g. 


Chapter VIII 


Asymmetrization 


e An automorphism of a map is an isomorphism from the map to 
itself. All automorphisms of a map form a group called its au- 
tomorphism group. of a map is, in fact, the trivialization of its 
automorphism group. 


e A number of sharp upper bounds of automorphism group orders 
for a variety of maps are provided. 


e The automorphism groups of simplified butterflies and those of 
simplified barflies are determined. 


e The realization of of a map is from rooting an element of the ground 
set. 


VIIL1 Automorphisms 


An isomorphism of a map to itself is called an automorphism . 
Let 7 be an automorphism of map M = (X, P). If for x € X, T(x) =y 
and x Æ y, then two elements x and y play the same role on M, or 
say, they are symmetric. Hence, an automorphism of a map reflects 
the symmetry among elements in the ground set of the map. 


Lemma 8.1 Suppose 7; and 72 are two automorphisms of map 
M, then their composition 7,7» is also an automorphism of map M. 


VIIL1 Automorphisms 191 


Proof Because 7, is an automorphism of M = (Xag, P), from 
[62) 
qr So mm ed ups eem. 


Similarly, for 75, 
ed E = =] 
T20T, =Q, ToDT, =p, PTa =P. 
Therefore, for 7175, 


(n)a(ri2) = (TaT2)a(73 ^n) 
= (mar; ri | 


= IQT] = a, 


(7172) 8 (7172) = (7172) 8 (Ty Ti ") 


= n (Trz) 


= nmn =P, 
and 
(ris)P(ri2) = (T172)P (ra^) 
= n(r4Pr; ri! 
=P =P. 
This implies that for 7,72, (7.2) is commutative. From Theorem 7.1, 
T|T2 is an automorphism of M as well. [] 


On the basis of the property on permutation composition, auto- 
morphisms satisfy the associate law for composition. 

Because an automorphism 7 is a bijection, it has a unique inverse 
denoted by 7~!. Because 


T 'ar=r (ror!) —(r rar ir) «a, 
and similarly, 


y peg PD 


from Theorem 7.1, 7^! is also an automorphism. 
If an element x € ¥ has r(x) = x for a mapping(particularly, an 
automorphism) 7, then z is called a fixed point of r. If every element 


192 Chapter VIII Asymmetrization 


is a fixed point of 7, then 7 is called an identity . Easy to see that an 
identity on ¥ is, of course, an automorphism of M, usually said to be 
trivial . By the property of a permutation, an identity is the unity of 
automorphisms, always denoted by 1. 

In summary, the set of all automorphisms of a map M forms 
a group, called the automorphism group of M, denoted by Aut( M). 
Its order is the cardinality of the set aut( M) = |Aut(M)]|, i.e., the 
number of elements in Aut(M) because of the finiteness. 


Theorem 8.1 Let 7 be an automorphism of map M = (&,P). 
If 7 has a fixed point, the 7 = 1, i.e., the identity. 


Proof Suppose x is the fixed point, ie. , r(x) = x. Because 7 
is an isomorphism, From (7.1), 
Tau) ewr(w) eu, 
(8x) = Br(2) = Ba 
T(Px) = P(T(£)) = Pz, 
i.e., ax, Dx and x are all fixed points. 
Then for any V € Via gp}, 


T(v(z)) = v(r(z)) = v(z). 
Therefore, from transitive axiom, every element on ¥ is a fixed 
point of 7. This means that 7 is the identity. [] 


In virtue of this theorem, the automorphism induced from T(x) = 
y can be represented by 7 = (x — y). 


Example 8.1 Let us go back to the automorphisms of the maps 
described in Pattern 7.1 and Pattern 7.2. 
If Mı and M» in Pattern 7.1 are taken to be 
M = (Kz + Ky, (x,y, By)(yz)) = Mı, 


then it is seen that only one nontrivial automorphism 7 = (x — az) 
exists. Thus, its automorphism group is 


Aut(M) = (1, (x ^ ax)}, 


VIII.2 Upper bound of group order 193 


i.e., a group of order 2. 
Then, maps M; and M» in Pattern 7.2 are taken to be 


M = (Kz 4 Ky, (xz, y, yy)(yz)) = Ms, 


it has also only one nontrivial automorphism 7 = (x — ox). So, its 
automorphism group is 


Aut(M) = (1, (x ^ ax)}, 


a group of order 2, as well. 

However, maps M; and M» here are not isomorphic. In fact, it 
is seen that M, is nonorientable with relative genus —1. and M^» is 
orientable of relative genus 1. 


VIIL2 Upper bounds of group order 


Because the automorphism group of a combinatorial structure 
with finite elements is an finite permutation group in its own right, its 
order must be bounded by an finite number. And, because there are 
n! permutations on a combinatorial structure of n elements, the order 
of its automorphism group is bounded by n!. 

However, n! is an exponential function of n according to the 
Stirling approximate formula, it is too large for determining the auto- 
morphism group in general. 

Now, it is asked that is there an constant c such that the order of 
automorphism group is bounded by n°, or denoted by O(n°), if there 
is, then such a result would be much hopeful for the determination of 
the group efficiently. 

In matter of fact, if the order of automorphism group is O(n*), c 
is independent of n for a structure with n elements, then an efficient 
algorithm can be designed for justifying and recognizing if two of them 
are isomorphic in a theoretical point of view. 


Lemma 8.2 For any map M = (4,P), the order of its auto- 
morphism group is 


aut(M) < |X| = 4e(M) (8.1) 


194 Chapter VIII Asymmetrization 


where e(M) = 1| | is the size of M. 
Proof From Theorem 8.1, M has at most |X| = 4e(M) auto- 
morphism, i.e., (8.1). O 


The bound presented by this lemma is sharp, i.e., it can not be 
reduced any more. For an example, the link map L = (Ka, (x)(»yx)). 
The order of its automorphism group is 4 = |Kz| = e(L). 


Lemma 8.3 For an integer i > 1, let v;( M) be the number of 
i-vertices(vertex incident with i semi-edges) in map M = (X, P), then 


aut(M) | 2ivi(M), (8.2) 
i.e., aut( M) is a factor of 2iv;( M). 


Proof Let r € Aut(M) be an automorphism of M. For x € X, 
(x)p is an i-vertex, assume T(x) = y. From the third relation of 
(7.1), (y)p is also an i-vertex. Then, the elements of X; = [x|Vx € 
X,|{x}p| = i) can be classified by the equivalent relation 


£ Aut Y 4> dr € Aut(M)x = Ty 


induced from the group Aut(M ). 
From Theorem 8.1, Aut(G) has a bijection with every equivalent 
class. This implies that each class has aut(M) elements. Therefore, 


aut(M) | ||. 
Because |X;| = 2iv;( M), (8.2) is soon obtained. O 


This lemma allows to improve, even apparently improve the bound 
presented by Lemma 8.1 for a map not vertex-regular(each vertex has 
the same valency). 


Lemma 8.4 For an integer j > 1, let ġ;(M) be the number of 
j-faces of map M = (X, P), then 


aut(M) | 2j6;(M), (8.3) 


VIII.2 Upper bound of group order 195 


i.e., aut(M) is a factor of 27¢;(M). 


Proof Let r € Aut(M) be an automorphism of M. For x € X, 
(x)p, is a j-face, assume T(x) = y. From the first two relations of 
(7.1), (yz) = yy. Then from this and the third relations, r((Py)x) = 
a Thus, (y)p4 is also a j-face. And, the elements of X; = {x|Vx € 

X,|{x}p,| = j} can be classified by the equivalence 


£ ~aut €— dr € Aut(M)x = Ty 


induced from the group Aut(M ). 

Further, from Theorem 8.1, Aut(G) has a bijection with every 
equivalent class. This leads that each class has aut(M) elements. 
Therefore, 

aut(M) | |: 


Because |X;| = 2j¢;(/), (8.3) is soon obtained. O 


This lemma allows also to improve, even apparently improve the 
bound presented by Lemma 8.1 for a map not face-regular(each face 
has the same valency). 


Theorem 8.2 Let v;(M) and ġ;(M) be, respectively, the num- 
bers of i-vertices and j-faces in map M = (X, P), i,j > 1, then 


aut(M) | (2iv;, 26; | Vi, i» 1, Vj, j > 1), (8.4) 


where (2iv; 276; | Vi, i > 1, Vj, j > 1) represents the greatest 


common divisor of all the numbers in the parentheses. 


Proof From Lemma 8.3, for any integer i > 1, 
aut( M) | 2iv;(M). 
From Lemma 8.4, for any integer j > 1, 
aut(M) | 2i$;(M). 


By combining the two relations above, (8.4) is soon found. O 


196 Chapter VIII Asymmetrization 


Based on this theorem, the following corollary is naturally de- 
duced. 


Corollary 8.1 Let v;(M) and ¢;(M) be, respectively, the num- 
bers of i-vertices and j-faces in map M = (X, P), i, j > 1, then 


aut(M) < (2iv;, 2jó; | Vi, i> 1, Vj, j > 1). (8.5) 
Proof A direct result of (8.4). O 
Corollary 8.2 For map M = (4,P), e(M) is its size, then 


aut( M) | 4e(M). (8.6) 


Proof Because 


4e(M) 2 25 ini(M) =2X_jġ;(M) 


i>1 jel 
= 5 2ivi(M) = 5 2jġ;(M), 
i21 j21 
we have 
(2iv;, 26; | Vi, i> 1, Vj, j > 1) | 4e(M). 
Hence, from Theorem 8.2, (8.6) is soon derived. El 


VIII.3 Determination of the group 


In this section, the automorphism groups of standard maps, t.e., 
simplified butterflies and simplified barflies, are discussed. 
First, observe the orientable case. For 


Oj= (fi) = (Aart AM, (1, Yi TY) 


VIIL3 Determination of the group 197 


by Algorithm 7.1, 


Sa (01) = Lys YL yw = 1, 2, 71, 32; 
Sas, (O1) = 1, Byi, 71, ayı = 1,2, 1, 72; 
Say, (O1) = L oy 71, 8j = 1,2, 31, 72; 
S45,(01) = 1, 71,71, y1 = 1,2, y1, 2; 

Bul CO) = Ly a= T9, eye 
Dag 1) e, gy. T a m 1,2 1,52: 
Ban tO) em T o Loa = T, 2 eu 

Bay (Oi) ec, Lue 12 01, 59. 


Uea 


Sx, (O1) = Sas (O1) 2s Sex, (O1) = S45, (01) = SO) 
= Say, (O1) = S54, (O1) — S44,(01) 
—L2,y1,32. 


Thus, O; has its automorphism group of order 8, t.e., 
aut(Qi) m4x(2x1l)2zs 


A map with a non-trivial automorphism group is said to be sym- 
metrical. If a map with its automorphism group of order 4 times its 
size, the it is said to be completely symmetrical. It can be seen that O4 
is completely symmetrical. However, none of Og, k > 2, is completely 
symmetrical although they are all symmetrical. 


Theorem 8.3  Forsimplified butterflies(orientable standard maps) 
Ox = (A, Je), k = 1, where 


k 


Ay = X Ci - Kyi) 
i=1 


and 
k 


Jk = (LIGA Vi, yi, yi), 


1=1 


198 Chapter VIII Asymmetrization 


we have 
2k, if k > 2; 


8.7 
8, ifk=1. o0 


aut(O,) = i 


Proof From the symmetry between (zi, yi, Y£i, Yyi), 1 > 2, and 
(z1, Uy, Y21, "Yy1) in Jk, k > 2, only necessary to calculate Sz, (Ox), 
Sa (Ok), Sy(Ox), S44 (0x), Sarı (Or), San(Ox) Sayı (Ok), and 
Say (Ox) by Algorithm 7.1. 

From Algorithm 7.1, 


k 
S3, (OX) = Lys Y1, vu | [ea vo veo vyd 
i=l 
k 
= 1,2, 71,72, [ [Gi — 1), 2i, 52i — 1), 72i) 
i=2 


k 
-H« ((2i — 1), 2i, (2i — 1), 21), 


k 
Saxil (Or) = = L; Q (T [I (xi, Yis YLi VY) BY, Y1, ayı 
i—1 


+ Sr (Ok), 
k 


Dn (Ox) = 1, YY, [i Vi; Yi, Vs); yl, yı 
=l 


* S5, (Ox), 
k 
Sa, (Ox) = 1, QUA, yl, at] [(z: Yi, VX; vyi)), 1 
i=l 
# S, (Ox), 
k 


S5 (Ox) = 1.92471; | [ener toyga 


i=1 
* S3, (Ox), 


VIIL3 Determination of the group 199 


k 
Say (Ox) =1 , QT], Q (ften yi yen 099) 5 01,6 


* Dei (Ox), 


k 
Bul (Ox) = um Sog b y. £1, V1, yz 
j=l 


Se (0x); 


k 
Say (Ox) = L Bun al, QT], a(] [ (ai. Vi; "Yi, qu)) | 
=l 
k 
-:1,251,92, [ [ (i= 1), 24,922 =1),;72) 
22 


= utt 


Because two automorphisms are from S%,,(Ox) = Sr (Or), Oy 
have 2 x k = 2k automorphisms altogether. Hence, when k > 2, 


aut(O;) = 2k. 
When k = 1, aut(O1) = 8 is known. O 
Then, observe the nonorientable case. for 
Qi = (45,23) = (IK, (21, 821)), 
by Algorithm 7.1, 


Sx, (Q1) = LG Don (1) = LoL 
S85, (Q1) x LOR Pais (Qi) = L 8l 


i.e., aut(Q1) = 


Theorem 8.4 For simplified barflies Qı = (45,71), | > 1, where 


l 
i=1 


200 Chapter VIII Asymmetrization 


and 
Tı = Tic: zi, 
i=l 
we have 
ato s a (8.8) 


Proof From the symmetry of (xj, Oxi), à > 2, and (x1, 92) in 
Tı, l > 2, only necessary to calculate 


S5 (Qi); Sazı (Qi); Spa (Qi) and S54, (Qi) 


by employing Algorithm 7.1. 
From Algorithm 7.1, 


l 
Sali) = 1, 81, | [ (4 Baa) 
i=2 
l 
ES 1, 61,2, 82, | | (wi, Bai) 
i=3 
l 
= | [G6 
i=1 


l 
Dads (Qi) = = 1 Q qK Tpu) me 
1—2 
* Dai G1); 


l 

Out Qi) — um da. 
i1—2 
Sz (Q1), 


VIILA4 Rootings 201 


l 


S291) = 1,81, a( e 


1—2 


2 


= 1, 61, L, BI, li (yi, o) 


= ] [G0 


i=1 
= Oy, (Qi). 


Because two automorphisms are from Sys, (Qi) = Sa (Q1), Qi has 
2 x | = 2l automorphisms altogether. Hence, when / > 2, 


aut(Qi) e. 
When l = 1, aut(Q1) = 4 is known. O 


Similarly, the two theorems can also be proved by employing 
Algorithm 7.2. 


VIIL.4 Rootings 


For a given map M = (X, P), if a subset R C X is chosen such 
that an automorphism of M with R fixed, i.e., an element of R does 
only correspond to an element of R, then M is called a set rooted map. 
The subset R is called the rooted set of M, and an element of R is 
called a rooted element . 


Theorem 8.5 For a set rooted map MÈ = (X,7), R is the 
rooted set, 


aut(M®) | |R]. (8.9) 


Proof Assume that all elements in R are partitioned into equiv- 
alent classes under the group Aut( MP). From Theorem 8.1, each class 
has aut( MÈ) elements. Therefore, (8.9) is satisfied. O 


202 Chapter VIII Asymmetrization 


Corollary 8.3 For a set rooted map MÈ? = (X,7), R is the 
rooted set, 


aut(MP) < |R]. (8.10) 
Proof A direct result of (8.9). E 


For a given map M = (X, P), if a vertex v,, x € X, is chosen such 
that an automorphism of M has to be with v, fixed, t.e., an element 
incident with v, has to correspond to an element incident with vz, then 
M is called a vertex rooted map. The vertex v, is called the rooted 
vertex of M, and an element incident with v,, rooted element . 


Corollary 8.4 For a vertex rooted map M" = (X, P), v, is 
the rooted vertex, 


aut( M") | 2|{a}p]. (8.11) 
Proof This is (8.9) when R = (xp U {ax}p. O 


For a given map M = (&, P), if face fr, x € X, is chosen such 
that an automorphism of M has f, fixed, i.e. , an element incident 
with f, should be corresponding to an element incident with f,, then 
M is said to be a face rooted map. The face f, is called the rooted 
face of M. An element in rooted face is called an rooted element. 


Corollary 8.5 Fora face rooted map M" = (X, P) with rooted 
face fr, 


aut(M^) | 2] (x1, ]. (8.12) 
Proof This is (8.9) when R = {x}p,U (ox) p,. O 


For given map M = (X, P), if edge e,, x € X, is chosen such 
that an automorphism of M is with e, fixed, t.e., an element in e, is 
always corresponding to an element in e;, then M is called an edge 
rooted map. Edge e; is the rooted edge of M. An element in the rooted 
edge is also called a rooted element. 


VIILA4 Rootings 203 


Corollary 8.6 For an edge rooted map M* = (X, P) with the 
rooted edge ez, 
aut( M*^) | |Kz|. (8.13) 


Proof The case of (8.9) when R= Ka. O 


For a given map M = (4,P), an element x € X is chosen such 
that an automorphism of M is with x as a fixed point, then M is 
called a rooted map . The element x is the root of M. The vertex, the 
edge and the face incident to the root are, respectively, called the root 
vertex, the root edge and the root face. 


Corollary 8.7 For a rooted map M" = (X, P) with its root z, 
aut( M") = 1. (8.14) 


Proof The case of (8.9) when R = {x}. E 


This tells us that a rooted map does not have the symmetry at all. 
The way mentioned above shows such a general clue for transforming 
a problem with symmetry to a problem without symmetry and then 
doing the reversion. 


Example 8.2 Map 


Mı = (Kz + Ky, (2)(y2, y, yy)) 


has 4 distinct ways for choosing the root. Because M; has the following 
4 primal trail codes 


Ox = lo, y1, 2,72) = Dar Du = 1, 2, 32, 114 = S Bx 
Sy = 1,71, 2,92, = Say Soy = 1, 2, 1,32, = Soy, 


the 4 ways of rooting are shown in Fig.8.1(a-d) where the root is 
marked at its tail. 


Example 8.3 Map M» = (Kx + Ky, (x)(y2, y, Gy)) has 4 dis- 
tinct ways for choosing the root. Because Mə has the following 4 


204 Chapter VIII Asymmetrization 


primal trail codes 


Sz = 19, Y1, 2, 82, = Sar, Sys = 1,2, 82,, 1, = Sg; 


S ex LL 52. = og Hil elus 


the 4 ways of rooting are shown in Fig.8.2(a-d) where the root is 
marked at its tail. 


| | 


(c) (d) 


Fig.8.1 Rootings in Example 1 


VIILA Rootings 





NG Aan NNO 


14 


(c) (d) 


Fig.8.2 Rootings in Example 2 


Activities on Chapter VIII 


VIIL5 Observations 


O8.1 Observe that how many embeddings the complete graph 
K4 of order 4 has. How many non-isomorphic maps are there among 
them?. 


O8.2 Consider the automorphism of the cube. Observe that 
what happens to the automorphism of those obtained by deleting, 
contracting, splitting and appending an edge on the cube. 


O8.3 Consider the automorphism of the octahedron. Observe 
that what happens to the automorphism of those obtained by deleting, 
contracting, splitting and appending an edge on the octahedron. 


O8.4 Consider the automorphism of the dodecahedron. Ob- 
serve that what happens to the automorphism of those obtained by 
deleting, contracting, splitting and appending an edge on the dodeca- 
hedron. 


O8.5 Consider the automorphism of the icosahedron. Observe 
that what happens to the automorphism of those obtained by deleting, 
contracting, splitting and appending an edge on the icosahedron. 


O8.6 Observe the duality between O8.2 and O8.3 and between 
O8.4 and O8.5. 


O8.7 Consider how to justify is there an edge in a map so that 
the order of the automorphism group of what is obtained by deleting 
the edge on the map does not change. 


O8.8 Consider how to justify is there an edge in a map so 


VIII.6 Exercises 207 


that the order of the automorphism group of what is obtained by 
contracting the edge on the map does not change. 


O8.9 Consider how to justify is there an edge in a map so that 
the order of the automorphism group of what is obtained by splitting 
the edge on the map does not change. 


OS8.10 Consider how to justify is there an edge in a map so that 
the order of the automorphism group of what is obtained by appending 
the edge on the map does not change. 

O8.11 Provide a map whose automorphism group is the cyclic 
group. 


O8.12 List all maps of size 3 with their automorphisms of order 


VIII.6 Exercises 


From the last two chapters, it is seen that the automorphism 
of a map provides a foundation for justifying is the map isomorphic 
to another. This is an pattern example for the automorphism of a 
general combinatorial structure. For instance, a graph, a network, a 
combinatorial design, a lattice, a group, a ring, a field etc. 


E8.1 Observe whether, or not, an automorphism of a map in- 
duces an automorphism of its under graph. 


E8.2 Determine the automorphism group of the map (X, P) 


where 
k 


X=) (Koi Ky) 
i=1 


and 


k k 
P= (LI Ti; Yi SIK is VYi)) 
i=l i=l 


for k > 2. 


208 Activities on Chapter VIII 


E8.3 Determine the automorphism group of the map (X, P) 


where : 
Ay KS 
ge] 
and 
Peru E xa, Dai) 
lor k > 2. 


A primal matching of a map is defined to be such a set of edges 
that any pair of its edges have no common end. If a primal matching is 
incident to all the vertices on the map, then it is said to be perfect. A 
dual matching of a map is such a set of edges that any pair of its edges 
have no incident face in common. If a dual matching is incident to all 
the faces on the map, then it is said to be perfect as well. By no means 
any map has a perfect primal matching. One having a prefect primal 
matching is called a primal matching map. By no means any map has 
a perfect dual matching either. One having a prefect dual matching is 
called a dual matching map. A map which is both of primal matching 
and of dual matching is said to be of bi-matching. 


E8.4 Suppose M is a primal matching map and P, a perfect 
primal matching of M. Let n;;(P) be the number of (i, j)-edges in P. 
Prove that for any i, j, ni;(P) #0, 


E8.5 Suppose M is a dual matching map and P*, a perfect dual 
matching. Let n; ;(P*) be the number of (i, 7)*-edges in P*. Prove that 
for any i, j, ni ;(P*) * 0, 


aut(M) | 2(i + j)n; ;(P"). 
E8.6 Design an algorithm for justifying does a primal matching 


map, a dual matching map, or a bi-matching map have an non-trivial 
automorphism. Explain their efficiency. Provide a condition for the 


VIII.7 Researches 209 


automorphism group of a primal matching map, a dual matching map, 
or a bi-matching map with, respectively, a primal matching, a dual 
matching, or a bi-matching rooted as the same as without rooting. 


For a map M, if a set of its faces is pairwise without common 
edge, then it is said to be independent. If an independent face set is 
pairwise without common vertex, it is called a cavity . If a cavity is 
spanning, i.e., all vertices of M are incident with the cavity, then it is 
called a full cavity . 


E8.7 Recognize if a vertex 3-regular(i.e., cubic) map has a full 
cavity. 


E8.8 For a vertex 3-regular map M with a full cavity H, let 
h;(H) be the number of i*-faces, à > 1, in this cavity. Prove that for 
any integer i > 1, A;(H) > 0, 


aut(M) | 6ih;(H). 
E8.9 Recognize whether, or not, a map has a full cavity. 


For a face f in a cavity H, its H-valency is the number of its 
incident primal semi-edges except for those in the boundary. 


E8.10 For a map M, let h;(H;j) be the number of j-faces of 
H-valency i. Prove that 


aut(M) | 2(¢ + 27)h;(H; j). 


E8.11 For a given integer k, find the number of non-isomorphic 
butterflies of relative genus k(k > 0), or of barflies of relative genus 
k(k « 0). 


VIII.7 Researches 


R8.1 Given 3 integers m > 1, g and s > 1, determine the 


210 Activities on Chapter VIII 


number of primal matching maps of relative genus g with size 7 and 
the order s of their automorphism groups. 


R8.2 Given 3 integers m > 1, g and s > 1, determine the 
number of dual matching maps of relative genus g with size m and the 
order s of their automorphism groups. 


R8.3 Given 3 integers m > 1, g and s > 1, determine the 
number of bi-matching maps of relative genus g with size m and the 
order s of their automorphism groups. 


The first three problems should be considered for starting from 
g = 0, 1, and then —1. Particularly, the three problems for self-dual 
maps should be firstly studied before the general cases. 


RS8.4 Find the cubic maps of size m > 7 with a given relative 
genus and the maximum order of their automorphism groups 


R8.5 Find the maps of size m > 1 with a given relative genus 
and the order 1 of their automorphism groups. 


R8.6 Prove, or disprove, the conjecture that for a given relative 
genus, almost all maps have their automorphism groups of order 1. 


R8.7 Given three integers m > 1, g and s > 1, determine the 
full cavity maps of size m with relative genus g and the order of their 
automorphism groups s. 


If a map has a set of edges inducing a Hamiltonian circuit on its 
under graph, then it is called a primal H-map . If a map has e set of 
edges inducing a Hamiltonian circuit on the under graph of its dual, 
then it is called a dual H-map . If a map is both a primal H-map and 
a dual H-map, then it is called a double H-map. 


R8.8 Given three integers m > 1, g and s > 1, determine the 
primal H-maps of size m with relative genus g and their automorphism 
group of order s. 


R8.9 Given three integers m > 1, g and s > 1, determine the 


VIII.7 Researches 211 


dual H-maps of size m with relative genus g and their automorphism 
group of order s. 

R8.10 Given three integers m > 1, g and s > 1, determine the 
double H-maps of size m with relative genus g and their automorphism 
group of order s. 


Chapter IX 
Rooted Petal Bundles 


e A petal bundle is a map which has only one vertex, or in other 
words, each edge of self-loop. 


e From decomposing the set of rooted orientable petal bundles, a lin- 
ear differential equation satisfied by the enumerating function with 
size as the parameter is discovered and then an explicit expression 
of the function is extracted. 


A quadratic equation of the enumerating function for rooted petal 
bundles on the surface of orientable genus 0 is discovered and then 
an explicit expression is extracted. 


e From decomposing the set of rooted nonorientable petal bundles, 
a linear differential equation satisfied by the enumerating function 
with size as the parameter is discovered in company with the ori- 
entable case and then a favorable explicit expression of the function 
is also extracted. 


e The numbers of orientable, nonorientable and total petal bundles 
with given size are, separately, obtained and then calculated for 
size not greater than 10. 


IX.1 Orientable petal bundles 


IX.1 Orientable petal bundles 213 


A single vertex map is also called a petal bundle, its under graph 
is a bouquet. In this section, the orientable rooted petal bundles are 
investigated for determining their enumerating function with size as a 
parameter by a simple form. 

Let D be the set of all non-isomorphic orientable rooted petal 
bundles. For convenience, the trivial map 0 is assumed to be in D. 

Now, D is divided into two classes: Dj and Dy, i.e., 


D = Di + Dy (9.1) 


where (D); = (9) and Dy, of course, consists of all petal bundles in 
D other than 9. 


Lemma 9.1 Let Day = (D — a|VD € Dy}. Then, 


Din = D. (9.2) 


Proof For any D = (X, P) € Dim, there exists a D' = (4",P’) € 
Dy such that D = D'—a'. Because D' is orientable, group Y = Yi pn 
has two orbits 


{r'}p and {ar }p 
on X". Because yr’ € {r'}p, D has also two orbits 


{r}p = {r'}p — {r yr} and far}p = {ar’}p — far’, 8r), 


and hence D is orientable as well. From Theorem 3.4, petal bundle 
D' leads thatD is a petal bundle. This implies that Diy) C D. 

Conversely, for any D = (X,P) € D, P = (r,Pr,--.,P r), 
D' = (X + Kr', P^) where 


T' =(r' r'r,Pr,- Pr ; 
20) 


Because D is orientable, group V = Y; p} has two orbits {r}w and 
{ar}y on X. Thus, group V' = V, has two orbits 


(r'hw = {r}y + {r yr'd and {ar tw = (or) + (ar, 8r) 


214 Chapter IX Rooted Petal Bundles 


on X’. This means that D’ is also orientable. Because D’ has only one 
vertex, D' € D. And, from D' Æ 0, it is only possible that D' € Dg. 
Therefore, in view of D = D' — a', D € Dap. [] 


From the last part in the proof of this lemma, for any D = (r)7 € 

D, D' has 2m(D) 4-1 distinct choices such that D' = D; = D+e; € Dy, 
0 € i € 2m(D), and hence D = D' — a! where e; = Kr’ and 

Do — (r^, P. T, JT, MEE Jm DEM i= 0; 

Dj — Um Ttt, gm qr’, Tr A oe, 

1<i<2m(D)-1; 

Dom(D) m (r; T, Ir, my ge. yr’), i = 2m(D), 

for ^y = aß. 


Lemma 9.2 Let H(D) = (Dj;|i = 0,1,2,---,2m(D)} for D € 
D. Then, 
Dy — 5 H(D). (9.3) 


DED 


Proof Because of Lemma 9.1, it is easily seen that the set on 
the left hand side of (9.3) is a subset of the set on the right. 

Conversely, from H(D) C Dy, for any D € D, the set on the 
right hand side of (9.3) is also a subset on the left. O 


The importance of Lemma 9.2 is that (9.3) provides a 1—to-1 
correspondence between the sets on its two sides. This is seen from 
the fact that for any two non-isomorphic petal bundles Dı and D», 
(D) NH(D2) = 0. 

On the basis of Lemmas 9.1-2, the enumerating functions of sets 
Dı and Dy can be evaluated as a function of D's as 


fol) y 7 (9.4) 


where m(D) is the size of D. 
Because Di only consists of the trivial map ? and v has no edge, 


folz) 7 1. (9.5) 


IX.1 Orientable petal bundles 215 


Lemma 9.3 For Dy, 
2dfp 


foal) = £ fp + 2x T (9.6) 
Proof From Lemma 9.2, 
Tola) = ` a) 
DED 
= ť (Zemo + gen) 
DED 
=g ` Po 4 On ` m(D)aztP) 
DED DED 
= £t fp + 2,287» 
dx 
where fp = fp(x). This is (9.6). [] 
Theorem 9.1 The differential equation about h 
dA 
gu cs E , 
l 2a? = l + (1—2)h; (9.7) 
ho = h|z=0 = 1 


is well defined in the ring of infinite series with integral coefficients and 
finite terms of negative exponents. And, the solution is h = fp(x). 


Proof Suppose h = Ho + Hix + Hox? + -+ Hz" +--+, for 
H; € Z,, i 2 0. According to the first relation of (9.7), via equating 
the coefficients of the terms with the same power of z on its two sides, 


the recursion 
—1 + Ho = 0; 


Hı — Ho = 0; (9.8) 
Hm = (2m — 1) Aya, M22 
is soon found. From this, Hp = 1(the initial condition!), Hı = 1, ---, 
and hence all the coefficients of h can be determined. Because only 
addition and multiplication are used in the evaluation, all Hm, m > 1, 
are integers from integrality of Ho. This is the first statement. 


216 Chapter IX Rooted Petal Bundles 


As for the last statement, from (9.1) and (9.5-6), it is seen that 
h = fp(x) satisfies the first relation of (9.7). And from the initial 
condition hy = fp(0) = 1, we only have that h = fp(x) by the first 
statement. L 


In fact, from (9.8), 


i (2m)! 
eet) = eur 
E: mm! 


where m > 0. 
Further, from Theorem 9.2, 


-14 xo TEN Pimp (9.9) 


Example 9.1 From (9.9), there are 3 orientable rooted petal 
bundles of size 2. 

However, there are 2 orientable non-rooted petal bundles as shown 
in (a) and (b) of Fig.9.1. 

In (a), based on primal trail code(or dual trail code), only 1 
rooted(r; as the root) element. In (b), 2 rooted(ro and r3 as the 
roots) elements. 














Fig.9.1 Petal bundles of size 2 


IX.2 Planar pedal bundles 217 


IX.2 Planar pedal bundles 


Petal bundles are here restricted to those of genus 0, i.e., planar 
pedal bundle. Rooted is still considered. Because orientable petal 
bundles can be partitioned into classes by genus as 

Dew (9.10) 
k>0 
where D; is the set of rooted petal bundles with orientable genus k. 
What is discussed in this section is Dp. For convenience, the trivial 
map ¥V is included in Dy. 

For this, Dp should be partitioned by the valency of root-face 

into classes as 
Dy > (9.11) 
s>0 
where F,, s > 0, is planar rooted petal bundles with the root-face of 
valency s. 


Lemma 9.4 Let S(the trivial map 0 is included) and T(V is 
excluded) be two set of rooted maps. If for any S = (X,P) € S — 0, 
there exist an integer k > 1 and maps S; = (X, P) € T, 1 €i k, 
such that 


k 
w=) x, (9.12) 
i=l 
and P is different from P;, 1 < i < k, only at vertex 
(r)p = ((r1) Py (72) Past tts (Tk) Px) (9.13) 


where r = ri. Then, 1 
fs(x) = 1- f) (9.14) 


where fs(x) and fr(x) are the enumerating functions of, respectively, 
S and 7. 


Proof First, S is classified based on k mentioned above, k > 0, 


SUY Ss 


k>0 


1.8: 


218 Chapter IX Rooted Petal Bundles 


Naturally, So = (0). Then, because any M; = (Vk, Qk) E Sk, k > 1, 
has the form as shown in (9.12) and (9.13) (4 and P are, respectively, 
replaced by Vy and Q;), we have 


fs, (x) = qe) 


My €Sy 
5 q m1) m(S2) + +m(Sk) 


($1,583.94) 
S,€T, 1<i<k 


= (fr(2))*. 
Therefore, by considering fs,(x) = 1, 


fs(x) = X fs,(2) 


k>0 


=14+ Y ro) 


k>1 


E 1 
1— fr(x) 
Notice that since x is an undeterminate, it can be considered for the 
values satisfying | fr(x)| < 1. This lemma is proved. O 


If S and 7 are, respectively, seen as Dp and F}, it can be checked 
that the condition of Lemma 9.4 is satisfied, then 


1 
= 1f) 
Further, another relation between fp,(r) and fz,(x) has to be 
found. 


(9.15) 


Lemma 9.5 Let Fay = {D — a|VD € 7j. Then, 
Fa = Do. (9.16) 
Proof Because 0 Z F,, for any D € 7, from the planarity of 


D, D' = D—ais planar and from D with a single vertex, D' = D—a 
is with a single vertex. Hence, D' € Do. This implies that Fa; C Do. 


IX.2 Planar pedal bundles 219 


Conversely, for any D' = (X', P") € Do, in view of a single vertex, 

P= (r)p. Let 
D= D'az-(X'- Kr,P) 

where P = (r,(rp,^r). Naturally, D is of single vertex. Because 
D is obtained from D' by appending an edge, from Corollary 4.2 and 
Lemma 4.6, the planarity of D' leads that D is planar. And, from 
(r)p, = (r), D € Fy. Since D' = D — a, D' € Fy). This implies that 
Dy ta [] 


Because this lemma provides a 1-to-1 correspondence between 
Fı and Dy, it is soon obtained that 


faz) = 3e pou 


DEF, 
=r »» pO) (9.17) 
Dc€Dg 
eeu 
In virtue of (9.17) and (9.15), 
= ER (9.18) 


Theorem 9.2 Let h® = fp,(x) be the enumerating function 
of planar rooted petal bundles with the size as the parameter, then 


A?) = Sn (9.19) 


m (m 4- 1)! 


Proof From (9.18), it is seen that h) satisfies the quadratic 
equation about h as 
zh? — h 4-1 — 0. 
It can be checked that only one of its two solutions is in a power series 
with all coefficients non-negative integers. That is 


(). 1— vl-—4v 


h 
2x 


220 Chapter IX Rooted Petal Bundles 


By expanding \/1 — 4x into a power series, (9.19) is soon found via 
rearrangement. El 


From the quadratic equation, a non-linear recursion can be de- 
rived for determining the coefficients of h. However, a linear recursion 
can be extracted for getting a simple result. This is far from an uni- 
versal way. 


Example 9.2 From known by (9.19), the number of planar 
rooted petal bundles of size 3 is 5. However, there are 2 planar non- 
rooted petal bundles altogether, shown in (a) and (b) of Fig.9.2. In 
(a), by primal trail codes(or dual trail codes), 3. Their roots are r1, 
r and r3. In (b), 2. Their roots are r4 and rs. 


TQ] =2 TA-— c 


wd 7 xL 
ze NEP 2n 


z 


(a) (b) 


Fig.9.2 Planar petal bundles of size 3 


IX.3 Nonorientable pedal bundles 


The central task of this section is to determine the enumerat- 
ing function of nonorientable rooted petal bundles with size as the 
parameter. 


Let U be the set of all nonorientable rooted petal bundles. Be- 
cause the trivial map is orientable, 0 is never in U. In other words, 
any map in U does have at least one edge. Now, & is partitioned into 


IX.3 Nonorientable pedal bundles 221 


two classes: Uy = (M|VM € U, M — a orientable} and 
Uy = (M|VM € U, M — a nonorientable], 


lEs 
U = ug. (9.20) 
First, the decomposition of the two sets Ur and My should be 
investigated. 


Lemma 9.6 Let Um = (M — a|VM € U}. Then, 
Hay =D (9.21) 
where D is the set of all orientable rooted petal bundles given by (9.1). 
Proof For M = (X,P) € Uy, if M 7 92, then M' = (XP^) 
where 4’ = X + Kr’ and P’ is different from P only at the vertex 
(r')p = (r', Br’, r, Pr, P?r, ..., Pr) 
such that M = M' — a’. From M' € Mı, M € M. If M = 9, then 
M = (Kr',(r', Br’) € ut 


such that M = M' — a'. Meanwhile, M € D. Hence, U € D. 
Conversely, for M = (4,P) € D, let M' = (X', P') such that 
X' = X + Kr’. Because M has a single vertex, 


Pi = (r)m — (r^, Br. r, Pr, P’r, Eds po. 
Therefore, M’ has a single vertex as well. And, since r^, Gr’ € {r’}w, 


M' € U. By reminding that M = M' — a’ is orientable, M’ € (f. 
Thus, M € Uy. This implies that D C U1. E 


Lemma 9.7 For any D = (X, F) € D, r = r(D), let B(D) = 
{ B;|0 € i € 2m(D)) where m(D) is the size of D, and 
Br vim. der 
(r'r, Br, F'r,---), 
1€ ix 2m(D) — 1; 
(r', v, , Fr,- , 8r), i = 2m(D). 


B;(D) = (9.22) 


222 Chapter IX Rooted Petal Bundles 


'Then, 
Uu — 5 B(D). (9.23) 


DED 
Proof For any M = (Z, P) € Uh, because 
Deu 


M is only some B;, 1 € i € 2m(D) in (9.22) such that Pr’ = r, 
or P?r' = r(Here, r' and r are, respectively, the roots of M and D). 
Therefore, M is also an element of the set on the right hand side of 
(9.23). 

