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    AI verified a years-old polynomial proof in Lean, a coauthor says

    The coauthor says the method evaluates a degree-n polynomial using ⌈(n+1)/2⌉ multiplications, a count they say they proved optimal.

    TA
    1 Source, 19d ago, first seen 19d ago

    TLDR

    A coauthor says the team proved the result years ago without AI, but the proof ran to 100 pages and they weren't confident enough to publish. They say AI has since verified it in Lean, a proof assistant. Their method reportedly needs ⌈(n+1)/2⌉ multiplications—rounding (n+1)/2 up to a whole number—to evaluate a degree-n polynomial, and they claim that count is optimal. The author highlights the potential benefit for finite fields, where they say multiplication is particularly expensive, and shares a tool for trying the method.

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    1 Source, first seen 19d ago

    Combined views

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    1 Source, first seen 19d ago

    566 likes
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    1 Source

    @thomasahleCompute Polynomials Twice as Fast - Here's a fun result, done entirely without AI. That because we proved it years ago, but it ran a hundred pages and we weren't quite sure enough to publish. Now AI verified it in Lean. A fun fact about polynomials is that P(x) = x⁴ + a₃x³ + a₂x² + a₁x + a₀ can be evaluated in just two(!) multiplications! The trick is to write y = (x + b₀)x + b₁ P(x) = (y + x + b₂)y + b₃ where b₀…b₃ are easy to calculate from a₀…a₃. Polynomials are everywhere from computing exp/sin, to cryptographic hashes and codes. Over finite fields multiplications are particularly expensive, so a 2x speedup matters. Donald Knuth and others showed ~n/2 multiplications suffice for any degree n, but their preprocessing needs complex roots: numerically unstable and useless over finite fields. Rabin & Winograd fixed that with rational preprocessing, but 2logn extra multiplications, which hurts at the small n we most care about. Our new method solves this. It uses ⌈(n+1)/2⌉ multiplications, and we prove this is optimal. You can try it out on your own polynomials at http://thomashale.com/fast-polynomials

    1 Source

    @thomasahleCompute Polynomials Twice as Fast - Here's a fun result, done entirely without AI. That because we proved it years ago, but it ran a hundred pages and we weren't quite sure enough to publish. Now AI verified it in Lean. A fun fact about polynomials is that P(x) = x⁴ + a₃x³ + a₂x² + a₁x + a₀ can be evaluated in just two(!) multiplications! The trick is to write y = (x + b₀)x + b₁ P(x) = (y + x + b₂)y + b₃ where b₀…b₃ are easy to calculate from a₀…a₃. Polynomials are everywhere from computing exp/sin, to cryptographic hashes and codes. Over finite fields multiplications are particularly expensive, so a 2x speedup matters. Donald Knuth and others showed ~n/2 multiplications suffice for any degree n, but their preprocessing needs complex roots: numerically unstable and useless over finite fields. Rabin & Winograd fixed that with rational preprocessing, but 2logn extra multiplications, which hurts at the small n we most care about. Our new method solves this. It uses ⌈(n+1)/2⌉ multiplications, and we prove this is optimal. You can try it out on your own polynomials at http://thomashale.com/fast-polynomials