How fast is binary search?

b.length no.of iterations log(b.length+1)

0 = 20 - 1 0 0

1 = 21 - 1 1 1

3 = 22 - 1 2 2

7 = 23 - 1 3 3

15 = 24 - 1 4 4

32767 = 215 -1 15 15

1048575= 220 -1 20 20

j is the �base 2 logarithm of 2j.

m is the base 10 logarithm of 10m.

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