# Unit Distance
**作者**: Sebastien Bubeck
**日期**: 2026-05-20T19:53:29.000Z
**来源**: [https://x.com/SebastienBubeck/status/2057187978720719114](https://x.com/SebastienBubeck/status/2057187978720719114)
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Claim: AI can make scientific breakthroughs.
Proof: An internal OpenAI model resolved the most famous conjecture in discrete geometry, namely the optimality (or lack thereof) of the grid for the unit distance problem. This conjecture had seen no progress, despite a lot of interest, since its inception 80 years ago. (There was a lot of activity and progress AROUND it though!)
Let me use this thread to explain concretely what happened. You can also find explanations at varying levels of complexity in our blogpost, in the companion paper written by world leading mathematicians (to appear on arxiv later today), in the report with the original AI proof, and in the (rewritten) chain of thought of the model solving the problem.
Okay so what are we talking about: the question is stupidly simple; if I put n points in the plane how many distances between those points can be the same? (By rescaling you can also just ask how many of those distances can be equal to 1, hence the name “unit” distance problem). Well, certainly you could put one point in the center of a circle and all the other ones on a circle centered at this point, which would result in n-1 distances being the same. And obviously there are at most n^2/2 distances. So what is the truth, is the best one can do of order n or of order n^2?
When Erdos introduced the problem in 1946 he analyzed the most natural construction for this problem: putting points on a simple grid. Okay so a point now has 4 neighbors on this grid, so certainly there are at least of order 2*n distances that are the same (2n and not 4n because of double counting). But let’s be a tiny bit more clever, instead of looking at distance 1 vertices (say the grid has unit length edges) we could look at vertices that are at distance sqrt(5) = sqrt(1+2^2). Just draw a little picture and you will see that there are 8 points at that distance! Indeed, you basically move along a L shape turned any way you want (and there 8 ways to do that). What Erdos proved (and I will give the proof below) is that you can keep going like this in powers of 2, up to about u(n) = 2^{log(n)/loglog(n)}. So it means the grid has at least about u(n)*n distances that are the same, and in fact this calculation is optimal for the grid. Note that u(n)*n = n^{1+o(1)} (specifically, n^{1+cst/loglog(n)}).
What Erdos conjectured is that the grid is essentially optimal: any configuration of points should have at most n^{1+o(1)} equal distances. This is the problem that saw zero progress in the past 80 years, again despite a lot of interest given how basic and natural this question is. My understanding is that Erdos strongly believed that the grid is optimal, and in fact in the closely related problem (introduced in the same 1946 paper!) of distinct distances he was vindicated. The distinct distances problem is simply the opposite version of the question, where one asks what is the minimal number of distinct distances that n points can form? The grid gives you n/sqrt(log(n)) distinct distances, and a breakthrough paper by Guth and Katz 10 years ago showed that this is indeed essentially optimal with a lower bound of n/log(n). In other words: everything was pointing to the grid being also an optimal candidate for the unit distance problem.
This is where the OpenAI internal comes in. It actually STRONGLY disproved this long-held belief and found a new (mind blowing) construction with a number of equal distances of order n^{1+delta} for some delta>0. To say a few words about how this breakthrough was achieved by the model I first need to tell you a little bit more about Erdos’ proof and where the 2^{log(n)/loglog(n)} comes from. Turns out the primes are lurking around!
We will assume two things about prime numbers: first the prime number theorem that says that there are about n/log(n) primes below n (well actually we need a slightly more refined version but it doesn’t matter for the level of this exposition). Second, that if a prime is equal to 1 modulo 4, then it factorizes over the Gaussian integers (which are integers of the form a+ib with a and b integers), namely in this case p = z bar{z}. For example 5=(1+2i)(1-2i), and this should remind you above when we counted 8 vertices at distance sqrt(5) = sqrt(1+2^2). Okay so now take the first k primes that equal to 1 modulo 4, p_1, …, p_k, and consider the number R=p_1…p_k = z_1 bar{z_1} … z_k \bar{z_k}. The key point is that we get 2^k Gaussian integers out of this with modulus equal to sqrt{R}, by selecting for each prime p_i to take either take z_i or bar{z_i} and then take their product (crucially we use that the modulus is multiplicative and that conjugation preserves the modulus). In other words we have found 2^k points at distance sqrt{R} from the origin on the grid! (To be precise we also have to prove that these points are distinct, which is where unique factorization in Z[i] becomes important, and something that will be key in the new proof, but let's ignore that here.) So now we just need to see how big we can take k while keeping sqrt{R}<sqrt{n} (the latter being the side length of a grid with n points). We have log(R) = sum_{i=1}^k log(p_i) which by the prime number theorem is roughly sum_{i=1}^k log(ilog(i)) which is basically klog(k). So we need klog(k) less than log(n), so k should be like log(n)/loglog(n), and we get the claimed 2^k = 2^{log(n)/loglog(n)}.
