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reference: 

Problem Definition:

The key idea is to incorporate the branch into one equation. 

Solution:

Minimum of x and y will be

y ^ ((x ^ y) & -(x < y))

It works because if x < y, then -(x < y) will be all ones, so r = y ^ (x ^ y) & ~0 = y ^ x ^ y = x. Otherwise, if x >= y, then -(x < y) will be all zeros, so r = y ^ ((x ^ y) & 0) = y.On some machines, evaluating (x < y) as 0 or 1 requires a branch instruction, so there may be no advantage.

To find the maximum, use

x ^ ((x ^ y) & -(x < y));

Code:

/*Function to find minimum of x and y*/int min(int x, int y){  return y ^ ((x ^ y) & -(x < y));} /*Function to find maximum of x and y*/int max(int x, int y){  return x ^ ((x ^ y) & -(x < y)); }

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