1
$\begingroup$

I was looking at solutions for a problem but I noticed it said $\langle2\rangle$ and was unsure of what that meant.

screenshot of relevant part

From Homework 5 (PDF)

$\endgroup$
  • $\begingroup$ That's a very strange notation to be used here. But the instructor as noted at the top corner of the paper, @chris-peikert who happens to be a member here, maybe we should ask him? $\endgroup$ – DannyNiu Mar 13 '18 at 7:32
  • $\begingroup$ I've started a chat room with him, I'm not sure if you could join. $\endgroup$ – DannyNiu Mar 13 '18 at 7:40
  • 2
    $\begingroup$ It appears from context that $\langle x \rangle$ is a canonical representation of the integer $x$ as an $n$- bit bitstring. Otherwise those inputs and outputs are not in the correct donation/range. $\endgroup$ – Maeher Mar 13 '18 at 7:44
  • $\begingroup$ @Maeher So basically the solution demonstrated an attack on a artifical hash compression function that is collision resistant but not pseudo-random? $\endgroup$ – DannyNiu Mar 13 '18 at 7:51
  • $\begingroup$ You also see this notation in quantum mechanics, as in part 1 of Why should one model an entropy source in order to build a TRNG?. Don't ask me, I just copied it but you see if often in photonic TRNG papers. It may have multiple meanings subject to domain. $\endgroup$ – Paul Uszak Mar 22 at 16:59
2
$\begingroup$

From the context of the exercise, it appears that $\langle x\rangle$ is meant to denote a canonical representation of the integer $x$ as a bitstring of length $n$.

If this were not the case, those inputs and outputs would not be in the correct domain and range respectively.

Likely this notation was defined somewhere in the lecture notes, but I cannot find them at the moment.

$\endgroup$
  • $\begingroup$ Sounds like a very logical answer, in that case it is likely defined as I2OSP as defined in RSA (a statically sized, unsigned big endian / network order integer representation. $\endgroup$ – Maarten Bodewes Mar 22 at 22:04

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.