Let $A$ be an array of $n$ integers containing the numbers $\{1, 2, \dots , n\}$ in some arbitrary order. For integers $i$ and $j$ such that $1 ≤ i < j ≤ n$, let $\mathrm{Reverse}(A, i, j)$ be a procedure that reverses the subarray $A[i], A[i + 1], \dots , A[j]$ of the array $A$ while leaving the remaining elements of the array unaffected. Prove that the following algorithm sorts the array $A$ and also terminates.

for i := 1 to n − 1
  while A[i] ≠ i do
    Reverse(A, i, A[i])

I am able to prove this if the elements are in descending order. Why does this algorithm work for an arbitrary order?


If the algorithm terminates at all then $A[i] = i$ for $i \leq n-1$, and so $A$ is sorted. Hence it suffices to show that the algorithm terminates. Furthermore, when reaching $i = i_0$, we know that $A[j] = j$ for $j < i_0$, and so $A[i_0],\ldots,A[n]$ is a permutation of $i_0,\ldots,n$, which is a smaller instance of the same problem. Hence it suffices to show that the inner while loop terminates when $i = 1$.

Suppose that the inner while loop never terminates when $i = 1$. Let $j$ be the maximal value which appears infinitely many times as $A[1]$. After finitely many iterations, $A[1] \leq j$ always. Since $A[1] = j$ infinitely many times, there must be some subsequent iteration in which $A[1] = j$. After that iteration, $A[j] = j$. In all subsequent iterations, $A[1] \leq j$, which ensures that $A[j]$ stays $j$, guaranteeing that $A[1]$ never equals $j$ again. However, this contradicts the definition of $j$.

The maximal number of iterations of the inner loop is A000375. There is no known formula!

I couldn't find the sequence for the maximal total number of iterations of the inner loop (over all $i$) on OEIS.

  • $\begingroup$ Its a good answer, but I think you will need to explain more why $j$ will stay at its place. We can prove (by induction on the time $t$ after we saw $j$ for the first time) that for any $A[1]=k$ which came after the iteration for which $A[1]=j$, we have $k<j$ (and not only $k\le j$). This is followed by that $j$ is maximal and is unique (there is only one $j$ in the array). $\endgroup$ – nir shahar May 4 at 20:39
  • $\begingroup$ It's a simple induction. $\endgroup$ – Yuval Filmus May 4 at 20:40
  • $\begingroup$ Can u please explain it with an example like taking a small array e.g [9,3,1,8]?? $\endgroup$ – Danjing_Chaw May 5 at 11:36
  • $\begingroup$ Your example is not a permutation of $1,\ldots,n$. Also, you can work out any specific example by hand, or by programming the algorithm. $\endgroup$ – Yuval Filmus May 5 at 12:02
  • $\begingroup$ Okay got it!. It is clear now. $\endgroup$ – Danjing_Chaw May 5 at 14:42

If you're familiar with cycle sort, you can start to see that this algorithm also operates in a similar way (although it's super inefficient).

Essentially, the inner loop doesn't terminate until the element in the current position $i$ is $i$. So when the outer loop moves on to the next element, you are guaranteed that the subarray to the left is sorted and in the final state ($1,2,3,\dots,i-1$)

Now to prove that the inner loop terminates... assume that it never terminates. So at a certain point, the value at $A[i]$ will loop through some values $t_1,t_2,\dots,t_k$, none of which are $i$. Let $T=\{\text{all values $A[i]$ loops through infinitely}\}$. Clearly $|T|<n-i$.

When $t_2$ is at $i$, then $t_1$ will be at position $t_1$. For $t_1$ to ever return to position $i$ (which it must as it's an infinite loop), there must be a $t_p>t_1$ in $T$. Otherwise $A[t_1]$ will never be affected by a reversal.

By symmetry this is true of all $t_j \in T$. Hence $T$ must contain an element which is strictly larger than every element in $T$... a contradiction. Hence the inner loop terminates.

You could also say that this finite set $T$ must contain an infinite sequence of strictly increasing numbers, which is not possible.

  • $\begingroup$ @Yuval Filmus Thanks for the edits $\endgroup$ – Hannah W. May 5 at 10:08

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