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A Simple Competitive Graph Coloring Algorithm
Journal of Combinatorial Theory, Series B 78, 5768 (2000) doi:10.1006jctb.1999.1927, available online at http:www.idealibrary.com on
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 chromatic number, of a planar graph is at most 18. This is a slight improvement of the current upper bound of 19. Perhaps more importantly, we bound the game coloring number of a graph G in terms of a new parameter r(G). We use this result to give very easy proofs of the best known upper bounds on game coloring number for forests, interval graphs, chordal graphs, outerplanar graphs, and line graphs and to give a new upper bound on the game coloring number of graphs embeddable on orientable 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 vertex set V. Let L # 6(G). The orientation G L =(V, E L ) of G with respect to L is obtained by setting E L =[(v, u): [v, u] # E and v>u in L]. (If this definition seems backwards, think of > as the head of an arrow.) For a vertex u # V, we denote the neighborhood of u in G by N G (v), the outneighborhood of u in G L by N + GL(u), and the inneighborhood of u in G L by + N& GL(u). We denote the various degrees of v by d G (v)= |N G |, d GL(v)= + & & + |N GL |, and d GL(v)= |N GL |. Let V L (u)=[v # V : vu]. The maximum outdegree of a digraph G is denoted by 2 + (G) and the maximum indegree of G is denoted by 2 & (G). When G and G L are clear from the context we will drop the subscripts. Finally let N[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 parameters chromatic number and coloring number. The chromatic number of a graph G is denoted by /(G). The coloring number col(G) of a graph G is defined by col(G)=1+ min 2 + (G L ). L # 6(G)
57 0095-895600 35.00 Copyright 2000 by Academic Press All rights of reproduction in any form reserved.
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Clearly /(G) col(G), since if col(G)=1+2 + (G L ), then First-Fit applied to 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 colors C, by two players Alice and Bob with Alice playing first. The players take turns coloring the vertices of G with colors from C so that no two adjacent vertices have the same color. Bob wins if at some time one of the players has no legal move; otherwise Alice wins when the players eventually create a 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 game is played on G using t colors. The game chromatic number was first introduced by Bodlaender [1]. The ordering game is also played on a graph G with a target t by Alice and Bob with Alice playing first. In this game the players take turns choosing vertices from the shrinking set of unchosen vertices. This creates a linear ordering L on the vertex set of G with x< y iff x is chosen before y. The score of the game is 2 + (G L )+1. Alice wins if the score is at most t; otherwise Bob wins. The game coloring number col g (G) of G is the least cardinal t such that Alice has a winning strategy for the ordering game played on G with target t. The game coloring number was formally introduced by Zhu in [8], although the notion had already been treated informally in [4]. Clearly /(G) / g (G) col g (G) 2(G)+1 and /(G) col(G) col g (G) 2(G)+1. It is also clear that if H is a spanning subgraph of G, then col g (H) col g (G). However, this does not hold for game chromatic number. It is not difficult to check that if G=K t, t and H=G&M, where M 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, planar graphs, and.line graphs. In all these cases the best known bounds on the game chromatic number are obtained by bounding the game coloring number. However this is not always the case. We have already noted that / g (K t, t )=3 while even col(K t, 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 col g (F ) 4 for all forests F. They also showed that col g (G) 3|(G)&2 for all interval graphs G and that there exist interval graphs with col g (G)=2|(G)&2, although these results were actually stated in terms of game chromatic number. Kierstead and Tuza [6] proved that col g (G) 6|(G)&8 for all chordal
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graphs G, although again the result was formally stated in terms of game chromatic number. This was proved by generalizing a new and simpler proof that col g (F) 4 for all forests F. Kierstead and Trotter [7] proved that / g (G) 33 for all planar graphs G. Their proof did not give bounds on 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 argument based on the acyclic chromatic number of planar graphs. Again their argument did not give bounds on the game coloring number of planar graphs. Later Kierstead remarked that the argument in [7] could be improved to also yield an upper bound of 30. Zhu made a major breakthrough when he realized that the techniques used for forests in [4] could also be applied to planar graphs in a much more complicated way. He proved [8] that col g (G) 19 for all planar graphs G. He also proved that if G is embeddable on an orientable surface of genus g then col g (G) w(3 - 1+48g+23)2x=(1+o(1)) - 108g. He used the same technique with Guan [5] to show that col g (G) 7 for all outerplanar graphs G and with Cai [2] he showed that if G is a k-degenerate graph with line graph L(G) then col g (L(G)) 2(G)+3k&1. Finally in [9] he used the technique to show that col g (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 play the ordering game. This strategy is based on a preordering of the vertices of the graph G on which the game is played. As in the proof in [7] we define a rank function on such orderings and then bound the score of the game in terms of the rank of the preordering. This approach allows us to very quickly obtain all the upperbounds discussed above by showing that the various graphs all have preorders with small enough rank. We also obtain the new result that col g (G) 18 for every planar graph G. In addition we show that if G is embeddable on an orientable surface of genus g then col g (G) (3 - 73+96g+41)4=(1+o(1)) - 54g. While our approach is based on Zhu's approach, it is simpler for two important reasons. First it requires only one strategy for Alice, while Zhu uses at least two distinct strategies. Second, the analysis of the score is completely different. While our analysis is based on the rank of the preordering of the vertices, Zhu's analysis 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 the strategy S(L, G) for Alice to use to play the ordering game on G with respect to L as follows.
