\documentclass[10pt]{article}
\usepackage{amsfonts}
\newtheorem{define}{Definition}
%\newcommand{\Z}{{\bf Z}}
\usepackage{psfig}
\oddsidemargin=0.15in
\evensidemargin=0.15in
\topmargin=-.5in
\textheight=9in
\textwidth=6.25in
\begin{document}
\input{preamble.tex}
\lecture{12}{October 20, 2004}{Madhu Sudan}{Anastasios Sidiropoulos}
%%%% body goes in here %%%%
\section{Overview}
This lecture is focused in comparisons of the following
properties/parameters of a code:
\begin{itemize}
\item
List decoding, vs distance.
\item
Distance, vs rate.
\item
List decoding, vs rate.
\end{itemize}
\section{The Plotkin's Bound}
Recall that for two binary strings $x,y\in \{0,1\}^n$, we denote by
$\Delta(x,y)$ the number of positions that $x$ and $y$ differ.
\begin{theorem}[Plotkin's Bound]\label{theorem:plotkin_bound}
If there exist codewords $c_1,c_2,\ldots,c_m\in \{0,1\}^n$, such that
for each $i,j$, with $i\leq j$, $\Delta(c_i,c_j)\geq n/2$, then $m\leq
2n$.
\end{theorem}
\begin{proof}
Assume that $m>2n$. We define vectors
$\tilde{c}_1,\ldots,\tilde{c}_m\in \{-1,1\}^n \subset \mathbb{R}^n$,
such that for each $i$, with $1\leq i\leq n$, $\tilde{c}_i$, and for
each $i$, with $1\leq j\leq n$, the $j$th coordinate of $\tilde{c}_i$
is $-1$, iff the $j$th bit of $c_i$ is $1$. Note that if
$\Delta(c_i,c_j)\geq n/2$, then this implies $\langle \tilde{c}_i,
\tilde{c}_j \rangle \leq 0$. Intuitively, this means that if two
codewords $c_i$, and $c_j$ have large Hamming distance, then the angle
between the corresponding vectors $\tilde{c}_i$, and $\tilde{c}_j$,
should be large.
Pick a random unit vector $x\in \mathbb{R}^n$. We have that w.h.p.,
$\langle x, \tilde{c}_i \rangle\neq 0$, for all $i$, with $1\leq i\leq
m$. Moreover, since there are $m$ codewords, either $x$, or $-x$ has
strictly positive inner product with at least $m/2$ of the
$\tilde{c}_i$s. We can assume w.l.o.g., that this holds for $x$.
Since $m>2n$, it follows that there exist $n+1$ vectors having
strictly positive inner product with $x$. W.l.o.g., assume that these
are the vectors $\tilde{c}_1,\ldots,\tilde{c}_{n+1}$.
Observe that a set of $n+1$ vectors in an $n$-dimensional space,
cannot be linear independent. Thus, we can assume that there exist
$\lambda_1,\ldots,\lambda_{n+1}\in \mathbb{R}$, with $\lambda_i>0$, for
each $i$, with $1\leq i\leq j$, and $\lambda_i\leq 0$, for each $i$,
with $j* 0$.
On the other hand,
\begin{eqnarray*}
\langle z, z \rangle & = & \left\langle \sum_{i=1}^{j}\lambda_i \tilde{c}_i, \sum_{i=j+1}^{n+1}\lambda_i \tilde{c}_i \right\rangle\\
& = & \sum_{i\leq j,i'>j}\lambda_i\lambda_{i'}\langle c_i,c_{i'}\rangle\\
& \leq & 0
\end{eqnarray*}
Thus, we obtain a contradiction.
\item{Case 2, $z=0$:}
We have
\begin{eqnarray*}
\sum_{i=1}^j \lambda_i\tilde{c}_i & = & 0,
\end{eqnarray*}
and thus
\begin{eqnarray*}
\langle z,x\rangle & = & \left\langle \sum_{i=1}^j \lambda_i \tilde{c}_i,x\right\rangle\\
& = & \sum_{i=1}^j \lambda_i \langle \tilde{c}_i, x\rangle\\
& > & 0
\end{eqnarray*}
The last inequality follows from the fact that $\lambda_i>0$, for
$1\leq i\leq j$, and that $\langle \tilde{c}_i,x\rangle > 0$. This
however is a contradiction, since $z=0$, which implies that $\langle
z,x\rangle = 0$.
\end{description}
\end{proof}
\section{The Johnson's Bound}
\begin{theorem}[Johnson's Bound]\label{theorem:johnson_bound}
For any $\epsilon$, with $0<\epsilon<1$, if $C$ is a
$\left[n,?,\left(\frac{q-1}{q}\right)(1-\epsilon)n\right]_q$-code,
then $C$ corrects less than
$\left(\frac{q-1}{q}\right)(1-\sqrt{\epsilon})n$ errors, with lists of
size $(q-1)n$.
