Timeline for One-product free sequences for $A_n$
Current License: CC BY-SA 4.0
16 events
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| Jul 16 at 12:05 | comment | added | Peter Taylor | No, that's clearly false. Otherwise considering $k=1$ you would have $D(G)$ equals the largest order of a cyclic subgroup. | |
| Jul 16 at 11:24 | comment | added | Mikel Martinez Puente | @PeterTaylor Thanks a lot! I am trying Visual Code using Python and Sage environment. As you said, it takes some time but it works! I am now looking for one-free sequences in $A_6$ of length 12 and 4 distinct elements, it is taking long time (maybe there are not). By the way, I've just came up with an interesting question: do you think that proving there is NO one-product free sequence with $k$ distinct elements and length $n$, implies there is also NO one-free sequence of length $n$ and $l$ distinct elements, for any $l>k$? Intuitively, the less distinct elements, more likely is but... | |
| Jul 14 at 14:11 | comment | added | Peter Taylor | If you're running it online, it aborts after about a minute. I have a local Sage installation, and it took more than a few minutes but less than 8 hours. | |
| Jul 14 at 7:27 | comment | added | Mikel Martinez Puente | @PeterTaylor Okay, thanks a lot! It seems I have to wait bit more then! The point is that the symbol of "loading" disappears after some time, which makes me think the computation is over. It seems it is not, or my computer does not wanna work ;) | |
| Jul 14 at 6:25 | comment | added | Peter Taylor |
It takes a while, but it gives me {(1,6,5,4,3,2): 1, (2,6,5,4,3): 3, (1,4,6,2,5): 4, (1,6,3,5,2): 4}
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| Jul 13 at 22:25 | comment | added | Mikel Martinez Puente | @PeterTaylor I am literally taking the Sage code you posted in the answer of the other question for $D(S_5)$, and changing the last parameters instead of oneFreeProduct(S5, 10, 3) now I write oneFreeProduct(S6, 12. 4). I just change that line. | |
| Jul 13 at 21:57 | comment | added | Peter Taylor | I'm not sure how you're setting up the check. Can you post your code somewhere? | |
| Jul 13 at 12:38 | comment | added | Mikel Martinez Puente | @PeterTaylor Yes, I agree! I asked you this since there is some contradiction I do not understand why it fails, maybe you could help me or my computer does not compute it correctly. As you previously mentioned in one comment above there is some one-free sequence of length 11 in $A_6$ with three distict elements, thus adding any odd permutation in $S_6$ it should be one-free sequence of length 12 in $S_6$ with 4 distinct elements. However, in the Sage code it seems not to exist such a sequence, when I look for it. Do you know why could that be? | |
| Jul 13 at 8:44 | comment | added | Peter Taylor | Yes, that's the bottleneck. I can't think of a way to represent the partial products in such a way that you can add one element and generate all of the partial products with it in arbitrary positions, so it recalculates a lot. | |
| Jul 12 at 15:02 | comment | added | Mikel Martinez Puente | @PeterTaylor If I understood well, the code runs thrugh all possible subsequence products, with all possible ordering right? Because that is so important point, for a fixed subsequence taking the different orders for the product into account! | |
| Jul 12 at 14:30 | comment | added | Mikel Martinez Puente | @PeterTaylor Thanks a lot once again! I am definitely going to use your Sage code from now on since it is very useful for the calculations I need! I really appreciate your help ;) | |
| Jul 11 at 23:32 | history | edited | Mikel Martinez Puente | CC BY-SA 4.0 |
edited body
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| Jul 11 at 15:37 | comment | added | Peter Taylor | Minor point: the statements of known values of $D$ look to actually be known values of $d$. More usefully, the Sage code I posted earlier elsewhere can't find a set of three elements of $A_6$ which give a one-product-free multiset of 12 elements. The best is 11 with e.g. $(1,5,4,3,2)^3, (1,6,5,2,3)^4, (1,3,6,2,4)^4$. That rather spoils the idea of finding four $5$-cycles to give a multiset $c_1^4, c_2^4, c_3^4, c_4^4$. | |
| Jul 11 at 10:38 | history | edited | Mikel Martinez Puente | CC BY-SA 4.0 |
deleted 166 characters in body; edited title
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| Jul 3 at 10:32 | history | edited | Mikel Martinez Puente | CC BY-SA 4.0 |
added 24 characters in body
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| Jul 2 at 15:16 | history | asked | Mikel Martinez Puente | CC BY-SA 4.0 |