deletions | additions       diff --git a/bits3.tex b/bits3.tex  index ab08014..86eaa21 100644  --- a/bits3.tex  +++ b/bits3.tex 
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\end{center}  \end{Example}  \begin{Example}\label{Buffo}  Let us consider a polynomial which is \emph{not} of the form $f=x^n+\ldots+1$, i.e. $f=x^3+x^2$, using it to construct a LFSR.  \\  We take the following initial state  \begin{center}  \begin{tabular}{ |c ||c|| c | c | c |}  \hline  ? & $\longrightarrow \longrightarrow$ & 0&1 & 1 \\  \hline  \end{tabular}  \end{center}  and we take the bit in the position represented by $x^2$, i.e. $0$ as mysterious bit:  \begin{center}  \begin{tabular}{ |c ||c|| c | c | c |}  \hline  0 & $\longrightarrow \longrightarrow$ & 0&1 & 1 \\  \hline  \end{tabular}  \end{center}  After a shift to the right, we obtain   \begin{center}  \begin{tabular}{ |c ||c|| c | c | c |}  \hline  ? & $\longrightarrow \longrightarrow$ & 0&0 & 1 \\  \hline  \end{tabular}  \end{center}  The mysterious bit turns out to be $0$ again, so we have  \begin{center}  \begin{tabular}{ |c ||c|| c | c | c |}  \hline  0 & $\longrightarrow \longrightarrow$ & 0&0 & 1 \\  \hline  \end{tabular}       \end{center}  and, after a shift to the right, we get    \begin{center}  \begin{tabular}{ |c ||c|| c | c | c |}  \hline  ? & $\longrightarrow \longrightarrow$ & 0& 0 & 0 \\  \hline  \end{tabular}  \end{center}    From now on, we are stuck; indeed the mysterious bit will always be equal to $0$ and so, after the shift to the right, we will always obtain $(0,0,0)$  \end{Example}  \begin{Theorem}  If $p\in \Fb[x]$ is primitive then is irreducible.  \end{Theorem}