problem
Let $a_4a_3a_2a_1a_0$ be a 5- bits binary numbers (each $a_i=0$ or $1$). Write an expression in Boolean algebra that evaluates to $1$ when odd number of bits of the number is $1$ and $0$ otherwise.
given example:
(1) $01101$ evaluates to $1$
(2) $00110$ evaluates to $0$
(3) $00010$ evaluates to $1$
My question: "when odd number of bits of the number is $1$"
This sentence was interpreted as "the number as decimal system is odd", but it is contrary to the example (3). I tried many other interpretations, but they didn't fit the examples given.
I don't know how to interpret it. Thank you for letting me know.
(I am a beginner in English.I'm sorry that it's not a math question, but it's related to English interpretation, but I'm leaving a question here because I think someone who's good at math can read the problem. )
This is an example of a parity function. More generally, the $n$-variable parity function is the Boolean function $f:\{0,1\}^n \to \{0,1\}$ defined by $f(x) = 1$ if and only if the number of ones in the vector $x \in \{0,1\}^n$ is odd. Using exclusive or, this function can be defined by $$ f(x) = x_1 \oplus x_2 \oplus \dotsm \oplus x_n $$ This expression can be converted into a Boolean expression by using the dictionary $x \oplus y = x\bar y + \bar x y$. Note that the size of a Boolean formula computing the parity function of $n$ variables is known to be at least $O(n^2)$.