This is implemented using a uniformly random

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Unformatted text preview: 2013, Midterm #2 4 Name: SID: 3. (15 points) Secret Sharing Alice is part of a secret sharing scheme in which n = 10 people have shares of a secret (which you may think of as a random number modulo 11) such that any k = 3 of them can reconstruct the secret. This is implemented using a uniformly random polynomial over GF (11) of degree (at most) 2. Suppose Alice’s share of the secret, P(5), is equal to 2. Suppose she happens to learn that Bob’s share of the secret, which is P(4), is not 10. For reference: Property 1: Polynomial of degree d has at most d roots. Property 2: There is a unique polynomial of degree d that goes through d + 1 points. Note: Please state explicitly where you use properties 1 and/or 2 in your solution (a) (5 points) Given her information, what is the probability that the secret P(0) is equal to 3? Justify your answer (1 point for the correct answer without proper justification). Solution: Assume Alice has both her share of the secret and Bob’s share. By Property 2, a polynomial of degree 2 is uniquely defined by 3 points. Since Alice only has 2 points, the secret P(0) can still take on any value. Since the polynomial is randomly chose, the probability that 1 the secret P(0) is equal to 3 is 11 . Since this is true for every possible value of Bob’s share...
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This document was uploaded on 01/28/2014.

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