Timeline for How many $40$-vertex cubic bipartite graphs have determinant $\pm 3$?
Current License: CC BY-SA 3.0
11 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Mar 15, 2018 at 14:02 | answer | added | Peter Heinig | timeline score: 2 | |
| Mar 15, 2018 at 11:32 | comment | added | Peter Heinig | (Technical comment @GerhardPaseman: needless to say, it is implied by, severally, two conditions in the OP that the number of black vertices equals the number of white vertices: (0) every regular bipartite graph necessarily is balanced, (1) the OP is about determinants of the biadjacency matrix, and determinants are only defined for square matrices.) | |
| Mar 15, 2018 at 5:19 | comment | added | Gordon Royle | @Brendan I have added the detail about matrices with side $3k$ into the main question. | |
| Mar 15, 2018 at 5:19 | history | edited | Gordon Royle | CC BY-SA 3.0 |
Added more information in answer to a question
|
| Mar 15, 2018 at 4:01 | answer | added | Gerhard Paseman | timeline score: 1 | |
| Mar 15, 2018 at 2:48 | answer | added | Brendan McKay | timeline score: 2 | |
| Mar 15, 2018 at 1:54 | comment | added | Brendan McKay | Since it is easy to generate random labelled bipartite graphs with a uniform distribution, what you are doing already should give a good estimate provided it is true that almost all of them with det=3 have trivial automorphism group. Is there any reason to suspect that having det=3 makes a non-trivial group more likely? In any case, if you compute the groups as well as the determinants you can compensate for any such bias. | |
| Mar 15, 2018 at 1:47 | comment | added | Brendan McKay | Can you explain the zeros in the table when n is a multiple of 3? | |
| Mar 15, 2018 at 1:32 | comment | added | Gordon Royle | Yes, if all row-sums are equal to some constant $r$ and all column sums are equal to a constant $s$, and the matrix is square then the total number of 1s is $nr$ and also $ns$. | |
| Mar 15, 2018 at 1:06 | comment | added | Gerhard Paseman | I believe this is implied by your conditions, but just to make sure: the bipartite graphs have the same number of black vertices and white vertices? And the resulting adjacency matrix is square with row sums and column sums all three? If so, I will dig up some old ideas for dealing with the kind of computation you propose. However Will Orrick is the person I would ask for this. Gerhard "Determinants Low On Priority List" Paseman, 2018.03.14. | |
| Mar 15, 2018 at 0:49 | history | asked | Gordon Royle | CC BY-SA 3.0 |