Coloring edges of embedded graphs

Coloring Edges ofEmbedded Graphs

Daniel P. Sanders1 and Yue Zhao2*

123 CLIFF ROAD

BELLE TERRE, NY 08544

2DEPARTMENT OF MATHEMATICS,

UNIVERSITY OF CENTRAL FLORIDA

ORLANDO, FL 32816

Received December 3, 1998

Abstract: In this paper, we prove that any graph G with maximum degree��G� � �11� �p

49ÿ24��=2, which is embeddable in a surface � ofcharacteristic �� 1 and satis®es jV�G�j > 2��G�ÿ5�ÿ2

�p6��G�, is

class one. ß 2000 John Wiley & Sons, Inc. J Graph Theory 35: 197±205, 2000

Keywords: edge colorings; class one and class two

1. INTRODUCTION

An edge coloring of a graph is a function assigning values (colors) to the edges ofthe graph in such a way that any two adjacent edges receive different colors. Agraph is edge k-colorable, if there is an edge coloring of the graph with colorsfrom f1; . . . ; kg. A ®nite simple graph G of maximum degree � is class one if itis edge �-colorable. Otherwise, Vizing's Theorem [6] guarantees that it is edge�� 1�-colorable, in which case, it is said to be class two.

In 1965. Vizing [7] (also see [1]) proved that any planar graph of maximumdegree � 8 is class one. In 1970, by using a similar method. Mel'nikov [4]generalized Vizing's result and proved that any projective planar graph ofmaximum degree � 8 is class one. In 1995 by using the discharging method,

ÐÐÐÐÐÐÐÐÐÐÐÐÐÐÐÐÐÐ

*Contract grant sponsor: National Science FoundationContract grant numbers: DMS 9896292, DMS 0096160.

ß 2000 John Wiley & Sons, Inc.

Hind and Zhao [3] proved that any graph of maximum degree � 8 which can beembedded in a surface of characterisitc � � 0 is class one and any graph ofmaximum degree � 9 which can be embedded in a surface of characteristic� � ÿ1 is class one. Similar results on some other surfaces were obtained in [5].Summarizing the previous results, it was proved that if a graph G can beembedded in a surface � of characteristic �� with ÿ5 � �� 1 and if��G� � H�� 2, where

H�� 7� ��49ÿ 24��p

2

$ %;

then G is class one. In general, Mel'nikov [4] proved that if a graph G can beembedded in a surface of characteristic �� 0 and if

��G� > max11� ��

25ÿ 24��p2

$ %;

8� 2��52ÿ 18��p

3

$ %( );

then G is class one. One can easily check that

lim��!ÿ1

8� 2��52ÿ 18��p

3ÿ H��

!� 1:

This raises the following question:Is it true that there exists an integer k � 0 such that if a graph G can be

embedded in a surface � of characteristic �� 0 and ��G� � H�� k, thenG is class one?

In this paper, by applying the discharging method, we will give a simple proofof the following theorem:

Theorem. Any graph G with � � �11� ��49ÿ 24��p �=2, which is embed-

dable in a surface � of characteristic �� 1 and satis®es jV�G�j >2��ÿ 5�ÿ 2

��6��p

, is class one.

This theorem answers the above question in the af®rmative with k � 3 underthe additional assumption that jV�G�j is suf®ciently large as a function of �.

Before proceeding, we introduce the following notation. A critical graph G isa connected graph such that G is class two Gÿ e is class one for each edge e

of G. Surfaces in this paper are compact, connected two manifolds withoutboundary. All embeddings considered in this paper are 2-cell embeddings. Givenan embedded graph G, let V(G), E(G), and F(G) be the vertex set, edge set, andface set of G, respectively. A k-face of an embedded graph is a face with k edgeson its boundary. A k-vertex, � k-vertex or � k-vertex is a vertex of degree k, atleast k or at most k, respectively. We use d(x) to denote the degree of x if x is a

198 JOURNAL OF GRAPH THEORY

vertex or the number of edges on the boundary of x if x is a face. We de®ne dk�x�to be the number of k-vertices adjacent to the vertex x, and similarly with � k and� k in place of k. Let ��G�; ��G� (or �; �) be the maximum degree and minimumdegree of G, respectively.

2. LEMMAS

Lemma 1 (Vizing's Adjacency Lemma). Let G be a critical graph of maximum

degree � and x 2 V�G�. Then

(a) if dk�x� � 1, where k 6� �, then d��x� � �ÿ k � 1:

(b) any vertex of G is adjacent to at least two �-vertices.

Lemma 2 (Mel'nikov [4]). Any graph with � >� 11�

��25ÿ24��p

2

�, which is

embeddable in a surface � of characteristic ��, where ÿ53 � �� 0, is

class one.

