Warning: foreach() argument must be of type array|object, bool given in /var/www/html/web/app/themes/studypress-core-theme/template-parts/header/mobile-offcanvas.php on line 20

Let \(x, a_{0}, \ldots, a_{\ell-1} \in R,\) and let \(k\) be an integer with \(0

Short Answer

Expert verified
Question: Given a collection of elements \(x, a_0, \ldots, a_{\ell - 1} \in R\) and an integer \(k\) with \(0 < k \leq \ell\), show how we can compute the values \(g_0(x), \ldots, g_{\ell - k}(x)\), where \(g_i(X) = \sum_{j=i}^{i + k - 1} a_j X^j \in R[X]\), using \(O(\ell)\) operations in the ring \(R\). Answer: We can compute the values \(g_0(x), \ldots, g_{\ell - k}(x)\) using the following algorithm: 1. Initialize \(g_0(x) = a_0 + a_1 x + a_2 x^2 + \cdots + a_{k - 1} x^{k-1}\). 2. For each \(i\) from \(0\) to \(\ell - k - 1\), compute \(g_{i+1}(x) = g_i(x) - a_i + a_{i + k} x^{k-1}\). 3. Output the computed functions \(g_0(x), g_1(x), \ldots, g_{\ell - k}(x)\). This algorithm requires \(O(\ell)\) operations in the ring \(R\), satisfying the constraint.

Step by step solution

01

Define the first function \(g_0(x)\)

We start by writing down the first function in the sequence \(g_0(X) = \sum_{j=0}^{k - 1} a_j X^j\). By evaluating this function at \(x\in R\), we have: $$ g_0(x) = a_0 + a_1 x + a_2 x^2 + \cdots + a_{k - 1} x^{k-1}. $$
02

Derive a general formula for computing the functions \(g_i(x)\)

Now, we need a general formula to compute the rest of the functions \(g_1(x), \ldots, g_{\ell - k}(x)\) in the sequence. To do this, we notice that each function can be computed from the previous one by simply updating two coefficients: $$ g_{i+1}(x) = g_i(x) - a_i + a_{i + k} x^{k-1}. $$ Using this formula, we can compute all the functions in the sequence iteratively.
03

Implement the algorithm for computing the functions \(g_i(x)\)

We are now ready to implement the algorithm for computing the functions \(g_0(x), \ldots, g_{\ell - k}(x)\). The algorithm consists of the following steps: 1. Initialize \(g_0(x) = a_0 + a_1 x + a_2 x^2 + \cdots + a_{k - 1} x^{k-1}\). 2. For each \(i\) from \(0\) to \(\ell - k - 1\), compute \(g_{i+1}(x) = g_i(x) - a_i + a_{i + k} x^{k-1}\). 3. Output the computed functions \(g_0(x), g_1(x), \ldots, g_{\ell - k}(x)\).
04

Analyze the complexity of the algorithm

We now analyze the complexity of the algorithm presented in Step 3. First, we initialize \(g_0(x)\), which requires \(k\) operations in total. For each iteration in steps 2-3, we do a constant number of ring operations (addition, subtraction, and multiplication). Since the loop runs for \(\ell - k\) iterations, the total number of operations performed in these steps is \(O(\ell-k)\). Thus, the overall complexity of the algorithm is \(O(\ell)\), as required.

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 Vaia!

Key Concepts

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

Algorithm Complexity
Understanding algorithm complexity is essential to evaluate a program’s efficiency. In this exercise, we need to compute multiple polynomial evaluations efficiently.
The algorithm developed in the solution utilizes a combination of initialization and iterative computation.
The first step involves calculating the function \(g_0(x)\). This requires \(k\) operations, including additions and multiplications. Subsequent calculations of \(g_i(x)\) from \(g_{i+1}(x)\) leverage previous results, ensuring that only a constant number of operations are needed per iteration.
Since this is done for each of the \(\ell-k\) possible functions, the algorithm’s complexity is described as \(O(\ell)\). In algorithm complexity, \(O(...)\) notation helps describe how the time to run an algorithm increases with the size of the input. An \(O(\ell)\) complexity means that the time grows linearly with \(\ell\), making this method efficient for large inputs.
Ring Theory
Ring theory provides the mathematical framework that underpins polynomial computations. A 'ring' in mathematics is a set equipped with two binary operations that generally resemble addition and multiplication. Each element within the ring retains closure, associativity, and distributive properties.

In this context, polynomial evaluation occurs within the ring \(R[X]\), where \(R[X]\) denotes the ring of polynomials over \(R\). The coefficients \(a_i\) belong to \(R\), and the variable \(X\) forms part of the polynomial operations. These operations obey the rules of ring arithmetic, ensuring reliable and consistent computation.
By understanding the ring structure, we can appreciate how polynomial operations and transformations maintain the essential properties needed for consistent results. This structure provides the reliability necessary for using algorithms to solve complex mathematical problems, like polynomial evaluations.
Iterative Methods
Iterative methods are key for efficiently computing sequences of values without recalculating unnecessary steps. The exercise demonstrates an iterative approach to compute the sequence of polynomial evaluations \(g_0(x), g_1(x), \ldots, g_{\ell-k}(x)\).

