Question

I found by chance that

int a = (h/2)*w+ (  (h+1)/2-h/2   )  *  (w+1)/2 ;

is equal to

int b = (w * h + 1) / 2 ;

when w and h are positive integers (assume no overflow).

Can you show me why these 2 are the same?

edit : integer -> positive integer.

Answer 1

Actually this is a math problem: (integer)/2 should be interpreted as floor. So, the problem is:

Show that floor(h/2)*w + ( floor((h+1)/2) - floor(h/2) ) * floor((w+1)/2) is equivalent to floor((w*h+1)/2)

Proof:

1. h = 2k, w = 2l: (both numbers are even) …
2. h = 2k + 1, w = 2l: …
3. h = 2k, w = 2l + 1: …
4. h = 2k + 1, w = 2k + 1: …

A hint: floor((2k+1)/2) == k. You can easily show the equivalence.

For example, the case 4:

a) floor(2k+1/2)*(2l+1) + ( floor((2k+2)/2) - floor((2k+1)/2) ) * floor((2l+2)/2) = 2kl+k + (k+1 - k)*(l+1) = 2kl + k + l + 1

b) floor(((2k+1)*(2l+1)+1)/2) = floor((4kl+2k+2l+2)/2) = 2kl + k + l + 1

Therefore, the two equations are equivalent.