Lemma 3. If � � 11� ��49ÿ24�p

2and t 2 �8;��, then

2��ÿ 5� ÿ 2��6��p

� ��ÿ t��ÿ t � 2�t ÿ 6

� ÿ6�

t ÿ 6:

Proof. Let f �t� � ÿ��ÿt��ÿt�2�ÿ6�tÿ6

. Since

f 0�t� � ÿt2 � 12t ��2 ÿ 10�ÿ 12� 6�

�t ÿ 6�2

� ÿt2 � 12t ÿ 12� 42:5� 5:5��49ÿ 24�p ÿ 6�ÿ 55ÿ 5

��49ÿ 24�p � 6�

�t ÿ 6�2

� ÿt2 � 12t ÿ 24:5� 0:5��49ÿ 24�p

�t ÿ 6�2

�ÿt2 � 12t ÿ 24:5� 11� ��

49ÿ 24�p ÿ 11

2

�t ÿ 6�2

� ÿt2 � 12t ÿ 30��

�t ÿ 6�2 � ÿ�t ÿ 6�2 �� 6

�t ÿ 6�2 :

Hence we have f 0�t� � 0 if t � 6� ��6��p

. It is clear that if8 � t < 6� ��

6��p

; f 0�t� > 0 and if 6� ��6��p

< t � �; f 0�t� < 0. Hence

maxt2�8;��

f �t� � f �6��6��p

�:

COLORING EDGES OF EMBEDDED GRAPHS 199

Since

f �6��6��p

�

� ÿ�6��6��p �2 � 2�� 1��6� ��

6��p � ÿ�2 ÿ 2�ÿ 6��

6��p

� ÿ36ÿ 12��6��p ÿ 6ÿ�� 12�� 2�

��6��p � 12� 2

��6��p ÿ�2 ÿ 2�ÿ 6��

6��p

� ÿ30ÿ 10��6��p � 9�� 2�

��6��p ÿ�2 ÿ 6��

6��p

� ÿ23ÿ 10��6��p � 2�

��6��p ÿ ��

49ÿ 24�p��

6��p

� 2��6��p ��ÿ 5� ÿ 2�� 6��

6��p

� 2��ÿ 5� ÿ 2��6��p

:

our lemma follows. &

3. DISCHARGING RULES

Let G be a critical graph with � � 24. Assume that G is embedded in a surface �of characteristic ��. By Euler's formula jV�G�j ÿ jE�G�j � jF�G�j � ��, wehave X

x2V�G� [F�G��4ÿ d�x�� 4��: �1�

We call M�x� � 4ÿ d�x� the initial charge of x. We will reassign a new chargedenoted by M0�x� (or M0) to each x 2 V�G�SF�G� according to the discharging

rules below.

R1. For each 2-vertex v, and for each �-vertex u adjacent to v, send 1 from vto u.

R2. For each 3-vertex v, and for each � ��ÿ 1�-vertex u adjacent to v, send 13

from v to u.R3. For each 3-face f that is incident with a � 4-vertex v and vertices u, w, send

12

from f to each of u, w. It will be convenient later to think of f sending 13

of this 12

directly to each of u, w, and sending the remaining 16

via v.R4. For each 3-face f that is not incident with any � 4-vertex, send 1

3from f to

each vertex incident with it.R5. For each 5-vertex v, send 2

15from v to each vertex adjacent to it.

Claim 1. For each x 2 V�G� with d�x� � 6;M0�x� � 0.


Proof. If x is a 2-vertex, d��x� � 2 by Lemma 1, and so it sends out 2 by R1.Thus M0�x� � 0. If x is a 3-vertex, then d��ÿ1��x� � 3, and x sends out 1 by R2.Thus M0�x� � 0. If x is a 4-vertex, then M0�x� � M�x� � 0. If x is a 5-vertex, thenx receives no charge from vertices, but it sends out 2

3to the vertices adjacent to it

by R5. Since x is incident with at most ®ve 3-faces, x receives at most 53

by R4 andM0�x� � M�x� ÿ �2=3� � �5=3� � 0. If x is a 6-vertex, then x receives no chargefrom vertices, but it receives 1

3from each incident 3-face by R4. Thus x receives at

most 63, and M0�x� � 0. &

Claim 2. For each f 2 F�G�;M0� f � � 0.

