Running time/time complexity for while loop with square root

Question

This question looks relatively simple, but I can\'t seem to find the running time in terms of n.

Here is the problem:

j = n;
while(j >= 2) {
    j = j         


        

Answer 1

Work backwards to get the number of time units for line 2:

                                   time
n              n       log_2(n)    units
1              1        0          0
2              2        1          1
4              4        2          2
16             16       4          3
16^2           256      8          4
(16^2)^2       65536    16         5
((16^2)^2)^2)  ...      32         6

In other words, for the number of time units t, n is 2^(2^(t-1)) except for the case t = 0 in which case n = 1.

To reverse this, you have

t = 0 when n < 2

t = log₂(log₂(n)) + 1 when n >= 2

where log₂(x) is known as the binary logarithm of x.