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\author{Thibaut Horel}
\title{CS 225 Problem Set 4 -- Solutions}

\begin{document}
\maketitle

\section*{Problem 5.1} I use a simple counting argument. If $\mathcal{C}$ is
$(\delta, L)$ list-decodable, then for all $r\in \Sigma^{\hat{n}}$ we have
$|B(r,\delta)\cap \mathcal{C}|\leq L$. Hence:
\begin{displaymath}
    \sum_{r\in \Sigma^{\hat{n}}} |B(r,\delta)\cap\mathcal{C}|\leq Lq^{\hat{n}}
\end{displaymath}
Furthermore, each code-word $c$ of $\mathcal{C}$ is counted several times in
the sum: once for each $r\in B(c,\delta)$. But $B(c,\delta)$ is of size
$q^{\hat{n}H_q(\delta, \hat{n})}$, so each $c\in\mathcal{C}$ appears
$q^{\hat{n}H_q(\delta, \hat{n})}$ times in the sum:
\begin{displaymath}
    |\mathcal{C}|q^{\hat{n}H_q(\delta, \hat{n})}=
    \sum_{r\in \Sigma^{\hat{n}}} |B(r,\delta)\cap\mathcal{C}|
\end{displaymath}
Putting the previous two equations together:
$ |\mathcal{C}|q^{\hat{n}H_q(\delta, \hat{n})} \leq Lq^{\hat{n}}$. Taking the
$\log$ both sides and reordering the terms (and using $\rho = \frac{\log
|\mathcal{C}|}{\hat{n}\log q}$) concludes the proof.

\section*{Problem 5.2}

\paragraph{1.} For two different messages $m_1$ and $m_2$, $\Enc_1(m_1)$ and
$\Enc_1(m_2)$ differ on at least $\delta_1 n_1$ letters by definition of the
minimum distance of $\Enc_1$. For each one of those $\delta_1 n_1$ positions
where the letters differ, the images by $\Enc_2$ of the letters at this
position differ on at least $\delta_2 n_2$ letters by definition of the
minimum distance of $\Enc_2$. By summing over all such positions, we obtain
$\delta_1n_1\delta_2n_2$ positions where $\Enc(m_1)$ and $\Enc(m_2)$ differ.
This proves that the minimum distance of $\Enc$ is at least $\delta_1\delta_2$.

\paragraph{2.} \emph{(Thanks Thomas for helping me late at night two days after
the homework was due!)} Let us consider a received word $r$, and a message $m$
such that $d(\Enc(m), r)\leq (1-\eps_1l_2)\delta_2$. We can view $r$ as the
concatenation of $n_1$ blocks, each of size $n_2$. Denoting by $r_i$ the $i$th
block, I first note that the fraction of $i$s such that $d(r_i,
\Enc_2(\Enc_1(m)_i)$ is at most $(1-\eps_1l_2)$. Otherwise, by summing over $i$
we would have $d(Enc(m), r)> (1-\eps_1l_2)\delta_2$, a contradiction.

Hence, by $l_2$ list-decoding each of the blocks $r_i$, we obtain $n_1$ lists
of $l_2$ characters from $\Sigma_1$, such the $i$s such that the $i$th list
does not contain $\Enc_1(m)_i$ are a fraction at most $(1-\eps_1l_2)$.

I now construct $l_2$ words $c_1,\dots, c_{l_2}$ such that $c_j$ is obtained by
taking the $j$th character in each of the $n_1$ lists of length $l_2$ obtained
by $l_2$ list-decoding in the previous paragraph.

I claim that one of those words $c_j$ is such that $d(c_j, Enc_1(m))\leq
1-\eps_1$. Otherwise, by summing over $j$, and denoting by $f$ the fraction of
$i$s such that the $i$th list does not contain $\Enc_1(m)_i$:
\begin{align*}
    (1-\eps_1)l_2&< \sum_{j=1}^{l_2} d(c_j, Enc_1(m))
    = \frac{1}{n_1}\sum_{i=1}^{n_1}\sum_{j=1}^{l_2} \left[(c_j)_i\neq
    Enc_1(m)_i\right]\\
    &= fl_2 + (1-f)(l_2-1) = l_2 - 1 + f\leq l_2-1+(1-\eps_1l_2)
    = (1-\eps_1)l_2
\end{align*}
where the second equality was obtained by splitting the sum over $i$ depending
on whether or not $\Enc_1(m)_i$ belongs to the $i$th list, and the last
inequality was obtained by using the bound on $f$ found in the first paragraph
of this answer. Looking at the two extremal terms of the above computation, we
see a contradiction.

Hence, one of the $c_j$ is such that $d(c_j, Enc_1(m))\leq
1-\eps_1$. By $l_1$-decoding all the $c_j$s, we obtain $l_2l_1$ words in
$\{1,\dots, N\}$ such that one of them is $m$ (since $\Enc_1$ is $l_2$
list-decodable at distance $1-\eps_1$). This proves that $\Enc$ is $l_1l_2$
list-decodable at distance $(1-\eps_1\l_2)\delta_2$.

\paragraph{3.} We consider:
\begin{itemize}
    \item $\Enc_1$: the Reed-Solomon code consisting of all polynomials of
        degree $d = \frac{n}{m}-1$ over $\mathbb{F}_{2^m}$.
    \item $\Enc_2$: the Hadamard code of all linear functions from
        $\mathbb{F}_2^m$ to $\mathbb{F}_2$.
\end{itemize}
Since $\mathbb{F}_2^m$ can be identified with $\mathbb{F}_{2^m}$ these two
codes are compatible for concatenation. Both codes have block length $2^m$ so
the resulting block length after concatenation is $2^{2m}$. The bit length of
the code is $(d+1)m = n$ as requested.

