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I have three boolean values. I need to return false if all three are true or if all three are false. I will return true in every other situation. Based on my research, in some specifications this is called a three-variable exclusive-or.

Edit: Some specifications claim a three variable XOR involves the only true result would come from a set where only one parameter is true. The XOR I am referring to here is of another specification where multiple values may be true, but not all.

  • What's the fastest way to perform this operation? a xor b xor c doesn't work

  • What's if it wasn't three but n parameters?

Here's the truth table for my desired operation (xor with three params).

A   B   C   -
T   T   T   F
T   T   F   T
T   F   T   T
T   F   F   T
F   T   T   T
F   T   F   T
F   F   T   T
F   F   F   F
Hatefiend
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    Possible duplicate of [XOR of three values](https://stackoverflow.com/questions/3466452/xor-of-three-values) – subdeveloper Aug 31 '18 at 05:25
  • @subdeveloper Unfortunately not. Edited my question to explain further why. In that question he asks `I want a statement that evaluates to true IFF only one of the three values is true.` This is not the same as mine. – Hatefiend Aug 31 '18 at 05:28

4 Answers4

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To make an algorithm for that, you need to know how to use karnaugh map in three variable. See sample karnaugh map here

Ok. First, to make things easier replace T as 1 and F as 0 in your truth table.

At first glance, it is just an increasing 3-bit binary. So arranging it in an increasing way is a good idea. Take a look below.

A   B   C       F(A,B,C)
0   0   0       0
0   0   1       1
0   1   0       1
0   1   1       1
1   0   0       1
1   0   1       1
1   1   0       1
1   1   1       0

By using karnaugh-map, you will get a boolean expression below. For the first expression we get A'B .

see image 1

For the second expression AB' .

see image 2

For the third expression B'C .

see image 3

For the fourth expression BC' .

enter image description here

To simply understand karnaugh-map, if all 1's are inside the straight sight towards the table of a variable then one term of expression will contain only that variable. But if 1's are outside the straight sight of that variable, then, it is a compliment of that variable.

F(A,B,C) = A'B + AB'+ B'C + BC'

but since

A XOR B = AB'+ A'B
B XOR C = BC'+ B'C

then our simplified form will be

F(A,B,C) = A XOR B + B XOR C

for pseudo code programming, it is equivalent to

result = (A XOR B) OR (B XOR C)
//other else
result = (A ^ B) | (B ^ C)
Leandro Keen Zapa
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4

Use this :

(A XOR B) OR (B XOR C)

Works for n inputs as well :

(A XOR B) OR (B XOR C) OR ...(n XOR n+1)

subdeveloper
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1

Hear is my python implementation of it

def xor_three(a, b, c):
    return (a ^ b) or (b ^ c)

A = True
B = False
C = True

print(xor_three(A, B, C))
abhinit21
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0

If all 3 are true or all 3 are false, I think the best way is:

if (a AND b AND c) or (not a AND not b AND not c):
    # Do something

If you only want it returns true when one and only one of the values is true:

if int(a) + int(b) + int(c) == 1:
    # Do something

Many thanks!

Xuan Quyen Nguyen
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