Conversely, for an element M in the set on the right of (9.23), 
from Lemma 9.6, M € Uy. This is to say that M is also an element of 
the set on the left hand side of (9.23). [] 


Example 9.3 Let D = (Kz,(z,yx)) € D. Then, D is of size 
l, be mi) 1, 

Three rooted petal bundles Bo(D), B4(D), and B»(D) € ttj are 
produced from D and shown as, respectively in (a), (b), and (c) of 
Fig.9.4 where r' = y and r = z. 






































Fig.9.3 Nonorientable petal bundles from orientable ones 


Lemma 9.8 Let Unn = (M — a|VM € Up}. Then, 
Un) =U. (9.24) 
Proof For M = (X,P) € Um, let M’ € Uy such that M = 


M' — a. Because M' is a nonorientable petal bundle and M’ € Uy, M 
is a nonorientable petal bundle as well, i.e., M € U. 


IX.3 Nonorientable pedal bundles 223 


Conversely, for any M = (X, P) € U, there exists M' = (X',P^) € 
Uy such that M = M' — d', e.g., X! = X + Kr’, P! = (r' yr, (rp). 
Therefore, M € Ui. [] 


Further, observe that for a map M € U, how many non-isomorphic 
maps M’ € Uyare there such that M = M’ — a’. Two cases should be 
considered: (1) r', yr’ and r are in the same orbit of P’; (2) r', Br’ 
and r are in the same orbit of P’. 

(1) Based on the rule of rooting, because yr’ only has 2m( M) 4-1 
possible positions, 7.e., 

yr! = Pir! P'r, P'(Pr), T: Ppp, 
then 
Ce Ae (r)p), i = 0; 
TT, r! P'r,---), 
HM) = E (9.25) 
1<i<2m(M) —-1; 
(r', (r)p, yr), i = 2m(M). 

(2) Based on the rule of rooting, because 8r’ also has 2m(M) + 

lpossible positions, t.e., 


Br! 2 lp! P'r, P'(Pr), I Pipe, 
then —- 
(r OF , (r)p), 1 = 0; 
m: POTE BF. P'r, -Js 
1<i<2m(M) -1; 
(r', (r)p, 8r), i 2 2m(M). 


Ji(M) = (9.26) 


Example 9.4 Let M = (Kz,(x,0x)) € U. The map M has 
only one edge, i.e., m( M) = 1. 

Six nonorientable petal bundles Jo(M), (M), and I2(M), with 
Jo(M), (M), and Jo(M) € Uy are produced for M and shown as, 
respectively, in (a), (b), and (c), with (d), (e), and (f) of Fig.9.4 where 
p andres. 


224 Chapter IX Rooted Petal Bundles 










































































Fig.9.4 Nonorientable petal bundles from nonorientable ones 


Lemma 9.9 For any M €U, let 


Er = U(M)fo < i € 2m(M)}; (9.27) 
J(M) = {Jj(M)|0 < j € 2m(M)}. 
Then, 

Un = X_(T(M) + J(M)). (9.28) 


Meu 


Proof For any M' = (4',P’) € Un, because M = M'—d' EU, 
M' is only some J;, 0 < i € 2m(M) — 1 in (9.25), or some J;, 0 < 
j € 2m(M) — 1 in (9.26) such that Pr’ = r, or P?r' = r(Here, r' and 
r are, respectively, the roots of M’ and M). Therefore, M is also an 
element of the set on the right hand side of (9.28). 

Conversely, for an element M in the set on the right of (9.28), 
from Lemma 9.8, M € Uy. This is to say that M is also an element 
of the set on the left hand side of (9.28). E 


Lemma 9.10 Let S and 7 be two sets of maps. If for any 
T € T, there exists a set C(T) C S such that 


IX.3 Nonorientable pedal bundles 225 


(i) for any T € 7, |£C(T)| = am(T) + b and for S € S, m(T) = 
m(S) — c, where a b and c are constants and m(T) is an isomorphic 
invariant, e.g., the size; and 

(y S= S UD 

TET 


then 


fs(x) = z°(bfr + art) (9.29) 


Proof Because £L(T) provides a mapping from a map in 7 toa 
subset of S and the cardinality of the subset is only dependent on the 
parameter of the enumerating function(by (i)), and (ii) means that 
the mapping provides a partition on S, then 


x spe 3 (am(T) + b)a™) 
TET 


=b >», LU) -- ax ` mT 


TET TET 


d 
= bfr + ur 
dg 


This is (9.29) by multiplying x^ to the two sides. E 


Theorem 9.3 The enumerating function g = fulx) of nonori- 
entable rooted petal bundles in the set U with size as the parameter 
satisfies the equation as 


dh 
4,238 = (1 — 2x)g — z(h + 22 —); 
o da (9.30) 
dg 
Jz 2=0 E 1, 
k 


where h = fp(x) is the enumerating function of orientable rooted petal 
bundles given by (9.9). 


Proof Because (9.23) and (9.28) provides the mappings from 
maps D € D and U € YU to, respectively, a subset of Ut with 2m(U)+1 
elements and a subset of Uy with 2(2m(U) +1) = 4m(U) +4 elements, 


226 Chapter IX Rooted Petal Bundles 


where D and U are 1 edge less than their images, from Lemma 9.10, 


dA dh 
= 2p—)- 2g? — 
fu(x) 9 x(h + emt th + 2x qm 
and 


— dg dg 
fu, (x) = «(2g + Ar) = 2xg 1,778 
By (9.20) again, 


dh d 
g = zh 4- 227 — 4 2zg 4+ Ag? 
dx dx 


Via rearrangement, (9.30) is soon obtained. [] 


IX.4 The number of pedal bundles 


Because (9.9) provides the number of orientable rooted petal bun- 
dles with size m, m > 0, t.e., 


(2m — 1)! 


Hg ee a7 a 
2m-l(m, — 1)! 


(9.31) 
for m > 1. Of curse, Hp = 1. 

Here, the number of nonorientable rooted petal bundles with size 
m is evaluated only by the equation shown in (9.30). Let Gm be the 
number of nonorientable rooted petal bundles with size m, m > 1. 

In fact, Gm, m > 1, are determined by the recursion as 


Gm = (4m — 2)Gm-1 + Hm, mM > 2; 
i (4m — 2) 88-1 m (9.32) 
Gil. 
The solution of the recursion (9.32) is obtained, i.e., 
Muriel Z 
Cm = Saim I] mt He-» 
" = (9.33) 


Mu 25 Ie 


=2 Ju 


IX.4 The number of pedal bundles 227 


Example 9.5 When m = 1, there are one orientable rooted 
petal bundle of 1 edge, i.e., M = (Kx, (x, yx)) and one nonorientable 
rooted petal bundle of 1 edge, i.e., N = (Ka, (x, Gx)) € U. 

By appending an edge Ky on M, 3 non-isomorphic nonorientable 
rooted petal bundles of 2 edges are produced and shown in (a-c) of 
Fig.9.3. 

By appending an edge Ky on N, 6 non-isomorphic nonorientable 
rooted petal bundles of 2 edges are produced and shown in (a-f) of 
Fig.9.4. Then, Gə = 9 which is in agreement with that provided by 
(9.32), or (9.33). 


Now, Hm, Gm, and HO for m < 10 are listed in Table 9.1. 





1 1 1 1 
2 3 9 2 
3 15 105 5 
4 105 1575 14 
5 945 29295 42 
6 10395 654885 132 
7 135135 17162145 A29 
8 2027025 516891375 | 1430 
9| 34459425 17608766175 | 4862 
10 | 654729075 669787843725 | 16796 


Table 9.1 Numbers of rooted petal bundles in 10 edges 


Lemma 9.11 For an integer m > 1, the number of non-isomorphic 
nonorientable rooted petal bundles with size m is 


Gm = (2? —1) Hm (9.34) 


where Hm is given by (9.31). 


Proof By induction. When m = 1, from H; = 1, Gi = 1. (9.34) 
is true. 
Assume Gi satisfies (9.34) for any 1 € k < m — 1, m > 2. Then, 


228 Chapter IX Rooted Petal Bundles 


from (9.32), 


Gm = (4m ES 2) Gi Suas 
= (4m - 2)((27 7 = 1) Hm) + Hm. 


Since it can, from (9.31), be seen that 


Hz = (2m = 1)Hm-1, 








we have H 
Ca = lám =a] = H,, 
(4m - 2)( > + 
Am — 2 
— 90-1 DET |) Hn 
mam) 
= (2(27^ — 1) + 1)Hm 
= (27 — 1)Hm 
This is (9.34). O 


Theorem 9.4 For an integer m > 1, the number of non- 
isomorphic rooted petal bundles with size m is 


2" (2m — 1)!! (9.35) 
where " 
(2m — 1)!! = | [Q: - 1). (9.36) 


Proof Because of (9.34), the number of non-isomorphic petal 
bundles with m edges is 


Hm + Gm = 2™ Hn: (9.37) 
By substituting (9.31) into (9.37), we have 
(2m — 1)! 

2m H. — 9m NO eye O 

* 9m-T(m — 1)! 
zc 
Ll (2m — 1)! 
(m — 1)! 


= 2" (2m — 1)!!. 


IX.4 The number of pedal bundles 229 
This is (9.35). O 


The theorem above reminds the number of embeddings of the 
bouquet of size m 
2™(2m — 1)! 


derived from (1.10) as a special case. This is (m— 1)! times the number 
of rooted petal bundles with m edges. 


Activities on Chapter IX 


IX.5 Observations 


A map with only one face is called a unisheet. 


O9.1 Observe that there is a 1—to-1 correspondence between 
the set of petal bundles and the set of unisheets. 


O9.2 Think that what types of graphs can be as the under 
graph of a unisheet and what type of graphs are not as the under 
graph of a unisheet. 


O9.3 Is any planar graph a under graph of a planar unisheet? 


O9.4 Think about three ways to justify a map which is a planar 
petal bundle. 


O9.5 Discuss how to determine that a petal bundle is on the 
projective plane. 


O9.6 Discuss how to determine that a petal bundle is on the 
torus. 


O9.7 Discuss how to determine that a petal bundle is on the 
Klein bottle. 


O9.8 Discuss how to determine that a unisheet is on the plane. 


O9.9 Discuss how to determine that a unisheet is on the pro- 
jective plane. 


O9.10 Discuss how to determine that a unisheet is on the torus. 


O9.11 Discuss how to determine that a unisheet is on the Klein 


IX.6 Exercises 231 


bottle. 


IX.6 Exercises 


E9.1 Show that for any graph G, there exists a unisheet U such 
that G is the under graph of U. 


E9.2 Prove that the number of rooted unisheet with size m is 


TIL 


(2m - 1)! = [ [Gi - 1). 


i=l 
E9.3 Prove that a unisheet is planar if, and only if, its under 
graph is a tree. 


E9.4 For an integer m > 1, determine the number of rooted 
petal bundles of size m on the torus. 


E9.5 For an integer m > 1, determine the number of rooted 
petal bundles of size m on the projective plane. 


A graph is said to be unicyclic if it has only one cycle. 


E9.6 Prove that a unisheet is on the projective plane if, and 
only if, its under graph is unicyclic. 


A graph is said to be eves-cyclic if it has two fundamental circuits 
incident. 


E9.7 Prove that a unisheet is on the torus if, and only if, its 
under graph is eves-cyclic. 


E9.8 For an integer m > 1, determine the number of non- 
isomor- phic rooted petal bundles with size m on the projective plane. 


E9.9 For an integer m > 1, determine the number of non- 
isomor- phic rooted petal bundles with size m on the torus. 


E9.10 For an integer m > 1, determine the number of non- 
isomorphic rooted petal bundles with size m on the Klein bottle. 


232 Activities on Chapter IX 


E9.11 For an integer m > 1, determine the number of non- 
isomorphic rooted unisheets with size m on the projective plane. 


E9.12 For an integer m > 1, determine the number of non- 
isomorphic rooted unisheets with size m on the Klein bottle. 


A map of order 3 is also called tri-pole map. 


E9.13 For a given integer m > 1, determine the number of all 
non-isomorphic rooted tri-pole maps with size m in the plane. 


IX.7 Researches 


R9.1 For two integers m > 1 and p > 2, determine the number 
of rooted petal bundles with size m on the surface of orientable genus 
p. 

R9.2 For two integers m > 1 and q > 3, determine the number 
of rooted petal bundles with size m on the surface of nonorientable 
genus q. 


R9.3 For two integers m > 1 and p > 2, determine the number 
of rooted unisheets with size m on the surface of orientable genus p. 


R9.4 For two integers m > 1 and q > 3, determine the number 
of rooted unisheets with size m on the surface of nonorientable genus 


q. 
A map of order 2 is also called bi-pole map. 
R9.5 For a given integer m > 1, determine the number of all 


non-isomorphic orientable rooted bi-pole maps with size m 


R9.6 For a given integer m > 1, determine the number of all 
non-isomorphic nonorientable rooted bi-pole maps with size m 


R9.7 For two integers m > 1 and p > 0, determine the number 
of all non-isomorphic orientable rooted bi-pole maps with size m of 


IX.7 Researches 233 


the surface of genus p. 


R9.8 For two integers m > 1 and q > 1, determine the number 
of all non-isomorphic nonorientable rooted bi-pole maps with size m 
on the surface of genus q. 


R9.9 For a given integer m > 1, determine the number of all 
non-isomorphic orientable rooted tri-pole maps with size m 


R9.10 For a given integer m > 1, determine the number of all 
non-isomorphic nonorientable rooted tri-pole maps with size m 


R9.11 For two integers m > 1 and p > 1, determine the num- 
ber of all non-isomorphic orientable rooted tri-pole maps with size m 
of the surface of genus p. 


R9.12 For two integers m > 1 and q > 1, determine the num- 
ber of all non-isomorphic nonorientable rooted tri-pole maps with size 
m on the surface of genus q. 


R9.13 For two integers n > 5 and p > s(n) where 


(n — 3)(n — 4) 
s(n) = E 


i.e., the least integer not less than the fractional (n — 3)(n — 4)/12(or 
called the up-integer of the fractional), determine the number of rooted 
maps whose under graph is the complete graph of order n with ori- 
entable genus p. 


R9.14 For two integers n > 5 and p > t(n) where 


t(n) = [EAE 


determine the number of rooted maps whose under graph is the com- 
plete graph of order n with nonorientable genus q. 
R9.15 For two integers n > 3 and p > c(n) where 
c(n) = (n — 4)2 5 +1, 


determine the number of rooted maps whose under graph is the n-cube 
with orientable genus p. 


234 Activities on Chapter IX 


R9.16 For two integers n > 3 and q > d(n) where 
dn) (n-—4)29-*4 
determine the number of rooted maps whose under graph is the n-cube 


with nonorientable genus q. 


R9.17 For three integers m, n > 3 and p > r(n) where 


(m — 2)(n — 2) 
r(n) — p[—— 92 


determine the number of rooted maps whose under graph is the com- 
plete bipartite graph of order m + n with orientable genus p. 
R9.18 For three integers m, n > 3 and q > l(n) where 


i(n) = [E th. 


determine the number of rooted maps whose under graph is the com- 
plete bipartite graph of order m + n with nonorientable genus q. 


Chapter X 


Asymmetrized Maps 


e From decomposing the set of rooted orientable maps, a quadratic 
differential equation satisfied by the enumerating function with 
size as the parameter is discovered and then a recursion formula is 
extracted for determining the function. 


A quadratic equation of the enumerating function in company with 
its partial values for rooted maps on the surface of orientable genus 
0 is discovered with an extra parameter and then an explicit ex- 
pression of the function with only size as a parameter is via char- 
acteristic parameters extracted for each term summation free. 


e From decomposing the set of rooted nonorientable maps, a nonlin- 
ear differential equation satisfied by the enumerating function with 
size as the parameter is discovered in company with the orientable 
case and then a recursion formula is extracted for determining the 
function. 


e The numbers of orientable, nonorientable and total maps with 
given size are, in all, obtained and then calculated for size not 
greater than 10. 


X.1 Orientable equation 


It is from Corollary 8.7 shown that a map with symmetry be- 


236 Chapter X Asymmetrized Maps 


comes a map without symmetry whenever an element is chosen as the 
root. Such a map with a root is called a rooted map. 

Rooting is, in fact, a kind of simplification in mathematics, par- 
ticularly in recognizing distinct combinatorial configurations for re- 
ducing the complexity. 

As soon as the rooted case is done, the general case can be re- 
covered by considering the symmetry in a suitable way. 

For maps, the estimation of the order of the automorphism group 
of a map in the last chapter and the efficient algorithm for justifying 
and recognizing if two maps are isomorphic in Chapter VII provide a 
theoretical foundation for transforming rooted maps into non-rooted 
maps. This will be seen in the next chapter. 

The main purpose of this chapter is to present some methods for 
investigating non-planar rooted maps as appendix to the monograph 
Enumerative Theory of Maps|Liu7] in which most pages are for planar 
maps, particularly rooted. 

Let M be the set of all orientable rooted maps. For M = 
(&,P) € M, let 


Uz = (x)p = (2, Pr; Pla) (10.1) 


be the vertex incident with x € X. The root is always denoted by r. 
The rooted edge which is incident with r is denoted by a = Kr. The 
rooting of M — a is taking P?r as its root where 


ô = min(i|P'r g Kr,i > 1). (10.2) 
In fact, 
1, if P ; 
"a A (10.3) 
2, othewise. 


In virtue of Theorem 3.4, M — a is a map if, and only if, a is not 
a harmonic loop except terminal link (or segmentation edge) of M. 

Now, let us partition M into three parts: Mr, My and Mp, 
oa 


M = Mı + Mu + Mir (10.4) 


X.1 Orientable equation 237 


where Mr = {V}, i.e., consisted of the trivial map, My and Mr are, 
respectively, consisted of those with a as a segmentation edge and not. 


Lemma 10.1 Let May — (M — a|VM € My}. Then, 

May — M x M (10.5) 
where x stands for the Cartesian product of two sets. 

Proof For any M = (X,P) € M x M, let M = Mi + Ms, 
Mj = (4, Pi), i = 1,2. Assume M' = (X', P^) such that X' = X - Kr' 
and P’ is different from P; or Pı only at, respectively, 

Up — (r)p = a, T2, Pore, ubi , Px ro) 
or 
UBr! = (Gr’)p = (a Br, T1, Pira, DI Piin). 

Since M’ € M and its rooted edge a’ = Kr’ is a segmentation 
edge, M' € My. It is checked that M = M' — a’. Therefore, M € 
Map. 

Conversely, for any M € May, we have M' € My such that 


M = M' — qd! where a’ = Kr’. From M' € My, M = M, + Mə where 
Mi, M» € M. This implies that M € M x M. [] 


It is seen from this lemma that there is a 1—to-1 correspondence 
between M(€ Muy, or M x M) and M'(€ My). Hence, 

For M = (X, P) € My, because M — a is a map (Theorem 3.4), 
from (10.3), the root r((M — a) of M — a has two possibilities: when 
Pr(M) Z yr(M), r(M — a) = Pr(M); otherwise, 

r(M — a) = P?r(M). 

Let M = (¥,P) = M —a where X = X — Kr and P are different 

from P only at 
(Fis = (Pr, DE ive Por), 


if a is not a loop; 
(P4r)s = (Pyr, T?wyr, e sal II à, 18 not a loop 


238 Chapter X Asymmetrized Maps 


otherwise, 7.e., when a is a loop, 


(P?r, P3r, ..., P lr), 34 yr = Pr; 
(T)s = (Pr, E pe pg TS P), 
if yr = Pu ss 


Since M(X, P) is orientable, group V = Vy, p; has two orbits 
{r}w and {ar}y on 4. For M = (X,P) = M —a, group V = V5, 
also has two orbits 

Ug = {r}w — tr, yr} 
and 
{ar}; = (lore — (or, Gr}. 
So, M is also orientable. 

Furthermore, for every element y € {f}y, is there exactly one 
position of a = Kr, i.e., yr is in the angle (ay, P), for M € vm such 
that M = M — a. This means that in Mj, there are 


{Fal = 514] = 2m) 


where m(M ) is the size of M, non-isomorphic maps for producing M. 


By considering for the case Pr = yr, in My, there are 2m(M) +1 
non-isomorphic maps for M altogether. 


Example 10.1 Let 


M = (Kz + Ky + Kz, (x,y, yz)(z, yx, yy)) 


where f = x is the root shown in Fig.10.1(a). Since it is orientable, the 
orbit of group V which 7 is in can be written as a cyclic permutation 
as 


(f)g = (x, y, Y2, Z, YT, VY). 
Then, Fig.10.1(b-h) presents all the 2m(M) +1 = 2x 3+1=7 maps 
in Mm, obtained by appending a = Kr on map M where (b) is for 
Pr = yr and (c-h) are for those obtained by appending a = Kr in 


the order of (#)g from M. 


239 


X.1 Orientable equation 








(b) 


(a) 





2 
E] S 2 


SN 





(d) 


(c) 











(f) 


(e) 


240 Chapter X Asymmetrized Maps 


Ne AE 
LIN 4 


(g) (h) 











EM 


Fig.10.1 New maps obtained by appending an edge 
Lemma 10.2 Let Mam = (M — a|VM € Mm}. Then, 


Many = M. (10.7) 


Proof Because for any M € Myr, M —a is also a map (Theorem 
3.4), then Mam € M. 

Conversely, for any M € M, any one, e.g., M' of the 2m(M) +1 
maps obtained by appending a’ from M in the above way is with 
M' € Mym. Because M = M' — d', then M € Many. [] 


For convenience, let H(M) be the set of all the 2m(M) + 1 maps 
in Mın, obtained from M by appending an edge in the above way. 
From Theorem 8.1, they are all mutually nonisomorphic in the sense 
of rooting. 


Lemma 10.3 For Mın, 


Mu = » H(M). (10.8) 


MEM 


Proof For any M = (X,P) € Mm, let M = (X,P) = M — a. 
Because a is not a segmentation edge, from Theorem 3.4 and Corollary 
3.1, M € .M. By orientability, because r € {7}% where v= Vy py 
there exists y € {7} such that Py = yr, or Pr = yr. Because 
l{7}¢| = 2mM, the former has 2m(M) possibilities and the latter, 
only one. This is the 2m(M) + 1 possibilities in H(M). Further, 


X.1 Orientable equation 241 


because M € M, M is an element of the set on the right hand side of 
(10.8). 

Conversely, for any M € H(M), M € M, Since M = M —a € 
M, by Theorem 3.4 and Corollary 3.1, a is not a segmentation edge. 
Therefore, M € Myr. [] 


Furthermore, (10.8) provides a 1-to-1 correspondence between 
the sets on its two sides. This enables us to construct all orientable 
maps with the rooted edge not a segmentation edge from general ori- 
entable maps with smaller size. 

In order to determine the number of non-isomorphic orientable 
rooted maps in M with size m > 0, the enumerating function of set 


M 
jui)e y qu (10.9) 
MEM 
has to be investigated for a simpler form in infinite power series where 
m(M) is the size of M. In the series form of (10.9), the coefficient 
of the term with 7”, m > 0, is just the number of non-isomorphic 


orientable rooted maps with size m. 
From (10.4), 


f(x) = faux) + f(x) + fux). (10.10) 


Lemmas above enable us to evaluate f(x), fq, (x) and fy (x) 
as functions of f = f(z). 

First, because M, contains only one map V and m(V) = 0, fm, (£) 
contributes the constant term 1 of f, i.e., 


fon) = 1 EUH 
Lemma 10.4 For My, 


Proof According to the 1-to-1 correspondence between My 
and M and that the former is with its size 1 greater than the latter 


242 Chapter X Asymmetrized Maps 


in the correspondence, by (10.6), 


didus =g » x a 


MEM (11) 


=i ` iM) 


MEMxM 
MEM 
= of’. 
This is (10.12). o 


Lemma 10.5 For Mm, 


2 E 


Satale )= Bf +22 (10.13) 


Proof From the 1—to-1 correspondence between My and Mm 
and that the former is with its size 1 greater than the latter in the 
correspondence, and then by Lemma 10.3 and Lemma 9.10, 


m(M) 
Emin (x = a 


MEM yy) 


df 
= Iy 
o(f + 2L) 
=axf+ og 
dx 

This is (10.13). O 


Theorem 10.1 The differential equation about f 


df 

Di S - 

l " dz Goana (10.14) 
fo = flz-o = 1 

is well defined in the ring of infinite power series with all coefficients 

nonnegative integers and the terms of negative powers finite. And, 

the solution is f = f(x). 





X.2 Planar rooted maps 243 


Proof Suppose f = Fo + Fix + Fave + Eme" +--+, Be 
Z,, i 2 0. Based on the first relation of (10.14), by equating the 
coefficients on the two sits with the same power of x, the recursion 


—1 + fo = 0; 
P HER 0 

ma (10.15) 
Fm = (2m — 1)Fm-1 + ` FiFm-i-i 

i—0 


qmi 2 


is soon extracted. Then, Fp = 1(the initial condition), Fi = 2, ---, all 
the coefficients of f can uniquely found from this recursion. Because 
only addition and multiplication are used for evaluating all the coef- 
ficients from the initial condition that Fo is an integer, Fm, m > 1, 
must all be integers. This is the first statement. 

For the last statement, from (10.10) and (10.11-13), it is seen 
that f = f(a) satisfies the first relation of (10.14). And, fo = 
fm(0) = 1 is just the initial condition. By the uniqueness in the first 
statement, the only possibility is f = f(x). E 


Although the form of the equation in Theorem 10.1 is rather 
simple, because of the occurrence of f?, it is far from getting the 
solution directly. In fact, it is an equation in the Riccati's type. It has 
no analytic solution in general. 


X.2 Planar rooted maps 


Let 7 bethe set of all planar rooted maps. Because it looks hard 
to decompose 7 into some classes so that each class can be produced 
by 7 with only size as the parameter. Now, another parameter for 
a map M, i.e., the valency of the rooted vertex n( M), is introduced. 
The enumerating function of 7 is 


gaa) ay (10.16) 
MeT 


244 Chapter X Asymmetrized Maps 


where m(M ) is still the size of M. 
Assume that 7 is partitioned into three classes: 79, 7; and 75, 
i.e., 

T=HtTh+h (10.17) 
where Jp = (0), Ti and J, are the sets of planar rooted maps with 
the rooted edge, respective a loop and a link(not loop). 

For M € T, let a= Kr be the rooted edge of M with the root 
r — r(M). For maps M; € 7 (1-1 and 2), let a; = Kr; be the rooted 
edge of M; with the root r; = r(M;j). 

The 1-addition of two maps Mı = (A, Pı) and M» = (A5, Po) 
is to produce the map Mı +- My = Mı U Ms with the root r = ri 
provided M; N Mə = (v, ] where v, = ((r1)p,, (r2)»,). 


Lemma 10.6 Let Ta) = {M — a|VM € Ti}, then 
des sT (10.18) 


where 7 x 7 = (M4 + M3|VM;, Mə € T Y is called the 1-product of T 
with itself. 


Proof For any M = (4,P) € Ti), let MW = (X', P) € 1; such 
that M' — a/ = M. Because a’ = Kr’ is a loop, 


(r)p — Us pue. Dentes yr, E Ptr), 


From the planarity, M’—a’ = M,+- Mo, Mj = (Xi, Pi), i = 1,2, where 
X = X +% = X' — Kr’, Pi and P» are different from P only at (r)p 
becoming, respectively, 


(rijp, = (P^r', P(P'r’), «+, PY yr!) 
where y = af and 
(rap, = (P'yr', P(P'yr’), ++, P' y. 


This implies M € 7 x: 7. 
Conversely, for M € T x- T, because M = Mı + Ms, let M' = 
M +a’, a’ = Kr’, such that 


(r)p = (r5 (rp Yr, (2) Pa) 


X.2 Planar rooted maps 245 


then M' € T and M = M' — q'. Since a’ is a loop, M' € Ti and hence 
MET). O 


Because this lemma presents a 1—to-1 correspondence M = Mı + 
M» between M € Ta) and Mj, Mo € T with m(M) = m(Mij) +m(M2) 
and n(M) = n( Mı) + n(M»), the enumerating function of Tay 


fnm y)- ` g M) mM) 

MeTx T 
gta) mE), ei Mp a 

pos (10.19) 
Tn 1VÀ n( M 2 

( > qM ROM) 


MeT 
= rae y). 


Then, from the 1-to-1 correspondence between 7; and Za) with 
the former of size 1 greater than the latter and the former of the rooted 
vertex valency 2 greater than the latter, the enumerating function of 
T; is 

fa sg) eu Pa bay) ec wy T Gs) (10.20) 


However, for 72, the correspondence between 7; and 7» = (M e 
a|lVM € To} with the former of size 1 greater than the latter is not of 
1-to-1 where M e a is the contraction of the rooted edge a on map 
M. The root on M ea is defined to be Pyr when Pyr 4 yr; or 
(P4)?r otherwise. Because a is a link, this is a basic transformation. 
According to Chapter V, M — a is planar if, and only if M is. Hence, 


Toy = T. (10.21) 


Further, observe what a correspondence between 75 and 7 is. 

For M = (X,P) € T, let (r)p = (r, Pr, --, P" )-17) where 
n(M) is the valency of the rooted vertex v, on M. By splitting a link 
at v, all those obtained are still planar because this operation is a 
basic transformation. For doing this, there are n(M) 4- 1 possibilities 
altogether. 


246 Chapter X Asymmetrized Maps 


Let M; = (Xi, Pi), 0 € i € n(M) be all the n(M) + 1 maps 
obtained from M by splitting an edge at the rooted vertex v, = (r)p 
as 


(ro, (r)p), while vari = (770); 
(ri, Pr,- , p OD715). while vgn, = (yr1, r); 
(ra, P?r, . .., p 719). while Un ceu T IPI 
CD IE (10.22) 
(Tr(M)-15 pat). 
while var 45, = (YTs(my-5 ^7 Jal M) 


(Tr(a), while Ubrany = (Yrn(m), (T)P). 


Lemma 10.7 For a map M € 7, let 
K(M) = (Mili 20,1,2,---, n(M)] 
where M;, 0 € i € n(M), are given by (10.22). Then, 


T; = M KM). (10.23) 


MET 


Proof For any M € T5, because a = Kr is a link, from (10.21), 
Mea c T and from (10.22, M € K(M ea). Therefore M is an 
element of the set on the right hand side of (10.23). 

Conversely, for M is an element of the set on the right of (10.23), 
there exists a map M’ € T such that M € K(M"). Because all maps 
in K(M) are planar if, and only if, M' is planar, M € 7 as well. 
Moreover, from (10.22), the rooted edge a is a link, M € 75. This 
meas that M an element of the set on the left hand side of (10.23).0 


Because (10.23) presents a 1—to-1 correspondence between maps 
with the same size on its two sides and the valency of rooted vertex 
of M;(0 € à € n(M)) is n(M) — i for any M € T, the enumerating 


X.2 Planar rooted maps 247 


function of 75 is 


IuG)e yey 





MEJ 
n(M) 
=a DD re 
MeT  i-0 (10.24) 
ff yn M) — 
MeT 
zy 
= to — yt 
{j= " o — yt) 


where t = t(x, y) and to = t(x, 1). 


Theorem 10.2 The enumerating function t = t(x, y) of planar 
rooted maps satisfies the equation as 


zy^(1— yt? — (1— y - zy))t +rytot (1— y) =0 (10.25) 








where ty = t(x, 1). 
Proof From (10.17), 


t= fax.) + fÍn(z.vy) + fIa(*,y). 


Because 79 = {J} and V has no edge, fz,(z,y) = 1. From (10.20) an 


(10.24), 
zy 





t=1+ry t + lto — yt). 


Via rearrangement of terms, (10.25) is soon found. O 


Although (10.25) is a quadratic equation, because the occurrence 
of tọ which is also unknown and the equation becomes an identity when 
y = 1, complication occurs in solving the equation directly. 

The discriminant of the equation (10.25), denoted by D(z, y), is 


D(z,y) = (zy^ — y + 1} — 4(y — 1) (y — 1 — xyto) 


—1-2gy 4 (1— 2z)y? + (a? + 2x 





248 Chapter X Asymmetrized Maps 


—H(x))y® + H(z)y* (10.26) 


where 
H (x) = Ax?tg + x? — 4a. (10.27) 


Assume that D(z, y) has the form as 
D(x,y) = (1 — 0y) (1 ay + by’) 
= 1— (20 — a)y + (6? — 2a0 + by? 
J-0(a0 — 2b)y? + 0?by. (10.28) 
By comparing with (10.26), 








Qe 2: (10.29) 
and 
] — 2x = 0(4 — 30) + b; 
a” + 2x — H (x) = 0(2(0 — 1)0 — 2b); (10.30) 
H(x) = 0?b. 
Then, an equation about b with 0 as the parameter is found as 
1 
7 (1 — 46 c 36 b) + 1— 40 4- 36? — b — 6b 


= 2(0 — 1)0? — 20b. 
By rearrangement, it becomes 


b? — (100? — 160 + 6)b + (96% — 320? 











-420? — 240 4- 5) — 0. (10.31) 
The discriminant of (10.31) is 





(100? — 160 + 6)? — 4(90* — 320? + 420? — 240 + 5) 
= 640* — 1920? + 2080? — 960 + 16 
= (80? = 120 + 4?. 
Therefore, 


b= (@—1)’,or 96° —140 +5 = (90 —5)(0 — 1). 


X.2 Planar rooted maps 249 


The latter has to be chosen in our case. By the last relation of 
(10.30), 
H(x)«6^(00 —5)(0 —1). 


From the first relation of (10.30) and (10.27), the expressions for 
x and bọ with parameter 0 are extracted as 


x = (30 — 2)(1— 0); 


T A0 — 3 (10.32) 
0^ 130 = 2)? 
This enables us to get bọ as a power series of x as 
2x9" 2m)! 
b = — r". 10.33 
(e) 2. m!(m + 2)! E ( ) 


by eliminating the parameter 0 via Lagrangian inversion. More about 
this method can be seen in the monograph[Liu8}. 


Example 10.2 For m = 2, it is known from (10.33) that the 
number of non-isomorphic planar rooted maps is the coefficient of x. 
That is 9. 

Because there are 4 non-isomorphic planar maps of size 2 as 
shown in Fig.10.2. The arrows on the same map represent the roots 
of non-isomorphic rooted ones. Such as there are, respectively, 2, 2, 1 
and 4 non-isomorphic rooted maps in (a), (b), (c) and (d) of Fig.10.2 
to get 9 altogether. 


V 
615 


(a) (c) (d) 


Fig.10.2 Planar rooted maps of two edges 


250 Chapter X Asymmetrized Maps 


X.9 Nonorientable equation 


Let Nm be the set of nonorientable rooted maps of size m. Of 
course, m > 1. And, let Nm be partitioned into ND and AD. i.e., 


Nm = NO + ND (10.34) 
where Ni) = {NIN € Nm, N — a orientable} and 
NO = Nm — N® = (N[N € Nn, N — a nonorientable}, 
a = e,(N) is still the rooted edge. 


Lemma 10.8 Let Mẹ’ = {N — a|VN € NP), then 
ALD e Ma (10.35) 
where M,,,_1 is the set of orientable rooted maps of size m — 1, m > 1. 


Proof Because of the nonorientability of N € ND and the ori- 
entability of N — a, from Corollary 3.1, a is not a segmentation edge. 
Based on Theorem 3.4, N — a is always an orientable map. So, for 
any N € VT, N € Mai, m > 1. This implies MP C Mmi. 

Conversely, for any N = (X, P) € M41, by appending an edge 
a’ on N, N' = (X',P") is obtained where X' = X + Kr' and P’ is 
different from P only at the vertex 

(r)p ES ge B (rip): 
Since Kr’ is not a segmentation edge, from Theorem 3.7, N' is a 
map. And since Gr’ € {r'}w, V' = Vipa}, from Theorem 4.1, N’ 


is nonorientable. Further, because N = N’ — a’ is orientable and 
m(N) +1=m(N’) =m, N € NR. This implies Mm- CMP. O 


For any M = (X, P) € Mm, since M is orientable, assume 


[re = (rar, oscar] 


V; € V = Vp, i = 1,2,---,2m(M) — 1 = 2m — 1. By appending 
the edge r’, 


A(M) LAo( M), A1( M), UO -, Aom(M)} 


X.3 Nonorientable equation 251 


is obtained where A;(M) = M+e,, = (£i, Pi) such that A; = X + Kr; 
and P; is determined in the following manner: 

Bro in the angle (o/P !r, r) 

brili = 1,2,---, 2m(M) — 1) in the angle (o/P- ir, vir), 

Dro, (w in the angle (r, aP'r). 

Because Br; € (ri) w, where V; = Vy, 53, à = 0,1,---,2m(M), 

from Theorem 4.1, A; are all nonorientable. Because A;(M) — e,, € 
Mm, Ai(M) € NO, 0 € i € 2m(M). From Lemma 10.8, 


NO:- M^ AM) (10.36) 
MEM, 


m > 0. Of course, Mo consists of only the trivial map. 
For MP, two cases should be considered: M® and NW), ie., 


ND — NOY 4 VD) (10.37) 
where 
N® = {NWN € N™® a =e, is a terminal, 
or segmentation edge] 
and 


AD = {NWN € N a is neither terminal 


nor segmentation edge]. 


Of course, MIP = MP — NO, 
Lemma 10.9 Let MY = {N - aNN € AG. Then, 
NO = M Max Nat XY Nux Mu 


nytng=m—-1 nytng=m—-1 
n1,n920 n1,n920 


+ > Nig X Na 


nytng=m-1 
n1,n920 


(10.38) 


where x represents the Cartesian product of sets. 


Proof Easy to see except for noticing that N — a has a transitive 
block which is the trivial map when a is a terminal edge. [] 


252 Chapter X Asymmetrized Maps 


Lemma 10.10 Let NS) ={N-alVNeE ND}. Then, 
NO = Ne (10.39) 
where m > 2. 


Proof Because a = e, is neither terminal nor segmentation edge, 
from Theorem 3.4, N € ND. By the nonorientability and the size 
m—1,NeE Mni; dB 

N e A ad 

On the other hand, for any N = (X, P) € N41, we have N' = 
(X', P^) such that 4’ = A + Kr' and P’ is different from P only 
at the vertex (rp. = (r’, yr’, (r)p). Since N is nonorientable and 


a’ = Kr’ is neither terminal nor segmentation edge, N' € ND. Thus, 
N = N'—a' € NAP. This implies 


Nm-1 C ACD. 
In consequence, the lemma is proved. L 


One attention should be paid to is that when m = 1, there is only 
one nonorientable map (Kr, (r, 8r)), and (Kr, (r, 8r)) € ND. Thus, 
(10.39) is meaningful only for m > 2. 

On the basis of this lemma, it is necessary to see how many 
N' E€ NO can be produced from one N € Nm such that N = N'— d. 

Because N is nonorientable, let I = {r, Vir, Vor, - -- , Vo 1r] be 
consists of half the elements in {r}y = A, V = V, p}, such that for 
any x € I, Kx 1I = (z,»yz). Two cases are now considered. 


Case 1 For any N = (X, P) € Nn, let 
B(N) — LBo(N), PY(N), Ba(N),- -+ , Ba (N)] 
where Bj(N) = (Xi, Pi) =N + e, j — 0,1,2,---, 2m, have 


Gro in the angle (o/P^ ir, r), 
Brj(j — 1,2,---,2m — 1) in the angle (o/P ^ y;r, jr), 
Brom in the angle (r, aP tr). 