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Strategy S(L, G). Let U denote the set of unchosen vertices. Alice maintains a subset A/V of active vertices. Initially A :=<. When a new vertex x is put into A we say that x is activated. Once a vertex is activated it will remain active forever. On her first turn Alice activates and chooses the least vertex in the ordering L. Now suppose that Bob has just chosen the vertex b. Alice uses the following algorithm to update A and choose the next vertex. v x :=b; v while x  A do A :=A _ [x]; s(x) :=min L N +[x] & (U _ [b]); x :=s(x) od; v if x{b then choose x else y :=min L 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 in N +[x] & (U _ [b]). If s(x) is inactive she activates it and continues. Eventually she chooses s(x) such that s(x) is active (possibly s(x)=x). If s(x){b, then she chooses s(x); otherwise she chooses the least unchosen vertex, after activating it if it is inactive. We shall show that when Alice uses strategy S(L, G), the score of the coloring game is bounded in terms of the following parameters. Suppose A and B are (not necessarily disjoint) subsets of V. We say that a matching M is a matching from A to B if M saturates A and B"A contains a cover of M. For u # V(G) the matching number m(u, L, G) of u with respect to L in G is defined to be the size of the largest set Z/N &[u] such that there exists a partition [X, Y] of Z and there exist matchings M from X/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 G+L(u)+m(u, L, G) r(L, G)=max r(u, L, G) u#V
r(G)= min r(L, G). L # 6(G)
Theorem 1. For any graph G=(V, E) and ordering L # 6(G), if Alice uses the strategy S(L, G) to play the ordering game on G, then the score will be at most 1+r(L, G). In particular, col g (G) 1+r(G). Proof. Suppose that Alice uses the strategy S(L, G) to play the ordering game on G. We shall show that at any time t any unchosen vertex u is adjacent to at most r(u, L, G) active vertices. Since every vertex chosen by Bob
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immediately becomes active and any vertex chosen by Alice is already active, 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. Since s(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 activated before 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 was activated, s(x) was chosen immediately after x was activated and so before y was activated. Thus when y is activated, s(x) is not available to be assigned 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 in N + (x) & U. Thus s(x) u and so s(x) # V +[u]. Also s(x){x. It follows that [P$, Q$] is a partition of N & (u) & A. So |N & (u) & A| = |P$| + |Q$|. Using our previous claim S P =[(x, s(x)): x # P]
and
S Q =[(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 setting X=P$ and Y=Q$ we have that [X, Y] witnesses (1). Otherwise we have a problem since the definition of m(u, L, G) only allows u to be the head 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 that s(x)=u>s(u). If u # P$ then set X=(P$&[x]) _ [u]
and
Y=Q$;
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otherwise set Y=P$
and
X=(Q$&[ y]) _ [u].
In either case, S X =[(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 of various classes of graphs. Corollary 2 (Faigle, Kern, Kierstead, and Trotter [4]). (V, E) is a forest then col g (G) 4.
If G=
Proof. Let L be an ordering of V such that |N + L (u)| 1, for every vertex u # V. Then r(L, G) 3. K Corollary 3 (Zhu [9]). If G is a chordal graph then col g (G) 3|(G)&1. Proof. Since G is chordal there exists an ordering L # 6(G) such that N+ 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] and s(z) # V + (u) for all z # X. Since N + (z) is a clique and u # N + (z), s(z) is adjacent 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] and s(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]). (V, E) is an interval graph then col g (G) 3|(G)&2.
