Question

Stuck with me HW – Need to try complexity

time=0;
for (i=n; i>=1; i = sqrt(i))
 for (j=1; j<=i; j++)
 time++;

What I did – First loop going like this:
i=n, n^(1/2), n^(1/4)…1
than we get:

n^(1/2)^k = 1
if I log both sides one side get 0… what should I do?

Answer 1

I suppose there is a typo somewhere because otherwise it’s Θ(∞) if the input n is not smaller than 1. (For i == 1, the update i = sqrt(i) doesn’t change i, so that’s an infinite loop.)

So let us suppose it’s actually

time = 0;
for (i = n; i > 1; i = sqrt(i))
    for (j = 1; j <= i; j++)
        time++;

Then, to get the complexity of nested loops, you need to sum the complexity of the inner loop for each iteration of the outer loop. Here, the inner loop runs i times, obviously, so we need to sum the values i runs through in the outer loop. These values are n, n^0.5, n^0.25, ..., n^(1/2^k), where k is characterised by

n^(1/2^(k+1)) < 2 <= n^(1/2^k)

or, equivalently,

2^(2^k) <= n < 2^(2^(k+1))
2^k <= lg n < 2^(k+1)
k <= lg (lg n) < k+1
k = floor(lg(lg n))

Now it remains to estimate the sum from above and below to get the Θ of the algorithm. This estimate is very easy if you start writing down the sums for a few (large) values of n.