Notes: 02-02 Stooge Sort

Stooge Sort

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 function stoogesort(array L, i = 0, j = length(L)-1){
     if L[i] > L[j] then       // If the leftmost element is larger than the rightmost element
         swap(L[i],L[j])       // Then swap them
     if (j - i + 1) > 2 then   // If there are at least 3 elements in the array
         t = floor((j - i + 1) / 3)
         stoogesort(L, i, j-t) // Sort the first 2/3 of the array
         stoogesort(L, i+t, j) // Sort the last 2/3 of the array
         stoogesort(L, i, j-t) // Sort the first 2/3 of the array again
     return L
 }

We can build a recurrence relation here:

T(n) = 3T(2n/3) + \Theta(1)

• 3T because 3 recursions.
• 2n/3 because each recursion is 2/3 as big.

Recursion Tree

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• Level 0 (Root): 1 node, non-recursive O(1) work
• Level 1: 3 nodes, each working on 2/3n iput, non-recursive O(1) work = 3
• Level 2: 3^2 nodes, each working on (2/3)^2 items. 9 work
• Level i: 3^i nodes, each working on (2/3)^i items. 3^i work

To get height, take n = (3/2)^h and solve for h:

• log(3/2) both sides
• So height is log_{3/2}(n).

So we end up with: \sum_{i=1}^{log_{3/2}(i)} of 3^{i} as the total amount of work.

The last level ends up dominating:

• Draw the tree.
• Show how we can line up all the other levels next to the last level and the sum is shorter.

So total work is \Theta(3^h).

• = 3^{log(3/2)}
• We’ve got a log identity: a^{log_{b}(c)} = c^{log_{b}(a)}.
• That gets us n^{log_{3/2}(3)}.
• Python: math.log(3, 1.5)
• So \Theta(n^{2.71})

So we end up with 3^{log_{3/2}(n)} ~ \Theta(n^2.71)

New Topic: Subset Sum

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Given a number K and list of integers xs, find a subset of xs that sums to K.

Simple solution:

• First, assume xs[0] is in the solution.
• Recursively try to find a subset of xs[1..] that adds up to K - xs[0].
• If not, xs[0] isn’t in the solution, recurse on xs[1..] and K.

What’s the runtime?