\end{theorem}
We will give a proof of Theorem \ref{theorem:johnson_bound}, for the
special case of $q=2$.
\begin{proof}
We will prove the contrapositive. That is, we assume that there exist
$r,c_1,\ldots,c_m\in \{0,1\}^n$, such that for each $i$, with $1\leq i\leq m$,
\begin{eqnarray*}
\Delta(r, c_i) & \leq & \frac{1-\tau}{2}n,
\end{eqnarray*}
and for each $i\neq j$,
\begin{eqnarray*}
\Delta(c_i, c_j) & \geq & \frac{1-\epsilon}{2}n.
\end{eqnarray*}
Define vectors $\tilde{r}, \tilde{c}_1,\ldots,\tilde{c}_m\in
\{0,1\}^n\subset\mathbb{R}^n$, as in the proof of Theorem
\ref{theorem:plotkin_bound}.
We have that for each $i$, with $1\leq i\leq m$,
\begin{eqnarray*}
\langle \tilde{r}, \tilde{c}_i \rangle & \leq & \tau n,
\end{eqnarray*}
and for each $i\neq j$,
\begin{eqnarray*}
\langle \tilde{c}_i, \tilde{c}_j \rangle & \geq & \epsilon n.
\end{eqnarray*}
We want to show that is $\tau > \sqrt{e}$, then $m\leq n$.
We have that the projection of each $\tilde{c}_i$ into $r$ is
``large'', and that the angle between each pair of $\tilde{c}_i$,
$\tilde{c}_j$ is also ``large''. Intuitively, the main idea of the
proof is that these two properties cannot be satisfied simultaneously,
if the number of the vectors $\tilde{c}_i$ is too large. We will
verify this argument by considering the vectors $\tilde{c}_i-\alpha
r$, for carefully chosen $\alpha$, and show that the angle between
each pair of such vectors is at least $90^\circ$. Thus, we will
obtain a bound on the number of such vectors.
Formally, we have
\begin{eqnarray*}
\langle c_i-\alpha r, c_j-\alpha \rangle & = & \langle c_i,c_j\rangle - \alpha \langle c_i,r\rangle -\alpha \langle c_j,r\rangle+\alpha^2\langle r,r\rangle\\
& \leq & (\epsilon -2\alpha\tau +\alpha^2)n
\end{eqnarray*}
By setting $\alpha=\sqrt{\epsilon}$, we obtain that the inner product
between each pair of vectors $\tilde{c}_i-\alpha r$, and
$\tilde{c}_j-\alpha r$ is
\begin{eqnarray*}
2\sqrt{\epsilon}(\sqrt{\epsilon}-\tau)n
\end{eqnarray*}
Thus, for any $\tau<\sqrt{\epsilon}$, the inner product is negative,
and the assertion follows by applying the Plotkin's Bound.
\end{proof}
We note that for the case $q>2$, the proof of Theorem
\ref{theorem:johnson_bound} becomes more technical. More specifically,
one needs to map each bit of a codeword $c_i$, into more than one
coordinates of the corresponding vector $\tilde{c}_i$. For example,
if we have codewords in $\{0,1,2\}^n$, we can map each symbol of a
vector in $\mathbb{R}$, such that the angle between each vector is at
lest $90^\circ$.
\section{Relating $R$ with $\delta$}
\subsection{Improving the Singleton Bound}
\begin{lemma}\label{lemma:suffix}
If there exists a $(n,k,d)_2$-code, then there also exists a
$(2d,k+2d-n,d)_2$-code.
\end{lemma}
\begin{proof}
Let $C$ be a $(n,k,d)_2$-code. $C$ contains $2^k$ codewords, of
length $n$. Thus, if we project each codeword into the first $n-2d$
coordinates, there are at least $2^{k+2d-n}$ codewords, that are
mapped into the same string. Since all these $2^{k+2d-n}$ codewords
have the same prefix of length $n-2d$, and since their distance is at
least $d$, it follows that their pairwise distance in the last $2d$
bits should be at least $d$. Thus, the suffixes of these codewords
form a $(2d,k+2d-n,d)_2$ code.
\end{proof}
It follows by Lemma \ref{lemma:suffix} that for any $(n,k,d)_2$-code,
with $k+2d-n\leq \log{4d}$, we have
\begin{eqnarray*}
R+2\delta-1 & \leq & 0.
