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Patrick2788
Silver Contributor
Jul 18, 2023

Formula Challenge: The most efficient way to generate a Number Spiral (Ulam spiral) ?

The goal: Ulam spiral - Wikipedia   The trick is creating a function capable of producing the largest matrix of numbers possible (this may rule out the recursive approach).   The approach ...
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Formulas and Functions
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Patrick2788's avatar
Patrick2788
Silver Contributor
Aug 31, 2025

One way to speed up the spiral is to use thunks/lazy evaluation all the way down:

SpiralĪ» = LAMBDA(n, [start_at],
        LET(
            start_at, IF(ISOMITTED(start_at), 1, start_at),
            i, LAMBDA(SEQUENCE(n * 2 + 1, , n, -1)),
            j, LAMBDA(SEQUENCE(, n * 2 + 1, -n, 1)),
            x, LAMBDA(j() * SEQUENCE(n * 2 + 1, , 1, 0)),
            y, LAMBDA(i() * SEQUENCE(, n * 2 + 1, 1, 0)),
            a_, LAMBDA((ABS(x()) > ABS(y())) * (x() > 0)),
            b_, LAMBDA((ABS(x()) > ABS(y())) * (x() <= 0)),
            c_, LAMBDA((ABS(x()) <= ABS(y())) * (y() > 0)),
            d_, LAMBDA((ABS(x()) <= ABS(y())) * (y() <= 0)),
            a, LAMBDA(4 * x() ^ 2 - 3 * x() + y()),
            b, LAMBDA(4 * x() ^ 2 - x() - y()),
            c, LAMBDA(4 * y() ^ 2 - y() - x()),
            d, LAMBDA(4 * y() ^ 2 - 3 * y() + x()),
            result, start_at + IF(a_(), a(), IF(b_(), b(), IF(c_(), c(), d()))),
            result
        )
    );

This generates a spiral with 51,854,401 elements in about 78 seconds.

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