-In Algorithm~\ref{algo:bbs_gpu}, $n$ is for the quantity
-of random numbers that a thread has to generate.
-The operation t<<=4 performs a left shift of 4 bits
-on the variable $t$ and stores the result in $t$, and
-$BBS1(bbs1)\&15$ selects
-the last four bits of the result of $BBS1$.
-Thus an operation of the form $t<<=4; t|=BBS1(bbs1)\&15\;$
-realizes in $t$ a left shift of 4 bits, and then puts
-the 4 last bits of $BBS1(bbs1)$ in the four last
-positions of $t$.
-Let us remark that to initialize $t$ is not a necessity as we
-fill it 4 bits by 4 bits, until having obtained 32 bits.
-The two last new shifts are realized in order to enlarge
-the small periods of the BBS used here, to introduce a variability.
-In these operations, we make twice a left shift of $t$ of \emph{at most}
-3 bits and we put \emph{exactly} the 3 last bits from a BBS into
-the 3 last bits of $t$, leading possibly to a loss of a few
-bits of $t$.
-
-It should be noticed that this generator has another time the form $x^{n+1} = x^n \oplus S^n$,
+In Algorithm~\ref{algo:bbs_gpu}, $n$ is for the quantity of random numbers that
+a thread has to generate. The operation t<<=4 performs a left shift of 4 bits
+on the variable $t$ and stores the result in $t$, and $BBS1(bbs1)\&15$ selects
+the last four bits of the result of $BBS1$. Thus an operation of the form
+$t<<=4; t|=BBS1(bbs1)\&15\;$ realizes in $t$ a left shift of 4 bits, and then
+puts the 4 last bits of $BBS1(bbs1)$ in the four last positions of $t$. Let us
+remark that the initialization $t$ is not a necessity as we fill it 4 bits by 4
+bits, until having obtained 32-bits. The two last new shifts are realized in
+order to enlarge the small periods of the BBS used here, to introduce a kind of
+variability. In these operations, we make twice a left shift of $t$ of \emph{at
+ most} 3 bits, represented by \texttt{shift} in the algorithm, and we put
+\emph{exactly} the \texttt{shift} last bits from a BBS into the \texttt{shift}
+last bits of $t$. For this, an array named \texttt{array\_shift}, containing the
+correspondance between the shift and the number obtained with \texttt{shift} 1
+to make the \texttt{and} operation is used. For example, with a left shift of 0,
+we make an and operation with 0, with a left shift of 3, we make an and
+operation with 7 (represented by 111 in binary mode).
+
+It should be noticed that this generator has once more the form $x^{n+1} = x^n \oplus S^n$,