Timeline for numbering the squares of a rectangular grid, was: counting sequences of pairs
Current License: CC BY-SA 3.0
10 events
| when toggle format | what | by | license | comment | |
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| Jun 12, 2012 at 9:32 | comment | added | Douglas Zare | It's good form to clean up questions if a better way to present them is found, although it is not good to edit a question too frequently. | |
| Jun 11, 2012 at 21:37 | comment | added | tortortor | @Barry Cipra Actually, maybe I should scrap my whole convoluted question and just ask it in your way. Would that be good mathoverflow style? | |
| Jun 11, 2012 at 21:26 | comment | added | tortortor | @Barry Cipra. Added the $(1,m)$ and $(2,2)$ cases as examples to the question. And yes, your last addendum (6/11/12) sounds like what I am looking for. And it actually sounds like maths! | |
| Jun 11, 2012 at 13:48 | history | edited | Barry Cipra | CC BY-SA 3.0 |
added 578 characters in body
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| Jun 11, 2012 at 0:37 | history | edited | Barry Cipra | CC BY-SA 3.0 |
correction
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| Jun 11, 2012 at 0:23 | comment | added | Barry Cipra | @tortortor, I see what you mean. I'll try to give it some additional thought. In the meantime, can you provide some total counts for a few small pairs $(m,n)$, as Gerry suggested? | |
| Jun 10, 2012 at 21:14 | comment | added | tortortor | Note that your formula turns out to be the same as $N!/(n! m!)$ which is something I tried before, but it gave me too many non-equivalent sequences. I think the cyclic permutations of the order of pairs, is not really accounted for in this formula, and I fail to see how it should be done. This is also a reason why my question is formulated in such a convoluted way, with the infinite sequences and all. | |
| Jun 10, 2012 at 21:05 | comment | added | tortortor | Thank you for the answer. But it gives too many. Take for example as @Gerry Myerson has suggested $N=4, n=2, m=2$ this gives $4$ splittings according to the formula, but I can get two of them to coincide due to a cyclic permutation. Think of the $2 \times 2$ grid you suggested. We get four arrangements, take the one with top row $1,2$ and bottom row $4,3$ as well as the one with top row $1,4$ and bottom $2,3$. Write the sequences out. Then in the latter, rename the top row and then move the last pair to the first position. The result is the former sequence. So the two should be equivalent. | |
| Jun 10, 2012 at 1:56 | comment | added | tortortor | The $N$ even requirement was wrong, I wanted it to not be prime because if $n$ or $m$ are $1$ it is boring for me. This is unnecessary though, I removed it. | |
| Jun 9, 2012 at 12:57 | history | answered | Barry Cipra | CC BY-SA 3.0 |