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See detailSkalpel: A Constraint-Based Type Error Slicer for Standard ML
Rahli, Vincent UL; Wells, Joe; Pirie, John et al

in Journal of Symbolic Computation (2016)

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See detailCubic Sieve Congruence of the Discrete Logarithm Problem, and fractional part sequences
Venkatesh, Srinivas Vivek UL; C. E. Veni Madhavan

in Journal of Symbolic Computation (2014), 64

The Cubic Sieve Method for solving the Discrete Logarithm Problem in prime fields requires a nontrivial solution to the Cubic Sieve Congruence (CSC) $ x^3 \equiv y^2 z (mod p) $, where 'p' is a given ... [more ▼]

The Cubic Sieve Method for solving the Discrete Logarithm Problem in prime fields requires a nontrivial solution to the Cubic Sieve Congruence (CSC) $ x^3 \equiv y^2 z (mod p) $, where 'p' is a given prime number. A nontrivial solution must also satisfy $ x^3 \neq y^2 z $ and $ 1 <= x,y,z < p^a $, where 'a' is a given real number such that $ 1/3 < a <= 1/2 $. The CSC problem is to find an efficient algorithm to obtain a nontrivial solution to CSC. CSC can be parametrized as $ x \equiv v^2 z (mod p) $ and $ y \equiv v^3 z (mod p) $. In this paper, we give a deterministic polynomial-time ($ O(ln^3 p) $ bit-operations) algorithm to determine, for a given 'v', a nontrivial solution to CSC, if one exists. Previously it took $ õ(p^a) $ time in the worst case to determine this. We relate the CSC problem to the gap problem of fractional part sequences, where we need to determine the non-negative integers 'N' satisfying the fractional part inequality $ {\theta N} < \phi $ ($ \theta $ and $ \phi $ are given real numbers). The correspondence between the CSC problem and the gap problem is that determining the parameter 'z' in the former problem corresponds to determining 'N' in the latter problem. We also show in the $ a = 1/2 $ case of CSC that for a certain class of primes the CSC problem can be solved deterministically in $ õ(p^{1/3}) $ time compared to the previous best of $ õ(p^{1/2}) $. It is empirically observed that about one out of three primes is covered by the above class. [less ▲]

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