multiple xor (sum of parities)

Question

If we have:

$b_1 \oplus b_2 = b_1 (1 - b_2) + b_2 (1 - b_1)$

what is (or are, if there are different versions) the compact general formula for a multiple "summation":

$b_1 \oplus b_2 \oplus \dotsb \oplus b_n$

[PS. Possibly in terms of ordinary addition/multiplications, avoiding modulus]

Answer 1

I think this will work:-

$$b_1 \oplus b_2 \oplus \cdots \oplus b_n = \frac{1 - \prod_{i=1}^n (1 - 2b_i)} 2 $$

Answer 2

You can think of the sum as an OR, the multiplications as AND and $1-x$ as NOT $x$, that is you can read: $b_1 (1 - b_2) + (1 - b_1) b_2$ as "($b_1$ AND NOT $b_2$) OR (NOT $b_1$ AND $b_2$)".

For any fixed $n$, you can write a similar logical statement for $b_1 \oplus b_2 \oplus \cdots \oplus b_n$ and convert it back into a formula, but it will get messy quickly. For example, when $n=3$ it is: $b_1 b_2 b_3 + b_1 (1 - b_2) (1 - b_3) + (1 - b_1) b_2 (1 - b_3) + (1 - b_1) (1 - b_2) b_3$.

If you want you can write this as $$b_1 \oplus b_2 \oplus \cdots \oplus b_n = \sum_{(x_1, \ldots, x_n) \in X} \prod_{i=1}^n b_i x_i + (1 - b_i) (1 - x_i)$$ where `$X = \{ (x_1, \ldots, x_n) \in \{0, 1\}^n : \sum x_i \textrm{ is odd} \}$.