2019
DOI: 10.1137/18m117488x
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The Maker--Breaker Rado Game on a Random Set of Integers

Abstract: Given an integer-valued matrix A of dimension ℓ × k and an integer-valued vector b of dimension ℓ, the Maker-Breaker (A, b)-game on a set of integers X is the game where Maker and Breaker take turns claiming previously unclaimed integers from X, and Maker's aim is to obtain a solution to the system Ax = b, whereas Breaker's aim is to prevent this. When X is a random subset of {1, . . . , n} where each number is included with probability p independently of all others, we determine the threshold probability p0 f… Show more

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Cited by 3 publications
(4 citation statements)
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“…Suppose A and B have dimensions A×kA and B×kB respectively. By Proposition 12 in [8], there exists vectors a,b such that every kA‐distinct solution x=(x1,,xkA) to A contains as an ordered subvector x=(xi1,,xik) (where i1<...<ik), a k ‐distinct solution to C(A)x=a and also every kB‐distinct solution y=(x1,,xkB) to B contains as an ordered subvector y=(yj1,,yjk) (where j1<<jk), a k ‐distinct solution to C(B)…”
Section: Proof Of Theorem 5 and The Ka=kb Case Of Theoremmentioning
confidence: 94%
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“…Suppose A and B have dimensions A×kA and B×kB respectively. By Proposition 12 in [8], there exists vectors a,b such that every kA‐distinct solution x=(x1,,xkA) to A contains as an ordered subvector x=(xi1,,xik) (where i1<...<ik), a k ‐distinct solution to C(A)x=a and also every kB‐distinct solution y=(x1,,xkB) to B contains as an ordered subvector y=(yj1,,yjk) (where j1<<jk), a k ‐distinct solution to C(B)…”
Section: Proof Of Theorem 5 and The Ka=kb Case Of Theoremmentioning
confidence: 94%
“…Suppose A and B are irredundant matrices that satisfy ( * ). Then Proposition 4.3(iv)-(v) in [9] implies that the definitions of m(A) and m(A, B) are well-defined (i.e., have positive denominator); Proposition 12 in [8] implies that C(A) is also well-defined and satisfies ( * ) itself. Theorem 6.…”
Section: A Unifying Theoremmentioning
confidence: 99%
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“…Maker wins the game if he is able to select all the vertices of one of the hyperedges from E, while Breaker wins if she is able to prevent Maker from doing so. We refer to the books of Beck [3] and of Hefetz et al [17] for more information on this game as well as to papers [15,24] for recent related developments.…”
Section: Introductionmentioning
confidence: 99%