Divide and conquer algorithm applied in finding a

问题:

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.

回答1:

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".