A Simple Competitive Graph Coloring Algorithm

Journal of Combinatorial Theory, Series B 78, 57�68 (2000)

A Simple Competitive Graph Coloring Algorithm

H. A. Kierstead

Department of Mathematics, Arizona State University,Main Campus, P.O. Box 871804, Tempe, Arizona 85287-1804

Received January 13, 1999

We prove that the game coloring number, and therefore the game chromaticnumber, of a planar graph is at most 18. This is a slight improvement of the currentupper bound of 19. Perhaps more importantly, we bound the game coloringnumber of a graph G in terms of a new parameter r(G). We use this result to givevery easy proofs of the best known upper bounds on game coloring number forforests, interval graphs, chordal graphs, outerplanar graphs, and line graphs and togive a new upper bound on the game coloring number of graphs embeddable onorientable surfaces with bounded genus. � 2000 Academic Press

1. INTRODUCTION

For a graph G=(V, E) let 6(G) be the set of linear orderings on the ver-tex set V. Let L # 6(G). The orientation GL=(V, EL) of G with respect toL is obtained by setting EL=[(v, u): [v, u] # E and v>u in L]. (If thisdefinition seems backwards, think of > as the head of an arrow.) For avertex u # V, we denote the neighborhood of u in G by NG(v), the out-neighborhood of u in GL by N +

GL(u), and the inneighborhood of u in GL by

N&GL

(u). We denote the various degrees of v by dG(v)=|NG |, d +GL

(v)=|N +

GL|, and d &

GL(v)=|N &

GL|. Let V +

L (u)=[v # V : v<u] and V &L (u)=

[v # V : v>u]. The maximum outdegree of a digraph G is denoted by2+ (G) and the maximum indegree of G is denoted by 2& (G). When G andGL are clear from the context we will drop the subscripts. Finally letN[u]=N(u) _ [u], N +[u]=N+ (u) _ [u], N&[u]=N & (u) _ [u],V +[u]=V + (u) _ [u], and V &[u]=V & (u) _ [u].

In this article we consider competitive versions of the graph parameterschromatic number and coloring number. The chromatic number of a graphG is denoted by /(G). The coloring number col(G) of a graph G is definedby

col(G)=1+ minL # 6(G)

2+ (GL).

doi:10.1006�jctb.1999.1927, available online at http:��www.idealibrary.com on

570095-8956�00 �35.00

Copyright � 2000 by Academic PressAll rights of reproduction in any form reserved.

Clearly /(G)�col(G), since if col(G)=1+2+ (GL), then First-Fit appliedto the vertices of G in the order L will use at most col(G) colors.

The chromatic game in played on a finite graph G, using a set of colorsC, by two players Alice and Bob with Alice playing first. The players taketurns coloring the vertices of G with colors from C so that no two adjacentvertices have the same color. Bob wins if at some time one of the playershas no legal move; otherwise Alice wins when the players eventually createa proper coloring of G. The game chromatic number of G, denoted /g (G),is the least cardinal t such that Alice has a winning strategy when the gameis played on G using t colors. The game chromatic number was firstintroduced by Bodlaender [1].

The ordering game is also played on a graph G with a target t by Aliceand Bob with Alice playing first. In this game the players take turns choos-ing vertices from the shrinking set of unchosen vertices. This creates alinear ordering L on the vertex set of G with x< y iff x is chosen beforey. The score of the game is 2+ (GL)+1. Alice wins if the score is at mostt; otherwise Bob wins. The game coloring number colg (G) of G is the leastcardinal t such that Alice has a winning strategy for the ordering gameplayed on G with target t. The game coloring number was formallyintroduced by Zhu in [8], although the notion had already been treatedinformally in [4]. Clearly

/(G)�/g (G)�colg (G)�2(G)+1

and

/(G)�col(G)�colg (G)�2(G)+1.

