User talk:Wonderstruck

Hello Wonderstruck and

Re this edit of yours: Could you please explain the link? I'm having trouble seeing the connection. Thanks. -- Jitse Niesen (talk) 00:15, 14 February 2007 (UTC)[reply]

Sure. A PRNG is an algorithm that takes a finite number of bits as input (the "seed") and produces a much larger (though finite) number of output bits. Therefore, every PRNG can be represented as a function, ${\displaystyle f(x)}$, where ${\displaystyle x}$ is the seed. More formally:

${\displaystyle f:\{0,1\}^{c}\rightarrow \{0,1\}^{n}}$

One of the requirements of a cryptographically secure PRNG is that it has to pass the "next-bit test". One way to formulate the next-bit test is to define a function ${\displaystyle g_{m}(x)}$ that evaluates to the leftmost ${\displaystyle m}$ bits of ${\displaystyle f(x)}$. A PRNG passes the next-bit test if and only if for every ${\displaystyle m<n}$, the probability of correctly guessing ${\displaystyle g_{m+1}(x)}$, given only ${\displaystyle g_{m}(x)}$, is no higher than 50%.

Imagine that Alice wants to convince her colleague, Bob, that a given PRNG ${\displaystyle f(x)}$ fails the next-bit test. Bob chooses a random 128-bit secret ${\displaystyle x}$, computes ${\displaystyle a_{200}=g_{200}(x)}$, and reveals ${\displaystyle a_{200}}$ to Alice. If Alice can predict the next 100 bits of PRNG output (that is, if she can correctly guess the value of ${\displaystyle g_{300}(x)}$) without being given ${\displaystyle x}$, then Bob will be convinced that ${\displaystyle f(x)}$ fails the next-bit test.

Let ${\displaystyle h(y)=g_{200}(y)-a_{200}}$. If Alice can use a root-finding algorithm to find ${\displaystyle y}$ such that ${\displaystyle h(y)=0}$, then it is likely that ${\displaystyle y=x}$, and, therefore, that ${\displaystyle g_{300}(y)=g_{300}(x)}$.

In other words, if a PRNG can be solved using a root-finding algorithm (in polynomial time), then it is not cryptographically secure. So, cryptographically secure PRNGs are designed so that they can't be solved (in polynomial time) by any known root-finding algorithms.

Does that make sense?

P.S. I just added the link as an interesting bit of trivia, so if you think it should be removed, that's fine.

-- Wonderstruck 05:40, 14 February 2007 (UTC)[reply]

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talk) 00:09, 8 March 2008 (UTC)[reply

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Your input is needed on the SOPA initiative

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