Mathematicians Harness Randomness To Crack a 55-Year-Old Conjecture

The late Ronald Graham wore two hats.

已故的罗纳德·格雷厄姆戴着两顶帽子。

He was a renowned mathematician, at one time president of the American Mathematical Society.

他是一位著名的数学家，曾任美国数学会主席。

He was also a serious juggler, and president of the International Jugglers’ Association.

他也是一位认真的杂耍演员，也是国际杂耍演员协会的主席。

“He loved tricks,” said Fan Chung , a mathematician at the University of California, San Diego, who was married to Graham.

“他喜欢把戏，”格雷厄姆的妻子、加州大学圣地亚哥分校的数学家范钟 (Fan Chung) 说。

“You know, spinning a ball, spinning a coat hanger, spinning several balls together, throwing pens against the wall.”

“你知道，旋转一个球，旋转一个衣架，一起旋转几个球，把笔扔到墙上。”

Sometimes Graham wore both hats at once.

有时格雷厄姆同时戴上两顶帽子。

“It’s interesting, in fact, that many mathematicians and computer scientists have an interest in juggling,” he said in a 1980 television interview .

“事实上，有趣的是，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道。

“I think it’s the search for patterns and structure that is responsible for this.”

“我认为这是对模式和结构的探索造成的。”

He would go on to write numerous papers, some with Chung , on the mathematics of juggling.

他继续写了许多关于杂耍数学的论文，其中一些是与钟一起写的。

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.

但早在 1971 年，也就是他明确表达这种联系的几十年前，他就提出了一个问题，一些数学家现在说这个问题也可能是受到杂耍的启发。

Start with a random set of different integers, not including zero.

从一组随机的不同整数开始，不包括零。

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?

你是否可以重新排列它们，以便如果你将前两个数字相加，然后将前三个数字相加，然后是前四个数字，依此类推，每个“部分和”的结果都会不同？

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.

用杂耍的语言来说，这意味着如果每个球在空中停留的时间不同，你总是可以找到一个将它们扔进去的命令，这样两个球就不会在同一节拍上坠落——这会破坏表演。

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.

如果数字都是正数，那么格雷厄姆问题的答案显然是肯定的：当你添加更多数字时，总和总是会变大。

Similarly, if there are both positive and negative numbers in the mix, the answer is also known to be yes.

同样，如果混合中同时存在正数和负数，则答案也是肯定的。

But what if the numbers live in a finite world — like numbers wrapped around a clock, which repeat after a certain count?

但是，如果这些数字存在于一个有限的世界中——就像缠绕在时钟上的数字一样，在一定的计数后会重复呢？

The late mathematician Ronald Graham was also a skilled juggler.

已故数学家罗纳德·格雷厄姆也是一位熟练的杂技演员。

That’s what Graham wanted to know.

这就是格雷厄姆想知道的。

He conjectured that the answer should still be yes.

他猜想，答案应该还是肯定的。

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.

他认为，这种情况经常发生，即使在处理严格的约束时，你仍然可以找到足够的灵活性来构造特殊的模式或结构——就像通常可以找到有效的数独板或拉丁方（另一种拼图），尽管它们有很多规则。

“It fits nicely in all these questions about designs and about very symmetric structures,” said Noga Alon , a mathematician at Princeton University.

普林斯顿大学数学家诺加·阿隆 (Noga Alon) 表示：“它非常适合所有这些有关设计和非常对称结构的问题。”

But for decades, no one could prove Graham’s intuition to be true.

但几十年来，没有人能够证明格雷厄姆的直觉是正确的。

That changed recently, when several young mathematicians picked up the balls.

最近，当几位年轻的数学家拿起球时，情况发生了变化。

In a proof that spanned four papers and various fields of mathematics, they finally resolved Graham’s rearrangement conjecture.

在跨越四篇论文和数学各个领域的证明中，他们最终解决了格雷厄姆的重排猜想。

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.

最终论文由波恩大学的 Lisa Sauermann 和芝加哥大学的 Huy Tuan Pham 于 2026 年 2 月发表，正式解决了该问题。

Across the papers, one theme prevailed: the power of randomness to draw out patterns.

在所有论文中，一个主题盛行：随机性绘制模式的力量。

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.

正如阿隆所说，“正是协作的力量、年轻一代的力量、概率方法的力量”解决了这个问题。

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.

牛津大学数学家 Alp Müyesser 经常发现自己被一些问题所吸引，这些问题的解决方案需要两个要素：随机过程和额外的东西。

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.

