20200903, 15:12  #12 
"Tilman Neumann"
Jan 2016
Germany
111011001_{2} Posts 
Guys, you are worrying me...
Shouldn't the permutation (0 2) lead to the rules Code:
2 > 0 1 > 1 0 > 2 Code:
2 > 1 1 > 0 0 > 2 
20200903, 15:34  #13 
Oct 2017
11^{2} Posts 

20200903, 15:38  #14 
Oct 2017
11^{2} Posts 

20200903, 15:41  #15 
"Tilman Neumann"
Jan 2016
Germany
11×43 Posts 
I see... Compare that to http://www.research.ibm.com/haifa/po...ember2020.html Just wondering how long it takes... 
20200905, 20:48  #16 
Oct 2017
171_{8} Posts 
Has anyone found an RPS(11) game with more than 55 automorphisms?

20200906, 21:03  #17 
Sep 2017
2·59 Posts 

20200907, 01:23  #18 
Oct 2017
11^{2} Posts 

20200907, 05:43  #19  
Oct 2017
171_{8} Posts 
Quote:
When I fix a0,...e0,a1,...,e1,a2,...,e2, the code is able to make an exhaustive search of a3,...,e10. Usually it finds 40000...50000 valid games, needing 7 hours (one thread). The time is needed for checking the 11! permutations of the games. One of these a0,...e0,a1,...,e1,a2,...,e2combinations, chosen at random, yielded three games with 55 automorphisms. Pure luck! I have tested some of the permutations with pencil and paper, and they were correct. So I hope that the code works correctly. 

20200907, 13:19  #20  
Sep 2017
2×59 Posts 
Quote:


20200910, 12:38  #21 
"Walter S. Gisler"
Sep 2020
Switzerland
11 Posts 
I have some solutions, but it felt way too easy, so I am pretty sure I am missing something and would like to check my understanding of the problem:
First of all, for the RPS(5) game, is the "permutation" that changes none of the labels also counted as an automorphism or not? I only found 4 permutations that change at least one of the labels for the given example. I am also confused by how the permutations are defined and what permutations are actually allowed in this case. Let's assume I have a permutation that relabels the numbers as follows: 0 to 1 1 to 2 2 to 0 3 to 4 4 to 3 I can't define this permutation in a single list. If we draw this in a graph, we get a disconnected graph with two cycles. Are we limited to permutations that result in a single cycle? 
20200911, 10:10  #22  
Sep 2017
2·59 Posts 
Quote:
And the mapping can have two disconnected cycles, but I doubt that it would yield a solution. Here is how you should define the mapping: [1,2,0,4,3] where the indices are the original numbers and the entries are the result of the permutation. You just have to be careful in applying this permutation to the game: do not apply one by one otherwise you will mess things up. It has to be applied all at once. Let's apply your mapping to the game given in the problem: 0 > 1, 3 1 > 2, 4 2 > 0, 3 3 > 1, 4 4 > 0, 2 Here is the result: 1 > 2, 4 2 > 0, 3 0 > 1, 4 4 > 2, 3 3 > 1, 0 when sorted by the left hand side this yields the game in canonical form: 0 > 1, 4 1 > 2, 4 2 > 0, 3 3 > 1, 0 4 > 2, 3 I hope that it is clear now. 

Thread Tools  
Similar Threads  
Thread  Thread Starter  Forum  Replies  Last Post 
September 2019  Xyzzy  Puzzles  10  20191008 13:47 
September 2018  Xyzzy  Puzzles  2  20181011 15:31 
September 2017  R. Gerbicz  Puzzles  21  20180317 13:19 
September 2016  Batalov  Puzzles  8  20161004 14:10 
Anyone going to Vienna in September?  fivemack  Factoring  1  20070907 00:29 