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\centerline{\textbf{HW 10 CMSC 456. Morally DUE Nov 30}}


\begin{enumerate}

\item
(25 points)
(You might want to write a program on your own for this problem,
but not for autograding.) 
Alice and Bob are doing LWE-private key over mod 101. 
They agree on secret key $(2,3,5,7)$. 
We give the random numbers Alice picks, the bit she wants to send,
and the value of $e$ she picks. YOU will supply what she sends Bob. 


\[
\begin{tabular}{|c|c|c|c|}
\hline
Rand Numb & bit & $e$ &  What Alice Sends Bob \cr
\hline
(1,8,7,90) & 1   & 0    &                         \cr
(1,8,7,90) & 0  & 1   &                         \cr
(1,8,7,90) & 1   & -1    &                         \cr
(1,8,7,90) & 0  & 0   &                         \cr
(10,81,71,91) & 1   & 1    &                         \cr
(10,81,71,91) & 0   & -1    &                         \cr
(10,81,71,91) & 1   & 0    &                         \cr
(10,81,71,91) & 0   & 1    &                         \cr
\hline
\end{tabular} 
\]



\vfill\eject

\item
(25 points) Alice and Bob do LWE-private with $p=1001$, vectors of length 3. 
(1001 is not a prime but one can still do LWE-private using it.) 
\begin{enumerate}
\item
(25 points) 
Describe a variant of LWE-private where Alice can send Bob one of
THREE possibilities, which we will call $0,1,2$. 
\item
(0 points) DO NOT HAND IN- do this for your own enlightenment. 
Describe a variant of LWE-private where Alice can send Bob one of
FOUR possibilities, which we will call $0,1,2,4$. 
\item
(0 points) 
Do parts 1 and 2 of this question with a general $p$. 
The key is to find $\gamma$ that works. 
\end{enumerate}

\newpage

\item 
(25 points) 
A1, A2, A3, A4 have cards similar to those used in the Alice-Bob-Cards-Dating lecture.
(e.g., hearts, spades, uparrows, make them clear, make them opaque, make them
fit into pez dispensers).
A1 has a bit $a_1$,
A2 has a bit $a_2$,
A3 has a bit $a_3$,
A4 has a bit $a_4$.
They want to compute $a_1\wedge a_2\wedge a_3\wedge a_4$ such that
\begin{enumerate}
\item
At the end they ALL know $a_1\wedge a_2\wedge a_3\wedge a_4$.
\item
At the end A1 only knows $a_1$ (of course), $a_1\wedge a_2\wedge a_3\wedge a_4$,  and what can be deduced from these.
So
\begin{enumerate}
\item
If $a_1=0$ and $a_1\wedge a_2\wedge a_3\wedge a_4=0$ 
then A1 knows nothing about $a_2$ or $a_3$ or $a_4$.
\item
If $a_1=0$ and $a_1\wedge a_2\wedge a_3\wedge a_4=1$ THIS CANNOT HAPPEN.
\item
If $a_1=1$ and $a_1\wedge a_2\wedge a_3\wedge a_4=0$ then A1 knows $a_2=0$ or $a_3=0$ or $a_4=0$, but does not know which of those happens. $a_1$ DOES NOT even know how many of $a_2,a_3,a_4$ said 0. 
\item
If $a_1=1$ and $a_1\wedge a_2\wedge a_3\wedge a_4=1$ then A1 knows $a_2=1$ and $a_3=1$ and $a_4=1$.
\end{enumerate}
\item
Similar for $a_2$, $a_3$, $a_4$.
\end{enumerate}

And now {\bf finally} the problem: Give a protocol for $A_1,A_2,A_3,A_4$ to
use that achieves the above conditions. Recall that they can use cards.

{\it Hint:} Use a variant of one of the schemes discussed in the  Alice-Bob-Cards-dating lecture

\vfill\eject 


\item
(25 points) 
In this problem we only deal with messages that are strings of bits. 
DO NOT do just part b and say is applies to Part a- just do two different
methods for Part a and Part b. 

In this problem when I ask for an error-detection or error-correction code I mean
you have to tell me:

If Alice wants to send $m_1\cdots m_L$, what does she send?

When Bob gets $b_1\cdots b_n$ what does he do to detect or correct? 

\begin{enumerate}
\item
(15 points) 
Give an error-detecting code such that the following happens.

\begin{itemize}
\item
If there are 0 errors, Bob is confident there are 0 errors.
\item
If there is 1 error, Bob is confident there is 1 error.
\item
If there are 2 errors, Bob is confident there are 2 error.
\item
If there are $\ge 3$ errors then Bob is confident of something but he is wrong. 
\end{itemize} 

(Hint: Use a variant of the Parity Check.) 


\item 
(10 point) 
Give an error-correcting code such that the following happens (note also that there is a
question on the fourth point). 

\begin{itemize}
\item
If there are 0 errors, Bob is confident there are 0 errors.
\item
If there is 1 error, Bob is confident there is 1 error and he knows where it is so he can correct it. 
\item
If there are 2 errors, Bob is confident there are 2 errors and he knows where it is so he can correct them. 
\item
If there are $\ge 3$ errors then Bob is confident of something but he MIGHT BE wrong. 
Give an example where there are 3 errors but Bob is right. 
Give an example where there are 3 errors but Bob is wrong. 
\end{itemize} 

(Hint: Use a variant of the repetition code.) 

\end{enumerate}






\end{enumerate}

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