2022 年，当他还是一名研究生时，他解决了一个这样的问题，之后遇到了格雷厄姆猜想，并意识到他刚刚完成的证明也能对此有所帮助。

Alp Müyesser enjoys thinking about problems that require him to combine randomness with something else.

Alp Müyesser 喜欢思考需要他将随机性与其他​​事物结合起来的问题。

The conjecture is set in the world of clock arithmetic.

该猜想以时钟算术世界为背景。

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 .

首先将所有数字放在数轴上，然后将线绕在钟面上，以便数字在某个素数 p 之后重复。

Say p is 7, for instance.

例如，假设 p 是 7。

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.

在此设置中，0、7、14 和 7 的所有其他倍数都是等价的 — 这意味着您可以将两个正数（如 3 和 4）相加并得到零。

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?

格雷厄姆提出了以下问题：如果您从该数轴上选取任何一组非零数字（对于任何 p ），您是否可以始终重新排列它们，以便您得到的部分和都不同？

The challenge depends on how big your set is compared to p .

挑战取决于您的集合与 p 相比有多大。

The more numbers you pick, the more sums there are to manage.

您选择的数字越多，需要管理的金额就越多。

But if you choose fewer numbers, there will be fewer ways to rearrange them.

但如果你选择的数字较少，重新排列它们的方法就会较少。

These different cases inspire different approaches.

这些不同的案例激发了不同的方法。

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 .

Müyesser 和他的前顾问、伦敦大学学院的 Alexey Pokrovskiy 一起解决了这样的情况：你的集合几乎包括直到 p 的所有可能数字。

With sets this large, it can be extremely hard to construct a valid ordering.

对于这么大的集合，构建有效的排序可能非常困难。

But it turned out that starting with a random ordering can bring you most of the way there.

但事实证明，从随机排序开始就可以完成大部分任务。

It’s embarrassing for humanity that we don’t know this.

我们不知道这一点，这对人类来说是尴尬的。

This situation just had to be rectified.

这种情况必须得到纠正。

Noah Kravitz, University of Oxford

诺亚·克拉维茨，牛津大学

“Computer scientists often call this a ‘ finding the hay in the haystack ’ problem,” Müyesser said.

“计算机科学家经常将此称为‘大海捞针’问题，”穆耶瑟说。

You might know that lots of good orderings are out there, but actually finding one is hard.

您可能知道有很多好的订单，但实际上找到一个很难。

“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.”

“如果你随机地做，它可能会起作用，但很难明确地描述解决方案应该是什么样子。”

Müyesser and Pokrovskiy needed to ensure that no sequence of numbers anywhere in the ordering added up to zero.

Müyesser 和 Pokrovskiy 需要确保排序中任何位置的数字序列加起来都不为零。

Otherwise, adding those numbers to the previous partial sum would repeat that sum.

否则，将这些数字添加到之前的部分总和中将重复该总和。

A completely random ordering might have a few of these troublesome sequences.

完全随机的排序可能会出现一些麻烦的序列。

So Müyesser and Pokrovskiy first set aside a few specially chosen numbers from the set, then randomly scrambled the rest.

因此，Müyesser 和 Pokrovskiy 首先从集合中专门挑选了一些数字，然后随机打乱其余的数字。

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.

他们扫描了随机排序是否存在任何问题；如果他们遇到加起来为零的间隔，他们可以插入一个备用数字来更改它。

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.

2022 年，他们发布了他们的解决方案，尽管它隐藏在一篇专注于将相同技术应用于更普遍问题的论文中。

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.

几年后，牛津大学的诺亚·克拉维茨 (Noah Kravitz) 不知道 Müyesser 和 Pokrovskiy 的解决方案，在未解决问题的在线档案中偶然发现了格雷厄姆的猜想。

“I saw there was an open problem, and I was like, it’s embarrassing for humanity that we don’t know this,” Kravitz said.

克拉维茨说：“我看到有一个悬而未决的问题，我想，我们不知道这一点，这对人类来说是尴尬的。”

“This situation just had to be rectified.”

“这种情况必须得到纠正。”

Noah Kravitz is one of several young mathematicians who recently revived the decades-old Graham conjecture.

诺亚·克拉维茨是最近复兴了已有数十年历史的格雷厄姆猜想的几位年轻数学家之一。

He decided to approach the conjecture from the opposite end.

他决定从相反的一端来探讨这个猜想。

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.

他与牛津大学的本杰明·贝德特 (Benjamin Bedert) 一起考虑了与 p 相比这组数字很小的情况——例如，阿隆说，如果你有一组 100 个数字，其中 p 是 10 亿。

Kravitz and Bedert solved Graham’s conjecture for those cases and posted their proof in September 2024.

