How many comparisons are needed for a binary search in a set of 64 elements?

Let f (n) be the number of comparisons required to search for an element in a search sequence of size n. Then from the binary search algorithm we can have f (n) = f (n/2) + 2, where n is even.

The number of comparisons needed for a binary search in a set of 64 elements is f (64).

f (64) = f (32) + 2

= f (16) + 2 + 2

= f (8) + 2 + 2 + 2

= f (4) + 2 + 2 + 2 + 2

= f (2) + 2 + 2 + 2 + 2 + 2

= f (1) + 2 + 2 + 2 + 2 + 2 + 2

= 0 + 2 + 2 + 2 + 2 + 2 + 2 + 2

f (64) = 14