It is also clear that if H is a spanning subgraph of G, thencolg (H)�colg (G). However, this does not hold for game chromaticnumber. It is not difficult to check that if G=Kt, t and H=G&M, whereM is a perfect matching in G, then /g (G)=3, but /g (H)=t.

We shall consider the problem of bounding the game coloring number,and therefore the game chromatic number, of various classes of graphs,including forests, subgraphs of chordal graphs, outerplanar graphs, planargraphs, and.line graphs. In all these cases the best known bounds on thegame chromatic number are obtained by bounding the game coloringnumber. However this is not always the case. We have already noted that/g (Kt, t)=3 while even col(Kt, t)=t+1. Bodlaender [1] showed that/g (F )�5 for all forests F and that there exist trees T with /g (T )=4.Faigle, Kern, Kierstead, and Trotter [4] proved that colg (F )�4 for allforests F. They also showed that colg (G)�3|(G)&2 for all interval graphsG and that there exist interval graphs with colg (G)=2|(G)&2, althoughthese results were actually stated in terms of game chromatic number.Kierstead and Tuza [6] proved that colg (G)�6|(G)&8 for all chordal

58 H. A. KIERSTEAD

graphs G, although again the result was formally stated in terms of gamechromatic number. This was proved by generalizing a new and simplerproof that colg (F )�4 for all forests F. Kierstead and Trotter [7] provedthat /g (G)�33 for all planar graphs G. Their proof did not give boundson the game coloring number. It did use a variant of the coloring number.Dinski and Zhu [3] improved this upper bound to 30 with an argumentbased on the acyclic chromatic number of planar graphs. Again their argu-ment did not give bounds on the game coloring number of planar graphs.Later Kierstead remarked that the argument in [7] could be improved toalso yield an upper bound of 30. Zhu made a major breakthrough when herealized that the techniques used for forests in [4] could also be applied toplanar graphs in a much more complicated way. He proved [8] thatcolg (G) � 19 for all planar graphs G. He also proved that if G isembeddable on an orientable surface of genus g then colg (G)�w(3 - 1+48g+23)�2x=(1+o(1)) - 108g. He used the same techniquewith Guan [5] to show that colg (G)�7 for all outerplanar graphs G andwith Cai [2] he showed that if G is a k-degenerate graph with line graphL(G) then colg (L(G))�2(G)+3k&1. Finally in [9] he used the techni-que to show that colg (G)�3|(G)&1 for all chordal graphs G. In [6]Kierstead and Tuza show that there exists an outerplanar graph G with/g (G)=6 and a planar graph with /g (G)=8.

In this article we present a single simple strategy for Alice to use to playthe ordering game. This strategy is based on a preordering of the verticesof the graph G on which the game is played. As in the proof in [7] wedefine a rank function on such orderings and then bound the score of thegame in terms of the rank of the preordering. This approach allows us tovery quickly obtain all the upperbounds discussed above by showing thatthe various graphs all have preorders with small enough rank. We alsoobtain the new result that colg (G)�18 for every planar graph G. In addi-tion we show that if G is embeddable on an orientable surface of genus gthen colg (G)�(3 - 73+96g+41)�4=(1+o(1)) - 54g. While our approachis based on Zhu's approach, it is simpler for two important reasons. Firstit requires only one strategy for Alice, while Zhu uses at least two distinctstrategies. Second, the analysis of the score is completely different. Whileour analysis is based on the rank of the preordering of the vertices, Zhu'sanalysis is based on a certain partitioning of the edges of the graph.

2. ALICE'S STRATEGY AND THE RANK OF A GRAPH

Fix a graph G=(V, E) and a linear ordering L of V. We first define thestrategy S(L, G) for Alice to use to play the ordering game on G withrespect to L as follows.

