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Output: A value x satisfying =. m ← Ceiling(√ n) For all j where 0 ≤ j < m: Compute α j and store the pair (j, α j) in a table. (See § In practice) Compute α −m. γ ← β. (set γ = β) For all i where 0 ≤ i < m: Check to see if γ is the second component (α j) of any pair in the table. If so, return im + j.
For k = 0, the kth power is the identity: b 0 = 1. Let a also be an element of G. An integer k that solves the equation b k = a is termed a discrete logarithm (or simply logarithm, in this context) of a to the base b. One writes k = log b a.
Pollard gives the time complexity of the algorithm as (), using a probabilistic argument based on the assumption that acts pseudorandomly. Since , can be represented using () bits, this is exponential in the problem size (though still a significant improvement over the trivial brute-force algorithm that takes time ()).
Let be a cyclic group of order , and given ,, and a partition =, let : be the map = {and define maps : and : by (,) = {() + (,) = {+ ()input: a: a generator of G b: an element of G output: An integer x such that a x = b, or failure Initialise i ← 0, a 0 ← 0, b 0 ← 0, x 0 ← 1 ∈ G loop i ← i + 1 x i ← f(x i−1), a i ← g(x i−1, a i−1), b i ← h(x i−1, b i−1) x 2i−1 ← ...
In mathematics, the logarithm of a number is the exponent by which another fixed value, the base, must be raised to produce that number. For example, the logarithm of 1000 to base 10 is 3, because 1000 is 10 to the 3 rd power: 1000 = 10 3 = 10 × 10 × 10. More generally, if x = b y, then y is the logarithm of x to base b, written log b x, so ...
Here’s another problem that’s very easy to write, but hard to solve. All you need to recall is the definition of rational numbers. Rational numbers can be written in the form p/q, where p and ...
The second stage solves the system of linear equations to compute the discrete logs of the factor base. A system of hundreds of thousands or millions of equations is a significant computation requiring large amounts of memory, and it is not embarrassingly parallel, so a supercomputer is typically used. This was considered a minor step compared ...
Clearly, a #P problem must be at least as hard as the corresponding NP problem, since a count of solutions immediately tells if at least one solution exists, if the count is greater than zero. Surprisingly, some #P problems that are believed to be difficult correspond to easy (for example linear-time) P problems. [ 18 ]