The above one paragraph argument (clever one I will give you that) has remained SOTA for 80 years. Now what AI did is pretty crazy in my opinion. First of all, as can be seen in the CoT, it almost immediately decided to try to improve the grid construction, which is the opposite of what most mathematicians had been trying to do so far. In my limited understanding the strategy it came to (and which it executed perfectly) is roughly like this: wouldn’t it be great if there was more ways to the split the primes? Maybe if we were to consider another field than Q, one of higher degree, then this could work with the integers Z replaced by the ring of integers of that field? Maybe instead of 2^k we could get 2^{f k} where f is the degree of the field? First guess would be to look at cyclotomic extensions but the model does that first in its CoT and quickly realizes that this won’t work. It keeps working hard and eventually brings in the language of ideals where there can be non-unique factorization that are handled by a class group. Now you need to start thinking about how you will construct high degree fields with all the parameters controlled (first the class number, but also this will be a higher dimensional lattice, so it will need to be projected back to the complex plane, and this projection will induce some collapsing that needs to be controlled, and on and on). That’s where the model uses a hammer from class field theory, the infinite towers from Golod-Shafarevich. At this point it’s probably better for you to head to the companion paper by actual experts on the subject for further details!
Okay let me take a step back: basically what the AI did is that it was able to use its vast knowledge of all of mathematics, to see a connection between discrete geometry and algebraic number theory, and then crucially it was able to masterfully chain together the argument, with expert level calculations at every step. It is truly a breakthrough result, yet at the same time it is also true that the model didn’t “invent” any “new mathematics” (say it didn’t invent some alternative class field theory, whatever that would mean). But this is the crucial point: merely being able to know deeply all the results in a scientific field, and being able to use all known arguments expertly and with just the right choice of parameters, that alone can lead to a ton of breakthroughs, and this is not just limited to mathematics, this type of (extremely) solid expert execution is the bread and butter of many many scientific advances.
