Ryan O'Donnell retweeted

Ryan Williams @rrwilliams.bsky.social@rrwilliams55m
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Here's a direct link to OpenAI's PDF claiming a refutation of ETH. In particular, it's claimed that 3SAT on n variables is in about exp(n/(log n)^{1/2+o(1)}) time
github.com/openai/math/blob/…
Ryan O'Donnell retweeted
Happy to share that our better-than-2 approximation results for k-median are now on arXiv:
arxiv.org/abs/2610.18144
arxiv.org/abs/2610.17172
It has been a pleasure working with this fantastic group: @rares_buhai , Vincent Cohen-Addad, Henry Fleischmann, @anupamg , Antoni "Toni" Jubés Monforte, Madhusudhan Reddy Pittu, @mstou73, and @LukasBirdyy
A special shout-out to Toni, who started working on this problem for his Master's thesis almost two years ago. His work led to the impressive analysis of the factor-revealing LP!
Now on to other things...
Sorry for spelling @buyolitsez2's name wrong; obviously it is 'Nikolai'.
I looked forward to this cleanup! Nikloai Chukhin (eccc.weizmann.ac.il/report/2…) significantly cleans and extends the OpenAI 2^(ω(sqrt(n))) lower bound for depth-3 circuits:
The (4n-1)-bit function sum_ij a_{i+j}x_iy_j mod 2 needs depth-d circuits of size 2^(Ω~(n log n)^{1/(d-1)}).
Replying to @TCzajka
Yeah, I just have no intuition about whether the algorithm can be made oblivious. But if you think it seems likely... that would be cool. Hopefully someone will work it out. Because this result seems noticeably more compelling to me if it also works for circuits.
Replying to @TCzajka
Seems to me that Boolean circuits might be the most natural model for investigating the problem? Has anyone discovered anything for circuits, post-OpenAI paper?
Replying to @AcerFur
It depends on the model, of course. In the standard model of algorithms you use in an algorithms class (namely, the RAM model with log n size registers, with multiplying two registers in time O(1) allowed), the complexity is O(n). This is not trivial, but known since the 1970s.
Replying to @rrwilliams
Also exponent 2d/(3d+2) for one d-dim tape.
Replying to @rrwilliams
My GPT claims the techniques can give exponent 2/3 for RAM model. I'll try to check if this is accurate.
Ryan O'Donnell retweeted
this is great: a unified and transparent exposition of the recent breakthroughs on the decoding of Reed-Solomon codes.
eccc.weizmann.ac.il/report/2…
Ryan O'Donnell retweeted

𝖬𝖺𝗁𝖽𝗂 𝖢𝗁. (bluesky:@mahdi.ch)@mahdi_tcs_Oct 6
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The what now?!
arxiv.org/abs/2610.06783v1
Ryan O'Donnell retweeted
oh that's interesting! The authors share the draft generated by Astra, as a companion to their arxiv submission.
arxiv.org/abs/2609.28028
STOC 2027 Call for Papers is out:
acm-stoc.org/stoc2027/stoc20…
Note quite a lot of changes from previous STOC CfPs.
Replying to @mahdi_tcs_ @OpenAI
Naw, SETH is true, GISO requires n^log(n) time, and no strongly poly algorithm for LP ;-)
Ryan O'Donnell retweeted
An amazing result in online algorithms. I worked on this for over a decade and it is fantastic to see a solution that looks very clean! Also great to see an expert behind the wheels. Congratulations @sahilsingla81 arxiv.org/abs/2609.14555
Replying to @BooleanAnalysis @henryquantum
... one might strive further to get the best of both worlds: "Rich" 2-to-1 with perfect completeness (eccc.weizmann.ac.il/report/2…). That would imply both results. Perhaps the rumors might concern this?
Replying to @henryquantum
I guess this result is morally close to 2-to-1 Conjecture with perfect completeness. (Maybe with tweaking you can get 4->2, who knows.)
Then it's true that UGC is somewhat different, being stronger in one sense (1-to-1) and weaker in another (imperfect completeness).
But...




