数学家利用随机性破解了一个55年前的猜想
内容总结：
数学家破解55年未解猜想：随机性成为关键工具
已故数学家罗纳德·格雷厄姆拥有双重身份——他既是著名数学家、曾任美国数学学会主席，也是一位严肃的杂耍艺人、国际杂耍者协会主席。他曾说，许多数学家和计算机科学家对杂耍感兴趣，根源在于对模式和结构的追寻。1971年，他提出了一个可能与杂耍有关的数学问题：从一组不含零的随机整数出发，能否总能重新排列它们，使得第一个数、前两个数之和、前三个数之和等所有“部分和”互不相同？用杂耍的语言说，如果每个球在空中停留不同时长，是否总能找到一种抛掷顺序，使两个球不会在同一拍落下？
当数字全为正数或正负混合时，答案显然或已知为“是”。但如果数字存在于有限世界中——如同钟面上的数字，超过某个素数p后循环——格雷厄姆猜想答案仍然是肯定的。然而数十年来无人能证明这一直觉。
近期，多位年轻数学家通过四篇论文最终解决了格雷厄姆的排列猜想。最后一篇论文由波恩大学的丽莎·绍尔曼和芝加哥大学的范辉俊于2026年2月发表，正式终结了这个问题。贯穿这些论文的核心主题是：随机性在揭示模式方面的力量。
牛津大学的阿尔普·穆耶塞尔在2022年解决另一个问题时意识到，他的方法也可用于格雷厄姆猜想。他与前导师、伦敦大学学院的阿列克谢·波克罗夫斯基合作，处理集合几乎包含p以内所有数字的情况。他们先随机打乱排列，再预留几个特殊数字用于修正零和序列，于2022年发布了解法。
与此同时，牛津大学的诺亚·克拉维茨独立接触到这一猜想，决定从相反方向入手，与牛津的本杰明·贝德特合作解决了集合远小于p的情况，于2024年9月发表证明。穆耶塞尔看到后主动联系，三人及两位同事组队扩展了原有方法，于2025年8月发表论文，处理了集合相对较大的更多情况。但中等规模集合的情况仍无人能解。
2026年2月，惊喜出现。绍尔曼和范辉俊是斯坦福大学旧识，2025年9月在德国一场会议上听到关于格雷厄姆猜想的报告后产生兴趣。绍尔曼高中时曾在国际数学奥林匹克竞赛中遇到过密切相关的问题并正确解答，后来四次夺得金牌。会议后范辉俊随绍尔曼到波恩短期访问，三天内两人便制定了攻克方案。
他们的方法核心是一种技术要求极高的“反集中”方法。两人先随机重排数字，再通过交换末尾数字修正零和序列。他们识别出三类可能导致失败的“坏事件”，并利用傅里叶分析证明这些坏事件发生的总概率低于100%，从而证明猜想成立。几个月后，他们发表了27页的证明，不仅表明总能找到满足条件的重排，还证明随机排列至少有90%的概率可通过调整消除坏事件。
此前研究该问题的数学家对这一剩余情况被如此迅速解决感到惊讶。穆耶塞尔说：“他们的方法完全不同。”四篇论文共同证明了格雷厄姆猜想对所有规模的集合都成立，但均假设p非常大（约10的100次方量级）。对于所有p的情况，技术上仍未解决。不过，证明证实了即使在这些奇怪有限的数字设定中，“总存在一些美好的结构”，总能通过重新排列数字来避免重复的部分和。
钟韵说：“提出一个好问题真是一门艺术。我想罗恩会非常高兴看到这个问题被解决。”
中文翻译：
数学家借助随机性破解55年未解猜想
引言
已故的罗纳德·格雷厄姆身兼两重身份。他是一位著名数学家，曾担任美国数学学会会长。他也是一位认真的杂耍艺人，曾任国际杂耍艺人协会会长。“他喜欢各种把戏，”加州大学圣迭戈分校数学家、格雷厄姆的妻子金芳蓉说，“你知道，转球、转衣架、同时转好几个球、往墙上扔笔。”
有时格雷厄姆会同时戴着这两顶帽子。“有意思的是，事实上许多数学家和计算机科学家都对杂耍感兴趣，”他在1980年的一次电视采访中说，“我认为，正是对模式和结构的追寻造就了这种兴趣。”
他后来撰写了大量论文，其中一些与金芳蓉合作，主题是杂耍的数学。但早在1971年，在他明确建立这种联系之前几十年，他就提出了一个问题——一些数学家现在认为，这个问题可能也受到了杂耍的启发。
从一组随机的、不含零的不同整数开始。你总能重新排列它们，使得当你把前两个数相加、再把前三个数相加、再把前四个数相加，依此类推时，每一个“部分和”都各不相同吗？用杂耍的语言来说，这意味着如果每个球在空中的停留时间都不同，你总能找到一种抛接顺序，使得两个球不会在同一拍同时落下——那样就会毁掉整个表演。
如果所有数字都是正数，那么格雷厄姆问题的答案显然是对的：随着你加入更多的数，和会不断增大。同样，如果数字中既有正数也有负数，答案也已知是对的。但如果这些数字生活在一个有限的世界里呢——就像数字绕着一个钟面转，达到某个数之后就会循环？
