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Game chromatic number of graphs with locally bounded number of cycles
Information Processing Letters 110 (2010) 757–760
Contents lists available at ScienceDirect
Information Processing Letters www.elsevier.com/locate/ipl
Game chromatic number of graphs with locally bounded number of cycles Konstanty Junosza-Szaniawski, Łukasz Rozej ˙ ∗ Faculty of Mathematics and Information Science, Warsaw University of Technology, Pl. Politechniki 1, 00-661 Warsaw, Poland
a r t i c l e
i n f o
a b s t r a c t
Article history: Received 9 February 2010 Received in revised form 9 June 2010 Accepted 10 June 2010 Available online 17 June 2010 Communicated by M. Yamashita
We consider the coloring game and the marking game on graphs with bounded number of cycles passing through any edge. We prove that the game coloring number of a graph G is at most c + 4, if every edge of G belongs to at most c different cycles. This result covers two earlier bounds on the game coloring number: for trees (c = 0) and for cactuses (c = 1). © 2010 Elsevier B.V. All rights reserved.
Keywords: Combinatorial problems Competitive algorithms Game coloring number Game chromatic number
1. Introduction The following graph coloring game was introduced independently by Brams (see [1]) and Bodlaender [2]. Two players, Alice and Bob, take turns by properly coloring vertices of a graph G with one of k available colors. Alice makes the first move, and she wins when all the vertices get colored. Bob wins if he can prevent it. The game chromatic number of a graph G, denoted by χ g (G ), is the smallest k such that Alice has a winning strategy in the coloring game on G with k colors. The graph coloring game is an extensively studied topic with many nice results, intriguing connections, and challenging open problems (see [1] for the recent survey). Not surprisingly the most challenging are problems concerning planar graphs. It is known that the game chromatic number is bounded for this class, but the optimal estimate still remains a mystery. The best result so far is due to Zhu [7], who proved recently that χ g (G ) 17 for every planar graph G. Curiously in the winning strategy Alice
*
Corresponding author. E-mail address: [email protected] (Ł. Rozej). ˙
0020-0190/$ – see front matter doi:10.1016/j.ipl.2010.06.004
© 2010
Elsevier B.V. All rights reserved.
cares only on choosing the right vertex—not on the color she actually uses. Indeed, consider the marking game in which Alice and Bob alternately mark vertices of a graph G without any restrictions. The goal of Alice is to keep the number of marked vertices around any unmarked vertex as low as possible (at any moment of the game). Bob is making efforts in the opposite direction. Denote by col g (G ) the game coloring number of a graph G, defined as the minimum number k such that Alice can play so that the number of marked vertices around any unmarked vertex is always strictly less than k. Hence, χ g (G ) col g (G ) holds for every graph G. It is somewhat surprising that all currently best bounds for the game chromatic have been obtained via the marking game, in which Alice acts actually like a color-blind person. In this paper we extend one of the earliest results in the area, due to Faigle et al. [3], asserting that col g ( T ) 4 for every tree T . Namely we prove the following theorem. Theorem 1.1. Let G be a graph such that every edge of G lies in at most c different cycles. Then col g (G ) c + 4. Our theorem extends also a recent result of Sidorowicz [6], which says that col g (G ) 5 if G is a cactus (a graph
758
K. Junosza-Szaniawski, Ł. Rozej ˙ / Information Processing Letters 110 (2010) 757–760
in which each edge lies in at most one cycle). We use activation strategy for the appropriate linear order on G (Section 2). The proof of Theorem 1.1 is given in Section 3, while Section 4 is devoted to a brief discussion and comparison with recent result by West et al. [5]. 