\documentclass[10pt]{article} \newtheorem{define}{Definition} %\newcommand{\Z}{{\mathbb{Z}}} %\usepackage{psfig} \usepackage{amsfonts} \oddsidemargin=0.15in \evensidemargin=0.15in \topmargin=-.5in \textheight=9in \textwidth=6.25in \newcommand{\pmone}{\{-1,1\}} \newcommand{\dist}{\mathop{\rm dist}} \begin{document} \input{preamble.tex} %{lecture number}{lecture date}{Ronitt Rubinfeld}{Your name} \lecture{3}{February 13, 2008}{Ronitt Rubinfeld}{Megumi Ando} %%%% body goes in here %%%% % \section*{Announcements} % \begin{itemize} % \item HW1 posted % \item Sign up to scribe % \end{itemize} \section*{Last Time} In the previous lecture, we made a notational switch from using Boolean functions of the form $f:\{0,1\}^n \rightarrow \{0,1\}$ to functions of the form $f:\pmone^n \rightarrow \pmone$. We defined linearity over this form, and what it meant to be $\epsilon$-close to linear. \begin{definition}$f: \pmone^n \rightarrow \pmone$ is \emph{linear} if \thinspace$\forall x, y \in \pmone^n$, $f(x) f(y) f(x \cdot y) = 1$, \\ where $x\cdot y = (x_1... x_n) \cdot (y_1 ... y_n) = (x_1y_1, x_2y_2, ... , x_ny_n)$. \end{definition} There are $2^n$ linear functions over $\pmone^n \rightarrow \pmone$. Each of them can be written as $\chi_S(x) = \prod_{i\in S} x_i$, where $S \subseteq \{1, ..., n\}$. \begin{definition} A function $f$ is \emph{$\epsilon$-close to linear} if \thinspace$\exists$ linear $g$ such that $\mbox{Pr}_x[f(x)\not=g(x)] \leq \epsilon$. Otherwise, $f$ is \emph{$\epsilon$-far}. \end{definition} Finally, we proposed the following linearity tester: \begin{itemize} \item Repeat $O\left(\frac{1}{\rho} \log \frac{1}{\beta}\right)$ times: \begin{itemize} \item Pick $x, y \in \pmone^n$. \item If $f(x)f(y)f(x\cdot y) \not= 1$, output ``FAIL" and halt. \end{itemize} \item Output ``PASS." \end{itemize} The rejection probability of one pass through the loop is $\delta \equiv E_{x,y} \left[ \frac{1-f(x)f(y)f(x\cdot y)}{2} \right]$. \section{Fourier Analysis (Basics)} We will use a few times the following simple fact about linear functions. \begin{fact}\label{fact0} $$\chi_S(x) \cdot \chi_T(x) = \prod_{i\in S} x_i \cdot \prod_{j\in R} x_j = \prod_{i\in S\triangle T} x_i,$$ where $S\triangle T$ is the symmetric difference of $S$ and $T$, i.e., the set of elements that appear in exactly one of the sets $S$ and $T$. \end{fact} % \subsection{Definitions} % % % % For two Boolean functions $f,g: \pmone^n \to \pmone$, we define their \emph{inner product} as % % $$\langle f,g \rangle \equiv \frac{1}{2^n}\sum_{x\in\pmone^n} f(x)g(x).$$ % % Furthermore, for any function % $f: \pmone^n \to \pmone$ and any subset $S\subset \{1,\ldots,n\}$, we define $\hat{f}(S)$ as $$\hat f(S) \equiv % %\langle f,\chi_S \rangle = % \frac{1}{2^n}\sum_{x\in\pmone^n}f(x)\chi_S(x).$$ % \subsection{Some Facts} % % The following Fourier analysis facts were presented in this lecture. % % % % % % \begin{fact}\label{fact4} % $E[f] = \hat{f} (\emptyset)$ % \end{fact} \subsection{Vector Space of Functions $g:\pmone^n \rightarrow \mathbb R$} The set $G = \{g | g:\pmone^n \rightarrow \mathbb R\}$ is a vector space of dim $2^n$. \begin{definition} \emph{Indicator functions} are functions of the form: If $x=a$, then $e_a(x) = 1$. Otherwise, $e_a(x) = 0$. \end{definition} Note that the indicator functions