Download Black-Box Models of Computation in Cryptology by Tibor Jager PDF

By Tibor Jager

Generic team algorithms remedy computational difficulties outlined over algebraic teams with out exploiting homes of a selected illustration of staff components. this is often modeled through treating the gang as a black-box. the truth that a computational challenge can't be solved by means of a pretty constrained category of algorithms could be obvious as aid in the direction of the conjecture that the matter can be demanding within the classical Turing computing device version. furthermore, a reduce complexity certain for yes algorithms is a worthy perception for the quest for cryptanalytic algorithms.

Tibor Jager addresses a number of basic questions bearing on algebraic black-box types of computation: Are the accepted crew version and its editions a cheap abstraction? What are the restrictions of those versions? do we chill out those types to carry them toward the reality?

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3 We say that the (N, , e)-Hensel-RSA problem is (εHRSA ,t)-hard, if Pr[A (N, , e, xe mod N) = xe mod N ] ≤ εHRSA for all algorithms A running in time t. As shown by Catalano et al. in [CNS02], the following two problems reduce to solving Hensel-RSA. • RSA PROBLEM . Let e be an integer such that gcd(e, φ (N)) = 1, where φ denotes Euler’s Phi function. The (N, e)-RSA problem is to compute x ∈ ZN $ on input xe mod N, where x ← ZN is chosen uniformly random. A special case is the (N, N)-RSA problem, where the exponent e equals the modulus N.

Note that the above corollary holds even if the reduction algorithm may query an −1 additional “inverting” Diffie-Hellman oracle returning gab on input ga , gb . Such an oracle does not to seem to be implied by a Diffie-Hellman oracle, if the order φ (N) of the group Z∗N is unknown. Recall here that computing φ (N) on input N is as hard as factoring N [Mil76, RSA78, May04]. However, our results say nothing if factoring N is easy, for instance if N is prime. Thus, there may still exist a generic reduction from the discrete logarithm to the Diffie-Hellman problem which works in any group whose order can be factored efficiently.

The list L is initialized with L1 = 1 and L2 = x = XN + x0 . Note that the variable X is used instead of x1 (x1 is not used throughout the game, but it is useful to have it defined in order to compare Game 2 to Game 1 in the analysis below). 2. Whenever the algorithm asks to perform a computation ◦ ∈ {+, −, ·} on two list elements Li , L j , the oracle computes Lk = Li ◦ L j . Note that each list element Li can be written as a polynomial Li (X) = (ai X + bi )N + ci , where ai , bi ∈ ZN −1 and ci ∈ ZN .

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