The Berman-Hartmanis conjecture discusses one-way functions (functions with hard to compute inverse functions).

As a step to solving the conjecture, if one-way functions could be reduced to a canonical or universal one-way function from which all one-way functions could be derived, this would be a major plus...

In a similar way, Turing devised a universal machine and Cook NP-completeness.

The question is then, are there universal one-way functions (discussed in the literature)? Can they be defined through NP-completeness?

  • $\begingroup$ note the "conventional wisdom" in the field is that the conjecture is false as pointed out by wikipedia $\endgroup$
    – vzn
    Commented Aug 2, 2014 at 15:52

1 Answer 1


That is an interesting question since there is no proof that one way functions exist. There is a lecture about this from Cornell's Cryptography course by Rafael Pass which can be found here

To summarize the lecture:

  1. If one way functions exist then $P\neq NP$.

  2. There exists a function $f_{UNIV}$ such that if there exists any one-way function, then $f_{UNIV}$ is a one-way function. Thus $f_{UNIV}$ can be considered a universal one-way function.

  3. $f_{UNIV}$ works by acting on all Turing machines after $n^2$ steps of execution. It is defined as

    $$f_{UNIV}=M_1^{(|x|^2)}(x)||M_2^{(|x|^2)}(x)||\dots M_{|x|}^{(|x|^2)}(x)$$

    where $M_{i}^{(|x|^2)}(x)$ denotes the output of Turing machine $M_i$ on input $x$ right after $|x|^2$ steps of computation.

  4. If a one way function $f$ exists it can be solved by a Turing machine $M$, which will be polynomial solvable by $f_{UNIV}$.

Caution: the construction of $f_{UNIV}$ is in no way practical. This is about asymptotic complexity, not practical security.

  • $\begingroup$ What do you mean by "$\hspace{.04 in}f_{UNIV}$ exists"? $\:$ What do you mean by "solved"? $\;\;\;\;$ $\endgroup$
    – user12859
    Commented Jul 31, 2014 at 21:37
  • $\begingroup$ By exists it means that there may be no one way functions they may not exists. $\endgroup$
    – lPlant
    Commented Jul 31, 2014 at 21:40
  • $\begingroup$ I've tried to clean up your answer, but I don't understand what you are saying in point 4 at all. Would you like to try rephrasing that? What do you mean when you say "$f$ can be solved by a Turing machine $M$"? If $f$ is one-way, there is no polynomial-time algorithm to invert it. What do you mean by "polynomial solvable by $f_{UNIV}$"? $\endgroup$
    – D.W.
    Commented Jul 31, 2014 at 21:57
  • 1
    $\begingroup$ The constructed function is not necessarily one-way, even if $M_k$ is one-way for some $k$. The reason is that all $M_i$ are evaluated on the same input $x$. Hence, if $M_1$ is easy to invert, the whole function is easy to invert. This problem can be fixed by first partioning the input $x$ into $x_1, \ldots, x_{\sqrt{|x|}}$ and then running $M_i$ on input $x_i$. $\endgroup$ Commented Jul 14, 2018 at 5:29
  • $\begingroup$ Liu and Pass have another universal one-way function with a more natural construction (time-bound Kolmogorov Complexity) arxiv.org/abs/2009.11514 $\endgroup$
    – Quantum7
    Commented Apr 25, 2022 at 9:14

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.