Timeline for Sets of symbols contained in contiguous substrings
Current License: CC BY-SA 4.0
19 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| May 27, 2021 at 21:58 | history | edited | Jukka Kohonen | CC BY-SA 4.0 |
OOPS. Fix the condition on substring length: not |Sigma| but |S| for each subset S.
|
| May 26, 2021 at 17:49 | history | edited | Jukka Kohonen | CC BY-SA 4.0 |
oops
|
| May 26, 2021 at 16:24 | history | edited | Jukka Kohonen | CC BY-SA 4.0 |
previous research found
|
| May 25, 2021 at 7:13 | history | edited | Jukka Kohonen | CC BY-SA 4.0 |
clarify; superdiagonal formula valid for m>=2, not for m=1
|
| May 25, 2021 at 0:47 | history | edited | Jukka Kohonen | CC BY-SA 4.0 |
date of update
|
| May 25, 2021 at 0:42 | history | edited | Jukka Kohonen | CC BY-SA 4.0 |
call them digits
|
| May 23, 2021 at 8:33 | history | edited | Jukka Kohonen | CC BY-SA 4.0 |
fix wrong word + shorter
|
| May 23, 2021 at 8:25 | history | edited | Jukka Kohonen | CC BY-SA 4.0 |
shorter
|
| May 23, 2021 at 7:43 | history | edited | Jukka Kohonen | CC BY-SA 4.0 |
fix typo
|
| May 23, 2021 at 7:19 | comment | added | Per Alexandersson | Right, that's a conjecture. But also, one can assume no two adjacent letters are equal, which makes the search a bit smaller. I wonder if there is a recursive way to combine two extremal words, to produce one containing all subsets for one letter more. This can be used to create an upper bound at least for T(m)... | |
| May 23, 2021 at 6:37 | comment | added | Jukka Kohonen | At least one can, without loss of generality, start with increasing numbers 1,2,... because whenever you pick an unseen letter, it does not matter which unseen letter you pick, so you can as well pick the smallest unseen letter. However there is always also the choice of picking a seen letter again, and I don't know if one can continue in order up to 1,2,...,m ? | |
| May 23, 2021 at 6:32 | comment | added | Jukka Kohonen | @Per: True, and also the question is somehow similar to (and different from) de Bruijn sequences, which contain all possible substrings of a given length. But now we are seeking all subsets of the alphabet. | |
| May 23, 2021 at 6:32 | comment | added | Per Alexandersson | Also, instead of 12314234, one can take 12343142, so that the extremal word starts with 1,2...m. Perhaps this choice is always possible? | |
| May 23, 2021 at 6:28 | comment | added | Per Alexandersson | The extremal words are reminicient of Super permutations, wikiwand.com/en/Superpermutation | |
| May 23, 2021 at 6:06 | history | edited | Jukka Kohonen | CC BY-SA 4.0 |
more values, 3rd and 4th diagonals OEIS, easier 1-based alphabet
|
| May 22, 2021 at 18:11 | history | edited | Jukka Kohonen | CC BY-SA 4.0 |
more values + second superdiagonal
|
| May 22, 2021 at 16:52 | history | edited | Jukka Kohonen | CC BY-SA 4.0 |
more values in table + fix text
|
| May 22, 2021 at 15:41 | history | edited | Jukka Kohonen | CC BY-SA 4.0 |
oops
|
| May 22, 2021 at 15:30 | history | answered | Jukka Kohonen | CC BY-SA 4.0 |