Question

Suppose we’re given a colored graph. Let G be its automorphism group. How do we test whether there is a permutation p in G such that p(x_i) = y_i for a given set of (x_i,y_i)? Concretely suppose that G is a permutation group over 6 elements. Then an example test that I’d like to be able to do is whether G contains any permutation that sends 2 to 2 and 3 to 5 and I don’t care where 1,4,5,6 end up.

You can assume that a program like saucy can efficiently compute a set of generators for the group G.

Answer 1

If G is connected and you can run saucy on a different graph, prepare a colored graph H consisting of the disjoint union of G with itself and, for each i, recolor vertex x_i in the first copy of G and y_i in the second copy of G to be color i distinct from the existing colors in G. There is a suitable automorphism of G if and only if there exists a generator of Aut(H) that maps a vertex in the first copy to a vertex in the second copy.

Perhaps there is a more direct method based on Schreier-Sims.

EDIT: here’s one way an SS-based method could go. Let there be p pairs x₁, y₁, …, x_p, y_p. If p = 0, the answer is of course yes. Otherwise, determine whether there exists a permutation g that carries x_p to y_p, i.e., x_p^g = y_p. If not so, the answer is no. If so, the permutation you’re looking for exists if and only if it can be written in the form h g (g followed by h), where h belongs to the stabilizer of y_p. Use the SS machinery to compute generators for the stabilizer and return the answer computed recursively for x₁^g, y₁, …, x_p-1^g, y_p-1.