Timeline for How large a subset of $\mathbb{F}_q^d$ can determine all determinants?
Current License: CC BY-SA 4.0
5 events
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| Nov 26, 2019 at 20:38 | comment | added | Gerhard Paseman | We don't . I am trying for not quite logarithmic representation. In the case of q=1009, I only need plus minus 504, which saves on about 70 primes and a few products. In the asymptotic picture, a multiplicative factor of two is easily ignored, since the work is in dividing up the factors. Gerhard "Is Having More Fun Dividing" Paseman, 2019.11.26. | |
| Nov 26, 2019 at 19:57 | comment | added | LSpice | One thing that I can't figure out how to make use of is that we get sign changes for free, by swapping rows, so we don't really have to cover (in the $q$ prime case) $0, \dotsc, q - 1$ but only, say, $0, \dotsc, \tfrac1 2(q - 1)$. | |
| Nov 26, 2019 at 18:54 | comment | added | Gerhard Paseman | I was rather loose in my estimations above. There are less than one hundred twenty products of two primes involved, and less than ten of three primes, so less than three hundred fifty vectors needed. With d=5 we can involve powers of seven and get away with a smaller number of vectors. Gerhard "This Is Just The Beginning" Paseman, 2019.11.26. | |
| Nov 26, 2019 at 18:31 | history | edited | Gerhard Paseman | CC BY-SA 4.0 |
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| Nov 26, 2019 at 18:22 | history | answered | Gerhard Paseman | CC BY-SA 4.0 |