XOR operation on three values

Question

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

Answer 1

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 .

For the second expression AB' .

For the third expression B'C .

For the fourth expression BC' .

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)

Answer 2

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)