XOR operation on three values

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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
xor logical-operators boolean-logic boolean
3 Answers

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)
user_50011070

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)

user_15846230

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))
user_133347780
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