X.3 Nonorientable equation 253 


Case 2 For any N = (X, P) € Nn, let 


C(N) = {Co(N), C (IN), C(N), ++, Com(N) } 
where C;(N) = (5, Qj) = N + er, j =0,1,2,---, 2m, have 
yro in the angle (o/P^!r, r}, 
yrj(j = 1,2,---,2m — 1) in the angle (o/P! y;r, jr), 
rom in the angle (r, o/P^!r). 


On the basis of Lemma 10.10, from the conjugate axiom, 
Ni = V (G(N) + C(N)) (10.40) 
NENm 


for m > 1. 
Because N = M +M 4- ---, the enumerating function 


—- 5.6, De” = fuo (2) + faz) 
m>1 NeN (10.41) 
= fox) + fym (z) + fyo (x). 


Lemma 10.11 For NMO = NO LAG 
adf 
dx 


where fm = f(x) is the enumerating function of orientable rooted 
maps determined by equation (10.14). 


fyo(x) = afm + 2x (10.42) 


Proof On the basis of (10.35-36), from Lemma 9.10, the lemma 
is obtained . O 


Lemma 10.12 For A) = NI +N) 
fon (x) = 2xfuty + eff (10.43) 
where fm = f(x) as in (10.10) and fy = f(x) as in (10.41). 
Proof A direct result of Lemma 10.9. [] 


254 Chapter X Asymmetrized Maps 


Lemma 10.13 For NV“) = NO +N? SEE 


fno (a) = 2z fu + ad (10.44) 


Proof On the basis of Lemma 10.10 with its extension (10.40), 
from Lemma 10.11, (10.44) is soon obtained. O 


Theorem 10.3 The following equation about f 


art — a(x) f — xf? — 2xb(x); 


df 
dx 


(10.45) 
=1 


y=0 





where 


hz) = f - 20M 


is well defined in the ring of power series with all coefficients non- 
negative integers and negative powers finite. And, the solution is 
f = fw). 

Proof Let f = Nyx + Nox? + Naz? 4- ---, then from (10.45) all 
the coefficients can be determined by the recursion 


on = 1-27 — 2z fm; 


Ny = (4m = 2)Nm-1 T (2m — LP 


m-—1 m—2 
2 Nos —i NiNm- —i 
E 2 dd D ! (10.46) 
m > 2; 


N=, 


where Fm, m > 0, are known in (10.15). Because all Nm, m > 1, 
determined by (10.46) are positive integers, the former statement is 
true. The latter is directly deduced from (10.41-44). [] 


X.4 Gross equation 255 


X.4 Gross equation 


Let Rm be the set of general(orientable and nonorientable) rooted 
maps with size m, m > 0. Of course, Rp consists of only the trivial 
map. 

For m > 1, Rm is partitioned into two subsets RIN ) and «RU ) 


1. €., 
Rm = RY + RDP (10.47) 
where 
RN) = {RIVR € Rm, e, (R) is a terminal link 
or segmentation edge} 
and 


RID) = (R|VR € Rm, e,(R) is neither terminal 


nor segmentation edge]. 
Of course, qi = Rm- RI 
Lemma 10.14 Let RẸ = {R — aVR € RO, then 


R= CN Rakha (10.48) 


ny+tng=m—1 
n1,n920 


m. 1. 


Proof Forany R € RW, because a = e,(R) is a terminal link 
or a segmentation edge, R — a has two transitive block (when a is a 
terminal link, the trivial map is seen as a transitive block in its own 
right), R— a = R4 + Rə and R4 € Rm, Ro € Rm. In other words, the 
set on the left hand side of (10.48) is a subset of the set of its right. 

Conversely, for any Ry = (44,71) € Rn, and Ry = (X2, P2) € 
Rn, by appending a = e,, R = (X,P) is obtained where ¥ = X, + 
A5 -- Kr and P is different from Pı and P2 only at the vertices (r)p = 
(r, (r1)p,) and (yr)p = (yr, (r2)p,). It is easily checked that R € Rm, 
m = nı +n2+ 1. In other words, the set on the right hand side of 
(10.48) is a subset of the set on the left. O 


256 Chapter X Asymmetrized Maps 





Since R = Ro + 4 + Ra» - ---, the enumerating function 
fra) = 9 C, Da" 
m»0 RERm (10.49) 
= fr (x) + from (at) + fgo(x), 


where RY = RMN 9... and RY = RID 4 RD +- 
First, because Ro consists of only the trivial map, 


fus) m 1. (10.50) 
Then, from Lemma 10.14, 
fg (x) = z fR, (10.51) 


where fr = fr(£). 
In order to evaluate fpg (x), R® has to be decomposed. 


Lemma 10.15 Let Ri = LR — avVR € RD}, then 
RIO = Rina, (10.52) 
where m > 1. 


Proof For any R' € qr ) because a’ = e, is neither terminal 
link nor segmentation edge, from Theorem 3.4, R = R' — a! € Ry_1. 
This implies 

RoC Ree: 

Conversely, for any R = (X, P) € Rm-1, by appending the edge 
a’ = Kr’, R' = (X', P’) is obtained where X' = X + Kr' and P’ is dif- 
ferent from P only at the vertex (r’)p: = (r', yr’, (r)p). From Theorem 


3.7, R! € Rm. Because a’ is neither terminal link nor segmentation 
edge and R = R' — a^, R € RIP. This implies 


Rast de 


'The lemma is proved. [] 


Based on this, what should be further considered for is how many 
Rc RID, can be produced from one R € Rm such that R = R — q'. 


X.4 Gross equation 257 


Because R is a map(orientable, or nonorientable), let 


I = {r, yir, bar, ++, amar) 


be the set of elements in correspondence with a primal trail code, or 
dual trail code. For any x € I, Kx NI = {x, yx} has two possibilities 
as cases. 


Case 1 For any R= (X, P) € Rm, let 
D(R) = {Do(R), D,(R), D»(R), uS Do, (R)] 
where D,(R) = (4), Pj) = RF e,, j =0,1,2,---,2m, have 


Gro in the angle aP r), 
Br;(j — 1,2,---,2m — 1) in the angle (aP7ty;r, jr), 
Brom in the angle (r, aP tr). 


Cases 2 For any R= (X,P) € Rm, let 
E(R) = {Eo(R), A(R), EX(R), +++, E (R)] 
where Ej(R) = (Yj, Qj) = RF erns J =0,1,2,---,2m, have 
yro in the angle (aP~'r,r), 


Yr; = 1,2,--.,2m — 1) in the angle (oP- sr, jr), 
Vom in the angle (r, o/P !r). 


Based on Lemma 10.15, from the conjugate axiom, 


T 
Riu = >, (D(R) + E(R) (10.53) 
RERm 
for m > 1. 
Because R = RP + RY +--+, from Lemma 10.15 with its 
extension (10.53) and Lemma 9.10, the enumerating function 


fram (x) = 2z fg + s lR (10.54) 


258 Chapter X Asymmetrized Maps 


Theorem 10.4 The equation about f 


d 
[o = 14 (1-22)f - xf’, (10.55) 
fo = f(0-1 


is well defined in the ring of power series with coefficients all nonneg- 
ative integers and terms of negative power finite. And, the solution is 


f = fr(z). 

Proof In virtue of the initial condition of equation (10.55), as- 
sume f = Ro + Rix + Rox? +--+. Of course, Ro = fo = 1. Further, 
from equation (10.55), the recursion 


m-1 
Rm = (4m — 2 Fd + Ss ee eee 
i=0 (10.56) 
mi: 
Hy 1 


is soon found for determine all the coefficients Rm, m > 0. It is easily 
checked that all of them are positive integers and hence the former 


statement is true. 
The latter is a direct result of (10.50—51) and (10.53). O 


X.5 The number of rooted maps 


First, let Om = (Fo, Fi, -++ , Fm), m => 0, be the m+1 dimensional 
vector where Fm, m > 0, are the number of non-isomorphic orientable 
rooted maps with size m. And, oÈ, = (Fm, Fm-1, +, Fo) called the 
reversed vector of the vector om. Easy to check that 


Om = ((0w) )^ = (em) ) = om (10.57) 


TIL m 


where T the transposition of a matrix. 
The recursion (10.15) for determining Fm, m > 0, becomes 


Fy, = (2m — 1) Fi + Onit 
B! (10.58) 
Fy=1. 


X.5 The number of rooted maps 259 


By (10.58), the number of non-isomorphic orientable rooted maps 
with size m, m > 1, can be calculated. In the first column of Table 
10.1, Fin, m < 10, are listed. 


Then, let ôm = (N1, No, --- , N1) where Npn is the number of non- 
isomorphic nonorientable rooted maps with size m for m > 1. 


The recursion (10.46) for determining Nm, m > 1, becomes 


Nm = (4m — 2) Ng.1 + (2m — 1) Fs 
iina atoma os 
m. 2: 
N,=1 


(10.59) 


where 2,4,» is given in (10.58). 

By (10.59), the number Nn can be calculated for m > 1. In the 
second column of Table 10.1, Nm, m < 10, are listed. 

Finally, let pm = (Ro, Ri,*--, Rm) where Rm is the number of 
non-isomorphic general maps with size m for m > 0. 


The recursion (10.56) becomes 


Rm = (4m — 2)Rm-1 + Pm—1P met 
m > 1; (10.60) 
m=i 


By (10.60), the number Rm can be calculated for m > 0. In the 
third column of Table 10.1, Rm, m < 10, are listed. 

From those numbers in Table 10.1, it is also checked that the 
enumerating functions f(x), f(z) and fr(x) of, respectively, non- 
isomorphic orientable, nonorientable and general(orientable and nonori- 
entable) rooted maps with size as the parameter satisfy the relation 
as 


fr(x) = f(x) + fw(x). (10.61) 


260 


mL 


0 
1 
2 
3 
4 
5 
6 
7 
8 
9 
0 


Chapter X Asymmetrized Maps 





1 0 1 

2 1 3 

10 14 24 

74 223 297 

706 4190 4896 
8162 92116 100278 
110410 2339894 2450304 
1708394 67825003 69533397 


29752066 2217740030 2247492096 
576037442 80952028936 | 81528066378 
12277827850 3268104785654 | 3280382613504 


Table 10.1 Numbers of rooted maps with size less than 11 


Activities on Chapter X 


X.6 Observations 


O10.1 Because a surface can be seen as a polygon with even 
edges pairwise identified in the plane, think that whether, or not, a 
map on a surface rather than plane can always represented by a planar 
one. If it can, explain the reason, or provide an example otherwise. 


O10.2 For a rooted map M, let m(M) and l(M) be, respec- 
tively, the size and the valency of root-face in M, observe the numbers 
of non-isomorphic planar rooted maps for m( M), I(M) < 3. 


Similarly, do the same with the valency of root-vertex s(M) in- 
stead of (M). 


A map with all its faces of 3-valent except for the root-face is 
called a near triangulation. 


O10.3 Fora near triangulation T, let m(T) and I(T) be, respec- 
tively, the size and the root-face valency of T. Observe the numbers 
of non-isomorphic planar rooted near triangulations for m, l < 4. 


Similarly, do the same with the valency of root-vertex s instead 
of l. 


A map with all its faces of valency 4 except for the root-face is 
called a near quadrangulation. 


O10.4 . For a near triangulation Q, let m(Q) and l(Q) be, 
respectively, the size and the root-face valency of T'Q. Observe the 
numbers of non-isomorphic planar rooted near quadrangulation for m, 
| 


262 Activities on Chapter X 


Similarly, do the same with the valency of root-vertex s instead 
of l. 


A planar map from which the result of deleting all the edges on 
a face is a tree is called a Halin map. The face is said to be specific. 


O10.5 Evaluate the number of non-isomorphic rooted Halin 
maps with the root on the specific face by the parameters: the size m 
and the valency l of the specific face for m > 6 and l > 3. 


010.6 ‘Try, by edge contraction, to determine the enumerating 
function of general rooted maps with the size as the parameter. 


O10.7 ‘Try, by edge contraction, to determine the enumerating 
function of rooted petal bundles with the size an the root-face valency 
as the two parameters. 


010.8 Try, by directly solving the equation (9.7), to determine 
the enumerating function of orientable rooted petal bundles h. 


010.9 ‘Try, by directly solving the equation (9.30), to deter- 
mine the enumerating function of nonorientable rooted petal bundles 
g. 

O10.10 Observe the numbers of general Eulerian rooted maps 
with the size smaller. 


X.7 Exercises 


For a map, if the result of deleting all inner vertices(not articulate 
vertex, i.e., a vertex of valency 1, or a terminal) of a spanning tree is, 
itself, a travel with only one vertex of valency probably greater than 
2, then it is called a pan-Halin map. Because this travel is still a map 
and becomes a petal bundle via decreasing subdivision, such a petal 
bundle is called the base map of the pan-Halin map. 

A pan-Halin map of which the base map is of size 2p and ori- 
entable genus p > 0, or of size q and nonorientable genus q > 1, is 
said to be pre-standard . If a pre-standard pan-Halin map has its base 


X.7 Exercises 263 


map with each edge incident with at least one terminal of the tree, 
then it is said to be standard. 

Let pg be the set of all pre-standard pan-Halin rooted maps. 
The root rg for H € Hy is chosen be an element incident with the 
vertex and the face of the base map of H. 

For any H = (X,P) € f, the tree T on H is seen as a 
planted tree(a plane tree with the root-vertex of valency 1) with its 
root rr = P(Py)'rx where 


t = min{s|(Py)*ry incident with a terminal of T}. 


E10.1 Given the partition of vertices according to their va- 
lencies on a planted tree j = (j1,js,-:::,), Le, ji, i > 1, is the 
number of unrooted vertices of valency i, prove that the number of 
non-isomorphic planted trees with the partition is 

(n — 1)! 
g! 


n=14+)S j; 


i>1 


where 


i.e., the order, and j! = [[;., j! 


E10.2 Given the vertex partition s = (55,53,::-), prove that 
the number of non-isomorphic pre-standard pan-Halin rooted maps 
with the partition and their base maps of size m on a surface of ori- 


entable genus p is 
pm ( " + 2p — 1 \ /s3\ nlm 
2p— 1 mj s! 


n+2=% s; and s! = [ [ si. 


i>2 s>2 


where 


E10.3 Given the vertex partition s = (s2, 53,::-), Prove that 
the number of non-isomorphic pre-standard pan-Halin rooted maps 


264 Activities on Chapter X 


with the partition and their base maps of size m on a surface of nonori- 


9m m+@q—1)\ (83\ nln 
q—1 mj s! 


n+2=%_ s; and s! = | [ si. 


i2 S22 


entable genus q is 


where 


E10.4 Given the vertex partition s = (s2,53,---), prove that 
the number of non-isomorphic standard pan-Halin rooted maps with 
the partition and their base maps of size m on a surface of orientable 


genus p is 
pm (M — 1\ /s3\n!m 
2p—1/\m/ s! 


n+2=%_ s; 


i22 


where 


and s > 0,s #0, m > 2p 7 1. 


E10.5 Given the vertex partition s = (55, 53,::-), prove that 
the number of non-isomorphic standard pan-Halin rooted maps with 
the partition and their base maps of size m on a surface of nonori- 


9m m — 1N (s3\n!m 
q—1/NmJ/ s! 


n+2=%_ s; 


i22 


entable genus q is 


where 


ands >0,s40,m>q2>1. 


E10.6 Evaluate the number of near triangulations of size m on 
the projective plane. 


E10.7 Evaluate the number of near triangulations of size m on 
the Klein bottle. 


X.8 Researches 265 


E10.8 Evaluate the number of rooted quadrangulations of size 
m on the projective plane. 


E10.9 Evaluate the number of near quadrangulations of size m 
on the Klein bottle. 


E10.10 Determine the enumerating function of rooted petal 
bundles with the size as the parameter on the torus. 


E10.11 Determine the enumerating function of rooted petal 
bundles with the size as the parameter on the projective plane. 


E10.12 Determine the enumerating function of orientable two 
vertex rooted maps with size as the parameter. 


E10.13 Determine the enumerating function of nonorientable 
two vertex rooted maps with size as the parameter. 


E10.14 Establish an equation satisfied by the enumerating func- 
tion of general non-separable rooted maps. 


E10.15 Establish an equation satisfied by the enumerating func- 
tion of general Eulerian rooted maps. 


A map is said to be loopless if its under graph has no self-loop. 


E10.16 Establish an equation satisfied by the enumerating func- 
tion of general loopless rooted maps. 


A map is said to be simple if its under graph has neither self-loop 
nor multi-edge. 


E10.17 Establish an equation satisfied by the enumerating func- 
tion of general simple rooted maps. 


X.8 Researches 


R10.1 Given a relative genus g Z O(the case g = 0 is solved), 
determine the number of rooted near triangulations of size m > |g| on 


266 Activities on Chapter X 


a surface of genus g. 


A rooted map with all vertices of the same valency except for 
probably one vertex is said to be near regular. Among them, near 
3-regular and near 4-regular are often encountered in literature. 

Although near triangulations or near quadrangulations are, re- 
spectively, the dual maps of rear 3-regular, or near 4- regular maps, 
they are still considered for most convenience from a different point 
of view. 


R10.2 Given a relative genus g # O(the case g = 0 is solved), 
determine the number of rooted near 3-regular maps of size m > |g| 
on a surface of genus g. 


R10.3 Given a relative genus g 4 O(the case g = 0 is solved), 
determine the number of rooted near quadrangulations of size m > |g| 
on a surface of genus g. 


R10.4 Given a relative genus g 4 0(the case g = 0 is solved), 
determine the number of rooted near 4-regular maps of size m > |g| 
on a surface of genus g. 


R10.5 Given a relative genus g Z O(the case g = 0 is solved), 
determine the number of non-separable rooted maps of size m > |g| 
on a surface of genus g. 


R10.6 Given a relative genus g Z O(the case g = 0 is solved), 
determine the number of Eulerian rooted maps of size m > |g| on a 
surface of genus g. 


R10.7 Given a relative genus g Z O(the case g = 0 ia solved), 
determine the number of non-separable Eulerian rooted maps of size 
m 2 |g| on a surface of genus g. 


For the problems above, another parameter / > 1 is absolutely 
necessary in almost all cases. it is the valency of the extra vertex, or 
face according as the regularity is for vertices, or faces. 


R10.8 Given a relative genus g Z O(the case g = 0 is known), 


X.8 Researches 267 


find a relation between general maps and quadrangulations on a sur- 
face of genus g. 


R10.9 Given a relative genus g Z 0(the case g = 0 is known), 
find a relation between general maps and triangulations on a surface 
of genus g. 


R10.10 Given a relative genus g 4 O(the case g = 0, a 1-to- 
1 correspondence between loopless planar rooted maps of size m — 
1 and 2-connected planar rooted triangulations of 2m — 1 unrooted 
faces should be found, but now unknown yet), find a relation between 
loopless rooted maps and triangulations on a surface of genus g. 


R10.11 Present an expression of the solution for equation (10.14) 
by special functions, particularly the hyperbolic geometric function. 


Chapter XI 


Maps with Symmetry 


e A relation between the number of rooted maps and the order of 
the automorphism group of a map is established. 


e A general procedure is shown for determining the group order dis- 
tribution of maps with given size via an example as an application 
of the relation. 


e A principle for counting unrooted maps from rooted ones is pro- 
vided. 


e Dased on the principle, a general procedure is shown for determin- 
ing the genus distribution of unrooted maps with given size via two 
examples. 


e Conversely, rooted maps can be also determined via unrooted maps. 


XI.1 Symmetric relation 


First, observe how to derive the number of non-isomorphic un- 
rooted maps from that of non-isomorphic rooted maps when the auto- 
morphism group is known, or in other words, how to transform results 
without symmetry to those with symmetry. 


XL1 Symmetric relation 269 


Theorem 11.1 Let no(U; I) be the number of non-isomorphic 
rooted maps with a given set of invariants including the size in the 
set of maps U considered. If the order of automorphism group of each 
map M in U is independent of the map M itself, but only dependent 
on U and I, denoted by aut(U; I), then the number of non-isomorphic 
unrooted maps with J in M is 


aut(U; D)no(Ut; I) 


m(U; I) = Ae 


(11.1) 


where e € Í is the size. 


Proof Let map M = (X,7P) € U. From Theorem 8.1, for any 
rca, 


|X.| = Hyl dr € Aut(M), y = 7x)| = aut(M). (11.3) 


In view of Corollary 8.2, M itself produce 


|.|]X| | 4e 
— |X,|  aut(M) 








no(M) (11.3) 


non-isomorphic rooted maps. Therefore, there are 


4e 
u; I) = — 
m1) = à, M 
Meu 
4e 
= — —n,(U;I 
aut(U; I ji 
non-isomorphic rooted maps in U. Via rearrangement, (11.1) is soon 
obtained. [] 


In chapter VIII, efficient algorithms are established for finding 
the automorphism group of a map, this enables us to get how many 
non-isomorphic rooted maps from a unrooted map by (11.1). 

However, from Chapter IX and Chapter X, it is unnecessary to 
know the automorphism group for counting rooted maps. This en- 
ables us to enumerate unrooted maps via automorphism groups by 
employing (11.1). 


270 Chapter XI Maps with Symmetry 


Problem of type 1 For a set of maps M known the number 
of non-isomorphic rooted maps with a given size, determine the num- 
ber of non-isomorphic unrooted maps with the given size according 
to the orders of their automorphism groups, or in other words, the 
distribution of unrooted maps on the orders of their automorphism 
groups. 


Although this problem does not yet have general progress in 
present, a great amount of results for rooted case have already pro- 
vided reachable conditions for the problem. 


XL2 An application 


In what follows, provide a general procedure for solving the prob- 
lem of type 1 via the determination of the distribution of rooted petal 
bundles on the orders of the automorphism groups of corresponding 
unrooted maps on the basis of Chapter IX. 

From Table 9.1 at the end of Chapter IX, the number of non- 
isomorphic planar rooted petal bundles with size 4 is H ©) = 14, shown 

n (a-n) of Fig.11.1. 

In virtue of Corollary 8.2, the orders of their automorphism 

groups are possibly 1, 2, 4, 8 and 16 only 5 cases. 





Case 1 aut(M) 21, M = (Kz 4- Ky + Kz + Kt, J). None. 


Case 2 aut(M) = 2. 8 planar rooted petal bundles: 


ida = (By IR 2,20, yt, "yy; 
; Sa = (2,9, yy, 2, "92, yt, o yt; 
| Se = (2592, 9s 2 "2," ys o yos 
ig en us do ^y d, 6 et ey 


DW. YY, YL, Z, YZ, t, yt 
T, Y, Z, YZ, t, Yt, YY, YT); 
T Y; Z, YZ; YY, t, yt, YT); 
T, Y, YY, Z, t, YZ, Yt, YT); 


Tou M omm MP om P c 


Si 
Iz 
Js = 
Jr 


Tee eed eei eri 


shown in (a-h) of Fig.11.1. 


XL2 An application 271 


Case 3 aut(M) = 4. 4 planar rooted petal bundles: 


Jo = (t, Y, VY, YT, Z, t, yt, yz); Jio = ras YT, y, 2, t, yt, YF, yy); 
Ju -(yzt,ytyz,yy,yx); Ju = (2, Y2 Va TE ts t), 


shown in (i-l) of Fig.11.1. 


Case 4 aut(M) = 8. 2 planar rooted petal bundles: 


Jig (55935 09. 2599, 05 y8)5 lace (5,35 09 2,38, 0; 8,9). 


shown in (m, n) of Fig.11.1. 
Case 5 aut(M) = 16. None. 


This procedure can be done for determining the automorphism 
groups of a unrooted maps via their primal trail codes, or dual trail 
codes, by computers and then via the collection of the same class of 
them according to the orders of the groups. 


NIZ. NIZ NZ 
Lx AN AN 


(a) (b) (c) 


NZ NIS NZ. 
AIX AIN VN 


(d) (e) (f) 









































262 Chapter XI Maps with Symmetry 

























































































Fig.11.1 Planar petal bundles of size 4 


XL3 Symmetric Principle 


Whenever the distribution of rooted maps on the orders of au- 
tomorphism groups is given for a set of maps, the number of non- 
isomorphic unrooted maps can be soon extracted 


Theorem 11.2 Let no;(.Mt; I) be the number of non-isomorphic 
rooted maps with the set of invariants / and the order of their auto- 
morphism groups 7 in a set of maps M for i|4e, 1 < i € 4e, where e is 


XL3 Symmetric Principle 273 


the size, then the number of non-isomorphic unrooted maps in M is 


m(M;1)= M: TRU (11.4) 


4e 
i|4e 


1<i<4e 


Proof Let ny;(M; I) be the number of non-isomorphic unrooted 
maps with the set of invariants J and the order of their automorphism 
groups i in the set of maps M for i|4e, 1 < i < 4e, where e is the size, 
then 

nil MET) ` ny; (Mt; I). (11.5) 


i|4e 
1<i<4e 


From Theorem 11.1, each unrooted map M € M, Aut(M) = i, 
produces 
4e 
d 
non-isomorphic rooted maps. Therefore, 
ingil M; I) 


nai (M; I) = = (11.6) 


By substituting (11.6) into (11.5), (11.4) is soon obtained. O 


On the choice of the set of invariants J, two types should be 
mentioned. One is that the set J consists of only the size and the 
genus for determining the genus distribution of non-isomorphic maps 
in a set of maps M. The other is that the set J consists of only the 
size and the orders of automorphisms for determining the symmetric 
distribution of non-isomorphic maps in a set of maps M. 


Problem of type 2 For a set of maps M with the number of 
non-isomorphic maps given, determine the number of non-isomorphic 
under graphs of maps in M. 


Although the justification of whether, or not, two graphs are 
isomorphic is much far from easy, a feasible approach to it is presented 
from the above discussion. Because the under graphs are isomorphic 


274 Chapter XI Maps with Symmetry 


if the two maps are isomorphic, the only thing we have to do is to 
classify non-isomorphic maps by their isomorphic under graphs. 

On the other hand, for a graph, it is also possible to discuss how 
many non-isomorphic rooted maps are with the graph as their under 
graph, and then to discuss how many non-isomorphic unrooted maps 
are with the graph as their under graphs, and finally to classify maps 
according to the isomorphism of their under graphs. 


XL4 General examples 


On the basis of the 15 orientable rooted petal bundles of size 3 
and the 9 nonorientable rooted petal bundles of size 2(in Table 9.1 at 
the end of Chapter IX), a general procedure is established for deter- 
mining the genus distribution of them. 


Orientable case Let M = (Ka+Ky4+ Kz, J), 1<i< 15. 
genus 0 5 orientable rooted petal bundles shown in (a-e) of 
Fig.11.2. Here, 
d = (£, YT, Y, YY, Z, YZ); Jr = [35:05 99) b mor es) 


with the order of its automorphism group aut(M) = 6 are one un- 
rooted map; 


Js = (zx, yx, yz, yz, vu) Ja = (2, y, WY, VE, 2, V2); 
Is = (CY eye VY.) 
with the order of its automorphism group aut(M) = 4 are one un- 


rooted map. 


Genus 1 10 orientable petal bundles shown in (f-o) of Fig.11.2. 
Here, 


Jes GG eoa a = E qr ave yarn 
da ec ndo mou y ce pop d ses 
Jio = Dish x m t var s Jı = (£, Y, YT, YY, £;^y«) 


XL4 General examples 275 


with the order of its automorphism group aut(M) = 2 are one un- 
rooted map; 


Sia = (2,5, 2) Vs VE) 72); Sia = (2, Y; YE, 2, Vs V2); 

Fis = (2, Y, 2, YE, 72, VY) 
with the order of its automorphism group aut(M) = 4 are one un- 
rooted map; and 

Jis = (£, Y, 2, YL, VY, YZ) 
with the order of its automorphism group aut(M) = 12 is one unrooted 
map itself. 
All are listed in Table 11.1 shown the genus distribution, group 

order distribution as well, of orientable unrooted maps. 









































276 Chapter XI Maps with Symmetry 
























































Fig.11.2 Orientable petal bundles of size 3 


Nonorientable case Let N = (Kx + Ky, Ji), 1<i< 9. 
Genusl 5 nonorientable rooted petal bundles shown in Fig.9.4(e, 
a,c) and in Fig.9.3(a,c). Here, 
eh = Ut By, pu. y) 


with the order of its automorphism group aut( N) = 8 is one unrooted 


map itself; 
Jo = (x, Bz,y, vy); Ja = (a, Bx, yy, y); 


Ja = (2,2, y, By); Js = (2, yz, By, y) 


XL4 General examples 277 


with the order of its automorphism group aut( N) = 2 are one unrooted 
map. 


Genus? 4 nonorientable rooted petal bundles shown in Fig.9.3(b) 
and Fig.9.4(b,d,f). Here, 


Je = (x, By, yz, y); Fr = (2, 7y, Bz, y); 
Js = 0, By); Jo m (2, D. D, y) 


with the order of its automorphism group aut(N) = 4 are 2 unrooted 
maps. 

All are listed in Table 11.2 shown the genus distribution, group 
order distribution as well, of nonorientable unrooted maps. 





Table 11.2 Distributions of nonorientable petal bundles 


Activities on Chapter XI 


XI.5 Observations 


O11.1 Given all the 54 planar rooted maps of size 3, find their 
distribution according to the orders of automorphism groups. 


O11.2 Given all the 40 outerplanar rooted maps(the root inci- 
dent with the outer face) of size 3, find their distribution according to 
the orders of automorphism groups. 


O11.3 Observe how many non-isomorphic rooted maps whose 
under graph is 4, i.e., the complete graph of order 4. 


O11.4 Observe the distribution of all the rooted maps whose 
under graph is Ky according to the orders of their automorphism 
groups. 


O11.5 Observe the genus distribution of all orientable maps 
whose under graph is K4. 


O11.6 Observe the genus distribution of all nonorientable maps 
whose under graph is K4. 


O11.7 Given all the 56 planar Euler rooted maps of size 4, find 
how many non-isomorphic unrooted maps among them. 


O11.8 Given all the 27 planar 4-regular rooted maps of co- 
order 4, find how many non-isomorphic unrooted maps among them. 


O11.9 For m > 5, try to determine the number of planar un- 
rooted petal bundles of size m. 


O11.10 For m 2 4, try to determine the number of orientable 


XI.G Exercises 279 


unrooted petal bundles of size m. 


O11.11 For m > 3, try to determine the number of nonori- 
entable unrooted petal bundles of size m. 


XI.6 Exercises 


E11.1 Prove that the number of non-isomorphic outerplanar 
rooted maps(the root is on the outer face) of size m is 
2m Jin)! 
(m+ 1)!m! 
for m > 1. 


E11.2 Prove that the number of planar 4-regular rooted maps 


of co-order n -- 1 is 
2(2n — 2)! 


(n+ (n — 1)! 


n—1 


forn > 1. 


E11.3 Prove that the number of non-isomorphic planar rooted 
maps of size m is 
gg 6) 
(n+1)(n+2)\m 


E11.4 Prove that the number of planar loopless rooted trian- 
culations of size 3m is 


for m > 0. 


27H (35m 

m!(2m + 2)! 
for m > 1. 

E11.5 Prove that the number of non-isomorphic planar Euler 
rooted maps of size m is 
8 x2?-l0m)l 

mI(m + 2)! 

for m > 1. 


280 Activities on Chapter XI 


E11.6 Prove that the number of non-isomorphic planar non- 
separable rooted maps of order p and co-order q is 


(2p 4- q — 5)!(2q 4- p — 5)! 
(p — 1)!(g — 1)! (2p — 3)! (2g — 3)! 
where p, q > 2. 


E11.7 Prove that the number of non-isomorphic planar simple 
rooted maps of size m is 


Y A(2m + 1) (2m — i — 4)! 
es CL ea 2)! (2m — i + L)!m! 
where m > 2. 

E11.8 Prove that the number of non-isomorphic planar 
3-connected rooted maps of size m > 6 is 

(—1)2 + fa 
where Rm, m > 2, are determined by the recursion 
(7m — 22) Rra + 200m. — 1) Rms 
2m 


with the initial conditions R; = —1 and R» = 2. 


R= m>3 


2 


E11.9 For m > 4, determine the genus distribution of ori- 
entable rooted petal bundles with size m. 


E11.10 For m > 5, determine the genus distribution of nonori- 
entable rooted petal bundles with size m. 


E11.11 For m > 4, determine the number of non-isomorphic 
outerplanar unrooted maps with size m. 


XI.7 Researches 


R11.1 Determine the number of non-isomorphic planar 4-regular 
unrooted maps of co-order n 4- 1 for n > 1. 


XI.7 Researches 281 


R11.2 Determine the number of non-isomorphic planar loop- 
less unrooted triangulations of size 3m, m > 2. 


R11.3 Determine the number of non-isomorphic planar Euler 
unrooted maps of size m > 2. 


R11.4 Determine the number of non-isomorphic planar non- 
separable unrooted maps of size m > 2. 


R11.5 Prove, or disprove, the conjecture that almost all trees 
have the order of their automorphism group 1 when the size is greater 
enough. 


R11.6 Prove, or disprove, the conjecture that almost all maps 
with a given relative genus have the order of their automorphism group 
1 when the size is large enough. 


R11.7 Prove , or disprove, the conjecture that for a positive 
integer gle, g > 2, almost no orientable map is with the order of 


automorphism group g when e is large enough. 


R11.8 Prove , or disprove, the conjecture that for a positive 
integer gle, g > 2, almost no nonorientable map is with the order of 
automorphism group g when e is large enough. 


R11.9 Determine the genus distribution of 4-regular rooted 
maps of co-order n+ 1, n > 1. 


R11.10 Determine the genus distribution of loopless rooted tri- 
angulations of size 3m, m > 2. 


R11.11 Determine the genus distribution of Euler rooted map 
with size m > 2. 


R11.12 Determine the genus distribution of nonseparable rooted 
map with size m > 2. 


Although corresponding problems about genus distribution can 
also posed for unrooted case, they would be only suitable after the 
solution of rooted case in general. 

Moreover, the genus distributions of maps with under graphs in 
some chosen classes can also be investigated. 


Chapter XII 


Genus Polynomials 


e The set of associate surfaces of a graph are constructed to deter- 
mine all of its distinct embeddings, or its super maps as well. 


e A layer division of an associate surface of a graph is defined for 
establishing an operation to transform this surface into another 
associate surface. A procedure can be constructed for listing all 
other associate surfaces from an associate surface by this operation 
without repetition. 


e A principle of determining the genus polynomial, called handle 
polynomial, of a graph is provided for the orientable case. 


e The genus polynomial of a graph for nonorientable case, also called 
crosscap polynomial, is derived from the handle polynomial of the 
graph. 


XII.1 Associate surfaces 


Given a graph G — (V, E) and a spanning tree T', the edge set 
E is partitioned into Er(tree edge) and Er(cotree edge), i.e., E = 
Er + Er. Let Er = [ili = 1,2,-..,0), 8 = B(G) be the Betti 
number(or cyclic number) of G. If i = (u[i], v[i]), then iu and 2, are, 
respectively, meant the semi-edges of i incident with u[i] and vļi]. 


XL1 Symmetric relation 283 


Write @ = (V + Vj, Er + Ej), where V, = (v, v;1 € i € 8) and 
FE, = {(uli], vi), (vli], vj) 1 < i € 8). Because G” is a tree itself, G” is 
called an expanded tree of T' on G, and denoted by To, or T in general 
case[Liu13-14]. 

Let ô = (01,605, ---, 0g) be a binary vector, or as a binary number 
of 2 digits. Denoted by T? that T, edges (u[i],v;) and (v[i], v;) are 
labelled by i with indices: +(always omitted) or —, 1 < i € B, where 
0; = 0 means that the two indices are the same; otherwise, different. 
Then, ô is called an assignment of indices on T 

For v € V, let o, be a rotation at v and og = {0,|Wv € V}, the 
rotation of G, then Î, determine an embedding of T on the plane. 


Theorem 12.1 For any o asa rotation and ô as an assignment 
of indices, T? determines a joint tree. 


Proof By the definition of a joint tree, it is soon seen. [] 


According to the theory described in Chapter 1, the orientabil- 
ity and genus are naturally defined to be that of its corresponding 
embedding. 


Lemma 12.1 Joint tree T5 is orientable if, and only if, ô = 0. 


Proof Because 6 = 0 implies each label with its two occurrences 
of different indices, the lemma is true. [] 


On a joint tree T$ , the surface determined by the boundary of 
the infinite face on the planar embedding of T, with ô on label indices 
is said to be an associate. 


Lemma 12.2 The genus of a joint tree Tu is that of its associate 
surface. 


Proof Only from the definition of orientability of a joint tree.L] 


Two associate surfaces are the same is meant that they have the 
same assignment with the same cyclic order. Otherwise, distinct. Let 


284 Chapter XII Genus Polynomials 


F(@) be the set of distinct surfaces on Ig = {1,2,---, 8}. 

For a surface F € F((3) and a tree T on a graph G, if there exists 
an joint tree 7? such that F is its associate surface, then F is said to 
be admissible . Let Fr(8) be the set of all distinct associate surfaces. 

Given two integers p, p > 0, and q, q > 1, let Fr(8;p)(or 
Fr(0;q),q = 1), p = 0, be all distinct admissible surfaces of ori- 
entable genus p(or nonorientable genus q). 


Theorem 12.2 For any integer p > 0( or q > 1), the cardinal- 
ity |Fr(G; p)|(or |-Fr(8; q)|) is independent of the choice of tree T on 
G. Further, it is the number of distinct embeddings of G on a surfaces 
of orientable genus p(or nonorientable genus q). 


Proof According to O1.14, a 1-to-1 correspondence between two 
sets of embeddings generated by two distinct spanning trees can be 
found such that same embeddings are in correspondence. This implies 
the theorem. [] 


Because of 


\Fr(8)| = X Er p) +X |Fr(6; a)l, 


p20 q>1 
the following conclusion is found from the theorem. 


Corollary 12.1 The cardinality |Fr(8)| is independent of the 
choice of tree T on G. Further, it is the number of distinct embeddings 
of G. 


From Lemma 12.1, the nonorientability of an associate surface 
can be easily justified by only checking if it has a label 7 with the 
same index, i.e., ó(i) = 1. 


Theorem 12.3 There is a 1-to-1 correspondence between as- 
sociate surfaces and embeddings of a graph. 


Proof First, we can easily seen that each embedding determines 
an associate surface. Then, we show that each associate surface is 


XIL2 Layer division of a surface 285 


determined by an embedding. Because of Theorem 12, this statement 
is derived. go 


From what is mentioned above, it is soon seen that the problem 
of determining the genus distribution of all embeddings for a graph is 
transformed into that of finding the number of all distinct admissible 
associate surfaces in each elementary equivalent class and the prob- 
lem on minimum and maximum genus of a graph is that among all 
admissible associate surfaces of the graph. All of them are done on a 


polygon. 


XIL2 Layer division of a surface 


Given a surface S — (A). it is divided into segments layer by 
layer as in the following. 

The Oth layer contains only one segment, i.e., A(— Ay). 

The 1st layer is obtained by dividing the segment Ag into l 
segments, i.e., S = (A1, Ao,---, Ai), where Aj, As, ---, Aj, are called 
the 1st layer segments. 

Suppose that on k — 1st layer, the k — 1st layer segments are 
where m,_1) is an integral k — 1-vector satisfied by 


1 


Ana- 


Tipi) < (nı, N2, M ny) < Aag 


with Ig = (1, 1, "ied. 1 


asit. 


