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Actually, why make a permutation matrix (mostly zeros and ones) and multiply by A (cost of O(n^3)). You can directly compute A with permuted rows/columns, then compute the error in (permuted) A. This has cost O(n^2) (computing the error) instead of O(n^3). IF the error is reduced, you can return the permutation, or even the permutation matrix (which can be constructed at the very end).
Actually, why make a permutation matrix (mostly zeros and ones) and multiply by A (cost of O(n^3)). You can directly compute A with permuted rows/columns, then compute the error in (permuted) A. This has cost O(n^2) (computing the error) instead of O(n^3). IF the error is reduced, you can return the permutation, or even the permutation matrix (which can be constructed at the very end).
Originally posted by @PaulWAyers in #62 (comment)
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