彼得·维多尔
这正是格雷厄姆想知道的。他猜想答案仍然是对的。他推断，即便面对严格的约束，你通常仍能找到足够的灵活性来构造特殊的模式或结构——就像尽管数独棋盘或拉丁方阵（另一种谜题）规则繁多，通常仍能找到有效的构造一样。“它很好地融入了所有这些关于设计和高度对称结构的问题中，”普林斯顿大学数学家诺加·阿隆说。但几十年来，没有人能证明格雷厄姆的直觉是正确的。
这种情况最近发生了变化，几位年轻数学家接过了球。在一项横跨四篇论文和多个数学领域的证明中，他们最终解决了格雷厄姆的重新排列猜想。最后一篇论文由波恩大学的莉萨·绍尔曼和芝加哥大学的范辉端撰写，于2026年2月发表，正式终结了这个问题。
贯穿这些论文的一个主题是：随机性揭示模式的力量。正如阿隆所说，“是合作的力量、年轻一代的力量、概率方法的力量”解决了这个问题。
级联
牛津大学数学家阿尔普·穆耶塞尔常常被那些需要两个要素才能解决的问题所吸引：一个随机过程，外加一些额外的东西。2022年，当他还是研究生时解决了一个这样的问题后，他遇到了格雷厄姆的猜想，意识到自己刚完成的证明在那里也能派上用场。
陈康欣
这个猜想设定在时钟算术的世界中。你先把整数放在一条数轴上，然后把数轴绕到一个钟面上，使得数字在某个素数p之后循环。比如说p是7。在这个设定中，0、7、14以及所有7的倍数都是等价的——这意味着你可以把两个正数（比如3和4）相加得到零。
格雷厄姆问的是：如果你从这条数轴上选取任意一组非零数字（对任意p），你总能重新排列它们，使得得到的所有部分和都各不相同吗？
这个挑战的难度取决于你的集合相对于p有多大。你选的数字越多，需要处理的和就越多。但如果你选的数字越少，重新排列它们的方式就越少。这些不同的情况需要不同的方法。
穆耶塞尔和他以前的导师、伦敦大学学院的阿列克谢·波克罗夫斯基处理的是你的集合包含几乎所有直到p的可能数字的情况。对于这么大的集合，构造一个有效的排列可能极其困难。但事实证明，从一个随机排列出发可以带你走完大部分路程。
“计算机科学家常把这叫做‘大海捞针’问题，”穆耶塞尔说。你可能知道有很多好的排列存在，但真正找到一个却很难。“如果你随机来做，很可能会成功，但很难明确描述解应该长什么样。”
穆耶塞尔和波克罗夫斯基需要确保排列中没有任何一段连续数字的和为零。否则，把这些数字加到之前的部分和上就会重复那个和。
一个完全随机的排列可能会有几个这样麻烦的序列。于是穆耶塞尔和波克罗夫斯基先从集合中留出几个特别挑选的数字，然后随机打乱其余的。他们扫描随机排列寻找任何问题；如果遇到一个和为零的区间，他们可以插入一个备用数字来改变它。2022年，他们发布了他们的解答，不过它隐藏在一篇专注于将同样技术应用于更一般问题的论文中。
几年后，牛津大学的诺亚·克拉维茨在不知穆耶塞尔和波克罗夫斯基已有解答的情况下，在一个未解问题在线档案中偶然发现了格雷厄姆的猜想。“我看到有一个未解问题，我当时想，人类竟然不知道这个，真是太丢人了，”克拉维茨说，“这种情况必须被纠正。”
他决定从相反的方向入手。他与牛津大学的本杰明·贝德尔特合作，考虑了数字集合相对于p非常小的情况——例如，阿隆说，如果你有一个100个数字的集合，而p是10亿。
克拉维茨和贝德尔特解决了这些情况下的格雷厄姆猜想，并于2024年9月发布了他们的证明。穆耶塞尔看到后主动联系，分享了自己的工作；他们三人（加上另外两位同事）随后合作扩展了穆耶塞尔最初的方法。
“这是一个相当不可能的人员组合，”克拉维茨说。他和穆耶塞尔来自组合数学中两个通常不合作的领域。“不同方向有完全不同的技术，”他说。
他们于2025年8月发布的论文处理了更多集合相对于p较大的情况。但在这些情况和克拉维茨与贝德尔特已覆盖的小集合情况之间，仍然存在一个缺口。没有人能想出如何处理中等大小的集合，比如那些包含大约p的一半数量的数字的集合。“我们的方法在那里行不通，而且有明确的理由说明它们为什么行不通，”穆耶塞尔说。
看起来对这个问题的研究可能又要进入一段漫长的停滞期。
然后，在2026年2月，一个惊喜出现在网上。
喷泉
莉萨·绍尔曼和范辉端是老朋友。这两位数学家于2015年在斯坦福大学相识，当时绍尔曼是研究生，范辉端是本科生。如今他们住在不同的大洲——绍尔曼在德国波恩，范辉端在芝加哥。但2025年9月在德国举行的一次会议为他们提供了一个难得的机会，让他们再次共用一块黑板，之后范辉端跟随绍尔曼到波恩做了短暂访问。他们只需要一个可以研究的问题。
