{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

# hw309 - A to calculate inverses Explain 4(8 points Suppose...

This preview shows page 1. Sign up to view the full content.

CS346 Cryptography, Fall 2009 Homework 3, Due October 1 1. (8 points) Suppose that for using RSA, Bob has chosen a large public modulus n for which the factorization cannot be found in a reasonable amount of time. Suppose Alice sends a message to Bob representing each alphabetic character as an integer between 0 and 25, and encrypting each as a separate plaintext character. Describe how Oscar can easily decrypt a message which is encrypted this way. 2. (8 points) If an RSA user’s public key is n = 17 · 43 and b = 29, what is the private exponent a ? Explain what you do and include the partial results of your calculations. HINT: Use the extended Euclidean algorithm. It will only take a few steps, you can do it by hand. 3. (8 points) Charlie doesn’t like the Extended Euclidean Algorithm, but instead has found an algorithm A to calculate the multiplicative inverse of a modulo m in time O ( m ). Can Charlie use a secure RSA public key and use
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: A to calculate inverses? Explain. 4. (8 points) Suppose that Oscar intercepts a message encoded with the RSA encryption, but he does not know the private key. Assume that n = p · q is the public modulus, and b is the public exponent. Suppose someone tells Oscar, that one of the plaintext blocks has a common factor with n . Explain how Oscar can use this information to decrypt the message. 5. (8 points) Prove that ( x b mod n ) a mod n = x ab mod n HINT: The binomial theorem will be helpful: ( s + t ) k = k X i =0 k i ! s i t k-i EXTRA CREDIT (10 points) Describe the most eﬃcient algorithm you can for computing x b mod n , if b is an arbitrary integer. Your solution should be eﬃcient enough to use for RSA encoding. HINT: We have already seen in class how to do this if b is of the form b = 2 l for some integer l ....
View Full Document

{[ snackBarMessage ]}

### What students are saying

• As a current student on this bumpy collegiate pathway, I stumbled upon Course Hero, where I can find study resources for nearly all my courses, get online help from tutors 24/7, and even share my old projects, papers, and lecture notes with other students.

Kiran Temple University Fox School of Business ‘17, Course Hero Intern

• I cannot even describe how much Course Hero helped me this summer. It’s truly become something I can always rely on and help me. In the end, I was not only able to survive summer classes, but I was able to thrive thanks to Course Hero.

Dana University of Pennsylvania ‘17, Course Hero Intern

• The ability to access any university’s resources through Course Hero proved invaluable in my case. I was behind on Tulane coursework and actually used UCLA’s materials to help me move forward and get everything together on time.

Jill Tulane University ‘16, Course Hero Intern