Lecture 10: Mental Poker

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Lecture 10: Mental Poker. Wayne Patterson SYCS 654 Spring 2010. Multi-Party Computation. The mental poker protocol is a kind of multi-party computation, and is based on a public-key encryption scheme with the following commutative composition property:

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Lecture 10: Mental Poker

Wayne Patterson

SYCS 654

Spring 2010

Multi-Party Computation
• The mental poker protocol is a kind of multi-party computation, and is based on a public-key encryption scheme with the following commutative composition property:
• EB(EA(M)) = EA(EB(M));
• where EXdenotes encryption using X's public key. Likewise, we use DXto denote decryption using X's private key.
• So, for example, an RSA-computation, a Diffie-Hellman, or an Elliptic Curve Computation.
Alice, Bob and Carol
• In the mental poker problem, three players, Alice, Bob, and Carol, wish to play poker using networked communication (e.g., email). The dealer for this hand is assumed to be Alice, but Bob and Carol both want to make sure they are dealt a fair hand. A possible protocol for this problem is as follows:
The Deal
• 1. Alice generates 52 distinct random numbers x1, x2, …, x52, where we assume that a sorted listing of these numbers corresponds to a sorted deck of cards. That is, the smallest xi is the Ace of spades, the second smallest is the Ace of clubs, and so on.
• 2. Alice encrypts each of the xi's using her own public key and sends the list of encrypted values, EA(x1), EA(x2), …, EA(x52), to Bob.
Bob’s Hand
• 3. Bob picks five of these encrypted numbers as his hand, say at indices {i1; i2; i3; i4; i5} and computes EB(EA(xi1)), EB(EA(xi2)), EB(EA(xi3)), EB(EA(xi4)), and EB(EA(xi5)), and sends these encrypted numbers to Alice. Bob also sends the remaining unchosen 47 “cards" to Carol.
Carol’s Hand
• 4. Carol picks five of these encrypted numbers as her hand, say at indices {j1; j2; j3; j4; j5} and computes EC(EA(xj1)), EC(EA(xj2)), EC(EA(xj3)), EC(EA(xj4)), and EC(EA(xj5)), and sends these encrypted numbers to Alice. In addition, Carol picks five of the remaining 42 “cards" as Alice's hand, and sends these to Alice.
Alice’s Hand
• 5. For each ik, k = 1, 2, …, 5, Alice computes DA(EB(EA(xik ))), which is the same as EB(xik ); and she sends these values to Bob. She also computes, for each ik, k = 1, 2, …, 5, DA(EC(EA(xjk ))), which is the same as EC(xjk); and she sends these values to Carol.
Who Wins
• 6. Bob and Carol decrypt the values of the cards they picked, Alice reveals the list of numbers x1; x2; : : : ; x52, and all the players learn their hands (and who is the winner).