\end{eqnarray*}
\subsection{The Elias-Bassalygo Bound}
The main argument in the proof of the Hamming bound is that if we have
$k$ non-intersecting balls of radius $\frac{d-1}/2$, in $\{0,1\}^n$,
then the sum of their volumes cannot exceed $2^n$. We will show how to
extend this idea in the case of intersecting balls, by bounding the
overlap.
Assume that we have a binary code of distance $\frac{1-\epsilon}{2}$.
For each codeword $c\in \{0,1\}^n$, we consider the ball in
$\{0,1\}^n$ of radius $\frac{1-\sqrt{\epsilon}}{2}$ around $c$. We
have
\begin{eqnarray*}
2^k \mbox{Vol}\left(n, \frac{1-\sqrt{\epsilon}}{2}n\right) & \leq & n2^n,
\end{eqnarray*}
and thus
\begin{eqnarray*}
2^{Rn} 2^{H\left(\frac{1-\sqrt{\epsilon}}{2}\right)n} & \leq & 2^{n+o(n)}.
\end{eqnarray*}
This implies
\begin{eqnarray*}
R + H\left(\frac{1-\sqrt{\epsilon}}{2}\right) & \leq & 1
\end{eqnarray*}
So, if $\delta=\frac{1-\epsilon}{2}$, then $R+H\left(\frac{1}{2} -
\frac{1}{2}\sqrt{1-2\delta}\right) \leq 1$.
\subsection{The Case $\delta \rightarrow 0$}
An interesting question is what are the best possible codes, when
$\delta\rightarrow 0$.
The Hamming bound gives
\begin{eqnarray*}
R & \leq & 1-H\left(\frac{\delta}{2}\right)\\
& \approx & 1-\frac{1}{2}(1+o(1))\delta \log_2\frac{1}{\delta}.
\end{eqnarray*}
On the other hand, we know that there exist codes satisfying
\begin{eqnarray*}
R & \geq & 1-(1+o(1))\delta \log_2\frac{1}{\delta}.
\end{eqnarray*}
\subsection{The Case $\delta \rightarrow 1/2$}
Another interesting question is what is the best possible value for
$R$, in the case where $\delta=\frac{1-\epsilon}{2}$, with
$\epsilon\rightarrow 0$. The Plotkin bound gives $R\leq 2\epsilon$.
Also, the EB-bound gives $R=O(\epsilon)$.
On the positive side, we can show (even for the case of random codes),
that there exist codes with $R=\Omega(\epsilon^2)$.
We also note that the Linear-Programming bound gives
$R=\tilde{O}(\epsilon^2)$ (also known as MRRW-bound, or JPL-bound).
\section{Relating $R$ with $p$}
We would like to know what is the best possible values for $R$, and
$p$, such that for infinitely many $n$, we have $(n,Rn,?)_2$-codes,
that are $(pn,n)$-error-correcting.
The Shannon bound gives
\begin{eqnarray*}
R & \leq & 1-H_2(p)
\end{eqnarray*}
We will next prove that this bound is tight.
\begin{lemma}
There exist codes, satisfying $R\geq 1-H_2(p)$.
\end{lemma}
Before we state the proof, we note that the same result can be
obtained by using random codes in $\{0,1\}^n$, but the proof is rather
technical.
\begin{proof}
We will show that there exists a linear code of rate $R$, that is
$(pn,n+1)$-error-correcting. We begin with an empty basis for the
code, and we repeatedly increase the basis, by greedily adding one
base-vector at a time.
More specifically, assume that we have already added the vectors
$b_1,b_2,\ldots,b_k\in \{0,1\}^n$ in the basis. Let
$C_i=\mbox{span}(b_1,\ldots,b_i\}$. We pick $b_{i+1}$, so as to
minimize the value $\Phi_{i+1}$, where for each $i$, the value
$\Phi_i$ is given by the following potential function:
\begin{eqnarray*}
\Phi_i & = & \mathbf{E}\left[2^{|B(x,pn)\cap C_i}\right],
\end{eqnarray*}
where the expectation is taken over the random choices of $x$, when
$x$ is distributed uniformly in $\{0,1\}^n$.
We have
\begin{eqnarray*}
\mathbf{E}[\Phi_{i+1}] & \leq & \Phi_i^2
\end{eqnarray*}
Thus, we can conclude that there exist base vectors $b_1,\ldots,b_k$,
such that
\begin{eqnarray*}
\Phi_k & \leq & \Phi_0^{2^k}
\end{eqnarray*}
Note that
\begin{eqnarray*}
\Phi_0 & = & 1 + \frac{\mbox{Vol}(n,pn)}{2^n}
\end{eqnarray*}
To be continued in the next lecture \ldots
\end{proof}
\end{document}
*