If G=
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
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and |X| |(G)&1 whenever M=[zs(z): z # X] is a matching with X/N &[u] and s(z) # V + (u) for all z # X. Consider a matching N=[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 i by 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 the claim, |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 outerplanar graph, then col g (G) 7. Proof. Without loss of generality we may assume that every interior face of G has degree 3. Then there exists an ordering L of V such that N+ L [u] is a clique of size at most 3 for every vertex u # V. It suffices to show 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, since otherwise contracting N + (u) to p and considering [ p, u] _ (X _ Y$) we obtain a minor of K 2, 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 oriented graph L(G)=(E9 , F), where two oriented edges (a, b) and (c, d) of G are adjacent vertices of L(H) iff b=c. Similarly, the line graph L(G) of a simple graph G=(V, E) is the simple graph L(G)=(E, F ), where two edges 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 that 2 + (G L )=k. Let H be the directed line graph (shift graph) of G L . Then col g (H) 3k+1. Proof. Let L$ order the vertex set E L of H lexicographically with respect to L, i.e., (a, b)<(c, d ) in L$ if and only if a
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b
Next let N be a matching from Y/N & ((a, b)) to V +[(a, b)]. Then every edge e # N has the form e=(( y, a), (a, c)). Thus |N| d G+L(a) k. Finally let M be a matching from X/N &[(a, b)] to V + ((a, b)). Then every edge e # M has the form e=((x, a), (a, c)), c{b or the form e=((a, b), (b, c)). Thus |M| (d G+L(a)&1)+1 k. Thus r((a, b), L$, H) 3k.
K
Corollary 7 (Cai and Zhu [2]). Let G=(V, E) be a graph with 2(G)=2 and let L # 6(G) be such that 2 + (G L )=k. Let H be the line graph of G L . Then col g (H) 2+3k&1. Proof. Identify the vertex set V(H) with E L . Let L$ order E L lexicographically with respect to L. Fix (a, b) # V(H)=E L . It suffices to show that r((a, b), L$, H ) 3k&2. Let M be a matching from X/N &[(a, b)] to V + ((a, b)), N be a matching from Y/N & ((a, b)) to V +[(a, b)], X and Y be disjoint, and m((a, b), L$, H)= |M| + |N|. Let P=N + ((a, b)). Partition M, N, and P by M a =[(e, f ) # M : a # e]
and
M b =[(e, f ) # M "[(a, b)] : b # e]
N a =[(e, f ) # N : a # e]
and
N b =[(e, f ) # N : b # e]
P a =[e # P : a # e]
and
P b =[e # P : b # e].
Then, noting that if (e, f ) # M b _ N b then e{(a, b), |P b | + |M b | + |N b | d G (b)&1 2&1. We claim that C=[e # E L : e=(a, d )] covers M a _ N a . To see this consider (e, f ) # M a _ N a . If e  C then e has the form e = (c, a). Since f # V +[(a, b)], f must be in C. So |M a | + |N a | 2d G+ (a) 2k.
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Finally |P a | =d G+ (a)&1 k&1. Thus r((a, b), L$, H) |P| + |M| + |N| |P a | + |M a | + |N b | + |P b | + |M b | + |N b | 2+3k&2.
K
Corollary 8. If G=(V, E) is a planar graph, then col g (G) 18. Proof. Fix a planar drawing of G. It suffices to construct a linear ordering L # 6(G) such that r(L, G) 17 as follows. Initially we have a set of chosen vertices C=< and a set of unchosen vertices U=V. At any stage of the construction we choose a vertex u # U, declare that u is bigger than all elements of U&[u] and smaller than all elements of C, and replace both U by U&[u] and C by C _ [u]. Note that at this point we will be able to evaluate r(u, L, G) to check that it is at most 17. Let H be the planar graph obtained from G by deleting all edges between vertices in C, deleting each vertex x # C such that |N G (x) & U| 3, and adding edges between any two nonadjacent vertices of U that are adjacent to the same deleted vertex x. Clearly H is still a planar graph. Let S=[xy # E(H) : x, y # U] A=[xy # E(H) : x # C, y # U, and 4 d H (x) 5] B=[xy # E(H) : x # C, y # U, and d H (x) 6]. Initially charge each vertex v # V(H) with a charge c(v)=d H (v). Next we redistribute the charges as follows: For each edge xy # A move a charge of 1 2 from the endpoint in U to the endpoint in C. Let the new charge of a vertex v be c$(v). Then the total charge is unchanged, but each vertex in C & V(H) has a charge of at least 6. Since : v # V(H)
c$(v)= :
d(v)<6 |V(G)|,
v # V(H)
there exists a vertex u # V(H) such that c$(u) 5.5. Note that u is not in C, since we have carefully arranged that each vertex in C & V(H) has charge at least 6. We choose u. Let _= |[ y # U : uy # S]|, := |[x # C : ux # A]|, and ;= |[x # C : ux # B]|. Then c$(u)=_+ 12 :+;. We claim that r(u, L, G) 3_+:+;+1. Clearly d G+ (u) _. Thus it suffices to show that m(u, L, G) 2_+:+;+1. Consider a set Z/N &[u]
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with |Z| =m(u, L, G) such that there exists a partition [X, Y] of Z and there exist matchings M from X/N &[u] to V + (u) and N from Y/N & (u) to V +[u]. Let N$=N&[xy # N : y=u]. Suppose that zy # M _ N$ with z # Z. If d + (z) 3, then uy # S. Otherwise uz # A _ B. It follows that m(u, L, G)= |Z| |M| + |N$| +1 2_+:+;+1. Finally, we check that 3_+:+;+1 17, provided that _, :, and ; are integers satisfying _+ 12 :+; 5.5
and
0 _, :, ;.