在会议上，他们听到了两场关于格雷厄姆猜想的报告，演讲者是曾尝试但未能弥合缺口的数学家。他们产生了兴趣。而后来发现，绍尔曼高中时在国际数学奥林匹克竞赛中遇到过一道密切相关的问题。她正确地解决了它，到高中毕业时，她已在这项著名竞赛中四次获得金牌。（很可能正是金芳蓉把那道题放进了当年的考试，因为她是在出题委员会中，而且她经常从格雷厄姆的众多谜题中汲取灵感。）
在三天的访问结束时，绍尔曼和范辉端已经有了破解这个问题的计划。
关键在于一种技术要求很高的方法，叫做反集中。在这里，一个反集中命题断言某个事件发生的概率特别低。但证明这类命题的机制如此复杂，以至于尽管克拉维茨和其他人知道这种反集中方法可能会成功，“我们就是没有胆量真正去尝试，”他说。
不过，绍尔曼和范辉端首先以前人的方式开始。他们随机重新排列数字集合，并设计了一个程序来修复任何问题——也就是任何和为零的序列。每当他们发现一个零和序列，就把序列中最后一个数字换成另一个。
这个程序通常进行得很顺利。但三类“坏事件”会导致它失败。第一：一个零和序列可能出现在整个排列的末尾附近；那样就没有其他数字可以换进来了。第二：许多零和序列可能出现得太靠近，使得无法全部修复。第三：修复一个坏序列可能会在之后制造出另一个零和序列。
绍尔曼和范辉端希望用反集中来证明，这些坏事件每一个都足够不可能发生。那么就一定存在一种重新排列数字集合的方式满足猜想。
为此，两人使用了傅里叶分析——一个让你把函数重写为简单波之和的数学领域——来证明一般来说，当你把随机数字集合相加时，没有哪个和特别容易出现。然后他们利用这一洞见仔细估计每个坏事件发生的概率，最终证明出现坏事件的总概率小于100%。这就足以解决猜想了。
在德国待了几个月后，绍尔曼和范辉端在网上发布了他们27页的证明。他们不仅证明了令人满意的重新排列总是可能的，而且证明了随机排列至少90%的情况下可以被重新排列以消除坏事件——这是一个巨大的成功率。
之前研究过这个问题的数学家们惊讶地看到剩余的情况被如此迅速地解决了。“他们的方法完全不同，”穆耶塞尔说。
这四篇论文共同证明了格雷厄姆猜想对所有大小的集合都成立。但它们都假设p非常大；虽然没有人计算过它的确切值，但可以想象成10的100次方那个量级。对数学家来说，这没问题——关键点在于你是在时钟算术的设定下工作。但从技术上讲，问题的另一个方面仍未解决——你可能还想对所有p解决这个猜想。如果你想用这个结果来编排一个真正的杂耍节目，那你就没戏了：要对应这么大的p，节目必须长得离谱。
这个证明确认了即使在这些奇怪、有限的数字设定中，“也总存在一些美好的结构，”阿隆说。你总能获得某种程度的灵活性，通过打乱集合中的数字来避免重复相同的部分和。
“提出一个好问题真的是一门艺术，”金芳蓉说。“我想罗恩看到这个问题被解决会非常高兴。”
英文来源：
Mathematicians Harness Randomness To Crack a 55-Year-Old Conjecture
Introduction
The late Ronald Graham wore two hats. He was a renowned mathematician, at one time president of the American Mathematical Society. He was also a serious juggler, and president of the International Jugglers’ Association. “He loved tricks,” said Fan Chung, a mathematician at the University of California, San Diego, who was married to Graham. “You know, spinning a ball, spinning a coat hanger, spinning several balls together, throwing pens against the wall.”
