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Considering that to prove a recursive algorithm we should refer to mathematical induction. Given the following algorithm (which sort an Array of size r) I found that base cases are for array size of 0 and 1 because an empty or 1-element array is already sorted. How can I prove mathematically that the algorithm behaviour holds for bigger sizes of the array?

1: MYSTERY(A,l,r) 
2:   range := r − l + 1 
3:   subrange := d(2 · range/3) // d() I assumed d() as the ceiling function 
4:   if range = 2 and A[l] > A[r] then 
5:     swap A[l] ↔ A[r] 
6:   else if range ≥ 3 then 
7:     MYSTERY(A,l,l + subrange − 1) 
8:     MYSTERY(A,r − (subrange − 1),r) 
9:     MYSTERY(A,l,l + subrange − 1) 
10:  end if 
11: end
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  • $\begingroup$ You could provide some description of what is going on. It appears MYSTERY is supposed to sort the elements from $l$ through $r$ of array A. Note that the use of $l$ makes the program very hard to read because it looks like $1$. Where do d2 and 3e come from? Clearly you should think about the subarrays that are sorted in lines 7-9. $\endgroup$ Commented Apr 19, 2013 at 17:21

2 Answers 2

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Given an array $A$, a call to MYSTERY(A, left, right) will sort the subarray $A[\text{left}, \dots,\text{right}]$ in increasing order. The algorithm is commonly known as stooge sort. Here's an outline of a correctness proof by induction on the size of the array, $s=right-left+1$.

Base. Clearly, when $s\le 3$, MYSTERY sorts the relevant part of the array.

Induction(strong). Let $s\ge 3$ and assume that MYSTERY sorts all arrays of size less than $s$. The algorithm makes three recursive calls:

  1. Sort the first 2/3 of the array.
  2. Sort the last 2/3 of the array.
  3. Sort the first 2/3 of the array.

For convenience, call the first third of the array to be sorted $P$, the second third $Q$ and the last third $R$. We have the following sequence of steps:

  1. After the first recursive call, the $[P, Q]$ portion of the array is sorted (by the inductive hypothesis), so in particular every element in $Q$ will be greater than or equal to everything in $P$.
  2. After the second call, sorting the $[Q,R]$ subarray, we'll have everything in $R$ in its correct (sorted) place in the array, so we're done with that part.
  3. After the last pass, everything in the $[P,Q]$ portion will be in its correct sorted place in the array, so we're done: the portion of the array from positions $left$ to $right$ will be sorted.

It's worth noting that this is even less efficient than, say, Bubblesort.

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This seems to have a number of problems.

First of all, what are d2 and 3e? These are not specified anywhere.

When range = 2, you are sorting the array. So it looks like the routine wants to sort its input.

Otherwise you process the first subrange elements, then the last subrange elements, then the first subrange elements again. If subrange is less than half the size of the array, there will be a middle part that will never get processed.

Disregarding all this, you have to state what the induction hypothesis is. In this case it would seem to be: For all arrays of size less than n, this routine sorts the array.

Assuming this is true, and assuming that the input array is of size n, prove that the routine then sorts that array.

However, it does not seem to me that this routine will do that.

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  • $\begingroup$ This seems better suited to a comment. $\endgroup$ Commented Apr 20, 2013 at 9:09

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