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Number of permutations to obtain identity

  1. Nov 17, 2011 #1
    1. The problem statement, all variables and given/known data

    Let [itex]s*(f)[/itex] be the minimum number of transpositions of adjacent elements needed to transform the permutation [itex]f[/itex] to the identity permutation. Prove that the maximum value of [itex]s*(f)[/itex] over permutations of [itex][n][/itex] is [itex]{n \choose 2}[/itex]. Explain how to determine [itex]s*(f)[/itex] by examining [itex]f[/itex].

    2. Relevant equations

    [itex]{n \choose 2} = \frac{n!}{k!(n-k)!}[/itex]

    Perhaps...
    Definition: The identity permutation of [n] is the identity function from [n] to [n]; its word form is 1 2 ... n. A transposition of two elements in a permutation switches their positions in the word form.

    A permutation f of [n] is even when P(f) is positive, and it is odd when P(f) is negative. When n = 1, there is one even permutation of [n] and no odd permutation. For n >= 2, there are n!/2 even permuatations and n!/2 odd permutations.

    My book solves a problem that counts the number of exchanges of entries needed to sort the numbers into the order 1,2,...,n (given a list of n numbers 1 through n). The solution defines the nth iterate of f: A --> A.
    Definition: The nth iterate of f: A --> A is the function [itex]f^{n}[/itex] obtained by composing n successive applications of f.
    Consequence: Since composition of functions is associative, we also have [itex]f^{k} o f^{n-k}[/itex] whenever 0 <= k <= n.

    3. The attempt at a solution

    The number of subsets of length 2 of a permutation of length n is n choose 2. I can see that from the definition, but how does one formulate a formal proof from just the definition?

    I can guess that the scenario requiring the most trials is when f is the permutation n...2 1.

    Any help would be greatly appreciated!
     
    Last edited: Nov 17, 2011
  2. jcsd
  3. Nov 18, 2011 #2

    Deveno

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    Science Advisor

    note that n choose 2 is (n-1)n/2.

    imagine that f takes 1→k.

    then (k-1 k)f takes 1→k-1.

    similarly (k-2 k-1)(k-1 k)f takes 1→k-2

    after applying at most k-1 such adjacent transpostions, we arrive a product:

    (1 2)(2 3).....(k-1 k)f which takes 1→1.

    ask yourself: what the maximum value of k?

    now we have a permutation g that takes 1→1.

    suppose g takes 2→m, and repeat.

    won't you eventually wind up with the identity?

    now sum the number of adjacent transpositions used, and use

    a well-known formula for this sum.
     
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