The paper focuses on the deterministic complexity of factoring polynomials over finite fields assuming the extended Riemann hypothesis (ERH). By the works of and , the general problem reduces deterministically in polynomial time to finding a proper factor of any squarefree and completely splitting
Factoring Polynomials and the Knapsack Problem
β Scribed by Mark van Hoeij
- Publisher
- Elsevier Science
- Year
- 2002
- Tongue
- English
- Weight
- 222 KB
- Volume
- 95
- Category
- Article
- ISSN
- 0022-314X
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β¦ Synopsis
For several decades the standard algorithm for factoring polynomials f with rational coefficients has been the Berlekamp-Zassenhaus algorithm. The complexity of this algorithm depends exponentially on n, where n is the number of modular factors of f . This exponential time complexity is due to a combinatorial problem: the problem of choosing the right subsets of these n factors. In this paper, this combinatorial problem is reduced to a type of Knapsack problem that can be solved with lattice reduction algorithms. The result is a practical algorithm that can factor polynomials that are far out of reach for previous algorithms. The presented solution to the combinatorial problem is different from previous lattice-based factorizers; these algorithms avoided the combinatorial problem by solving the entire factorization problem with lattice reduction. This led to lattices of large dimension and coefficients, and thus poor performance. This is why lattice-based algorithms, despite their polynomial time complexity, did not replace Berlekamp-Zassenhaus as the standard method. That is now changing; new versions of computer algebra systems such as Maple, Magma, NTL and Pari have already switched to the algorithm presented here.
π SIMILAR VOLUMES
The tree center problems are designed to find a subtree minimizing the maximum distance from any vertex. This paper shows that these problems in a tree network are related to the bottleneck knapsack problems and presents linear-time algorithms for the tree center problems by using the relation.