Binary Search Algorithm - Performance

Performance

With each test that fails to find a match at the probed position, the search is continued with one or other of the two sub-intervals, each at most half the size. More precisely, if the number of items, N, is odd then both sub-intervals will contain (N - 1)/2 elements, while if N is even then the two sub-intervals contain N/2 - 1 and N/2 elements.

If the original number of items is N then after the first iteration there will be at most N/2 items remaining, then at most N/4 items, at most N/8 items, and so on. In the worst case, when the value is not in the list, the algorithm must continue iterating until the span has been made empty; this will have taken at most ⌊log₂(N) + 1⌋ iterations, where the ⌊ ⌋ notation denotes the floor function that rounds its argument down to an integer. This worst case analysis is tight: for any N there exists a query that takes exactly ⌊log₂(N) + 1⌋ iterations. When compared to linear search, whose worst-case behaviour is N iterations, we see that binary search is substantially faster as N grows large. For example, to search a list of one million items takes as many as one million iterations with linear search, but never more than twenty iterations with binary search. However, a binary search can only be performed if the list is in sorted order.