Kravitz 和 Bedert 解决了 Graham 对这些情况的猜想，并于 2024 年 9 月发布了他们的证明。

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.

Müyesser 看到了，伸出手来分享他自己的作品；他们三人（加上另外两名同事）随后联手扩展了 Müyesser 最初的方法。

“It was a pretty unlikely combination of people,” Kravitz said.

“这是一个不太可能的人组合，”克拉维茨说。

He and Müyesser come from two areas of combinatorics that don’t typically collaborate.

他和 Müyesser 来自两个通常不合作的组合数学领域。

“Different sections have completely different techniques,” he said.

“不同的部门有完全不同的技术，”他说。

Their paper, which they posted in August 2025 , handled more cases where the set of numbers is relatively large compared to p .

他们于 2025 年 8 月发布的论文处理了更多与 p 相比，数字集相对较大的情况。

But between those cases and the small-set cases that Kravitz and Bedert had covered, a gap remained.

但这些案例与克拉维茨和贝德特报道的小型案例之间仍然存在差距。

No one could figure out what to do about medium-size sets, such as those that include roughly half as many numbers as p .

没有人能弄清楚如何处理中等大小的集合，例如那些包含大约 p 一半数量的集合。

“Our methods didn’t work there, and there were clear reasons that they would not have worked,” Müyesser said.

“我们的方法在那里不起作用，而且有明显的理由表明它们不起作用，”穆耶塞尔说。

It seemed as though research on the problem might enter another long hiatus.

似乎对这个问题的研究可能会进入另一个长期的间歇期。

Then, in February 2026, a surprise appeared online.

然后，2026 年 2 月，网上出现了一个惊喜。

Lisa Sauermann and Huy Tuan Pham were old friends.

丽莎·索尔曼 (Lisa Sauermann) 和范辉俊 (Huy Tuan Pham) 是老朋友。

The two mathematicians had met in 2015 at Stanford University, where Sauermann was a graduate student and Pham an undergraduate.

这两位数学家于 2015 年在斯坦福大学相识，当时索尔曼是该校的研究生，范是该校的本科生。

Today they live on different continents — Sauermann in Bonn, Germany, and Pham in Chicago.

如今，他们生活在不同的大陆——索尔曼住在德国波恩，范住在芝加哥。

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.

但2025年9月在德国举行的一次会议为他们再次共享黑板提供了难得的机会，随后范跟随索尔曼前往波恩进行了短暂访问。

All they needed was a problem to work on.

他们所需要的只是一个需要解决的问题。

At the conference, they heard two talks on Graham’s conjecture by mathematicians who had attempted but failed to bridge the gap.

在会议上，他们听到了数学家关于格雷厄姆猜想的两次演讲，他们曾试图弥合这一差距，但未能成功。

They were intrigued.

他们很感兴趣。

And as it later turned out, Sauermann had encountered a closely related problem in the International Mathematical Olympiad as a high school student.

后来事实证明，索尔曼在高中时的国际数学奥林匹克竞赛中遇到了一个密切相关的问题。

She solved it correctly, and by the time she finished high school, she’d won a gold medal in the prestigious competition four times.

她正确地解决了这个问题，当她高中毕业时，她已经四次在这项享有声望的比赛中获得金牌。

(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.)

（最有可能的是，是钟把这个问题放在了当年的考试中，因为她是编写这些问题的委员会的成员，而且她经常从格雷厄姆的许多谜题中获得灵感。）

Lisa Sauermann (top) and Huy Tuan Pham have been friends for the past decade.

Lisa Sauermann（上）和 Huy Tuan Pham 在过去十年中一直是朋友。

Their recent collaboration on the Graham conjecture finally settled the problem.

他们最近在格雷厄姆猜想上的合作终于解决了这个问题。

Barbara Frommann/University of Bonn; Courtesy of Huy Tuan Pham

芭芭拉·弗洛曼/波恩大学；由范辉俊 (Huy Tuan Pham) 提供

By the end of their three-day visit, Sauermann and Pham had a plan for how to crack the case.

在为期三天的访问结束时，索尔曼和范已经制定了如何破案的计划。

It hinged on a technically demanding method called anti-concentration.

它取决于一种技术要求很高的方法，称为反集中。

Here, an anti-concentration statement asserts that some event has a particularly low chance of happening.

在这里，反集中声明声称某些事件发生的可能性特别低。

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.