are basis functions of $G$. However, we will not be using them. Instead we will be using the parity functions, $\{\chi_S\}_{S\subseteq [n]}$, described in the previous lecture. \begin{definition} For $f,g : \pmone \rightarrow \pmone$, the ``inner product" $$\langle f,g\rangle \mbox = \frac{1}{2^n} \sum_{x\in \pmone^n} f(x)g(x),$$ \end{definition} where the sum $\sum f(x)g(x)$ is the ``correlation," a measure of how often $f$ and $g$ agree. \\ \noindent Note that: \begin{enumerate} \item $<\chi_S, \chi_S> = \frac{1}{2^n} \sum_{x\in\pmone^n} \chi_S^2 (x) = 1.$ (Absolute correlation.) \item If $S\not= T$, \end{enumerate} \begin{eqnarray*} <\chi_S, \chi_T> &=& \frac{1}{2^n} \sum_{x\in\pmone^n} \chi_S (x) \chi_T (x) \\ &=& \frac{1}{2^n} \sum_{x\in\pmone^n} \prod_{i\in S} x_i \prod_{j\in T}x_j \mbox{\qquad(by definition)} \\ &=& \frac{1}{2^n} \sum \prod_{i \in S\triangle T} x_i \mbox{\qquad(by Fact \ref{fact0})} \end{eqnarray*} where $S\triangle T$ is non-empty. Therefore, there exists a $j \in S\triangle T$. Let $x^{\oplus j}$ equals $x$ with the $j$-th bit flipped. \begin{eqnarray*} &=& \frac{1}{2^n} \sum_{\mbox{pairs } (x, x^{\oplus j})} \left( x_j \prod_{i \in S\triangle T \setminus\{j\}} x_i + \overline{x_j} \prod_{i \in S\triangle T \setminus\{j\}} x_i \right) \\ &=& 0 \end{eqnarray*} From Notes 1 and 2 above, we see that every parity function $X_S$ is normal to the others, and thus, the parity functions form an orthonormal basis. \bigskip % \subsection{Digression: Matrix Multiplication Tester (Freivald's Algorithm)} % For one of the non-handin problems, we were asked to figure out a randomized matrix multiplication tester that runs in $O(n^2)$ time. If matrix $C$ is equivalent to the product of matrices $A$ and $B$, then the tester should always output ``$A\cdot B \equiv C$." Otherwise, the tester should correctly output ``$A\cdot B \not\equiv C$" at least three-fourths of the time. \\ % % \noindent The answer is Freivald's Algorithm. \\ % On input $$, where $A, B, C$ are $n$-by-$n$ matrices in $Z_2$: % \begin{enumerate} % \item Let $r \in_R \{0,1\}^n$. % \item Compute $r' \leftarrow A\times (B\times r) - C\times r \mbox{ (mod }2)$. % \item If $r '= \vec{0}$, repeat Steps 1-2. Else, output ``$A\times B \not\equiv C$.'' % \item If $r '= \vec{0}$ again, output ``$A\times B \equiv C$.'' Else, output ``$A\times B \not\equiv C$.'' % \end{enumerate} % % Running Time. Instead of multiplying $A$ and $B$ first, we multiply $B$ and $r$ first. So the total running time is $O(n^2)$ rather than $O(n^3)$. \\ % % \noindent Correctness: \\ % Case 1: $A\times B \equiv C$. When $A\times B \equiv C$, $[A\times (B\times r)] - (C\times r)$ is equivalent to zero because of matrix associativity. Thus, the algorithm always correctly outputs ``$A\times B \equiv C$." \\ % % \noindent Case 2: $A\times B \not\equiv C$. % \begin{eqnarray} % && \mbox{Let } D = (A\times B) - C \mbox{ (mod $2$)} \nonumber\\ % && \mbox{Let } D_{i,j} \mbox{ be the element of $D$ in the $i$-th row, $j$-th column.