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Solving and Verifying the Boolean Pythagorean Triples Problem via Cube-and-Conquer

机译:通过多维数据集和征服解决和验证布尔毕达哥兰三星问题

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The boolean Pythagorean Triples problem has been a longstanding open problem in Ramsey Theory: Can the set N = {1, 2, . . . } of natural numbers be divided into two parts, such that no part contains a triple (a, b, c) with a~2 + b~2 = c~2 ? A prize for the solution was offered by Ronald Graham over two decades ago. We solve this problem, proving in fact the impossibility, by using the Cube-and-Conquer paradigm, a hybrid SAT method for hard problems, employing both look-ahead and CDCL solvers. An important role is played by dedicated look-ahead heuristics, which indeed allowed to solve the problem on a cluster with 800 cores in about 2 days. Due to the general interest in this mathematical problem, our result requires a formal proof. Exploiting recent progress in unsatisfiability proofs of SAT solvers, we produced and verified a proof in the DRAT format, which is almost 200 terabytes in size. From this we extracted and made available a compressed certificate of 68 gigabytes, that allows anyone to reconstruct the DRAT proof for checking.
机译:Boolean Pythagorean Triples问题在Ramsey理论中是一个长期的开放问题:可以设置n = {1,2,。 。 。自然数分为两部分,使得没有部分含有三倍(a,b,c),具有〜2 + b〜2 = c〜2?二十年前罗纳德格雷厄姆提供了该解决方案的奖品。我们解决了这个问题,实际上证明了不可能的,通过使用立方体和征服范例,混合SAT方法用于难以解决的难题,采用展望前方和CDCL求解器。专用的展望前提发仪扮演一个重要的作用,这确实允许在大约2天内用800个核心的群集解决问题。由于对此数学问题的一般兴趣,我们的结果需要正式证明。利用近期宿用索引的不可起点证明的最新进展,我们在DRAT格式中生产和验证了证据,差少约为200吨。由此,我们提取并提供了68千兆字节的压缩证书,允许任何人重建DRAT证明进行检查。

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