# Proof of correctness of reversal algorithm for array rotation

We are given an array A of size n and we have to rotate it in left direction by d positions. So e.g. if A = {1, 2, 3, 4, 5, 6, 7} then for d = 2, the resultant rotated array is {3, 4, 5, 6, 7, 1, 2}.

One algorithm which does this goes as follows:

1. Reverse A[0..d-1] (0-indexing)
2. Reverse A[d..n-1]
3. Reverse A[0..n-1]

These three steps surprisingly rotate the array correctly. What is the math behind this algorithm? Why does it give a correct solution? It feels magical to me. I am not able to put up a formal proof of its correctness.

Here is a proof by picture, which follows the steps of the algorithm: $$0,\ldots,d-1,d,\ldots,n-1 \\ d-1,\ldots,0,d,\ldots,n-1 \\ d-1,\ldots,0,n-1,\ldots,d \\ d,\ldots,n-1,0,\ldots,d-1$$ You can easily turn this into a formal proof by giving a formula for the permutation after each step, which you can easily prove correct.

• This is a good starting point but not sure how to give formula for the permutation? Aug 12, 2018 at 14:04
• Take it as an exercise. Your formula could involve two separate cases. Aug 12, 2018 at 16:52

First, we can see the rotation problem (as described in the question) as equivalent to: interchange of two blocks of elements which are at index-ranges $$[0, d-1]$$ and $$[d, n-1]$$. Let's call these blocks, with elements exactly as the original array $$A$$, as $$B_1$$ and $$B_2$$. As result of this interchange, the order of these blocks in $$A$$ should change from $$(B_1,B_2)$$ to $$(B_2,B_1)$$.

In the provided algorithm, after the first two steps, array $$A$$ would contain:

$$B_1$$(reversed), followed by, $$B_2$$(reversed).

Now, if this modified array $$A$$ is read/iterated from end to start, it should return elements in order:

$$[B_2$$(reversed)(reversed) $$\leftrightarrow B_2]$$, followed by, $$[B_1$$(reversed)(reversed) $$\leftrightarrow B_1]$$.

which is the intended order. The third step modifies $$A$$ to reflect such order, by reversing.