If we choose $m$ large enough such that $\frac{n}{m2^m}\leq \eps$, then the
minimum distance of the RS code is $1-\frac{n/m-1}{2^m}\geq
1-\frac{n/m}{2^m}\geq 1-\eps$. Hence, by \textbf{1.}, the minimum distance of
the concatenated code is at least $\frac{1}{2}(1-\eps)$ as requested.
Finally, if $\frac{n}{m2^m}\leq \eps$, the block length of the concatenated
code, $2^{2m}$, is $O(\frac{n^2}{\eps^2})$ as requested.

Since $m = O(\log \frac{n}{\eps})$ and $m(d+1) = n$, it is clear that the code
is fully explicit.

\paragraph{} To get list-decoding, we need a different setting of $m$. We know
that the Hadamard code can be $\frac{1}{2}-\frac{\eps}{2}$ list-decoded for any
$\eps$ with list length $l_2 = \Theta(\frac{1}{\eps^2})$. If we set $m$ such
that:
\begin{displaymath}
2l_2\sqrt{\frac{n}{m2^m}} = \eps
\end{displaymath}
i.e $\frac{n}{m2^m} = \Theta(\eps^6)$,  then applying \textbf{2.} the
concatenated code will have distance
$(1-\eps)(\frac{1}{2}-\frac{\eps}{2}) \geq (\frac{1}{2}
-\eps)$.  The resulting list length will be $\sqrt{\frac{m2^m}{d}}l_2
=\Theta(\frac{1}{\eps^5})$.

The inner Hadamard code as $2^m = \Theta(\frac{n}{\eps^6})$ code-words, so the
list-decoding can be done in polynomial time by brute force testing all
code-words. Finally, since $m = \Theta(\log \frac{n}{\eps^6})$, The outer
RS code can be list-decoded in polynomial time using the algorithm seen in
class. Note that by \textbf{2.} we only need to apply this algorithm $l_2
= \Theta(\frac{1}{\eps^2})$ times, so the overall resulting running time is
still polynomial.

\section*{Problem 5.6}

\paragraph{1.} I follow the hint by allowing the bivariate polynomial $Q$ in
the proof of Theorem 5.19 to have as many monomials as possible. More
precisely, we allow all monomials of degree $(d_1, d_2)$ such that:
\begin{displaymath}
    d_1 + d\cdot d_2\leq \lfloor\eps q\rfloor \eqdef D
\end{displaymath}
and we will be fine as long as the number of monomials is strictly larger than
$q$ (to make sure we can find such a $Q$). Hence, all we have to do is count
the number of pairs $(d_1,d_2)$ such that $d_1 + d\cdot d_2\leq \eps q$ and
make sure that it is strictly larger than $q$. We see
that $d_2$ can be at most $A\eqdef\lfloor D/d\rfloor$ (in which case
$d_1$ has to be zero). If $d_2 = A - k$ for $k\in\{0,\dots, A\}$, then $d_1$
can be as large as $D -kd$. Hence, the total number of monomials is:
\begin{displaymath}
    N = \sum_{k=0}^A (D-kd + 1) = (D+1)(A+1) - \frac{d}{2} A(A+1)
    = (A+1)\left(D+1-\frac{d}{2}A\right)
\end{displaymath}
For $\eps=\sqrt{2d/q}$, we have $A=\lfloor D/d\rfloor = \lfloor
\sqrt{2q/d}\rfloor$ and $D=\lfloor
\sqrt{2qd}\rfloor$. This leads to:
\begin{displaymath}
    N > \sqrt{2q/d}\left(\sqrt{2qd} - \frac{d}{2}\sqrt{2q/d}\right) = q
\end{displaymath}
as desired.

\section*{Problem 5.7}

\paragraph{1.} Let us denote by $r\in\{0,1\}^{\hat{n}}$ the answers given by
the mean oracle trying to hide the secret from us. First observe that, denoting
by $[A]$ the Iverson bracket of event $A$ and $c_i = \Enc(s_i), i\in\{1,2\}$:
\begin{displaymath}
    \hat{n}d(c_1, c_2) = \sum_{l=1}^{\hat{n}} [(c_1)_i\neq (c_2)_i]
    = \sum_{i=1}^{\hat{n}}[(c_1)_i\neq r_i] + [(c_2)_i\neq r_i]
    = \hat{n} d(c_1, r) + \hat{n}d(c_2, r)
\end{displaymath}
where the second equality follows from the fact that $r_i$ is either equal to
$(c_1)_i$ or $(c_2)_i$  when $(c_1)_i\neq (c_2)_i$. The other equalities simply
use the definition of the relative distance.

I claim that one of the two secrets $s\in\{s_1, s_2\}$ is such that the
relative distance $d(\Enc(s), r)\leq \frac{1}{4} + \eps$. Otherwise, using the
identity proven above:
\begin{displaymath}
    d(c_1, c_2) = d(c_1, r) + d(c_2, r)> \frac{1}{2} + 2\eps
\end{displaymath}
This implies that $\Enc(s_1)$ and $\Enc(s_2)$ agree on less than a $\frac{1}{2}
- 2\eps$ fraction of positions and contradicts the assumption on $\Enc$.

Since one of the two secrets $s\in\{s_1,s_2\}$ is such that
$d(\Enc(s),r)\leq\frac{1}{4}+\eps$, by list-decoding the received answer $r$ at
distance $\frac{1}{4}+\eps$, we are guaranteed that the list $L$ of length $l$
returned by the decoder contains one of the two secrets. By assumption, the
decoding takes polynomial time.

\paragraph{2.}

\end{document}