59COMPETITIVE GRAPH COLORING

Strategy S(L, G). Let U denote the set of unchosen vertices. Alicemaintains a subset A/V of active vertices. Initially A :=<. When a newvertex x is put into A we say that x is activated. Once a vertex is activatedit will remain active forever. On her first turn Alice activates and choosesthe least vertex in the ordering L. Now suppose that Bob has just chosenthe vertex b. Alice uses the following algorithm to update A and choose thenext vertex.

v x :=b;

v while x � A do A :=A _ [x]; s(x) :=minL N +[x] & (U _ [b]);x :=s(x) od;

v if x{b then choose x

else y :=minL U; if y{A then A :=A _ [ y] fi; choose y fi;

So after Bob chooses a vertex b, Alice first activates b if b is inactive.Whenever she activates a vertex x she defines s(x) to be the least vertex inN+[x] & (U _ [b]). If s(x) is inactive she activates it and continues. Even-tually she chooses s(x) such that s(x) is active (possibly s(x)=x). Ifs(x){b, then she chooses s(x); otherwise she chooses the least unchosenvertex, after activating it if it is inactive.

We shall show that when Alice uses strategy S(L, G), the score of thecoloring game is bounded in terms of the following parameters. Suppose Aand B are (not necessarily disjoint) subsets of V. We say that a matchingM is a matching from A to B if M saturates A and B"A contains a coverof M. For u # V(G) the matching number m(u, L, G) of u with respect to Lin G is defined to be the size of the largest set Z/N&[u] such that thereexists a partition [X, Y] of Z and there exist matchings M fromX/N&[u] to V + (u) and N from Y/N& (u) to V +[u]. The rank r(L, G)of L with respect to G and rank r(G) of G are defined by

r(u, L, G)=d +GL

(u)+m(u, L, G)

r(L, G)=maxu # V

r(u, L, G)

r(G)= minL # 6(G)

r(L, G).

Theorem 1. For any graph G=(V, E) and ordering L # 6(G), if Aliceuses the strategy S(L, G) to play the ordering game on G, then the score willbe at most 1+r(L, G). In particular, colg (G)�1+r(G).

Proof. Suppose that Alice uses the strategy S(L, G) to play the orderinggame on G. We shall show that at any time t any unchosen vertex u is adja-cent to at most r(u, L, G) active vertices. Since every vertex chosen by Bob

60 H. A. KIERSTEAD

immediately becomes active and any vertex chosen by Alice is alreadyactive, this will prove the theorem. The main task is to show that

|N& (u) & A|�m(u, L, G). (1)

Let

P=[x # A : x is activated before s(x)]

and

Q=[x # A : x is activated after s(x)].

We claim that s restricted to P and s restricted to Q are both one-to-one.First consider two vertices x, y # P, where x was activated before y. Sinces(x) was activated immediately after x was activated, either y=s(x) or s(x)was activated before y. Since y was activated before s( y), s(x) was activatedbefore s( y) and thus s(x){s( y). Similarly, consider two vertices x, y # Q,where x was activated before y. Since s(x) was already active when x wasactivated, s(x) was chosen immediately after x was activated and so beforey was activated. Thus when y is activated, s(x) is not available to beassigned to s( y).

Let

P$=N& (u) & P

and

Q$=N& (u) & Q.

Note that for any vertex x # N& (u) & A, the unchosen vertex u is inN+ (x) & U. Thus s(x)�u and so s(x) # V +[u]. Also s(x){x. It followsthat [P$, Q$] is a partition of N& (u) & A. So |N& (u) & A|=|P$|+|Q$|.Using our previous claim

SP=[(x, s(x)): x # P] and SQ=[(x, s(x)): x # Q]

are matchings from N& (u) to V +[u].If there do not exist x # P$ and y # Q$ such that s(x)=u=s( y), then set-

ting X=P$ and Y=Q$ we have that [X, Y] witnesses (1). Otherwise wehave a problem since the definition of m(u, L, G) only allows u to be thehead of an edge in one of the two matchings M and N. Fix x # P$, y # Q$,with s(x)=u=s( y). Since x is activated before u and y is activated after u,x is activated before y. Since u is still unchosen, it must be thats(x)=u>s(u). If u # P$ then set

X=(P$&[x]) _ [u] and Y=Q$;


otherwise set

Y=P$ and X=(Q$&[ y]) _ [u].