但证明此类陈述的机制是如此复杂，以至于尽管克拉维茨和其他人意识到这种反集中方法可能会成功，但“我们只是没有勇气真正尝试它，”他说。

First, though, Sauermann and Pham began the way their predecessors had.

不过，索尔曼和范首先采用了他们前辈的做法。

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.

他们随机地重新排序了一组数字，并提出了一个解决任何问题的程序——即任何加起来为零的序列。

Any time they found a zero-sum sequence, they swapped out the last number in the sequence with another one.

每当他们发现零和序列时，他们就会用另一个数字替换序列中的最后一个数字。

This procedure often went without a hitch.

这个过程通常都很顺利。

But three types of “bad events” would cause it to fail.

但三种类型的“坏事件”会导致它失败。

One: A zero-sum sequence might occur toward the end of the entire arrangement; then there would be no other numbers to swap in.

一：在整个安排结束时可能会出现零和序列；那么就没有其他号码可以交换了。

Two: Many zero-sum sequences might appear too close together, making it impossible to fix them all.

第二：许多零和序列可能看起来太接近，从而无法全部修复。

And three: Fixing one bad sequence might create another zero-sum sequence down the line.

第三：修复一个错误的序列可能会产生另一个零和序列。

Sauermann and Pham hoped to prove, using anti-concentration, that each of these bad events was sufficiently unlikely.

绍尔曼和范希望通过反集中证明，这些不良事件中的每一个都不太可能发生。

Then there would have to be a way to rearrange the set of numbers to satisfy the conjecture.

那么就必须有一种方法来重新排列这组数字以满足猜想。

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.

为了做到这一点，两人使用了傅立叶分析（一种数学领域，可以让你将函数重写为简单波的和）来表明，一般来说，当你将随机的一组数字相加时，不会特别有可能出现任何一个和。

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%.

然后，他们利用这一洞察仔细估计了每个不良事件发生的概率，最终表明发生不良事件的总几率小于 100%。

That was enough to settle the conjecture.

这足以解决这个猜想。

A few months after their stint in Germany, Sauermann and Pham posted their 27-page proof online.

在德国工作几个月后，Sauermann 和 Pham 在网上发布了 27 页的证明。

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.

他们不仅证明了令人满意的重新排列总是可能的，而且可以重新排列随机排序以消除至少 90% 的不良事件——这是一个巨大的成功率。

The mathematicians who had previously worked on the problem were surprised to see the remaining case closed so quickly.

之前研究过这个问题的数学家们惊讶地发现剩下的案例如此之快就结案了。

“Their approach is just completely different,” Müyesser said.

“他们的方法完全不同，”穆耶瑟说。

Together, the four papers prove Graham’s conjecture for sets of all sizes.

这四篇论文共同证明了格雷厄姆对各种大小集合的猜想。

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.

但他们都假设 p 非常大；虽然没有人计算出它的确切值，但可以按照 10 的 100 次方来思考。

To mathematicians, that’s fine — the salient point is that you’re working in the setting of clock arithmetic.

对于数学家来说，这很好——重点是你正在研究时钟算术的设置。

But another aspect of the problem technically remains unsolved — you might still try to resolve the conjecture for all p .

但问题的另一个方面在技术上仍未解决——您可能仍然尝试解决所有 p 的猜想。

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.

如果你想用这个结果来编排一个真正的杂耍动作，那你就不走运了：为了对应这么大的 p ，这个动作就必须太长了。

The proof confirms that even within these strange, limited number settings, “there are some nice structures that always exist,” Alon said.

证据证实，即使在这些奇怪的、有限的数量设置中，“仍然存在一些不错的结构，”阿隆说。

You can always achieve some degree of flexibility, shuffling the numbers in your set around to avoid revisiting the same partial sums.

您始终可以实现一定程度的灵活性，对集合中的数字进行洗牌以避免重新访问相同的部分和。

“To pose a good problem is really an art,” Chung said.

“提出一个好问题确实是一门艺术，”钟说。

“I think Ron would be extremely happy to see the problem solved.”

“我认为罗恩会非常高兴看到问题得到解决。”

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《Quanta》杂志对评论进行审核，以促进知情、实质性、文明的对话。

Abusive, profane, self-promotional, misleading, incoherent or off-topic comments will be rejected.

辱骂、亵渎、自我推销、误导、语无伦次或偏离主题的评论将被拒绝。

Moderators are staffed during regular business hours (New York time) and can only accept comments written in English.

版主在正常工作时间（纽约时间）有工作人员，并且只能接受英文评论。