The Coloring Game on Matroids 3

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This number is known as the arboricity of the underlaying graph G. The chromatic number of a matroid can be easily expressed in terms of its rank function. Edmonds [4] gave an explicit formula, extending a theorem of Nash- Williams [10] for graph arboricity. Seymour [12] (see also [7]) proved that the chromatic number χ(M) of a matroid M is equal to the list chromatic number ch(M) – another well-known concept from graph theory, initiated by Vizing [13], and independently by Erd˝os, Rubin and Taylor [5]. In this note we study a game-theoretic variant of the chromatic number of a matroid (for other variants see ex. [8, 9]). It is defined by a game beetwen two players – Alice and Bob. They alternately color elements of the ground set E of a matroid M using a fixed set of colors C. The only rule that both players have to obey is that at any moment of the play the partial coloring is proper. Who starts does not influence our results, but to make the definition strict suppose that Alice makes the first move. The game ends when the whole matroid has been colored, or if they arrive to a partial coloring that cannot be further extended (what happens when trying to color any uncolored element with any possible color results in a monochromatic circuit of M). Alice wins in the first case, while Bob in the second. The game chromatic number of a matroid M, denoted by χg(M), is the minimum size of the set of colors C for which Alice has a winning strategy. The above game is a matroidal analog of a well-studied graph game coloring, which was introduced by Brams (cf. [6]) for planar graphs with a motivation to give an easier proof of the four color theorem. The game was independently reinvented by Bodlaender [3], and since then the topic developed into several directions leading to interesting results and challenging open problems (see a recent survey [1]). The first step in studying game chromatic number of matroids was made by Bartnicki, Grytczuk, and Kierstead [2]. They showed that for every graphic matroid M inequality χg(M) ≤ 3χ(M) holds. In Theorem 1 we improve and extend this result by proving that for every loopless matroid M we have χg(M) ≤ 2χ(M). This gives a nearly tight bound, since in Theorem 5 we provide a family of matroids Mk (for k ≥ 3) satisfying χ(Mk) = k and χg(Mk)=2k − 1. Our bounds remain true also for the fractional parameters, as well as for list version of the game chromatic number.

2. A strategy for Alice To achieve our goal we shall need a more general version of the matroid coloring game. Let M1,...,Md be matroids on the same ground set E, and let {1,...,d} be the set of colors. As before, the players alternately color elements of E, but now for all i the set of elements colored with i must be independent in the matroid Mi. As before, Alice wins if at the end of the game the whole set E is colored, Bob wins otherwise. We call this game coloring game on M1,...,Md. The initial game on M with d colors coincides with the coloring game on M1 = ··· = Md = M.

Theorem 1. Let M1,...,Md be matroids on a ground set E. If there exist sets V1,...,Vd, with Vi independent in Mi, such that V1 ∪···∪ Vd = E ∪ E as multisets, then Alice has a winning strategy in the coloring game on M1,...,Md. In particular, for every loopless matroid M we have χg(M) ≤ 2χ(M). THE COLORING GAME ON MATROIDS 3

Proof. Let us ﬁx a 2-covering of E by sets V1,...,Vd independent in corre- sponding matroids. Let Ui be the set of elements colored (at a ﬁxed moment of the play) with i. Then C = U1 ∪···∪ Ud is the set of all colored elements. Alice will try to keep the following invariant after each of her moves:

(2.1) for each i, the set Ui ∪ (Vi \ C) is independent in Mi.

Moreover, element e ∈ Vi ∩ Vj will be colored by Alice only with i or j. Observe first that if the condition (2.1) holds and there is an uncolored element e (e ∈ Vi for some i), then the player can always make an ‘obvious’ move, namely to color e with i. After this move the condition (2.1) remains true. To prove that Alice can keep the condition (2.1) assume that the condition held after her previous move (let Ui and C be defined at this moment) and then Bob colored an element e with color j. If (Uj ∪ e) ∪ (Vj \ C) is independent in Mj, then (2.1) still holds and we use the above observation. When (Uj ∪ e) ∪ (Vj \ C) is dependent in Mj, then by the augmentation axiom we can extend the independent set Uj ∪ e from the independent set Uj ∪ (Vj \ C) in Mj. The extension equals to (Uj ∪ e) ∪ (Vj \ C) \ f for some f ∈ Vj \ C. Since sets V1,...,Vd form a 2-covering we know that f ∈ Vl for some l 6= j. Now Alice has an admissible move, and her strategy is to color f with color l. It is easy to observe that after her move the condition (2.1) is preserved. To get the second part of the assertion suppose that χ(M) = k. By the first part of the assertion applied to M1 = ··· = M2k = M we infer that Alice has a winning strategy with 2k colors, and therefore χg(M) ≤ 2k. As usually in this kind of games the following question seems to be natural and non trivial (for graph coloring game it was asked by Zhu [14]). Question 2. Suppose Alice has a winning strategy in the coloring game on a matroid M with k colors. Does she also have a winning strategy with l > k colors? We generalize our result to list version. The rules of the game between Alice and Bob do not change, except that now each element has its own list of available colors. So, instead of a fixed set of colors C, each element e ∈ E can be colored by Alice or Bob only with a color from its list L(e). The minimum number k for which Alice has a winning strategy for every assignment of lists of size k is called the game list chromatic number of M, and denoted by chg(M). Clearly, chg(M) ≥ χg(M) for every loopless matroid M, however the same upper bound as for χg(M) works also for list parameter.

