# Odd Parity Function [closed]

I am trying to define a Odd Parity Function that takes three 1 bit inputs and will output a 1 if the 3 bits are odd as a Boolean function.

1 1 0 = 0
1 0 0 = 1
0 0 0 = 0
1 1 1 = 1


I understand this has a relationship to XOR as I can define this with 2 parameter as

X xor Y = (XY')+(X'Y)


My assumption is the function will look like this

(X xor Y) xor Z = (((XY')+(X'Y))Z')+(((XY')+(X'Y))'Z)


Can this function be simplifed?

## closed as unclear what you're asking by David Richerby, Rick Decker, Yuval Filmus, D.W.♦, vonbrandApr 23 '14 at 16:48

Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.

Your question is addressed (for the parity of $n$ bits) by Troy Lee, who shows in his paper The formula size of PARITY that the (optimal) formula size (number of literals) of parity on $n = 2^\ell + k$ bits (where $0 \leq k < 2^\ell$) is $2^\ell (2^\ell + 3k)$. In your particular case, $\ell = k = 1$ and so the formula size is $10$, matching your formula, and showing that it is tight under this complexity measure.