What Is the Boolean Expression for a 3-Input XOR Gate?

A 3-input XOR gate combines three binary inputs, usually named $A$, $B$, and $C$. The key rule is parity: its output is 1 when an odd number of the inputs are 1, and 0 when the number of 1s is even.

The compact Boolean expression is $A \oplus B \oplus C$. You can also write it as two cascaded XOR operations, $(A \oplus B) \oplus C$. To express the same function using only AND, OR, and NOT, use the sum-of-products expression derived below.

The Boolean expression

XOR, or exclusive OR, of two inputs is 1 when the inputs differ. For three inputs, apply that rule twice:

$$Y = (A \oplus B) \oplus C = A \oplus B \oplus C.$$

XOR is associative, so the parentheses do not change the result. The output is 1 for input combinations with one or three 1s. In canonical sum-of-products form, this is:

$$Y = A'B'C + A'BC' + AB'C' + ABC.$$

Here a prime means NOT, adjacency means AND, and plus means OR. Each product term identifies one input combination that makes the output 1. For instance, $A'BC'$ is true when $A=0$, $B=1$, and $C=0$.

Truth table

$A$	$B$	$C$	Output $Y$
0	0	0	0
0	0	1	1
0	1	0	1
0	1	1	0
1	0	0	1
1	0	1	0
1	1	0	0
1	1	1	1

The rows with output 1 contain either one 1 or three 1s. The rows with output 0 contain zero or two 1s. This is the odd-parity rule.

Deriving the sum-of-products form

A direct way to derive the expression is to list every row of the truth table where the output is 1, then write a product term for each row. Use a complemented variable for an input that is 0 and an uncomplemented variable for an input that is 1.

For example, the row $A=0$, $B=0$, $C=1$ gives $A'B'C$. The row $A=1$, $B=1$, $C=1$ gives $ABC$. Doing this for all four output-1 rows gives:

$$\begin{aligned} Y &= A'B'C + A'BC' + AB'C' + ABC. \end{aligned}$$

This form is useful when you need an expression built from AND, OR, and NOT gates. If XOR is an allowed operation, $A \oplus B \oplus C$ is shorter and usually makes the parity behavior clearer.

Worked example

Suppose the inputs are $A=1$, $B=0$, and $C=1$. Evaluate the cascaded form:

$$\begin{aligned} Y &= (1 \oplus 0) \oplus 1 \\ &= 1 \oplus 1 \\ &= 0. \end{aligned}$$

There are two input 1s, which is even, so the output is 0. In the sum-of-products expression, none of the four terms is true for $(1,0,1)$, giving the same result.

For the input $(1,1,1)$, the chained expression is $(1\oplus1)\oplus1=0\oplus1=1$. The $ABC$ term is also true, so the expanded expression gives 1.

Common mistakes

• Treating the gate as “1 if any input is 1.” That describes a 3-input OR gate. XOR returns 0 for inputs such as $(1,1,0)$ because there are two 1s.
• Including only the one-1 cases. The three rows with exactly one 1 produce three minterms, but the row $(1,1,1)$ also produces 1 and contributes $ABC$.
• Assuming XOR means exactly one input is 1. That is a different function for three or more inputs. A multi-input XOR is 1 for any odd number of 1s, including three.
• Changing the result by regrouping. XOR is associative, so $(A\oplus B)\oplus C=A\oplus(B\oplus C)$. It is not generally valid to replace XOR with OR or to omit an input.

A 3-input XOR can therefore be described either as an odd-parity function, as a chain of two 2-input XOR gates, or by its four sum-of-products minterms.