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Cake day: February 6th, 2024

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  • If the growth is superexponential, we make it so that each successive doubling takes 10% less time.

    (From AI 2027, as quoted by titotal.)

    This is an incredibly silly sentence and is certainly enough to determine the output of the entire model on its own. It necessarily implies that the predicted value becomes infinite in a finite amount of time, disregarding almost all other features of how it is calculated.

    To elaborate, suppose we take as our “base model” any function f which has the property that lim_{t → ∞} f(t) = ∞. Now I define the concept of “super-f” function by saying that each subsequent block of “virtual time” as seen by f, takes 10% less “real time” than the last. This will give us a function like g(t) = f(-log(1 - t)), obtained by inverting the exponential rate of convergence of a geometric series. Then g has a vertical asymptote to infinity regardless of what the function f is, simply because we have compressed an infinite amount of “virtual time” into a finite amount of “real time”.










  • Yes - on the theoretical side, they do have an actual improvement, which is a non-asymptotic reduction in the number of multiplications required for the product of two 4x4 matrices over an arbitrary noncommutative ring. You are correct that the implied improvement to omega is moot since theoretical algorithms have long since reduced the exponent beyond that of Strassen’s algorithm.

    From a practical side, almost all applications use some version of the naive O(n^3) algorithm, since the asymptotically better ones tend to be slower in practice. However, occasionally Strassen’s algorithm has been implemented and used - it is still reasonably simple after all. There is possibly some practical value to the 48-multiplications result then, in that it could replace uses of Strassen’s algorithm.