Proof. If f is a 3-face, then M� f � � 1. Since f sends out 1 to vertices incidentwith it by R3 and R4, M0� f � � 0. If f is a k-face with k � 4, thenM0�f � � M�f � � 0. &

Claim 3. If x is a k-vertex with 7 � k � 20, then M0�x� � �2�6ÿ k�=3�.Proof. Since x is not adjacent to any � 5-vertices, x receives nothing by R1,

R2, R3, and R5. By R4, x receives 1/3 directly from each incident 3-face. Hencewe have M0�x� � M�x� � �k=3� � �2�6ÿ k�=3�. &

Claim 4. If x is a 21-vertex, then M0�x� � M�x� � R�x�, where

R�x� � 21

3� 2d5�x�

15:

The proof of Claim 4 is easy. Hence we omit it.


R�x� � 22

3� d4�x�

3� 2d5�x�

15: �2�

Proof. Since x is not adjacent to any � 3-vertices, x receives nothing byR1, R2. By R3, for each edge xy joining x to a 4-vertex y, x receives at most 1

3

in total via y from the two faces on either side of xy. By R4, x receives13

directly from each 3-face incident with it and by R5, x receives 215

from

each 5-vertex adjacent to it. Hence we have M0�x� � M�x� � 223� d4�x�

3� 2d5�x�

15.

&


R�x� � 23

3� 2d3�x�

3� d4�x�

3� 2d5�x�

15:


Claim 7. If x is a k-vertex with k � 24, then M0�x� � M�x� � R�x�, where

R�x� � k

3� 5d2�x�

6� 2d3�x�

3� d4�x�

3� 2d5�x�

15: �3�

Since the proof of Claim 6 is not only similar to, but easier than, the one ofClaim 7, we only give the proof of Claim 7.

Proof of Claim 7. The ®nal term in (3) is an upper bound for the contributionto x by 5-vertices. The previous term arises as in (2). For the term in d3�x�, notethat for each edge xy joining x to a 3-vertex y, x receives 1

3from y by R2 and at

most 13

in total via y from the two faces on either side of xy, by R3. For the term ind2�x�, note that d2�x� � 1 by Lemma 1 and if y is a 2-vertex adjacent to x then,since G is simple, edge xy lies in the boundary of at most one 3-face, and so x

receives 1 from y by R1 and at most 12

from the two faces on either side of edgexy by R3. Hence x receives at most kÿ2

3� 1

2� 1 � k

3� 5

6from 3-faces and the

2-vertex y. &

4. PROOF OF THE MAIN RESULT

Proof of the Theorem. Suppose that G is a counter-example to our theoremwith the minimum number of edges. Then G is a critical graph. By [3], [4], [5],and Lemma 2, we can assume that �� ÿ54 and ��G� � 24. By (1), we haveX

x2V�G� [F�G��4ÿ d�x��

Xx2V�G� [F�G�

M�x� � 4��:

For each x 2 V�G�SF�G�, we consider M0�x� which is obtained according tothe discharging rules in Section 3.

If x is a � 6-vertex or a face, by Claim 1 and Claim 2, M0�x� � 0.If x is a k-vertex with 7 � k � 20, by Claim 3, M0�x� � 2�6ÿk�

3.

Let x be a 21-vertex. Since d4�x� � 0 and since x is adjacent to at most one 5-vertex, R�x� � 107

15. Hence we have M0�x� � ÿ 148

15.

Let x be a 22-vertex. By Claim 5, M0�x� � M�x� � R�x�, whereR�x� � 22

3� d4�x�

3� 2d5�x�

15. We will prove that M0�x� � ÿ 31

3.

If d4�x� > 0, then d��x� � 21, and so R�x� � 233

. If d4�x� � 0 and d5�x� > 0,then d��x� � 20, again R�x� � 23

3. Hence we have M0�x� � ÿ 31

3.

Let x be a 23-vertex. By Claim 6, M0�x� � M�x� � R�x�, whereR�x� � 23

3� 2d3�x�

3� d4�x�

3� 2d5�x�

15. We will prove that M0�x� � ÿ 32

3.

If d3�x� > 0, then d��x� � 22, and so R�x� � 233� 2

3� 25

3. If d3�x� � 0 but

d4�x� > 0, then d��x� � 21, and so d4�x� � d5�x� � 2; thus again R�x� � 253

. If


d�4�x� � 0 but d5�x� > 0, then d��x� � 20 and so R�x� � 233� 6

15< 25

3. Since in

all cases R�x� � 253

, we have M0�x� � ÿ 323

.Let x be a k-vertex with k � 24. By Claim 7, we have M0�x� � M�x� � R�x�,

where R�x� � k3� 5d2�x�

6� 2d3�x�

3� d4�x�

3� 2d5�x�

15.

If d2�x� > 0, then d��x� � �ÿ 1, so R�x� � k3� 5

6. Thus M0�x� � 29ÿ4k

6.

