1
$\begingroup$

Let's say we have a randomly encrypted message and I pasted the contents of that message on a blog for some reason I do not share the key.

Would you be able to figure out the key?

This is the message not encrypted:

This is the message.

This is the encrypted message:

 -----BEGIN OneTime MESSAGE-----

Version: OneTime 1.122

Pad ID: b1255b5389588da3c0e00b87213afb6bd79a0bbb

Offset: 0

QlpoOTFBWSZTWUyTjxYAAAJ+gAgCA4FKYABwSAEoAQAQIAAigA0xMEKBpoZGTEHGQdDBjUXXyo7R
HxdyRThQkEyTjxY=

-----END OneTime MESSAGE-----

The package will not let you use keys twice. Its protocol follows this rule especially for communicating to two computers over telnet, nc, ssh, etc. The protocol can be found here in this link https://www.red-bean.com/onetime/ .

$\endgroup$
3
  • $\begingroup$ A quick search shows that OneTime is some kind of Linux package. I'm however not aware of any particular protocol that it adheres to. Do you have any reference to the protocol used? Note that analyzing ciphertext is off topic; we need to have some idea of the protocol used to give a meaningful answer. Without that we'll have to put the question on hold. $\endgroup$
    – Maarten Bodewes
    Jan 29 '19 at 2:33
  • 1
    $\begingroup$ I'm not asking to analyze the ciphertext. I'm asking if its possible for an attacker could get the key for some unpractical arbitrary reason just by already knowing the contents of the message. The package will not let you use keys twice. Its protocol follows this rule especially for communicating to two computers over telnet, nc, ssh, etc. The protocol can be found here in this link red-bean.com/onetime $\endgroup$ Jan 29 '19 at 3:32
  • $\begingroup$ Thanks for the followup and the protocol indication. Exactly what was needed. One minor remark: additional info is best placed in the question. I'll do that for now and then I'll delete these comments tomorrow. $\endgroup$
    – Maarten Bodewes
    Jan 29 '19 at 23:19
4
$\begingroup$

Knowing the plaintext and ciphertext will allow you to recover (a portion of) the key.

The OneTime package referenced in the comments implements a simple one-time pad. Because the plaintext and ciphertext are related by a simple XOR it is trivial to calculate the portion of the one-time pad that is used for that message. If all of the rules for using a one-time pad are followed (e.g. no re-use, truly independent random values for each bit, etc), then that portion of the key can be calculated, but it wouldn't provide any value beyond what was already known.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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