2. Activation strategy and the order of vertices To prove our theorem we use activation strategy which has been previously used for many classes of graphs [4]. In this strategy vertices are ordered linearly: v 1 , . . . , v n . Vertices can be active or inactive. If i < j, we call v i a backward neighbour of v j and v j a forward neighbour of v i . At the beginning of the game all vertices are inactive. In her first move Alice marks and activates vertex v 1 . Each time Bob marks a vertex v Alice activates it. If v doesn’t have any unmarked backward neighbours, then Alice marks and activates the least unmarked vertex. Suppose on the other hand that v has unmarked backward neighbours and that w is the smallest of them. If w is active, then Alice marks it. If w is not active, then Alice activates it and checks if it has unmarked backward neighbours like she did with vertex v. She continues this process until she stops at a vertex which is active or doesn’t have any unmarked backward neighbours. After Alice’s move any marked vertex is active. Alice visits each vertex at most twice—first time she activates it, second time she colors it. Now we define the linear order which will be used in the activation strategy to prove Theorem 1.1. The order will be constructed iteratively. We start with the empty graph G 0 . In each step we add some vertices and edges of the graph G to the already ordered subgraph until the whole graph is ordered. After k iterations we have a subgraph G k of the graph G, and one of the following situations: 1. There is a cycle C with some vertices in G k which is not a subgraph of G k . Then we put the vertices of V (C ) \ V (G k ) in one of the two possible linear orders on the cycle and we add them to G k , together with all edges between them and V (G k ). In this way we get G k+1 . 2. Situation 1 does not hold. There is a unique edge e connecting a vertex v ∈ V (G ) \ V (G k ) with a vertex in G k . Then we put G k+1 = G k + e + v. 3. Situations 1, 2 do not hold. If there is a vertex v ∈ V (G ) \ V (G k ) we add v to G k together with all edges from v to G k obtaining G k+1 . Let k( v ) := max{k ∈ N: v ∈ / V (G k )}. With G ( v ) we denote the maximal common subgraph of graphs G k( v ) and the component of G containing v. Our construction of the order of vertices guarantees that the graph G ( v ) is connected for any v. 3. Proof of Theorem 1.1 We use activation strategy for the marking game. We will find an upper bound for the number of active neighbours of any unmarked vertex after any of Alice’s moves.
Fig. 1.
Fig. 2.
Let v be any unmarked vertex (Fig. 1). Let N 1 be the set of backward neighbours of v. Let N 2 be the set of forward neighbours of v having some neighbours before v. Let N 3 be the set of forward neighbours of v having no neighbours before v. So, the vertex v is the smallest backward neighbour of vertices from N 3 . After activating any vertex in N 3 Alice visits v. If there were more than one active vertex in N 3 , v would be marked. So, there is at most one active vertex in N 3 . Now, one of the following holds: 1. The vertex v was added in point 3. Then | N 1 | = 0 and | N 2 | = 0. So, v has at most one active neighbour in N 3 . 2. The vertex v was added in point 2. Then | N 1 | = 1 and | N 2 | = 0. Hence, v has at most one active neighbour in N 3 , and at most one active neighbour in N 1 . 3. The vertex v was added in point 1 as a vertex of cycle C . Then v has two neighbours on C . Let w be a backward neighbour of v on C . We have w ∈ N 1 . Let n1 = | N 1 | and n2 = | N 2 |. All the vertices in N 1 and N 2 can be active. We will find upper bound for the number of cycles containing the edge w v. We will show that for each vertex in ( N 1 \ { w }) ∪ N 2 there is a different cycle containing the edge w v. First consider cycles corresponding to vertices in N 1 \ { w }. Let x be any vertex in N 1 \ { w }. If x ∈ V (C ) (Fig. 2) then in C there is a path P xv from v to x containing edge / V (C ) w v and P xv ∪ vx is a cycle containing edge w v. If x ∈ (Fig. 3), then x ∈ V (G ( v )). Let y be the first vertex from V (G ( v )) which appears on cycle C when it is passed from v through w. Let P v y be a path in C from v to y, which contains the edge w v. Graph G ( v ) is connected, so in G ( v ) there is a path P yx from y to x. Thus we get a cycle P v y ∪ P yx ∪ xv containing the edges w v and vx. Thus for every
K. Junosza-Szaniawski, Ł. Rozej ˙ / Information Processing Letters 110 (2010) 757–760
759
Fig. 6. A tree T (c ( T ) = 0) with col g ( T ) = 4 [2].
Fig. 3.
Cycles corresponding to vertices in N 2 are different from cycles corresponding to vertices in N 1 \ { w }. So the number of cycles, to which the edge w v belongs, is equal to n1 + n2 − 1 and it is at most c. We have n1 + n2 c + 1. After Alice’s move v has active neighbours in N 1 , N 2 and at most one in N 3 . The number of v’s active neighbours is at most | N 1 | + | N 2 | + 1 = n1 + n2 + 1 c + 2. After Alice’s move there are at most c + 2 marked neighbours of v. After Bob’s move there are at most c + 3. So col g (G ) c + 4 and the proof is complete. 4. Concluding remarks
Fig. 4.