If d2�x� � 0 but d3�x� > 0, then d��x� � �ÿ 2, and R�x� � k3� 4

3. Thus

M0�x� � 16ÿ2k3

.If d�3�x� � 0 and d4�x� > 0, then d��x� � �ÿ 3. Hence M0�x� � 15ÿ2k

3.

If d�4�x� � 0 and d5�x� > 0, then d��x� � �ÿ 4 and so R�x� � k3� 8

15. Hence

M0�x� � 68ÿ10k15

< 14ÿ2k3

.Finally, if d�5�x� � 0;M0�x� � 2�6ÿk�

3.

Hence we have M0�x� � 0 for each x 2 V�G�SF�G�. According to �, weconsider the following cases.

If � � 2, by Lemma 1, one can obtain that G has at least � �-vertices. HenceXx2V�G�[F�G�

M�x� �X

x2V�G�[F�G�M0�x�

�Xx2V�

M0�x� � �16ÿ 2�

3

� �� ÿ 1

3�2�2 ÿ 16�� 4�ÿ

��49ÿ 24��

p� 1 < 4�;

a contradiction.If � � 3, then G has at least ��ÿ 1� �-vertices. HenceX

x2V�G�[F�G�M�x� �

Xx2V�G�[F�G�

M0�x�

�Xx2V�

M0�x� � ��ÿ 1� 16ÿ 2�

3

� �� ÿ 1

3�2�2 ÿ 18�� 16� � 4�ÿ 1

3�2� 2

��49ÿ 24��

p� < 4�;

a contradiction.If � � 4, then G has at least ��ÿ 2� �-vertices. HenceX



M0�x�

�Xx2V�

M0�x� � ��ÿ 2� 15ÿ 2�

3

� �� ÿ 1

3�2�2 ÿ 19�� 30� � 4�ÿ 1

3

21

2� 3

2

��49ÿ 24��

p� �< 4�;

a contradiction.


If � � 5, then G has at least ��ÿ 3� �-vertices. Hence


M�x� �X


�Xx2V�

M0�x� � ��ÿ 3� 14ÿ 2�

3

� ÿ 1

3�2�2 ÿ 20�� 42� � 4�ÿ 1

3�17�

��49ÿ 24��

p� < 4�;

a contradiction.If � � 6, then G has at least ��ÿ 4� �-vertices. Hence


M�x� �X


�Xx2V�

M0�x� � ��ÿ 4� 12ÿ 2�

3

� �� ÿ 1

3�2�2 ÿ 20�� 48� < 4�;

a contradiction.If � � 7, then G has at least ��ÿ 5� �-vertices. Hence


M�x� �X


�Xx2V�

M0�x� � ��ÿ 5� 12ÿ 2�

3

� �� ÿ 1

3�2�2 ÿ 22�� 60� � 4�ÿ 8 < 4�;

a contradiction.Hence we assume that 8 � � � �. Since jV�G�j > 2��ÿ 5� ÿ 2

��6��p

, wehave X



M0�x�

� 2�6ÿ ��3

�jV�G�j ÿ�� ÿ 2� � 2�6ÿ��3

��ÿ � � 2�

� 2

3fÿ�� ÿ 6�jV�G�j ÿ ��ÿ ��ÿ � � 2�g


� 2

3ÿ�� ÿ 6� jV�G�j � ��ÿ ��ÿ � � 2�

� ÿ 6

� �� <

2

3ÿ�� ÿ 6� 2��ÿ 5� ÿ 2

��6��p

� ��ÿ ��ÿ � � 2�� ÿ 6

� �� :

� 2

3ÿ�� ÿ 6� ÿ6��

� ÿ 6

� �� 4��;

a contradiction. Hence our theorem is true. &

References

[1] S. Fiorini and R. J. Wilson, Edge-colorings of graphs, Pitman, 1977.

[2] A. G. Chetwynd and A. J. W. Hilton, Regular graphs of high degree are 1-factorizable, Proc London Math Soc 50 (3) (1985), 193±206.

[3] H. Hind and Y. Zhao, Edge colorings of graphs embeddable in a surface of lowgenus, Discrete Math 190 (1998), 107±114.

[4] L. S. Mel'nikov, The chromatic class and location of a graph on a closedsurface. Mat Zametki 7 (1970), 671±681 / Math Notes 7 (1970), 405±411.

[5] Z. Yan and Y. Zhao, Edge colorings of embedded graphs, Graphs and Combin16 (2000), 245±256.

[6] V. G. Vizing, On an estimate of the chromatic class of a p-graph. DiskretAnaliz 3 (1964), 25±30.

[7] V. G. Vizing, Critical graphs with given chromatic class, Diskret Analiz 5(1965), 9±17.


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Coloring edges of embedded graphs