Fast permutation-word multiplication and the simultaneous conjugacy problem
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Given a finite sequence a_1, a_2,..., a_d of d permutations in the symmetric group S_n, and a permutation word k_1k_2... k_m over the alphabet {1,2,..., d}, computation of the product a_k_1a_k_2... a_k_m in a straightforward manner takes O(n m) time. However, it appears that this multiplication is such an elementary operation that, surprisingly enough, it went on unquestioned. We show that the above product can be computed in time O(min{ n m, n m log d / log m}) using O(m + n m^ϵ) space, where 0 < ϵ < 1. Consequently, this computation takes o(n m) time whenever log d = o(log m), which is a reasonable assumption in practice. The above result is used to solve the transitive simultaneous conjugacy problem in O(n^2 log d / log n + dnlog n) time and O(n^1+ ϵ + dn) space, where 0 < ϵ <1. This problem asks whether there exists a permutation τ∈ S_n such that b_j = τ^-1 a_j τ holds for all j = 1,2, ..., d, where a_1, a_2, ..., a_d and b_1, b_2, ..., b_d are given sequences of d permutations in S_n, each of which generates a transitive subgroup of S_n. As from mid 70' it has been know that the problem can be solved in O(dn^2) time. An algorithm with running time O(dn log(dn)), proposed in late 80', does not work correctly on all input data.
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