Timeline for Davenport constant $D(S_5)=10$ or $11$?
Current License: CC BY-SA 4.0
17 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Jul 2 at 22:55 | vote | accept | Mikel Martinez Puente | ||
| Jul 2 at 22:54 | vote | accept | Mikel Martinez Puente | ||
| Jul 2 at 22:55 | |||||
| Jul 2 at 16:19 | history | edited | LSpice | CC BY-SA 4.0 |
Typos; deleted "thanks"
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| Jul 2 at 16:16 | comment | added | Daniel Weber | $120 \cdot \binom{120}{11}$ isn't that large, this might be doable in a few GPU days using $f(S) = \bigcup_{x \in S} (f(S \setminus \{x\}) \cup f(S \setminus \{x\}) x)$, although storage is a problem. Perhaps $f$ contains the identity fairly quickly so the actually search space is much smaller than expected? It could also be possible to only compute this up to length 4 and then test all possibilities of using that. | |
| Jul 2 at 15:26 | answer | added | Peter Taylor | timeline score: 3 | |
| Jul 2 at 14:44 | comment | added | Mikel Martinez Puente | @PeterTaylor you mean the order of the elements in the sequence or in the product? If you refer to the order in the sequence, you are absolutely right. However, "sequence" is the name used in all papers about the Erdös-Ginzburg-Ziv constant and the Davenport constants, it is somehow a convention. | |
| Jul 2 at 8:45 | comment | added | Peter Taylor | Am I correct in understanding that the order is nowhere relevant? If so, it would be clearer to talk about multisets instead of sequences. | |
| Jun 30 at 12:57 | comment | added | Mikel Martinez Puente | @BrendanMcKay yeah, that's true! However, I have been trying to find such a sequence using GAP computer algebra system and it seems not to be such a sequence! Moreover, I am pretty convinved that $D(S_5)$=10 and hence every sequence of 10 elements is one-product. But I don't know how to prove it... | |
| Jun 30 at 12:48 | comment | added | Gerry Myerson | "I have already deleted those questions in Math Stack." math.stackexchange.com/questions/4937972/… is not deleted. | |
| Jun 30 at 11:25 | comment | added | Brendan McKay | If $(12)$ is in the sequence then it isn't inverse-free. | |
| Jun 30 at 10:45 | comment | added | Mikel Martinez Puente | @GerryMyerson I have already deleted those questions in Math Stack. I think there were not appropiated so I tried with overflow. Thanks for your advise! | |
| Jun 30 at 10:38 | comment | added | Mikel Martinez Puente | @BrendanMcKay Thanks for the rwsponse! Yes exactly, I allow non-contiguous subsequences. In particular, we allow any subsequence $g_{i1}, ..., g_{ir}$ so that the product of elements in some order gives identity. I see what you idea is, however I believe is not enough. Imagine that we find sich a sequence with some transposition, wlog $(12)$. Then, when we take the copy we have two times $(12)$ hence we get identity. | |
| Jun 30 at 8:21 | comment | added | Gerry Myerson | And another unlinked question by this same user, mathoverflow.net/questions/474161/… | |
| Jun 30 at 7:32 | comment | added | Gerry Myerson | Also posted to math.stack, math.stackexchange.com/questions/4937972/… without notification to either site, a violation. | |
| Jun 30 at 6:09 | comment | added | Brendan McKay | Can you find a sequence of length 5 which is both one-free and inverse-free (i.e. $g$ and $g^{-1}$ don't both appear as subsequence products)? Then you can join two copies together to make a sequence of length 10. | |
| Jun 30 at 5:45 | comment | added | Brendan McKay | I think that by "subsequence" you allow non-contiguous subsequences, right? | |
| Jun 29 at 22:23 | history | asked | Mikel Martinez Puente | CC BY-SA 4.0 |