In either case, SX=[(x, s(x)): x # X] is a matching from N&[u] to V + (u).So [X, Y] witnesses (1).

Finally note that

|N(u) & A|�d + (u)+|N & (u) & A|�d + (u)+m(u, L, G)=r(u, L, G). K

3. APPLICATIONS

We can now obtain a series of corollaries by calculating the ranks ofvarious classes of graphs.

Corollary 2 (Faigle, Kern, Kierstead, and Trotter [4]). If G=(V, E) is a forest then colg (G)�4.

Proof. Let L be an ordering of V such that |N +L (u)|�1, for every ver-

tex u # V. Then r(L, G)�3. K

Corollary 3 (Zhu [9]). If G is a chordal graph then colg (G)�3|(G)&1.

Proof. Since G is chordal there exists an ordering L # 6(G) such thatN+

GL(u) is a clique for every vertex u. It suffices to show that

r(u, L, G)�3|(G)&2, for any vertex u. First note that d + (u)�|(G)&1.Consider a matching M=[zs(z): z # X], where X/N &[u] ands(z) # V + (u) for all z # X. Since N+ (z) is a clique and u # N+ (z), s(z) isadjacent to u and so [s(z): z # X]/N+ (u). Thus |X|=|[s(z): z # X] |�|(G)&1. Also, if N=[zs(z): z # Y] is a matching with Y/N&[u] ands(z) # V + (u) for all z # Y, then [s(z): z # Y]/N +[u]. Thus|Y |=|[s(z): z # Y] |�|(G). So

r(u, L, G)�3|(G)&2. K

In the special case of interval graphs we can do slightly better.

Corollary 4 (Faigle, Kern, Kierstead, and Trotter [4]). If G=(V, E) is an interval graph then colg (G)�3|(G)&2.

Proof. Let L be the ordering of the intervals in V by left endpoints.Then N +

GL(u) is a clique for every vertex u. It suffices to show that

r(u, L, G)�3|(G)&3, for every vertex u # V. As above d + (u)�|(G)&1

62 H. A. KIERSTEAD

and |X|�|(G)&1 whenever M=[zs(z): z # X] is a matching withX/N&[u] and s(z) # V + (u) for all z # X. Consider a matchingN=[zs(z): z # Y], where Y/N& (u) and s(z) # V +[u] for all z # Y.Choose y # Y such that s( y) has the smallest right endpoint. We claim that[ y] _ [s(z): z # Y] is a clique: Denoting the left endpoint of an interval iby l(i) and the right endpoint by r(i) we have

l(s(z))�l(u)�l( y)�r( y)�r(s(z)).

Thus the right endpoint of y is in every s(z). This proves the claim. By theclaim, |Y |=|[s(z): z # Y]|�|(G)&1 So

r(u, L, G)�3|(G)&3. K

Corollary 5 (Guan and Zhu [5]). If G=(V, E) is an outerplanargraph, then colg (G)�7.

Proof. Without loss of generality we may assume that every interiorface of G has degree 3. Then there exists an ordering L of V such thatN+

L [u] is a clique of size at most 3 for every vertex u # V. It suffices toshow that r(u, L, G)�6, for every vertex u # V. Suppose that M=[xs(x): x # X] is a matching from X/N&[u] to V + (u), N=[ ys( y):y # Y] is a matching from Y/N& (u) to V +[u], and X and Y are disjoint.Since s(z), u # N+[z] and N+[z] is a clique for all vertices z # X _ Y, [s(z):z # X _ Y]/N+[u]. For any x, y # V, |M"[ux]|+|N"[ yu]|�2, sinceotherwise contracting N+ (u) to p and considering [ p, u] _ (X _ Y$) weobtain a minor of K2, 3 , which is impossible in an outerplanar graph. Thus

r(u, L, G)�d + (u)+|M"[ux]|+|N"[ yu]|+2�6. K

The line graph L(G) of an oriented graph G=(V, E9 ) is the orientedgraph L(G)=(E9 , F ), where two oriented edges (a, b) and (c, d) of G areadjacent vertices of L(H) iff b=c. Similarly, the line graph L(G) of asimple graph G=(V, E) is the simple graph L(G)=(E, F ), where twoedges ab and cd of G are adjacent vertices of L(G) iff |[a, b] & [c, d]|=1.

