CSCI 7000 - Cryptography - Spring 2003

Problem Set #1

Due: Feb 4th, 2001 at 11am



1. Assume Pr[E] > 0 and Pr[F] > 0. Prove Pr[E | F] = Pr[E] iff Pr[F | E] = Pr[F].

2. Recall the definition of mutually independent events. In class, we mentioned the defintion of pairwise independence as well, but didn't write it on the board. Here it is again: Let X1, ..., Xn be n events (from the same sample space). We say that the set of events {X1, ... , Xn} is pairwise independent iff for any distinct i,j in {1, ..., n}, Xi is independent of Xj.

Give an example of a collection of events which are pairwise independent, but not mutually independent. Clearly explain why your example exhibits these properties.

3. "The Drunken Professor"
Your professor shows up drunk to class (typical, eh?). And proceeds to hand back n corrected homework assignments to the n students in the class. But he's so intoxicated that instead of doing this correctly, he hands them back to random people (ie, the mapping of papers to students is a uniform random permutation). What is the expected number of students who will receive his own homework back?

(Hint: consider breaking into smaller RVs and using the linearity of expectation to help you.)

4. A Dealer shuffles a deck of cards perfectly (ie, each of the possible 52! orderings of the deck are equally likely). Then he deals 26 of these cards to Alice and the remaining 26 cards to Bob. Eavesdropper Eve is listening in, but cannot see any of the cards.

  1. Alice has a message M of 48 bits. Give a method whereby Alice can send M to Bob by saying something out loud (Eve can hear this!) such that Eve obtains zero information about M.
  2. Now suppose Alice's message M is 49 bits instead of 48 in the preceeding scenario. Prove that no protocol exists which allows Alice to communicate M to Bob in such a way that Eve obtains no information about M. (To say that Eve learns nothing means that the probability of Alice saying S is independent of the message. More precisely, for all messages M1, M2, we have that
        Pr[ Alice says S | M = M1 ]  =  Pr[ Alice says S | M = M2]
    
    where the probability is taken over the shuffles of the cards.)