Timeline for Enumerating certain types of permutation polynomials
Current License: CC BY-SA 3.0
15 events
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| Jan 24, 2015 at 5:26 | history | edited | Anurag |
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| May 27, 2014 at 1:19 | comment | added | Anurag | @MichaelZieve: I have edited the question details again. Hopefully this is the last edit. | |
| May 27, 2014 at 1:18 | history | edited | Anurag | CC BY-SA 3.0 |
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| May 26, 2014 at 23:49 | comment | added | Anurag | $31555584/2^{13}$ = 3852. I think for $q = 3$ each perfect matching corresponds $2^{(1+3+3^2)}$ such polynomials because for every pair $U$, $V$ of $1$-dimensional subspaces with $f(U) = f(V)$ we can take any permutation of the non-zero elements of $V$ to define another permutation polynomial corresponding to the same perfect matching and since there are $13$ $1$-dimensional subspaces we have that factor. In general I'll have to be careful about this and I think I should re-formulate the conditions. (Thank you for pointing it out and I apologise for so many errors) | |
| May 26, 2014 at 22:43 | comment | added | Michael Zieve | I want to clarify that the current (7th) version of this question is internally inconsistent. For $q=3$ there are 31555584 permutations $f$ of $GF(q^3)$ which satisfy conditions (1) and (2). The OP's assertion that there are only 3852 such permutations is wrong. It appears that the OP did not compute these permutations, but got the number 3852 from the number of perfect matchings in the incidence graph of $PG(2,3)$. So the real question is to find the mistake in the OP's proof (given in version 7 of the question) that the number of such permutations equals the number of perfect matchings. | |
| May 26, 2014 at 22:04 | history | edited | Anurag | CC BY-SA 3.0 |
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| May 26, 2014 at 21:50 | review | Close votes | |||
| May 27, 2014 at 20:45 | |||||
| May 26, 2014 at 21:27 | history | edited | Anurag | CC BY-SA 3.0 |
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| May 26, 2014 at 21:24 | comment | added | Anurag | The number of perfect matchings in the incidence graph of $PG(2,3)$ is certainly $3852$. You can also check it here: oeis.org/A000794. So, there might be some error in the interpretation of a perfect matching as a permutation polynomial. I would check it again and see if I can find the error. | |
| May 26, 2014 at 20:29 | comment | added | Anurag | Edited. Is it clear now? | |
| May 26, 2014 at 20:29 | history | edited | Anurag | CC BY-SA 3.0 |
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| May 26, 2014 at 20:12 | history | edited | Anurag | CC BY-SA 3.0 |
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| May 26, 2014 at 20:06 | history | edited | Anurag | CC BY-SA 3.0 |
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| May 26, 2014 at 19:55 | history | edited | Anurag | CC BY-SA 3.0 |
clarified the trace function, it is not the absolute trace.
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| May 26, 2014 at 19:40 | history | asked | Anurag | CC BY-SA 3.0 |