I know that there has not been a proof against or for, the existence for a true one-way function. But i was wondering has such a thing been proven for collision-less (injective) one-way functions.


closed as off-topic by Adam Hughes, Daniel W. Farlow, Claude Leibovici, JonMark Perry, Ben Jun 17 '17 at 9:51

This question appears to be off-topic. The users who voted to close gave this specific reason:

  • "This question is not about mathematics, within the scope defined in the help center." – Adam Hughes, Daniel W. Farlow, JonMark Perry
If this question can be reworded to fit the rules in the help center, please edit the question.

  • 1
    $\begingroup$ You mean, injective? $\endgroup$ – Clement C. Jun 16 '17 at 19:09
  • $\begingroup$ yes i will edit the question $\endgroup$ – Alonzo C Jun 16 '17 at 19:24

Comment: be careful with the terms you use. "Injective" is common, but "collision-less" is non-standard and may be confused with "collision-free," which is something else.

First, note that since what you ask is a subset of one-way functions, proving their existence would imply the existence of one-way functions. Thus, you already know that existence of injective one-way functions hasn't been established.

Now, their existence has not been disproven either: the existence of one-way permutations (i.e., one-way functions that are bijective, and thus a fortiori injective) is a common cryptographic assumption. To this day, it has not been proven nor disproven, and their existence is not known to be implied by that of one-way functions.

  • $\begingroup$ thank you. i knew there existence had not been proven, but i had know idea if they had been disproved. $\endgroup$ – Alonzo C Jun 16 '17 at 19:34
  • $\begingroup$ sorry did not mean to unclick with be compressing the whole bitcoin block-chain & running a vm it gets a bit laggy. $\endgroup$ – Alonzo C Jun 16 '17 at 19:50
  • $\begingroup$ Oh, I see. No worries. $\endgroup$ – Clement C. Jun 16 '17 at 19:51

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