IN k-i) = (Ni, No, 1ta Ny), 


Ni = l4 = Nay); No = lana N3 = lan eux m Ny. a4 = lAng 9? then the 
kth layer segments are obtained by dividing each k— 1st layer segment 
as 


Ay An ity Ån (12.1) 


1p Rk) (1) Ang, jj 


where 





lg = (1g 1,1) € Qs i) € Nay = QN o ay Ne) 


286 Chapter XII Genus Polynomials 


and N; = lA, ay 1 €i < Ny. Segments in (L1) are called sons of 
Ap, .," Conversely, Aj, is the father of any one in (12.1). 

A layer segment which has only one element is called an end 
segment and others, principle segments. 

For an example, let 


S = (1, —7,2, —5,8, —1,4, —6,5, —2,6,7, —3, —4). 





Fig.12.1 shows a layer division of S and Tab.12.1, the principle seg- 
ments in each layer. 


Layers Principle segments 


Oth layer 
1st layer 


Asi 7959 146 567 4) 





2nd layer ; 
H = (—2;6), I = (-3; 4) 





Tab.12.1 Layers and principle segments 


(1, —7,2 — 5;3, 1,4, —6, 5; —2, 6, 7, —3 — 4) 





(;—7,2;-5) | (3, —1;4, —6; 5) (—2, 6; 7; —3, —4) 





Fig.12.1 A layer division of S 


For a layer division of a surface, if principle segments are dealt 
with vertices and edges are with the relationship between father and 
son, then what is obtained is a tree denoted by T'. On T, by adding 
cotree edges as end segments, a graph G = (V, E) is induced. For 
example, the graph induced from the layer division shown in Fig.12.1 
is as 

Y= 14, B,C, D, EJ GA (12.2) 


XIL2 Layer division of a surface 287 


and 
E =1{6, 6,c,0,¢, 7,9, h;1,2,9,4, 5,0, T]; (12.3) 
where 
a= (A,B), b= (A, C),c = (A, D),d = (B, E), 
e — (C,F), = (0,G),g = (ID, H),h — (D,I), 
and 


1 = (B, F),2 = (E, H),3 = (F,D,4 = (G,I), 
5 = (B,C),6 = (G, H),7 = (D, E). 


By considering Er = {a,b,c,d,e, f, g, h, Er = (1,2,3,4,5,6,7),0; = 
0,2 = 1,2,---,7, and the rotation o implied in the layer division, a 
joint tree a is produced. 


Theorem 12.4 A layer division of a surface determines a joint 
tree. Conversely, a joint tree determines a layer division of its associate 
surface. 


Proof From the procedure of constructing a layer division, a 
joint tree is determined. Conversely, it is natural. [] 


Then, an operation on a layer division is discussed for trans- 
forming an associate surface into another in order to visit all associate 
surfaces without repetition. 

A layer segment with all its successors is called a branch in the 
layer division. The operation of interchanging the positions of two 
layer segments with the same father in a layer division is called an 
exchanger. 


Lemma 12.3 A layer division of an associate surface of a graph 
under an exchanger is still a layer division of another associate surface. 
Conversely, the later under the same exchanger becomes the former. 


Proof From the correspondence between layer divisions and as- 
sociate surfaces, the lemma can be obtained. [] 


On the basis of this lemma, exchanger can be seen as an operation 
on the set of all associate surfaces of a graph. 


288 Chapter XII Genus Polynomials 


Lemma 12.4 The exchanger is closed in the set of all associate 
surfaces of a graph. 


Proof From the correspondence between joint trees and layer 
divisions, the conclusion of the lemma is seen. [] 


Lemma 12.5 Let .A(G) be the set of all associate surfaces of 
a graph G, then for any S1, S9 € A(G), there exist a sequence of 
exchangers on the set such that Sı can be transformed into So. 


Proof By considering the joint trees and layer divisions, the 
lemma is right. E] 


If .A(G) is dealt as the vertex set and an edge as an exchanger, 
then what is obtained is called the associate surface graph of G, and 
denoted by H(G). From Theorem 12.3, it is also called the surface 
embedding graph of G. 


Theorem 12.5 In H(G), there is a Hamilton path. Further, 
for any two vertices, H(G) has a Hamilton path with the two vertices 
as ends. 


Proof By arranging an order, an Hamiltonian path can be ex- 
tracted based on the procedure of the layer division. [] 


First, starting from a surface in A(G), by doing exchangers at 
each principle in one layer to another, a Hamilton path can always be 
found in considering Theorem 12.3. This implies the first statement. 

Further, for chosen $1, $5 € A(G) = V(H(G)) adjective, starting 
from from S1, by doing exchangers avoid Ss except the final step, on 
the basis of the strongly finite recursion principle, a Hamilton path 
between Sı and $5 can be obtained. This implies that H(G) has a 
Hamilton circuit and hence the last statement. 

This theorem tells us that the problem of determining the mini- 
mum, or maximum genus of graph G has an algorithm in time linear 


on H(G). 


XIL3 Handle polymomials 289 


XIL3 Handle polynomials 


Let S(G) be the set of associate surfaces of a graph G and S,(G), 
the subset of S(G) with genus g. The the enumerating function 


Jmax 


(Giz) = So |S (Ale (12.4) 


9=Jmin 


is called the genus polynomial of G where gmin and gmax are, respec- 
tively, the minimum and maximum genus of G for orientable, or 
nonorientable case. In orientable case, u(G;x) = y(G; x) is called 
the handle polynomial. In nonorientable case, v(G; y) = 7(G; y) is the 
crosscap polynomaal. 

On the basis of the theory described in 12.1 and 12.2, (12.4) 
is in fact the genus distribution of embeddings of G. Because the 
enumerating function of super rooted maps of G is a constant times 
the genus polynomial y(G; z), for the enumeration of naps by genus it 
is enough only to discuss y(G; z). 


Lemma 12.6 An orientable associate surface of a graph with- 
out two letters interlaced has a letter x such that zx! is a segment 
of the surface. 


Proof Let (x,x~') be a segment of the surface with minimum of 
letters. If it does not contain only the letter x, then there is another 
letter y in it. Because of x and y noninterlaced, the segment (y, y^ 1), 
or (y !,y), is a subsegment of (x,x~'). However, it has at least one 
letter less than the minimum. [] 


Lemma 12.7 An orientable associate surface of a graph is with 
genus 0 if, and only if, no two letters are interlaced. 


Proof On the basis of Lemma 12.6, by the finite recursion prin- 
ciple the lemma can soon be found. [] 


Theorem 12.6 If an orientable associate surface of a graph 
has two letters interlaced, i.e., in form as ArByCxz- ! Dy ! E, then its 


290 Chapter XII Genus Polynomials 


genus is k, k > 1, if, and only if, the orientable genus of ADC'BE is 
k — 1. 


Proof On the basis of Relation 1 in L2, the theorem is soon 
found. [] 


According this theorem, a linear time algorithm can be designed 
for classifying the orientable associate surfaces of a graph G by their 
genus. Let N;(G) be the number of orientable associate surfaces of G 
with genus 7, 2 > 0 


Theorem 12.7 The handle polynomial of G is 


u(G; x) = ` Ni(G)z* (12.5) 


0xix 2| 
where 0 is the Betti number of G. 
Proof From (12.4), the theorem follows. O 


XII.4 Crosscap polynomials 


Let £5, = (S55;]1 € j < si} where s; is the number of orientable 
20-gons (surfaces) with genus i, then 


Fog = X Fig. (12.6) 


0s, 


Given a surface 5 € Fog, S induces 228 1 nonorientable surfaces. 
Let Ng be the set of all nonorientable surfaces induced by S. then the 
polynomial 
ós(y) = X. INS) (12.7) 
1<j<6 
is called the nonorientable form of S where N;(S) is the subset of Ns, 
se 
For a graph G with Betti number 5, the set of all associate ori- 
entable surfaces of determined by joint trees of G is denoted by S(G). 


XIL4 Crosscap polymomials 291 


Let S5;(G) for 6 € As be the subset of S(G) with nonorientable form 
ô where As is the set of all nonorientable forms of surfaces in S(G). 


Theorem 12.8 The crosscap polynomial of a graph G is 


v(G;y) = X. |Ss(G)|d(y). (12.8) 


EAs 


Proof From (12.4) and (12.7), the theorem is deduced. O 


Theorem 12.9 Ifa nonorientable associate surface of a graph 
is in form as ArBzxC, then its genus is k, k > 1, if, and only if, the 
genus of AB-1C is 

l k — 1, if AB-!C is nonorientable; 


k—1 12.9 
, otherwise. 





2 


Proof From Relation 2 in I.2, the theorem is soon found. [] 


According to Theorem 12.9, a linear time algorithm can also 
be designed for determining the genus of a surface and then classify 
nonorientable associate surfaces of a graph by genus. Hence, the cross- 
cap polynomial expressed by (12.8) can soon be found. 


Activities on Chapter XII 


XII.5 Observations 


O12.1 For the bouquet of size 3, list all its associate surfaces 
and then classify them by genus, orientable and nonorientable. 


O12.2 For K,, list all of its associate surfaces and then classify 
them by genus, orientable and nonorientable. 


O12.3 Compare the two sets of associate surfaces obtained in 
012.1 and 012.2. 


O12.4 Observe how many layer divisions of the surface 


(aa bb lec). 


O12.5 List all orientable surfaces of 4-gons. 


O12.6 For each orientable surface obtained in O12.5, find its 
nonorientable form. 


O12.7 List all orientable surfaces of 6-gons. 


O12.8 For each orientable surface obtained in O12.7, find its 
nonorientable form. 


012.9 Find the nonorientable form of surface (aa !bb cc). 
O12.10 Find the nonorientable form of surface (abca 1b 1c 5). 


O12.11 Find the nonorientable form of surface (abcc b ta~t). 


XII.6 Exercises 293 


XILO Exercises 


A graph is called a necklace if it is Hamiltonian and the result of 
deleting all 2-edges is a perfect matching. 

E12.1 Find the handle polynomial of a necklace with order 2n, 
no. 


E12.2 Find the crosscap polynomial of a necklace with order 
2n, n > 2. 


E12.3 Show that the nonorientable form of surface 
(a,b, +++ dnbna b an n) 
for n > 1 is 
a((1+ z)?^ — z?"), 
E12.4 Show that the nonorientable form of surface 


(a1a9 os aiu ts a,') 


for n > 1 is 
z((1-4 zr)" — x"). 
E12.5 Show that the nonorientable form of surface 


(aya; asaz eH ana) 


for n > 4 is 
(1-F x)" — 1. 


E12.6 For all surfaces of 20-gons, 6 > 4, with genus 0, Show 
that their nonorientable forms are 


(1-4 x)? — 1. 


A graph is called a ladder if it has a Hamiltonian circuit and all 
edges not on the circuit are geometrically parallel. 


E12.7 Find the handle polynomial of a ladder with m edges 
not on the Hamiltonian circuit. 


294 Activities on Chapter XII 


E12.8 Find the crosscap polynomial of a ladder with m edges 
not on the Hamiltonian circuit. 


A graph is called a Ringel ladder if it is cubic without multi- 
edge and consists of a ladder with the two 2-edges each of which is 
subdivided into 3 edges by two vertices and the other two edges are 
interlaced on the hamiltonian circuit 


E12.9 Find the handle polynomial of a ladder with m edges 


not on the Hamiltonian circuit. 


E12.10 Find the crosscap polynomial of a ladder with m edges 
not on the Hamiltonian circuit. 


XII.7 Researches 


R12.1 Find the handle polynomial of the bouquet B, of size 
m, m > 1 by joint tree model. 


R12.2 Find the crosscap polynomial of the bouquet of size m, 
m > 1 by joint tree model. 

R12.3 Find the handle polynomial of the wheel W,, of order n, 
n 2 4 by joint tree model. 

R12.4 Find the crosscap polynomial of the wheel W,, of order 
n, n 2 4 by joint tree model. 

R12.5 Find the handle polynomial of the complete graph Kn 
of order n, n > 4 by joint tree model. 

R12.6 Find the crosscap polynomial of the complete graph Kn 
of order n, n > 4 by joint tree model. 

R12.7 Find the handle polynomial of the complete bipartite 
graph Km,n of order m+n, m,n > 3 by joint tree model. 


R12.8 Find the crosscap polynomial of the complete bipartite 
graph Km,n of order m+n, m,n > 3 by joint tree model. 


R12.9 Find the handle polynomial of the n-cube of order n, 


XII.7 Researches 295 


n > 3 by joint tree model. 


R12.10 Find the crosscap polynomial of the n-cube of order n, 
n > 3 by joint tree model. 


R12.11 For the n-cube Q,, n > 3, prove that the minimum 
genus Yn = Jmin(Qn) with 7,_1 satisfies the relation 


Grint Qn) — 2" (n — 3) T Jmin(Qn-1) 


from an associate surface of Qn-ı with genus 7y,_; to get an associate 
surface of Qn with genus Yn. 


R12.12 For the complete bipartite graph Am», M > n > 4, 
prove that the minimum genus Ym,n = Jmin(Km,n) with Ymn—1 satisfies 
the relation 


ool Pas) = 2 T T TE, = 
where 
EE, m= 00 fn), 1(2 Mn: 2n, 2117/2), 
: 3(2 /n,2 f[n/2]; 2/n, 2] |n/2 |); 
[um = 2 m = 2(mod 4); 


4 
[== |, m - 02n), 12h. 2 Mn/21), 
3(2 Jn, 2|Lm/2];2In,2 Jln/2]) 


from an associate surface of Km n-1 with genus Ym,n-1 to get an asso- 
ciate surface of Km, n with genus Ym,n- 


R12.13 For the complete graph K,, n > 5, prove that the 
minimum genus ^; = Jmin( Kn) with *, 4 satisfies the relation 


n— 4 


Gal Ka) E ( 6 





) ale Grint A usd) 


296 Activities on Chapter XII 


where 


P5 n= 230 fLn/6)),3(2ILn/6]), 5Q f1n/6]) 
pes Ed n = 4(mod 6); 
|^ e zn n = 0,1(2|[n/6]), 3(2 J|n/6]), 5(2]|n/6]) 


from an associate surface of KK, , with genus *,.., to get an associate 











surface of K, with genus Yn. 


Chapter XIII 


Census with Partitions 


e The planted trees are enumerated with vertex partition vector in 
an elementary way instead as those methods used before. 


e A summation free form of the number of outerplanar rooted maps 
is derived from the result on planted trees. 


e On the basis of the result for planted outerplanar maps, the num- 
bers of Hamilton cubic rooted maps is determined. 


e The number of Halin rooted maps with vertex partition is gotten 
as a form without summation. 


e Biboundary inner rooted maps on the sphere are counted by, an 
explicit formula with vertex partitions. 


e On the basis of joint tree model, the number of general rooted 
maps with vertex partition can also expressed via planted trees in 
an indirected way. 


e The pan-flowers which have pan-Halin maps as a special case are 
classified according to vertex partition and genus given. 


XIIL.1 Planted trees 


A plane tree is such a super planar rooted map of a tree. A 
planted tree is a plane tree of root-vertex valency 1. In Fig.13.1, (a) 
shows a plane tree and (b), a planted tree. 


298 Chapter XIII Census with Partitions 


Let T be a planted tree of order n with vertices vo, v1, U2, --, 
Un, Nn > 1, where vo is the rooted vertex. The segment recorded as 
travelling along the face boundary of T from vp bach to itself and 
then vg left off is called a V-code of T when v; is replaced by i for 
i—1,2,---,n as shown in Fig.13.2 and Fig.13.3. 





(a) A plane tree (b) A planted tree 


Fig.13.1 Plane tree and planted tree 


A sequence of numbers is said to be polyhedral if each adjacent 
pair of numbers occurs twice. It is easily seen that a V-code of a 
planted tree is a polyhedral segment. 


The vector n = (n4,n5,::-,ni,:--), where n;(? > 1) is the number 
of unrooted vertices of valency 2, is called the vertex partition of a 
planted tree. 


For a sequence of nonnegative integers n4, no, * -:, ni, --- denoted 
by a vector n = (n1, n2,- , Ni, +), if 


3. 2- ini = 1, (13.1) 


i>1 
then n is said to be feasible. Let 


n = (nj, n5, ma) 


XIII.1 Planted trees 299 


where 
nı — 1, when: = l; 
ng—-ı + 1, when; = k — 1; 
yg um | (13.2) 
ny —1, when t = k; 
nj, otherwise 
for a k > 2 and ni, ng > 0, then n' is called a reduction of n. 


Lemma 13.1 A reduction n’ of a sequence of nonnegative in- 
tegers n is feasible if, and only if n is feasible. 


Proof By considering (13.2), we have the eqaulity as 
3. Q-)ni- M (2-)n = -1 — (2 — k) + (2- k1) 20. 
i>1 i21 


' his leads to the lemma. [] 





The sequence n = (1, 1) is feasible but no reduction can be done. 
So, it is called irreducible. 


Lemma 13.2 Any feasible sequence n has nı > 0. 


Proof By contradiction. Suppose n is feasible but n; — 0. Be- 


cause of 
3. 2-ini2 (2- ini x 0. 
i21 i22 
This contradicts to (13.1), the feasibility. O 


Lemma 13.3 Any feasible sequence n # ngo can always be 
transformed into n, only by reductions. 


Proof Because of n Z ny, Lemma 13.2 enables us to get a reduc- 
tion. Whenever the reduction is not ng, another reduction can also be 
done from Lemma 13.1. By the finite recursion principle, the lemma 
is done. [] 


Theorem 13.1 For a nonnegative integer sequence n = (n4, n, 
s Ni, +++), there exists such a planted tree that n; unrooted vertices 
are of valency i(i > 1) if, and only if, n is feasible. 


300 Chapter XIII Census with Partitions 


Proof Necessity. Suppose T' is such a planted tree. Because n; 
is the number of unrooted vertices with valency i, i > 1 in T, the size 


of T' is 
Y 


i21 


1c ini 22V 7n. 


i21 i21 


and hence 


This means that n satisfies (13.1), i.e., n is feasible. 

Sufficiency. First, it is seen that the irreducible sequence is the 
vertex partition of the planted tree whose under graph is a path of two 
edges. Then, by following the inversion of the procedure in the proof 
of Lemma 13.3, a planted tree with a given feasible sequence can be 
found. [] 


For a polyhedral segment L with 1 as both starting and ending 
numbers on the set N = {1,2,3,---,n}$, n > 1, let the vector be the 
point partition of L where n; be the number of occurrences of i in L, 
i >l. 

In a polyhedral segment L, if vuv is a subsegment of L, then u is 
said to be contractible. The operation of deleting u and then identify- 
ing v, or in other words vuv is replaced by v, is called contraction. If L 
can be transformed into a single point, then L is called a celluliform. 

If the point partition of L satisfies (13.1), then L is said to be 
feasible as well. 

It can be seen that any celluliform is a feasible segment but con- 
versely not necessary to be true. 

In what follows, the notation bellow is adopted as 


O- Cm) HC) o enm 


where s > 2, n; > 0 are all integers and 


8 i-1 
n= ) Ni, Oj-1 = X nj. 
i=1 j=1 


XIII.1 Planted trees 301 


Notice that when s — 2, it becomes the combination of choosing 
ni from m. 


Example 13.1 In Fig.13.2, two distinct planted trees of order 
5 are with vertex partition n = (3,0,2) satisfying (13.1). (a) is with 
sequence 123242151 and (b), 121343531. Here, 


1 5 1 5! 
5 \3,0,2 ~ 531021 


ang 


(a) 123242151 ) 121343531 
Fig.13.2 Trees with n = (3,0, 2) 


In Fig.13.3, six distinct planted trees of order 5 shown by (a-f) 
are with vertex partition n = (2,2, 1) satisfying (13.1). Here, 


lf 5 1 5! 
5221) 522m1 


For a feasible segment of numbers on N, the occurrences of i € N 
divides the segment into sections in number equal to that of times of 
its occurrences. Each of the sections is called an i-section. 

If a feasible segment on N is with the property that all numbers 
less than 7 have occurred before the first occurrence of i, 1 <i € n, 
then it is called favorable. Denote by 11 a 1-to-1 correspondence 
between two sets. 


Lemma 13.4 Let 7, be the set of all planted trees of order n+1 
with vertex partition n and £n, the set of all favorable celluliforms on 
N with point partition n, then J, 1&1 Lp. 


302 Chapter XIII Census with Partitions 


Proof Necessity. For T € 7,, it is easy to check that its V-code 
A(T) is uniquely a favorable celluliform, i.e., u(T) € Ln. 

Sufficiency. Let u € £n. Because of the uniqueness of the great- 
est point which is contractible, a point can be done by successfully 
contracting the greatest points. By reversing the procedure, a tree 
T(u) € 1, is done. O 


Theorem 13.2 The number of nonisomorphic planted trees of 
order n + 1 with vertex partition n is 


where 
n! = | |m! (13.5) 





(a) 123432521 (b) 1232345421 (c) 123435321 





(d) 123214541 (e) 121345431 (f) 123432151 


Fig.13.3 Trees with n = (2, 2, 1) 


XIII.1 Planted trees 303 


Proof On the basis of Lemma 13.4, it suffices to discuss the set 
of all favorable celluliforms £,. Since each favorable celluliform has 
n possibilities to choose the minimum point and different possibilities 
correspond to different ways of choosing n from n elements, the set of 
all ways is partitioned into 


classes. A way is represented by number sequence of length n with 
repetition as occurrence in the natural order. Two ways A and B 
are equivalent if, and only if, there exists a number i € N such that 
A + i(mod n) is B in cyclic order. A way starting from 1 is said to 
be standard. Because of each class with n ways in which only the 
standard way enables us to form the V-code of a planted tree, the 
theorem is soon obtained. [] 


In Example 13.1, Fig.13.2 and Fig.13.3 show two cases of (13.4). 
Only take n = (3,0, 2) as an example. There are 10 ways of combina- 
tions of choosing 2 points with 3 occurrences each and 3 points with 
1 occurrence each from 5 points numbered by 1,2,3,4 and 5 as 


1) 111222345; (2) 111233345; (3) 111234445; 
4) 111234555; (5) 122233345; (6) 122234445; 
7) 122234555; (8) 123334445; (9) 123334555; 
10) 123444555 


"m LOS LS OS 


in which 2 classes are divided as C = {(1), (5), (8), (10), (4)} and C5 = 
{(2), (6), (9), (3), (7)} because of (5) = 222333451, (8) = 333444512, 
(10) = 444555123 and (4) = 555111234 as (1) = 111222345 for Ci, 
and the like for C5. 

For a general outerplanar rooted map M = (Xa B(X), P) with 
(r)p, on the specific circuit where r is the root and y = ag and its 
dual M* = (Xa4(.X), Py) with root r as well without loss of generality, 
let Hy be the map obtained from M* by transforming the vertex (r)p, 
of M* into vertices (r), ((Py)r), +++, (Py) !r). Such an operation 
is called articulation. The root rg of Hm is taken r as shown in 


304 Chapter XIII Census with Partitions 


Fig.13.4 in which bold lines are on M and dashed lines, on Hm. Here, 
multiedges are permitted. 





Fig134 M ana Hy 


Further, it is easily checked that Hy), is a planted tree of size 
which is equal to the size of M. 


Lemma 13.5 An outerplanar rooted map M of order n 4- 1 
with face partition s is 1-to-1 corresponding to a planted tree Hm 
with vertex partition t = s + nl,. 


Proof By the procedure of getting Hm from M, it is seen that 
the number of 7-faces in M is the same as that in Hy for i > 1. 

For i = 1, Hy has n — 1 articulate vertices greater than sı. In 
virtue of the nonseparability of M, sı = 0. 

Conversely, it is still true and hence the lemma. [] 


An attention which should be paid to is that all articulate edges 
in Hymy are 1—to-1 corresponding to all edges on the root-face boundary 
of M. 


Theorem 13.3 The number of nonisomorphic outerplanar rooted 
maps of order n with face partition s is 


ee en aso 


where s, +n — n, i.e., 5; = 0, because of no articulate vertex and s is 
the number of unrooted faces. 





Proof On the basis of Lemma 13.5, the theorem is obtained 
from Theorem 13.2. [] 


XIIL2 Hamiltonian cubic maps 305 


Since a bipartite map has all of its faces of even valency, its face 
partition s is of all s; = 0 when 7 is even. 


Corollary 13.1 The number of nonisomorphic outerplanar rooted 
bipartite maps of order 2m with face partition s is 





ji —1 
————— iui (13.7) 
2m 4- s 1 Ns 4- (2m — 1)1, 
where s, +n — n, i.e., 5; = 0, because of no articulate vertex and s is 
the number of unrooted faces. [] 


A map is said to be simple if it has neither selfloop nor multiedge. 


Corollary 13.2 The number of nonisomorphic outerplanar rooted 
simple maps of order n with face partition s is 





1 ire — 1 
—_ 13.8 
m Me nm Es 
where s, +n — n, i.e., 5; = 0, because of no articulate vertex and s is 
the number of unrooted faces. [] 


Corollary 13.3 The number of nonisomorphic outerplanar rooted 
bipartite maps of order 2m with face partition s is 





1 —1 
————— E (13.9) 
2m 4- s — 1 Xs 4- (2m — 1)1, 
where s, +n — n, i.e., 5; = 0, because of no articulate vertex and s is 
the number of unrooted faces. [] 


XIII.2 Hamiltonian cubic map 


For saving the space occupied, this section concentrate to discuss 
on Hamiltonian planar rooted quadregular maps as super maps of a 
Hamiltonian planar graph, and then provide a main idea for general 


306 Chapter XIII Census with Partitions 


such maps. A map is said to be quadregular if each of its vertices is of 
valency 4. 

A Hamiltonian planar rooted quadregular map with the two 
edges not on the Hamiltonian circuit not success in the rotation at 
each vertex is called a quaternity. 

Let Mi = (Xa vel X1), J1) and Mz = (Xa vel X2), J2) be two rooted 
maps with their roots, respectively, rı and ra. Assume (rı)zy and 
(72) my are with the same length. 

The map obtained by identifying rı and arg with Kr; = Karo 
as well as (J1y)'ri and (aJey)'r2 for i > 1 is called the boundary 
identification of Mj and M», denoted by I(Mi, Mz). The operation 
from Mı and M» to I(Mi, Mə) is called boundary identifier. 

A boundary identification of two outerplanar cubic rooted maps 
is a quaternity because of Mı and Mə both outerplanar and cubic with 
its root rı = rə. 


Lemma 13.6 Let O, and Z, be the sets of all, respectively, 
quaternities and boundary identifiers with face partition s, then there 
is a 1-to-1 correspondence between Q, and Z;. 


Proof By considering the inverse of a boundary identifier, a qua- 
ternity becomes two cubic outerplanar maps whose boundary identi- 
fication is just the quaternity with the same face partition s. This is 
the lemma. [] 


From the proof of this lemma, it is seen the identity 


Lemma 13.7 The number of nonisomorphic outerplanar cubic 
rooted maps of order n with face partition s is 


crei (non) sa 


for s € Scu», the set of all the vectors available as the face partition of 
an outerplanar cubic map. 





XIIL3 Halin maps 307 
Proof From Theorem 13.3, the conclusion is true. [] 


Theorem 13.4 The number of nonisomorphic quaternities of 
order n with face partition s is 


p OMGE) v. 


$1:52€ cub 
8—8]-82 


where a; — |s;|, called the absolute norm of s;, i.e., the sum of all the 
absolute values of entries in s; for i = 1,2. 


Proof Sincethe set of all quaternities of order n is the Cartesian 
product of the set of all cubic outerplanar rooted maps and itself, the 
formula (13.12) is soon obtained. [] 


'This method can be also employed for the case when the bound- 
ary is cubic and further for others with observing boundary combina- 
torics. 


XIII.3 Halin maps 


If a graph can be partitioned into a tree and a circuit whose 
vertex set consists of all articulate vertices of the tree, then it is called 
a Halin graph. A planar Halin map is a super map of a Halin graph 
on the surface of genus 0 such that the circuit forms a face boundary. 

Let H = (X,,5(X), J) be a planar Halin rooted map with (r) gy, 
y = af, as the face formed by the specific circuit where r is the root. 
The associate planted tree denoted by Ty is obtained by deleting all 
the edges Kr, K(Jy)r, +++, K(Jy) 1r on the circuit. 


Lemma 13.8 A planar Halin rooted map with vertex partition 
u of the specific circuit with length n is 1-to-1 corresponding to a 
planted tree with vertex partition v = u + (n — 1)(1, — 1). 


Proof By considering the procedure from a Halin map H to a tree 


308 Chapter XIII Census with Partitions 


Ty, carefully counting the numbers of vertices with the same valency 
and comparing them of H with those of Ty, the lemma is found. OU 


Theorem 13.5 The number of nonisomorphic planar Halin 
rooted map with vertex partition u of the specific circuit with length 


"C 1 lu 4-n—1 
AH) = Tena " (n — 1 a) Hey 


where |u| is the absolute norm of u. 


n is 


Proof On the basis of Lemma 13.8, by Theorem 13.2, the con- 
clusion of the theorem is done. [] 


Let Hy = (%o,9(X1), A) and Hə = (Xo,3(X2), J2) be two planar 
Halin rooted maps with |{ri}7,4| = {r2} nyl, the boundary identifica- 
tion of Hı and H^» is called a double leaf. 

A graph with a specific circuit of all vertices of valency 4 is called 
a quadcircularity. A super map of a quadcircularity is a quadcircula- 
tion. 


Lemma 13.9 A planar rooted quadcirculation M is a double 
leaf if, and only if, the map obtained from M by deleting all edges on 
the specific circuit can be partitioned into two trees such that each of 
vertices on the circuit is articulate of both the trees. 


Proof Since a double leaf is obtained by boundary identifier 
from two Halin maps, the conclusion is directly deduced. [] 


Lemma 13.10 A planar rooted quadcirculation with vertex 
partition u of the specific circuit of length n is 1-to-1 corresponding 
to a pair of planar Halin rooted maps H, and Hə with vertex partitions, 
respectively, s and £ such that 


u =s +t-— (n — 1)(21, — 14). (13.14) 
where 1; is the vector of all entries 0 but the i-th 1 for i = 3,4. 


Proof By considering that u does not involve n — 1 unrooted 


XIIL3 Halin maps 309 


3-vertices in s and £ each and involves n — 1 unrooted 4-vertices, the 
formula (13.14) holds. O 


Theorem 13.6 The number of nonisomorphic double leafs with 
vertex partition u of the specific circuit of valency n is 


` A(HS) ACHL) (13.15) 
S-t—u-(n—1)(213—14) 
8,t€Sq] 
where Sq is the set of vectors available as vertex partitions of planar 
Halin maps. 


Proof On account of Lemma 13.10, the theorem is soon derived 
from Theorem 13.2 for the Cartesian product of two sets. [] 


Given a nonseparable graph G with a cocircuit C* of an orien- 
tation defined, if G is planar in companion with such a orientation 
then G is said to have the C*-oriented planarity, or cocircuit oriented 
planarity. A planar super map of such a graph is called a cocircuit 
oriented map. If each edge on the cocircuit is bisectioned and then 
snip off each new 2-valent vertex as two articulate vertices in a cocir- 
cuit oriented map M so that what obtained is two disjoint plane trees, 
then M is called a cocircular map. The root is always chosen to be an 
element in an edge on the cocircuit in a cocircular map. 


Lemma 13.11 A cocircular map with the oriented cocircuit 
of n+ 1 edges and the vertex partition u is 1—to-1 corresponding to 
a pair of planted trees (T1, T5) with vertex partitions u and us such 
that uj; = uU», = n. 


Proof By considering the uniqueness of a cocircular map com- 
posed from two planted trees, the coclusion is directly deduced. UO 


Let U, be the set of all integer vectors feasible to a planted tree 
with n unrooted articulate vertices. 


Theorem 13.8 The number of nonisomorphic cocircular maps 


310 Chapter XIII Census with Partitions 


with the oriented cocircuit of n edges and given vertex partition u is 


iua IY /lul 
(a) m 


Uy 42 €Un uy | [us] 


uj tug=u 


Proof Based on Lemma 13.11, the formula (13.16) is derived 
from Theorem 13.2. L 


A cocirculation is such a planar rooted map which has a cocircuit 
oriented. For this type of planar maps, the number of nonsepsrable 
ones can be determined from maps with cubic boundary of root-face. 

More interestingly, maps with cubic boundary of root-face can 
be transformed into maps with root-vertex valency as a parameter. 

In view of this, many types of planar maps with cubic boundary 
can be known from what have been done for counting maps with size 
and root-vertex valency as two parameters. 


XIIL4 Biboundary inner rooted maps 


A map is said to be biboundary if it has a circuit C that two 
trees are obtained by deleting all the edges on C. In view of this, a 
Hamilton cubic map is a uniboundary map because it is not necessary 
to have two connected components as all the edges on the Hamiltonian 
circuit are deleted. Here, only planar case is considered. 

LetM = (X, 7)be a biboundary map, r = r(M) is its root. The 
length of boundary is m and the vertex partition vector of nonbound- 
ary vertices is n = (ni,n5,-:--), nj, i > 1, the number of i-vertices not 
on the boundary. 

Assume M; = (4, 41) and My = (A5, J) are two submaps of 
M. Denote byC = (Kr, Kgr,-:-,Ky™ 'r) the boundary circuit of 
M where 


i J^yx;, if yx; not incident with an inner edge of Mo; 
pti = 


aT Dxz;, otherwise. 


XIII.4 Biboundary inner rooted maps 311 


To= TIm =T and 2z;=y'r,i=1,2,---,m—1. 

Let rı = r(Mi) =r and ra = r(M2) = aq" tr. This means the 
root-vertex of Mı adjacent to that of M» in M. Such a root is said to 
be inner rooted. 

First, denoted by Bm, m > 6, the set of all biboundary rooted 
maps with the boundary lengthm. 


Lemma 13.12 Let Wm, and Wm, are, respectively, unibound- 
ary maps of boundary lengths mı > 3 and mə > 3, then a pair of 
(WA, Wo}, Wi € Wm, and Ws € Wma, m = m; + Mo, composes of 


mı 


Sm (Mı) = M (13.17) 


biboundary maps in B,,. And, this combinatorial number is deter- 
mined by the recursion as 


Tn» os Mə — 1 
[ims [=|] mz 
d (13.18) 


1 
Ma =1, m 21; FA =1, m = 0. 
Proof By induction on m», m» > 2, for any m; > 1. 
First, check the case of my = 2, for m, > 1, that 


[2] -[] [o] [2 Jem 


From the fact that if o vertices in the first segment, then the second 
segment has to have mj, vertices; if 1 vertex in the first segment, then 
the second segment has to have m, — 1 vertices; ---; if mı vertices in 
the first segment, then the second segment has to have 0 verticesm;. 
They are all together m; 4- 1. Thus, (13.18) is true for m» — 2. 
Then, assume £4, 1(mij), m2 > 3, have been determined by 
(13.18). To prove tm,(mi) is determined by (13.18). Because of 
mı + 1 occurrences for putting mı vertices in to m» segments as 


312 Chapter XIII Census with Partitions 


when m vertices in the first segment, then no vertex in all other 
m» — 1 segments and hence 5,,, 1(0) ways; when mı — 1 vertices in 
the first segment, then i vertex in all other m; — 1 segments and hence 
Sm,-1(1) ways; ---; when 0 vertices in the first segment, then m; ver- 
tices in all other m» — 1 segments and hence s,,, 1(mm1i) ways. They 
are Sm (M1) = 357 9 Sm—1 (i) ways all together. That is 


My 
mə | _ Mə — 1 
[m SE r | mza 
i=0 

By the induction hypothesis, Sm (mı) is determined. The lemma is 
true. g 


Then, denote by Dm, m > 6, the set of all biboundary rooted 
maps with the boundary length m. 


Lemma 13.13 Let Mm, and Mm, be the uniboundary rooted 
maps of boundary lengths, respectively, mı > 3 and mə > 3, then a 
pair {M), Mo}, Mı € Mm, and M» € Mm, composes 


Ls (mi) = ( ri ) (13.19) 


biboundary rooted maps in Dm, Mm = m44-m». And, this combinatorial 
number is determined by 


m \ _ = my —1 
(m)-X| ; i; (13.20) 
i=0 
where the terms on the right hand side of (13.19) are given in Lemma 
19.13. 


Proof By induction on mg for any mı > 1. 
First, when m» = 2, By considering for assigning rm vertices in 
M» edges on the boundary in the order determined that when 1, 2, 
-+, my, vertices in the first edge(incident with the root), the second 
edge has to have, respectively, mı — 1, m4 — 2, ---, 0 vertices, we have 
to(m) = s1(0) + si(1) Tec si(mi — 1) = mı. 


XIII.4 Biboundary inner rooted maps 313 


Then, assume £4, 1(mij), m2 > 3, have been determined by 
(13.20). To prove that 


tma (m1) = $m4-1(0) + Sma—1(1) = Pee $m;-1(mà 2 b 


is determined by (13.20) as well. Because of the first edge in M» edges 


allowed to have mi, mı — 1, ---, 1 vertex, the other mə — 1 edges only 
allowed to have, respectively, 0, 1, ---, mı — 1 vertex. This implies 
mM — 1 mM — 1 mM — 1 
m epe pem mi] 
Hence, (13.20) is right. [] 


Denote by Qi, 1 € i € tm (Mı) the t4,(mi) biboundary inner 
rooted maps mentioned in this lemma with M (Mi, M5) = (Qi, Qs, +>, 
Qi, (m)]- 


Lemma 13.14 Let Hm = ( M(Mi, M3)|IV(Mi, M») € Mm, x 
M miz m + M2 = my}, then 


Des wee (13.21) 
HEH 


i.e., Hm is a partition of Dm. 


Proof For any D € Dm, it is known from biboundary maps that 
there exist mı and m2, Mı + Mə = m, such that Mı € Mm, and 
M» € Mm, compose of D. Thus, D = (Mi, M3) € Hin. 

Conversely, for any Q € H, H € Hm, because of Q composed 
from two uniboundary maps M; € Mm, and M» € Mma, m, + m2 = 
m, there exist D € Dm such that D = Q. 

In summary, the lemma is obtained. [] 


In what follows, observe how many nonisomorphic uniboundary 
maps of boundary length m with vertex partition vector n. 


Lemma 13.15 The number of uniboundary rooted maps of 
boundary length m, m > 3, and nonboundary vertex partition vector 


314 Chapter XIII Census with Partitions 


(m t n — 1)! 


"im — Tin (13.22) 


n(m, n) = 


where n = |n| 2 n1 +n +--+.. 


Proof Let M = (X, J) bea uniboundary rooted maps of bound- 
ary length m, m > 3, and nonboundary vertex partition vector n. Its 
root is r. Because of cubicness on the boundary, Jr is incident with 
an articulate vertex of the tree. Let Jr be the root to make the tree 
planted. Because of all 1-vertices of the planted tree on the boundary, 
its vertex partition vector is n+ (m — 1)1,, where 1, is the vector of all 
entries 0 but the first entry 1. Since a planted tree with vertex parti- 
tion is 1-to-1 corresponding to a uniboundary rooted map of boundary 
length m, m > 3, and vertex partition vector n, from Theorem 13.2, 
the number of nonisomorphic uniboundary rooted maps f boundary 
length m, m > 3, and vertex partition vector n is 


(om (m t n — 1)! 

nh, bh) = 2 aaa. S XT 

ý (m +n — 1)!(n — m1)! 

where n = |n| = ni + n3 4- ---. By considering that nı = 0 in n, the 
lemma is done. C 


On the basis of the above tree lemmas, the main result of this 
section can be gotten. 


