diff --git a/cw/cw.tex b/cw/cw.tex
index 89ad2bb..8c5c332 100644
--- a/cw/cw.tex
+++ b/cw/cw.tex
@@ -145,10 +145,22 @@
 	where $R$ is the list of random values generated for each pair
 
 	\subsection*{4.3}
+	Our random value is  $R = u-bit long digit$ which means that it has $2^u$ possible values.
+	And since we know that $q$ balls into $p$ holes a colision is bound to happend at the probability of $\frap{q^2}{2p}$ we can calculate:
+	$$\frac{q^2}{2(2^u)}\gt\frac{1}{2}\iff q\gt 2^{\frac{u}{2}}$$
+
+	\subsection*{4.4}
+	The size of TripleDES is 64 bit long which makes $u=64/2=32$ making the q
+	$q\gt 2^\frac{32}{2}\iff q \gt 65536$
+	\subsection*{4.5}
+	The size of AES is 128 bit long which makes $u=128/2=64$ making the q
+	$q\gt 2^\frac{64}{2}\iff q \gt 4294967296$
+	\subsection*{4.6}
+	Since in in both 4.4 and 4.5 the value of $q$ is not large enough the scheme is not CPA secure
 
 	\section*{5}
 	\subsection*{5.1}
-	The hash function is collision resistante for $n=1$, since if the block size is one the hash function is the encryption. Therefore:
+	The hash function is collision resistanteeee for $n=1$, since if the block size is one the hash function is the encryption. Therefore:
 	if the message is only one block long:
 	$$H=E$$
 	$$m\ne m'$$