The process begins with calculating \(g_0(x)\), which serves as a basis. Subsequent functions \(g_i(x)\) are computed using the simple iterative formula:
  • \(g_{i+1}(x) = g_i(x) - a_i + a_{i+k} x^{k-1}\).
Instead of recalculating each polynomial from scratch, each function builds upon the previous one. This method dramatically reduces the computational effort, improving efficiency. Iterative methods often reveal elegant solutions to problems by breaking them into smaller, more manageable steps.
In practice, this approach not only saves time but also resources, which can be crucial in applications that involve large-scale data or computations.

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

Given polynomials \(g, h \in R[X],\) show how to compute their composition \(g(h) \in R[X]\) using \(O\left(\operatorname{len}(g)^{2} \operatorname{len}(h)^{2}\right)\) operations in \(R\).

This exercise develops a fast algorithm, called the fast Fourier transform or FFT, for computing the function \(\mathcal{E}_{n, \omega} .\) This is a recursive algorithm 17.6 Faster polynomial arithmetic (*) FFT \(\left(n, \omega ; a_{0}, \ldots, a_{2^{n}-1}\right)\) that takes as input an integer \(n \geq 0,\) a primitive \(2^{n}\) th root of unity \(\omega \in R,\) and elements \(a_{0}, \ldots, a_{2^{n}-1} \in R,\) and runs as follows: if \(n=0\) then return \(a_{0}\) else $$ \begin{array}{l} \left(\alpha_{0}, \ldots, \alpha_{2^{n-1}-1}\right) \leftarrow \mathrm{FFT}\left(n-1, \omega^{2} ; a_{0}, a_{2}, \ldots, a_{2^{n}-2}\right) \\\ \left(\beta_{0}, \ldots, \beta_{2^{n-1}-1}\right) \leftarrow \mathrm{FFT}\left(n-1, \omega^{2} ; a_{1}, a_{3}, \ldots, a_{2^{n}-1}\right) \\\ \text { for } i \leftarrow 0 \text { to } 2^{n-1}-1 \mathrm{do} \\ \quad \gamma_{i} \leftarrow \alpha_{i}+\beta_{i} \omega^{i}, \gamma_{i+2^{n-1}} \leftarrow \alpha_{i}-\beta_{i} \omega^{i} \\ \text { return }\left(\gamma_{0}, \ldots, \gamma_{2^{n}-1}\right) \end{array} $$

Let \(g, h \in R[X, Y]\) with \(g=\sum_{i=0}^{m-1} g_{i} Y^{i}\) and \(h=\sum_{i=0}^{m-1} h_{i} Y^{i},\) where each \(g_{i}\) and \(h_{i}\) is a polynomial in \(X\) of degree less than \(k .\) The product \(f:=g h \in R[X, Y]\) may be written \(f=\sum_{i=0}^{2 m-2} f_{i} Y^{i},\) where each \(f_{i}\) is a polynomial in \(X .\) Show how to compute \(f,\) given \(g\) and \(h,\) using \(O(M(k m))\) operations in \(R\). Hint: for an appropriately chosen integer \(t>0,\) first convert \(g, h\) to \(\tilde{g}, \tilde{h} \in R[X],\) where \(\tilde{g}:=\sum_{i=0}^{m-1} g_{i} X^{t i}\) and \(\tilde{h}:=\sum_{i=0}^{m-1} h_{i} X^{t i} ;\) next, compute \(\tilde{f}:=\tilde{g} \tilde{h} \in R[X] ;\) finally, "read off" the \(f_{i}\) 's from the coefficients of \(\tilde{f}\).

Let \(f \in R[X]\) be a polynomial of degree \(\ell>0\) with \(\operatorname{lc}(f) \in R^{*},\) and let \(E:=R[X] /(f) .\) Suppose that in addition to \(f,\) we are given a polynomial \(g \in R[X]\) of degree less than \(k\) and an element \(\alpha \in E,\) and we want to compute \(g(\alpha) \in E .\) This is called the modular composition problem. (a) Show that a straightforward application of Horner's rule yields an algorithm that uses \(O\left(k \ell^{2}\right)\) operations in \(R,\) and requires space for storing \(O(\ell)\) elements of \(R\). (b) Show how to compute \(g(\alpha)\) using just \(O\left(k \ell+k^{1 / 2} \ell^{2}\right)\) operations in \(R,\) at the expense of requiring space for storing \(O\left(k^{1 / 2} \ell\right)\) elements of \(R\). Hint: first compute a table of powers \(1, \alpha, \ldots, \alpha^{m},\) for \(m \approx k^{1 / 2}\)

Suppose you are given three polynomials \(f, g, h \in \mathbb{Z}_{p}[X],\) where \(p\) is a large prime, in particular, \(p \geq 2 \operatorname{deg}(g) \operatorname{deg}(h) .\) Design an efficient probabilistic algorithm that tests if \(f=g(h)\) (i.e., if \(f\) equals \(g\) composed with \(h\) ). Your algorithm should have the following properties: if \(f=g(h),\) it \(\begin{array}{ll}468 & \text { Polynomial arithmetic and applications }\end{array}\) should always output "true," and otherwise, it should output "false" with probability at least \(0.999 .\) The expected running time of your algorithm should be \(O\left((\operatorname{len}(f)+\operatorname{len}(g)+\operatorname{len}(h)) \operatorname{len}(p)^{2}\right)\) EXERCISE 17.6. Let \(x, a_{0}, \ldots, a_{\ell-1} \in R,\) and let \(k\) be an integer with \(0

See all solutions

Recommended explanations on Math Textbooks

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.

Sign-up for free