# Talk:Hard-core predicate

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## Recent Discoveries

There's a new paper The security of all RSA and discrete log bits whose abstract states that:

• all bits of RSA and discrete log are hard core
• every block of log |x| bits is simultaneously hard core

which is very relevant and very exciting; I haven't read the paper yet; the article needs to be updated. Arvindn 07:35, 18 Nov 2004 (UTC)

Yep, this has actually been known for awhile, it's just now appearing in journal form. I can try to take a crack at some point in the near future. --Chris Peikert 20:36, 18 Nov 2004 (UTC)
• Yep. Took a look + added what I understood! RobertHannah89 (talk) 15:07, 27 January 2011 (UTC)

This question concerns blackbox one-way functions and computable one-way functions. [I use slightly different notation - ${\displaystyle g}$ is used to indicate the hard-core predicate instead of ${\displaystyle B}$]

Let ${\displaystyle f:\{0,1\}^{*}\mapsto \{0,1\}^{*}}$ be a length-preserving one-way function, and

Due to the Goldreich-Levin theorem, such a hardcore predicate exists for all length-preserving one-way functions.

We define the probabilities

Here: ${\displaystyle f(.)}$ represents the algorithm for computing} ${\displaystyle f}$

${\displaystyle B(f(.))}$ represents a blackbox for computing ${\displaystyle f}$

${\displaystyle x}$ is a finite length string chosen uniformly from ${\displaystyle \{0,1\}^{*}}$, and

${\displaystyle A}$ is any PPT algorithm.

The Goldreich-Levin theorem says that if ${\displaystyle f}$ is one-way, then for all algorithms ${\displaystyle A}$, the probability ${\displaystyle u}$ is negligible function of ${\displaystyle |x|}$. Thus, ${\displaystyle v}$ is also a negligible function of ${\displaystyle |x|}$.

However, the Goldreich-Levin theorem does not say anything about the ratio ${\displaystyle k=v/u}$.

My question: Is ${\displaystyle k}$ a negligible function of ${\displaystyle |x|}$ too? According to my logic, ${\displaystyle k}$ should be close to ${\displaystyle 1}$, and should be independent of ${\displaystyle x}$.

Observation: If ${\displaystyle k=1}$ then having access to the algorithm of ${\displaystyle f}$ does not give any additional advantage over having blackbox access to ${\displaystyle f}$ (in computing the hard-core predicate)