Corollary 6. Let G=(V, E) be a graph and let L # 6(G) be such that2+ (GL)=k. Let H be the directed line graph (shift graph) of GL . Thencolg (H)�3k+1.

Proof. Let L$ order the vertex set EL of H lexicographically withrespect to L, i.e., (a, b)<(c, d ) in L$ if and only if a<c in L or a=c and


b<d in L. Fix (a, b) # V(H)=EL . It suffices to show that r((a, b),L$, H)�3k. Note that

d +HL$

((a, b))=|[(c, d ) # EL : c=b]|=d +GL

(b)�k.

Next let N be a matching from Y/N& ((a, b)) to V +[(a, b)]. Then everyedge e # N has the form e=(( y, a), (a, c)). Thus

|N|�d +GL

(a)�k.

Finally let M be a matching from X/N&[(a, b)] to V + ((a, b)). Thenevery edge e # M has the form e=((x, a), (a, c)), c{b or the forme=((a, b), (b, c)). Thus

|M|�(d +GL

(a)&1)+1�k.

Thus r((a, b), L$, H)�3k. K

Corollary 7 (Cai and Zhu [2]). Let G=(V, E) be a graph with2(G)=2 and let L # 6(G) be such that 2+ (GL)=k. Let H be the line graphof GL . Then colg (H)�2+3k&1.

Proof. Identify the vertex set V(H) with EL . Let L$ order EL

lexicographically with respect to L. Fix (a, b) # V(H)=EL . It sufficesto show that r((a, b), L$, H ) � 3k&2. Let M be a matching fromX/N&[(a, b)] to V + ((a, b)), N be a matching from Y/N& ((a, b)) toV +[(a, b)], X and Y be disjoint, and m((a, b), L$, H)=|M|+ |N|. LetP=N+ ((a, b)). Partition M, N, and P by

Ma=[(e, f ) # M : a # e] and Mb=[(e, f ) # M"[(a, b)] : b # e]

Na=[(e, f ) # N : a # e] and Nb=[(e, f ) # N : b # e]

Pa=[e # P : a # e] and Pb=[e # P : b # e].

Then, noting that if (e, f ) # Mb _ Nb then e{(a, b),

|Pb |+|Mb |+ |Nb |�dG(b)&1�2&1.

We claim that C=[e # EL : e=(a, d )] covers Ma _ Na . To see this con-sider (e, f ) # Ma _ Na . If e � C then e has the form e = (c, a). Sincef # V +[(a, b)], f must be in C. So

|Ma |+|Na |�2d+G (a)�2k.

64 H. A. KIERSTEAD

Finally |Pa |=d +G (a)&1�k&1. Thus

r((a, b), L$, H)�|P|+|M|+|N|

�|Pa |+|Ma |+|Nb |+|Pb |+|Mb |+|Nb |

�2+3k&2. K

Corollary 8. If G=(V, E) is a planar graph, then colg (G)�18.

Proof. Fix a planar drawing of G. It suffices to construct a linear order-ing L # 6(G) such that r(L, G)�17 as follows. Initially we have a set ofchosen vertices C=< and a set of unchosen vertices U=V. At any stageof the construction we choose a vertex u # U, declare that u is bigger thanall elements of U&[u] and smaller than all elements of C, and replaceboth U by U&[u] and C by C _ [u]. Note that at this point we will beable to evaluate r(u, L, G) to check that it is at most 17.