Theorem 13.9 The number of biboundary inner rooted maps 
of boundary length m > 6 and nonboundary vertex partition is 


Y Mə (n! s TI — 1)! 
plz y p Mao 13.23 
m (n? + m3 — 1)! E 


(mg = 1)!n?! 


where L = ((r4, m», nl, n?)| m; + ma = m, n! + n? = n, m4, m; > 3). 


Proof For any given m, and m», mı +M = m, with n! and n?, 


XIIL5 General maps 315 


nl +n? =n, from Lemma 13.14-15, Dm can be classified into 
5 (n! 4- m4 — 1)! (n? 4- m5 — 1)! 
(m; — l)mn!! (me — 1)!n?! 


(m; ,ma,n!,n?)e£ 


classes. From Lemma 13.13, each class has 


Cm) 


nonisomorphic biboundary inner rooted maps of boundary length m 
and nonboundary vertex partition vector n. Thus, the theorem is 
proved. L 


XIII.5 General maps 


Based on the joint tree model shown in Chapter XII, it looks 
general maps on surfaces in a closed relation with joint trees. In this 
section, only orientable case is considered as an instance. 

Because of the independence with a tree chosen, general maps 
with a cotree marked are particularly investigated. 

For the convenience for description, all maps are assumed to have 
no articulate edge. 

Let M = (X, J) be a map with cotree edges aj = Ka, a2 = 
Kx, ---, aj = Ka; marked where | = (M) is the Betti number of 
M. The root of M is chosen on a cotree edge, assume r = r(M) = x4. 

Another map Hy = (Xy, Jz) is constructed as 


l 
Xa =X +Y (Ksi + Kt; — ai) (13.24) 
i=1 
where Ks; = zi, oui, 95i, ysjsand Kt; = {yz;, 82i, Bt, yu}, 1 <1 < 
l, ; Jg is defined as 


T { (x)z, when x € X; 


(x), when z ¢ X. (13.25) 


316 Chapter XIII Census with Partitions 


Lemma 13.16 For any rooted map M with a cotree marked, 
the map Hy is a planted tree with the number of articulate vertices 
two times the Betti number of M. 


Proof Because of connected without circuit on Hy, Hy is a 
tree in its own right. Since the number of cotree edges is the Betti 
number of M, from the construction of Hm and no articulate edge on 
M, the number of articulate vertices on Hm is two times the Betti 
number of M. go 


Let M (l; n) be the set of all general rooted maps with a cotree of 
size | marked and vertex partition n including the root-vertex. And, 
let H(n) be the set of all planted trees with articulate vertices two 
times the number of cotree edges and vertex partition n excluding the 
root-vertex. 


Lemma 13.17 There is a 1-to-1 correspondence between M (l; 
n) and H(n + (21 — 1)1,) as the set of joint trees. 


Proof For M € M(l; n), it is easily seen that the corresponding 
Hy is just a joint tree of M and hence Hy € H(n + (21 — 1)1,). 

Conversely, for H € H(n+(2l—1)1,), in virtue of a joint tree with 
its articulate vertices are pairwise marked as cotree edges of the cor- 
responding map M as H — Hy, by counting the valencies of vertices, 
it is checked that M € M(l; n). 

Therefore, the lemma is true. [] 





This lemma enables us to determine the number of general rooted 
maps with a cotree marked with vertex partition given. 


Theorem 13.10 The number of rooted general maps with a 
cotree marked for a vertex(root-vertex included) partition n given is 


(n + 21 — 2)! 


ME n)| = “Ol iym ^ 


(13.26) 


where | is the Betti number(the size of cotree) and n = |n]. 


Proof A direct result of Lemma 13.17 and Theorem 13.2. OU 


XIII.G Pan-flowers 317 


XIIL.G Pan-flowers 


A map is called a pan-flower if it can be seen as a standard petal 
bundle added a tree such that only all vertices are in the inner parts of 
edges on the petal bundle. The petal bundle is seen as the boundary 
of a pan-flower. Because of not a circuit for the base graph of a petal 
bundle in general, a pan-flower is reasonably seen as a generalization 
of a map with a boundary, or a boundary map. A pan-Halin map is 
only a special case when the petal bundle is asymmetric. 

For convenience, the petal bundle in a pan-flower is called the 
base map. If a edge of a fundamental circuit on the underlying graph 
of base graph is allowed to have no articulate vertex of the tree, then 
the pan-flower is said to be pre-standard. If the edge has at least one 
articulate vertex of the tree, then the pan-flower is standard. 

This section is concerned with pan-flowers in the two classes foe 
a vertex partition vector given. 

Let Hpsu be the set of all rooted pre-standard pan-flowers, where 
the root is chosen to be an element incident to the vertex of base 
map. For any H = (X, J) € Hy, the tree Ty is always seen as a 
planted tree whose root is first encountered on the rooted face of H 
starting from the root of H. Otherwise, the first encountered at the 
root- vertex of H from the root. 


Lemma 13.18 Let Hpsu(p;s) be the set of all rooted pre- 
standard pan-flowers with vertex partition vector s = (s2, 53,-:-) ona 
surfaces of orientable genus p, then 


jı —Ópa jit 2p . 
|Hpsu(p; 8)| = 2/179» ie E P) A (13.27) 


where 71(j) is the set of planted trees with vertex partition vector 
j = (ju J2,°++), such that s; = ji, i A 3; s3 = Ja ja +1, pz 1,82 0, 
j > 0, but s # 0 and j z 0. Further, 6,1 is the Kronecker symbol, 
i.e., dpi = 1, when p = 1; 0p = 0, otherwise. 


Proof On the basis of pre-standardness, it is from the defini- 
tion of pan-flowers seen that an element in the set on the left hand 


318 Chapter XIII Census with Partitions 


side of (13.27) has an element in the set on the right hand side in 
correspondence. 
In what follows, to prove that each T = (Vv, J) € 71(j), produces 


| +2 
51-2 f d ) maps in Hpsu(p; s). 


951— 

Denied by (r); (v1), (2), +++, (2j) all the articulate vertices 
of T, where 0 < h < ly < --- « l4, such that x; = (Jar, i = 
1,,2,---, 91, r = r(T) is the root of T. 

First, by considering that the underlying graph of base map has 
2p loops, only one vertex, its embedding on the orientable surface of 
genus 2p has exactly one face. Because of the order of its automor- 
phism group 8 when p = 1, only one possible way; 2p when p > 2, two 
possible ways. 

Then, the assignment of the jı + 1 articulate vertices, (r), (x;), 
i = 1,2,--.,7]1 of T on the base map has the number of ways as 
choosing 2p — 1 from 7, + 2 intervals with repetition allowable. That 


1S 
A-2-t(2p-1)—-1Y _ fjit+2p 
2p—1 2p—1/- 


Finally, since each of elements r, x;, i = 1,2,---, jı, has 2 ways: 
one side (ax, ax), or the other (x, Gx}, they have 





gjı +1 
ways altogether. 


In summary of the three cases, The aim reaches at. [] 


On the basis of this lemma, by employing a result in 813.1, the 
following theorem can be deduced. 


Theorem 13.11 The number of pre-standard rooted pan-flowers 
with vertex partition vector s = (s2, 53,:--) and its base involving m 
missing vertices on an orientable surface of genus p, p > 1, is 


2p —1 | 
pete d à 0) E (13.28) 


where s! = [[,.5 5;! = 59!53! --- and n+ 2 = $5.5 si. 


XIII.G Pan-flowers 319 


Proof From Lemma 13.18, this number is 


- m=+2p-— 1 , 
9 a pi )n@ 


where j = (jı, j2, ja, -) such that 7; +1 = m, j3 = 53 — m and ji = si, 
i 3, 22. Then from Theorem 13.2, we have 


(a —1)! — n! 


OR T a Dlss mia 


S3 \ n!m 
Am) s^ 


where n = n' — 1 = 905,91 — 1 = $5255; — 2. By substituting this 
into the last, (13.28) is obtained. E 


Let Tii: s) be the set of all pre-standard rooted pan-flowers 
with vertex partition vector s = (s2, 53, : --) on a nonorientable surface 
of genus q, s > 0, but s #0. 


Lemma 13.19 For Hps(q; 8), q > 0, s > 0, but s Z 0, we have 


mt+q-1 


(Hpsu(q; s)| = z-ten[ ed 


Jm (13.29) 
where j = (m— 1, 52, 83—m, S4, 85, -) and m is the number of trivalent 
vertices(i.e., missing vertices on its base). 


Proof Lj Similarly to the proof of Lemma 13.18. However, an 
attention should be paid to that the size of the base is q, q > 1, instead 
of 2p, p > 1 and the order of automorphism group of the base is 2q 
when q > 2; 4 when q= 1. [] 


similarly to Theorem 13.11, we have 


Theorem 13.12 The number of pre-standard rooted pan-flowers 
with vertex partition vector s = (55,53,---) and its base involving m 
missing vertices on a nonorientable surface of genus q, g > 1, is 


jm-&i 1 (" Tue ) (5) nim. (13.30) 


q—-1 mj s! 


320 Chapter XIII Census with Partitions 


where «F2 = » sys and 8| e [pga 


Proof Similarly to the proof of Theorem 13.11, from Lemma 
13.19 and Theorem 13.2, the theorem is done. L 


For standard pan-flowers, let Hs be the set of all such maps. 
From the definition, if the base map has m missing vertices, then it is 
not possible on an orientable surface of genus greater then m/2, or on 
a nonorientable surface of genus greater than m. 


Lemma 13.20 Let Hsy(p;.s) be the set of all standard rooted 
pan-flowers with vertex partition vector s = (89, 83,---) on an ori- 
entable surface of genus p. If maps in H,y(p;s) have its base with 
m 2 2p missing vertices, then we have 


: — 9m—ôp 1+1 m= : 
Pax) zr img — asap 


where 7,(5), as above, is the set of panted trees with vertex partition 
vector j = (ji, Jo, ja.) for jj = m — 1, ja = 83 m, ji = si, i 3, 
$23 

Proof For any H € Hsy(p;s), from the definitions of standard- 
ness and pan-flowers, it is seen that there exists a planted tree in 71(7) 
corresponding to H. Thus, it suffices to prove that any planted tree 
in T = (X, J) € T(J) produces 


Qm— pitt m — 1 
2p—1 
maps in Hsy(p; s). 


First, an attention should be paid to that maps in Hsn(p; s) are 
with their base of size 2p on an orientable surface of genus p. Since 
the order of automorphism group of the base is 4p when p > 2; 8 when 
p = 1, the base has, respectively, 2 ways when p > 2; 1 way to choose 
its root. 

Second, since the number of missing vertices on the base is m, T' 
must have m — 1 unrooted articulate vertices. Let them be incident 
with (21), +++, (@m_1). From the standardness again, there are m — 


XIIL6 Pan-flowers 321 


1 intervals for choice in the linear order ((r), (x1), (£2), +++, (zi 1)). 
Thus, any 2p — 1 points insertion divides the linear order into 2p 
nonempty segments. This has 


m -—1 
2p— 1 
distinct ways. 


Third, notice that each of the m articulate edges including the 
root-edge in T' has 2 choices, and hence 
pL 
distinct choices altogether. 
In summary rom the three cases, the lemma is soon found. UO 
Based on this, we have 


Theorem 13.13 The number of standard rooted pan-flowers 
with vertex partition vector s = (s2, 53,:--) and their base map of m, 
m > 2p, unrooted vertices on an orientable surface Sp of genus p > 1, 


us m — 1X (/s3\nlm 
oF aa »" E ) (5) Fi (13.32) 


where n + 2 = 55.55; 

Proof Similarly to the proof of Theorem 13.11. However, by 
Lemma 13.20 instead of Lemma 13.8. [] 

At a look again for the nonorientable case. 

Lemma 13.21 Let H.y(q; s) be the set of standard rooted pan- 
flowers with vertex partition vector s = (s2, 53, --) on a nonorientable 
surface of genus q > 1. If each map in Hsy(q; s) has its base map of 
m unrooted vertices, then we have 


Han(q;s)| = 27-9 CNO (13.33) 


where n(j) zx A )L Jj = 63m Jaen A = m— L J3 = 63 — M, 
Jli = s 13, i22. 


322 Chapter XIII Census with Partitions 


Proof Similarly to the proof of Lemma XII.2.3. However, what 
an attention should be paid to is the base of size q instead of 2p for a 
surface of genus q. [] 


Thus, we can also have 
Theorem 13.14 The number of standard rooted pan-flowers 


with vertex partition vector s = (s2, 53,---) and their bases of m un- 
rooted vertices on a nonorientable surface of genus q, q > 1, is 


E ! 
ee (s P) (5) ma (13.34) 


where s 20, $20, m2q21andn-c2-—95/.5si. 


Proof Similarly to the proof of Theorem 13.13. However, by 
Lemma 13.21 instead of by Lemma 13.20. [] 


Activities on Chapter XIII 


XIIL.7 Observations 


O13.1 Observe the number of plane rooted trees of size n > 0 
given. 


O13.2 Observe the number of outerplanar rooted maps of size 
n 2 0 given. 
O13.3 Observe the number of wintersweets of size n > 0. 


O13.4 To show a relationship between outer planar maps and 
trees. 


O13.5 Observe how to evaluate the number of plane rooted 
trees with the number of articulate vertices m > 2 and size n > 1 
given. 


O13.6 Consider what relation have the enufunction with the 
number of nonrooted vertices, nonrooted faces and the enufunction of 
vertex partition and genus as parameters. 


O13.7 Observe the number of rooted plane tree with root- 
vertex valency and vertex partition. 


O13.8 Observe the number of rooted plane trees with the num- 
ber of articulate vertices and vertex partition of other vertices. 


O13.9 Observe the difference between boundary maps and non- 
boundary maps. 


O13.10 observe the automorphism groups of a map and one of 
its boundary map by . 


324 Activities on Chapter XIII 


XIII.S Exercises 


E13.1 Determine the enufunction of planted trees with vertex 
partition as parameters by establishing and solving a equation. 


A wintersweet is a rooted map with the property that it becomes 
a tree if missing circuits only at nonrooted terminal vertices of the 
tree. 


E13.2 Establish an equation satisfied by the vertex partition 
function of Wintersweets and then try to solve the equation. 


A rooted map is called unicyclic if it has only one circuit. 


E13.3 Establish an equation satisfied by the vertex partition 
function of unicyclic maps when the root is on a circuit and then try 
to solve the equation. 


E13.4 Establish an equation satisfied by the vertex partition 
function of Halin rooted maps and then try to solve the equation. 


E13.5 Establish an equation satisfied by the vertex partition 
function of outerplanar rooted maps and then try to solve the equation. 


E13.6 Establish an equation satisfied by the face partition func- 
tion of outerplanar rooted maps when the root is not on the circuit 
and then try to solve the equation. 


E13.7 Establish an equation satisfied by the face partition func- 
tion of planar rooted petal bundles and then try to solve the equation. 


E13.8 Establish an equation satisfied by the face partition func- 
tion of outerplanar rooted maps when the root is on the circuit and 
then try to solve the equation. 


E13.9 Establish an equation satisfied by the vertex partition 
function of unicyclic maps when the root is not on a circuit and then 
try to solve the equation. 


E13.10 Establish an equation satisfied by the face partition 
function of planar rooted supermaps of bouquets and then try to solve 


XIII.7 Researches 325 


the equation. 


E13.11 Establish an equation satisfied by the vertex partition 
function of planar rooted maps with two vertex disjoint circuits and 
then try to solve the equation. 


XIII.9 Researches 


R13.1 Determine the vertex partition function of general rooted 
maps on the sphere. 


R13.2 Determine the vertex partition function of general rooted 
maps on all surfaces. 


R13.3 Determine the vertex partition function of general Eu- 
lerian rooted maps on all surfaces. 


R13.4 Determine the vertex partition function of 2-edge con- 
nected rooted maps on the sphere. 


R13.5 Determine the vertex partition function of 2-edge con- 
nected rooted maps on all surfaces. 


R13.6 Determine the vertex partition function of nonseparable 
rooted maps on the sphere. 


R13.7 Determine the vertex partition function of nonseparable 
rooted maps on all surfaces. 


R13.8 Determine the vertex partition function of loopless rooted 
maps on the sphere. 


R13.9 Determine the vertex partition function of loopless rooted 
maps on all surfaces. 


R13.10 Determine the vertex partition function of rooted tri- 
angulations on all surfaces. 


Chapter XIV 


Super Maps of a Graph 


e A semi-automorphism of a graph is a bijection from its semiedge 
set to itself generated by the binary group sticking on all edges 
such that the partitions in correspondence. 


e An automorphism of a graph is a bijection from the edge set to 
itself such that the adjacency on edges in correspondence. 


e The semi-automorphism group of a graph is different from its au- 
tomorphism group if, and only if, a loop occurs. 


e Nonisomorphic super rooted and unrooted maps of a graph can 
be done from the embeddings of the graph via its automorphism 
group or semi-automorphism group of the graph. 


XIV.1 Semi-automorphisms on a graph 


A pregraph is considered as a partition on the set of all semiedges 
as shown in Chapter I. Let G = (4¥,6;7) be a pregraph where X, 
ô and 7 are, respectively, the set of all semiedges, the permutation 
determined by edges and the partition on X. 

Two regraphs Gi = (4, ĉi; m1) and G5 = (X2, ô2; 75) are said to 


XIV.1 Semiedge automorphisms on a graph 327 


be seme-isomorphic if there is a bijection 7 : YA; — A» such that 


X —————À A» 
"| |» (14.1) 
p) ———9 QA) 


are commutative for y = ô and m where m, is induced from 7 on 
^1(41). The bijection 7 is called a semi-automorphism between Gi 
and Go. 


Example 14.1 Given two pregraphs G = (4, 61; 7) where 
8 


= S210, (D) à = [ [(@:(0), 2:1); 
m= {X)1<i<8} i 
with 
Xi = (1(0), 26(0), ze(1)}, X2 = {x1(1), v2 (0), z3(0)}, 
Xs = (z2(1), 25(0), 77(0)}, X4 = {x4(0), v5 (1); v7 (1), va(0) 
Xs = {x4(1), v5(1), x8(1)} 
and H = (YV, 69; 72) where 


y= Suo (1)}, d = | [(v0) (2); 
micis) i 
with 
Yı = {y4(0), ye(0), y7(O)}, Yo = {y5(0), yo(1), y7(1), ys(0)}, 
Ys = {y3(0), ys(1), ys(1)}, Ya = {ye(0), ys(1), ya) J, 
Ys = {y1(0), (1), y2(1)}- 
Let T: A — Y be a bijection with (14.1) commutative for y; = ôi, 
i = ] and 2, as 


X1 X2. X3 X4 UH X6 XT Tg 
doy2 Ó2U4 Y3 Ys Yo Yı Y7 Ys 


328 Chapter XIV Super Maps of a Graph 


Because of 

TX; = {7x1(0), TX6(0), rzo(1)) = {y2(1), (0), y1(1)} = YS, 
), ya(1), y3(1)} = Y 
,7£5(0), 7£7(0)} = (y4(0), ye(0), v; (0)) = Yi, 


9 

8 

N 

Qut, 

© 

n>" 

4 

8 

w 

Qm. 

=) 

io od 
ww —— 

| 
mma m 

ea 

bo 

aet mss 

c 


4 
A 
| 
4 
—— 
8 
A 
c 
8 
A 
m 
nex” 
8 
p 
LL 
8 
a 
oO 
— 
—— 


= {yo(1), yr (1), ys (0), ys(O)} = Y2, 
TX5 = (7za(1), rz4(1), Teg(1)} = {y3(0), ys(1), us(1)) = Ys; 


we have 74,71 = TT, i.e., (14.1) is commutative for yi = 7;, i = 1 and 
2. Therefore, 7 is a semi-isomorphism between G and H. 


Lemma 14.1 If two pregraphs G and H are semi-isomorphic, 
then they have the same number of connected components provided 
omission of isolated vertex. 


Proof By contradiction. Suppose G and H are semi-isomorphic 
with a semi-isomorphism 7 : G — H but G = Gi + Ga with two 
components: G4, and Gj and H, a component itself. From the the 
commutativity of (14.1), H has two components as well. This contra- 
dicts to the assumption that H is a component itself. [] 


If G — H, then a semi-isomorphism between G and H is called 
a semi-automorphism of G. Lemma 14.1 enables us to discuss semi- 
automorphism of only a graph instead of a pregraph without loss gen- 
erality. 

Lemma 14.1 allows us to consider only graphs instead of pre- 
graphs for semi-automorphisms. 

Moreover for the sake of brevity, only graphs of order greater 
than 4 are considered as the general case in what follows. 


Theorem 14.1 The set of all semi-automorphisms of a graph 
forms a group. 


Proof Because of all semi-automorphisms as permutations act- 
ing on the set of semi-edges, the commutativity leads to the closedness 
in the set of all semi-automorphisms under composition with the as- 


XIV.2 Automorphisms on a graph 329 


sociate law. Moreover, easy to check that the identity permutation is 
a semi-automorphisms and the inverse of a semi-automorphism is still 
a semi-automorphism. This theorem holds. [] 


This group in Theorem 14.1 is called the semi-automorphism 
group of the graph. 


Example 14.2 In Example 14.1, the pregraph G = (X, 61; 77) 
is a graph. It is easily checked that 


_ ( T1 T2 T3 X4 X5 Te T7 Tg | _ 
= E fo U3 L4 Ly Ó]Xg T7 2» = (xe(0), xe(1)) 


is a semi-automorphism on G. It can also be checked that the semi- 
automorphism group of G is Autyi(G) = {7;|0 € i < 11} where 


To = 1, the identity; 


Ti = (£5, £7); 

que (£4, £8); 

T3 = (£5, £7) (24, £8); 

Ta = (£2, £3) (£4, 125) (zv, 01238); 
T5 = (£2, £3) (25, $118) (£7, 0174); 


XIV.2 Automorphisms on a graph 


Now, let us be back to the usual form of a graph G = (V, E) 
where V and E are, respectively, the vertex and edge sets. In fact, if 
X; as described in 8XIV.1 is denoted by v;, then V = (vo|i = 0.1.2. --) 
and E = (zj|j = 0,1,2,---}. 

An edge-isomorphism of two pregraphs G; = (Vj, Ei), i = 1 and 


330 Chapter XIV Super Maps of a Graph 


2, is defined as a bijection 7: E, —> E» with diagram 


FA —— , E» 
"| |” (14.2) 
V —“— VW 


commutative where 7;, i = 1 and 2, are seen a mapping 2^ — V;. 
When G = G4 = G^», an edge-isomorphism between G and G2 
becomes an edge-automorphism on G. 


Lemma 14.2 If two pregraphs G and H are edge-isomorphic, 
then they have the same number of components provided omission of 
isolated vertex. 


Proof Similar to the proof of Lemma 14.1. [] 


This lemma enables us to discuss only graphs instead of pre- 
graphs for edge-isomorphisms or edge-automorphisms. 


Theorem 14.2 All edge-automorphisms of a graph G forms a 
group, denoted by Autee(G). 


Proof Similar to the proof of Theorem 14.1. [] 


Example 14.3 The graph G In Example 14.2 has its Aut..(G) = 
{7;|0 € i < 5) where 


To = 1, the identity; 


An isomorphism, or in the sense above vertez-isomorphism, be- 
tween two pregraphs G; = (Vj, Ei), i = 1 and 2, is defined as a bijection 


XIV.3 Relationships 331 


T: Vj — V which satisfies that the diagram 


VW ———3À V2 


«| e (14.3) 


Téi 


BiG Wx Vi ——— eV Ve 


is commutative where é;(v;) = Ein, for vj € Vi, i = 1 and 2. 
When G = Gi = Go, a isomorphism between G4 and G^ is called 
an automorphism of G. 


Lemma 14.3 If two pregraphs G and H are isomorphic, then 
they have the same number of components. 


Proof Similar to the proof of Lemma 14.2. [] 


This lemma enables us to discuss only graphs instead of pre- 
graphs for isomorphisms or automorphisms. 


Theorem 14.3 all automorphisms of a graph G form a group, 
denoted by Aut(G). 


Proof Similar to the proof of Theorem 14.2. [] 


The group mentioned in this theorem is called the automorphism 
group of G. 


Example 14.4 The graph G In Example 14.2 has its Aut(G) = 
{7;|0 € i < 1) where 
i To = 1, the identity; 
Tı = (X3, X5). 


Because of no influence on the automorphism group of a graph when 
deleting loops, or replacing multiedge by a single edge, 7;, ? — 1,2 and 
3 in Example 14.3 are to the identity 7) and 74 and 75 in Example 14.3 
to 7, here. 


332 Chapter XIV Super Maps of a Graph 


XIV.3 Relationships 


Fundamental relationships among those groups mentioned in the 
last section are then explained for the coming usages. 


Theorem 14.4 Auty:(G) ~ Aute(G) if, and only if, G is loop- 
less. 


Proof Necessity. By contradiction. Suppose Autyr(G) ~ Autee(G) 
with an edge-automorphism 7 but G has a loop denoted by z — 
(z(0), z(1)). Assume 7(I) = l without loss of generality. However, 
both the semi-automorphisms: 7; and 7» corresponding to 7 are found 


as 
T(x), when x Z z; 
T(z) = 4 z(0), when x = z(0); 
1), when x = z(1) 


( 
and 
T(x), when x Z z; 
( 


0), when x = z(1); 
), when x = z(0). 
This implies Autyr(G) 2^ Autee(G), a contradiction. 
Sufficiency. Because of no loop in G, the symmetry between two 
ends of a link leads to Autyr(G) ~ Autec(G). E 


mlr)= 4 z 


From the proof of Theorem 14.4, the following corollary can be 
done. 
Corollary 14.1 Let l be the number of loops in G, then 
Autne(G) ~ S. x Autee(G) 
where Sə is the symmetric group of order 2. 
Proof Because of exact two semi-automorphisms deduced from 


an edge-automorphism and a loop, the conclusion is done. L 


From this corollary, we can soon find 
autne(G) = 2! x autelG). (14.4) 


XIV.3 Relationships 333 


Because of no contribution of a loop to the automorphism group 
of G, the graph G has its automorphism group Aut(G) always for that 
obtained by deleting all loops on G. 

Theorem 14.5 Aute.(G) ~ Aut(G) if, and only if, G is simple. 

Proof Because of no contribution of either loops or multiedges 


to the automorphism group Aut(G), the theorem holds. [] 


In virtue of the proof of Theorem 12.5, a graph with multi-edges 
G has its automorphism group Aut(G) always for its underlying simple 
graph, i.e., one obtained by substituting a link for each multi-edge on 


G. 


Lemma 14.4 Let G be a graph with i-edges of number nj, 
i > 2. Then, its edge-automorphism group 


Autee(G) = 5 m;S; x Aut(G) 


i»2 
where S; is the symmetric group of order i, i > 2. 


Proof In virtue of S,, as the edge-automorphism group of link 
bundle Pm of size m, m > 2, the lemma is done. [] 


On the basis of Lemma 14.4, we can obtain 


Corollary 14.2 Let l and m; be, respectively, the number of 
loops and 7-edges, | > 1,2 > 2 in G, then 


autne(G) = 2'nmeaut(G) (14.5) 


where 


Nme = Mne(G) = Y ilm; 


i>1 
which is called the multiplicity of G. 


Proof By considering Corollary 14.1, the conclusion is done. O 


334 Chapter XIV Super Maps of a Graph 


XIV.4 Nonisomorphic super maps 


For map M = (Xas, P), its automorphisms are discussed with 
asymmetrization in Chapter VIII. Let M(G) be the set of all noniso- 
morphic maps with underlying graph G. 


Lemma 14.5 For an automorphism Ç on map M = (X45(X), P), 
we have exhaustively 


lax) € Autne(G) and Caļsx) € Autur(G) 
where G = G(M), the under graph of M, and 6(X) = X + 6X. 


Proof Because Xal X) = (X + 6X) + (aX + aó X) = 6(X) + 
aó(X), by Conjugate Axiom each Ç € Aut(M) has exhaustively two 
possibilities: C|5.x) € Autye(G) and Ga|;(x) € Autyr(G). O 


On the basis of Lemma 14.5, we can find 


Theorem 14.6 Let £,(G) be the set of all embeddings of a 
graph G on a surface of genus g(orientable or nonorientable), then the 
number of nonisomorphic maps in £,(G) is 


1 
= — X 14. 


where ®(7) = (M € €,(G)|7(M) = M or ra(M) = M}. 


Proof Suppose X1, X», ---, Xm are all the equivalent classes of 
X = €,(G) under the group Autye(G) x (a), then m = m,(G). Let 


S(x) = {T € Autye(G) x (a)| T(x) = x} 
be the stabilizer at z, a subgroup of Autyr(G) x (a). Because 
|Autar(G) x (o)] = ISG)I [Xi], 


x; E X;,1=1,2,---,m, we have 


m|Autur(G) x (a)] = 25 ISI Xi]. (1) 


XIV.4 Nonisomorphic super maps 330 


By observing |S(x;)| independent of the choice of x; in the class X;, 
the right hand side of (1) is 


2/1525; 2 1 


rcx LEX TES (a 


E Y 2,1 (2) 


7€Autne(G)x (a) z—r(x) 


= P5, Ie». 


T€ Autyrc(G)x (o) 


From (1) and (2), the theorem can be soon derived. O 


The theorem above shows how to find nonisomorphic super maps 
of a graph when the semi-automorphism group of the graph is known. 


Theorem 14.7 Let G be a graph with / loops and m; multi- 
edges of multiplier i and €,(G), the set of all embeddings of G on a 
surface of genus g(orientable or nonorientable), then the number of 
nonisomorphic maps in €,(G) is 


1 
m,(G) = Uns) P |®(r)| (14.7) 
qn Uthf 


where ®(7) = (M € £,(G)|T(M) = M or To(M) = M} and nme is 
the multiplicity of G. 
Proof On the basis of Theorem 14.6, the conclusion is soon de- 


rived from Corollary 14.2. [] 


Corollary 14.3 Let G be a simple graph. Then, the number 
of nonisomorphic maps in £,(G) is 


my(G) = zr PN le) (14.8 


where ®(7) = (M € €,(G)|7(M) = M or rTa(M) = M} and nme is 
the multiplicity of G. 


336 Chapter XIV Super Maps of a Graph 


Proof A direct result of Theorem 14.7 via considering G with 
neither loop nor multi-edge. [] 


XIV.5 Via rooted super maps 


Another approach for determining nonisomorphic super maps of 
a graph is via rooted ones whenever its distinct embeddings are known. 


Theorem 14.8 Fora graph G, let R,(G) and €,(G) be, respec- 
tively, the sets of all nonisomorphic rooted super maps and all distinct 
embeddings of G with size e(G) on a surface of genus g(orientable or 
nonorientable). Then, 

2e(G) 


mRA(G)| ES TOL (14.9) 


Proof Let M,(G) be the set of all nonisomorphic super maps 
of G. By (11.3), we have 


RO=- D A 


MEM, (G) 





i.e., 


4e(G) 2x autne(G) 


2 x autne(G) MEM, (GC) aut(M) 


By considering that 2 x autn(G) = |Autr(G) x (a)| is 
|(Autne(G) x (0))|u] x [Autur(G) x (a) | 


and (Autne(G) x (0))|u = Aut(M) from Lemma 14.5, we have |R,(G)| 
is 


- 4O — : s 
|Autgr(G) x mi, 2 ^ tar(G) x (a) (M)| 


|. 2e(G) 
m autyr(G) |£4(G)]. 


XIV.5 Via rooted super maps 337 
This is (14.9). O 


This theorem enables us to determine all the super rooted maps 
of a graph when the semi-automorphism group of the graph is known. 


Theorem 14.9 For a graph G with l, | > 1, loops and m; 
multi-edges of multiplier 7, i > 2, let R,(G) and £,(G) be, respec- 
tively, the sets of all nonisomorphic rooted super maps and all distinct 
embeddings of G with size e(G) on a surface of genus g(orientable or 
nonorientable). Then, 

«(G) 

2! - 1n neaut(G) 


where Nme is the multiplicity of G. 


mRA(G)| — £5 (G)] (14.10) 


Proof A direct result of Theorem 14.8 from Lemma 12.2. O 


Corollary 14.4 For a simple graph G, let R,(G) and €,(G) 
be, respectively, the sets of all nonisomorphic rooted super maps and 
all distinct embeddings of G with size e(G) on a surface of genus 
g(orientable or nonorientable). Then, 

2e(G) 


RCI = ESI. (14.11) 





Proof The case of | = 0 and m; = 0, i > 2, of Theorem 14.9. O 


Corollary 14.5 The number of rooted super maps of a simple 
graph G with n; vertices of valency 2, i > 1, on orientable surfaces is 


m [ke-n (14.12) 


i22 





where e is the size of G. 
Proof Because of the number of distinct embeddings on ori- 


entable surfaces 
S > é(G) = [[(G - 09" (14.13) 


g20 i22 


338 Chapter XIV Super Maps of a Graph 


known, Corollary 14.4 leads to the conclusion. [] 
Corollary 14.6 The number of rooted super maps of bouquet 
Bm, m > 1, on orientable surfaces is 


(2m)! 
2m! 





(14.14) 


Proof Because of the number of all distinct embeddings of B,, 
on orientable surfaces (2m ,)! and the order of its semi-automorphism 
group 2"! known, the conclusion is deduced from Theorem 14.8. CO 


In virtue of petal bundles all super maps of bouquets, (14.14) is 
in coincidence with (9.9). 


The number of nonisomorphic super maps of a graph can also be 
derived from rooted ones. 


Theorem 14.10 Fora given graph G, let £&(G) be the set of all 
its nonequivalent embeddings with automorphism group order k. Then 
we have the number of all nonisomorphic unrooted supper maps of G 


is 
1 


nu(G) = TORTEK > ie (14.15) 


i|4e 
1<i<de 


where € = €(G) is the size of G and I(G) is the number of loops in G. 
Proof On the basis of Theorem 14.8, we have 


2e(G) 


[Ri(G)| = TIA) 


|E:(G)| 


where R;(G) is rooted super maps of G with automorphism group 
order i. By Theorem 14.4 and Corollary 14.1, 


«(G) 


I;(G)| = AC laut(G) 


iei (G)]. 


Because of 4€(G')/i rooted maps produced by an unrooted map 


XIV.5 Via rooted super maps 339 


in R;(G) as known in the proof of Theorem 11.1, we have 





] a e(G) 
acc; iO = iG) MO iau (aj O 
eG). 


~ 2/6 aut(G) 
Overall possible ;|4e(G) is the conclusion of the theorem. O 


Further, this theorem can be generalized for any types a set of 
graphs. 


Theorem 14.11 For a set of graphs G, the number of noniso- 
morphic unrooted super maps of all graphs in G is 


nu(9) = Y^ sert *. d&(G)l. (14.16) 


GEG i|4e(G) 
1<i<4e(G) 


Proof From Theorem 14.10 overall G € G, the theorem is soon 
done. Oo 


For a given genus g of an orientable or nonorientable surface, let 
E.(G;g) be the set of all nonequivalent embeddings of a graph G on 
the surface with automorphism group order k. 


Theorem 14.12 For a given genus g of an orientable or nonori- 
entable surface, the number of all nonisomorphic unrooted supper maps 
of a G on the surface is 


w(G: 9) = sperma È EG) (14.17) 


i|4e 
1<i<4e 


where € = €(G) is the size of G. 


Proof By classification of maps and embeddings as well with 
genus, from Theorem 14.11 the theorem is done. [] 


Furthermore, this theorem can also generalized for any types a 
set of graphs. 


340 Chapter XIV Super Maps of a Graph 


Theorem 14.13 For a given genus g of an orientable or nonori- 
entable surface, the number of nonisomorphic unrooted super maps of 
all graphs in a set of graphs Gp with a given property P is 


1 mn 
mel) = gy 2; il£;(G; g)]. (14.18) 


1<i<4e(G) 


Proof A particular case of Theorem 14.11. [] 


Activities on Chapter XIV 


XIV.6 Observations 


Let B, be the bouquet of n loops for n > 1. 
O14.1 Find the semi-automorphism group of B, for n > 1. 
O14.2 Find the edge-automorphism group of B, for n > 1. 


O14.3 Find the automorphism group of D, for n > 1. 


Let D,, be the dipole which is of order two and size m for m > 1 
without loop. 


O14.4 Show Autne(Dm) ~ Autee(Dm) ~ Sm where Sm is the 
symmetric group of order m for m > 1. 


O14.5 Find a condition for Aut(D,,) ~ Autyr(D,,), m > 1. 
O14.6 Determine the number of super maps of K4. 
O14.7 Determine the number of rooted super maps of K5. 


014.8 Determine the number of super maps of K3 3 rooted and 
unrooted. 


014.9 Determine the number of super maps of K4, rooted and 
unrooted. 


O14.10 Suppose A;,, be the number of distinct embeddings of 
Bı, | > 1, on the orientable surface of genus p > 0, determine the 
number of rooted and unrooted super maps of Bı on the orientable 


342 Activities on Chapter XIV 


surface of genus p > 0. 


XIV.7 Exercises 


E14.1 Show that the number of rooted super maps of Kn, the 
complete graph of order n, n > 4, is 


(n — 2)*-l. 


E14.2 Show that the number of rooted super maps of Km,n, 
the complete bipartite graph of order m+n, m,n > 3, is 


o= apes. 


E14.3 Let 7, be the set of non-isomorphic trees of order m, 
n 2 2. Show that 


I6-D" — Qn-1) 
y^ I (2n = 1) 


ed |Aut(T)| | n!(n4- 1)" 


the number of rooted plane trees of order n — 21 n;, Where n; is the 
number of vertices of valency i, i > 1. 


E14.4 Determine the distribution of rooted super maps of the 
complete graph Kn, n > 4, by automorphism group orders. 


E14.5 Determine the distribution of rooted super maps of the 
complete bipartite graphs K(m, n), m,n > 3, by automorphism group 
orders. 


E14.6 Determine the distribution of rooted super maps of su- 
per cube Q,, n > 4, by automorphism group orders. 


E14.7 Determine the distribution of rooted super maps of the 
complete tripartite graph Kọ, m,n, l,m,n > 2, by automorphism 
group orders. 


E14.8 Determine the distribution of rooted super maps of the 
complete equi-bipartite Ky», n > 3, by automorphism group orders. 


XIV.8 Reseaches 343 


E14.9 Determine the distribution of rooted super maps of the 
complete equi-tripartite /€(n,n,n), n > 2, by automorphism group 
orders. 


E14.10 Determine the distribution of rooted super maps of the 
complete quadpartite graph Kk lmn, k,l, m,n > 2, by automorphism 
group orders. 


E14.11 Determine the distribution of rooted super maps of the 
complete equi-quadpartite graph Knnnn, n > 2, by automorphism 
group orders. 


E14.12 Determine the distribution of rooted super maps of su- 
per wheel Wn, n > 4, by automorphism group orders. 


XIV.8 Researches 


R14.1 For given integer n > 1, determine the distribution of 
outer planar graphs of order n by the order of their semi-automorphism 
groups. 


