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This site contains various discussions of one-way functions and their relation to P versus NP.

Some of these discussions use a language $L=\{(x',y) ~\mid~ x'\le x \text{ and } f(x)=y \}$, where $f:\Sigma^*\to\Sigma^*$ is the one-way function and $x'\le x$ is the prefix relation. Now one central claim is that this language $L$ is contained in NP, since the word $x$ is a YES-certificate for $(x',y)\in L$.

I do not see why this claim is justified.
Why is the length of the certificate $x$ polynomially bounded in the length of $(x',y)$?

Couldn't it be possible that $x$ is exponentially long in $y$ and $x'$, but $f(x)$ is short and quickly computable from $x$?

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  • $\begingroup$ It is likely that a proof that P=NP is not an effective proof. Crypto works about the same if the effort to crack n bit keys is 2^n or n^65536. $\endgroup$ – Joshua Dec 17 '18 at 19:56
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Yes, it could be that in the language you give, $x$ is exponentially long in $(y,x')$, and $f$ is an efficiently computable one-way function (note that it only has to run in time polynomial in its input length, so $f(x)$ needs not be computable in time polynomial in $(y,x')$).

However, this is really a minor issue: the answers to this question that you read are simply a bit informal, and only give an intuition of the proof that OWF implies $P \neq NP$. Intuitively, to fix this, modify your language as follows:

$L=\{(1^n, x',y) ~\mid~ \exists x, |x| = n, x'\le x, \text{ and } f(x)=y \}$,

where $1^n$ means a sequence of $n$ consecutive one, which exactly allows to fix the issue you point out (note that here $x'\le x$ means $x'$ is a prefix of $x$).

Note: the second answer to the question you link to does provide a link to an exercise sheet which contains the more formal solution.

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  • $\begingroup$ Sorry, but with your modification it is not clear anymore how to invert function $f$ in polynomial time, in case $L$ is in $P$. For exploting $L$, it seems that now I need some a priori bound on $n$, but this might be exponentially large in the length of $y$. $\endgroup$ – Alexis Dec 17 '18 at 15:04
  • $\begingroup$ Inverting $f$ needs not be polynomial in the output size $y$, but still in the input size $x$, which is the case if $L$ is in $P$. You should check the detailed solution given on page 2-3 of the exercise sheet I link to (courses.cs.ut.ee/all/MTAT.07.004/2016_fall/uploads/solution/…). $\endgroup$ – Geoffroy Couteau Dec 17 '18 at 15:29
  • $\begingroup$ With polynomial runtime for f you can not generate an output f(x) with superpolynonial size - so actually any limitation on x also translates to y implicitly. $\endgroup$ – tylo Dec 18 '18 at 11:56

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