>>> from sympy.combinatorics import * >>> Cycle()*(1,2)*(2,3) [(1, 3, 2)] >>> _.as_list() [0, 3, 1, 2] >>> Permutation([[2,3]],size=4).array_form [0, 1, 3, 2] >>> Permutation([[1,2]],size=4).array_form [0, 2, 1, 3]
So using the (1,2) permutation to select from the (2,3) one gives the final answer of [0, 3, 1, 2] (applied the (1,2) last, hence R to L) -- You received this message because you are subscribed to the Google Groups "sympy" group. To post to this group, send email to [email protected]. To unsubscribe from this group, send email to [email protected]. For more options, visit this group at http://groups.google.com/group/sympy?hl=en.