K
Theorem 9. If G is a graph that is embeddable on an orientable surface of genus g 1 then col g (G)
3 - 73+96g+41 . 4
Proof. Let b=(3 - 73+96g+41)4. Fix an embedding of G. We shall construct a preorder L # 6(G) such that when Alice uses the strategy S(L, G) any unchosen vertex v has at most b&1 active neighbors. First note that if v is the i th vertex in L, i.e., |V + L [v]| =i and v is unchosen, then v has at most d + (v)+i active neighbors: Clearly v has at most d + (v) active outneighbors. Every time an inneighbor of v is activated, v is a candidate to be activated, so either v or some vertex in V + (v) is chosen. Thus if v has i active inneighbors, then v is chosen immediately after its i th active inneighbor is activated. Also, by Theorem 1, v has at most r(v, L, G) active neighbors. Thus it suffices to construct L so that if v is the i th vertex in L then either i+d + (v) b&1 or r(v, L, G) b&1. We construct L using the same approach as in the proof of Corollary 8. Suppose that we have a set C of chosen vertices that have already been ordered at the top of L and a set U of unchosen vertices that will be ordered at the bottom of L. Let i= |U|. We must choose the next largest vertex u from U so that either i+d + (u) b&1 or r(u, L, G) b&1. Let G$ be the graph obtained from G by deleting all edges between vertices in C. Next let G" be the graph obtained from G$ by adding edges where necessary so that for each c # C, N G$ (c) contains either a hamiltonian cycle B c or a hamiltonian path on at most two vertices in G". Finally let H=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 d G$ (c) 3 is contained in a face f c of H with boundary B c . Call the face f c the face of c and let F C =[ f c: c # C and d G$ (c) 4]. Notice that a face of H is the face of at
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COMPETITIVE GRAPH COLORING
most one chosen vertex. Let == |E(H)|, ,= |F (H)|, = C = f # FC | f |, and , C = |F C |. 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 # F C . Also : a(v)=2 : v#U
:
f # FC v # f
| f | &3 =2 : ( | f | &3)=2(= C &3, C ). |f| f # FC
Using 2== :
| f | 3,+(= C &3, C ),
f # F (H)
,
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) : (d H (v)+a(v)). v#U
So there exists a vertex u # U such that d H (u)+a(u) (12(g&1))i+6. We choose u. We claim that r(u, L, G)
\
12( g&1) 12( g&1) +6 +1+ 2 +6 i i
\\
+ .
First note that d + (u)=d H (u) w(12( g&1)i+6x, so it suffices to show that 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 and there exist matchings M from X/N &[u] to V + (u) and N from Y/N & (u) to V +[u]. Let N$=N&[xy # N : y=u]. Suppose that zy # M _ N$ with z # Z. If uy  E(H), then d G" (z) 4. So u # f z and f z contributes at least 12 to a(u). It follows that
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H. A. KIERSTEAD
m(u, L, G)= |Z| |M| + |N$| +1 2d H (u)+2a(u)+1 2
\
12( g&1) +6 +1, i
+
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. In some sense the parameters (a, b) measure how far a graph deviates from being a chordal graph. A (0, 0)-pseudo chordal graph is just a chordal graph. Zhu proved that (a, b)-pseudo chordal graphs with clique size k have game coloring number at most 3k+2a+b. We have not been able to duplicate this result with our techniques. This is a class that was defined specifically because Zhu's techniques applied to it. One could also define the class of graphs to which our techniques apply, i.e., graphs with rank at most 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 same important subclasses of graphs. It would be very interesting to clear up these connections. REFERENCES 1. H. L. Bodlaender, On the complexity of some coloring games, in ``Graph Theoretical Concepts in Computer Science'' (R. H. Mohring, Ed.), Lecture Notes in Computer Science, Vol. 484, pp. 3040, Springer-Verlag, New YorkBerlin, 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), 143150. 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), 569584. 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.