Let H be the planar graph obtained from G by deleting all edges betweenvertices in C, deleting each vertex x # C such that |NG(x) & U|�3, andadding edges between any two nonadjacent vertices of U that are adjacentto the same deleted vertex x. Clearly H is still a planar graph. Let

S=[xy # E(H) : x, y # U]

A=[xy # E(H) : x # C, y # U, and 4�dH(x)�5]

B=[xy # E(H) : x # C, y # U, and dH(x)�6].

Initially charge each vertex v # V(H) with a charge c(v)=dH(v). Next weredistribute the charges as follows: For each edge xy # A move a charge of12 from the endpoint in U to the endpoint in C. Let the new charge of a ver-tex v be c$(v). Then the total charge is unchanged, but each vertex inC & V(H) has a charge of at least 6. Since

:v # V(H)

c$(v)= :v # V(H)

d(v)<6 |V(G)|,

there exists a vertex u # V(H) such that c$(u)�5.5. Note that u is not in C,since we have carefully arranged that each vertex in C & V(H) has chargeat least 6. We choose u. Let _=|[ y # U : uy # S]|, :=|[x # C : ux # A]|,and ;=|[x # C : ux # B]|. Then

c$(u)=_+ 12:+;.

We claim that r(u, L, G)�3_+:+;+1. Clearly d +G (u)�_. Thus it suf-

fices to show that m(u, L, G)�2_+:+;+1. Consider a set Z/N&[u]


with |Z|=m(u, L, G) such that there exists a partition [X, Y] of Z andthere exist matchings M from X/N&[u] to V + (u) and N fromY/N& (u) to V +[u]. Let N$=N&[xy # N : y=u]. Suppose thatzy # M _ N$ with z # Z. If d + (z)�3, then uy # S. Otherwise uz # A _ B. Itfollows that

m(u, L, G)=|Z|�|M|+|N$|+1�2_+:+;+1.

Finally, we check that 3_+:+;+1�17, provided that _, :, and ; areintegers satisfying

_+ 12:+;�5.5 and 0�_, :, ;. K

Theorem 9. If G is a graph that is embeddable on an orientable surfaceof genus g�1 then

colg (G)�3 - 73+96g+41

4.

Proof. Let b=(3 - 73+96g+41)�4. Fix an embedding of G. We shallconstruct a preorder L # 6(G) such that when Alice uses the strategyS(L, G) any unchosen vertex v has at most b&1 active neighbors. Firstnote that if v is the i th vertex in L, i.e., |V +

L [v]|=i and v is unchosen, thenv has at most d + (v)+i active neighbors: Clearly v has at most d + (v)active outneighbors. Every time an inneighbor of v is activated, v is acandidate to be activated, so either v or some vertex in V + (v) is chosen.Thus if v has i active inneighbors, then v is chosen immediately after its i thactive inneighbor is activated. Also, by Theorem 1, v has at most r(v, L, G)active neighbors. Thus it suffices to construct L so that if v is the i th vertexin L then either i+d + (v)�b&1 or r(v, L, G)�b&1.

We construct L using the same approach as in the proof of Corollary 8.Suppose that we have a set C of chosen vertices that have already beenordered at the top of L and a set U of unchosen vertices that will beordered at the bottom of L. Let i=|U|. We must choose the next largestvertex u from U so that either i+d + (u)�b&1 or r(u, L, G)�b&1.

Let G$ be the graph obtained from G by deleting all edges between ver-tices in C. Next let G" be the graph obtained from G$ by adding edgeswhere necessary so that for each c # C, NG$ (c) contains either a hamiltoniancycle Bc or a hamiltonian path on at most two vertices in G". Finally letH=G"&C. Then H is embeddable on an orientable surface of genus g.Moreover this can be done so that each vertex c # C with dG$ (c)�3 is con-tained in a face fc of H with boundary Bc . Call the face fc the face of c andlet FC=[ fc: c # C and dG$ (c)�4]. Notice that a face of H is the face of at

66 H. A. KIERSTEAD

most one chosen vertex. Let ==|E(H)|, ,=|F (H)|, =C=�f # FC| f |, and

,C=|FC |. For a vertex v # U, let

a(v)=2 :v # f # FC

| f |&3| f |

.

