From 100ba825657c9031b9b527cfae06ce3f8f7ea732 Mon Sep 17 00:00:00 2001
From: Thibaut Horel <thibaut.horel@gmail.com>
Date: Mon, 6 Apr 2015 10:29:41 -0400
Subject: [ps4] hardest questions done, still a few gaps here and there

---
 ps4/main.tex | 136 ++++++++++++++++++++++++++++++++++++++++++++++++++++++++++-
 1 file changed, 134 insertions(+), 2 deletions(-)

(limited to 'ps4/main.tex')

diff --git a/ps4/main.tex b/ps4/main.tex
index 7799f32..ecda8e0 100644
--- a/ps4/main.tex
+++ b/ps4/main.tex
@@ -24,6 +24,8 @@
 
 
 \DeclareMathOperator{\E}{\mathbb{E}}
+\DeclareMathOperator{\Enc}{\mathrm{Enc}}
+\DeclareMathOperator{\Dec}{\mathrm{Dec}}
 \let\P\relax
 \DeclareMathOperator{\P}{\mathbb{P}}
 \DeclareMathOperator*{\argmin}{\arg\min}
@@ -59,14 +61,144 @@ $|B(r,\delta)\cap \mathcal{C}|\leq L$. Hence:
 \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 at most 
+$q^{\hat{n}H_q(\delta, \hat{n})}$, 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})}\leq 
+    |\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.
+
+\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}
-- 
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