Question

I have to find the recurrence equation from this algorithm:

    ALGO(n)
    if n <= 2 then return(0)
    else
        y = ALGO(n/3)
        i = 2^n
        while i >= 2 do
            j = (1/2) * log(i) //base 2
            while j > 0 do
                i = i/2
                j = j - 1
        z = ALGO(n/2)
        return(z+y) 

I reasoned about it and concluded that
T(n) = O(1) if n<=2

I think that the inner while is an O(n) (j is decreased by 1 at each iteration) while the outer while is an O(logn) (the i is halved at each iteration), giving an O(n*log(n)):

T(n) = T(n/3) + T(n/2) + O(n*log(n)) if n > 2

Is this a good argument?

Answer 1

The two while loop should be O(n):

1. i = 2^n;
2. j = (1/2) * log (i) = 1/2 * n = n/2
3. the inner while executes for n/2 times then j becomes 0, now i = i / 2^(n/2) = 2^(n/2);

Now this program will go back to step 1, only with i halved. So the two loops should be:

n/2+n/4+n/8+... = O(n)

In fact this could also be viewed from a simpler perspective. Since the loop won’t terminate until i == 1, and for every execution i = i / 2, the inner loop will run n times. Imagine we flatten the inner loop and the out loop, there will be n times of i = i / 2, hence the two loops are O(n).

In conclusion, T(n) = T(n/3) + T(n/2) + O(n).

Hope this could be helpful!