A parity function $f_s:\{0,1\}^{n}\rightarrow\{0,1\}$, for some $s\in \{0,1\}^n$, is a function of the form $f_s(x) = x \cdot s$, where the inner product is taken modulo 2.

Show that $f_s$ is a balanced function for all $s$

We have $f_s(x) =\sum_i x_is_i \mod 2$. If $s \neq 0^n$, then there exist $i$ such that $s_i \neq 0$. So, for all $x$, $f_s(x) \neq f_s(x^i)$, where $x^i$ the string obtained from by inverting bit $i$. Hence $f_s$ is balanced.

I really don't understand the second from last sentence. Why this does this imply the function is balanced?


Let's say that the first bit of $s$ $s_0=1$ (the argument will be exactly the same for any bit, just for convenience).

You can split the space of inputs $x \in \{0,1\}^n$ in two halves: one half where $x_0 = 0$ and the other half where $x_0 = 1$. For each bitstring $x$ from the first half you'll have a bitstring $\tilde{x}$ from the second half which will be equal to $x$ in all bits except the first one, and will differ from $x$ in the first bit.

Now consider $f_s(x)$ and $f_s(\tilde{x})$; you'll have $f_s(x) = x_0s_0 + F$ and $f_s(\tilde{x}) = \tilde{x}_0s_0 + F$ ($F$ is the sum of all terms except the first one). You know that $x_0 = 0$ and $\tilde{x}_0 = 1$, so you'll have $F = f_s(x)$ and $f_s(\tilde{x}) = 1 + F$, so you're guaranteed that $f_s(x) \neq f_s(\tilde{x})$.

This way you can split all possible $x$ in $2^{n-1}$ pairs, and each of the pairs will have different values of $f_s$ - which means that exactly $2^{n-1}$ values are 0 and $2^{n-1}$ are 1, and the function is balanced.

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  • $\begingroup$ Genius! Thanks! $\endgroup$ – Permian Jun 16 at 21:03

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