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Jan 21, 2018 at 17:52 history edited Turbo CC BY-SA 3.0
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Jan 21, 2018 at 1:57 comment added Josiah Park The size of the matrix $n$ may show up in the bound unlike the $2^{O(r)}\times 2^{O(r)}$ bound.
Jan 21, 2018 at 1:51 comment added Josiah Park A lower bound is achievable by looking at finite fields of order $p=2b+1$, when $p$ is prime. Is the bound independent of the size of the matrix $n$?
Jan 19, 2018 at 10:47 comment added Vincent Why $2b + 1$? In the $\pm 1$-matrices case this would evaluate to $3$ rather than $2$, right?
Jan 18, 2018 at 21:24 history edited Turbo CC BY-SA 3.0
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Jan 18, 2018 at 21:06 history edited Turbo CC BY-SA 3.0
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Jan 18, 2018 at 14:02 history edited Turbo CC BY-SA 3.0
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Jan 18, 2018 at 13:51 history asked Turbo CC BY-SA 3.0