The two previous results for c = 0 and c = 1 are known to be tight. Fig. 6 presents a tree for which Bob has a winning strategy with 3 colors. For a proof it is enough to consider just a few cases. Fig. 7 contains a cactus for which Bob has a winning strategy with 4 colors. The proof requires analysis of over 20 cases. It remains an open question if bound c + 4 is tight for graphs with c 1. Even when c = 2 answering this question needs a different approach or consideration of a lot of cases. A recent result by Montassier et al. [5] introduces a bound for game chromatic number for graphs with bounded maximum average degree defined as:
Mad(G ) = max d( H ): H G Fig. 5.
vertex in N 1 \ { w } there is a cycle containing w v. These cycles differ at least in edges between v and vertices from N 1 \ { w }. So we have n1 − 1 different cycles containing the edge w v. Next consider cycles corresponding to vertices in N 2 . Let x be any vertex in N 2 , and y one of its neighbours preceding v. If y ∈ V (C ) (Fig. 4), then in C there is path P yv from y to v and containing the edge w v. We have cycle P yv ∪ vx ∪ xy, so there is a cycle containing the edges w v and vx. If y ∈ / C (Fig. 5), then y ∈ V (G ( v )). Let z be the first vertex from G ( v ) which appears on the cycle C when it is passed from v through w. Let P v z be a path in C from v to z containing the edge w v. Graph G ( v ) is connected, so in G ( v ) there is a path P zy from z to y. We have cycle P v z ∪ P zy ∪ yx ∪ xv, so there is a cycle containing the edges w v and vx. Thus for every vertex in N 2 there is a cycle containing w v. These cycles differ at least in edges between v and vertices from N 2 . So we have n2 different cycles containing the edge w v.
where d(G ) is average degree of graph G. The authors proved that: Theorem 4.1 (Montassier, Pecher, Raspaud, West, Zhu). Any 4(k+1)(k+3) simple graph with maximum average degree k2 +6k+6 has game coloring number at most 4 + k. We introduce a class of graphs for which our bound is better then the one from Theorem 4.1. Let G k,l be a graph (see Fig. 8). It is obvious that c (G k,l ) = k, so from Theorem 1.1 col g (G k,l ) k + 4. We will show that the bound given by Theorem 4.1 is greater for any k > 1 and l > 1. It suffices 4(k+1)(k+3) to show that k2 +6k+6 < Mad(G k,l ). We have:
Mad(G k,l ) d(G k,l ) =
=
2kl + (l − 1)(2k + 2) + 2k + 2 kl + l + 1 4k + 4
4k + 4 k+1+
1 l
k+
3 2
.
)(k+3) It is easy to check that 4(kk2++16k < 4k+34 for any k > 1. +6 k+ 2
760
K. Junosza-Szaniawski, Ł. Rozej ˙ / Information Processing Letters 110 (2010) 757–760
Fig. 7. A cactus C (c (C ) = 1) with col g (C ) = 5 [6].
Fig. 8.
References [1] T. Bartnicki, J. Grytczuk, H.A. Kierstead, X. Zhu, The map coloring game, Amer. Math. Monthly (November, 2007). [2] H.L. Bodlaender, On the complexity of some coloring games, Internat. J. Found. Comput. Sci. 2 (1991) 133–147. [3] U. Faigle, U. Kern, H.A. Kierstead, W.T. Trotter, On the game chromatic number of some classes of graphs, Ars Combin. 35 (1993) 143–150.
[4] H.A. Kierstead, A simple competitive graph colouring algorithm, J. Combin. Theory Ser. B 78 (2000) 57–68. [5] M. Montassier, A. Pecher, A. Raspaud, D.B. West, X. Zhu, Decomposition of sparse graphs, with application to game coloring number, http://www.math.uiuc.edu/~west/pubs/publink.html. [6] E. Sidorowicz, The game chromatic number and the game colouring number of cactuses, Inform. Process. Lett. 102 (2007) 147–151. [7] X. Zhu, Refined activation strategy for the marking game, J. Combin. Theory Ser. B 98 (2008) 1–18.