R14.2 For given integer n > 1, determine the distribution of 
Eulerian planar graphs of order n by the order of their semi-automorphism 
groups. 


R14.3 For given integer n > 1, determine the distribution of 
general planar graphs of order n by the order of their semi-automorphism 
groups. 


R14.4 For given integer n > 1, determine the distribution of 
nonseparable planar graphs of order n by the order of their semi- 
automorphism groups. 


R14.5 For given integer n > 1, determine the distribution of 
cubic planar graphs of order n by the order of their semi-automorphism 
groups. 


R14.6 For given integer n > 1, determine the distribution 


344 Activities on Chapter XIV 


of 4-regular planar graphs of order n by the order of their semi- 
automorphism groups. 


R14.7 For given integer n > 1, determine the distribution of 
nonseparable graphs of order n by the order of their semi-automorphism 
groups. 


R14.8 For given integer n > 1, determine the distribution of 
general graphs of order n by the order of their semi-automorphism 
groups. 


R14.9 For given integer n > 1, determine the distribution of 
Elerian graphs of order n by the order of their semi-automorphism 
groups. 


R14.10 For given integer n > 1, determine the distribution of 
general graphs of order n by the order of their semi-automorphism 
groups. 


R14.11 For given integer n > 1, determine the distribution 
of cubic graphs of order n by the order of their semi-automorphism 
groups. 


R14.12 For given integer n > 1, determine the distribution of 
4-regular graphs of order n by the order of their semi-automorphism 
groups. 


R14.13 For given integer n > 1, determine the distribution of 
5-regular graphs of order n by the order of their semi-automorphism 
groups. 


Chapter XV 


Equations with Partitions 


e The meson functional is used for describing equations discovered 
from census of maps via vertex, or face, partition as parameters. 


e Functional equations are extracted form the census of general maps 
and nonseparable maps with the root-vertex valency and the vertex 
partition vector on the sphere. 


e Dy observing maps without cut-edge on general surfaces, a func- 
tional equation has also be found with vertex partition. 


e Functional equations satisfied by the vertex partition functions of 
Eulerian maps on the sphere and general surfaces are derived from 
suitable decompositions of related sets of maps. 


e All these equations can be shown to be well definedness. However, 
they are not yet solved in any way. 


XV.1 The meson functional 


Let fy) E Ry}, where y= (yi, TAT J be a function, and 
V(f,y) > 0, i—1,2,--. 
A transformation is established as 1 y! "TP 


y 
convinced y? = 1 > yo. 


346 Chapter XV Equations with Partitions 


Since J is a function from the function space F with basis 


y 
{1,y,y?, =+} to the vector space V with basis {yo, y1, ys, +-}, it is 
called the meson functional. i.e., the Blissard operator. For any 


Vi = > ay’, L= 1,2, 


j20 


it is easy to check that 


Je + v2) = $ (ay + az) E 


j20 
= ) Q1jYj + 1 02,8; 
j20 j20 


= fut fu 
y y 


Hence, the meson functional is linear. 


zii 
The inverse of the meson functional J is denoted by J Dye 
y y 


E 
y), i = 1,2,--., convinced J yo = 1, or simply yọ = 1. However, 1 
y 
is seen as a vector in V. 
Two linear operators called /eft an dright projection, denoted 


by, respectively, S, and Ry, are defined in the space V as: let v = 
2? 50 474; € V, then 


Syv = SoU ES Lig qus 


j20 


1 
Ryu = > j 4-197: 


j21 


(15.1) 


In other words, if y; is considered as the vector with all entries 0 but 
only the i-th 1, then the matrices corresponding to S, and R, are, 
respectively, as 


XV.1 The meson functional 347 


where 
te when j = 1; 
Le 


(j —1)1)1, when j > 2 
for 1; being the infinite vector of all entries 0 but only the (j — 1)-st 
1 and 


HL deos) (15.2b) 
where 
0, when 7 = 1; 
T. = . 
z jor when j > 2 


for the super index ‘T’ as the transpose. 
Easy to check that 


I 0° = 
n= (1), m~( 


where / is the identity. 


je o 


À 2), (15.3) 


-1 
Theorem 15.1 For v = v(yos yi,:::) € V, let f(y) = J v, 
y 


i [ss [roo f me so 


Proof M equating the coefficients of terms in same type on the 
two sides, the theorem is done. [] 


If f(x,y) is a function with two types of unknowns, and assume 
f(x, y) € V(z, y), a bilinear space, then it is easily checked that 


[ [ 1e» -| [| fe» (15.5) 


Denoted by F(z,y) the function in (15.5). Conversely, for F(z, y) 
€ 'R(x, y), we have 


fay =f | Fe) (15.6) 


DU dy 


because of interchangeable between J and J l 
z y 


348 Chapter XV Equations with Partitions 


Let f(z) € R{z}. The following two operators on f as 


r—y 
and 
any f = ——— (15.8) 


are, respectively, called the (x, y)- difference and (x, y)-difference of f 
with respect to z. 
Lemma 15.1 For any function f(z) € R{z}, let f = f(z), 
then 
Oba) eeu. (15.9) 
Proof Because of the linearity of the two operators O; , and 6; ,, 
this enables us only to discuss f(z) = z", n » 0. Then, it is seen 


On y2f = Ox eee v 





7 yx” n+l _ cyt! 
= mmm 
p" y” 
=> LY 
t—y 
= zyó,.yz^ 
= zyó,,f. 
This is what we want to prove. [] 
Theorem 15.2 For any f € R{z}, we have 
^82. ylz f) — 0b pl2f) = xa y epl’). (15.10) 


Proof From (15.7) and (15.8), the left hand side of (15.10) is 
wy" (^ f (x7) — v fü) - ay PG) - Ea) 
zi y? 
PREPO -PP 
= 
From (15.7), this is the right hand side of (15.10). O 


XV.1 The meson functional 349 


For a set of maps .A, let 


falz, y) = b» gm UD yn) (15.11) 
AEA 
where m(A) and n(A) are, respectively, the invariant parameter and 
vector on A. Let F4(x,y) be such a function of two unknowns that 


falz, y) = ] Pe» (15.12) 


The powers of x and y in FA4(x,y) are, respectively, called the 
first parameter and the second parameter. 


Theorem 15.3 Let S and 7 be two sets of maps. If there is a 
mapping A(T) = (51, Sn --, Smry4if such that 5; and {i m(T) +2— 
i} are with a 1-1 correspondence from 7 to S for any T € 7, where i 
and m(T) + 2 — i are the contributions to, respectively, the first and 


the second parameters, i = 1,2, ---,m(T) + 1, with the condition as 
S — S(T), 
TET 
then 
Fs(x,y) = £yôsy(z fr) (15.13) 


where fr = fr(z) = fr(z,y). 


Proof From the definition of A, we have 


(T)+1 
Fs(z, y) = ys > gy) He a n) 
TeT i=l 


m(T)+1 ,m(I)41 
x yY n 
= 2y 2: — 
TET y 
= usu fr). 
This is (15.13). [] 


Theorem 15.4 Let Sand 7 be two sets of maps. If there exists 
a mapping A(T) = (51,55, --, S4, 71) such that S; and (i, m(T) - i) 


350 Chapter XV Equations with Partitions 


are in a 1-1 correspondence for T € T, where i and m(T) + 2 — i are 
the contributions to, respectively, the first and the second parameters, 
i —1,2,---,m(T) — 1, with the condition 


S=) (T) 
TET 


then 
Fs(z, y) = Os y (fr) (15.14) 


where fr = fr(z) = fr(z,y). 


Proof From the definition of A, we have 


m(T)—1 
12-607) E > >, o 
TeT i=1 
op me) 
yx" ry" n 
-ayy = 
TET 
This is (15.14). D 


XV.2 General maps on the sphere 


A map is said to be general if both loops and multi-edges are 
allowed. Of course, the vertex map ? is also treated as degenerate. Let 
M p be the set of all rooted general planar maps. For any M € Msep; 
let a = e,(M) be the root-edge. Then, Mp can be divided into three 
classes: JM ass JV au and MW us 0:8 


IVA ao em dV us T VE T Mi (15.15) 
such that Mp, consists of a single map Ù, 
Ma, = (M|VM € Msep, a is a loop]. 
Of course, 


Ma, = UMIVM € M, ais a link) 


XV.2 General maps on the sphere 351 


in its own right. 


Lemma 15.2 Let M; 
have 


= (M — a|VM € Mep}. Then, we 


(gep)1 geP1 


Nb), = Meer OM ge (15.16) 
where © is the 1-production as defined in $2.1. 


Proof For a map M € Msp); because there is a map Me 
Mis) such that M = M — à,à = e,(M), the root-edge of M, by 
considering the root-edge à as a loop we see that M = M,+Mb, pro- 
vided Mı N Mə = o, the common root-vertex of M and M. Since Mi 
and M» are allowed to be any maps in M 
0, this implies that M € M,.. © M. 

Conversely, for any M € Mj © M, since M = Mj- M», Mi, 
M» € Mp, we may always construct a map M by adding a loop à 
at the common vertex of M; and M» as the root-edge of M such that 
Mı and M^» are in different domains of the loop. Of course, M isa 

general map. Because the root-edge of M isa loop added, M €. M 
However, it is easily seen that M — M — à. Therefore, M € Migs, 
In consequence, the lemma is proved. D 


sp Including the vertex map 


8gep1 * 


For Msp, because the root-edges are all links we consider the 
set M (gp) = {M ea|VM € Mas, a = e (M), the root-edge as usual. 
The smallest map in Mep, is the link map L = (Kr, (r)(agr)) and 
it is seen that Lea = 2. Thus, 0 € Miep, For any M € My, 
because the root-edge of M is not a loop we know that M ea € 
Nas. Conversely, for any M € Mep we may always construct a map 
Me Msp by splitting the root-vertex of M into two vertices with 
a new edge a as the root-edge connecting them. This implies that 
Mea=Me M gep),- Therefore we have 


M eep)y — IM: (15.17) 


352 Chapter XV Equations with Partitions 


Lemma 15.3 For Mep, we have 
Men = >, {ViM|0<i<m(M)} (15.18) 
MeMgep 


where m(M) is the valency of the root-vertex of M and V; is the 
operator defined in 87.1. 


Proof For any M € Msep, because the root-edge a is a link, we 
may assume a = (01, 02) such that 


01 = (r, S) and o» = (apr, T). 


Let M be the map obtained by contracting the root-edge a into a 
vertex 6 = (T, S) as the root-vertex of M. It is easily checked that 
M € Ms from (15.17) and that 


M = Vis M, 0 « |S| « m(M) 


where m(M) = |S| + |T|, and |Z|, Z = S or T, stands for the cardi- 
nality of Z. That implies M is a member of the set on the right hand 
side of (15.18). 

Conversely, for any M in the set on the right, because there 
exist a map Me M, and an integer 1,0 < i < m(M), such that 
M = V,M, we may soon find that M € Mep, by considering that the 
root-edge of M is always a link and that M € Mep as well. Thus, 
M EM as: 

Therefore the lemma follows. [] 


Bep 


From the two Lemmas above we are now allowed to determine 


the contributions of M,.,,,7 = 0, 1, 2, to the enufunction 


mana = >, anp (15.19) 
MEM gep 
where n(M) = (nı( M), na( M), --- , nj(M),---), ni(m) is the number 
of vertices of valency i in M and m(M), the valency of the root-vertex 
of M. 


XV.3 Nonseparable maps on the sphere 353 


First, since 9 has neither non-rooted vertex nor edge we soon see 
that 


IMgep, = L (15.20) 
Then, by Lemma 15.3, 
IMgep, = Tg. (15.21) 


where g = gu, (x, y) defined by (15.19). 
Further, from Lemma 15.3 


m(M)-1 
OM ep, =| ` pya Ma ya 
Y MEMgep i=1 
By Theorem 15.3, 
ÜMgep, — s | (5.0). (15.22) 
y 


Theorem 15.5 The enufunction g defined by (15..5) satisfies 
the following functional equation: 


g—-1-4zgLz J (uos (2g) ): (15.23) 
y 


Proof According to (15.15), from (15.20-22) the theorem is soon 
obtained. [] 


XV.3 Nonseparable maps on the sphere 


Let M,, be the set of all rooted nonseparable planar maps with 
the convention that the loop map L4 = (Kr, (r, o fir)) is included but 
the link map L = (Kr, (r)(ar)) is not for convenience. 

Then, M,, is divided into two parts Ms and M,,,,2.€., 


Mas = Masy + Mas; (15.24) 


such that M s consists of only the loop map Lı. 


354 Chapter XV Equations with Partitions 


Lemma 15.4 A map M € M,,, M z L4 if, and only if, its 
dual M* € Ms- 


Proof By contradiction. Assume M = (Xa B, P) € Mus, M z 
L; and its dual M* = (X54, PaB) Z Mns. Let 


v = (x, Paßz,---, Pal") 


be a cut-vertex of M*. Then we have a face f* = (x, Px,- --,P”x) 
on M* such that there exists an integer 7,1 < j € n, on f* satisfying 
Pix = (Paz for some i,1 € i € m, ie, v; = v5, = v*. However, 
f* is a vertex of M which has the face v* having the symmetry and 
hence f* is a cut-vertex of M. A contradiction to the assumption 
appears. The necessity is true. 

Conversely, from the duality the sufficiency is true as well. O 


For any M € M,,, let m(M) be the valency of the root-vertex 
and n(M) = (mi( M), na( M), ---), ni((:M) be the number of nonrooted 
vertices of valency 2,2 > 1. 

From the nonimputability, the root-edge a = (v1, vg) of any map 
M in M,,, is always a link. The map M e a obtained by contracting 
the root-edge a in M has the same number of faces as M does. 


Lemma 15.5 For any M € M,,, there is an integer k > 1 with 
k 

M ea — 3 Mi (15.25) 
i=1 


such that all M; are allowed to be any map in M,, and that M;, i = 
1,2,---,k, does not have the form (15.25) for k > 1. 


Proof In fact, from what were mentioned in 96.2, we see that 
k is the root-index of M and that all M;,1 < i € k, do not have the 
form (15.2) for k > 1. From the nonseparability of M, by considering 
that all vertices of M; except for the root-vertex are the same as those 
of M for i = 1,2,---,k, since M; does not have the form (15.2) for 


XV.3 Nonseparable maps on the sphere 355 


k > 1, the root-vertex is not a cut-vertex for i = 1,2,---,k. That 
implies all M;, 1 < i < k, are allowed to be any map in M,, including 
the loop map. The lemma follows. [] 


Now let us write 
k 
Mx = { X :Mi| VM; € Mus, 1 <i € k}, (15.26) 
i=1 


and 
M ns), = {M e a\VM € Masi}, (15.27) 


where a = e,, the root-edge of M. 


Lemma 15.6 For M,,,, we have 


Mors), = >> Mz; 
k>1 
My, = ME (15.28) 


where is the inner 1v-production. 


Proof By the definition of inner 1v-product, the last form of 
(15.28) is easily seen. 
From Lemma 15.5 we can find that 


Mins, = U M x. 
k>1 


Moreover, for any i, 7,2 Æ j, we always have 
Mi () Me= d. 
Therefore, the first form of (15.28) is true. O 


Based on the two lemmas above, we are allowed to evaluate the 
contributions of Mas and M,,, to the enufunction fm, of Mas with 
vertex partition, t.e., 


Telam >, ee. (15.29) 
M €.Mns 


356 Chapter XV Equations with Partitions 


where m(M) is the valency of root-vertex and 
n(M) = (m(M), ni(M), ---) 
with n;(M) being the number of nonroot-vertices of valency 7,7 > 1. 


Since M, consists of only the loop map, which has the root- 
vertex of valency 2 without nonrooted vertex, we have 


NEL (15.30) 
For M,,,, we have to evaluate the function 
f(z,2) = ` g 04) a aM) (15.31) 
MEM ys, 


where s(M) is the valency of the nonrooted vertex vg, incident with 
the root-edge e, of M. 

By considering that for M € M,,,, m(M ea) = (m(M) —1) + 
(s(M) — 1), we soon find 


M€ Miss, 


where s(M) is the contribution of the valency of the nonrooted end of 
the root-edge of M to the valency of the root-vertex of M — Mea, M € 


M. Because s(M) is allowed to be any number between 1 and 
m(M) — 1, from Lemma 15.2 we have 


m(M)-1 5 
od m(M Zi n( M 
jeje qp ox «UB Cy (2 yee 
k>1 \MeMns = 


By Theorem 15.4, 
TUB) = 2z > Ti 


k>1 
£20 p,2f 
i L- Ongar 
where f = f(u) = f, (u, y), and hence 


Ma, = [fen 


XV.4 Maps without cut-edge on surfaces 391 


YOryf 
=, /—— | (15.32) 


Theorem 15.6 The enufunction of M,, defined by (15.29) sat- 
isfies the following functional equation: 


nay 
matte | IL. 15.33 
f p ( ) 


Proof Since fuu, = fts, + fMns,» from (15.30) and (15.32) the 
theorem is obtained. [] 


XV.4 Maps without cut-edge on surfaces 


In this section, only maps without cut-edge(or, 2-connected)are 
considered. Let M be the set of all(including both orientable and 
nonorientable) rooted maps without cut-edge. Classify M into three 
classes as 


M = Mo 4 Mt Mo (15.34) 


where Mọ consists of only the vertex map 0, My, is of all with the 
root-edge self-loop and, of course, Mg is of all with the root-edge not 
self-loop. 


Lemma 15.7 The contribution of the set Mo to f = fm(z.y) 


J5 eL, (15.35) 


where fo = f(x, y). 
Proof Because of V neither cut-edge nor nonrooted vertex, m(0) = 
0 and n(2) = 0. Thus, the lemma is obtained. O 


In order to determine the enufunction of Mı, how to decompose 
Mı should first be considered. 


Lemma 15.8 For Mı, we have 


May =M, (15.36) 


358 Chapter XV Equations with Partitions 


where M = (M — a|VM € Mı}, a= Kr(M). 


Proof Because of Lı = (r, yr) € Mı, y = ag, we have Lı — a = 
0 € M. 

For any 5 € My), since there exists M € M such that 5 = 
M — a, by considering the root-edge of M not cut-edge, it is seen 
5 € M. Thus, M) € M. 

Conversely, for any M = (¥, 7) € M, a new edge a’ = Kr' is 
added to the root-vertex (r)7 for getting 5;, whose root-vertex is 

(r'r, Utt Le, am. ne MEE gos), 

where 0 <i < m(M) — 1. Because of S; — a’ = M, we have S; € Mı. 
Hence, M € Mq. [] 

From this lemma, it is seen that each map M = (X, J) in M 
not only produces S; € M,,0 € i € m(M) — 1 but also Sm € Mı 


nonisomorphic to them. Its root-vertex is (r^, (r).7, yr’). For M € M, 
let 


Lemma 15.9 The set Mı has a decomposition as 
M,= p» SM, (15.38) 
MEM 


where Sy is given from (15.37). 


Proof First, for M € Mı, because of M' = M — a € My), 
Lemma 15.8 enables us to have M' € M. Via (15.37), M € Sw is 
obtained. Thus, what on the left hand side of (7.4.5) is a subset of 
that on the right hand side. 

Conversely, for a map M on the left hand side of (15.38), because 
of the root-edge a self-loop,we have M € Mı. Thus, the set on the 
left hand side of (15.38) is a subset of that on the right hand side. O 


On the basis of this lemma, we have 


Lemma 15.10 For gı = gm: (x.y) = fn (22, y), we have 


Of 


ee (15.39) 


Rh sc e 


XV.4 Maps without cut-edge on surfaces 359 


where f = f, y). 
Proof From Lemma 15.9, 


= p» (m(M) + Yay. 


MEM 
By Lemma 9.10, we get 
Of 
— m2 ad 
This is the conclusion of the lemma. [] 


In what follows, Mə is considered. 


Lemma 15.11 For Mz, let Mi) = (M ea|VM € Mo}, then 
M (2) — M —23, (15.40) 


where Ù is the vertex map. 


Proof For any M € Mv), there is a map M’ € Mp such that 
M = M' ed'. Because of a’ neither cut-edge nor self-loop, M € M. 
And, sice the link map Lo = (Kr, (r)(yr)) € M», May CM — 9. 

Conversely, for any M = (X, J) € M —9, let U;,1 be obtained by 
splitting the root-vertex (r)5 of M with an additional edge a’ = Kr’ 
whose two ends are (7, r, --- , Jr) and 


(yr', girl e Vim 1 < i < m(M). 


Because of a’ not cut-edge, U; € Mə, 1 < i € m(M). And, because 
of M = U; e a Me Mo) Thus, M C Ü m M (o. L] 


For any M = (4,7) € M — 9, let 
Uy = (Uil x à € m(M)), (15.41) 
where U; is appeared in the proof of Lemma 15.11. 


Lemma 15.12 The set M? has the following decomposition: 
M:= M Um, (15.42) 


MEM-v 


360 Chapter XV Equations with Partitions 


where Um is given from (15.41). 


Proof First, for any M € Mo, from Lemma 15.5, M' = M èa € 
M — 2 and further M € Um. This implies that 


My = U Us. 


MEeEM-v 


Then, for any bMi, M» € M — 0, because of M, not isomorphic 
to M», 


Um, (um - 0. 
Thus, (15.42) is right. The lemma is obtained. O 


This lemma enables us to determine the contribution of Mə to 
Lemma 15.13 For f» = fm,(x, y), we have 


y 


where f = f(2) = fíu(z. y). 
Proof From Lemma 15.12, 


m(M) 
fz -| 2. ( `. qim 2i) yM), 


Y MeM-9 i=l 
By employing Theorem 15.4, (15.43) is obtained. E 


On the basis of those having been done, the main result can be 
deduced in what follows. 


Theorem 15.7 The functional equation about f 
o 
NET +=- s | ydes (15.44) 
ox y 


is well defined on the field (9t; z, y}. And, its solution is f = f(x) = 


XV.5 Eulerian maps on the sphere 361 


Proof The first statement can be proved in a usual way except 
for involving a certain complication. 

The second statement is derived from (15.34) in companion with 
(15.35), (15.39), and (15.43). [] 


XV.5 Eulerian maps on the sphere 


A map is called Eulerian if all the valencies of its vertices are 
even (or say, all vertices are even). Let U be the set of all the rooted 
planar Eulerian maps with the convention that the vertex map Ó is in 
U for convenience. 

Further, U is divided into 3 classes: Up, U and Wh, i.e., 


U = Ue o Us o Ub (15.45) 
such that Uy = (0), or simply write {J} = 0, and 


U, = {U|VU € U with a = e,(U) being a loop}. 


Lemma 15.14 Any Eulerian map (not necessarily planar) has 
no cut-edge. 


Proof By contradiction. Assume that a Eulerian map M has a 
cut-edge e = (u,v) such that M = Mi U e U M», Mı n M» = Ø, where 
Mı and M» are submaps of M with the property that Mi is incident 
tou and Ms, to v. From the Eulerianity of M, u and v are the unique 
odd vertex in Mı and Mo, respectively. This contradicts to that both 
Mı and M» are a submap of M because the number of odd vertices in 
a map is even. E 


Lemma 15.15 Let Uy) = {U — a|VU € Ui} where a = e, (U) 
is the root-edge. Then, we have 


Un; = Uu (15.46) 


where © is the 1v-production. 


362 Chapter XV Equations with Partitions 


Proof Because for U € M, the root-edge a is a loop, we see that 
U — a = U414-Us where U; and U; are in the inner and outer domain of 
a respectively. Of course, it can be checked that both U and U» are 
maps in U. Thus, the set on the left hand side of (15.46) is a subset 
of that on the right. 

On the other hand, for any U = U44-Uo, U1, Us € U, we may 
uniquely construct a map U’ by adding a loop at the common vertex 
of U; and Uy. The root-edge of U’ is chosen to be the loop such that 
U and Us are respectively in its inner and outer domains. It is easily 
checked that U’ is a Eulerian map and hence U’ € U. However, 
U = U'—a € Un. That implies the set on the right hand side of 
(15.46) is a subset of that on the left as well. O 


For any map U € 0f», we see that the root-edge a of U has to be 
a link. From Lemma 15.14, if U ea = U;+U> such that the root-vertex 
is the common vertex, then the valencies of the vertices in both U; and 
U» are odd. Further for any U € U, if U = U44-Us, then the valencies 
of the common vertex between U; and U»5 are both even as well. 


Lemma 15.16 Let Uo) = {U ea|VU € Lj where a is the 
root-edge of U. Then, we have 


Uo, =U — ù (15.47) 
where J? = U for simplicity. 


Proof From what has just been discussed, the set on the left 
hand side of (15.47) is a subset of that on the right. 

Conversely for any U € U — 9, we may always construct a map 
U' by splitting the root-vertex into o; and o9 with the new edge a’ = 
(01, 02) as the root-edge of U’ such that the valencies of o1 and os in 
U' are both even. However, U = U' ea' € Uy). That implies the set 
on the right hand side of (15.47) is a subset of that on the left. O 


For a map U € U — 0, assume the valency of the root-vertex 
ois 2k, k > l,without loss of generality. The map U’ obtained by 


XV.5 Eulerian maps on the sphere 363 


splitting the root-vertex into o; and o» with the new edge a’ = (01, 02) 
such that the valency p(o1; U') = 2i and hence p(o»; U") = 2k — 2i + 2 
is denoted by Upi, i = 1,2,---,k. 

From Lemma 15.14, we see that the procedure works and that 
all the resultant maps Uj, are also Eulerian maps. 


Lemma 15.17 For Lf», we have 


[U| = ` Ua] i=1,2,---,m(U)}| (15.48) 


Ucu-) 
where 2m(U) is the valency of the root-vertex of U. 


Proof First, we show that for any U € Uy, it appears in the set 
on the right hand side of (15.48) only once. Assume that a — (o, v) is 
the root-edge of U and that p(o) — 2s and p(v) — 2t. Let U' be the 
map obtained by contracting the root-edge a, i.e., U' — U ea. Then, 
there is the only possibility that U = U [2s] in the set on the right hand 
side of (15.48). 

Then, we show that for any map U in the set on the right hand 
side of (15.48), it appears also only once in U2. This is obvious from 
Lemma 15.16 because all elements are distinguished and they are all 
maps in U by considering the Eulerianity with the root-edges being 
links. [] 


In what follows, we see what kind of equation should be satis- 
fied by the enufunction u of rooted planar Eulerian maps with vertex 
partition. Write 

gy g ga (15.49) 
Ucu 
where 2m(U) is the valency of the root-vertex as mentioned above and 
n(U) = (na(U),---, na(U),---), na; (U) is the number of nonrooted 
vertices of valency 22, i > 1. 


Theorem 15.8 The function u defined in (15.5) satisfies the 


364 Chapter XV Equations with Partitions 


following functional equation: 
u — 1-4 z^) a J (ul Qu(/2))) (15.50) 
y 
where u(z) = u|;—; = u(z, y). 
Proof The contribution of Uo to u is 
ugo =1 (15.51) 


since 2m(0) = 0 and n(v) = 0. 
From Lemma 15.15, the contribution of Mı to u is 


uj = n b» gm yal) 
UU) 
= a^. (15.52) 
The contribution of U to u is denoted by us. Let 


ii(z) = ` AITE 
UEU 


where 2j(U) is the valency of the nonrooted end of the root-edge 
and n(U) = (fi3(U), R4(U), «fii (U),-- -), alU) is the number of 
vertices of valency 2i except for the two ends of the root-edge. It is 
easily seen that 

R(U) = n(U) — ea) 


where e»;(yj is the vector with all the components 0 except only for 
the j(U)-th which is 1. In addition, it can be verified that 


u» = fao (15.53) 


From Lemma 15.17, we have 


m(U) 
ü(z) E ` »» gi ,2m(U)-2i*2 ya. 


Ucu—o i=l 


XV.6 Eulerian maps on the surfaces 365 


By Theorem 15.3, 
ü(z) = a?220,» s (u( V/t)). 
Then by (15.53), we find that 


ug = x" J y ôy alul Vt). (15.54) 


From (15.51), (15.52) and (15.54), the theorem is obtained. UO 


XV.6 Eulerian maps on the surfaces 


Let Megu be the set of all orientable Euler rooted maps on sur- 
faces. Because of no cut-edge for any Eulerian map, Eulerian maps 
are classified into three classes as Mẹ ,, ME; and M, such that 
M, consists of only the vertex map J, Mi has all its maps with 
the root-edge self-loop and 


Ma = Meu Mia m Miny- (15.55) 


Naturally, MŽ; has all maps with the root-edge a link. 

The enufunction g = fmz,(2,y) is of the powers 2m and n = 
(n2, n4, -+ -) of, respectively, x and y as the valency of root-vertex and 
the the vertex partition operator. i 


Lemma 15.18 For MÌ „ we have 
go = 1, (15.56) 
where go = fmo (2^, y). 


Proof Because of V with neither root-edge nor nonrooted vertex, 
m(0) = 0 and n(0) = 0. The lemma is done. O 


In order to determine the enufunction of ML, a suitable decom- 
position of ML. should be first considered. 


Lemma 15.19 For ML, we have 


MJ = Meu; (15.57) 


366 Chapter XV Equations with Partitions 


where M, = (M — a|VM € Mi), a = Kr(M), the root-edge. 

Proof Because of Ly = (r, yr) € ML yo, L1—a = 9 € Meu 
is seen. 

For any S € Ma). since there is a map M € Megu such that 
S = M — a, from M as a Eulerian map, S € Megu is known. Thus, 
MS C Meu. 

Conversely, for any M = (X,J) € Meu, By adding a new 
root-edge a’ = Kr’ at the vertex (r)5 to get S;, whose root-vertex is 
(ries, ir, oyr!, fr, cs, g20007M), 0 € i € 2m(M) — 1. From 
S; — à! = M, S; € Ml, Thus, Mga € MË. Oo 


In the proof of this lemma, it is seen that each map M = (X, J) 
in Mg produces not only S; € Mt p0 < i € 2m(M) — 1, but also 
Som € ME, nonisomorphic to them. Its root-vertex is (r^, (r) z, yr’). 

For M € Megu, let 


Lemma 15.20 Set Mi, has the following decomposition: 
Meu = ` Sm, (15.59) 
M € Mngu 


where Sm is given from (15.58). 
Proof First, for any M € ML, from Lemma 15.19, M' = 
M — a € Megu, and hence M € Sw. Thus, 
Mia = U Sw. 


M € Mmgu 


Then, for any Mı, M» € Mey, because of nonisomorphic be- 
tween them, 


Sm ( Su = 9. 
Therefore, the conclusion of the lemma is true. [] 


On the basis of this lemma, the following lemma can be seen. 


XV.6 Eulerian maps on the surfaces 367 


Lemma 15.21 For gı = gMbu(z.y) = fm, (2°, y), we have 


ð 
gı = xg 2255) (15.60) 


where g — 9 Ms UU V) = anes y). 
Proof From (15.59), 


g= >) (2m(M) + 1)s" "Dy. 
M €.M gu 


By employing Lemma 9.10, (15.60) is obtained. [] 
In what follows, M3. is investigated. 


Lemma 15.22 For Mj,,, let MB = (M ea|VM € Mju}, 
then 
M, = Mea — 9, (15.61) 


where Ù is the vertex map. 


Proof Because of L4 € M 0 ¢ MC. (Q)). Then, M2, C 
M Eu E Ü. 

Conversely, for any M = (X,P) € Mru — 9, since Mo; = (A + 
Kra;,P3;) € M, where the two ends of as; = Kroj is obtained by 
splitting the root-vertex (r)p of M, i.e., 


(2j) Pa; — (raj; r, Pr, TT (p)? 


and | 
(9725) Po; — ("yro;, Ptr, ey (Pj umo, 


1<%t< m-—1. Because of M = Mo; è ao;, Mo; € MI s Thus, 


For any M = (X, J) € Mru — 2, let 
Muy = {Mp,;|1 € j € m(M)), (15.62) 


where Mo;, 1 € j € m(M) — 1, have appeared in the proof of Lemma 
15.22. 


368 Chapter XV Equations with Partitions 


Lemma 15.23 Set Mpy” has the following decomposition: 
Mau? = b Mm (15.63) 


MEMgu— Ü 


where M m is given from (15.62). 


Proof First, for any M € MS as because of M = Mea c 
Mru — v, Lemma 15.22 tells us that M € M(M). Thus, 


2 
Mew = |] Mu. 
M €.M gu Ü 


Then, for any Mi, Ma € Mguu — 9, because of nonisomorphic 
between M; and Mo, 


Mm (| Mas, # 9. 
This implies (15.63). O 
On the basis of this lemma, the following conclusion can be seen. 
Lemma 15.24 For go = fm (v, y), we have 
g2 = 2 f Pip (15.64) 
y 


where g = g(z) = fa@ou(2?, y)- 
Proof From Lemma 15.23, 


92 = ` Ji 


McMmgu-? j= 


m(M) 
x 2j, FM) (M)+2— ye, 
1 


By employing Theorem 15.3, 
E 2 
gs =T f» 042 2g( Vz). 
y 


This is the lemma. O 


Now, the main result of this section can be described. 


XV.6 Eulerian maps on the surfaces 369 


Theorem 15.9 The functional equation about g 


ont 98 _ 1 4 (1— z?)g pco (15.65) 


ðr? — / 
is well defined on the field L{R; x,y}. Further, its solution is g = 
9 Ms UU Y) E fT Mga (5 9): 
Proof The last conclusion is deduced from (15.55), in compan- 
ion with (15.56), (15.60), and (15.64). 
'The former conclusion is a result of the well definedness for the 


equation system obtained by equating the coefficients on the two sides 
of (15.65). [] 





Activities on Chapter XV 


XV.7 Observations 


O15.1 Let f = fuļx.y) be the vertex partition function of a 


set U of maps. Suppose U, = {U + a'| VU € U} where a’ = Kr’ is the 
root-edge of U' = U +a’. To evaluate f+ = fu, (x, y) from f. 


O15.2 Let A be the set of supermaps of K4, the complete graph 
of order 4. To evaluate f(z, y). 


O15.3 Let B; be the set of all bipartite rooted maps with the 
root-edge a cut-edge and B(1) = (B — a|VB € Bı}. Suppose the 
vertex partition function f of all bipartite rooted maps on surfaces is 
known. To evaluate fs, (x, y) from f. 


O15.4 Fora set of maps M, observe the relationship between 
vertex partition function f(x,y) and the enufunction fm(x, y) of two 
parameters: the valency of root-vertex and the size. 


O15.5 Fora set of maps M, observe the relationship between 
vertex partition function f(x,y) and the enufunction fm(x, y) of two 
parameters: the valency of root-vertex and the order. 


O15.6 Fora set of maps M, observe the relationship between 
vertex partition function f(x,y) and the enufunction fm(x, y) of two 
parameters: the valency of root-vertex and the coorder. 


O15.7 Consider the conditions satisfied by the face partition 
of petal bundle with size n > 1. 


O15.8 Consider the conditions satisfied by the face partition 
of a supermap of the complete graph K,, with order n > 4. 


XV.8 Exercises 371 


O15.9 Show a 1-to-1 correspondence between the two sets U’ 
and M where U’ = {U —alW = (4,7) € U,a = Kr(U),yr € 
(r)7,r =r(U)}, U consists of all rooted maps on the projective plane 
and M is of all rooted planar maps. 


O15.10 Observe the relationship between the vertex partition 
function of plane rooted trees and the face partition function of out- 
erplanar rooted maps. 


O15.11 Observe the existence of a tree for a given vector n as 
its vertex partition. 


XV.8 Exercises 


E15.1 Prove that the vertex partition function of planted trees 
satisfies the functional equation about f as 


n=0- f pr. (15.66) 





E15.2 Solve the functional equation (15.66) in a direct way. 


E15.3 Ifthe vertex partition function of Halin rooted maps is 
taken to have the root-face valency instead of root-vertex valency, then 
show that the function satisfies the functional equation about f as 





NC" T y^ f? 
featy to | TR (15.67) 


where f = f(x) = f(x, y) and f, = f(y). 


E15.4 Provide à method for solving the functional equation 
shown in (15.67). 


E15.5 Prove that the vertex partition function of Wintersweets 
with root not on a circuit satisfies the following functional equation 
about f as 

LY3 


(1-41 N =1 +a | viuo) (15.68) 


372 Activities on Chapter XV 


E15.6 Find a way for solving the functional equation shown in 
(15.68). 


E15.7 Prove that the vertex partition function of unicyclic 
maps with root not on the circuit satisfies the functional equation 
about f as 


f E LT + s | vdes, (15.69) 
y 


where 7, is the vertex partition function of planted trees, which is 
known in El15.1. 


E15.8 Solve the functional equation shown in (15.69). 


E15.9 Show that the following functional equation about f is 
satisfied by the vertex partition function of outerplanar rooted maps 
as 


acero ies | ybeo(2f) (15.70) 


y 
where 


1 
p= z:- V1 — 4z?). 
x 


E15.10 Solve the functional equation shown in (15.70). 


E15.11 Find a functional equation satisfied by the vertex par- 
tition function of general planar rooted maps. 


XV.9 Researches 


R15.1 For given orientable genus p Z 0, determine a functional 
equation satisfied by the vertex function of a set of maps on the surface 
of genus p. 


R15.2 For given nonorientable genus q > 1, determine a func- 
tional equation satisfied by the vertex function of a set of maps on the 
surface of genus q. 


R15.3 Determine a functional equation satisfied by the vertex 
partition function of nonseparable rooted maps on the Klein bottle. 


XV.9 Researches 313 


R15.4 Determine a functional equation satisfied by the vertex 
partition function of nonseparable rooted maps on the torus. 


R15.5 Determine a functional equation satisfied by the vertex 
partition function of bipartite rooted maps on the torus. 


R15.6 Solve the functional equation about f as 


sf Lo | viue (15.71) 
y 
R15.7 Solve the functional equation about f as 
Gt [1o | ierat. (15.72) 
y y 
R15.8 Solve the functional equation about f as 
(faeit -PA= | (fy +avbrslef)). (15.73) 
y y 
R15.9 Solve the functional equation about f as 
1 
f= [= 15.74 
y Le D ote 
R15.10 Solve the functional equation about f as 
1 — 0,55 z 
fe J l eei) O (15.75) 
y 1— 205 a (Ef. n) = vy Oo fo) 
R15.11 Solve the functional equation about f as 
face ; | Mel (15.76) 
y 1— Ony te 
R15.12 Solve the functional equation about f as 
fax E (15.77) 
y (1 ~~ 0,2. o fuz)? = (1052 yf yz)? 


R15.13 Solve the functional equation about f as 


(lg fel aè | ifo (15.78) 
y 


Appendix I 


Concepts of Polyhedra, Surfaces, 
Embeddings and Maps 


This appendix provides a fundamental of basic concepts of poly- 
hedra, surfaces, embeddings and maps from original to developed as a 
compensation for Chapters I-II. Only those available in the usage from 
combinatorization to algebraication are particularly concentrated on. 


Ax. Polyhedra 


A polyhedron P is a set (C;|1 < i € k}, k > 1, of cycles of letters 
such that each letter occurs exactly twice with the same power (or 
index) or different powers: 1(always omitted) and —1 and denoted by 
P = ((Ci1 € i € k}). It is seen as a set of all the cycles in any cyclic 
order. 

This is a general statement of Heffter's|Hefl| (and more than half 
a century later Edmonds’|Edm1] as dual case) which has the minimal- 
ity of no proper subset as a polyhedron for the convenience usages. 

