-In algorithm~\ref{algo:bbs_gpu}, t<<=4 performs a left shift of 4 bits
-on the variable t and stores the result in t. BBS1(bbs1)\&15 selects
-the last four bits of the result of BBS1. It should be noticed that
-for the two new shifts, we use arbitrarily 4 BBSs that have previously
-been used.
+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
+correspondence 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 $x^{n+1} = x^n \oplus S^n$,
+where $S^n$ is referred in this algorithm as $t$: each iteration of this
+PRNG ends with $x = x \wedge t$. This $S^n$ is only constituted
+by secure bits produced by the BBS generator, and thus, due to
+Proposition~\ref{cryptopreuve}, the resulted PRNG is cryptographically
+secure.
+