For 3 input XOR gate and XNOR gate, by solving the equations I got the result as in the picture.

So according to the solution the outputs of the 3 input XOR and XNOR gates are same. This solution holds good when number of inputs to the gates are odd.

For the case of even number of inputs XOR and XNOR are complement to each other.

With this assumption the answer to the circuit in the picture should be option A,B, and C. but the correct answer is D. I am confused as to how?

Thanks in advance for the help.

## Best Answer

The misunderstanding is that, given XOR as a logic gate, XNOR is

definedas being always its negation.Having defined your XOR-3 as an

odd parity checker(by accepting the minterm \$xyz\$ - otherwise it would be aone hot checker), the correct interpretation of a XNOR-3 would then be aneven parity checker(as pointed by Bradman175). This simply means that the expression for your algebraic XNOR-3 is not correct in this context.In other words, \$x \odot y \odot z \ne \text{XNOR-3}\$.

Let's observe an implementation through logic gates.

A three way XOR gate can be implemented with a XOR-2 itself XORed with the remaining input, and remember that

anyXNOR has then to be rendered as a XOR in series with a NOT gate. Therefore, a XNOR-3 implementation would be:^{simulate this circuit – Schematic created using CircuitLab}This yields a truth table coherent with the functionality expected above (1 when there are an even amount of input at 1). This also shows that a triple XNOR is algebraically rendered as

$$ \text{XNOR-3} \buildrel def\over = \text{NOT} (\text{XOR-3}) = \overline{(x \oplus y \oplus z)} = \overline{(xyz + \overline{x}\overline{y}z + \overline{x}y\overline{z} + x\overline{y}\overline{z})} $$

by carefully expounding all terms, you eventually get to the even parity checker expression which is

$$ \text{XNOR-3} = \overline{x}\overline{y}\overline{z} + \overline{x}yz + xy\overline{z} + x\overline{y}z $$

It follows that: $$\text{XNOR-3} =(x \oplus y)\odot z \ne x \odot y \odot z$$ as mentioned in the beginning.