# Tag Info

0

If you generate group elements at random as you suggest then you can indeed invoke the "Birthday Paradox" to find logarithms in time $O(\sqrt p)$. Unfortunately your storage requirements are the same and for cryptographically interesting group orders your method is therefore far from optimal. The fastest way for groups with (apparently) no exploitable ...

3

Let's say I choose $m = 8192$, $a = 4801$, and $c = 83$. By looking at the value of $x$, I can learn quite a lot about the value of $y$: for any $x$, there are only about nine or ten possible subsequent $y$s, and they're very unevenly distributed. With a LCG every $L_1$ has a single possible $L_2$, and for a maximal period the reverse is also true. ...

4

Since bits are independently generated, the entropy of the key is the sum over the entropy of the individual bits. The following calculations use the chance $P(x)$ of a zero or one bit. The first 5 bits are constant and thus have 0 entropy. The others are unbiased and have 1 bit of entropy each. $P(0)=P(1)=0.5$. \$-2\cdot(0.5 \cdot \log_2(0.5))= ...

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