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%{lecture number}{lecture date}{Ronitt Rubinfeld}{Your name}
\lecture{24}{May 3, 2007}{Ronitt Rubinfeld}{Ivaylo Riskov}
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\section{Recap}
$$f:\{0,1\}^n \rightarrow \{0,1\}$$
$x < y$ if $\forall i, x_i \le y_i$
$f$ is monotone if $\forall x \le y, f(x) \le f(y)$
We define a violating edge $(x,y)$ such that $y=x\oplus e_i, x < y, f(x) > f(y)$. We can use the following tester: Pick $O\left(\frac{n}{\epsilon}\right)$ edges and reject if any violate.
Finally, we started the proof of the following theorem:
\begin{theorem}
The above tester passes monotone functions and fails $\epsilon$-far from monotone functions (must chnage $f$ on at least $\epsilon$ fraction of the inputs) with probability at least $3/4$.
\end{theorem}
\section{Proof of Theorem}
\begin{claim}
If $f$ is $\epsilon$-far from monotone it has at least $\epsilon 2^n$ violating edges.
\end{claim}
\begin{proofof}{claim}
By contrapositive: assume $f$ has less than $\epsilon 2^n$ violating edges.
Let $f^{(i)}$ be $f$ after the $i$-th phase. That is, phase $i$ fixes edges in $i$-th direction without ruining lower direction edges.
Inductive assumption: $f^{(i - 1)}$ has no violating edges in directions $1\ldots i - 1$.
\begin{definition}
Consider the following definitions:
\begin{itemize}
\item $H_0 \leftarrow $ nodes $x$ such that $x_i = 0$
\item $H_1 \leftarrow $ nodes $x$ such that $x_i = 1$
\item $M_i \leftarrow $ violating edges in $i$-th direction.
\item $U_1 \leftarrow $ violating edges in $H_0$ having common endpoint with $M_i$
\item $U_2 \leftarrow $ other violating edges in $H_0$
\item $V_1, V_2 $ same as $U_1$ and $U_2$, but for $H_1$
\end{itemize}
\end{definition}
>From the above definition the set of violating edges in $f^{(i-1)}$ is
$M_i \cup U_1 \cup U_2 \cup V_1 \cup V_2$.
\emph{Notes}:
\begin{enumerate}
\item Edges in $i$-th direction, i.e. of the form $(x, x\oplus e_i)$, give perfect matching between $H_0$ and $H_1$.
\item Violating edges in $M_i$ have 1 in $H_0$ and 0 in $H_1$.
\item Need to change 1 to 0 in $H_0$ or 0 to 1 in $H_1$.
\end{enumerate}
If $|U_1| \le |V_1|$: modify $f^{(i - 1)}$ to get $f^{(i)}$ by setting $f^i(y) = 1, \forall (x, y)\in M_i$.
This fixes all edges in $M_i$ and $V_1$. All edges in $U_1, U_2$, and $V_2$ stay the same. However, this can create new violating edges.
We will show that $|V_1|$ is at most the number of new violating edges, so in $f^{(i)}$ we fixed all edges in $M_i$ and did not create more violating edges.
We also need to show that no violations occured in directions $1, \ldots, i - 1$. Note that only edges in $H_1$ can become violated.
\begin{claim}
At most $|U_1|$ violating edge are created. (and therefore at most $|V_1|$ violating edges, because in this case $|U_1| \le |V_1|$.
\end{claim}
\begin{proofof}{claim}
Let $(x, y)$ be a violating edge in $M_i$ and $(y,z)$ be a non-violating edge in $H_1$ that became violating.
$$ x = y \oplus e_i \quad z = y\oplus e_j = x \oplus e_i \oplus e_j$$
$f(x) = 1$ and $f(y)$ was 0, but is now 1. Consider the node $w= x\oplus e_j$ and so $z=w\oplus e_i$. However, $f(z)$ was not changed (we don't change nodes from 1 to 0) and so $(w,z)$ must not have been in $M_i$, i.e. $(w,z)\notin M_i$. Therefore, $f(w) = 0$ which means that $(x,w)$ is a violating edge. This gives a 1-to-1 mapping between new violating edges $(y,z)$ to old violating edges in $U_1$
\end{proofof}
The change of violating edges at $i$-th step is $-|M_i|-|V_1|+\text{\# new violations}$. From the above claim, the number of new violations is at most $|U_1| \le |V_1|$, so in total we got rid of at least $M_i$ violating edges. Therefore after all $n$ phases the total number of fixed violating edges is $|\bigcup M_i|$.
Note that no violations in previous directions are introduced. Suppose that $(z,y), z = y\oplus e_j, j < i$ is new violating in previous direction. Then $(x,w), w=x\oplus e_j, j < i$ was also violating, which contradicts the invariant for $f^{(i - 1)}$.
\end{proofof}
\section{Linearity testing}
We now turn our attention to the problem of linearity testing.
Consider a mapping $f: G \to H$.
\begin{definition}
$f$ is ``linear'' or a ``homomorphism'' if
$$ \forall x, y\quad f(x) + f(y) = f(x + y)$$.
\end{definition}
\begin{definition}
$f$ is $\epsilon$-linear if we can change it on at most $\epsilon$ fraction of the domain to make it linear.
\end{definition}
Let $G$ be a finite group aand
$$\forall a,y\in G\quad Pr_{x\in G} [y = a + x] = \frac{1}{|G|}$$
Then if pick $x\in_R G$, $a +x$ if distributed uniformly in $G$.
\subsection{Self-correcting}
Given is $f$ such that $\exists g, Pr_x[f(x) = g(x)] \ge 7/8$ and $g$ is linear. In other words, $g$ agrees with $f$ on at least $7/8|G|$ of the inputs).
The following algorithm computes $g(x)$:
\indent For $i = 1 \ldots c\log 1/\beta$ \\
\indent \indent Pick $y\in_R G$ \\
\indent \indent $answer_i \leftarrow f(y) + f(x - y)$ \\
\indent Output: most common answer
\begin{claim}
The probability that the above algorithm outputs $g(x)$ is at least $1-\beta$.
\end{claim}
\begin{proof}
$Pr[f(y) \neq g(y)] \le 1/8$ and so $Pr[f(x - y) \neq g(x - y)] \le 1/8$ by the observation that $x - y\in_R G$.
$$Pr[f(y) + f(x - y) \neq g(y) + g(x - y)] = Pr[f(y) + f(x - y) \neq g(x)] \le 1/4$$
By Chernoff bounds we repeat and take the majority.
\end{proof}
\subsection{Testing Linearity}
Consider the following algorithm:
\indent Do $O(?)$ times: \\
\indent \indent Pick random $x, y$ \\
\indent \indent If $f(x) + f(y) \neq f(x + y)$ fail and halt. \\
Example:
$$f(x) = \left\{
\begin{aligned}
1& \quad x\equiv 1 \text{ mod } 3 \\
0& \quad x\equiv 0 \text{ mod } 3 \\
-1& \quad x\equiv -1 \text{ mod } 3 \\
\end{aligned}\right.$$
Note that $f$ fails the test for $x\equiv y\equiv 1 \text{mod} 3$ or $x\equiv y \equiv 2 \mod 3$, otherwise it passes.
The probability of failure is then: $\delta_f = Pr[f(x) + f(y) \neq f(x + y)] = 2/9$. But $f$ is $2/3$-far from linear.
\end{document}