A polyhedron is orientable if there is an orientation of each cycle, 
clockwise or anticlockwise, such that the two occurrences of each letter 
with different powers; nonorientable, otherwise. 

The support of polyhedron P = ({C;|1 < i € k]) is the graph 
G = (Vp, Ep) with a weight w on Ep where Vp = {Cill € i € k}, 
(Ci, C;) € Ep if, and only if, C; and Cj, 1 € i, j € k, have a common 
letter, and 


(L1) 


(e) l 0, when two powers are different; 
wW — 


1, otherwise 


Ax.I.1 Polyhedra 375 


for e € Ep. 
The set of all the edges with weight 1 is called the 1-set of the 
polyhedron. 


Theorem I.1 A polyhedron P = ((C;|1 € i € k}) is orientable 
if, and only if, one of the following statements is satisfied: 

(1) What obtained by contracting all edges of weight 0 on the 
support is a bipartite graph; 

(2) No odd weight fundamental circuit is on the support; 


(3) No odd weight circuit is on the support; 
(4) The 1-set forms a cocycle; 
(5) The equation system about x; = zc;, C; € Vp, on GF(2) 


for (C;, Cj) € Ep has a solution. 


Proof Because P is orientable, the two occurrences of each let- 
ter are with different powers. Since the weights of all edges are the 
constant 0, the equation system (I.2) has a solution of x; = 0 for all 
C; € Vp, 1 € i € k. Further, by considering that the consistency of 
equation system (1.2) is not changed from switching the orientation 
of a cycle between ce clockwise and anticlockwise while interchanging 
the weights between 0 and 1 of all the edges incident with the cycle on 
the support, statement (5) is satisfied for any orientable polyhedron. 

On the basis of statement (5), from a solution of equation system 
(1.2) the vertices of Gp are classified into two classes by x; = 1 or 0: 
1-class or 0-class respectively. According to equation (1.2), each edge 
with weight 1 has its two ends in different classes and hence the 1-set 
is a cocycle. This is statement (4). 

On the basis of statement (4), since any circuit meets even num- 
ber of edges with a cocycle, all circuits are with even weight. This 
means no odd weight circuit. Therefore, statement (3) is satisfied. 

On the basis of statement (3), the statement (2) is naturally 
deduced because a fundamental circuit is a circuit in its own right. 


376 Appendix I: Concepts of Polyhedra and Maps 


On the basis of statement (2), by contracting all edges of weight 
0 in each fundamental circuit on the support, (1) is satisfied. 

On the basis of statement (1),the vertices are partitioned into two 
classes by the equivalence that two vertices are joined by even weight 
path. By switching the orientation of all vertices in one of the two 
classes and those in the other class unchanged, a polyhedron without 
weight 1 edge is found. This implies that P is orientable. 

In summary, the theorem is proved. [] 


On the support Gp — (Vp, Ep) of a polyhedron P, the operation 
of switching the orientations of all vertices in a subset of Vp between 
clockwise and anticlockwise and the weights of all edges incident with 
just one end in the subset interchanged between 0 and 1 is called a 
switch on P. 


Theorem I.2 The orientability of a polyhedron does not change 
under switches. 


Proof From the definition of orientability, the conclusion of this 
theorem is true. [] 


Let T' be a minimal set of edges having an edge in common with 
all cocycles in the support of a polyhedron. In fact, it can seen that 
T is a spanning tree. 

All the polyhedra obtained by switching on a polyhedron P are 
seen to be the same as P; different, otherwise. From Theorem I.2, in 
order to discuss all different polyhedra it enables us only to consider 
all such polyhedra of the support with weight 0 on all tree edge for 
a spanning tree chosen independently in any convenient way. Such a 
polyhedron is said to be classic. 


Theorem I.3 A classic polyhedron is orientable if, and only if, 
all edges as letters have their two occurrences with different powers. A 
classic polyhedron is nonorientable if, and only if, the set of letters each 
of which has its two occurrences with same power does not contain a 
cocycle. 


Ax.l2 Surfaces 377 


Proof The first statement is deduced from Theorem I.1(3). The 
second statement is by contradiction derived from Theorem I.1 and 
Theorem I.2. [] 


Now, a polyhedron(always summed to be classic below) P is con- 
sidered as a permutation formed by its cycles. Let ó be the permuta- 
tion with each cycle only consists of the two occurrences of each letter 
in P. Then, the dual, denoted by P*, of P is defined to be P* — Pó 
such that their supports are with the same weight. The cycles in P 
are called faces and those in P* are vertices. Cycles in 6 are edges. Let 
v(P), e(P) and ¢(P) be, respectively, the number of vertices, edges 
and faces on P, then v(P) — e(P) - G(P) is the Eulerian characteristic 
of P. The graph which is formed by vertices and edges of P is called 
a skeleton of P. Of course, the skeleton of P is the support of P*. 


Theorem I.4  P* isa polyhedron and P** = P. P* is orientable 
if, and only if, so is P with the same Eulerian characteristic. 


Proof It is easily checked that P* is a polyhedron from P as a 
polyhedron. Since 5? = 1, the identity, we have 


pete puqpiéeP umm 


This is the first statement. From Theorem I.3, the second statement 
is obtained. [] 


Ax.L2 Surfaces 


Surfaces seen as polyhedral polygons can be topologically classi- 
fied by a type of equivalence. Let P be the set of all such polygons. 

For P = (((Aj))à > 1}) € P, the following three operations 
including their inverses are called elementary transformation: 

Operation 0: For (A;) = (Xaa !Y), (Ai) & (XY) where at 
least one of X and Y is not empty; 

Operation 1: For (A;) = (XabY ab) (or (XabYb~'a~')), (Ai) & 
(XaYa)(or XaY a); 


378 Appendix I: Concepts of Polyhedra and Maps 


Operation 2: For (A;) = (Xa) and (A;) = (a !Y), i z j, 
({(A;), (4;)}) & (XY) where at least one of X and Y is not empty. 
Particularly, ({(A;), (A;)}) = (XaYa !) when both (X) and (Y) are 
polyhedra. 

If a polyhedron P can be obtained by elementary transformation 
into another polyhedron Q, then they are called elementary equiva- 
lence, denoted by P ~a Q. In topology, the elementary equivalence is 
topological in 2-dimensional sense. 


Lemma I.1 For P € P, there exists a polyhedron Q — (X) € 
P where X is a linear order such that P ~a Q. 


Proof Let P = ({(A,)|1 € i € k]). If k = 1, P is in the form 
as Q itself. If k > 2, by employing Operation 2 step by step to reduce 
the number of cycles 1 by 1 if any, the form Q can be found. [] 


Lemma I.2 For P € P, if P = ((A)(B)) with both (A) and 
(B) as polyhedra, then for any x ¢ AU B, P ~a ((A)z(B)c 1). 


Proof It is seen that 
P = (AB) ~a (Azz ^! B) (by Operation 0) 
^a (CAx) (z^! B)) (by Operation 2) 
= ((A)e(B)a7)). 7 


From Lemmas I.1-2, for classifying P it suffices to only discuss 
polygons as Q. 


Lemma L3 Let Q = (ArByCz !Dy |), then 
Q ~a (ADzyBzx !Cy |). (L3) 


Proof It is seen that 
Q ^a ((Azz)(z !ByCz ! Dy !)) (by Operation 2) 
~e (zADy !z !ByC) (by Operation 2) 
= (ADzyBzx Cy ). o 


Ax.L2 Surfaces 319 


Lemma L4 Let Q = (AxByCz !Dy |), then 


Q ~a (BAzxyzr |DCy )). (1.4) 


Proof It is seen that 


Q ^a ((x Dy Axz)(ByCz")) (by Operation 2) 
m~e (BAzzz |DCz !) (by Operation 2) 
= (BAzyz DOY"): n 


Lemma L5 Let Q = (AzByCz Dy b) then 
Q ~a (ADCBzyz y). (1.5) 


Proof From Lemma I.4 and then Lemma I.3, the lemma is soon 
done. gO 


According to Lemma I.5, if A is replaced by EA in polyhedron 
(ADC B), then the relation is soon derived as 


Relation 1: (ArByCx !Dy !E) ~a (ADCBEzyx y |). 
Lemma I.6 Let Q = (AzBz) € P, then Q ~a (AB ' rz). 
Proof It is seen that 


Q ~a ((Azz)(z ^ Bz)) = ((zAx) (x B™z)) (by Operation 2) 
~el (zAB 1z) = (AB xx) (by Operation 2). gO 


According to Lemma I.6, if A is replaced by C A in polyhedron 
(AB), then the relation is soon derived as 


Relation 2: (ArBzC) ~a (AB !Czz). 


Lemma I.7 LetQ = (Axyz M zz) € P, thenQ ~a (Aryzyzz). 


380 Appendix I: | Concepts of Polyhedra and Maps 


Proof It is seen that 
Q ~a ((zAzyt)(t 1x y !z)) (by Operation 2) 
~e CAxytyrt) (by Operation 2) 
= (Axyzyzz). go 


According to Lemma 1.7, then by Relation 2 twice for x and y, 
the relation is soon derived as 


Relation 3: (Aryx !y !zz) ea (Avxyyzz). 
Lemma L8 If Q € P orientable not as (AxByCx Dy 1E), 
then Q ~a (xxz~')(= Op). 


Proof Because Q is not in the above form, Q has to be in form as 
(Azz 1B). If both A and B are empty, them Q ~a (zx 3); otherwise, 
Q ~a (AB). Because (AB) still satisfies the given condition, by the 
finite recursion principle, (zz !) can be found. O 


Theorem I.5 For any Q € P orientable, if Q 4. (zx 1), then 
there exists an integer p > 1 such that 


Q ~el qI viyi; y; )(— Op). (1.5) 
i=l 


Proof Because Q ~a (AziByiCz, Dy, E), by Relation 1 we 
have 
Q ~a (ADCBExzqyizi yi |). 


If (ADOBE) ca (zr), the Q a (tiyr Yr): That is the case 
p= 1. Otherwise (ADCBE) = (A1z5B1yo C325 D1y3 E1). Because 


(ADCBEzyyizi yi!) = (AirsBiyyCizg Dy; Ej)mwii yi 


is still in the given condition. By the finite recursion principle, (1.5) 
is found. E 


Ax.L3 Embeddings 381 


Theorem I.6 For any Q € P nonorientable, there exists an 
integer q > 1 such that 


Q ~al qJI £;£;)(= Qq). (1.6) 


Proof Because Q is nonorientable, there is a letter x; in Q 
such that Q = (Az,Bz,C). By Relation 2, D ~a (AB !Czqz). 
If (AB^1C) ~a (xxl), then by Operation 0 we have Q ~a (2121). 
This is the case of q = 1. Otherwise, there exits an integer k > 1 such 
that 


k 
Q ~el aJI ze Qu 


and (A) %q (xz!) is orientable. By Theorem L5, there exists an 
integer s > 1 such that 


(A) ~a (T nud -w E= O,). 


Thus, by Relation 3 for s times, we have 
2s+k 


Q ~a (|| zx) Q2. 
i—1 
This is g=2s+k > 1. [] 


On the basis of Lemma I.8 and Theorems I.5-6, surfaces in topol- 
ogy are in fact the classes of polyhedra under the elementary equiv- 
alence. Surfaces Oo, Op, p > 1, are, respectively, orientable standard 
surfaces of genus 0, p, p > 1. Surfaces Q,, q È 1, are nonorientable 
standard surfaces of genus q. 


Ax.l3 Embeddings 


An embedding(i.e., cellular embedding in early references particu- 
larly in topology and geometry) of a graph is such a polyhedron whose 
skeleton is the graph. 


382 Appendix I: | Concepts of Polyhedra and Maps 


The distinction of embeddings are the same as polyhedra. Pre- 
cisely speaking, two distinct embeddings on a 2-dimensional manifold 
are not equivalent topologically in 1-dimensional sense. 

According to Ax.I.1, all embeddings always imply to be classic. 

For a graph G = (V, E), Heffter-Edmonds' model of an embed- 
ding of G by rotation system at vertices, in fact, only for orientable 
case [Hef1] and [Edm1]. 

Let c = {o,|v € V) be the rotation system on G where o, is the 
cyclic order of semi-edges at v € V. Then, by the following procedure 
to find an embedding of G: 


Procedure I.1 First, put different vertices in different position 
marked by a hole circle or a bold point on the plane. Draw lines for 
edges such that no interior point passes through a vertex and o; is in 
clockwise when v is a hole circle; in anticlockwise, otherwise. 

Then, by travelling along an edge in the rule: passing through 
on the same side when the two ends of the edges are in same type; 
crossing to the other side, otherwise. Find all cycles such that each 
edge occurs just twice. The set of cycles is denoted by Pg. 


Lemma I.9 Pg is a polyhedron. 


Proof Because it is easily checked from the definition of a poly- 
hedron. [] 


Lemma I.10 Fg is orientable. 


Proof Because the dual is orientable, from Theorem L4, the 
lemma is true. L 


Theorem I.7 The dual of Pg is an orientable embedding of G. 


Proof Because the support of the dual of Pg is G itself, the 
theorem is deduced. [] 


However, Pg in general is not classic except for all vertices are of 
same type. 


Ax.L3 Embeddings 383 


Theorem I.8 For a given rotation system c of a graph G, let 
Pa(a;0) be the polyhedron obtained by the procedure above for all 
vertices of same type, then P¢(a;0) is unique. 


Proof From the uniqueness of classic polyhedron in this case, 
the theorem is done. Ll 


On the basis of Theorem L8, it suffices only to make all vertices 
with the same type, e.g., in clockwise. Further, in order to extend 
to nonorientable case, on account of Theorem 1.3, edges in a set not 
containing cocycle are marked for crossing one side to the other in 
the Heffter-Edmonds' model. The marked edges are called twist. This 
model as well as the Procedure I.1 here is called an expansion. 


Theorem I.9 The dual of what is obtained in an expansion is 
a unique nonorientable embedding of G for twist edges fixed. 


Proof Because one obtained in an expansion is a classic poly- 
hedron, from the uniqueness of the dual of a polyhedron, the theorem 
deduced. [] 


Theorem I.10 All embeddings of a graph G obtained by ex- 
pansions for all possible rotation system and twist edges in a subset 
of the cotree T' of a given spanning tree T' on G are distinct. 


Proof Asa result of Theorem I.9. [] 


This theorem enables us to choose a spanning tree T' on a graph 
G for discussing all embeddings of G on surfaces. 

Let Tj and T5 be two spanning trees of a graph G. The sets 
of all embeddings of G as shown in Theorem 1.10 for Tj and T» are, 
respectively, denoted by €; and £». 


Theorem I.11 Let £f and £j be, respectively, the subsets of £ 
and Ez on surfaces of genus g(orientable g = p > 0, or nonorientable 
g=q2 1). Then €; =. 


384 Appendix I: | Concepts of Polyhedra and Maps 


Proof Because of Theorem L.10, it suffices only to discuss ex- 
pansions for T, and T». Since |T;| = |T5|, Theorems I.8-9 implies the 
theorem. [] 


For an embedding P € £7, if P ¢ £3, then there exits a twist 
edge e in T». By doing a switch with the fundamental cocircuit con- 
taining e for 7», an embedding P’ in the same distinct class with P is 
found. If no twit edge is in T», then P’ is the classic embedding in £7 
corresponding to P. Otherwise, by the finite recursion, a classic em- 
bedding Q € £3 in the same distinct class with P is finally found. In 
this way, the 1-to-1 correspondence between Ef and £3 is established. 

The last two theorems form the foundation of the joint tree model 
shown in |Liu13-14]. Related topics are referred to |Sta1-2]. 


Ax.l.4 Maps 


Maps as polyhedra or embeddings of its underlying graph had 
been being no specific meaning until 1979 when Tutte(William T., 
1917-2002) clarified that a map is a particular type of permutation 
on a set formed as a union of quadricells[Tut1—3]. All quadricells are 
with similar construction that four elements have the symmetry as a 
straight line segment with two ends and two sides. 

This idea would go back to Klein(Felix, 1849-1925) who consid- 
ered a triangulation of an embedding on a surface by inserting a vertex 
in the interior of each face and each edge and then connecting all line 
segments from a vertex in the interior of a face to all vertices on the 
boundary of the face. It is seen that each edge is adjacent to four 
triangles called flags as a quadricell. So, such a pattern of map used 
in this course can be named as Klein-Tutte's model. Related topics 
are referred to [Vin1-2]. 

Now, we have seen that a surfaces is determined by an elementary 
class of polyhedra, an embedding is by a distinct class of polyhedra and 
a map is by an isomorphic class of embeddings. The distinction of em- 
beddings is based on edges labelled by letters, or numbers. This is also 


Ax.L4 Maps 385 


a kind of asymmetrization. But edges on a map are without labelling. 
Isomorphic maps are combinatorially considered with symmetry. So, 
a map is an isomorphic class of embeddings of its underlying graph. 

Let G = (V, E) be a graph. As shown in Sect. 1.1, V = Par(X) 
and E = (Bx|r € X) where Par(X) is a partition on B(X) = 
UxzexBxz, Bx = {x(0), x(1)} for a set X. Two graphs G4 = (Vi, E) 
and G» = (Vo, E>) are isomorphic if, and only if, there exists a bijection 
t: X1 — Xə such that the diagrams 


. a A 


| ba (L7) 


X, ———— Xo 


for c; = Bj;Par; i = 1,2, are commutative. Let Aut(G) be the 
automorphism group of G. 

On the other hand, a semi-arc isomorphism between two graphs 
G, = (Vi, E4) and G2 = (Vo, E2) is defined to be such a bijection 7: 
By(X4) — B3( X3) that 


B(X) —————  Bs(X3) 


hy, it as) 


T 


for c; = B; Pary i = 1,2, are commutative. Let Aut;/5(G) be the 
semi-arc automorphism group of G. 


Theorem I.12 If Aut(G) and Aut;/5(G) are, resp., the auto- 
morphism and semi-arc automorphism groups of graph G, then 


Auty/2(G) = Aut(G) x Sj (1.9) 


where / is the number of self-loops on G and 55 is the symmetric group 
of degree 2. 


Proof Because each automorphism of G just induces two semi- 
arc isomorphisms of G for a self-loop, the theorem is true. [] 


386 Appendix I: | Concepts of Polyhedra and Maps 


For map M = (Xa, P), its automorphisms are discussed with 
asymmetrization in Chapter VIII. Let M(G) be the set of all noniso- 
morphic maps with underlying graph G. 


Lemmal.11 Foran automorphism Ç on map M = (A,5(X), P), 
we have exhaustively G|p(x; € Auty2(G) and Go|g(x; € Auty/2(G) 
where G = G(M), the underlying graph of M, and B(X) = X + BX. 


Proof Because Xa 5(X) = (X + BX) 4 
a B(X), by Conjugate Axiom each ¢ € Aut( 
possibilities: G[p(x) € Autj/5(G) and Go|np(x) 





(aX +a3X) = B(X)+ 
M) has exhaustively two 
c Auty/2(G). [] 
Theorem 1.13 Let €,(G) be the set of all embeddings of a 


graph G on a surface of genus g(orientable or nonorientable), then the 
number of nonisomorphic maps in £,(G) is 


™(O)= se o MO (1.10) 


TEAuty /2(G) 


where ®(7) = (M € &(G)|r(M) = M or rTa(M) = M}. 
Proof Suppose X;, X», --:, Xm are all the equivalent classes of 
X = €,(G) under the group Auty/2(G) x (o), then m = m,(G). Let 
S(z) = {7 € Autij(G) x (a)| T(z) = x) 


be the stabilizer at x, a subgroup of Aut; /;(G) x (a). Because |Aut4/(G) x 
(a)| 2 [S(2;)l Xi], gee X, t= 1,2, -*-, m, we have 


m|Auty/2(G) x (a)| = De IS (xi) || Xi]. (1) 


By observing |S(x;)| independent of the choice of x; in the class X;, 


Ax.L4 Maps 387 


the right hand side of (1) is 


2,5)25, 5,1 


rcx LEX TES (x) 


E 2 (2) 


TEAuty /2(G)x (a) z—T(x) 


», Ie. 


T€Aut4/2(G) x (a) 
From (1) and (2), the theorem can be soon derived. O 


The theorem above shows how to find nonisomorphic super maps 
of a graph when the automorphism group of the graph is known. 


Theorem 1.14 Fora graph G, let R,(G) and €,(G) be, respec- 
tively, the sets of all nonisomorphic rooted super maps and all distinct 
embeddings of G with size e(G) on a surface of genus g(orientable or 
nonorientable). Then, 


2e(G) 


RG)| = auty /2(G) 


Exa (1.11) 


Proof Let M,(G) be the set of all nonisomorphic super maps 
of G. By (11.3), 


IR (G)| ES 


4e(G) 2 x auti/5(G) 
7 2 x autı/2(G') 


aut( M) 
MeM,(G) 


By considering that 


2 x auty/2(G) = |Auty/2(G) x (a)l 
= [(Autijs(G) x (a))|u| x |Auti2(G) x (a) (M)| 
and (Aut;/5(G) x (o))|u = Aut(M), we have 
4e(G) 
[R (G)| = ERES » |Autij2(G) x (o)(M)| 
n2 MeM,(G) 
A). le (a) 


H auty /2 (G) 


388 Appendix I: | Concepts of Polyhedra and Maps 
This is (1.11). O 


This theorem enables us to determine all the super rooted maps 
of a graph when the automorphism group of the graph is known. How- 
ever, the problem of finding an automorphism of a graph is much more 
difficult than that of finding an automorphism of a map on the basis 
of Chapter VIII in general. For asymmetric graphs the two theo- 
rems above provide results much simpler. More results are referred to 
[MLW1]. 


Appendix II 


Table of Genus Polynomials for 
Embeddings and Maps of 
Small Size 


For a graph G, let pe(x), a(x) and utc (x) be, respectively, the 
orientable genus distributions of embeddings, super maps and rooted 
super maps of G, or called orientable genus polynomials. Similarly, 
let qc(zx .), vc(x !) and v5(x !) be, respectively, the nonorientable 
genus distributions of embeddings, super maps and rooted super maps 
of G, or called nonorientable genus polynomials. 


Ax.IL1  Triconnected cubic graphs 


First, list all nonisomorphic 3-connected cubic graphs from size 
6 through 15. 


Size 6 


390 Appendix Il: Table of Genus Polynomials for Embeddings and Maps 





Size 9 
C931 C92 
Size 12 
C121 C12,2 
C12.3 C124 
Size 15 
C353 C15,3 


C154 C15,5 C15,6 


Ax.IL1 Triconnected cubic graphs 391 


Q9 S V 


C45,7 C15,8 C15,9 
C35,10 Cis C152 
C1543 C1514 


In what follows, the orientable and nonorientable genus poly- 
nomials of embeddings, super maps and rooted super maps of 3- 
connected cubic graphs shown above are provided. 


Case of size 6: 


Orientable 
Po, (x) = 2+ 14z, 
Bos (z) = 1+ 2a, 
Mo, (@) = 1+ 72. 
Nonorientable 


qos, (z 1) = 14a + 422? + 562°, 
Ve. (eo) = 2a Ba" 432", 
VO, (2 *) = "x + 21a? + 282°. 


392 Appendix Il: Table of Genus Polynomials for Embeddings and Maps 


Case of size 9: 


Orientable 
Doy, () = 2 + 38x + 242”, 
Ho, (x) = 1-4 5x + 22”, 
Mo, (x) = 3+ 57a + 3627; 
Do, (x) = 40x + 2427, 
pos Qc) = 20 ae 
Ho, (x) = 10x + 6x”. 
Nonorientable 


Gcy,(a~*) = 22x + 1222? + 4242? + 39227, 
vc, (£t) = 3x + 12x? + 282? + 23x, 
Vo, (X 1) = 33x + 1832? + 6362? + 588^; 


qo, (x!) = 12x + 1082? + 4322? + 40824, 
ve, (x 1) = x + 2x? + 62? + 62^; 
Vo, (2°) = 3x + 272^ + 1082? + 102. 





Case of size 12: 
Orientable 
PC (@) = 2+ 70x + 1842”, 


Boc) = 1+ 15z- 2827, 
Ho, (2) = 12 + 420% + 110427; 


Ax.IL1 Triconnected cubic graphs 393 


Dc, (x) = 64x + 19227, 
I C13,5 (x) = 4z + 122, 
Hp, (x) = 128x + 3842”; 


Posle) = 56x + 2002”, 
ics (m) = 5x + 1327, 
Ho, (x) = 84x + 30027; 


Dc, (X) = 2 + 54x + 2002”, 
Jic, (m) = 1+ 52x + 82°, 
MO, (X) = 14+ 27x + 1002. 


Nonorientable 


Cia (X 7) = 30x + 2422? + 14482? + 32722 + 29442", 
Vc, (m 1) = Tx + Ada? + 217a? + 4522 + 382°, 
V... (£71) = 180x + 14522? + 8688x? + 1963224 + 176647; 


qc, (X 7) = 12x + 1802? + 13602? + 331227 + 30722”, 
Vc, (2 1) = £ + 92? + 642? + 14924 + 13725, 
Vp, (X 1) = 24a + 36027 + 2720x? + 6624x + 61447"; 














qc, (X 7) = 10x + 1582? + 12722? + 32962 + 32002”, 
Voss (X 1) = 2a + Ma? + 572? + 1332 + 11825; 
Vp, |) = 15a + 23727 + 19087? + 2944" + 48002”; 














qc, (7 .) = 24x + 192z* + 12882? + 326427 + 31682, 
Vc, (x 1) = £ + Tz? + 24r? + 58x + 405; 
Vo, (X 1) = 12z + 962? + 644r” + 16322 + 15842°. 





Appendix II: Table of Genus Polynomials for Embeddings and Maps 


Case of size 15: 


Orientable 


Dc, (x) = 2 + 102z + 664? + 25622, 

Jc, (m) = 1+ 27a + 17627 + 682°, 

Lc, (x) = 30 + 15302 + 996027 + 384025; 
( 


PCs, r) = 720 + 6642? + 288r, 
Hess (x) = 20x + 1802" + 67227, 
Hes, (m) = 1080x + 99602? + 43202; 


Dc, (t) = 56x + 64827 + 32027, 

HCis,3 (x) 12x + 96x? d 4425, 

Hc, (x) = 420x + 48602? + 240027; 
( 


Dc, (x) = 80x + 688z? + 25627, 
Kesal T] = liz + 9327 + 32x3, 
Ho, (£) = 600x + 51602? + 19202; 


Dc, (£) = 2 + 118x + 6482? + 25622, 
14 27z + 8827 + 362°, 
15 + 885x + 4860x? + 19202 


T 


Ho s 


(2) = 
(2) = 
Dc, (0) = 2+ 110x + 688x? + 22427, 
(£) = 1 + 142 + 602" + 202°, 
CL) = 
( 
( 
( 


Uc, (£) = 10 + 550x + 344027 + 11205. 


Dc, (5) = 2 + 78x + 65677 + 28827, 


,(v) 9 1-4 142 + 812? + 2425, 
r | \= 


Dc, (x) = 96x + 672x? + 25627, 
Hass) = 9r + 492? + 182°, 
His (m) = 3602 + 25202? + 9602°: 


Ax.IL1 Triconnected cubic graphs 395 


PoigaUt) = 48r + 656x? 4 3202, 
HC15,9 (x) = 8rd 59x? 4r 2523, 
Ho, (£) = 180x + 2460x? + 1200z°; 


DC, 4 (1) = 88x + 64827 + 2882°, 
HC45,10 (x) = 5x + 31r’ + 162°, 
Hr, (£) = 220x + 16202 + 1202; 


Pois (£) = 2 + 70x + 63227 + 3202°, 
HCis,11 (x) = = 1] + 5x + 28x" + 102°, 
MCs (£) = 3 + 1052 + 9482? + 48027; 


PCis, a= 72% + 6324? T 3202, 

UCs S (X) = 6x + 24a? + 142°, 

Ho, (©) = 108x + 948x? + 48055. 

PCs (£) = 48x + 72027 + 25627, 

HC, (x) = 2x + 152” SF 6x’ ; 

Hc, G0) = 302 + 45027 + 16025; 

PCs (£) = 40x + 6642? + 32027, 

jig) =L + Tx? e 229, 

He, uu) = 10x + 1662? + 802. 
Nonorientable 


qc, (x 7) = 38x + 39427 + 33362? + 1274424 + 270082? 
+ 209922°, 


Vc (x 1) = 10x + 1042? + 8382? + 322027 + 67682? + 53002; 
Vos 1) = 570a + 59102? + 500402? + 1911602* + 4051202? 
+ 314880z°: 


396 


Appendix Il: Table of Genus Polynomials for Embeddings and Maps 


qc, (£t) = 10x + 214x? + 25762? + 11664z* + 274242? 
+ 226247°, 
VCs (x 1) = 4v + 602? + 6761? + 29882 + 69524? + 56882; 
Vp, (x |) = 150a + 32102? + 386402? + 1749602* + 4113602? 
+ 3393602°; 


Ciz3(@') = 6x + 1582? + 21882? + 1091224 + 275042? 
+ 237442, 
Vc, (x 1) = 2x + 27a? + 3132? + 146624 + 35722? + 30442; 
Vox 1) = 45a + 11852? + 164102? + 818402* + 2062802? 
+ 17808025; 


qc, (07 .) = 12x + 244z* + 28162? + 12224z* + 274562? 
+ 217602, 
Vc, (x 1) = 2x + 33a? + 36827? + 15652 + 34802? + 27362; 
Voy, 4(@ 1) = 90x + 18302? + 211202? + 916802* + 2059202? 
+ 1632002°; 


qc, (0 |) = 38x + 410? + 34962? + 129522 + 268802? 
+ 207362°, 
Voy, (x1) = 8x + T6a? + 5242? + 17682 + 34602? + 26522; 
VO s (27) = 3857 + 30752? + 262202? + 971402* + 2016002? 
+ 15552025, 


Cizg(@ |) = 38x + 402? + 34482? + 1304024 + 270722? 
+ 205122°, 
Ve, (x 1) = 6r + 44a? + 3192? + 11572? + 23542? + 174425; 
Vox 1) = 190x + 20102? + 172402? + 652002* + 1353602? 
+ 102560; 


Ax.IL1 Triconnected cubic graphs 397 


qc, (0 |) = 32x + 312x? + 28002? + 1180024 + 272002? 
+ 223682°, 
VCs (x 1) = 5x + 35a? + 2672? + 10772 + 23582? + 18662; 
VO,,,(@ +) = 160x + 15602? + 140002? + 5900074 + 1360002? 
+ 1118402°; 


qc, (£t) = 12x + 260x? + 297621? + 124322 + 273282? 
+ 215042°, 
Vc, (x 1) = 2x + 212? + 2072? + 82824 + 1772x° + 13822°; 
vL, (a!) = 45x + 9752? + 111602? + 2662024 + 1924802° 
+ 806402°; 


QC, (x .) = 4a + 1322? + 20492? + 1072024 + 276162? 
+ 240002°, 
Vc, (m |) = £ + 16x? + 1522? + 753x* + 18112? + 155925; 
Voy, (X |) = 15x + 4952? + 16502? + 402002* + 1035602? 
+ 90000z°: 


QC, ao (0 7) = 12x + 2522? + 28642? + 121362 + 272642? 
+ 219846, 

Voss) = £ + Ma? + 1242? + 5172 + 11542? + 9412; 

Vern (0 1) = 30x + 63027 + T1602? + 30340x* + 6815602? 
+ 54960; 


QC, (£71) = 30a + 2822? + 25602? + 112402 + 271682? 
+ 232322°, 

Vos u (27t) = 2x + 17x? + 922? + 351z* + 1542? + 62425; 

VO (©) = 45a + 4232 + 38402? + 168602* + 407522? 
+ 348482°, 


398 Appendix Il: Table of Genus Polynomials for Embeddings and Maps 


QC,s ao (0 7) = 12x + 2202? + 24802? + 1124024 + 272642? 
+ 232969, 

Vo, s (t 1) = 2x + Ma? + 902? + 34324 + T5625 + 63825; 

VoU 1) = 18r + 3302? + 37202? + 1686074 + 408962? 
+ 34944x; 


dC, (2 1) = 120x? + 22322? + 11568x + 27936x" 
+ 226562°, 

Vo, (t 1) = 4a? + 28x? + 14474 + 3072? + 25925; 

Vos 1) = 1927 + 13952? + 72302 + 174602? 
+ 14160; 


dC, (xL) = 4x + 120x? + 19002? + 104402 + 276642? 
+ 243842, 
Vo, (X 1) = a + 22? + 16x? + 62 + 14225 + 11125; 
Ver, (#1) = a + 302? + 474r? + 261024 + 69162? 
+ 60392°. 


Ax.IL2 Bouquets 


Let Bm be the bouquet of size m, m > 1. 


Case of m — 1: 


Orientable 


Ax.IL2 Bouquets 399 


Nonorientable 


Case of m — 2: 


Orientable 
pp,(x) = 4-F 2x, 
HB» (x) =1+2, 
Hp, (x) = 2+ x. 
Nonorientable 


dp, (x t) = 10x + 827, 
vele) = 2g + 2z^, 


Vp, (£~!) = 5g + 42”. 


Case of m — 3: 


Orientable 
pp,(x) = 40 + 80x, 
lp (t) = 2+ 3x, 
Lp, (x) =5+10z. 
Nonorientable 


qp, (xz |) = 176x + 336z? + 3282°, 


400 Appendix Il: Table of Genus Polynomials for Embeddings and Maps 


vg,(x^l) = 5x + 8x? + 82%, 
Vg, (x 1) = 22x + 422? + 412°. 


Case of m — 4: 


Orientable 


pp,(x) = 672 + 3360x + 10082”, 
bop, (2) = 3 +102 + 42”, 
Uf, (m) = 14 + 70x + 212°. 


Nonorientable 


qp,(x |) = 44642 + 145922? + 331202? + 23424a**, 
vg,(z +) = 129 + 33z? + 642? + 4724, 
vp, (x 1) = 93x + 304z? + 6902? + 48827. 


Case of m — 5: 


Orientable 


pp,(x) = 16128 + 161280 + 18547227, 
Lp, (x) = 6 + 35x + 382”, 
UJ, (x) = 42 + 420x + 48327. 


Nonorientable 


qp,(x |) = 1482242 + 718080z? + 27456002? + 447744024 
+ 31599362°, 

vp,(z |) = 33x + 131z? + 4422? + 686z^ + 4732°, 

vg, (x 1) = 386x + 187077 + 71502? + 116602* + 82292”. 


Ax.II.3 Wheels 401 


Case of m — 6: 


Orientable 


PB (XL) = 506880 + 8870400x + 24837120x? + 570240027, 
La, (xz) = 124+ 1327 + 3282? + 822°, 
Lp, (x) = 132 + 2310x + 6468z? + 14852°. 


Ax.IL3 Wheels 


Let W, be the wheel of order n, n > 4, i.e., all vertices are of 
valency (or degree) 3 but one and all 3-valent vertices form a circuit. 


Case of n — 4: 


Orientable 
pw,(x) = 2 + 142, 
pw,(z) = 1+ 2z, 
by, (2) = 14+ Te. 
Nonorientable 


qw,(x 5) = 14x + 42x? + 562°, 
uma J= 2x + 8x? + 325, 
Viy,(@ |) = Tx + 212? + 282°. 


Case of n= 5: 
Orientable 
pw, (x) = 2 + 58x + 362°, 


pw, (2) = 1+ 8a + 42’, 
Uy, (x) = 4 + 1162 + 722°. 


402 Appendix Il: Table of Genus Polynomials for Embeddings and Maps 
Nonorientable 


qw, (£7}) = 28x + 1762? + 6402? + 59624, 
vy, (xt) = 4r + 18x? + 522? + 482$, 
viy, (x 1) = 56x + 352z? + 12802? + 1192z*. 


Case of n — 6: 


Orientable 
pw, (x) = 2 + 190x + 57627, 
ima = 1-- 14x + 412’, 
py, (0) = 4+ 380x 4-115227. 
Nonorientable 


qw,(x 1) = 52x + 5802? + 40802? + 9880z* + 92162, 
vw,(z +) = 6x + 38x? + 2272? + 539x + 4942, 
Vi, (x 7) = 104x + 11602? + 81602? + 197602* + 1843227. 


Case of n = T: 


Orientable 


pw, (x) = 2 + 550x + 4968z? + 216027, 
pw; (x) = 14 34x + 240z? + 1062, 
Uy (x) = 4 + 1100x + 99362? + 432027. 


Nonorientable 


qw,(x 1) = 94x + 1680z? + 194822? + 87536z* + 2054962? 
+ 1695526. 


Ax.II.4 Link bundles 403 


vw, (£71) = 8x + 89a? + 8782? + 3829x* + 8788x? + 72412°, 
Vi (x 7) = 188x + 33602? + 389642? + 175072x* + 4109922? 
+ 3391042. 


Case of order n — 8: 


Orientable 


pw, (x) = 2 + 1484x + 311782? + 5949627, 
uw, (£) = 1 + 63x + 11762? + 224627, 
Ly, (x) = 4 + 2968x + 623562? + 1189927. 


Ax.ILA4 Link bundles 


Let L,, be the link bundle of size m, m > 3. A link bundle is a 
graph of order 2 without loop. 


Case of size m = 3: 


Orientable 
Duls)e2-p2u, 
pn(x)-1-z, 
Bp, (£) = l- lm. 
Nonorientable 


ina) = 6x + 627, 
victa t = laos, 


v; (£71) = 32 + 32°. 


404 Appendix Il: Table of Genus Polynomials for Embeddings and Maps 


Case of size m = 4: 


Orientable 
pr,(x) = 6 + 30z, 
Hr, (z) = 1-4 2z, 
ui, (£) = 1+ 5z. 
Nonorientable 


qr, (x .) = 36x + 96z? + 12027, 
vns ess 35", 
v; (x |) = 6x + 162? + 202°. 


Case of size m = 5: 


Orientable 
pr,(x) = 24 + 360x + 1922”, 
ir.(2) = 14-32 + 32", 
ur, (£) = 1+ 15x + 82". 
Nonorientable 


qr, (x |) = 240x + 12002? + 38402? + 3360274, 

vp, (zx 1) = 2x + Tx? + Mz? + 142%, 

v; (x!) = 10x + 50x? + 1602? + 14014. 
Case of size m = 6: 


Orientable 


pr, (x) = 120 + 4200x + 100802, 
Dr, (x) = 1 + 6x + 102°, 
us, (x) = 1+ 35a + 842”. 


Ax.IL5 Complete bipartite graphs 405 


Nonorientable 


qn, (x ) = 1800x + 144002? + 841202? + 184320z* + 1617602, 
vp,(z |) = 3x + 142? + 482? + 9627 + 722°, 
vi (x |) = 15x + 1207? + 7012? + 1536x* + 13482”. 


Case of size m = T: 


Orientable 


pr,(x) = 720 + 50400 + 3376802? + 12960025, 
pz, (x) = 14 8x + 31x? + 162°, 
ur, (£) = 1 + 70x + 46927 + 18027. 





Ax.IL5 Complete bipartite graphs 


Let Km,n be the complete bipartite graph of order m+n, m,n > 


Case of order m+n — 6: 


Orientable 
DK,s(x) = 40x + 24x". 
Ux (t) = 2x + z?, 
Hk, (£) = 10x + 6x”. 
Nonorientable 


qK,,(') = 122 + 108z? + 4322? + 4082+, 
Vigna) — vd 227 | 6x? | 62, 
Vi. (1-1) = 3x + 27x? + 1082? + 1022. 





