Question

This is the question I’m trying to solve:

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The following divide-and-conquer algorithm is proposed for finding the simultaneous maximum and minimum:

• If there is one item, it is the maximum and minimum

• if there are two items, then compare them and in one comparison you can find the maximum and minimum.

• Otherwise, split the input in two halves, divided as evenly as possibly (if N is odd, one of the two halves will have one more element than the other).

• Recursively find the maximum and minimum of each half, and then in two additional comparisons produce the maximum and minimum for the entire problem.

(b) Suppose N is of the form 3 + 2k. What is the exact number of comparisons used by this algorithm?

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for this point (b), I tried to find a recurrence equation to solve but it didn’t work.
I’ve tried

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

where three is the minimum cost when I try 3 inputs.
any help?

Answer 1

Your recurrence equation should not have a term for the special case of n = 3. The algorithm gives you these facts:

• T(1) = 0
• T(2) = 1
• T(n) (n > 2) = T(floor(n/2)) + T(ceil(n/2)) + 2

That should be all you need to work out the answer.