Contents lists available at ScienceDirect
Information Processing Letters www.elsevier.com/locate/ipl
Game chromatic number of graphs with locally bounded number of cycles Konstanty Junosza-Szaniawski, Łukasz Rozej ˙ ∗ Faculty of Mathematics and Information Science, Warsaw University of Technology, Pl. Politechniki 1, 00-661 Warsaw, Poland
a r t i c l e
i n f o
a b s t r a c t
Article history: Received 9 February 2010 Received in revised form 9 June 2010 Accepted 10 June 2010 Available online 17 June 2010 Communicated by M. Yamashita
We consider the coloring game and the marking game on graphs with bounded number of cycles passing through any edge. We prove that the game coloring number of a graph G is at most c + 4, if every edge of G belongs to at most c different cycles. This result covers two earlier bounds on the game coloring number: for trees (c = 0) and for cactuses (c = 1). © 2010 Elsevier B.V. All rights reserved.
Keywords: Combinatorial problems Competitive algorithms Game coloring number Game chromatic number
1. Introduction The following graph coloring game was introduced independently by Brams (see [1]) and Bodlaender [2]. Two players, Alice and Bob, take turns by properly coloring vertices of a graph G with one of k available colors. Alice makes the first move, and she wins when all the vertices get colored. Bob wins if he can prevent it. The game chromatic number of a graph G, denoted by χ g (G ), is the smallest k such that Alice has a winning strategy in the coloring game on G with k colors. The graph coloring game is an extensively studied topic with many nice results, intriguing connections, and challenging open problems (see [1] for the recent survey). Not surprisingly the most challenging are problems concerning planar graphs. It is known that the game chromatic number is bounded for this class, but the optimal estimate still remains a mystery. The best result so far is due to Zhu [7], who proved recently that χ g (G ) 17 for every planar graph G. Curiously in the winning strategy Alice
*
Corresponding author. E-mail address: [email protected] (Ł. Rozej). ˙
0020-0190/$ – see front matter doi:10.1016/j.ipl.2010.06.004
© 2010
Elsevier B.V. All rights reserved.
cares only on choosing the right vertex—not on the color she actually uses. Indeed, consider the marking game in which Alice and Bob alternately mark vertices of a graph G without any restrictions. The goal of Alice is to keep the number of marked vertices around any unmarked vertex as low as possible (at any moment of the game). Bob is making efforts in the opposite direction. Denote by col g (G ) the game coloring number of a graph G, defined as the minimum number k such that Alice can play so that the number of marked vertices around any unmarked vertex is always strictly less than k. Hence, χ g (G ) col g (G ) holds for every graph G. It is somewhat surprising that all currently best bounds for the game chromatic have been obtained via the marking game, in which Alice acts actually like a color-blind person. In this paper we extend one of the earliest results in the area, due to Faigle et al. [3], asserting that col g ( T ) 4 for every tree T . Namely we prove the following theorem. Theorem 1.1. Let G be a graph such that every edge of G lies in at most c different cycles. Then col g (G ) c + 4. Our theorem extends also a recent result of Sidorowicz [6], which says that col g (G ) 5 if G is a cactus (a graph
758
K. Junosza-Szaniawski, Ł. Rozej ˙ / Information Processing Letters 110 (2010) 757–760
in which each edge lies in at most one cycle). We use activation strategy for the appropriate linear order on G (Section 2). The proof of Theorem 1.1 is given in Section 3, while Section 4 is devoted to a brief discussion and comparison with recent result by West et al. [5]. 2. Activation strategy and the order of vertices To prove our theorem we use activation strategy which has been previously used for many classes of graphs [4]. In this strategy vertices are ordered linearly: v 1 , . . . , v n . Vertices can be active or inactive. If i < j, we call v i a backward neighbour of v j and v j a forward neighbour of v i . At the beginning of the game all vertices are inactive. In her first move Alice marks and activates vertex v 1 . Each time Bob marks a vertex v Alice activates it. If v doesn’t have any unmarked backward neighbours, then Alice marks and activates the least unmarked vertex. Suppose on the other hand that v has unmarked backward