Finally a word about what this means for mathematics going forward. The companion paper has a lot of reflections on that from leading mathematicians, so better to just read what THEY have to say. But one thing interesting to note is that we are NOT submitting the model’s proof to arxiv. Indeed no human author can claim to have contributed in the traditional sense (although of course it’s really the fruit of all human researchers in OpenAI who have created this amazing model, as well as humanity in general developing mathematics for millennia ...). On the other hand the companion paper by humans goes beyond just reflections on the significance of the moment, it also digests the proof, puts it into broader context, and even simplifies it a bit. While the community still has a lot of work to do to fully adapt to these new developments, we believe that this principle of separating the AI proof from the human’s understanding of it will be an important piece of the puzzle.
## 相关链接
- [Sebastien Bubeck](https://x.com/SebastienBubeck)
- [@SebastienBubeck](https://x.com/SebastienBubeck)
- [our blogpost](https://openai.com/index/model-disproves-discrete-geometry-conjecture/)
- [the companion paper](https://cdn.openai.com/pdf/00191fd2-3b93-47a3-aff3-6e8bcf787959/unit-distance-proof.pdf)
- [the report](https://cdn.openai.com/pdf/ab811408-0d1b-473c-a7c3-6978a122e3d4/unit-distance-remarks.pdf)
- [the (rewritten) chain of thought](https://cdn.openai.com/pdf/1625eff6-5ac1-40d8-b1db-5d5cf925de8b/unit-distance-cot.pdf)
- [Upgrade to Premium](https://x.com/i/premium_sign_up)
- [3:53 AM · May 21, 2026](https://x.com/SebastienBubeck/status/2057187978720719114)
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*导出时间: 2026/5/21 10:35:28*
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## 中文翻译
# 单位距离
**作者**: Sebastien Bubeck
**日期**: 2026-05-20T19:53:29.000Z
**来源**: [https://x.com/SebastienBubeck/status/2057187978720719114](https://x.com/SebastienBubeck/status/2057187978720719114)
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结论：AI 可以取得科学突破。
证明：OpenAI 的一个内部模型解决了离散几何中最著名的猜想，即关于网格在单位距离问题中的最优性（或缺乏最优性）。尽管外界对此兴趣浓厚，但在该问题提出后的 80 年里，该猜想一直未能取得进展。（尽管其周边确实有很多活跃的研究和进展！）
让我用这个帖子来具体解释一下发生了什么。您也可以在我们的博客文章中、由世界顶尖数学家撰写的配套论文（将于今天晚些时候在 arxiv 上发布）中、包含原始 AI 证明的报告中，以及模型解决问题的（重写后的）思维链中，找到不同复杂程度的解释。
好吧，那么我们到底在讨论什么：这个问题其实简单得有些“蠢”；如果我在平面上放置 n 个点，这些点之间能有多少对距离是相等的？（通过重新缩放，你也可以直接问有多少对距离可以等于 1，这就是“单位”距离问题的名称由来）。嗯，当然你可以把一个点放在圆心，把其他所有点放在以该点为中心的圆周上，这样就会有 n-1 对距离相等。显然，距离的总数最多也就是 n^2/2。那么真相是什么？最好的结果是与 n 同阶（linear）还是与 n^2 同阶（quadratic）呢？
当 Erdos 在 1946 年提出这个问题时，他分析了针对该问题最自然的构造：将点放在简单的网格上。好吧，现在网格上的一个点有 4 个邻居，所以肯定至少有大约 2n 对距离是相等的（是 2n 而不是 4n，因为有重复计数）。但让我们稍微再聪明一点，除了看距离为 1 的顶点（假设网格的边长为 1），我们还可以看距离为 sqrt(5) = sqrt(1+2^2) 的顶点。只要画个小图你就会发现，那个距离有 8 个点！确实，你基本上是沿着任意方向旋转的“L”形移动（有 8 种方式可以做到这一点）。Erdos 证明的内容（我会在下面给出证明）是，你可以继续这样以 2 的幂次推进，直到大约 u(n) = 2^{log(n)/loglog(n)}。这意味着网格至少有大约 u(n)*n 对相等的距离，实际上对于网格而言，这个计算是最优的。注意 u(n)*n = n^{1+o(1)}（具体来说是 n^{1+cst/loglog(n)}）。
Erdos 猜想网格本质上是最优的：任何点构型最多都只有 n^{1+o(1)} 对相等距离。这就是在过去 80 年里零进展的问题，尽管鉴于这个问题多么基础和自然，外界对此有极大的兴趣。我的理解是，Erdos 坚信网格是最优的，事实上，在一个密切相关的问题（也在 1946 年的同一篇论文中提出！）——不同距离问题——上，他是被证明正确的。不同距离问题简单来说就是这个问题的反面，它问的是 n 个点能组成的最少的不同距离是多少？网格给出的不同距离是 n/sqrt(log(n))，而 Guth 和 Katz 10 年前的一篇突破性论文表明，通过 n/log(n) 的下界，这确实本质上是最优的。换句话说：一切迹象都表明网格也是单位距离问题的最优候选。
