This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
Unformatted text preview: 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.three possibilities: we could take the first branch of the if statement (i....
View
Full
Document
This note was uploaded on 09/20/2007 for the course COM S 211 taught by Professor Kozen during the Spring '06 term at Cornell.
 Spring '06
 KOZEN
 Mathematical Induction, Binary Search, Recursion, Inductive Reasoning, Control flow, recursive calls, binary search runs

Click to edit the document details