Divide and conquer algorithm applied in finding a peak in an array.

Question

For an array a: a₁, a₂, … a_k, … a_n, a_k is a peak if and only if a_k-1 ≤ a_k ≥ a_k+1 when 1 < k and k < n. a₁ is a peak if a₁ ≥ a₂, and a_n is a peak if a_n-1 ≤ a_n. The goal is to find one peak from the array.

A divide and conquer algorithm is given as follows:

find_peak(a,low,high):
    mid = (low+high)/2
    if a[mid-1] <= a[mid] >= a[mid+1] return mid // this is a peak;
    if a[mid] < a[mid-1] 
        return find_peak(a,low,mid-1) // a peak must exist in A[low..mid-1]
    if a[mid] < a[mid+1]
        return find_peak(a,mid+1,high) // a peak must exist in A[mid+1..high]

Why this algorithm is correct? I think it may suffer from losing half of the array in which a peak exists.

It really comes down to the assertions; is it true that "a peak must exist in A[low...mid-1]" if a[mid] < a[mid-1]? — Oliver Charlesworth, Jun 02 '13 at 21:07
This algorithm assumes an awful lot about the way the data is structured.. — christopher, Jun 02 '13 at 21:08
re "I think it may suffer from losing half of the array in which a peak exists." You said earlier you only wanted to find _one_ peak. So it doesn't matter if you lose half the array so long as there is a peak in the other half. Also, have you tried coding up the algorithm and running it in a debugger with some different test cases? Might give you some intuition as to what it is doing. — TooTone, Jun 02 '13 at 21:22
This is assuming that the array is holding a smoothing changing function. — Raedwald, Jul 12 '13 at 09:32

score 7 · Answer 1 · edited Jun 02 '13 at 22:14

7

The algorithm is correct, although it requires a bit of calculus to prove. First case is trivial → peak. Second case is a "half peak", meaning that it has the down slope, but not up.

We have 2 possibilities here:

The function is monotonically decreasing till a_mid → a₁ ≥ a₂ is the peak.
The function is not monotonically decreasing till a_mid → There is a 1≥k≥mid, for which a_k ≥ a_k-1. Let's choose the largest k for which this statement is true → a_k+1 < a_k ≥ a_k-1 → and that's the peak.

Similar argument can be applied for the third case with the opposite "half peak".

edited Jun 02 '13 at 22:14

icktoofay

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answered Jun 02 '13 at 21:57

SomeWittyUsername

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is it worth adding that this can break down if we have a run of constants at one side or the other? – TooTone Jun 02 '13 at 22:39
Thanks for your answer. I see your point and it's intuitively right. Can you specify how to prove it? – roland luo Jun 02 '13 at 23:41
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@TooTone Nope, it's still valid - 3 equal consecutive values also qualify for a peak according to the problem definition – SomeWittyUsername Jun 03 '13 at 06:29
@icepack I stand corrected: it's a bit of a non-intuitive definition of peak. – TooTone Jun 03 '13 at 06:46

Divide and conquer algorithm applied in finding a peak in an array.

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