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 chromatic number, of a planar graph is at most 18. This is a slight improvement of the current upper bound of 19. Perhaps more importantly, we bound the game coloring number of a graph G in terms of a new parameter r(G). We use this result to give very easy proofs of the best known upper bounds on game coloring number for forests, interval graphs, chordal graphs, outerplanar graphs, and line graphs and to give a new upper bound on the game coloring number of graphs embeddable on orientable 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 vertex set V. Let L # 6(G). The orientation G L =(V, E L ) of G with respect to L is obtained by setting E L =[(v, u): [v, u] # E and v>u in L]. (If this definition seems backwards, think of > as the head of an arrow.) For a vertex u # V, we denote the neighborhood of u in G by N G (v), the outneighborhood of u in G L by N + GL(u), and the inneighborhood of u in G L by + N& GL(u). We denote the various degrees of v by d G (v)= |N G |, d GL(v)= + & & + |N GL |, and d GL(v)= |N GL |. Let V L (u)=[v # V : v
57 0095-895600 35.00 Copyright 2000 by Academic Press All rights of reproduction in any form reserved.
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Clearly /(G) col(G), since if col(G)=1+2 + (G L ), then First-Fit applied to 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 colors C, by two players Alice and Bob with Alice playing first. The players take turns coloring the vertices of G with colors from C so that no two adjacent vertices have the same color. Bob wins if at some time one of the players has no legal move; otherwise Alice wins when the players eventually create a 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 game is played on G using t colors. The game chromatic number was first introduced by Bodlaender [1]. The ordering game is also played on a graph G with a target t by Alice and Bob with Alice playing first. In this game the players take turns choosing vertices from the shrinking set of unchosen vertices. This creates a linear ordering L on the vertex set of G with x< y iff x is chosen before y. The score of the game is 2 + (G L )+1. Alice wins if the score is at most t; otherwise Bob wins. The game coloring number col g (G) of G is the least cardinal t such that Alice has a winning strategy for the ordering game played on G with target t. The game coloring number was formally introduced by Zhu in [8], although the notion had already been treated informally in [4]. Clearly /(G) / g (G) col g (G) 2(G)+1 and /(G) col(G) col g (G) 2(G)+1. It is also clear that if H is a spanning subgraph of G, then col g (H) col g (G). However, this does not hold for game chromatic number. It is not difficult to check that if G=K t, t and H=G&M, where M 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, planar graphs, and.line graphs. In all these cases the best known bounds on the game chromatic number are obtained by bounding the game coloring number. However this is not always the case. We have already noted that / g (K t, t )=3 while even col(K t, 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 col g (F ) 4 for all forests F. They also showed that col g (G) 3|(G)&2 for all interval graphs G and that there exist interval graphs with col g (G)=2|(G)&2, although these results were actually stated in terms of game chromatic number. Kierstead and Tuza [6] proved that col g (G) 6|(G)&8 for all chordal
COMPETITIVE GRAPH COLORING
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graphs G, although again the result was formally stated in terms of game chromatic number. This was proved by generalizing a new and simpler proof that col g (F) 4 for all forests F. Kierstead and Trotter [7] proved that / g (G) 33 for all planar graphs G. Their proof did not give bounds on 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 argument based on the acyclic chromatic number of planar graphs. Again their argument did not give bounds on the game coloring number of planar graphs. Later Kierstead remarked that the argument in [7] could be improved to also yield an upper bound of 30. Zhu made a major breakthrough when he realized that the techniques used for forests in [4] could also be applied to planar graphs in a much more complicated way. He proved [8] that col g (G) 19 for all planar graphs G. He also proved that if G is embeddable on an orientable surface of genus g then col g (G) w(3 - 1+48g+23)2x=(1+o(1)) - 108g. He used the same technique with Guan [5] to show that col g (G) 7 for all outerplanar graphs G and with Cai [2] he showed that if G is a k-degenerate graph with line graph L(G) then col g (L(G)) 2(G)+3k&1. Finally in [9] he used the technique to show that col g (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 play the ordering game. This strategy is based on a preordering of the vertices of the graph G on which the game is played. As in the proof in [7] we define a rank function on such orderings and then bound the score of the game in terms of the rank of the preordering. This approach allows us to very quickly obtain all the upperbounds discussed above by showing that the various graphs all have preorders with small enough rank. We also obtain the new result that col g (G) 18 for every planar graph G. In addition we show that if G is embeddable on an orientable surface of genus g then col g (G) (3 - 73+96g+41)4=(1+o(1)) - 54g. While our approach is based on Zhu's approach, it is simpler for two important reasons. First it requires only one strategy for Alice, while Zhu uses at least two distinct strategies. Second, the analysis of the score is completely different. While our analysis is based on the rank of the preordering of the vertices, Zhu's analysis 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 the strategy S(L, G) for Alice to use to play the ordering game on G with respect to L as follows.
