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cw/cw.tex
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cw/cw.tex
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where $R$ is the list of random values generated for each pair
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\subsection*{4.3}
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Our random value is $R = u-bit long digit$ which means that it has $2^u$ possible values.
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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:
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$$\frac{q^2}{2(2^u)}\gt\frac{1}{2}\iff q\gt 2^{\frac{u}{2}}$$
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\subsection*{4.4}
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The size of TripleDES is 64 bit long which makes $u=64/2=32$ making the q
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$q\gt 2^\frac{32}{2}\iff q \gt 65536$
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\subsection*{4.5}
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The size of AES is 128 bit long which makes $u=128/2=64$ making the q
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$q\gt 2^\frac{64}{2}\iff q \gt 4294967296$
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\subsection*{4.6}
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Since in in both 4.4 and 4.5 the value of $q$ is not large enough the scheme is not CPA secure
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\section*{5}
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\subsection*{5.1}
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The hash function is collision resistante for $n=1$, since if the block size is one the hash function is the encryption. Therefore:
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The hash function is collision resistanteeee for $n=1$, since if the block size is one the hash function is the encryption. Therefore:
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if the message is only one block long:
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$$H=E$$
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$$m\ne m'$$
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