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So I think it's possible to create an infinite sequence of transpositions T = { ti, i ≥ 2 } satisfying ∀ i[2,n!]

  • a,b[1,n] s.t. ti = (a b)
  • n!|(j - i)tj = ti

such that if you consider the sequence of permutations Pn = { pi,n, i ∈ [1,n!] } defined by

  • p1,n is the identity permutation
  • pi+1,n = pi,n ti+1

that Pn contains every permutation of n elements exactly once.


For example, if T starts with

{ (1 2), (1 3), (1 2), (1 3), (1 2), (1 4),
  (1 2), (1 3), (1 2), (1 3), (1 2), (2 4),
  (1 2), (1 3), (1 2), (1 3), (1 2), (3 4),
  (1 2), (1 3), (1 2), (1 3), (1 2), ....

then P4 is given by

{ (1 2 3 4), (2 1 3 4), (3 1 2 4), (1 3 2 4), (2 3 1 4), (3 2 1 4),
  (4 2 1 3), (2 4 1 3), (1 4 2 3), (4 1 2 3), (2 1 4 3), (1 2 4 3),
  (1 3 4 2), (3 1 4 2), (4 1 3 2), (1 4 3 2), (3 4 1 2), (4 3 1 2),
  (4 3 2 1), (3 4 2 1), (2 4 3 1), (4 2 3 1), (3 2 4 1), (2 3 4 1) }

and P4 contains every permutation of 4 elements exactly once.


What I'd like to find is a constructive proof for the existence of T. I'm sure I'm retreading old ground here, I just don't seem to be searching for the right terms to turn up what I'm looking for.

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See for instance en.wikipedia.org/wiki/…. – Richard Stanley Mar 17 2012 at 21:35
@Richard Stanley: Thanks! The link you gave was broken, but it was enough for me to find [Wikipedia's page on the Steinhaus–Johnson–Trotter algorithm](en.wikipedia.org/wiki/…). If you'd like to upgrade your comment to an answer, I'd be happy to accept it. – rampion Mar 17 2012 at 22:43
O.k., I will upgrade my comment. I am glad it helped. – Richard Stanley Mar 17 2012 at 23:55

1 Answer

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See for instance http://en.wikipedia.org/wiki/Steinhaus-Johnson-Trotter_algorithm.

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