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diff --git a/bits3.tex b/bits3.tex
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...
\hline
\end{tabular}
\end{center}
so we have a new three-bit vector, that we call again
state.\\ \emph{state}.\\
We now need to compute the new mysterious value ?, and so we proceed
with the same algorithm, by taking the two bits at positions $x$ (i.e., $1$) and $1$ (i.e., $0$).\\
We will then get $1+0=1$ and so
...
\end{tabular}
\end{center}
%
which becomes, after another shift to the right, the following
state
%
\begin{center}
\begin{tabular}{ |c ||c|| c | c | c |}
...
\end{tabular}
\end{center}
%
with the last shift
we find the final state, which is the same as the initial state.
%
\begin{center}
\begin{tabular}{ |c ||c|| c | c | c |}
...
We notice the following facts:
\begin{enumerate}
\item[(1)] After having performed the above algorithm seven times, we get again the initial
vector state $(1,0,1)$.
\item[(2)] While performing the above algorithm, we find all seven nonzero $3$-bit strings
%only the numbers $1,2,3,4,5,6,7$ can be written using three bits not all zero and, performing the algorithm, we found all these numbers.
\end{enumerate}
...
any polynomial $f\in \Fb[x]$ of the form $f=x^n+\ldots+1$ to construct an LFSR
producing $n$-bit vectors and such that any initial state is got again after
a finite number of iterations.\\
Fact (2) is more difficult to generalize. If we want an LFSR that can start from any initial nonzero
vector state and obtain all other nonzero
vectors, vectors as states, then we must use very special polynomials, called \emph{primitive} polynomials.
%
\begin{Definition}\label{Primitive}
We call primitive any polynomial $f\in \Fb[x]$ of degree $n$ that, used as feedback polynomials of a LFSR, can generate all the non-zero
...
\end{tabular}
\end{center}
After a shift to the right, we obtain
the second state
\begin{center}
...
\end{center}
and, after a shift to the right, we get
the third state
\begin{center}
\begin{tabular}{ |c ||c|| c | c | c |}
...
\end{tabular}
\end{center}
From now on, we are
stuck; indeed stuck with a zero state. Indeed the mysterious bit will always be equal to $0$ and so, after the shift to the right,
we the state will always
obtain remain $(0,0,0)$.
\end{Example}
We recall the notion of irreducible polynomial (Definition \ref{REDIRR}). An irreducible polynomial cannot be written as product of two non-trivial factors. It turns out that there is a close links between primitive polynomials and irreducible polynomials, as shown in the following theorem.
\begin{Theorem}
