Fastest way of finding the middle value of a triple?

There's an answer here using min/max and no branches (https://stackoverflow.com/a/14676309/2233603). Actually 4 min/max operations are enough to find the median, there's no need for xor's:

median = max(min(a,b), min(max(a,b),c));

Though, it won't give you the median value's index...

Breakdown of all cases:

a b c
1 2 3   max(min(1,2), min(max(1,2),3)) = max(1, min(2,3)) = max(1, 2) = 2
1 3 2   max(min(1,3), min(max(1,3),2)) = max(1, min(3,2)) = max(1, 2) = 2
2 1 3   max(min(2,1), min(max(2,1),3)) = max(1, min(2,3)) = max(1, 2) = 2
2 3 1   max(min(2,3), min(max(2,3),1)) = max(2, min(3,1)) = max(2, 1) = 2
3 1 2   max(min(3,1), min(max(3,1),2)) = max(1, min(3,2)) = max(1, 2) = 2
3 2 1   max(min(3,2), min(max(3,2),1)) = max(2, min(3,1)) = max(2, 1) = 2

It's possible to answer the query without branches if the hardware can answer min and max queries without branches (most CPUs today can do this).

The operator ^ denotes bitwise xor.

Input: triple (a,b,c)
1. mx=max(max(a,b),c)
2. mn=min(min(a,b),c)
3. md=a^b^c^mx^mn
4. return md

This is correct because:

xor is commutative and associative
xor on equal bits produces zero
xor with zero doesn't change the bit

The appropriate min/max functions should be chosen for int/float. If only positive floats are present then it's possible to use integer min/max directly on the floating point representation (this could be desirable, since integer operations are generally faster).

In the unlikely scenario that the hardware doesn't support min/max, it's possible to do something like this:

max(a,b)=(a+b+|a-b|)/2
min(a,b)=(a+b-|a-b|)/2

However, this isn't correct when using float operations since the exact min/max is required and not something that's close to it. Luckily, float min/max has been supported in hardware for ages (on x86, from Pentium III and onwards).

If you are looking for the most efficient solution, I would imagine that it is something like this:

if (array[randomIndexA] > array[randomIndexB]) {
  if (array[randomIndexB] > array[randomIndexC]) {
    return "b is the middle value";
  } else if (array[randomIndexA] > array[randomIndexC]) {
    return "c is the middle value";
  } else {
    return "a is the middle value";
  }
} else {
  if (array[randomIndexA] > array[randomIndexC]) {
    return "a is the middle value";
  } else if (array[randomIndexB] > array[randomIndexC]) {
    return "c is the middle value";
  } else {
    return "b is the middle value";
  }
}

This approach requires at least two and at most three comparisons. It deliberately ignores the possibility of two values being equal (as did your question): if this is important, the approach can be extended to check this also.

Fastest way of finding the middle value of a triple?

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