Binary search runs in O(log n) time.

# Binary search runs in O(log n) time. - Binary search runs...

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Binary search runs in O (log n ) time. Michael George Tuesday March 29, 2005 This is a proof that binary search runs in O (log n ) time. Here is the code: binsearch ( A, x, a, b ) if b = a then return false m b - a 2 + a if A [ m ] > x then return binsearch ( A, x, a, m ) else if A [ m ] = x then return true else if A [ m ] < x then return binsearch ( A, x, m, b ) Let C be the amount of required to run all of the code in the procedure except for the two recursive calls, and let T ( n ) be the total amount of time required to run the procedure when b - a = n . I claim that T ( n ) C log n + T (1) for all n 1. I will prove this by strong induction. The base case (when n = 1) is clear: C log 1 + T (1) = 0 + T (1) = T (1) Now, choose a particular n > 1. For our inductive hypothesis we will assume that for all k < n , that T ( k ) C log k + T (1). How long does binsearch take to run if b - a = n ? Well, there are three possibilities: we could take the first branch of the if statement (i.e. A [ m ] > x ), we could take the second branch ( A [ m ] = x ), or we could take the third branch ( A [ m ] < x ).

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