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$f(0,0,0)=0,\, f(0,0,1)=0,\, f(0,1,0)=1,\,
f(0,1,1)=1,\, f(1,0,0)=1,\, f(1,0,1)=1,\,
f(1,1,0)=0,\, f(1,1,1)=0 \,.$\\
The absolute norm form We want to derive the ANF of
$f$ is $f$, We know that it something like
$$f (x, y, z) = axyz + bxy + cxz + dzy + ex + f y + gz + h,$$
where %
for some $a, b, c, d, e, f, g, h \in
\Fb$, so by \Fb$.
By evaluating the ANF in $(0,0,0)$ we see that $f (0, 0, 0) = h$.
Since the function $f$ is defined so that By definition of $f$, $f (0, 0, 0) = 0$,
so we
can conclude that have identified $h$ as $h = 0$.\\
Evaluating the ANF in the other $3$-tuples of bits and comparing with the definition of $f$
we can find all the other
coefficients coefficients.
\end{Example}
\begin{Exercise}
Determine the ANF of the function in the previous example.\\
(hint: start with
evaluating $f$ at the $3$-tuples s.t. the bit $1$ appears only once, then
evaluate on pass to evaluating $f$ at the $3$-tuples where $1$ appears twice and
conclude finish with
$(1,1,1)$) and we obtain evaluating at $(1,1,1)$).\\
The result is $f (x, y, z) = x + y$
\end{Example} \end{Exercise}
The following function, often used in cryptography, is called the \emph{majority function}:
$$f: (\Fb)^3 \rightarrow \Fb $$