} \nonumber\\ % && A\times B \not\equiv C \nonumber \nonumber\\ % &\longrightarrow& D \not= \matrix{0} \nonumber\\ % &\longrightarrow& \exists D_{i,j}\not= 0 \nonumber % \end{eqnarray} % Let $r_j$ be the $j$-th element of $r$. If the $r_j$ is one, then $a_i$ will not equal $c_i$, and $a$ will not equal $c$. However, if $r_j$ is zero, then $a$ could still be $c$ (if no other column in $D$ contains a non-zero element). Therefore, the probability that $a$ will not equal $b$ is bounded below by $\frac{1}{2}$, the probability that randomly picked $r_j$ is one. \\ % % \noindent Since the algorithm constructs two independent $r$'s, the probability of correctly outputting ``$A\times B\not\equiv C$'' is bounded below by $\frac{3}{4}$. \subsection{Fourier Coefficients} The following corollary follows. \begin{corollary} $$\forall f \mbox{, } f(x) = \sum_{S \subseteq [n]} \hat{f}(S) \chi_S(x),$$ where $\hat{f}(z)$ is the Fourier coefficient, which can be calculated as follows: \begin{eqnarray*} \hat{f}(S) &=& \langle f, \chi_S \rangle \\ &=& \frac{1}{2^n} \sum_{x\in \{0,1\}^n} f(x) \chi_S(x) \end{eqnarray*} \end{corollary} In particular, a parity function has all but one coefficients equal zero. \begin{fact}[Fourier Coefficients of Parity Functions $\chi_T$]\label{fact1} \begin{eqnarray*} f = \chi_T &\Longleftrightarrow& \hat{f}(T) = 1. \\ && \mbox{Furthermore, } \forall S\not=T, \hat{f}(S) = 0. \end{eqnarray*} \end{fact} \noindent \textbf{A few more examples of Fourier coefficients:} \\ \\ \begin{tabular}{| l | l |} \hline \begin{textbf}{Function}\end{textbf} & \begin{textbf}{Fourier Representation}\end{textbf} \\ \hline \hline $f(x) = 1$ & $1 \cdot \chi_\emptyset$ \\ \hline $f(x) = x_i$ & $1 \cdot \chi_{\{i\}}$ \\ \hline and$(x_1, x_2)$ & $\frac{1}{2}\chi_{\emptyset} + \frac{1}{2}\chi_{\{1\}} + \frac{1}{2}\chi_{\{2\}} - \frac{1}{2}\chi_{\{1,2\}}$ \\ \hline maj$(x_1, x_2, x_3)$ & $\frac{1}{2}\chi_{\{1\}} + \frac{1}{2}\chi_{\{2\}} + \frac{1}{2}\chi_{\{3\}} - \frac{1}{2}\chi_{\{1,2,3\}}$ \\ \hline \end{tabular} \bigskip \subsection{Fourier Coefficients and Distance to Linearity} Let $\dist(f,g)$ denote the fraction of inputs on which two Boolean functions $f,g : \pmone^n \to \pmone$ disagree. That is, $\dist(f,g) = \Pr_{x \in \pmone^n}[f(x) \ne g(x)]$. For instance, the distance between two different parity functions is $1/2$. \begin{fact}\label{fact3} For $S \not=T$, $\dist(\chi_S, \chi_T) = \frac{1}{2}$. \\ \end{fact} It turns out that Fourier coefficients can be used to express the distance of a function to a given linear function. \begin{fact}[Agreement of $f$ with Linear Functions]\label{fact2} For $f: \pmone^n \rightarrow \pmone$, $$\hat{f}(S) = 1 - 2 \dist(f_1, \chi_S).$$ \end{fact} \begin{proof} \begin{eqnarray*} 2^n \hat{f}(s) &=& \sum_x f(x)\chi_S(x) \\ &=& \sum_{\mbox{$x$ s.t. $f(x) = \chi_S(x)$}} f(x) \chi_S(x) + \sum_{\mbox{$x$ s.t. $f(x) \not= \chi_S(x)$}} f(x) \chi_S(x) \\ &=& 2^n - 2|\{x| f(x) \not= \chi_S(x)\}| \\ &=& 2^n \left(1 - 2\frac{|\{x| f(x) \not= \chi_S(x)\}|}{2^n} \right) \\ \hat{f}(s) &=&1 - 2 \dist(f_1\cdot \chi_S) \end{eqnarray*} \end{proof} \subsection{Plancherel's Theorem} The following simple theorem holds. \begin{theorem}[Plancherel's Theorem] For $f,g: \pmone \rightarrow \mathbb R$, $$\langle f,g \rangle = E_x[f(x)\cdot g(x)] = \sum_{S\subseteq [n]} \hat{f}(S) \hat{g}(S).