MathGroup Archive 1997

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All involutions of a permutation


bored?

then, for what it's worth, 

i built me a function that takes any permutation p as argument, and spews
out a list of "involutions" or decompositions (permutations q :
q^2==Identity, so that q r == p with r^2==Identity). Although the math is
simple (once you get the knack), the programming in Mathematica turned out
harder than 1 would guess.


A much simpler function counts the number of involutions for each permutation p.


Anyone who begs & screams long enough might (just might) get a copy. (;-)#


wouter.

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