fix spelling

cw/cw.tex

@@ -219,11 +219,24 @@
 
 	which means that
 
-	$$(((m^{r_{a\text{ alice}}})^{r_{a\text{ bob}}})^{r_{b\text{ alice}}})^{r_{b\text{ bob}}}) = m (\text{mod } p$$
+	$$(((m^{r_{a\text{ alice}}})^{r_{a\text{ bob}}})^{r_{b\text{ alice}}})^{r_{b\text{ bob}}} = m (\text{mod } p)$$
 
 	in this case $r_a$ from alice cancels $r_b$ from alice, and $r_a$ from bob cancels $r_b$ from bob
 	\subsubsection*{6.3.2}
-	To send an encrypted message using this system 
+	To send an encrypted message using this system:
+	\begin{enumerate}
+	\item Bob and Alice choose a prime $p$
+	\item The sender, let's say Alice, selects $m$ and two random values $r_{a1}$ and $r_{a2}$ such that $(m^{r_{a1}})^{r_{a2}} = m (\text{mod } p)$
+	\item Alice then calculates $t1 = m^{r_{a1}} (\text{mod } p)$, Alice sends $t1$ to bob.
+	\item Bob selects two random values $r_{b1}$ and $r_{b2}$ such that $(m^{r_{b1}})^{r_{b2}} = m (\text{mod } p)$
+	\item Bob then calculates $t2 = t1^{r_{b1}} (\text{mod } p)$, Bob sends $t2$ to Alice
+	\item Alice then calculates $t3 = t2^{r_{a2}} (\text{mod } p)$, this undoes step 3, then Alice sends $t3$ to bob
+	\item Bob then calculates $m = t3^{r_{b2}} (\text{mod } p)$, this undoes step 5
+	\end{enumerate}
+	\subsubsection*{6.3.3}
+	Bob and Alice exchange information 4 times, they choose the primes and then 3 exchanges during the encryption process.
+	While for ElGamal you need to exchange information only twice, once to exchange public keys and the second to exchange the encrypted message
+	\subsubsection*{6.3.4}
 	
 
 	\section*{7}