Timeline for The calculation of permanent of a matrix
Current License: CC BY-SA 4.0
10 events
| when toggle format | what | by | license | comment | |
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| Oct 4, 2021 at 12:46 | comment | added | Max Alekseyev | Also, if permanent of the 0-1 matrix with elements of $A$ taken by absolute value is zero (which is easy to test), then permanent of $A$ is zero as well. | |
| Oct 4, 2021 at 5:53 | history | edited | YCor | CC BY-SA 4.0 |
removed capitals from title, added tag
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| Oct 4, 2021 at 4:50 | comment | added | Anurag Sahay | Here's a relevant answer from the cstheory stack exchange. cstheory.stackexchange.com/questions/32885/… Of course, this doesn't answer your question. | |
| Oct 4, 2021 at 4:42 | comment | added | Jacob.Z.Lee | That is a good idea. Unfortunately, the permanent of my concerned (1,-1,0)-matrix is always even， I think. | |
| Oct 4, 2021 at 4:17 | comment | added | Max Alekseyev | We can sort out some cases by noticing that pemanent and determinant are congruent modulo 2. So, if determinant is odd, then permanent must be nonzero. | |
| Oct 4, 2021 at 4:04 | comment | added | Jacob.Z.Lee | yes, it is a (1.-1,0)-matrix. | |
| Oct 4, 2021 at 3:02 | comment | added | LSpice | $A(1, -1, 0)$ means all entries are $1$, $-1$, or $0$? | |
| Oct 4, 2021 at 2:42 | comment | added | Jacob.Z.Lee | Computing the permanent of a (0,1)-matrix is #P-complete. If given a matrix A(1,-1,0), can we detect it zero or nonzero without computing it out by some methods? | |
| Oct 4, 2021 at 2:34 | comment | added | Will Sawin | Isn't this #P-complete? | |
| Oct 4, 2021 at 2:27 | history | asked | Jacob.Z.Lee | CC BY-SA 4.0 |