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On the two problems in Ramsey achievement games

Published 2 Aug 2024 in math.CO | (2408.01479v1)

Abstract: Let p,qp,q be two integers with pqp\geq q. Given a finite graph FF with no isolated vertices, the generalized Ramsey achievement game of FF on the complete graph KnK_n, denoted by (p,q;Kn,F,+)(p,q;K_n,F,+), is played by two players called Alice and Bob. In each round, Alice firstly chooses pp uncolored edges e1,e2,...,epe_1,e_2,...,e_p and colors it blue, then Bob chooses qq uncolored edge f1,f2,...,fqf_1,f_2,...,f_q and colors it red; the player who can first complete the formation of FF in his (or her) color is the winner. The generalized achievement number of FF, denoted by a(p,q;F){a}(p,q;F) is defined to be the smallest nn for which Alice has a winning strategy. If p=q=1p=q=1, then it is denoted by a(F){a}(F), which is the classical achievement number of FF introduced by Harary in 1982. If Alice aims to form a blue FF, and the goal of Bob is to try to stop him, this kind of game is called the first player game by Bollob\'{a}s. Let a<sup>(F){a}<sup>*(F) be the smallest positive integer nn 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){a}(T) is realized when TT is a path and the maximum value of a(T){a}(T) is realized when TT is a star among all trees TT of order nn. He also asked which graphs FF satisfy a<sup>(F)=a(F)a<sup>*(F)=a(F)? In this paper, we proved that na(p,q;T)n+q(n2)/pn\leq {a}(p,q;T)\leq n+q\left\lfloor (n-2)/p \right\rfloor for all trees TT of order nn, and obtained a lower bound of a(p,q;K1,n1){a}(p,q;K_{1,n-1}), where K1,n1K_{1,n-1} is a star. We proved that the minimum value of a(T){a}(T) is realized when TT is a path which gives a positive solution to the first part of Harary's conjecture, and a(T)2n2{a}(T)\leq 2n-2 for all trees of order nn. We also proved that for n3n\geq 3, we have 2n2(4n8)ln(4n4)a(K1,n1)2n22n-2-\sqrt{(4n-8)\ln (4n-4)}\leq a(K_{1,n-1})\leq 2n-2 with the help of a theorem of Alon, Krivelevich, Spencer and Szab\'o. We proved that a<sup>(Pn)=a(Pn)a<sup>*(P_n)=a(P_n) for a path PnP_n.

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