neighbours and that w is the smallest of them. If w is active, then Alice marks it. If w is not active, then Alice activates it and checks if it has unmarked backward neighbours like she did with vertex v. She continues this process until she stops at a vertex which is active or doesn’t have any unmarked backward neighbours. After Alice’s move any marked vertex is active. Alice visits each vertex at most twice—first time she activates it, second time she colors it. Now we define the linear order which will be used in the activation strategy to prove Theorem 1.1. The order will be constructed iteratively. We start with the empty graph G 0 . In each step we add some vertices and edges of the graph G to the already ordered subgraph until the whole graph is ordered. After k iterations we have a subgraph G k of the graph G, and one of the following situations: 1. There is a cycle C with some vertices in G k which is not a subgraph of G k . Then we put the vertices of V (C ) \ V (G k ) in one of the two possible linear orders on the cycle and we add them to G k , together with all edges between them and V (G k ). In this way we get G k+1 . 2. Situation 1 does not hold. There is a unique edge e connecting a vertex v ∈ V (G ) \ V (G k ) with a vertex in G k . Then we put G k+1 = G k + e + v. 3. Situations 1, 2 do not hold. If there is a vertex v ∈ V (G ) \ V (G k ) we add v to G k together with all edges from v to G k obtaining G k+1 . Let k( v ) := max{k ∈ N: v ∈ / V (G k )}. With G ( v ) we denote the maximal common subgraph of graphs G k( v ) and the component of G containing v. Our construction of the order of vertices guarantees that the graph G ( v ) is connected for any v. 3. Proof of Theorem 1.1 We use activation strategy for the marking game. We will find an upper bound for the number of active neighbours of any unmarked vertex after any of Alice’s moves.
Fig. 1.
Fig. 2.
Let v be any unmarked vertex (Fig. 1). Let N 1 be the set of backward neighbours of v. Let N 2 be the set of forward neighbours of v having some neighbours before v. Let N 3 be the set of forward neighbours of v having no neighbours before v. So, the vertex v is the smallest backward neighbour of vertices from N 3 . After activating any vertex in N 3 Alice visits v. If there were more than one active vertex in N 3 , v would be marked. So, there is at most one active vertex in N 3 . Now, one of the following holds: 1. The vertex v was added in point 3. Then | N 1 | = 0 and | N 2 | = 0. So, v has at most one active neighbour in N 3 . 2. The vertex v was added in point 2. Then | N 1 | = 1 and | N 2 | = 0. Hence, v has at most one active neighbour in N 3 , and at most one active neighbour in N 1 . 3. The vertex v was added in point 1 as a vertex of cycle C . Then v has two neighbours on C . Let w be a backward neighbour of v on C . We have w ∈ N 1 . Let n1 = | N 1 | and n2 = | N 2 |. All the vertices in N 1 and N 2 can be active. We will find upper bound for the number of cycles containing the edge w v. We will show that for each vertex in ( N 1 \ { w }) ∪ N 2 there is a different cycle containing the edge w v. First consider cycles corresponding to vertices in N 1 \ { w }. Let x be any vertex in N 1 \ { w }. If x ∈ V (C ) (Fig. 2) then in C there is a path P xv from v to x containing edge / V (C ) w v and P xv ∪ vx is a cycle containing edge w v. If x ∈ (Fig. 3), then x ∈ V (G ( v )). Let y be the first vertex from V (G ( v )) which appears on cycle C when it is passed from v through w. Let P v y be a path in C from v to y, which contains the edge w v. Graph G ( v ) is connected, so in G ( v ) there is a path P yx from y to x. Thus we get a cycle P v y ∪ P yx ∪ xv containing the edges w v and vx. Thus for every
K. Junosza-Szaniawski, Ł. Rozej ˙ / Information Processing Letters 110 (2010) 757–760
759
Fig. 6. A tree T (c ( T ) = 0) with col g ( T ) = 4 [2].
Fig. 3.
Cycles corresponding to vertices in N 2 are different from cycles corresponding to vertices in N 1 \ { w }. So the number of cycles, to which the edge w v belongs, is equal to n1 + n2 − 1 and it is at most c. We have n1 + n2 c + 1. After Alice’s move v has active neighbours in N 1 , N 2 and at most one in N 3 . The number of v’s active neighbours is at most | N 1 | + | N 2 | + 1 = n1 + n2 + 1 c + 2. After Alice’s move there are at most c + 2 marked neighbours of v. After Bob’s move there are at most c + 3. So col g (G ) c + 4 and the proof is complete. 4. Concluding remarks
Fig. 4.
