How do I find the time complexity of this algorithm?

Question

I have been trying to calculate the time complexity of this algorithm but I don't think that I am right.

void unknownRuntime( FILE* input )
    {
    int temp;
    int size;
    int n;
    int i=0;
    int *array;

    if( fscanf( input, "%d", &size ) != 1 )
    {
        exit(-1);
    }
    array = (int *) malloc( size*sizeof(int) );
    if( array == NULL )
    {
        exit(-1);
    }

    while( fscanf( input, "%d", &temp ) == 1 && i<size)
    {
        array[i] = temp;
        i++;
    }

    for( i=0; i<size; i++ )
    {
        for( n=size-1; n>1; n/=1.01 )
        {
            array[n-1] = array[n];
        }
    }

    free(array);
}

I developed a program that tests different array sizes to establish a relationship between input size and runtimes of the given mystery function and here were my results:

N vs. seconds plot

From the running times that I plotted, it appears to be O( N log N ), but I'm not sure if I can see where the algorithm takes O(1) time to cut the problem size by a fraction. Any help would be appreciated.

Answer 1

Let's look at the inner loop. And let's pretend we're dealing with real numbers (as opposed to type-constrained numbers).

i=0:  n = (size-1)
i=1:  n = (size-1)/1.01
i=2:  n = (size-1)/1.01/1.01
i=3:  n = (size-1)/1.01/1.01/1.01
...

We could write that as

n = (size-1) / ( 1.01 ^ i )    // "^" is being used to denote exponentiation.

The loop will therefore stop when

1.01 ^ i >= size - 1

which is to say when

i >= log[1.01](size - 1)       // log[base](x)

The inner loop is therefore O(log N), so the whole is O(N log N). ^[1]

---

Now, the above was assuming real numbers. What the algorithm is really doing is

i=0:  n = (size-1)
i=1:  n = floor( (size-1) / 1.01 )
i=2:  n = floor( floor( (size-1) / 1.01 ) / 1.01 )
i=3:  n = floor( floor( floor( (size-1) / 1.01 ) / 1.01 ) / 1.01 )
...

Where floor represents the truncation caused from the implicit cast to int . (There could also be some error from limited floating-point precision not represented above.)

n decreases a little bit faster in this progression than the earlier progression, so this progression will approach 1 faster, but the difference is minor, and the number of step will still be O(log N).

---

1. log usually refers to base 2 logarithms in Computer Science, but it's irrelevant here because we ignore constant factors in Big-O notation, and you can convert logarithms from one base to another by multiplying by a constant factor.

   For example, log[1.01](x) = log[2](x) * 1/log[1.01](2) .