Question

I’m preparing for an exam and these are some of problems from last year’s tests. The task is to calculate both exact and asymptotic complexity. How would you solve it? Universally, if possible.

for ( i = j = 0; i < n; j ++ ) {
  doSomething ();
  i += j / n;
  j %= n;
}

for ( i = 0; i < 2 * n; i += 2 )
  for ( j = 1; j <= n; j <<= 1 )
    if ( j & i )
      doSomething ();

for (i = 0; i < 2*n; i++) {
  if ( i > n )
    for (j = i; j < 2 * i; j ++ ) doSomething();
  else
    for (j = n; j < 2 * n; j ++ ) doSomething();
}

Thanks in advance

Answer 1

My solution for the third loop is

t(n) = [ (n-1)*n +  ((n-1)*n)/2 ] *D  + [ n^2 +n ] *D + [ 2n ]*I

so it is in O(n^2) given that doSomething() has a constant time
and that i and j are integers.

The second term ( [ n^2 +n ] *D ) is fairly easy.

The loop

for (j = n; j < 2 * n; j ++ ) doSomething();

gets called while i <= n so it will be called n+1 times, since it starts from 0.

The loop for (j = n; j < 2 * n; j ++ ) calls doSomething() n times, so we have (n+1)*n*D = [n^2+n] *D. I assume that doSomething() has a constant time which is equal to D

The first term is a little bit more complex.

for (j = i; j < 2 * i; j ++ ) doSomething();

gets called when i>n so it will be called n-1 times.
The loop calls doSomething() i times.
The first time it gets called n+1, the second time ´n+2´ and so on until it is 2n-1 which is equal to n + (n-1).
So we get a sequence likes this {n+1, n+2, n+3, ... , n+(n-1)}.

If we sum up the sequence we get n-1 times n and the sum 1+2+3+...+ (n-1).
The last term can be solved with the “Gaußsche Summenformel” (sorry I don’t have the English name for it but you can see the formula in the German wiki link) so it is equal to ((n-1)*n)/2

So the first term is (n-1) * n + ((n-1)*n)/2 *D

And the last term is therefor the if statement which is called 2*n*I, where I is the time to execute the If statement.