2009
DOI: 10.1002/rsa.20279
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Asymptotic random graph intuition for the biased connectivity game

Abstract: ABSTRACT:We study biased Maker/Breaker games on the edges of the complete graph, as introduced by Chvátal and Erdős. We show that Maker, occupying one edge in each of his turns, can build a spanning tree, even if Breaker occupies bedges in each turn. This improves a result of Beck, and is asymptotically best possible as witnessed by the Breaker-strategy of Chvátal and Erdős. We also give a strategy for Maker to occupy a graph with minimum degree c (where c = c(n) is a slowly growing function of n) while playin… Show more

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Cited by 66 publications
(106 citation statements)
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“…Now we can describe the main tool of our proof, a recent result of Gebauer and Szabó, who analyzed in [6] the biased minimum degree game. For our goals, it will suffice to specialize their analysis to the game where Maker's goal is to reach a graph of minimum degree at least 12.…”
Section: Lemma 1 Let G Be a Connected Non-hamiltonian K-expander Thmentioning
confidence: 99%
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“…Now we can describe the main tool of our proof, a recent result of Gebauer and Szabó, who analyzed in [6] the biased minimum degree game. For our goals, it will suffice to specialize their analysis to the game where Maker's goal is to reach a graph of minimum degree at least 12.…”
Section: Lemma 1 Let G Be a Connected Non-hamiltonian K-expander Thmentioning
confidence: 99%
“…The proof of Lemma 3 is a straightforward modification of the proof of Theorem 1.2 of [6]. More specifically, the argument of [6] can be used to analyze a slightly different game in which Breaker wins if he accumulates at least (1 − δ)n edges at a vertex whose Maker degree is still less than 12.…”
Section: -Game Played On the Edge Set Of The Complete Graph K N On N mentioning
confidence: 99%
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“…For the connectivity game, it was shown by Chvátal and Erdős [9] and Gebauer and Szabó [11] that the threshold bias is b T = (1 + o(1)) n log n . The result of Krivelevich [16] gives the leading term of the threshold bias for the Hamilton cycle game, b H = (1 + o(1)) n log n .…”
Section: Introductionmentioning
confidence: 99%
“…Indeed, it was proved in [11] that, playing a (1 : 1) game on E(K n ), Maker can claim all edges of a Hamilton path of K n in n − 1 moves. Moreover, if Maker just wants to build a connected spanning graph, that is, he does not have to declare in advance which spanning tree he intends to build, then he can do so in n − 1 moves even in a (1 : (1 − ε)n/ log n) game (see [8]). On the other hand, it was conjectured by Beck [3] and subsequently proved by Bednarska [5] that, playing a (1 : q) game on E(K n ), where q ≥ cn for an arbitrarily small constant c > 0, Maker cannot build a complete binary tree on q vertices in optimal time, that is, in q − 1 moves.…”
Section: Introductionmentioning
confidence: 99%