How to find out of 4 numbers, where 3 are equal to each other, one is not equal, at a time?

Question

Programming

Tell me how, using the minimum number of conditional operators, to find out of 4 numbers, one is not equal, if the remaining 3 are equal to each other.
Only without loops and arrays, only conditional statements.
I was told that this is related to the problem of three coins, supposedly the same algorithm.

Answer 1

In 4 equal numbers, there is no one that is not equal.
0 comparison operators. I'm done?

Answer 2

It is necessary to clarify - what to count, comparison operations, or conditional operators. Count written operations/statements, or in progress.
Possible solutions:

int X[4];
return X[(X[0]==X[2])+2*(X[0]==X[1])];

int f(int p,int q,int r){
  return p==r ? q : p;
}
int g(int a,int b,int c,int d){
  return a==b ? f(c,d,a) : f(a,b,c);
}

int x=a^b,y=a^c;
x=(x|-x)>>31; y=(y|-y)>>31;
return ((a^b^c^d)&x&y)^((b^d)&x)^((c^d)&y)^d;

(the latter has no comparisons and conditional statements at all).

Answer 3

Without explicit conditions, but with floating arithmetic, therefore it is inaccurate:
With one condition for positive numbers:

x = x1+x2-x3-x4;
if (0 > x) x = -x;

Answer 4

I could only come up with two comparisons. For one thing I can’t imagine how =)
https://jsfiddle.net/mjdtfbkq/1/

Answer 5

JavaScript:

a1=3; a2=2; a3=3; a4=3;
a1*!(2*a2-a3-a4)+a2*!(2*a3-a4-a1)+a3*!(2*a4-a1-a2)+a4*!(2*a1-a2-a3)

0 conditional.

Answer 6

Compare any two, this comparison will give you a pair of known 'good' and a pair of fake in any outcome. The second comparison is to compare the obviously good with any of the pair with a fake. This will give the answer. Two comparisons.