The two previous results for c = 0 and c = 1 are known to be tight. Fig. 6 presents a tree for which Bob has a winning strategy with 3 colors. For a proof it is enough to consider just a few cases. Fig. 7 contains a cactus for which Bob has a winning strategy with 4 colors. The proof requires analysis of over 20 cases. It remains an open question if bound c + 4 is tight for graphs with c 1. Even when c = 2 answering this question needs a different approach or consideration of a lot of cases. A recent result by Montassier et al. [5] introduces a bound for game chromatic number for graphs with bounded maximum average degree defined as:
Mad(G ) = max d( H ): H G Fig. 5.
vertex in N 1 \ { w } there is a cycle containing w v. These cycles differ at least in edges between v and vertices from N 1 \ { w }. So we have n1 − 1 different cycles containing the edge w v. Next consider cycles corresponding to vertices in N 2 . Let x be any vertex in N 2 , and y one of its neighbours preceding v. If y ∈ V (C ) (Fig. 4), then in C there is path P yv from y to v and containing the edge w v. We have cycle P yv ∪ vx ∪ xy, so there is a cycle containing the edges w v and vx. If y ∈ / C (Fig. 5), then y ∈ V (G ( v )). Let z be the first vertex from G ( v ) which appears on the cycle C when it is passed from v through w. Let P v z be a path in C from v to z containing the edge w v. Graph G ( v ) is connected, so in G ( v ) there is a path P zy from z to y. We have cycle P v z ∪ P zy ∪ yx ∪ xv, so there is a cycle containing the edges w v and vx. Thus for every vertex in N 2 there is a cycle containing w v. These cycles differ at least in edges between v and vertices from N 2 . So we have n2 different cycles containing the edge w v.
where d(G ) is average degree of graph G. The authors proved that: Theorem 4.1 (Montassier, Pecher, Raspaud, West, Zhu). Any 4(k+1)(k+3) simple graph with maximum average degree k2 +6k+6 has game coloring number at most 4 + k. We introduce a class of graphs for which our bound is better then the one from Theorem 4.1. Let G k,l be a graph (see Fig. 8). It is obvious that c (G k,l ) = k, so from Theorem 1.1 col g (G k,l ) k + 4. We will show that the bound given by Theorem 4.1 is greater for any k > 1 and l > 1. It suffices 4(k+1)(k+3) to show that k2 +6k+6 < Mad(G k,l ). We have:
Mad(G k,l ) d(G k,l ) =
=
2kl + (l − 1)(2k + 2) + 2k + 2 kl + l + 1 4k + 4
4k + 4 k+1+
1 l
k+
3 2
.
)(k+3) It is easy to check that 4(kk2++16k < 4k+34 for any k > 1. +6 k+ 2
760
K. Junosza-Szaniawski, Ł. Rozej ˙ / Information Processing Letters 110 (2010) 757–760
Fig. 7. A cactus C (c (C ) = 1) with col g (C ) = 5 [6].
Fig. 8.
References [1] T. Bartnicki, J. Grytczuk, H.A. Kierstead, X. Zhu, The map coloring game, Amer. Math. Monthly (November, 2007). [2] H.L. Bodlaender, On the complexity of some coloring games, Internat. J. Found. Comput. Sci. 2 (1991) 133–147. [3] U. Faigle, U. Kern, H.A. Kierstead, W.T. Trotter, On the game chromatic number of some classes of graphs, Ars Combin. 35 (1993) 143–150.
[4] H.A. Kierstead, A simple competitive graph colouring algorithm, J. Combin. Theory Ser. B 78 (2000) 57–68. [5] M. Montassier, A. Pecher, A. Raspaud, D.B. West, X. Zhu, Decomposition of sparse graphs, with application to game coloring number, http://www.math.uiuc.edu/~west/pubs/publink.html. [6] E. Sidorowicz, The game chromatic number and the game colouring number of cactuses, Inform. Process. Lett. 102 (2007) 147–151. [7] X. Zhu, Refined activation strategy for the marking game, J. Combin. Theory Ser. B 98 (2008) 1–18.