Sometimes Graham wore both hats at once. “It’s interesting, in fact, that many mathematicians and computer scientists have an interest in juggling,” he said in a 1980 television interview. “I think it’s the search for patterns and structure that is responsible for this.”
He would go on to write numerous papers, some with Chung, on the mathematics of juggling. But back in 1971, decades before he made that connection explicit, he posed a question that some mathematicians now say might have been inspired by juggling, too.
Start with a random set of different integers, not including zero. Can you always rearrange them so that if you add up the first two numbers, then the first three, then the first four, and so on, every “partial sum” turns out different? In the language of juggling, this would mean that if each ball stays in the air for a different amount of time, you can always find an order to throw them in such that two balls won’t come crashing down on the same beat — which would ruin the act.
If the numbers are all positive, then the answer to Graham’s question is obviously yes: The sums will always grow larger as you add more numbers. Similarly, if there are both positive and negative numbers in the mix, the answer is also known to be yes. But what if the numbers live in a finite world — like numbers wrapped around a clock, which repeat after a certain count?
Peter Vidor
That’s what Graham wanted to know. He conjectured that the answer should still be yes. It often happens, he figured, that even when dealing with rigid constraints, you can still find enough flexibility to construct special patterns or structures — just as it’s usually possible to find a valid sudoku board or Latin square (another kind of puzzle) despite their many rules. “It fits nicely in all these questions about designs and about very symmetric structures,” said Noga Alon, a mathematician at Princeton University. But for decades, no one could prove Graham’s intuition to be true.
