Timeline for Davenport constant $D(S_5)=10$ or $11$?
Current License: CC BY-SA 4.0
10 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Jul 4 at 16:51 | comment | added | Mikel Martinez Puente | Thank you so much! I have already proved indeed that $D(S_5)=11$ by exploiting the automorphism group you mention. | |
| Jul 4 at 15:51 | comment | added | Peter Taylor | @MikelMartinezPuente, Sage code. | |
| Jul 4 at 13:36 | comment | added | Peter Taylor | @MikelMartinezPuente, yes, I checked all possibilities with 4 distinct elements. I found the inverse-free multiset of 4 elements by iterative deepening: extend inverse- and one-free multisets of $k$ elements to $k+1$ elements by trying all possibilities. I actually wrote that in Python and just converted to Sage for the answer because it shifts the burden of trusting correctness to Sage's implementation rather than mine. I could rewrite it in Sage, but it's quite naïve (and in particular doesn't exploit the automorphism group to gain efficiency by using representatives of cosets). | |
| Jul 4 at 11:38 | comment | added | Mikel Martinez Puente | I also have another request for you. How did you find the inverse-free multiset of 4 elements? I think this idea of getting inverse-free multisets of maximal length could be extremely useful for my master thesis. However, I've started few days ago programming in GAP and I struggle a bit. I wonder if you would like to share with me an easy program in GAP/SAGE which computes such sequences of maximal length. THANK YOU VERY MUCH!! | |
| Jul 4 at 11:32 | comment | added | Mikel Martinez Puente | Yes, it makes sense! Thank you so much for your suggestions and help!! I will try to work on that for further cases of $S_n$. By the way, I am not an expert in Gap but I am trying to prove that $D(S_5)=11$ by brute force (almost done), in the last comment you say that any one-free multiset of 11 elements contain at least 5 distinct elements, I guess you have checked all possibilities with only 4 distinct elements and all are one-product sequences, right? Or have you found some one-free multiset of 11 elements? | |
| Jul 3 at 18:47 | comment | added | Peter Taylor | By exhaustive search, any one-free multiset of 11 elements contains at least 5 distinct elements. | |
| Jul 3 at 8:28 | comment | added | Peter Taylor | It appears that the typical maximum one-free multisets contain few distinct elements each with multiplicity one less than its order, and that makes some intuitive sense because repeating an element as much as possible reduces the number of distinct products. Thus one strategic idea we can take is to test multisets of this form first in order to get lower bounds. | |
| 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 15:26 | history | answered | Peter Taylor | CC BY-SA 4.0 |