Recurrence Relation for a loop

The question is t开发者_如何学运维o set up a recurrence relation to find the value given by the algorithm. The answer should be in teta() terms.

foo = 0;

for int i=1 to n do
    for j=ceiling(sqrt(i)) to n do
        for k=1 to ceiling(log(i+j)) do
             foo++

Not entirely sure but here goes.

Second loop executes 1 - sqrt(1) + 2 - sqrt(2) + ... + n - sqrt(n) = n(n+1)/2 - n^1.5 times => O(n^2) times. See here for a discussion that sqrt(1) + ... + sqrt(n) = O(n^1.5).

We've established that the third loop will get fired O(n^2) times. So the algorithm is asymptotically equivalent to something like this:

for i = 1 to n do
    for j = 1 to n do
        for k = 1 to log(i+j) do
            ++foo

This leads to the sum log(1+1) + log(1+2) + ... + log(1+n) + ... + log(n+n). log(1+1) + log(1+2) + ... + log(1+n) = log(2*3*...*(n+1)) = O(n log n). This gets multiplied by n, resulting in O(n^2 log n).

So your algorithm is also O(n^2 log n), and also Theta(n^2 log n) if I'm not mistaken.