Notice that 2( | f |&3)�( | f | )� 12 for all f # FC . Also

:v # U

a(v)=2 :f # FC

:v # f

| f |&3| f |

=2 :f # FC

( | f |&3)=2(=C&3,C).

Using

2== :f # F (H)

| f |�3,+(=C&3,C),

,�2=&(=C&3,C)

3,

and Euler's formula, i&=+,=2(1& g), we have

i+2(g&1)�=+(=C&3,C)

3

6i+12(g&1)� :v # U

(dH(v)+a(v)).

So there exists a vertex u # U such that dH(u)+a(u)�(12(g&1))�i+6. Wechoose u.

We claim that

r(u, L, G)�\12(g&1)i �+6 +1+\2 \12(g&1)

i+6+� .

First note that d + (u)=dH(u)�w(12(g&1)�i+6x, so it suffices to showthat m(u, L, G)�1+w2((12(g&1))�i+6)x. Consider a set Z/N&[u]with |Z|=m(u, L, G) such that there exists a partition [X, Y] of Z andthere exist matchings M from X/N&[u] to V + (u) and N fromY/N& (u) to V +[u]. Let N$=N&[xy # N : y=u]. Suppose thatzy # M _ N$ with z # Z. If uy � E(H), then dG" (z)�4. So u # fz and fz con-tributes at least 1

2 to a(u). It follows that


m(u, L, G)=|Z|� |M|+ |N$|+1

�2dH(u)+2a(u)+1

�2 \12(g&1)i

+6++1,

and so m(u, L, G)�w2((12(g&1))�i+6)x+1.It is easy to check that if i�(13+- 73+96g)�2, then i+d + (u)�b&1

and if i�(13+- 73+96g)�2, then r(u, L, G)�b&1. K

4. REMARKS

In [9] Zhu defined the new class of (a, b)-pseudo chordal graphs. Insome sense the parameters (a, b) measure how far a graph deviates frombeing a chordal graph. A (0, 0)-pseudo chordal graph is just a chordalgraph. Zhu proved that (a, b)-pseudo chordal graphs with clique size khave game coloring number at most 3k+2a+b. We have not been able toduplicate this result with our techniques. This is a class that was definedspecifically because Zhu's techniques applied to it. One could also definethe class of graphs to which our techniques apply, i.e., graphs with rank atmost r. It is not clear what the relationship between these two classes are.However, they must be closely related since they include many of the sameimportant subclasses of graphs. It would be very interesting to clear upthese connections.

REFERENCES

1. H. L. Bodlaender, On the complexity of some coloring games, in ``Graph TheoreticalConcepts in Computer Science'' (R. H. Mo� hring, Ed.), Lecture Notes in Computer Science,Vol. 484, pp. 30�40, Springer-Verlag, New York�Berlin, 1991.

2. L. Cai and X. Zhu, Game coloring index and game chromatic index of graphs, manuscript,1998.

3. T. Dinski and X. Zhu, Game chromatic number of graphs, Discrete Math., in press.4. U. Faigle, U. Kern, H. A. Kierstead, and W. T. Trotter, On the game chromatic number

of some classes of graphs, Ars Combin. 35 (1993), 143�150.5. D. Guan and X. Zhu, The game chromatic number of outerplanar graphs, J. Graph Theory,

in press.6. H. A. Kierstead and Z. Tuza, Game chromatic number and treewidth, manuscript, 1996.7. H. A. Kierstead and W. T. Trotter, Planar graph coloring with an uncooperative partner,

J. Graph Theory 18 (1994), 569�584.8. X. Zhu, Game coloring number of planar graphs, J. Combin. Theory Ser. B, in press.9. X. Zhu, Game coloring number of pseudo partial k-trees, manuscript, 1998.

68 H. A. KIERSTEAD

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A Simple Competitive Graph Coloring Algorithm