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Jun 16, 2022 at 13:01 vote accept Rebecca J. Stones
Jun 15, 2022 at 6:25 comment added Brendan McKay @hoboonsuan Thanks, that saves me some computing time. I have $K_5^{(4)}$ down to 20 seconds and wonder if $K_5^{(6)}$ is plausible.
Jun 15, 2022 at 5:42 comment added ho boon suan @BrendanMcKay By Bosák's result in my answer, $K_5^{(2j+1)}$ for $j\ge0$ is ungraceful.
Jun 15, 2022 at 5:37 comment added Brendan McKay Actually, 2 seconds and 2 minutes once I realized that the reverse of a labelling is also a labelling. Now I wonder about $K_5^{(5)}$ — is anything known.
Jun 15, 2022 at 5:09 comment added Brendan McKay My little C program says that neither $K_5^{(3)}$ nor $K_5^{(4)}$ have graceful labellings. It took 3 seconds and 4 minutes, respectively. This confirms Mikhail's result so I think this is sufficient verification.
Jun 15, 2022 at 2:45 answer added ho boon suan timeline score: 7
Jun 15, 2022 at 1:10 comment added Timothy Chow I'd expect that a SAT solver should be able to confirm the nonexistence of a graceful labeling in these two cases. Introduce Boolean variables $x_{ij}$ to indicate that vertex $i$ gets label $j$ and Boolean variables $y_{ij}$ to indicate that edge $i$ gets label $j$. It is not hard to write down all the necessary constraints as SAT clauses. I'd also recommend adding a few symmetry-breaking constraints to speed up the computation.
Jun 14, 2022 at 23:13 history edited Rebecca J. Stones CC BY-SA 4.0
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Jun 14, 2022 at 23:02 history edited Rebecca J. Stones CC BY-SA 4.0
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Jun 14, 2022 at 21:57 history edited Rebecca J. Stones CC BY-SA 4.0
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Jun 14, 2022 at 21:28 history edited Rebecca J. Stones CC BY-SA 4.0
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Jun 14, 2022 at 21:08 history edited Rebecca J. Stones CC BY-SA 4.0
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Jun 14, 2022 at 12:05 comment added Mikhail Tikhomirov My brute-force wasn't able to find a labelling for any of the two.
Jun 14, 2022 at 11:53 history edited Rebecca J. Stones CC BY-SA 4.0
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Jun 14, 2022 at 11:45 history edited Rebecca J. Stones CC BY-SA 4.0
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Jun 14, 2022 at 11:37 history asked Rebecca J. Stones CC BY-SA 4.0