这就是 OpenAI 内部模型介入的地方。它实际上有力地反驳了这一长期存在的信念，并找到了一种新的（令人震惊的）构造，其相等距离的数量达到了 n^{1+delta} 阶，其中 delta > 0。为了多谈谈这个突破是如何由模型实现的，我首先需要多讲一点 Erdos 的证明以及 2^{log(n)/loglog(n)} 是从哪里来的。事实证明，质数潜伏在其中！
我们将假设关于质数的两件事：首先是质数定理，它指出在 n 以下大约有 n/log(n) 个质数（嗯，实际上我们需要一个稍微精细一点的版本，但这对于本文的解释水平并不重要）。其次，如果一个质数模 4 余 1，那么它可以在高斯整数（即形如 a+ib 且 a、b 为整数的整数）上分解，即在这种情况下 p = z bar{z}。例如 5=(1+2i)(1-2i)，这应该能让你回想起上面我们计算距离 sqrt(5) = sqrt(1+2^2) 时有 8 个顶点的情况。好吧，现在取前 k 个模 4 余 1 的质数，p_1, …, p_k，并考虑数字 R=p_1…p_k = z_1 bar{z_1} … z_k \bar{z_k}。关键点在于，我们从中得到了 2^k 个模等于 sqrt{R} 的高斯整数，方法是对每个质数 p_i 选择取 z_i 或 bar{z_i}，然后取它们的乘积（关键在于我们使用了模的乘法性质以及共轭不改变模的性质）。换句话说，我们在网格上找到了距离原点 sqrt{R} 的 2^k 个点！（准确地说，我们还需要证明这些点是不同的，这在涉及 Z[i] 中的唯一分解定理时变得很重要，这也是新证明中的关键，但让我们在这里先忽略这一点）。所以现在我们需要看看在保持 sqrt{R}<sqrt{n}（后者是具有 n 个点的网格的边长）的情况下，我们可以把 k 取多大。我们有 log(R) = sum_{i=1}^k log(p_i)，根据质数定理，这大约是 sum_{i=1}^k log(ilog(i))，基本上就是 klog(k)。所以我们需要 klog(k) 小于 log(n)，所以 k 应该像 log(n)/loglog(n)，我们就得到了声称的 2^k = 2^{log(n)/loglog(n)}。
上述这一段论证（我会承认它很聪明）在 80 年里一直是最先进的水平（SOTA）。现在 AI 所做的在我看来非常疯狂。首先，正如在思维链（CoT）中看到的那样，它几乎立刻决定尝试改进网格构造，这与大多数数学家迄今为止一直尝试做的恰恰相反。以我有限的理解，它提出的策略（并且它完美地执行了）大致如下：如果有更多分解质数的方法岂不是很好？也许如果我们考虑 Q 以外的域，一个更高次数的域，那么这可以用该域的整数环代替整数 Z 来实现？也许我们得到的不是 2^k 而是 2^{f k}，其中 f 是域的次数？最初的猜想是看分圆扩张，但模型在其思维链中首先尝试了这一点，并很快意识到这行不通。它继续努力工作，最终引入了理想的语言，在那里存在由类群处理的非唯一分解。现在你需要开始思考如何构建所有参数都受控的高次域（首先是类数，但这将是一个高维格，所以它需要被投影回复平面，这种投影会引发一些需要控制的坍缩，等等）。这就是模型使用类域论的大锤——Golod-Shafarevich 无限塔——的地方。在这一点上，您最好去阅读该主题实际专家撰写的配套论文以获取更多细节！
好吧，让我退一步讲：基本上 AI 所做的是，它能够利用其对所有数学的庞大知识，看到离散几何和代数数论之间的联系，然后关键是它能够熟练地将论证串联起来，并在每一步进行专家级的计算。这确实是一个突破性的结果，但与此同时，模型并没有“发明”任何“新数学”（比如说它没有发明某种替代的类域论，不管那意味着什么）。但这正是关键点：仅仅能够深入了解一个科学领域的所有结果，并能够专家级地使用所有已知的论证，加上恰到好处的参数选择，仅仅这一点就能导致大量的突破，这不仅仅局限于数学，这种（极其）扎实的专家级执行是许多许多科学进步的衣食父母。
最后，关于这对未来数学意味着什么，我也想说几句。配套论文中有很多顶尖数学家对此的思考，所以最好直接阅读他们不得不说的话。但有趣的一点要注意，我们并没有将模型的证明提交到 arxiv。确实，没有人类作者可以声称在传统意义上做出了贡献（尽管当然这实际上是 OpenAI 所有创造这个惊人模型的人类研究人员的成果，以及数千年来发展数学的人类大众的成果……）。另一方面，人类撰写的配套论文不仅仅是对这一时刻意义的反思，它还消化了证明，将其置于更广泛的背景中，甚至对其进行了一些简化。虽然社区仍有很多工作要做来完全适应这些新发展，但我们相信，这种将 AI 证明与人类对它的理解区分开来的原则，将是拼图中的重要一块。
## 相关链接
- [Sebastien Bubeck](https://x.com/SebastienBubeck)
- [@SebastienBubeck](https://x.com/SebastienBubeck)
- [我们的博客文章](https://openai.com/index/model-disproves-discrete-geometry-conjecture/)
- [配套论文](https://cdn.openai.com/pdf/00191fd2-3b93-47a3-aff3-6e8bcf787959/unit-distance-proof.pdf)
- [报告](https://cdn.openai.com/pdf/ab811408-0d1b-473c-a7c3-6978a122e3d4/unit-distance-remarks.pdf)
- [（重写后的）思维链](https://cdn.openai.com/pdf/1625eff6-5ac1-40d8-b1db-5d5cf925de8b/unit-distance-cot.pdf)
- [升级到 Premium](https://x.com/i/premium_sign_up)
- [3:53 AM · May 21, 2026](https://x.com/SebastienBubeck/status/2057187978720719114)
- [168.4K Views](https://x.com/SebastienBubeck/status/2057187978720719114/analytics)
- [View quotes](https://x.com/SebastienBubeck/status/2057187978720719114/quotes)
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*导出时间: 2026/5/21 10:35:28*