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?

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

polygenelubricants · Accepted Answer · 2010-08-12T10:09:20.253

53

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

edited Aug 12 '10 at 10:09

answered Aug 12 '10 at 09:54

polygenelubricants

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Great, but the general solution does not short-circuit. – Ani Aug 28 '10 at 09:35
2

`a ? 1 : 0` can be simplified to `!!a` – kaspersky May 26 '14 at 07:46
4

@gg.kaspersky, only in JavaScript, C, and languages which have truthy/falsy tests via the `!` operator. For example, this wouldn't work in Java or C#. – Drew Noakes Sep 03 '14 at 14:50
@DrewNoakes, when I made the comment, I thought for some reason that the question was C specific. Indeed, certain syntax constructs are valid only in certain languages. – kaspersky Sep 03 '14 at 17:52

score 20 · Answer 2 · answered Aug 12 '10 at 09:54

20

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

answered Aug 12 '10 at 09:54

Hemant

19,486
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1

I love it. Very simple and understandable. – Aaron D Aug 20 '10 at 15:52
The simplest and must understandable when reading the code (well, I'd add few spaces) – gerardw Aug 20 '14 at 16:50
In Haskell: `list_xor xs = sum ( map (\x -> if x then 1 else 0) xs ) == 1` – Jivan Pal May 28 '22 at 01:26

score 9 · Answer 3 · answered Aug 12 '10 at 09:53

9

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)

answered Aug 12 '10 at 09:53

Aaron Digulla

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score 9 · Answer 4 · answered Aug 25 '10 at 12:41

9

Another possibility:

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

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

answered Aug 25 '10 at 12:41

Timwi

65,159
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Kinda glad this is not the accepted answer for readability reasons, but still, I had to upvote as shortness still has some kinky appeal to me. On university in C class, we had a contest in writing the shortest source code that generates a song similar to Heidis Kuken, and I was in TOP 5 in a contest. – Vlasec Nov 22 '19 at 22:23
@Vlasec I assume you know about [code golf](https://codegolf.stackexchange.com/), but if you didn't here you go. – ThomasPM Jun 07 '23 at 21:33

score 3 · Answer 5 · answered May 19 '19 at 23:57

3

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)

answered May 19 '19 at 23:57

Pier Carlo Cadoppi

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5

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python converts bool to int, so `result = bool(a) + bool(b) + bool(c) == 1` – Drdilyor May 18 '22 at 07:44
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and if a, b and c are already booleans, you can simply say `result = a + b + c == 1` – Pier Carlo Cadoppi May 19 '22 at 13:57

kaspersky · Answer 6 · 2014-06-20T14:18:32.453

2

You could also try (in C):

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

edited Jun 20 '14 at 14:18

answered May 20 '14 at 07:37

kaspersky

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This is only valid in some languages [eg. javascript], and it's not a simplification. In a language which auto-converts boolean to number for `+`, if `a`, `b`, and `c` are already boolean, you also don't need the double negation `!!`: just `a+b+c===1` will be equivalent. – blgt Jun 20 '14 at 10:23
1

@blgt, my mistake, I should specify that I meant to be an answer for C language. – kaspersky Jun 20 '14 at 14:18
1

Using `!!` to guarantee a C true is `1` is actually pretty neat. A bit paranoid in the context of a boolean logic question though. If the rest of your code is well-written `a+b+c==1` should still be sufficient – blgt Jun 20 '14 at 14:35

score 1 · Answer 7 · answered Nov 22 '12 at 09:11

1

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.

answered Nov 22 '12 at 09:11

nurettin

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score 1 · Answer 8 · answered Aug 28 '10 at 09:37

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;
}

Jivan Pal · Answer 9 · 2022-05-28T01:30:35.543

0

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;
}

edited May 28 '22 at 01:30

answered Apr 27 '22 at 20:58

Jivan Pal

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10

I'm not sure this works, nor is it simple. – Jaan May 05 '23 at 04:27
Define "simple". It definitely works, as what it's doing is checking that exactly one element of the array is `true`. It works for an arbitrarily large array of booleans, not just 3 elements, and short circuits as soon as it finds a second element that is `true` (hence `&& count < 2`). – Jivan Pal May 05 '23 at 11:36

XOR of three values

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