That changed recently, when several young mathematicians picked up the balls. In a proof that spanned four papers and various fields of mathematics, they finally resolved Graham’s rearrangement conjecture. The final paper, by Lisa Sauermann of the University of Bonn and Huy Tuan Pham of the University of Chicago, appeared in February 2026, officially closing the problem.
Across the papers, one theme prevailed: the power of randomness to draw out patterns. As Alon put it, “It’s the power of collaboration, the power of the young generation, the power of probabilistic methods” that solved the problem.
The Cascade
Alp Müyesser, a mathematician at the University of Oxford, often finds himself drawn to problems whose solutions need two ingredients: a random process, and something extra as well. After solving one such problem in 2022 while he was still a graduate student, he encountered Graham’s conjecture and realized that his just-finished proof could help there, too.
Kangxin Chen
The conjecture is set in the world of clock arithmetic. You start by placing the whole numbers on a number line, then you wrap the line around the face of a clock so that the numbers repeat after some prime number, p. Say p is 7, for instance. In this setting, 0, 7, 14, and all other multiples of 7 are equivalent — meaning that you can add two positive numbers (like 3 and 4) and get zero.
Graham asked the following: If you pick any set of nonzero numbers off this number line (for any p), can you always rearrange them so that the partial sums you get are all different?
The challenge depends on how big your set is compared to p. The more numbers you pick, the more sums there are to manage. But if you choose fewer numbers, there will be fewer ways to rearrange them. These different cases inspire different approaches.
Müyesser, along with his former adviser, Alexey Pokrovskiy of University College London, tackled the case where your set includes almost every possible number up to p. With sets this large, it can be extremely hard to construct a valid ordering. But it turned out that starting with a random ordering can bring you most of the way there.
“Computer scientists often call this a ‘finding the hay in the haystack’ problem,” Müyesser said. You might know that lots of good orderings are out there, but actually finding one is hard. “If you do it randomly, it’s likely going to work, but it’s hard to explicitly describe what the solution is supposed to look like.”
Müyesser and Pokrovskiy needed to ensure that no sequence of numbers anywhere in the ordering added up to zero. Otherwise, adding those numbers to the previous partial sum would repeat that sum.
A completely random ordering might have a few of these troublesome sequences. So Müyesser and Pokrovskiy first set aside a few specially chosen numbers from the set, then randomly scrambled the rest. They scanned their random ordering for any problems; if they came across an interval that added up to zero, they could insert one of the spare numbers to change it. In 2022, they posted their solution, though it was hidden in a paper that focused on applying the same technique to a more general problem.
A couple of years later, Noah Kravitz of Oxford, unaware of Müyesser and Pokrovskiy’s solution, stumbled on Graham’s conjecture in an online archive of unsolved problems. “I saw there was an open problem, and I was like, it’s embarrassing for humanity that we don’t know this,” Kravitz said. “This situation just had to be rectified.”
He decided to approach the conjecture from the opposite end. Together with Benjamin Bedert of Oxford, he considered the case where the set of numbers is tiny compared to p — for instance, Alon said, if you have a set of 100 numbers where p is 1 billion.
Kravitz and Bedert solved Graham’s conjecture for those cases and posted their proof in September 2024. Müyesser saw it and reached out, sharing his own work; the three of them (plus two other colleagues) then teamed up to extend Müyesser’s original approach.
“It was a pretty unlikely combination of people,” Kravitz said. He and Müyesser come from two areas of combinatorics that don’t typically collaborate. “Different sections have completely different techniques,” he said.
