On the two problems in Ramsey achievement games

Abstract

Let p,q be two integers with p≥ q. Given a finite graph F with no isolated vertices, the generalized Ramsey achievement game of F on the complete graph Kn, denoted by (p,q;Kn,F,+), is played by two players called Alice and Bob. In each round, Alice firstly chooses p uncolored edges e1,e2,...,ep and colors it blue, then Bob chooses q uncolored edge f1,f2,...,fq and colors it red; the player who can first complete the formation of F in his (or her) color is the winner. The generalized achievement number of F, denoted by a(p,q;F) is defined to be the smallest n for which Alice has a winning strategy. If p=q=1, then it is denoted by a(F), which is the classical achievement number of F introduced by Harary in 1982. If Alice aims to form a blue F, and the goal of Bob is to try to stop him, this kind of game is called the first player game by Bollob\'as. Let a*(F) be the smallest positive integer n for which Alice has a winning strategy in the first player game. A conjecture due to Harary states that the minimum value of a(T) is realized when T is a path and the maximum value of a(T) is realized when T is a star among all trees T of order n. He also asked which graphs F satisfy a*(F)=a(F)? In this paper, we proved that n≤ a(p,q;T)≤ n+q (n-2)/p for all trees T of order n, and obtained a lower bound of a(p,q;K1,n-1), where K1,n-1 is a star. We proved that the minimum value of a(T) is realized when T is a path which gives a positive solution to the first part of Harary's conjecture, and a(T)≤ 2n-2 for all trees of order n. We also proved that for n≥ 3, we have 2n-2-(4n-8) (4n-4)≤ a(K1,n-1)≤ 2n-2 with the help of a theorem of Alon, Krivelevich, Spencer and Szab\'o. We proved that a*(Pn)=a(Pn) for a path Pn.

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