### made suggested changes in lecture 2 notes.

parent be10f1cc
 ... ... @@ -133,10 +133,11 @@ We will define the function $\gt(x,y)$ to be \begin{marginfigure} \centering \includegraphics{figs/ge_matrix_2} \caption{Example matrix for $\gt(x,y)$ when $|x| = |y| = 5$ with fooling set elements of size 2 drawn.} \end{marginfigure} \sunote{Better to have just the figure as a PDF and use the caption for the text. This allows you to use the \gt command for the function name, to get a consistent look.} %\sunote{Better to have just the figure as a PDF and use the caption for the % text. This allows you to use the \gt command for the function name, to get a % consistent look.} According to the definition of $\gt(x,y)$ the lower triangle ... ... @@ -159,26 +160,28 @@ follows directly. % recording ~20:20 \paragraph{Example: \disjoint} \sunote{Didn't we define \disjoint in the last lecture notes?} We define $x$ and $y$ to be a vector of bits which define set membership. \begin{align*} & x,y \in [0,1]^n \\ \text{and } & x_i \Leftrightarrow i \in S_x \\ & y_i \Leftrightarrow i \in S_y \end{align*} the function $\disjoint(x,y)$ is defined as follows \begin{equation*} \disjoint(x,y) = \left\{ \begin{array}{l l} 1 & \quad \text{if $(x,y)$ are disjoint: } \left\{ \forall i, x_i \land y_i = 0 \right\} \\ 0 & \quad \text{otherwise} \end{array} \right. \end{equation*} %\sunote{Didn't we define \disjoint in the last lecture notes?} %\ronote{In response, we did define \disjoint, but in in lecture 1 % we did not reason about the definition with as much detail.} % %We define $x$ and $y$ to be a vector of bits %which define set membership. % %\begin{align*} % & x,y \in [0,1]^n \\ %\text{and } & x_i \Leftrightarrow i \in S_x \\ % & y_i \Leftrightarrow i \in S_y %\end{align*} % %the function $\disjoint(x,y)$ is defined as follows % %\begin{equation*} %\disjoint(x,y) = \left\{ % \begin{array}{l l} % 1 & \quad \text{if $(x,y)$ are disjoint: } \left\{ \forall i, x_i \land y_i = 0 \right\} \\ % 0 & \quad \text{otherwise} % \end{array} \right. %\end{equation*} \begin{marginfigure} ... ... @@ -191,9 +194,9 @@ $1$ & $0$ & $1$ & $1$ & $0$ & $0$ \\ $1$ & $1$ & $1$ & $0$ & $0$ & $0$ \\ \end{tabular} \\ \vspace{1em} Matrix for $DISJ(x,y)$ when $n=4$ showing which \caption{Matrix for $DISJ(x,y)$ when $n=4$ showing which combinations will result in a $1$ or a $0$. a $0$.} \end{marginfigure} Let us observe one of the items in the ... ... @@ -201,7 +204,7 @@ fooling set using our example. Using the pair $(x_1, y_1) = (01, 10)$ and $(x_2, y_2) = (10, 01)$ we can see this is part of the fooling set. This set is interesting because $(x_1,y_1) = (\overline{x_2}, \overline{y_2})$. In fact, the fooling st for $DISJ(x,y)$ are the bit patterns with the bits flipped; In fact, the fooling st for $\disjoint(x,y)$ are the bit patterns with the bits flipped; it is the set of all $A$ and $\overline{A}$. % recording ~33:20 ... ... @@ -348,7 +351,31 @@ rest we do not care about. We are choosing some subset of the $r$ bits to be fixed to $1$, the rest we can pick to be anything else. Doing this results in $2^n - 2^{n-r}$ \sunote{See the note that I provided in the discussion group and insert it here} % \sunote{See the note that I provided in the discussion group and insert it here} The statement that the rank of the IP matrix over GF($2$)$=2^n$ is not true. A proof of this can be found in Sherstov's notes\footnote{http://www.cs.ucla.edu/~sherstov/teaching2012-winter/docs/lecture03.pdf}, which we will briefly discuss here. The first observation is that the rank argument works over any field. So we can choose the field we wish to evaluate matrix rank over. In particular, the rank over the reals is larger than the rank over any other field. Recall that we would like to show that the inner product has deterministic communication complexity $n$ via the rank method. So we need to show that the rank of the induced matrix (where $M_{ij}=<\!\!x_i,y_j\!\!>$) is large: remember that IP is a function from $n$ bit string to $\{0, 1\}$. Firstly, consider the auxiliary matrix $M'=2M-1$ (where $1$ is the all ons matrix). All this does is change $0$ to $-1$ and $1$ to $1$, which is more convenient to work with. It is a nonsingular transformation, so $\text{rank}(M) \ge \text{rank}(M')-1$. Now it is easy to see that $M'_{\text{IP}}$ on one-bit strings is the Hadamard matrix\footnote{$\begin{bmatrix} 1 & 1 \\ 1 & -1 \end{bmatrix}$} and the IP matrix for $n$ bit strings is merely the $n$-th order version of this (or the $n$th tensor power of this matrix). But this matrix has full rank: in particular its rank is $2^n$. Therefor the rank of the IP matrix is at least $2^n-1$. %recording ~1:00:00 \section{Randomized Algorithms} ... ...
 \usepackage{amsmath} ... ... @@ -233,9 +233,7 @@ h Example matrix for $gt(x,y)$ when \\$|x| = |y| = 5$ with fooling set elements of size 2 drawn. \\ \\  \begin{bmatrix} 1 & 0 & 0 & 0 & 0 \\ 1 & 1 & 0 & 0 & 0 \\ ... ...
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