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A SAT Attack on the Erdos Discrepancy Conjecture

机译:SAT攻击erdos差异猜想

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In 1930s Paul Erdos conjectured that for any positive integer C in any infinite ±1 sequence (x_n) there exists a subsequence x_d, x_(2d), x_(3d),...,x_(kd), for some positive integers k and d, such that |∑_(i=1)~k x_(id)| > C. The conjecture has been referred to as one of the major open problems in combinatorial number theory and discrepancy theory. For the particular case of C = 1 a human proof of the conjecture exists; for C = 2 a bespoke computer program had generated sequences of length 1124 of discrepancy 2, but the status of the conjecture remained open even for such a small bound. We show that by encoding the problem into Boolean satisfiability and applying the state of the art SAT solver, one can obtain a discrepancy 2 sequence of length 1160 and a proof of the Erdos discrepancy conjecture for C = 2, claiming that no discrepancy 2 sequence of length 1161, or more, exists. We also present our partial results for the case of C = 3.
机译:在20世纪30年代,Paul Erdos猜测,对于任何无限±1序列(X_N)的任何正整数C,存在后续x_d,x_(2d),x_(3d),...,x_(kd),对于某些正整数k和D,使得|Σ_(i = 1)〜k x_(id)| > C.猜想被称为组合数字理论和差异理论中的主要开放问题之一。对于C = 1的特定情况,存在猜想的人类证明;对于C = 2,定制计算机程序已经产生了差异2的长度1124的序列,但即使对于这种小限制,猜想的状态也保持开放。我们认为,通过将问题进行编码到布尔满可取性并应用现有技术的状态,可以获得长度1160的差异2序列和用于C = 2的鄂尔多斯差异猜想的证据,声称没有差异2序列存在长度1161或更多。我们还为C = 3的情况提供了我们的部分结果。

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