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Strategy S(L, G). Let U denote the set of unchosen vertices. Alice maintains a subset A/V of active vertices. Initially A :=<. When a new vertex x is put into A we say that x is activated. Once a vertex is activated it will remain active forever. On her first turn Alice activates and chooses the least vertex in the ordering L. Now suppose that Bob has just chosen the vertex b. Alice uses the following algorithm to update A and choose the next vertex. v x :=b; v while x  A do A :=A _ [x]; s(x) :=min L N +[x] & (U _ [b]); x :=s(x) od; v if x{b then choose x else y :=min L 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 in N +[x] & (U _ [b]). If s(x) is inactive she activates it and continues. Eventually she chooses s(x) such that s(x) is active (possibly s(x)=x). If s(x){b, then she chooses s(x); otherwise she chooses the least unchosen vertex, after activating it if it is inactive. We shall show that when Alice uses strategy S(L, G), the score of the coloring game is bounded in terms of the following parameters. Suppose A and B are (not necessarily disjoint) subsets of V. We say that a matching M is a matching from A to B if M saturates A and B"A contains a cover of M. For u # V(G) the matching number m(u, L, G) of u with respect to L in G is defined to be the size of the largest set Z/N &[u] such that there exists a partition [X, Y] of Z and there exist matchings M from X/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 G+L(u)+m(u, L, G) r(L, G)=max r(u, L, G) u#V
r(G)= min r(L, G). L # 6(G)
Theorem 1. For any graph G=(V, E) and ordering L # 6(G), if Alice uses the strategy S(L, G) to play the ordering game on G, then the score will be at most 1+r(L, G). In particular, col g (G) 1+r(G). Proof. Suppose that Alice uses the strategy S(L, G) to play the ordering game on G. We shall show that at any time t any unchosen vertex u is adjacent to at most r(u, L, G) active vertices. Since every vertex chosen by Bob
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immediately becomes active and any vertex chosen by Alice is already active, 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. Since s(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 activated before 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 was activated, s(x) was chosen immediately after x was activated and so before y was activated. Thus when y is activated, s(x) is not available to be assigned to s( y). Let P$=N & (u) & P and Q$=N & (u) & Q. Note that for any vertex x # N & (u) & A, the unchosen vertex u is in N + (x) & U. Thus s(x) u and so s(x) # V +[u]. Also s(x){x. It follows that [P$, Q$] is a partition of N & (u) & A. So |N & (u) & A| = |P$| + |Q$|. Using our previous claim S P =[(x, s(x)): x # P]
and
S Q =[(x, s(x)): x # Q]
are matchings from N & (u) to V +[u]. If there do not exist x # P$ and y # Q$ such that s(x)=u=s( y), then setting X=P$ and Y=Q$ we have that [X, Y] witnesses (1). Otherwise we have a problem since the definition of m(u, L, G) only allows u to be the head 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 that s(x)=u>s(u). If u # P$ then set X=(P$&[x]) _ [u]
and
Y=Q$;
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otherwise set Y=P$
and
X=(Q$&[ y]) _ [u].
In either case, S X =[(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 of various classes of graphs. Corollary 2 (Faigle, Kern, Kierstead, and Trotter [4]). (V, E) is a forest then col g (G) 4.
If G=
Proof. Let L be an ordering of V such that |N + L (u)| 1, for every vertex u # V. Then r(L, G) 3. K Corollary 3 (Zhu [9]). If G is a chordal graph then col g (G) 3|(G)&1. Proof. Since G is chordal there exists an ordering L # 6(G) such that N+ 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] and s(z) # V + (u) for all z # X. Since N + (z) is a clique and u # N + (z), s(z) is adjacent 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] and s(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]). (V, E) is an interval graph then col g (G) 3|(G)&2.
