3
$\begingroup$

This question already has an answer here:

How is it possible to do an exhaustive key search (DES algorithm) and only search for 2^55 instead of 2^56 keys ?

This is a slide from my course if this helps you. I don't understand it.

1

$\endgroup$

marked as duplicate by Maarten Bodewes, otus, fgrieu, e-sushi Sep 21 '16 at 13:15

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

  • $\begingroup$ I've voted to close this question as this has already been explained in an answer. If you need more detail, please specify what you don't understand or want detailed from the given answer in the duplicate. $\endgroup$ – Maarten Bodewes Sep 20 '16 at 22:14
  • $\begingroup$ In the question's quote, underline (rather than the usual overline as in $\overline K$) is used for bitwise complement. $\endgroup$ – fgrieu Sep 21 '16 at 5:54
  • $\begingroup$ @fgrieu Maybe they wanted to mirror the usual notation :P $\endgroup$ – Maarten Bodewes Sep 21 '16 at 7:04
1
$\begingroup$

What you want follows from the DES complementation property so only the remaining 55 key bits need to be brute-forced.

$\endgroup$
  • $\begingroup$ Sorry this doesn't answer the question in the detail I am looking for. I can understand how we can do an exhaustive search with 2^56 tries. I want to know how exactly is it possible to do 2^55 tries. In detail. $\endgroup$ – dimitris93 Sep 20 '16 at 22:00

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