找出以下重现的复杂性：T（n）= T（n / 2）+ log（n）

Question

How can I estimate the run time of two loops, where each runs in a logarithmic time, as shown here:

for(int i=n; i>=0; i /= 2)
{
    for( int j=i; j>=0; j /= 2)
    {
        count++;
    }
}

My analysis:

i = n  , 1  => j = lg n  | 1+lg (n) 
i = n/2, 1  => j = lg n/2 | 1+lg (n/2)
i = 1  , 1  => j = 0      | 1

How to sum up the logarithmic term in this case?

Answer 1

i = n.      | Inner loop runs log(n) iterations.       | O(log(n))
i = n/2.    | Inner loop runs log(n/2) iterations.     | O(log(n))
i = n/4.    | Inner loop runs log(n/4) iterations.     | O(log(n))
.           |                                          |
.           |                                          |
i = log(n). | Inner loop runs log(log(n)) iterations.  | O(log(log(n)))

What are we noticing ? that for each n we are adding O(log(n))
This recurrence is T(n) = T(n/2) + log(n) + 1 Which can be expanded the following way:

T(n) = log(n) + log(n/2) + log(n/4) + ... + 1
     = log(n) + [log(n)-1] + [log(n)-2] + [log(n) - 3] + .... + 1
     = log(n)*log(n) - (2 + 3 + .... log(n))
     = log(n)*log(n) - ([(2+log(n))*log(n)]/2) 
     = log(n)*log(n) - ~0.5log(n)*log(n)
     = ~0.5log(n)*log(n)

T(n) = O(log(n)*log(n))