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Old 2018-02-02, 02:38   #19
CRGreathouse
 
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Aug 2006

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Quote:
Originally Posted by science_man_88 View Post
I do see ways to cut down the number to search. Because, you can use symmetries picking (a,b) and (b',a') where apostrophies mean reversed permutations, will not change much at the same length, it can in theory reverse the superpermutation.
That cuts it down to a little more than
\[n^{n! + (n-1)! + (n-2)! + n-3}/2.\]

You can do better: reassign the variables so the first one you use is 1, the first one you use other than that is 2, and so on. This cuts it to about
\[n^{n! + (n-1)! + (n-2)! + n-3}/n! \approx n^{n! + (n-1)! + (n-2)! - 3}.\]

But these numbers are huge, we need to do much, much better.
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