1 / 4

Announcements: HW4 – DES due midnight So far the record is less than 15 sec on 1 million iters

This announcement provides practical considerations for using RSA encryption, including the process of sending messages and the security measures involved. It also addresses common issues and provides examples using larger numbers. The announcement concludes with information about using Maple's worksheet mode and a note on the lambda expression myConcatenator.

ellaw
Download Presentation

Announcements: HW4 – DES due midnight So far the record is less than 15 sec on 1 million iters

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  1. DTTF/NB479: Dszquphsbqiz Day 20 • Announcements: • HW4 – DES due midnight • So far the record is less than 15 sec on 1 million iters • Quiz on ch 3 postponed until after break • Term project groups and topics due end of week after break • Use ch 10 – 19 as inspiration • Today • Using RSA: practical considerations • Questions?

  2. RSA (Rivest – Shamir – Adelman) For Alice to send a message to Bob. • Bob chooses primes p,q (large, ~100 digits each) • He publishes his public key (n,e): • n = pq • e, a large number such that gcd(e, (p-1)(q-1)) = 1 • Alice has a message m < n. • Otherwise (if m > n), break message into chunks < n • Alice sends c = me(mod n) • Bob computes cd (mod n) = (me)d = m (mod n). • What does he use for d?

  3. Alice – (m)  Bob Bob’s key: n = pq = (13)(17) = 221 e = 35: gcd(e, (p-1)(q-1)) = 1 d=e-1 mod 192 exists: d = __11__ m = 20 (letter t) 1-based, so leading ‘a’ = 1 not ignored c = me(mod n) = _197___ cd (mod n) = _20__ Issues: How to compute 2035(mod 221)? Efficiency is O(log e) How to compute d? Extended Euclidean alg. And why is this secure? Why can’t Eve calculate d herself? Toy example

  4. Example with larger numbers • Maple’s worksheet mode • For some reason, inert power (&^) only works for me when entering in the red (single-line exponents) entry-mode; press F5 to toggle. • myConcatenator is a lambda expression.

More Related