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Oct 17, 2022 at 20:34 comment added Michael Hardy For an orthogonal matrix, is there such a thing as the permutation matrix that is closest to it? (If so, that should do it.)
Oct 17, 2022 at 19:29 history edited Ben Deitmar CC BY-SA 4.0
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Oct 17, 2022 at 18:46 answer added Christophe Leuridan timeline score: 1
Oct 17, 2022 at 17:26 comment added YCor Using a Voronoi tiling you get 1,2,3 for free (define the projection arbitrarily on boundaries. However it's easily definable but maybe not easily computable. No idea about 4.
Oct 17, 2022 at 17:26 comment added Ben Deitmar Thank you! Up to a measure zero is fine.
Oct 17, 2022 at 15:59 answer added Neil Strickland timeline score: 3
Oct 17, 2022 at 14:58 comment added David E Speyer Does it really have to be defined for all $g \in O(n)$, or are you happy working up to a set of measure zero? If the latter, send any $g \in O(n)$ to the permutation $\sigma$ which maximizes $\sum_j g_{\sigma(j) j}$, discarding the matrices where ties occur. This is even rapidly computable, since it is an instance of maximum weight matching en.wikipedia.org/wiki/Maximum_weight_matching .
Oct 17, 2022 at 14:27 history asked Ben Deitmar CC BY-SA 4.0