20180127, 22:04  #12  
Aug 2006
3×1,993 Posts 
Quote:


20180127, 22:14  #13 
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Jul 2009
Dumbassville
2^{6}·131 Posts 

20180131, 23:32  #14 
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Jul 2009
Dumbassville
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Superpermutations
Still can't believe the lowest number of instruction needed to have all permutations of n instructions present in a code isn't well known. 
20180201, 03:29  #15  
Aug 2006
3×1,993 Posts 
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20180201, 22:02  #16  
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Jul 2009
Dumbassville
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20180201, 22:26  #17  
Aug 2006
1011101011011_{2} Posts 
Quote:
\[n^{n! + (n1)! + (n2)! + n3}\] possibilities to check in the best case, it's hard to reduce the options to something reasonable. Googol to the sixth power and all that. 

20180201, 23:28  #18 
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Jul 2009
Dumbassville
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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.
Last fiddled with by science_man_88 on 20180201 at 23:46 
20180202, 02:38  #19  
Aug 2006
3×1,993 Posts 
Quote:
\[n^{n! + (n1)! + (n2)! + n3}/2.\] 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 \[n^{n! + (n1)! + (n2)! + n3}/n! \approx n^{n! + (n1)! + (n2)!  3}.\] But these numbers are huge, we need to do much, much better. 

20180202, 03:25  #20  
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Jul 2009
Dumbassville
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Quote:
Last fiddled with by science_man_88 on 20180202 at 03:34 

20180203, 18:49  #21 
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Jul 2009
Dumbassville
20C0_{16} Posts 
At least we can use derangement math to show as n gets larger about 37% of orderings will have at least 1 permutaion placed correctly.

20180204, 03:23  #22 
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Jul 2009
Dumbassville
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. Not sure it wouldn't be solved already, I found a solution for the equation for n=14 just take the coeffcients on 1001 to add to 391 Mod 432 and the coefficient on 432 is 906 mod 1001.
Last fiddled with by science_man_88 on 20180204 at 03:34 
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