$$ \end{theorem} \begin{proof} \begin{eqnarray*} \langle f,g \rangle &=& \left\langle\sum_S \hat{f} (S) \chi_S (x), \sum_T \hat{g}(T) \chi_T (x)\right\rangle \\ &=& \sum_S \sum_T \hat{f} (S) \hat{g}(T) \left\langle\chi_S(x), \chi_T(x) \right\rangle \\ &=& \sum_{S=T} \hat{f}(S) \hat{g}(T) \cdot 1 = \sum_{S} \hat{f}(S) \hat{g}(S) \end{eqnarray*} \end{proof} The theorem yields multiple useful properties. \begin{corollary}[Parseval's identity] For $f: \pmone^n \rightarrow R$, $\langle f,f\rangle = \sum \hat{f}^2 (S)$. \end{corollary} \begin{corollary} For $f: \pmone^n \rightarrow \pmone$, $\sum \hat{f}^2 (S) = \langle f,f\rangle =1$. \end{corollary} \begin{corollary} \begin{eqnarray*} E_x[\chi_S(x)] = \left \{ \begin{array}{ll} 1 & \mbox{ if $S = \emptyset$, } \\ 0, & \mbox{ otherwise. } \end{array} \right. \end{eqnarray*} \end{corollary} \section{Analysis of the Proposed Linearity Tester} Recall that $\delta$ is the probability that a single pass through the loop detects that the input function $f$ is not linear, and it can be expressed as $$\delta = E_{x,y} \left[ \frac{1-f(x)f(y)f(x\cdot y)}{2} \right].$$ \begin{lemma}[Main Lemma] $1-\delta = \frac{1}{2} + \frac{1}{2} \sum_{S\subseteq [n]} \hat{f}^3 (S)$ \\ \end{lemma} \begin{proof} \begin{eqnarray*} 1 - \delta &=& E_{x,y} \left[ \frac{1 + f(x)f(y)f(xy)}{2} \right] \\ &=& \frac{1}{2} + \frac{1}{2} E_{x,y} [f(x)f(y)f(xy)] \\ \\ E_{x,y} [f(x)f(y)f(xy)] &=& E_{x,y}[(\sum_S \hat{f} (S) \chi_S (x)) (\sum_T \hat{f}(T) \chi_T (y))(\sum_U \hat{f}(U) \chi_U(x\cdot y)] \\ &=& \sum_{S,T,U} \hat{f}(S) \hat{f}(T) \hat{f}(U) E_{x,y}[\chi_S(x) \chi_T(y) \chi_U (x\cdot y)] \\ \\ E_{x,y}[\chi_S(x) \chi_T(y) \chi_U (x\cdot y)] &=& E_{x,y}[\prod_{i \in S} x_i \prod_{j\in T} y_j \prod_{k \in U} x_ky_k] \\ &=& E_{x,y}[\prod_{i \in S\triangle U} x_i \prod_{j \in T\triangle U} y_j] \\ &=& E_x[\chi_{S\triangle U} (x)] E_y[\chi_{T\triangle U} (y)] \\ E_x[\chi_{S\triangle U} (x)] &=& \left \{ \begin{array}{ll} 1 & \mbox{ if $S = U$,} \\ 0, & \mbox{otherwise } \end{array} \right. \\ E_y[\chi_{T\triangle U} (y)] &=& \left \{ \begin{array}{ll} 1 & \mbox{ if $T = U$,} \\ 0, & \mbox{ otherwise } \end{array} \right. \end{eqnarray*} So, $E_{x,y}[\chi_S(x) \chi_T(y) \chi_U (x\cdot y)]$ is non-zero if and only if $S=T=U$. If $S=T=U$, then the expectation is 1. Hence, $$E_{x,y} [f(x)f(y)f(xy)] = \sum_{S,T,U} \hat{f}(S) \hat{f}(T) \hat{f}(U) E_{x,y}[\chi_S(x) \chi_T(y) \chi_U (x\cdot y)] = \sum_S \hat{f}^3 (S)$$ and $$1-\delta = \frac{1}{2} + \frac{1}{2} E_{x,y} [f(x)f(y)f(xy)] = \frac{1}{2} + \frac{1}{2}\sum_S \hat{f}^3 (S).$$ \end{proof} \begin{theorem} If $f $ is $\epsilon$-far from linear, then $\delta=\Pr_{x,y}[f(x)f(y)f(x\cdot y) \not= 1] \geq \epsilon$. \end{theorem} \begin{proof}\\ We will prove Theorem 10 by proving its contrapositive; we will assume that $\delta < \epsilon$, and demonstrate that this assumption implies that $f$ is $\epsilon$-close. \\ The Main Lemma implies that \begin{eqnarray*} 1 - \delta &\leq& \frac{1}{2} + \frac{1}{2} \sum_S \hat{f}^3 (S) \\ 1 - 2\delta &\leq& \sum_S \hat{f}^3 (S) \\ &\leq& \left( \max_S \hat{f}(S) \right) \sum_S \hat{f}^2 (S) = \max_S \hat{f}(S), \end{eqnarray*} Let $T = \mathop{\arg\max}\limits_S \hat{f}(S)$. We have $$1 - 2\delta \le \hat{f}(T),$$ and by Fact \ref{fact2}, \begin{eqnarray*} \dist(f,\chi_T) = \frac{1}{2} - \frac{1}{2}\hat{f}(T) < \frac{1}{2} - \frac{1}{2}(1 - 2\delta) = \delta < \epsilon. \end{eqnarray*} Therefore, $f$ is $\epsilon$-close to a linear function; an impossibility. \end{proof} \end{document}