From 96941cad0f409c3f14a460c2a592786789d05472 Mon Sep 17 00:00:00 2001 From: Andre Henriques Date: Tue, 7 Nov 2023 22:06:05 +0000 Subject: [PATCH] fix spelling --- cw/cw.tex | 14 +++++++++++++- 1 file changed, 13 insertions(+), 1 deletion(-) 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: $$H=E$$ $$m\ne m'$$