Thread: September 2020
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Old 2020-09-11, 10:10   #22
SmartMersenne
 
Sep 2017

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Quote:
Originally Posted by Walter View Post
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?
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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