Corollary 3. For every loopless matroid M we have chg(M) ≤ 2χ(M). Proof. Let L be a list assignment of size 2χ(M). Without loss of generality we can assume that lists L(e) are contained in the set {1,...,d} for some d. Bya theorem of Seymour [12], there is a 2-covering of E by sets V1,...,Vd, such that elements of Vi have color i on their lists and Vi is independent in M. Let Mi be the matroid M restricted to the set Vi with the ground set trivially extended to E. By Theorem 1 Alice has a winning strategy in the coloring game on lists L. In view of Seymour’s Theorem [12] we ask the following natural question. Question 4. Does for every matroid M the game list chromatic number equals to the game chromatic number chg(M)= χg(M)? 4 MICHALLASO N´

3. A strategy for Bob

We present a family of transversal matroids Mk for k ≥ 3, with χ(Mk) = k and χg(Mk) ≥ 2k − 1. This slightly improves the lower bound from the paper of Bartnicki, Grytczuk and Kierstead [2]. They gave an example of graphic matroids Hk satisfying χ(Hk)= k and χg(Hk) ≥ 2k − 2 for every k ≥ 1. Fix k ≥ 3. Let D1,...,D3k(2k−1) be disjoint sets such that each set Di = {d1,i,...,dk,i} has exactly k elements. Let C = {c1,1,...,ck,2k−1} be a set with k(2k − 1) elements, disjoint from sets Di. Let E be the union of sets C and D1,...,D3k(2k−1). Let Mk be the transversal matroid (see [11] for a deﬁnition) on a ground set E with multiset of subsets A consisting of sets D1,...,D3k(2k−1) and (2k − 1) copies of E. In other words, a subset I ⊂ E is independent in Mk if there is some subset J ⊂ I with |J|≤ 2k − 1, such that I \ J contains no elements of C and at most one element of each Di.

Theorem 5. For k ≥ 3 matroid Mk satisfies χ(Mk)= k and χg(Mk) ≥ 2k −1. Proof. To prove the first part observe that we can partition the ground set E into k independent sets Vi = {ci,1,...,ci,2k−1, di,1,...,di,3k(2k−1)}. To prove the second part notice that the rank of C ∪ Di equals to 2k, so if there are t elements from Di colored with i, then there are at most 2k − t elements from C colored with i. This suggests that Bob should try to color each Di with one color. Suppose {1,...,h} is the set of colors in the game, and h ≤ 2k − 2. We will describe a winning strategy for Bob. Alice will always loose the game because she will be not able to color all elements of the set C. Assume first that Alice colors only elements from the set C, and her goal is to color all of them. It is the main case to understand. Denote by di and ci the number of elements colored with i in Di and C respectively (at some moment of the play). Bob wants to keep the following invariant after each of his moves:

(3.1) di ≥ ci, for every color i.

It is easy to see that he can always do it, and in fact there is an equality di = ci for every i. Bob just mimics Alice’s moves. Whenever she colors some c ∈ C with i he responds by coloring an element of Di with i. He can do it, because when after Alice’s move ci = di +1 for some i, then ci + di ≤ 2k. But, then also ci + (di + 1) ≤ 2k, so elements colored with i are independent, and di +1 ≤ k, so there was an uncolored element in Di. Observe that when Bob plays with this strategy, then for each color i we have ci + di ≤ 2k and ci ≤ di, so as a consequence ci ≤ k. This means that Alice can color only hk elements of C, thus she looses. It remains to justify that coloring elements of D1 ∪···∪ D3k(2k−1) by Alice can not help her in coloring elements of C. To see this we have to modify the invariant that Bob wants to keep. We assume k ≥ 4, because for k = 3 more careful case analysis is needed. Denote by di,fi the number of elements in Di ∪ Di+h ∪ Di+2h colored with i and with some other color respectively, and ci as before. Now the invariant that Bob wants to keep after each of his moves is the following:

(3.2) di ≥ ci + fi or di ≥ k +2, for every color i. Analogously to the previous case one can show that Bob can keep this invariant. He just have to obey one more rule. Whenever Alice colors an element of Di ∪ Di+h ∪ THE COLORING GAME ON MATROIDS 5

Di+2h, then Bob colors another element of this set with i always trying to keep ǫi, the number of sets among Di,Di+h,Di+2h with at least one element colored with i, as low as possible. This completes the description of Bob’s strategy. Notice that fi ≥ ǫi − 1. Condition (3.2) gives the same consequence as (3.1). Indeed, let Di be a the union of those of Di,Di+h,Di+2h, which have an element colored with i. Matroid Mk restricted to the set C ∪ Di has rank 2k − 1+ ǫi. If di ≥ ci + fi, then

2k − 1+ ǫi ≥ ci + di ≥ 2ci + fi ≥ 2ci + ǫi − 1.

Otherwise, if di ≥ k + 2, then

2k +2 ≥ 2k − 1+ ǫi ≥ ci + di ≥ ci + k +2.

In both cases ci ≤ k, so there can be at most k elements in C colored with i. Our results lead naturally to the following question.

Question 6. Does the inequality χg(M) ≤ 2χ(M) − 1 hold for every loopless matroid M?

Acknowledgements I would like to thank Jarek Grytczuk for many inspiring conversations, in particular for introducing me to game arboricity and posing a question for arbitrary matroids. Additionally, I want to thank him for the help in preparation of this manuscript. References

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