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For a permutation $\pi$, let $S(\pi$) be the minimum number of transpositions required to sort the elements in increasing order. Show that for a random permutation $\pi \in S_{10001}$ $E[S(\pi)] ≥ 5000$, and deduce that $P(S(\pi) ≤ 4500) < 1/20$.

By using the probability space suggested in the answer to this question: https://mathoverflow.net/questions/120163/concentration-bounds-for-sums-of-random-variables-of-permutations it can be easily seen that the random variable S is 2-Lipschitz. If $E[S(\pi)] ≥ 5000$ then $P(S(\pi) ≤ 4500) < 1/20$ follows immediately by using the Azuma-Hoeffding inequality.

My question is: How can I show that $E[S(\pi)] ≥ 5000$?

mathcounterexamples.net
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Andreas Dahlberg
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  • I guess that given $\pi$ and its reverse permutation $\overline{\pi}$, $S(\pi)+S(\overline{\pi})$ has to fulfill some simple lower bound. Maybe it is enough to consider permutations with long runs of elements in decreasing order: in order to be sorted out, these runs should require a number of transpositions which is roughly half their length. – Jack D'Aurizio Jan 18 '19 at 10:15
  • @JackD'Aurizio What do you name the reverse permutation? – mathcounterexamples.net Jan 18 '19 at 10:55
  • @mathcounterexamples.net: because I suppose that in sorting $\pi$, then $\overline{\pi}$, one needs roughly the same number of transpositions, close to the length of $\pi$. In other terms I am trying to spot a hidden symmetry in the problem. – Jack D'Aurizio Jan 18 '19 at 11:15
  • @JackD'Aurizio In fact my question was just how do you define the reverse permutation? – mathcounterexamples.net Jan 18 '19 at 12:25
  • $$\overline{\pi}(m) = \pi(n+1-m)$$ assuming $\pi\in S_n$. – Jack D'Aurizio Jan 18 '19 at 12:30
  • I suppose the lower bound is 10000, then it follows from linearity of expectation and the fact that the expected value of π and its reverse is the same that E[S(π)]≥5000. However, I don't know how to prove this either. – Andreas Dahlberg Jan 18 '19 at 14:12
  • Maybe you can do something with cycles. You can partition a permutation into cycles. A cycle of length $k$ needs $k-1$ transpositions to be fixed. (I think, I tried it manually with some examples) So the number of transpositions needed for a permutation is $10001 - c$, where $c$ is the number of cycles (including trivial ones) – Paul Jan 18 '19 at 15:58
  • Combine this with this question and you've got a simple expression for $E[S(\pi)]$. – Paul Jan 20 '19 at 02:56
  • Although I think there is an easier solution to this question, I will accept this answer. Thanks! – Andreas Dahlberg Jan 20 '19 at 07:24
  • @AndreasDahlberg I turned it into an answer – Paul Jan 20 '19 at 18:04

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You can partition a permutation into cycles. For example, the permutation $\begin{pmatrix} 1 & 2 & 3 & 4 & 5 \\ 5 & 1 & 3 & 4 & 2\end{pmatrix}$ has cycles $(5, 2, 1)$, $(3)$ and $(4)$.

Now let's do a transposition. When swapping two elements that belong to the same cycle, the cycle is split into two smaller cycles. When swapping two elements that belong to different cycles, the two cycles merge into one cycle.

Let $n$ be the number of elements and $c$ be the number of cycles in the permutation. As soon as we have $n$ cycles, the permutation is sorted. So we need $n$ - $c$ transpositions.

The expected number of cycles in a permutation is $H_n$, the $n$-th harmonic number. This makes the expected minimum number of transpositions to sort a permutation equal to $n - H_n$.

Paul
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