
Professor @MITEECS and @MIT_CSAIL. Complexity, algorithms, and related math. I'll let you know when P != NP (and when it's not)
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![We show that for all k, there exists εₖ > 0 such that graph k-coloring can be solved by a randomized algorithm with one-sided error in time O((2-εₖ)ⁿ). Prior to this work and independent concurrent work of Zamir [arXiv, 2026], exponential improvements over the 2ⁿ ⋅ poly(n)-time algorithm of Björklund, Husfeldt, and Koivisto [SIAM Journal on Computing, 2009] were only known for k ≤ 6.](https://pbs.twimg.com/media/HOfM0-mXcAA9yg6.png?name=small&format=webp)
ALT We show that for all k, there exists εₖ > 0 such that graph k-coloring can be solved by a randomized algorithm with one-sided error in time O((2-εₖ)ⁿ). Prior to this work and independent concurrent work of Zamir [arXiv, 2026], exponential improvements over the 2ⁿ ⋅ poly(n)-time algorithm of Björklund, Husfeldt, and Koivisto [SIAM Journal on Computing, 2009] were only known for k ≤ 6.
![We show that for all k, there exists εₖ > 0 such that graph k-coloring can be solved by a randomized algorithm with one-sided error in time O((2-εₖ)ⁿ). Prior to this work and independent concurrent work of Zamir [arXiv, 2026], exponential improvements over the 2ⁿ ⋅ poly(n)-time algorithm of Björklund, Husfeldt, and Koivisto [SIAM Journal on Computing, 2009] were only known for k ≤ 6.](https://pbs.twimg.com/media/HOfM0-mXcAA9yg6.png?name=small&format=webp)
ALT We show that for all k, there exists εₖ > 0 such that graph k-coloring can be solved by a randomized algorithm with one-sided error in time O((2-εₖ)ⁿ). Prior to this work and independent concurrent work of Zamir [arXiv, 2026], exponential improvements over the 2ⁿ ⋅ poly(n)-time algorithm of Björklund, Husfeldt, and Koivisto [SIAM Journal on Computing, 2009] were only known for k ≤ 6.

ALT Fable chain of thought: "4nln²p/p ≤ n/(4p³): 16p²ln²p ≤ 1: NEVER!!!! THE LADDER IS NEVER THE BEST?!?! Let me recheck the earlier example: p = √(log n) hmm earlier I computed p≈√log n, L≈n/log n: new ≈ √(p³nL′) = (((log n)^{3/4}·n/log n)^{1/2}? hmm: p³ = log^{1.5}n; nL′ = n²/log n: product p³nL′ = n²log^{0.5}n: sqrt: nlog^{0.25}n: TIMES 2: ≈ 2n log^{1/4}n > n!!! WORSE than product!! Earlier I wrote √(p³nL) = n/log^{1/4}n” — ARITHMETIC ERROR (I divided instead of multiplied). Let me recompute: p^{3/2}= log^{3/4}n; √(nL′) = √(n²/log n) = n/log^{1/2}n: product: n·log^{3/4-1/2} = n log^{1/4}n. Yes: WORSE than product. So the ladder construction is DOMINATED. Damn!! OK let me recheck the optimization — maybe I set up"













