/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 2 (Lenstra 1990) Consider the foll... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 19: Problem 2

(Lenstra 1990) Consider the following special polynomial factorization task: input is a prime \(p\) and \(f \in F_{p}|x|\) of degree \(n\) and dividing \(x^{p}-x_{1}\), so that all monic irreducible factors of \(f\) in \(\mathrm{F}_{p}|x|\) are linear and distinct. Adapt the Pollard and Strassen method to find a deterministic algorithm for factoring \(f\) with \(O^{*}(n \sqrt{p})\) operations in \(\mathrm{E}_{p}\) if \(p^{2}>n\).

Short Answer

Expert verified
Use roots of \( x^p - x \) and iterate across linear factors \( (x - a) \) to factor \( f \).

Step by step solution

01

Understand the Setup

We have a polynomial \( f \) of degree \( n \) that divides \( x^p - x_1 \). This means each monic irreducible factor of \( f \) is a linear factor, which implies that \( f \) can be expressed as a product of degree-one polynomials. Since \( f \) divides \( x^p - x \), the roots of \( f \) must be among the roots of \( x^p - x \).
02

Recall Pollard-Strassen Method

The Pollard and Strassen method for polynomial factorization involves using Frobenius and powers of a random element to build the factorization cycle. Specifically, with \( f(x) \) dividing \( x^p - x \), the roots of \( f \) are exactly the distinct roots of \( x^p - x \).
03

Set Up the Factorization

We create a series of polynomials using the roots of \( x^p - x \). We use Fermat's Little Theorem to decipher states of these roots: \( x^p \equiv x \pmod{f(x)} \). Simultaneously, we construct linear factors iteratively evaluating \( x \) values in \( F_p \).
04

Implement Iterative Algorithm

We must iterate through possible linear factors \( (x - a) \) where \( a \/in F_p \). For each root \( a \), test \( f(a) = 0 \) to confirm it as a root. Use efficient root-finding methods to lower computational cost from testing every linear polynomial.
05

Complexity and Conditions

We verify that this aligns with the \( O^*(n \sqrt{p}) \) operations due to the constraint \( p^2 > n \). This access to enough elements in \( F_p \) ensures convergence speed thereby meeting desired complexity.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91影视!

Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Pollard-Strassen method
The Pollard-Strassen method is a fascinating technique for factoring polynomials, especially useful in situations where working in finite fields like \( \mathrm{F}_p \) is necessary. Originally proposed for integer factorization, it has been adapted for use in polynomial factorization. The method leverages the mathematical properties of cyclic groups and polycyclic groups. In the context of factoring polynomials like \( f \), which divide expressions such as \( x^p - x \), Pollard-Strassen utilizes algorithms based on Frobenius maps and cyclic testing.
  • Start with a known polynomial \( f \) and relate it to cycles formed by the iterates of random elements from \( \mathrm{F}_p \).
  • These cycles help in revealing the linear factors by grouping similar elements.
This technique is useful in trying to manage the complexity of computations, especially aimed at reducing operations within large finite fields settings. Instead of brute-force checking, the method's genius is efficiently navigating through potential roots, extracting just those that contribute to a complete factorization.
Fermat's Little Theorem
Fermat's Little Theorem is a wonderful mathematical tool that simplifies computations in modular arithmetic, particularly in finite fields like \( \mathrm{F}_p \). This theorem states that if \( p \) is a prime and \( a \) is an integer not divisible by \( p \), then \( a^{p-1} \equiv 1 \pmod{p} \). In polynomial factorization, particularly involving expressions such as \( x^p - x \), Fermat's Little Theorem is pivotal.
  • It guarantees that \( x^p \equiv x \pmod{f(x)} \) for all \( x \) in \( \mathrm{F}_p \), simplifying many expressions.
  • This implication directly assists in confirming if a particular \( x \) value is a root of \( f \).
This provides a reliability check across the calculations, ensuring that the polynomial division holds true within finite fields, considerably speeding up the factoring process.
Deterministic algorithm
A deterministic algorithm assures a specific output given a particular input, always following the same sequence of computational steps. This certainty is crucial when factoring polynomials in contexts requiring stringent accuracy. In adapting the Pollard-Strassen method for polynomial factorization, the deterministic approach means that the outcome鈥攆actoring \( f \)鈥攊s predicable and reproducible.
  • In the exercise, such an algorithm involves breaking down the polynomial into its linear factors.
  • Every step, from testing linear polynomials to verifying roots, follows a definitive road without randomness.
This is particularly important given the condition \( p^2>n \), which allows the algorithm to operate efficiently within a finite but large field, ensuring that every possible factor is considered.
Finite fields
Finite fields, denoted as \( \mathrm{F}_p \), where \( p \) is a prime number, play a vital role in many areas of mathematics and cryptography. These fields contain a finite number of elements, exactly \( p \), forming a complete system for arithmetic operations such as addition, subtraction, multiplication, and division (except by zero).
  • Polynomials over finite fields, particularly those like \( f \) dividing \( x^p - x \), follow unique properties due to the field's modular arithmetic constraints.
  • Due to these properties, tests for factorization can exploit the cyclical nature of multiplicative groups within the field.
When factoring polynomials, especially in contexts needing deterministic methods like in the exercise, finite fields offer a structured environment where operations and solutions are bound by the field's fixed size, influencing the complexity and feasibility of potential solutions.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Three integers \((x, y, z) \in \mathbb{N}^{3}\) form a Pythagorean triple if \(x^{2}+y^{2}=z^{2}\), and a primitive Py. thagorean triple if furthermore gcd \((x, y, z)=1\). (i) Show that any Pythagorean triple is of the form \((\lambda x, \lambda y, \lambda z)\) for a primitive Pythagorean triple \((x, y, z)\) and some \(\lambda \in \mathbb{N}\), (ii) Let \(s, t \in N\) be coprime with \(s>t\) and \(s t\) even. Show that \(\left(s^{2}-t^{2}, 2 s t, s^{2}+t^{2}\right)\) is a primitive Pythagorean triple, (iii) Let \((x, y, z)\) be a primitive Pythagorean triple. Show that \(z\) is odd and either \(x\) or \(y\) is odd, but not both. Hint: Calculate modulo 4. Assume that \(x\) is odd. Prove that \((z+x) / 2\) and \((z-x) / 2\) are coprime squares, not both odd, and conclude that \((x, y, z)\) is of the form as in (ii). (iv) Use (ii) and (iii) to generate all primitive Pythagorean triples \((x, y, z)\) with \(z \leq 100\).

Fuctor the integer \(N=23802996783967\) using Pollard's \(\rho\) method, and also with the Pollard and Strassen method.

Let \(E\) be an elliptic curve and \(P, Q \in E\). Explain why \(P+Q=S\), where \(S\) is the third intersection point of \(E\) with the line through \(P\) and \(Q\) (sec Figure 19.6), is not a group operation.

(Guy 1975) Let \(x_{0}=2\) and \(x_{i}=x_{i-1}^{2}+1\) for \(i \geq 1\). For \(p \in N_{\text {, we let }} e(p)=\min \left\\{i \in N_{\geq 1}\right.\) : \(\left.x_{i} \equiv x_{2 i} \bmod p\right\\}\). (i) Calculate \(e(p)\) for the primes \(p \leq 11\). (ii) Calculate \(e(p)\) for the primes \(p \leq 10^{6}\). You should find \(e(p) \leq 3680\) for all these \(p\). (Guy (1975) notes that \(e(p)\) seems to grow like \((p \ln p)^{1 / 2}\).) (iii) Let \(N\) be a number to be factored, run Pollard's \(\rho\) method on it with initial value \(x_{0}=2\), and assume that \(\operatorname{gcd}\left(x_{i}-x_{2} . N\right)=1\) for \(0 \leq i \leq k\). Show that \(e(p)>k\) for all prime divisors \(p\) of \(N\). (iv) Conclude that if the ged in (iii) is trivial for 3680 steps, then \(N\) has no factor up to \(10^{6}\).

Check that the curve \(E=\left\\{(x, y) \in \mathbb{F}_{7}^{2}: y^{2}=x^{3}+x+3\right\\}\) over E7 is nonsingular. Compute all points on it, and verify that it is cyclic and generated by \((4.1)\).
See all solutions

Recommended explanations on Computer Science Textbooks

Cloud Services

Read Explanation

Big Data

Read Explanation

Blockchain Technology

Read Explanation

Computer Organisation and Architecture

Read Explanation

Problem Solving Techniques

Read Explanation

Theory of Computation

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.