Timeline for A specific collection of subgraphs in $K_{70, 70}$
Current License: CC BY-SA 4.0
18 events
| when toggle format | what | by | license | comment | |
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| Sep 18, 2019 at 1:30 | history | edited | RobPratt | CC BY-SA 4.0 |
tightened formulation
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| Sep 14, 2019 at 20:15 | comment | added | Aaron Meyerowitz | From your first list (left hand side) I see that 24 appears with every other number except $1,2,22,23.$ (That was one eyeballing of the answer so I might be off a bit.) To find $70$ lists for the right hand side which mesh with these would require (along with much more) $24$ partitions of $70$ using just $1,22,23,24.$ That won’t work with just a single $1$. If there is a solution it would need an abundance of multiple lists (vertices) on each side with exactly the same members (neighbors on the other side). | |
| Sep 14, 2019 at 1:15 | comment | added | user44143 | Can you show the lists of vertices? | |
| Sep 14, 2019 at 0:33 | history | edited | RobPratt | CC BY-SA 4.0 |
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| Sep 14, 2019 at 0:26 | comment | added | RobPratt | Updated just now and will rerun the solver. | |
| Sep 14, 2019 at 0:25 | history | edited | RobPratt | CC BY-SA 4.0 |
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| Sep 14, 2019 at 0:15 | comment | added | Ilya Bogdanov | Notice that you can use the same partition of $70$ several times (for several vertices in one part); so I do not understand why you put $x_i\in\{0,1\}$. Moreover, by these reasons, the second half of your inequalities should be more complicated. | |
| Sep 13, 2019 at 14:54 | history | edited | RobPratt | CC BY-SA 4.0 |
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| Sep 13, 2019 at 3:50 | comment | added | RobPratt | Not yet, just proposing a formulation that captures the problem. | |
| Sep 13, 2019 at 3:43 | comment | added | Brendan McKay | I'm puzzled. Did you find a solution? | |
| Sep 13, 2019 at 3:09 | history | edited | RobPratt | CC BY-SA 4.0 |
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| Sep 12, 2019 at 21:42 | comment | added | RobPratt | I am using SAS. | |
| Sep 12, 2019 at 20:28 | comment | added | dvitek | @RobPratt What software are you using to do the integer linear programming? | |
| Sep 12, 2019 at 18:27 | vote | accept | Chain Markov | ||
| Sep 13, 2019 at 6:32 | |||||
| Sep 12, 2019 at 6:15 | comment | added | Brendan McKay | Yes, that should do it. | |
| Sep 12, 2019 at 4:40 | comment | added | RobPratt | Hmm, I guess after the edits to @dvitek's answer we need two such sets of partitions, with no pair of parts appearing together on both sides? | |
| Sep 12, 2019 at 4:29 | comment | added | Brendan McKay | I see the 70 partitions corresponding to the vertices on one side of the $K_{70,70}$, but I don't see how to make a partition of $K_{70,70}$ out of them. | |
| Sep 12, 2019 at 3:41 | history | answered | RobPratt | CC BY-SA 4.0 |