406 Appendix Il: Table of Genus Polynomials for Embeddings and Maps 


Case of order m -4- n — T: 


Orientable 


Dk, (x) = 156x + 22442? + 105622, 
pcs (0) = 3x + 162? + 102°, 
Heale) = 26x + 314a? + 17627. 


Nonorientable 


qi, (x |) = 12x + 432? + 68522? + 3628827 + 933602? 
+ 807842°, 
Vk, (X 3) 





Y) = g + Az? + 3325 + 15624 + 35825 + 31726, 
Vic, (@ 1) = 2x + 72x? + 11422? + 6048x* + 155602? 
+ +134642°. 


Case of order m+n = 8: 


Orientable 


PK,,(@) = 108x + 249842? + 5650202? + 108950414, 
HK, (T) = 2x + 252? + 3182? + 53027, 
Hic, (£) = 3x + 6942? + 156952? + 302642. 


Dk,,(x) = 240x + 315842? + 2908802? + 1136642", 
pcs (x) = x + 3327 + 2252? + 10527, 
Hg, (x) = 10x + 156627 + 121202? + 473627. 


Ax.IL6 Complete graphs 407 


Ax.IL6 Complete graphs 
Let , be the complete graph of order n, n > 4. 


Case of order n = 4: 


Because of K4 = Wi, seen from Case of order n = 4 in Ax.II.3. 


Case of order n = 5: 


Orientable 


PK,(@) = 462x + 49742? + 324027, 
ju, (£) = 6x + 312? + 132, 
Up (m) = 77x + 8292? + 3902°. 


Nonorientable 


qx, (x .) = 54x + 1320x? + 174902? + 8466024 
+ +208776x° + 17758829, 
v,(x 1) = Qe + 112? + 992? + AVI + 9552? + 7962, 
Vy. (xL) = 9x + 220x7 + 29152? + 14110z* + 347962? 
+ 4-295982. 


Case of order n = 6: 


Orientable 


Dk, (x) = 1800x + 6545762? + 24613800.X? 
+41242502087 + 415825922°. 


408 Appendix Il: Table of Genus Polynomials for Embeddings and Maps 


Nonorientable 


qx,(x 1) = 24a + 45602? + 3709202? + 1082844027 
+ 1922645762° + 1927543032x5 + 119055609602" 
+ 42386101920? + 798313881604? 
+ 592442818562”, 


Appendix III 


Atlas of Rooted and Unrooted Maps 
for Small Graphs 


In the symbol X : a,b,c for a map appearing under a figure 
below, X is the under graph of the map, a — oy or qy are, respectively, 
orientable or nonorientable genus y, b is the series number with two 
digits and c is the number of ways to assign a root. And, r on a 
surface is for zl, or —z, x = 1,2,--+. 


Ax.III.1 Bouquets Bm of size 4> m > 1 
Case m = 1: 
Orientable genus 0 


1 


CE» 


1 


B1:00—01—01 


Nonorientable genus 1 


1 


B; : q1 — 01 — 01 


410 Appendix III: | Atlas of Rooted and Unrooted Maps 


Case m — 2: 


Orientable genus 0 





Bə : o0 — 01 — 02 


Orientable genus 1 





Bə : ol — 01 — 01 


Nonorientable genus 1 





Ax.III.1 Bouquets Bm of size4 2 m 2-1 


Nonorientable genus 2 


H 


By: q2 — 01 — 02 


Case m = 3: 


Orientable genus 0 


2 


Bz : 00 — 01 — 02 


Orientable genus 1 





All 


EM 


Bə : q2 — 02 — 02 





B3 : o0 — 02 — 03 





1 


B; : ol — 01 — 04 B; : ol — 02 — 03 B3 : ol — 03 — 03 


412 Appendix III; | Atlas of Rooted and Unrooted Maps 


Nonorientable genus 1 





B; : q1 — 01 — 06 Bs : q1 — 02 — 06 B3 : q1 — 03 — 06 





3 


B3:q1—04—-03  B3:q1—05— 01 


Nonorientable genus 2 





Bz: q2 — 02 — 12 Bz : q2 — 03 — 06 


1 1 





Ax.III.1 Bouquets Bm of size 4>m> 1 413 





3 


B; : q2 — 07 — 03 Bs : q2 — 08 — 03 


Nonorientable genus 3 





3 


B; : q3 — 04 — 03 B; : q3 — 05 — 06 B3 : q3 — 06 — 06 





1 





B3 : q3 — 08 — 03 


414 Appendix III: Atlas of Rooted and Unrooted Maps 


Case m — 4: 


Orientable genus 0 






B; : 00 — 01 — 02 


Orientable genus 1 


Bg: o1 — 02 — 08 Bag: 01 — 03 — 08 


1 1 





B; : o1 — 04 — 08 B4:01— 05 — 16 B4: 01 — 06 — 08 
1 1 1 





B4:01 — 07 — 04 B4:01— 08 —04 B; : ol — 09 — 04 


Ax.IIL2 Link bundles L,,,7 > m > 3 415 





B4:01 — 10 — 02 


Orientable genus 2 


eI 





B4:02—01— 04 Bg : 02 — 02 — 08 B4: 02 — 03 — 08 


wi 





By: 02 — 04 — 01 


Ax.IIL2 Link bundles Lm, 6 > m > 3 


Case m = 3: 


Orientable genus 0 


416 Appendix III: Atlas of Rooted and Unrooted Maps 


hol 


Ls : o0 — 01 — 01 


Orientable genus 1 


L3 : ol — 01 — 01 


Nonorientable genus 1 


Ls : q1 — 01 — 03 


Nonorientable genus 2 


Ax.IIL2 Link bundles L,,,7 > m > 3 417 


Ls : q2 — 01 — 03 


Case m = 4: 


Orientable genus 0 





L4:00—01— 01 


Orientable genus 1 





L4:01—01—01  L4:01—02—04 


418 Appendix III: Atlas of Rooted and Unrooted Maps 


Nonorientable genus 1 





La :q1—01-02 L4 :q1-— 02 = 04 


Nonorientable genus 2 





Lg: q2 — 01 — 02 Lg: q2 — 02 — 08 L4: q2 — 03 — 02 





L, : q2 — 04 — 04 


Nonorientable genus 3 


Ax.IIL2 Link bundles L,,,7 > m > 3 419 





La : q3 — 01 — 08 La : q3 — 02 — 08 La : q3 — 03 — 04 


Case m = 5: 


Orientable genus 0 


wi 





L5:01 — 01 — 05 Ls; : ol — 02 — 05 Ls; : ol — 03 — 05 


Orientable genus 2 


420 Appendix III: Atlas of Rooted and Unrooted Maps 


px] 





L5:02—01 — 02 L5:02— 02 — 05 L5:02—03—01 


Nonorientable genus 1 





Ls : q1 — 01 — 05 Ls : q1 — 02 — 05 


Nonorientable genus 2 





Ls; : q2 — 01 — 05 Ls : q2 — 02 — 10 Ls : q2 — 03 — 10 





Ls : q2 — 04 — 05 Ls : q2 — 05 — 10 Ls : q2 — 06 — 05 


Ax.IIL2 Link bundles L,,,7 > m > 3 421 





Ls : q2 — 07 — 05 


Nonorientable genus 3 


bol 





Ls : q3 — 01 — 10 Ls : q3 — 02 — 20 Ls : q3 — 03 — 20 





L5:43 — 07—10 L5:q3 — 08 — 10 Ls : q3 — 09 — 20 


422 Appendix III: | Atlas of Rooted and Unrooted Maps 


wi 





Ls :q3— 10 — 10 L5:q03—11— 10 Ls : q3 — 12 — 05 





Ls :q3—13—-05  Lg:g8—14— 05 


Nonorientable genus 4 





Ls :q4— 01 — 10 Ls : q4 — 02 — 10 Ls :q4— 03 — 10 





L5:q4 — 04 — 10 L5:q4 —05 — 10 L5:q4 — 06 — 10 


Ax.IIL2 Link bundles L,,,7 > m > 3 423 





L5 :94 — 07 — 05 Ls : q4 — 08 — 20 Ls :q4— 09 — 10 





Ls :g4 — 10 — 05 Ls :q4— 11—05 Ls : q4 — 12 — 20 





Case m = 6: 


Orientable genus 0 


424 Appendix III: Atlas of Rooted and Unrooted Maps 





Lg:01—04—06  Lg:01—05—06 . Lg:01 — 06 — 06 


Orientable genus 2 


Ax.IIL2 Link bundles L,,,7 > m > 3 425 







Le : 02 — 03 — 03 


E 5 
3 





2 


2M 


: 02 — 09 — 06 


Íargd-07-19- Le:02—08—12 Le 
3 


2 4 





Case m = T: 


Orientable genus 0 


5 2 
6 i 
6 1 
5 2 
Ly : o0 — 01 — 01 
Orientable genus 1 


Orientable genus 2 


SEE 


— 28 L; : 02 — 15 — 28 
= 4 3 
1 3 1 2 
6 1 6 1 
NEA ; $ 


L7 : 02 — 14 
1x \ 
SEDAN 


A 
= 


oye 
CAN 5 


m" 


i'd 
AM 
HUE 
0 S: 
AR: 


1 
6 


= 


Al Ms 
it 
AL Mn 


y 


V 
px 


[ine] gent 


A xb 


AA 


N 


o 

9 

Rd 

S 

© 
i 


: Af NE 
OW : 


Orientable genus 3 


a 


5 
3 


2 
5 
4 
o3 


i 


A Ne 


k 
- 
ut 
S| 
| 


CN 

z j=) 

| 

eo 

IN 10 = 
bs 

I Re) -] 

KO o H 

ho, N = 

Oo 


3 - 
2 

I 

6 
5 
L7 : 03 — 01 — 


AX. 
WM | 


V 


Im 
ho vet 


sy 
Ly 


Im 


Ist iml 


Xf M : 
QU : 


r- 


J 
<H 
- 
| 
Aum 
e 
| 
ine) 
o 


r- 


= 


dy. oe. gh. 
QAP PAN PUAJY | 
Ax :dx:a: gx ! 
AP QP PAP AP: 
Ui ap: wa: 
pO 


5 
2 
1 
2 


XN 
€ 


1 
6 
5 
L7:03—10— 
2 
6 
5 
Lz: 


432 Appendix III: | Atlas of Rooted and Unrooted Maps 
Ax.III.3 Complete bipartite graphs Ky, 7,4 > m,n > 3 


Case m+n — 6: 


Orientable genus 1 





K35:0l— 1-109 K33:01 —02—01 


Orientable genus 2 





K33:02— 01 — 06 


Nonorientable genus 1 





K33:g1—01-—03 


Nonorientable genus 2 


Ax.IIL3 Complete bipartite graphs Km n,4 > m,n > 3 433 





K33:q02— 01 — 18 K33:q2— 02 — 09 


Nonorientable genus 3 





K35:q3— 01— 18 Kost 09 — 18 K33:q3— 03 — 36 





K35 : q3 — 04— 09 K3 : q3 — 05 — 18 K3 3 : q3 — 06 — 09 


Nonorientable genus 4 


434 Appendix III: | Atlas of Rooted and Unrooted Maps 





K33:g4 — 01 — 03 K3 3 :q4— 02—18 K3 3 : q4 — 03 — 36 





K3 3: q4— 04— 18 K3 3 : q4 — 05 — 09 K3 3 :q4— 06 — 18 


Case m -- n — t: 


Orientable genus 1 





K43:01-— 01 — 06 K43:01— 02 — 12 K43:01— 03 — 08 


Orientable genus 2 


435 


Ax.III.3 Complete bipartite graphs Km n,4 > m,n > 3 





— 03 — 12 


Ka3 : 02 


K45:00—02—94 


K43:02—01— 02 





K43:02— 05 — 24 K4,3 : 02 — 06 — 24 


K43 : 02 — 04 — 24 





K43:02— 08 — 24 K4,3 : 02 — 09 — 04 


K43:02— 07 — 12 





48 


K4,3 : 02 — 12 — 


K43:02— 11 — 24 


— 48 


K4,3 : 02 — 10 


436 Appendix III: | Atlas of Rooted and Unrooted Maps 





K43:02— 13 — 24 K43:02— 14 — 48 K43:02— 15 — 08 





K43:02—16—24 


Orientable genus 3 





K4,3 : 03 —01 — 24 K43:03 — 02 — 24 K43:03 — 03 — 08 


Ax.III.4 Wheels W, of order 6 > n > 4 437 





K43:08 — 04 — 24 K4,3 : 03 — 05 — 48 K4,3 : 03 — 06 — 08 





K43 : 03 — 07 — 08 Ka, : 03 — 08 — 24 K43 : 03 — 09 — 06 





K4,3 : 03 — 10 — 02 


Ax.IIL.A4 Wheels W,,5 >n > 4 


Case n = 4(i.e., the complete graph K, of order 4): 


Orientable genus 0 


438 Appendix III: | Atlas of Rooted and Unrooted Maps 





W4:00—01-— 01 


Orientable genus 1 





W3 : ol — 01 — 03 W3 : ol — 02 — 04 


Nonorientable genus 1 





W, : q1 — 01 — 06 W, : q1 — 02 — 01 


Nonorientable genus 2 


Ax.III.4 Wheels W, of order 6 > n > 4 439 





W3 : q2 — 01 — 06 W3 : q2 — 02 — 12 W3 : q2 — 03 — 03 


Nonorientable genus 3 





Wz : q3 — 01 — 12 W3 : q3 — 02 — 12 W3 : q3 — 03 — 04 


Case n = 5: 


Orientable genus 0 


wi 





Ws : 00 — 01 — 04 


440 Appendix III; | Atlas of Rooted and Unrooted Maps 


Orientable genus 1 





W; : ol — 02 — 16 W; : ol — 03 — 04 





W; : ol — 04 — 32 Ws : ol — 05 — 16 Ws : ol — 06 — 16 





W; : o1 — 07 — 08 Ws : ol — 08 — 08 


Orientable genus 2 


Ax.III.4. Wheels W, of order 6 > n > 4 441 





Ws : 02 — 01 — 16 Ws : 02 — 02 — 08 Ws : 02 — 03 — 16 


1 


eI 





Ws : 02 — 04 — 32 


Case n = 6: 


Orientable genus 0 





We : 00 — 01 — 04 


Orientable genus 1 


442 Appendix III: Atlas of Rooted and Unrooted Maps 





We : ol — 03 — 40 





We : ol — 06 — 40 





We : ol — 07 — 20 We : ol — 08 — 20 





Wa : o1 — 10 — 40 We : o1 — 11 — 40 We : ol — 12 — 20 


Ax.III.4 Wheels W, of order 6 > n > 4 443 





Ws : o1 — 13 — 20 Ws : 01 — 14 — 20 


Orientable genus 2 





We : 02 — 01 — 20 We : 02 — 02 — 20 We : 03 — 03 — 20 





We : 02 — 04 — 20 We : 02 — 05 — 04 We : 02 — 06 — 40 


444 Appendix III: Atlas of Rooted and Unrooted Maps 





We : 02 — 07 — 40 We : 02 — 08 — 40 We : 02 — 09 — 20 





Wa : 02 — 10 — 20 We : 02 — 11 — 40 We : 02 — 12 — 40 





Ws : 02 — 13 — 40 Ws : 02 — 14 — 40 Ws : 02 — 15 — 40 





Ws : 02 — 16 — 20 We : 02 — 17 — 40 We : 02 — 18 — 20 


Ax.III.4 Wheels W, of order 6 > n > 4 445 





Ws : 02 — 19 — 20 Ws : 02 — 20 — 20 Ws : 02 — 21 — 40 





We : 02 — 22 — 40 We : 02 — 23 — 40 Ws : 02 — 24 — 40 





We : 02 — 25 — 40 We : 02 — 26 — 40 We : 02 — 27 — 20 





We : 02 — 28 — 20 We : 02 — 29 — 40 We : 02 — 30 — 40 


446 Appendix III: Atlas of Rooted and Unrooted Maps 





We : 02 — 31 — 40 We : 02 — 32 — 40 We : 02 — 33 — 20 





We : 02 — 34 — 20 We : 02 — 35 — 20 We : 02 — 36 — 20 





Ws : 02 — 37 — 20 We : 02 — 38 — 20 We : 02 — 39 — 20 





We : 02 — 40 — 04 We : 02 — 41 — 04 


Ax.IIL5 Complete graphs K„ of order 5 > n > 4 44T 
Ax.III.5 Complete graphs of order 5 > n > 4 

Size 4: (K4 = Wa, W4 is known in Ax.IIL4). 

Size 5: 


Orientable genus 1 





K; : o1 — 02 — 20 Ks : ol — 03 — 10 





Ks : ol — 04 — 02 We : ol — 05 — 20 We : ol — 06 — 05 


Orientable genus 2 





K; : 02 — 02 — 20 Ks : 02 — 03 — 20 


448 Appendix III: Atlas of Rooted and Unrooted Maps 





K; : 02 — 04 — 20 K; : 02 — 05 — 40 Ks : 02 — 06 — 20 





K; : 02 — 07 — 05 K; : 02 — 08 — 40 K; : 02 — 09 — 20 





K; : 02 — 10 — 20 Ks : 02 — 11 — 40 Ks : 02 — 12 — 20 





K; : 02 — 13 — 20 K; : 02 — 14 — 40 K; : 02 — 15 — 40 


Ax.III.5 Complete graphs Kn of order 5 > n > 4 449 





Ks : 02 — 16 — 40 K; : 02 — 17 — 40 Ks : 02 — 18 — 20 


3 1 





ou 


wi 





6 2 


Ks : 02 — 19 — 40 K; : 02 — 20 — 40 Ks : 02 — 21 — 20 





K; : 02 — 22 — 40 K; : 02 — 23 — 40 K; : 02 — 24 — 20 





K; : 02 — 27 — 20 


Ks : 02 — 25 — 10 K; : 02 — 26 — 40 


450 Appendix III: Atlas of Rooted and Unrooted Maps 





Ks : 02 — 28 — 20 Ks : 02 — 29 — 40 Ks : 02 — 30 — 10 





K; : 02 — 31 — 20 


Orientable genus 3 





K; : 03 — 01 — 40 K; : 03 — 02 — 20 K; : 03 — 03 — 40 


Ax.III.5 Complete graphs Kn of order 5 > n > 4 451 





K; : 03 — 04 — 40 K; : 03 — 05 — 20 K; : 03 — 06 — 40 





K; : 03 — 07 — 40 K; : 03 — 08 — 20 K; : 03 — 09 — 10 





K; : 03 — 10 — 40 Ks : 03 — 12 — 20 





452 Appendix III: Atlas of Rooted and Unrooted Maps 


Ax.III.6 Triconnected cubic graphs of size in [6, 15] 
Size 6: (C51 = K4 = W4, W4 is known above). 


Size 9: (Co 2 = K33) 


C91: 


Orientable genus 0 


E 





C9, : 00 — 01 — 03 


Orientable genus 1 





Co, : o1 — 01 — 09 C9, : ol — 02 — 18 C94 :01— 03 — 18 


Ax.III.6 "Iriconnected cubic graphs of size in |6, 15] 453 


EI 





Co, : ol — 04 — 09 C9, : 01 — 05 — 18 


Orientable genus 2 





Ü53 :02— d — 18 C51 : 02 — 02 — 18 


C2: 


Orientable genus 1 





Co,2 : 01 — 01 — 01 Co,2 : o1 — 02 — 09 


Orientable genus 2 


454 Appendix III: Atlas of Rooted and Unrooted Maps 





Co,9 : 02 — 01 — 06 


Size 12: (C124 is the cube) 


C121: 


Orientable genus 0 





Ci24 : 00 — 01 — 12 


Orientable genus 1 





Ci24 : ol — 01 — 24 Ci24 : ol — 02 — 24 Ci24 : o1 — 03 — 12 


Ax.III.6 "Iriconnected cubic graphs of size in |6, 15] 455 





Ci24 of ol = 04 — 24 Ci24 :ol— 05 = 48 Ci24 H ol = 06 — 24 





Ci24 : ol — 07 — 48 Ci24 : ol — 08 — 24 Ci24 : ol — 09 — 48 





Ci24 : ol — 10 — 24 Ci24 : ol — 11 — 24 Ci24 : o1 — 12 — 48 


456 Appendix III; | Atlas of Rooted and Unrooted Maps 





Ci24 : ol — 13 — 12 Ci24 : ol — 14 — 24 Ci24 : o1 — 15 — 12 


Orientable genus 2 





Ci24 :02 — 01 — 28 Ci24 : 02 — 02 — 48 Ci24 : 02 — 03 — 48 





Ci24 : 02 — 04 — 48 Ci24 : 02 — 05 — 48 Ci24 : 02 — 06 — 24 


Ax.III.6 "Iriconnected cubic graphs of size in |6, 15] 457 





Ci24 : 02 — 0T — 24 Ci24 : 02 — 08 — 24 Ci24 : 02 — 09 — 48 





Ci24 :02— 10 — 48 Ci24 : 02 — 11 — 48 Ci24 : 02 — 12 — 48 





Ci24 : 02 — 15 — 48 





Ci24 B 02 — 16 — 48 Ci24 : 02 — 17 — 48 Ci24 : 02 — 18 — 48 


458 Appendix III: Atlas of Rooted and Unrooted Maps 


1 





Ci24 : 02 — 25 — 48 Ci24 : 02 — 27 — 24 


Ci24 : 02 — 28 — 24 


Ax.III.6 "Iriconnected cubic graphs of size in |6, 15] 459 


C123: 


Orientable genus 1 


2 





C2 :01 — 01 — 08 C2 : ol — 02 — 24 C122 : ol — 03 — 48 





C123 :ol— 04 = 48 


Orientable genus 2 





C2 :02 — 01 — 24 C2 : 02 — 02 — 48 C122 :02 — 03 — 24 


460 Appendix III; | Atlas of Rooted and Unrooted Maps 


1 





C23 $ 02 = 04 — 24 C2 :02— 05 = 48 C123 H 02 = 06 — 24 





C123 : 02 — 07 — 24 C2 : 02 — 08 — 08 C122 : 02 — 09 — 48 





C12 $ 02 — 10 — 48 C2 :02— 11— 16 C12,2 : 02 —12— 48 


Orientable genus 1 


Ax.III.6 "Iriconnected cubic graphs of size in |6, 15] 461 





Ci2,3 :01 — 01 — 12 C12,3 : ol — 02 — 24 Ci2,3 : ol — 03 — 24 





Ci2,3 : o1 — 04 — 12 C123 :01 — 05 — 12 


Orientable genus 2 





Ci2,3 :02 — 01 — 24 C12,3 : 02 — 02 — 03 Ci2.3 :02 — 03 — 24 


462 Appendix III; | Atlas of Rooted and Unrooted Maps 





Ci2,3 $ 02 = 04 — 24 C12,3 : 02 — 05 = 48 C123 ` 02 =i 06 — 24 





Ci2,3 : 02 — 07 — 48 C12,3 :02 — 08 — 24 C123 : 02 — 09 — 24 





C2,3 : 02 — 10 — 06 Ci2,3 : 02 — 11 — 24 Ci»,3 :02 — 12 — 24 


Ax.III.6 "Iriconnected cubic graphs of size in |6, 15] 463 


3 





C12,3 :02— 13 = 03 


C124: 


Orientable genus 0 





Ci24 :00— 01— 01 


Orientable genus 1 





Ci24 :01 — 01 — 03 Ci24 : ol — 02 — 12 Ci» : ol — 03 — 03 


464 Appendix III; | Atlas of Rooted and Unrooted Maps 





Ci24 : o1 — 04 — 08 Ci24 :01 — 05 — 01 


Orientable genus 2 





Ciz x 02 = 01 = 04 Ci24 :02 — 02 — 24 Ci24 : o2 az 03 —12 





C24 : 02 — 04 — 04 Cua : 02—05 — 12 054 : 02 — 06 — 24 


Ax.III.6 "Iriconnected cubic graphs of size in |6, 15] 465 





Ci24 : 02 — 07 — 08 Ci24 : 02 — 08 — 12 


Size 15: Two chosen where C4514 is the Petersen graph. 


Cis qi: 


Orientable genus 0 


wi 





6 5 


Cii : 00 — 01 — 03 


Orientable genus 1 





C1511 :01 — 01 — 30 C1511 : ol — 02 — 30 Cikin :02 — 03 — 15 


466 Appendix III; | Atlas of Rooted and Unrooted Maps 





C15,11 :01 — 04 — 15 Cisai :01 — 05 — 15 


Orientable genus 2 


BI 








6 6 5 






: 02 — 03 — 60 


Cis; : 02 — 04 — 30 Cisa1i :02— 05 — 15 Cisai : 02 — 06 — 30 


1 5 


bol 





ou 


I 


C1511 : 02 — 07 — 60 C1511 : 02 — 08 — 30 Cikin : 02 — 09 — 30 


Ax.III.6 "Iriconnected cubic graphs of size in |6, 15] 467 


C1511 : 02 — 10 — 30 Cisai :02 — 11 — 30 





Cisi : 02 — 13 — 60 Cis. : 02 — 14 — 30 Cisai :02 — 15 — 30 





1 6 


Cisi : 02 — 16 — 03 C15,11 :02 — 17 — 60 Cisai : 02 — 18 — 30 





Cis, : 02 — 19 — 30 Cis ii : 02 — 20 — 30 Cikin : 02 — 21 — 30 


468 Appendix III: Atlas of Rooted and Unrooted Maps 





Orientable genus 3 


bol 
NI 


Ol 
I 
Ol 





ou 


I 5 I 


C1511 : 03 — 01 — 60 Cis ii : 03 — 02 — 60 Cikin : 03 — 03 — 60 


Ax.III.6 "Iriconnected cubic graphs of size in |6, 15] 469 


[x] 





1 6 





C1511 : 03 — 05 — 60 Cisai : 03 — 06 — 60 





Cisai : 03 — 07 — 60 Ci5,11 : 03 — 08 — 30 Cisai : 03 — 09 — 30 





C15,11 : 03 — 10 — 30 


C1514: 


Orientable genus 1 


470 Appendix III; Atlas of Rooted and Unrooted Maps 





C15,14 :ol — 01 = 10 


Orientable genus 2 






C15,14 : 02 — 02 — 30 Cis,4 : 02 — 03 — 30 


C15,14 : 02 — 04 — 20 C15,14 : 02 — 05 — 30 Cis,4 : 02 — 06 — 10 


Ax.III.6 "Iriconnected cubic graphs of size in |6, 15] AT] 





C15,14 : 02 — 07 — 06 


Orientable genus 3 


I 
ou 





6 2 


Cis,14 : 03 — 01 — 60 Cis,4 : 03 — 02 — 20 


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9. 


Terminology 


(i, j)-edge, 186 

(i, j)-map, 58 

(i, j) s-map, 123 

(i*, 7*)-map, 58 

(l, s)*-edge, 186 

(x, y)-difference, 348 
(x, y)-difference, 348 
1-addition, 244 
1-product, 244 
C*-oriented planarity, 309 
H-valency, 209 
i-connected, 35 

i-cut, 35 

i-map, 57 

i-section, 301 
i-vertex, 194 

j-face, 194 

j'-map, 58 

n-cube, 141 
N-standard map, 149 
O-standard map, 126 
s-manifold, 29 
V-code, 298 

1-set, 375 

1st level segment, 285 
2-partition, 34, 122 
3-map, 123 


absolute genus, 158 
absolute norm, 307 
admissible, 284 
appending an edge, 86 


arc, 2 

articulate vertex, 187, 262 
articulation, 303 

assignment, 283 

associate net, 121 

associate surface, 283 
associate surface graph, 288 
automorphism, 190, 331 
automorphism group, 192, 331 


balanced, 121 

barfly, 146 

Base map, 317 

base map, 262 

basic adding, 96 

basic appending, 96 

basic contracting, 96 

basic contracting irreducible, 101 
basic deleting, 96 

basic deleting edge irreducible, 101 
basic equivalence, 112 

basic partition, 45 

basic permutation, 43 

basic set, 41 

basic splitting, 96 

basic subtracting, 96 

basic subtracting irreducible, 102 
basic transformation, 96 

Betti number, 17 

bi-matching map, 208 

bi-pole map, 232 

biboundary, 310 


480 


Terminology 


bijection, 42, 101 

bipartite, 100 

bipartite graph, 6 

bipole, 37 

Blissard operator, 346 
boundary identification, 306 
boundary identifier, 306 
bouquet, 37, 213 

branch, 287 

butterfly, 125 


cavity, 209 

cellular embedding, 381 
celluliform, 300 

chromatic number, 40 
circuit, 5 

circuit partition, 7 

class of basic equivalence, 113 
co-order, 113 

cocircuit oriented map, 309 
cocircuit oriented planarity, 309 
cocircular map, 309 
cocirculation, 310 

cocycle, 34, 122 
commutative, 168 
completely symmetrical, 197 
composition, 4 

conjugate axiom, 46 
conjugate), 46 

connected, 5 

contractible, 10 

contractible point, 300 
contracting, 76 

contraction, 300 

coorder, 19, 169 


481 


corank, 17 
cosemiedge, 56 
crosscap polynomial, 289 
crosscap, 13 

crossing number, 40 
cubic, 209 

cut-vertex, 59 
cuttable, 68, 69 
cutting, 68 

cutting face, 69 
cutting graph, 69 
cutting vertex, 68 
cycle, 43 

cyclic number, 17 
cyclic permutation, 43 


decomposition, 37 
decreasing duplition, 112 
decreasing subdivision, 112 
degree, 7 

deleting, 72 

different indices, 19 
different signs, 187 
digraph, 5 

dipole, 341 

directed pregraph, 2 
distinct, 283 

distinct, 376 

double edge, 70 

double H-map, 210 
double leaf, 308 

double link, 70 

double loop, 70 

double side curve), 10 
down-embeddable, 20 


482 


dual, 49, 65, 377 

dual Eulerian map), 100 
dual H-map, 210 

dual map, 66 

dual matching, 208 
dual matching map, 208 
dual regular, 57 

dual trail code, 179 


edge, 2 

edge independent partition), 59 

edge rooted, 202 

edge-automorphism, 330 

edge-isomorphism, 329 

efficient, 176 

elementary equivalence, 378 

elementary transformation, 377 

embedding, 381 

embedding), 17 

empty pregraph, 2 

end segment, 286 

end, 1, 56 

enumerating function, 241 

equilibrious embedding, 64 

equilibrium, 64 

equivalent class, 113 

Euler characteristic, 20 

Euler characteristic(Euler charac- 
teristic), 113 

Euler graph, 7 

Eulerian, 361 

Eulerian characteristic, 377 

even, 361 

even assigned conjecture, 102 

even assigned map, 100 


Terminology 


even pregraph, 7 
eves-cyclic, 231 
exchanger, 287 
expanded tree, 283 
expansion, 383 


face, 17, 49, 377 

face regular, 57 

face representative, 121 

face rooted, 202 

face-algorithm, 172 

face-regular, 195 

fan-flower, 317 

father, 286 

favorable embedding, 60 

favorable map, 60 

favorable segment, 301 

feasible segment, 300 

feasible sequence, 298 

finite pregraph), 2 

finite recursion principle), 6 

finite restrict recursion principle, 
6 

first operation, 47 

first parameter, 349 

fixed point, 191 

flag, 384 

full cavity, 209 


general, 350 

generalized Halin graph, 38 
generated group, 50 

genus polynomial, 289 
genus, 138 

graph, 5 


Terminology 


ground set, 2, 42 


Halin graph, 307 

Halin map, 262 
Hamilton map, 123 
Hamiltonian circuit, 122 
Hamiltonian graph, 122 
handle polynomial, 289 
handle, 13 

harmonic link, 70 
harmonic loop, 70 
hexagonalization), 102 
homotopic), 10 


identity, 192 

immersion, 33 

incidence equation, 34 
incident pair, 100 
included angle, 56 
increasing duplition, 112 
increasing subdivision, 112 
independent face set, 209 
independent pair, 100 
independent set, 100 
induced, 44, 99 

infinite pregraph, 2 
initial end, 2 

inner rooted, 311 

inner vertex, 262 
interlaced, 134 
interpolation theorem, 21 
irreducible, 140, 299 
isomorphic, 385 
isomorphic class, 168 
isomorphism, 165, 330 


483 


joint sequence, 187 
joint tree, 18, 187 


Klein group, 22 
Klein model, 31 


ladder, 293 

left projection, 346 
link, 70 

link bundle, 37, 403 
link map, 194 

loop, 70 

loop bundle, 37 
loop map, 67 
loopless map, 265 


map, 50 

map geometry, 29 

maximum genus, 21 

maximum genus embedding, 162 

maximum nonorientable face num- 
ber embedding, 63 

maximum nonorientable genus, 162 

maximum orientable face number 
embedding, 63 

meson functional, 346 

minimum genus, 21 

minimum genus embedding, 162 

minimum nonorientable genus, 162 

minimum nonorientable genus em- 
bedding, 63 

minimum orientable genus embed- 
ding, 63 

multiplicity, 333 


near quadrangulation, 261 


484 


near regular, 266 

near triangulation, 261 

necklace, 293 

network, 6 

node, 2 

noncuttable, 68, 69 

noncuttable block, 69 

nonorientable form, 290 

nonorientable, 10, 104, 374 

nonorientable favorable genus, 162 

nonorientable genus, 11, 158, 162 

nonorientable pan-tour conjecture, 
163 

nonorientable pan-tour genus, 162 

nonorientable pan-tour maximum 
genus, 162 

nonorientable preproper genus, 162 

nonorientable proper map conjec- 
ture, 163 

nonorientable rule, 153 

nonorientable rule 3, 155 

nonorientable single peak conjec- 
ture, 39 


nonorientable small face proper map 


conjecture, 163 
nonorientable tour genus, 162 
nonorientable tour map conjecture, 

163 
nonorientable tour maximum genus, 

162 
nonseparable, 58 


odd circuit, 34 
orbit, 23, 43 
order, 19, 40, 43, 169, 192 


Terminology 


orientable genus polynomial, 389 

orientable, 10, 104, 374 

orientable genus, 11, 138 

orientable minimum pan-tour genus, 
142 

orientable pan-tour conjecture, 142 

orientable pan-tour maximum genus, 
143 

orientable proper map conjecture, 
142 

orientable single peak conjecture, 
39 

orientable tour conjecture, 142 


pan-Halin map, 262 
pan-tour face), 142 
pan-tour map, 142 
parallel, 134 

partition, 2 

path, 5 

petal bundle, 125, 213 
planar graph, 35 

planar map, 141 

planar pedal bundle, 217 
plane tree, 297 

planted tree, 263, 297 
point partition, 300 
polygonal map, 62 
polyhedral sequence, 298 
polyhedron, 374 
pre-standard, 317 
pre-standard pan-Halin map), 262 
prefect dual matching, 208 
prefect primal matching, 208 
pregenus, 11 


Terminology 


pregraph, 2 

premap, 47 

preproper map, 60 
primal H-map, 210 
primal matching, 208 
primal matching map, 208 
primal regular, 57 
primal trail code, 179 
principle segment, 286 
problem of type 1, 270 
problem of type 2, 273 
proper circuit, 99 
proper cocircuit, 99 
proper embedding, 60 
proper map, 60 


quadcircularity, 308 
quadcirculation, 308 
quadrangulation, 57 
quadregular map, 306 
quadricell, 42 
quaternity, 306 
quinquangulation, 102 


rank, 17 

reduced rule, 134 
reduction, 299 
relative genus, 158 
reversed vector, 258 
right projection, 346 
Ringel ladder, 294 
root, 203 

root edge, 203 

root face, 203 

root vertex, 203 


485 


rooted edge, 202 

rooted element, 201, 202 
rooted face, 202 

rooted map, 203, 236 
rooted set, 201 

rooted vertex, 202 
rotation, 17 


same, 283, 376 

same sign, 187 

second operation, 47 
second parameter, 349 
segmentation edge, 75 
self-dual, 98 

semi-arc isomorphism, 385 
semi-automorphism, 328 
semi-automorphism group, 329 
semi-isomorphic, 327 
semi-isomorphism, 327 
semi-regular map, 61 
semiedge, 56 

separable, 35 

set rooted), 201 

sharp, 194 

shearing loop, 82 

side, 56 

simple map, 265, 305 
simplified barfly, 151 
simplified butterfly, 128 
single edge, 70 

single link, 70 

single loop, 70 

single peak, 39 

single side curve, 10 
single vertex map, 125 


486 


singular link, 70 

singular loop, 70 

size, 19, 40, 169 

skeleton, 377 

small face favorable embedding, 61 

small face proper embedding, 61 

Smarandache geometry, 29 

Smarandache multi-space, 28 

son, 286 

spanning cavity, 209 

spanning tree, 17 

specific face, 262 

splitting, 88 

splitting block, 35 

splitting edge, 88 

splitting pair, 35 

standard, 317 

standard, 303 

standard form, 12 

standard pan-Halin map), 263 

standard splitting block, 35 

standard splitting block decompo- 
sition, 36 

standard surface, 381 

straight line embedding, 17 

strong embedding, 60 

strong map, 60 

subsidiary graph, 121 

super premap, 50 

support, 374 

surface closed curve axiom, 10 

surface embedding graph, 288 

surface embedding, 17 

switch, 376 


Terminology 


symmetric, 190 
symmetrical map, 197 


terminal, 262 
terminal end, 2 
terminal link, 75 
terminal loop, 81 
thickness, 40 

tour, 5, 142 

tour map, 142 

trail, 5 

transitive, 5, 33, 50 
transitive axiom, 50 
transitive block, 66 
transitive decomposition, 66 
travel and traverse rule, 33 
travel, 5 

tree, 17 

tri-pole map, 232 
triangulation, 57 
trivial, 192 

trivial map, 127 
TT-rule, 33 

twist, 383 

twist loop, 81 


under pregraph, 50 
uniboundary, 310 
unicyclic, 324 

unisheet, 230 
up-embeddable, 20, 189 
up-integer, 233 


vertex, 2, 47, 377 
vertex partition, 298 
vertex regular, 57 


Terminology 


vertex rooted, 202 
vertex-algorithm, 172 
vertex-isomorphism, 330 
vertex-regular, 194 


walk, 5 
wheel, 38 
wintersweet, 324 


487 


Abstract: A Smarandache system (3; R) is such a mathematical system with at least 
one Smarandachely denied rule F in R such that it behaves in at least two different ways 
within the same set X, i.e., validated and invalided, or only invalided but in multiple dis- 
tinct ways. A map is a 2-cell decomposition of surface, which can be seen as a connected 
graphs in development from partition to permutation, also a basis for constructing Smaran- 
dache systems, particularly, Smarandache 2-manifolds for Smarandache geometry. As an 
introductory book, this book contains the elementary materials in map theory, including 
embeddings of a graph, abstract maps, duality, orientable and non-orientable maps, iso- 
morphisms of maps and the enumeration of rooted or unrooted maps, particularly, the 
joint tree representation of an embedding of a graph on two dimensional manifolds, which 
enables one to make the complication much simpler on map enumeration. All of these 
are valuable for researchers and students in combinatorics, graphs and low dimensional 


topology. 


In 
9 1 


QU Leu ss) 


| 
46