Guys, you are worrying me...
Shouldn't the permutation (0 2) lead to the rules [CODE] 2 > 0 1 > 1 0 > 2 [/CODE]instead of [CODE] 2 > 1 1 > 0 0 > 2 [/CODE]?? 
[QUOTE=Till;555885]Guys, you are worrying me...
Shouldn't the permutation (0 2) lead to the rules [CODE] 2 > 0 1 > 1 0 > 2 [/CODE]instead of [CODE] 2 > 1 1 > 0 0 > 2 [/CODE]??[/QUOTE] 2 becomes 0 and 2 becomes 0. Einfach nur die zweien und die Nullen austauschen! 
[QUOTE=Dieter;555888]2 becomes 0 and 2 becomes 0.
Einfach nur die zweien und die Nullen austauschen![/QUOTE] I wanted to say: 2 becomes 0 and 0 becomes 2, of course! 
[QUOTE=Dieter;555890]I wanted to say:
2 becomes 0 and 0 becomes 2, of course![/QUOTE] I see... Compare that to [url]http://www.research.ibm.com/haifa/ponderthis/challenges/September2020.html[/url] Just wondering how long it takes... 
Has anyone found an RPS(11) game with more than 55 automorphisms?

[QUOTE=Dieter;556171]Has anyone found an RPS(11) game with more than 55 automorphisms?[/QUOTE]
Did you find one with 55? I think the question required at least 50. 
[QUOTE=SmartMersenne;556274]Did you find one with 55? I think the question required at least 50.[/QUOTE]
There are many games with 5 or 11 and some with 9. I have found three games with 55. Seems again to be the search for the needle ... 
[QUOTE=SmartMersenne;556274]Did you find one with 55? I think the question required at least 50.[/QUOTE]
Needle in a haystack, computing times: 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. 
[QUOTE=Dieter;556305]Needle in a haystack, computing times:
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.[/QUOTE] This is not pure luck. After some point it is a skill, and you have proven yourself to have that problemsolving skill. 
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? 
[QUOTE=Walter;556630]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?[/QUOTE] Yes, the identity mapping is counted as a permutation. 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. 
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