If G=
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
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and |X| |(G)&1 whenever M=[zs(z): z # X] is a matching with X/N &[u] and s(z) # V + (u) for all z # X. Consider a matching N=[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 i by 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 the claim, |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 outerplanar graph, then col g (G) 7. Proof. Without loss of generality we may assume that every interior face of G has degree 3. Then there exists an ordering L of V such that N+ L [u] is a clique of size at most 3 for every vertex u # V. It suffices to show 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, since otherwise contracting N + (u) to p and considering [ p, u] _ (X _ Y$) we obtain a minor of K 2, 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 oriented graph L(G)=(E9 , F), where two oriented edges (a, b) and (c, d) of G are adjacent vertices of L(H) iff b=c. Similarly, the line graph L(G) of a simple graph G=(V, E) is the simple graph L(G)=(E, F ), where two edges 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 that 2 + (G L )=k. Let H be the directed line graph (shift graph) of G L . Then col g (H) 3k+1. Proof. Let L$ order the vertex set E L of H lexicographically with respect to L, i.e., (a, b)<(c, d ) in L$ if and only if a
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b
Next let N be a matching from Y/N & ((a, b)) to V +[(a, b)]. Then every edge e # N has the form e=(( y, a), (a, c)). Thus |N| d G+L(a) k. Finally let M be a matching from X/N &[(a, b)] to V + ((a, b)). Then every edge e # M has the form e=((x, a), (a, c)), c{b or the form e=((a, b), (b, c)). Thus |M| (d G+L(a)&1)+1 k. Thus r((a, b), L$, H) 3k.
K
Corollary 7 (Cai and Zhu [2]). Let G=(V, E) be a graph with 2(G)=2 and let L # 6(G) be such that 2 + (G L )=k. Let H be the line graph of G L . Then col g (H) 2+3k&1. Proof. Identify the vertex set V(H) with E L . Let L$ order E L lexicographically with respect to L. Fix (a, b) # V(H)=E L . It suffices to show that r((a, b), L$, H ) 3k&2. Let M be a matching from X/N &[(a, b)] to V + ((a, b)), N be a matching from Y/N & ((a, b)) to V +[(a, b)], X and Y be disjoint, and m((a, b), L$, H)= |M| + |N|. Let P=N + ((a, b)). Partition M, N, and P by M a =[(e, f ) # M : a # e]
and
M b =[(e, f ) # M "[(a, b)] : b # e]
N a =[(e, f ) # N : a # e]
and
N b =[(e, f ) # N : b # e]
P a =[e # P : a # e]
and
P b =[e # P : b # e].
Then, noting that if (e, f ) # M b _ N b then e{(a, b), |P b | + |M b | + |N b | d G (b)&1 2&1. We claim that C=[e # E L : e=(a, d )] covers M a _ N a . To see this consider (e, f ) # M a _ N a . If e  C then e has the form e = (c, a). Since f # V +[(a, b)], f must be in C. So |M a | + |N a | 2d G+ (a) 2k.
COMPETITIVE GRAPH COLORING
65
Finally |P a | =d G+ (a)&1 k&1. Thus r((a, b), L$, H) |P| + |M| + |N| |P a | + |M a | + |N b | + |P b | + |M b | + |N b | 2+3k&2.
K
Corollary 8. If G=(V, E) is a planar graph, then col g (G) 18. Proof. Fix a planar drawing of G. It suffices to construct a linear ordering L # 6(G) such that r(L, G) 17 as follows. Initially we have a set of chosen vertices C=< and a set of unchosen vertices U=V. At any stage of the construction we choose a vertex u # U, declare that u is bigger than all elements of U&[u] and smaller than all elements of C, and replace both U by U&[u] and C by C _ [u]. Note that at this point we will be able to evaluate r(u, L, G) to check that it is at most 17. Let H be the planar graph obtained from G by deleting all edges between vertices in C, deleting each vertex x # C such that |N G (x) & U| 3, and adding edges between any two nonadjacent vertices of U that are adjacent to the same deleted vertex x. Clearly H is still a planar graph. Let S=[xy # E(H) : x, y # U] A=[xy # E(H) : x # C, y # U, and 4 d H (x) 5] B=[xy # E(H) : x # C, y # U, and d H (x) 6]. Initially charge each vertex v # V(H) with a charge c(v)=d H (v). Next we redistribute the charges as follows: For each edge xy # A move a charge of 1 2 from the endpoint in U to the endpoint in C. Let the new charge of a vertex v be c$(v). Then the total charge is unchanged, but each vertex in C & V(H) has a charge of at least 6. Since : v # V(H)
c$(v)= :
d(v)<6 |V(G)|,
v # V(H)
there exists a vertex u # V(H) such that c$(u) 5.5. Note that u is not in C, since we have carefully arranged that each vertex in C & V(H) has charge at least 6. We choose u. Let _= |[ y # U : uy # S]|, := |[x # C : ux # A]|, and ;= |[x # C : ux # B]|. Then c$(u)=_+ 12 :+;. We claim that r(u, L, G) 3_+:+;+1. Clearly d G+ (u) _. Thus it suffices to show that m(u, L, G) 2_+:+;+1. Consider a set Z/N &[u]
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H. A. KIERSTEAD
with |Z| =m(u, L, G) such that there exists a partition [X, Y] of Z and there exist matchings M from X/N &[u] to V + (u) and N from Y/N & (u) to V +[u]. Let N$=N&[xy # N : y=u]. Suppose that zy # M _ N$ with z # Z. If d + (z) 3, then uy # S. Otherwise uz # A _ B. It follows that m(u, L, G)= |Z| |M| + |N$| +1 2_+:+;+1. Finally, we check that 3_+:+;+1 17, provided that _, :, and ; are integers satisfying _+ 12 :+; 5.5
and
0 _, :, ;.