Their paper, which they posted in August 2025, handled more cases where the set of numbers is relatively large compared to p. But between those cases and the small-set cases that Kravitz and Bedert had covered, a gap remained. No one could figure out what to do about medium-size sets, such as those that include roughly half as many numbers as p. “Our methods didn’t work there, and there were clear reasons that they would not have worked,” Müyesser said.
It seemed as though research on the problem might enter another long hiatus.
Then, in February 2026, a surprise appeared online.
The Fountain
Lisa Sauermann and Huy Tuan Pham were old friends. The two mathematicians had met in 2015 at Stanford University, where Sauermann was a graduate student and Pham an undergraduate. Today they live on different continents — Sauermann in Bonn, Germany, and Pham in Chicago. But a conference in Germany in September 2025 provided a rare chance for them to share a chalkboard again, and afterward Pham followed Sauermann to Bonn for a short visit. All they needed was a problem to work on.
At the conference, they heard two talks on Graham’s conjecture by mathematicians who had attempted but failed to bridge the gap. They were intrigued. And as it later turned out, Sauermann had encountered a closely related problem in the International Mathematical Olympiad as a high school student. She solved it correctly, and by the time she finished high school, she’d won a gold medal in the prestigious competition four times. (Most likely, it was Chung who placed the problem on that year’s exam, as she was on the committee that wrote the questions, and she frequently took inspiration from Graham’s many puzzles.)
By the end of their three-day visit, Sauermann and Pham had a plan for how to crack the case.
It hinged on a technically demanding method called anti-concentration. Here, an anti-concentration statement asserts that some event has a particularly low chance of happening. But the mechanics of proving these kinds of statements are so intricate that, though Kravitz and others were aware that such an anti-concentration approach might succeed, “we just hadn’t had the guts to actually try it,” he said.
First, though, Sauermann and Pham began the way their predecessors had. They randomly reordered their set of numbers and came up with a procedure to fix any problems — that is, any sequences that add up to zero. Any time they found a zero-sum sequence, they swapped out the last number in the sequence with another one.
This procedure often went without a hitch. But three types of “bad events” would cause it to fail. One: A zero-sum sequence might occur toward the end of the entire arrangement; then there would be no other numbers to swap in. Two: Many zero-sum sequences might appear too close together, making it impossible to fix them all. And three: Fixing one bad sequence might create another zero-sum sequence down the line.
Sauermann and Pham hoped to prove, using anti-concentration, that each of these bad events was sufficiently unlikely. Then there would have to be a way to rearrange the set of numbers to satisfy the conjecture.
To do this, the duo used Fourier analysis — an area of math that lets you rewrite functions as sums of simple waves — to show that in general, when you add up random sets of numbers, no one sum is especially likely to appear. They then used this insight to carefully estimate the probability that each bad event would occur, ultimately showing that the total chance of getting a bad event was less than 100%. That was enough to settle the conjecture.
A few months after their stint in Germany, Sauermann and Pham posted their 27-page proof online. They had shown not only that a satisfactory rearrangement was always possible, but that a random ordering could be rearranged to eliminate bad events at least 90% of the time — a massive success rate.
The mathematicians who had previously worked on the problem were surprised to see the remaining case closed so quickly. “Their approach is just completely different,” Müyesser said.
Together, the four papers prove Graham’s conjecture for sets of all sizes. But they all assume that p is very large; though no one has calculated its exact value, think along the lines of 10 raised to the 100th power. To mathematicians, that’s fine — the salient point is that you’re working in the setting of clock arithmetic. But another aspect of the problem technically remains unsolved — you might still try to resolve the conjecture for all p. And if you want to use the result to choreograph a real juggling routine, you’re out of luck: To correspond to such a large p, the routine would have to be much too long.
The proof confirms that even within these strange, limited number settings, “there are some nice structures that always exist,” Alon said. You can always achieve some degree of flexibility, shuffling the numbers in your set around to avoid revisiting the same partial sums.
“To pose a good problem is really an art,” Chung said. “I think Ron would be extremely happy to see the problem solved.”