XOR of three values

Question

What is the simplest way to do a three-way exclusive OR?

In other words, I have three values, and I want a statement that evaluates to true IFF only one of the three values is true.

So far, this is what I've come up with:

((a ^ b) && (a ^ c) && !(b && c)) || ((b ^ a) && (b ^ c) && !(a && c)) || ((c ^ a) && (c ^ b) && !(a && b))

Is there something simpler to do the same thing?

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Here's the proof that the above accomplishes the task:

a = true; b = true; c = true
((a ^ b) && (a ^ c) && !(b && c)) || ((b ^ a) && (b ^ c) && !(a && c)) || ((c ^ a) && (c ^ b) && !(a && b))
=> false

a = true; b = true; c = false
((a ^ b) && (a ^ c) && !(b && c)) || ((b ^ a) && (b ^ c) && !(a && c)) || ((c ^ a) && (c ^ b) && !(a && b))
=> false

a = true; b = false; c = true
((a ^ b) && (a ^ c) && !(b && c)) || ((b ^ a) && (b ^ c) && !(a && c)) || ((c ^ a) && (c ^ b) && !(a && b))
=> false

a = true; b = false; c = false
((a ^ b) && (a ^ c) && !(b && c)) || ((b ^ a) && (b ^ c) && !(a && c)) || ((c ^ a) && (c ^ b) && !(a && b))
=> true

a = false; b = true; c = true
((a ^ b) && (a ^ c) && !(b && c)) || ((b ^ a) && (b ^ c) && !(a && c)) || ((c ^ a) && (c ^ b) && !(a && b))
=> false

a = false; b = true; c = false
((a ^ b) && (a ^ c) && !(b && c)) || ((b ^ a) && (b ^ c) && !(a && c)) || ((c ^ a) && (c ^ b) && !(a && b))
=> true

a = false; b = false; c = true
((a ^ b) && (a ^ c) && !(b && c)) || ((b ^ a) && (b ^ c) && !(a && c)) || ((c ^ a) && (c ^ b) && !(a && b))
=> true

a = false; b = false; c = false
((a ^ b) && (a ^ c) && !(b && c)) || ((b ^ a) && (b ^ c) && !(a && c)) || ((c ^ a) && (c ^ b) && !(a && b))
=> false

Answer 1

For exactly three terms, you can use this expression:

(a ^ b ^ c) && !(a && b && c)

The first part is true iff one or three of the terms are true . The second part of the expression ensures that not all three are true .

Note that the above expression does NOT generalize to more terms. A more general solution is to actually count how many terms are true , so something like this:

int trueCount =
   (a ? 1 : 0) +
   (b ? 1 : 0) +
   (c ? 1 : 0) +
   ... // more terms as necessary 

return (trueCount == 1); // or some range check expression etc

Answer 2

bool result = (a?1:0)+(b?1:0)+(c?1:0) == 1;

Answer 3

a^b^c is only 1 if an uneven number of variables is 1 (two '1' would cancel each other out). So you just need to check for the case "all three are 1":

result = (a^b^c) && !(a&&b&&c)

Answer 4

Another possibility:

a ? !b && !c : b ^ c

which happens to be 9 characters shorter than the accepted answer :)

Answer 5

Better yet on Python:

result = (1 if a else 0)+(1 if b else 0)+(1 if c else 0) == 1

This can be used also on if statements!

It saved my day for CLI mutually exclusive arguments through Click (everyone hates click)

Answer 6

您也可以尝试（在 C 中）：

!!a + !!b + !!c == 1

Answer 7

f= lambda{ |a| [false, false, true].permutation.to_a.uniq.include? a }
p f.call([false, true, false])
p f.call([false, true, true])

$ true

$ false

Because I can.

Answer 8

Here's a general implementation that fails quickly when more than one bool is found to be true .

Usage :

XOR(a, b, c);

Code :

public static bool XOR(params bool[] bools)
{
    return bools.Where(b => b).AssertCount(1);
}

public static bool AssertCount<T>(this IEnumerable<T> source, int countToAssert)
{
    int count = 0;
    foreach (var t in source)
    {
        if (++count > countToAssert) return false;
    }

    return count == countToAssert;
}

Answer 9

In C:

#include <stdbool.h>

bool array_xor(size_t array_size, bool[] array) {
    int count = 0;

    for (int i = 0; i < array_size && count < 2; i++) {
        if (array[i]) {
            count++;
        }
    }

    return count == 1;
}