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I'm solving a practice quiz and came across the following question

Write down the recurrence corresponding to the below divide and conquer algorithm, labeling exactly the components for each of: dividing, conquering, and combining.

1. Foo (p, r):
2.     if p = r
3.          return (1)
4.     else
5.         s ← 1
6.         for i = p to r
7.             s ← s * i
8.         q ← Foo(p, r − 1) * s
9.     return (q)

My attempt at an answer.

  1. Let T(n) be the work done by Foo over p to r, so T(n) is equivalent of Foo(p, r) where n is r - p + 1.

  2. I get the following recurrence T(n) = T(n - 1) + Θ(n) + Θ(1)

  3. The dividing part would be a constant Θ(1) which corresponds to the r-1 operation.

  4. The conquering part would be the T(n - 1) which is recursively solving the sub-problem.

  5. The combining part is a constant Θ(1) for the multiplication operation of T(n - 1) * s.

But that seems wrong as I didn't mention the Θ(n). What part of dividing, conquering, combining should the Θ(n) of lines 6,7 fall into?

thiton
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ayh
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  • Umm, where does "Theta" come from in the code above? Or "T"? – Jim Clay Nov 17 '11 at 17:42
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    @JimClay I meant the theta of asymptotic notation, and edited my post to try and clarify what I meant by T(n) – ayh Nov 17 '11 at 18:03
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    Wow, nowadays even divide by 1 (calling itself only once) can be called "divide and conquer"... – kennytm Nov 17 '11 at 18:13
  • @KennyTM +1 LOL You're right it's a quite bad example of divide and conquer... also the algorithm itself does not make any sense, for Foo(p, r) it calculates the product r (r-1)^2 (r-2)^3 ... p^(r-p+1) ... why is that useful? The algorithm is also as fast as a trivial iterative implementation, so no benefit from "divide and conquer" here – Antti Huima Nov 19 '11 at 01:57

1 Answers1

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s accumulates from p to r so this would seem to fall into the "combining" part. So we have Θ(n) coming from combining.

As we combine the elements back together, we essentially have to zip the back up over the n elements.

Fahd A.
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