Maximum of Two Numbers Without Using if-else

Below are the methods to get minimum(or maximum) without using branching.

• Using XOR and comparison operator
  Minimum of x and y will be

  Min(x,y) = 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.

  Similarly the maximum of two integers can be found from the following:

  Max(x,y) = x ^ ((x ^ y) & -(x < y))

  Hence the code for caculating minimum and maximum is given by

  // 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));
  }

•  Using subtraction and shift
  Here is a another solution which uses the fact that a number is stored in its 2’s complement form where the msb (most significant bit) is 0, if the number is positive, otherwise its 1.

  int getMax(int x, int y) {
    int c = x - y;  
    int k = (c >> 31) & 0x1;  
    int max = x - k * y;  
    return max;  
  }

  Here, c stores the difference of x and y. Right shifting c by 31 times gives the msb of c; bit wise AND with 1 results in 1 if x < y, and 0 if x > y.

• Using middle value
  Consider a straight line with the two numbers as its end-points. Now the larger number, say y, is away from their midpoint M by half the distance between the two endpoints. If d be the distance between x and y and M be the midpoint, then larger number is

  // Large number
  y = M + d /2
  
  // Since
  M = (x+y)/2
  d = | x - y |