[Retros] happy prime new year
andrew buchanan
andrew at anselan.com
Mon Jan 30 02:49:53 EST 2017
Hi Noam,
>> [2017 is] the number of permutations of 7 elements which *don't* contain>> a double rise (e.g. 1342 contains a rise from 1 to 3, and then
>> another rise from 3 to 4, so is excluded from the count).
>> Equivalently, it's those permutations which *don't* have an embedded
>> increasing run (e.g. 1423 contains 123 (although these do not appear
>> adjacently) so is excluded from the count.
>This equivalence is not obvious, though I have not tried to
>confirm or refute it.
A string containing symbols 1..n can be regarded as a function from position i to value j (e.g. for 1342 1->1, 2->3, 3->4 & 4->2). But also it can be regarded as the inverse: moving value i to position j, which as a string is e.g. 1423. A "sequential rise" in one of these (e.g. 13 in 1342) corresponds to a "non-sequential run" of 2 in the other (e.g. 12 in 1423), and vice versa. Similarly, 1342 contains the rise 134 as 1423 contains the run 123.
Neither perspective seems easy to embed in chess. For example for the second perspective: to avoid the run 123 means that the position must distinguish between the game after 1 followed by 2 and the game after 2 followed by 1. Normally can say in chess: yes 1 must happen before 2, but if both are viable the resulting position doesn't distinguish between them.
>2017 is also the number of permutations of 1,2,...,16
>that, when arranged in a 4x4 array, are increasing by rows, columns,
>and diagonals. (Without the diagonal condition that's a special case
>of the hook-number formula, giving 24024.) This seems more amenable
>to a chess realization, though it would still take some work even once
>we remove 1,2,15,16 whose locations are forced.
Cool! Without the diagonals, a representation of the 12 moves e.g. shifts a queue xx.xx 3 cells to the right. But I can't imagine how to embed the diagonals.
Regards,Andrew
On Monday, January 30, 2017 3:05 AM, Noam Elkies <elkies at math.harvard.edu> wrote:
andrew buchanan <andrew at anselan.com> writes:
> (1) There is a naturally occurring combinatorial way to reach 2017.
> It's the number of permutations of 7 elements which *don't* contain
> a double rise (e.g. 1342 contains a rise from 1 to 3, and then
> another rise from 3 to 4, so is excluded from the count).
> Equivalently, it's those permutations which *don't* have an embedded
> increasing run (e.g. 1423 contains 123 (although these do not appear
> adjacently) so is excluded from the count.
This equivalence is not obvious, though I have not tried to
confirm or refute it.
> Several queue problems are already built around the Euler zigzag
> function, which is not unrelated. However I can't begin to figure out
> a way to represent this new approach in chess terms.
Me neither... 2017 is also the number of permutations of 1,2,...,16
that, when arranged in a 4x4 array, are increasing by rows, columns,
and diagonals. (Without the diagonal condition that's a special case
of the hook-number formula, giving 24024.) This seems more amenable
to a chess realization, though it would still take some work even once
we remove 1,2,15,16 whose locations are forced.
NDE
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