After seeing the Jacobian conjecture and the Maxwell conjecture proven false, I decided to try Kimi K3 on the 4chan problem. If you don’t know it: it’s the question that was born on 4chan’s /sci/ board in 2011, what’s the shortest string that contains every possible ordering of $n$ episodes as a contiguous binge-watch? People call them superpermutations; $s(n)$ is the answer. For $n = 6$ it’s been stuck between 867 and 872 for over a decade.
A video explains it a lot better.
Kimi did research and didn’t find any new improvements at the time, so I went ahead and tried to improve the bounds myself. And we actually found something: a neat, human-understandable lemma I’m calling the absorption lemma. It says every “2-loop” in the problem has exactly five doors in, so if your walk has $R$ segments you must have entered at least $\lceil (R-1)/5 \rceil$ of those loops. The two best constructions anyone has ever found both hit that bound exactly, which is how you know it’s the right invariant. Combined with a rigidity argument and an exhaustive check of all 10,068 possible covers, it proves $s(6) \ge 868$. The full writeup with the math is here, and it’s the one page of all this that I’d genuinely recommend reading.
Then I went looking for someone to tell about it, and that’s when I found the Superpermutators Google group, and my heart sank a little. Not only was $s(6) \ge 868$ essentially known since an unpublished 2019 draft by Zach Hunter, but in the two weeks right before we finished, a wave of AI-assisted work had hit the problem:
So my 868 is not a new lower bound. It’s an independent proof of something the field got three days earlier by three different methods. In computer-assisted math, independent confirmations genuinely matter, but let’s be honest about what happened here: AI ripped superpermutations wide open, and I got to watch it happen in real time from about three days behind the wave.
And since everyone loves the punchline: you need to watch 872 episodes of the first 6 Haruhi episodes back to back in order to have watched every way you could have arranged them during the binge itself. That’s what ten years of silence and two weeks of chaos were about.
| $n$ | lower bound | upper bound | status |
|---|---|---|---|
| 6 | 872 (preliminary, vlad-ds) | 872 (Houston 2014) | probably done |
| 7 | 5888 (Hunter & Raudvere, Lean) | 5906 (Coanda/Egan/Houston 2019) | the open frontier |
| 8 | 46103 (Hunter & Raudvere, Lean) | 46204 | wide open |
The n = 7 gap is where the game is now, and we’ve narrowed it to one question: can 141 of the 840 two-loops cover all 720 one-cycles? Yes+cheap = a 5905-string (new record); no = s(7) >= 5906 (settled). The 5905 question, explained. Our baseline notes already have one fun observation in them: the absorption lemma holds at $n = 7$ too, and the current 5906 champion is the only known string anywhere that doesn’t saturate it (it’s off by exactly one). Make of that what you will.
The repo with all the verification machinery: github.com/levkropp/superperm. The lemma, properly typeset: here.