What’s left over?

0:00The question

Start with n dots and every possible link between them. Now pick a triangle at random, among the triangles whose three links are all still there, and erase its three links. Repeat, until no triangle is left. The erased triangles never share a link, so they form a packing, and the links that survive are called its leave. How many survive? For tiny networks we can compute the answer exactly. On 7 dots, about 28% of the time nothing is left at all: the process stumbles onto a perfect packing, the Fano plane. Here is one run on 9 dots: there are 84 triangles to choose from at first, then 65, 50, and so on down to one, and 9 links remain. For large n, the question is how the leave grows, and with what constant. The paper tells the history this way. Bollobás and Erdős conjectured in 1990, as reported by Bohman, Frieze and Lubetzky, that the expected leave has order n^(3/2). Spencer’s branching-process analysis, and the nibble argument of Rödl and Thoma, showed that the leave is a vanishing fraction of all the links. Grable proved a bound of n^(11/6), plus any small amount, and outlined the calculations for 7/4. Bohman, Frieze and Lubetzky gave a short proof of n^(7/4), up to a logarithmic factor, and then, in 2015, proved that the leave is n^(3/2+o(1)), with high probability: the exponent, though not yet the exact order. Felix Joos and Marcus Kühn developed removal estimates for general, strictly balanced hypergraphs, and conjectured an exact constant; for triangles, it is 1/(2√2).

1:39The claim

Result 188 is one OpenAI preprint, 25 pages, dated September 25, 2026. It claims: for triangle removal starting from the complete graph, the number of links left at the end, divided by n^(3/2), converges in mean square to 1/(2√2), about 0.354. So the same holds in probability, and for the average. This is the triangle case of Joos and Kühn’s conjecture, which they stated in probability; the paper proves it in mean square. Our simulations of the process approach the constant from below: about 92% of it at 50 dots, 98% at 1,000, and 99% at 5,000, with the spread shrinking as n grows. Not claimed: any rate of convergence; the overall error is only shown to vanish. No fluctuation law. Only removal from the complete graph: the paper does not assert a sharp constant for other starting graphs, or for general hypergraph removal. The exponent 3/2 is not new; it is Bohman, Frieze and Lubetzky’s, up to that little o. And the proof is not self-contained: for most of the process, it relies on Joos and Kühn’s estimates, which the paper calls its only external proof input.

3:00How the proof goes

The proof hands off the early game. Joos and Kühn’s estimates show that, for most of the process, the graph looks like a random graph whose density shrinks on a predictable track. The paper stops their clock when the density is n^(−1/2+1/2000), so that each remaining link lies in about D triangles, where D is n^(1/1000). Everything after that is proved in the paper itself, though not all of it is new. Give every triangle a random priority, a standard trick, and scan them in order, accepting a triangle when its links are still there; this gives exactly the same process. A link survives up to some time exactly when no earlier triangle through it finds its other two links still there at its own priority. So survival becomes a recursive question, asked of lower-priority triangles, which ask further questions, and the paper tests the two links of each triangle together, with one jointly ordered list. Now imagine the questions never overlap: every occurrence of a triangle gets fresh randomness. In that idealized tree, a link survives to time t with probability close to Spencer’s, 1/√(1+2Dt). In our simulation on 300 dots, the real process follows that curve closely, though applying it from the start is our heuristic. The constant falls out: m links at the hand-off, each surviving with probability 1/√(1+2D), gives n²p/(2√(1+2np²)), which tends to n^(3/2)/(2√2), whatever the density p at the hand-off, as long as D is large. Joos and Kühn’s own heuristic reaches the same point from the triangle count, shown here for 1,000 dots. The work is to show that the real process matches the idealized tree accurately enough for second moments. That precision matters because the chance that two given links both survive is itself of order 1/D. The paper names two new estimates. First, stability in the maximum norm: small differences between links do not grow, which it proves from cycle counts in the graphs around each vertex and a bound on the resulting evolution. Second, collisions: the real process differs from the tree only when the same triangle turns up twice, and a repeat forces an extra link joining points on at most two query paths, a rare picture that the paper counts and prices, for an error smaller than 1/D. Then the mean and the second moment of the normalized leave both tend to one, so its variance tends to zero.

5:41Where it stands

The preprint is unrefereed and has not been revised. OpenAI’s Lean files state the main theorem exactly, with all three forms of convergence, and prove it in 287 files and about 34,000 lines, with no gaps we could find by search; we did not run the check. Notably, the Lean proof proves its own version of the early-process estimates that the paper takes from Joos and Kühn. The paper is not among those OSVerify reviewed, and we have seen no expert review. If it holds, random greedy triangle removal from the complete graph typically leaves, for large n, about 0.354 times n^(3/2) links, the constant Joos and Kühn predicted.