Running time/time complexity for while loop with square root

问题

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^(1/2)
}

I don't really need the total running time, I just need to know how to calculate the amount of times the second and third lines are hit (they should be the same). I'd like to know if there is some sort of formula for finding this, as well. I can see that the above is the equivalent of:

for(j = n; n >= 2; j = j^(1/2)

Please note that the type of operation doesn't matter, each time a line is executed, it counts as 1 time unit. So line 1 would just be 1 time unit, line 2 would be:

• 0 time units if n were 1,
• 1 time unit if n were 2,
• 2 time units if n were 4,
• 3 time units if n were 16, etc.

Thanks in advance to anyone who offers help! It is very much appreciated!

回答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.

来源：https://stackoverflow.com/questions/32798251/running-time-time-complexity-for-while-loop-with-square-root