We consider a ``word'' formed of 4 bits which we denote $b _ { 1 } , b _ { 2 } , b _ { 3 }$ and $b _ { 4 }$. We add to this list a check key $c _ { 1 } c _ { 2 } c _ { 3 }$ formed of three bits:
\begin{itemize}
  \item $c _ { 1 }$ is the remainder of the Euclidean division of $b _ { 2 } + b _ { 3 } + b _ { 4 }$ by 2;
  \item $c _ { 2 }$ is the remainder of the Euclidean division of $b _ { 1 } + b _ { 3 } + b _ { 4 }$ by 2;
  \item $c _ { 3 }$ is the remainder of the Euclidean division of $b _ { 1 } + b _ { 2 } + b _ { 4 }$ by 2.
\end{itemize}
We then call ``message'' the sequence of 7 bits formed of the 4 bits of the word and the 3 control bits.

\begin{enumerate}
  \item Preliminaries\\
a. Justify that $c _ { 1 } , c _ { 2 }$ and $c _ { 3 }$ can only take the values 0 or 1.\\
b. Calculate the check key associated with the word 1001.

  \item Let $b _ { 1 } b _ { 2 } b _ { 3 } b _ { 4 }$ be a word of 4 bits and $c _ { 1 } c _ { 2 } c _ { 3 }$ the associated key.\\
Prove that if we change the value of $b _ { 1 }$ and recalculate the key, then:
\begin{itemize}
  \item the value of $c _ { 1 }$ is unchanged;
  \item the value of $c _ { 2 }$ is modified;
  \item the value of $c _ { 3 }$ is modified.
\end{itemize}

  \item We assume that, during the transmission of the message, at most one of the 7 bits was transmitted incorrectly. From the first four bits of the received message, we recalculate the 3 control bits, and compare them with the received control bits.\\
Without justification, copy and complete the table below. The letter $F$ means that the received control bit does not match the calculated control bit, and $J$ means that these two bits are equal.

\begin{center}
\begin{tabular}{|l|l|l|l|l|l|l|l|l|}
\hline
\backslashbox{Calculated control bit}{Erroneous bit} & $b _ { 1 }$ & $b _ { 2 }$ & $b _ { 3 }$ & $b _ { 4 }$ & $c _ { 1 }$ & $c _ { 2 }$ & $c _ { 3 }$ & None \\
\hline
$c _ { 1 }$ & $J$ &  &  &  &  &  &  &  \\
\hline
$c _ { 2 }$ & $F$ &  &  &  &  &  &  &  \\
\hline
$c _ { 3 }$ & $F$ &  &  &  &  &  &  &  \\
\hline
\end{tabular}
\end{center}

  \item Justify briefly, using the table, that if a single received bit is erroneous, we can in all cases determine which one it is, and correct the error.

  \item Here are two messages of 7 bits:
$$A = 0100010 \quad \text { and } \quad B = 1101001 .$$
We admit that each of them contains at most one transmission error. Say whether they contain an error, and correct it if necessary.
\end{enumerate}