CS 259C/Math 250: Elliptic Curves in Cryptography Homework 3 Solutions
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چکیده
Thus, both decryption algorithms correctly recover M (b) Suppose there is an adversary A with advantage at least in the real-or-random game. Then we can construct the following algorithm B which solves the BDDH problem: * On input (P,Q,R, S, γ), send (P,Q,R) to A, where Q is interpreted as QA and R is interpreted as QB. * When A send the challenge plaintext M , respond with (S,Mγ) as the challenge ciphertext. * Output the output of A Since Q = QA = [a]P and R = QB = [b]P for random a, b, the public parameters seen by A are distributed as they would be in the real-or-random game. If γ = ê(P, P ) where S = [c]P , then the challenge ciphertext (S,Mγ) = ([c]P,Mγ) is a correctly distributed encryption of M since c is random. Hence,
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CS 259C/Math 250: Elliptic Curves in Cryptography Homework 3 Solutions
Hence, any party can convert a signature from Sign to one from Sign′ and vice versa. (a) According to the note above, Verify′(pk,M, σ′) = Verify(pk,M, σ) where σ = ([s]P − [e]Q, s). Specifically, Verify′(pk,M, σ) works as follows: 1. Compute R = [s]P − [e]Q 2. Compute e′ = H(M ||R) 3. Accept if R + [e′]Q = [s]P . Note that this condition is equivalent to e = e′ (b) Suppose we have an adversary ...
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تاریخ انتشار 2011