K
Theorem 9. If G is a graph that is embeddable on an orientable surface of genus g 1 then col g (G)
3 - 73+96g+41 . 4
Proof. Let b=(3 - 73+96g+41)4. Fix an embedding of G. We shall construct a preorder L # 6(G) such that when Alice uses the strategy S(L, G) any unchosen vertex v has at most b&1 active neighbors. First note that if v is the i th vertex in L, i.e., |V + L [v]| =i and v is unchosen, then v has at most d + (v)+i active neighbors: Clearly v has at most d + (v) active outneighbors. Every time an inneighbor of v is activated, v is a candidate to be activated, so either v or some vertex in V + (v) is chosen. Thus if v has i active inneighbors, then v is chosen immediately after its i th active inneighbor is activated. Also, by Theorem 1, v has at most r(v, L, G) active neighbors. Thus it suffices to construct L so that if v is the i th vertex in L then either i+d + (v) b&1 or r(v, L, G) b&1. We construct L using the same approach as in the proof of Corollary 8. Suppose that we have a set C of chosen vertices that have already been ordered at the top of L and a set U of unchosen vertices that will be ordered at the bottom of L. Let i= |U|. We must choose the next largest vertex u from U so that either i+d + (u) b&1 or r(u, L, G) b&1. Let G$ be the graph obtained from G by deleting all edges between vertices in C. Next let G" be the graph obtained from G$ by adding edges where necessary so that for each c # C, N G$ (c) contains either a hamiltonian cycle B c or a hamiltonian path on at most two vertices in G". Finally let H=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 d G$ (c) 3 is contained in a face f c of H with boundary B c . Call the face f c the face of c and let F C =[ f c: c # C and d G$ (c) 4]. Notice that a face of H is the face of at
67
COMPETITIVE GRAPH COLORING
most one chosen vertex. Let == |E(H)|, ,= |F (H)|, = C = f # FC | f |, and , C = |F C |. 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 # F C . Also : a(v)=2 : v#U
:
f # FC v # f
| f | &3 =2 : ( | f | &3)=2(= C &3, C ). |f| f # FC
Using 2== :
| f | 3,+(= C &3, C ),
f # F (H)
,
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) : (d H (v)+a(v)). v#U
So there exists a vertex u # U such that d H (u)+a(u) (12(g&1))i+6. We choose u. We claim that r(u, L, G)
\
12( g&1) 12( g&1) +6 +1+ 2 +6 i i
\\
+ .
First note that d + (u)=d H (u) w(12( g&1)i+6x, so it suffices to show that 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 and there exist matchings M from X/N &[u] to V + (u) and N from Y/N & (u) to V +[u]. Let N$=N&[xy # N : y=u]. Suppose that zy # M _ N$ with z # Z. If uy  E(H), then d G" (z) 4. So u # f z and f z contributes at least 12 to a(u). It follows that
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H. A. KIERSTEAD
m(u, L, G)= |Z| |M| + |N$| +1 2d H (u)+2a(u)+1 2
\
12( g&1) +6 +1, i
+
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. In some sense the parameters (a, b) measure how far a graph deviates from being a chordal graph. A (0, 0)-pseudo chordal graph is just a chordal graph. Zhu proved that (a, b)-pseudo chordal graphs with clique size k have game coloring number at most 3k+2a+b. We have not been able to duplicate this result with our techniques. This is a class that was defined specifically because Zhu's techniques applied to it. One could also define the class of graphs to which our techniques apply, i.e., graphs with rank at most 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 same important subclasses of graphs. It would be very interesting to clear up these connections. REFERENCES 1. H. L. Bodlaender, On the complexity of some coloring games, in ``Graph Theoretical Concepts in Computer Science'' (R. H. Mohring, Ed.), Lecture Notes in Computer Science, Vol. 484, pp. 3040, Springer-Verlag, New YorkBerlin, 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), 143150. 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), 569584. 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.