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Given a polynomial \(g \in R[X]\) and an element \(x \in R,\) a particularly elegant and efficient way of computing \(g(x)\) is called Horner's rule. Suppose \(g=\sum_{i=0}^{k-1} a_{i} X^{i},\) where \(k \geq 0\) and \(a_{i} \in R\) for \(i=0, \ldots, k-1 .\) Horner's rule computes \(g(x)\) as follows: $$ y \leftarrow 0_{R} $$ for \(i \leftarrow k-1\) down to 0 do $$ y \leftarrow y x+a_{i} $$ output \(y\) Show that this algorithm correctly computes \(g(x)\) using \(k\) multiplications in \(R\) and \(k\) additions in \(R\).

Short Answer

Expert verified
Question: Explain how Horner's rule computes the value of a polynomial g(x) using k multiplications and k additions. Answer: Horner's rule is an efficient algorithm to compute the value of a polynomial g(x) using k multiplications and k additions. The algorithm starts by initializing the result, y, with the zero element of the ring R. Then, it iterates from i = k-1 down to i = 0, updating y by multiplying the current value of y by x and adding the corresponding coefficient a_i of the polynomial g in each step. After the iteration, the final value of y is the result of the polynomial evaluation at x. The algorithm requires k multiplications and k additions, as each iteration involves one multiplication and one addition, and the algorithm iterates k times in total. The correctness of Horner's rule can be proven using mathematical induction.

Step by step solution

01

Horner's Rule Algorithm Initialization

According to Horner's rule, we initialize the result, y, with the zero element of the ring R: $$ y \leftarrow 0_{R} $$ This initialization is the starting point for the iterative process of evaluating the polynomial.
02

Iterate from k-1 to 0

Now, the algorithm iterates from i = k-1 down to i = 0 with the following expression: $$ y \leftarrow yx + a_{i} $$ This step updates y by multiplying the current value of y by x, and then adding \({a_i}\), where \({a_i}\) are the coefficients of the polynomial g.
03

Output the result

Upon completion of the iteration, the final value of y is the result of the polynomial evaluation at x: $$ g(x) = y $$
04

Efficiency of Horner's Rule

To prove that Horner's rule requires k multiplications and k additions, let's count the operations in the algorithm. In each iteration (Step 2), we perform one multiplication (y times x) and one addition (adding \({a_i}\)). The algorithm iterates k times (from i = k-1 to i = 0), so there are k multiplications and k additions in total.
05

Correctness of Horner's Rule

To prove the correctness of the algorithm, we will use mathematical induction. Base Case (i = k-1): $$ y = 0_{R}x + a_{k-1} = a_{k-1}x^{0} $$ Assume correctness for i = j: $$ y = a_{j+1}x^{0} + a_{j}x^{1} + \dots + a_{k-2}x^{k-j-1} + a_{k-1}x^{k-j} $$ Consider the next iteration (i = j-1): $$ y \leftarrow yx + a_{j-1} = (a_{j+1}x^{0} + a_{j}x^{1} + \dots + a_{k-2}x^{k-j-1} + a_{k-1}x^{k-j})x + a_{j-1} $$ $$ y = a_{j+1}x^{1} + a_{j}x^{2} + \dots + a_{k-2}x^{k-j} + a_{k-1}x^{k-j+1} + a_{j-1}x^{0} $$ The induction step shows that Horner's rule correctly computes the polynomial g(x) for every iteration. Therefore, the algorithm is correct. In conclusion, Horner's rule correctly computes the polynomial g(x) using k multiplications and k additions.

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Key Concepts

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

Polynomial Evaluation
Evaluating a polynomial efficiently is crucial in mathematics and computer science. Horner's rule is an elegant method for polynomial evaluation that is particularly beneficial due to its efficiency.
This method allows us to evaluate a polynomial such as \( g(x) = a_{0} + a_{1}x + a_{2}x^2 + \cdots + a_{k-1}x^{k-1} \) by restructuring it to minimize the computational cost. Using Horner's rule, the polynomial can be reformulated as:
  • \( g(x) = (((a_{k-1}x + a_{k-2})x + a_{k-3})x + \cdots + a_{1})x + a_0 \)
This format uses a sequence of nested operations, which reduces the number of multiplications needed to just \( k \), where \( k \) is the degree of the polynomial.
With this form, Horner's rule simplifies the steps:
  • Start with the highest degree coefficient and work to the constant term (\( a_0 \)).
  • Combine multiplication and addition iteratively, keeping calculations at a minimum.
This decrease in the number of operations speeds up the computation, making Horner's rule an extremely useful tool.
Mathematical Induction
Mathematical induction is a fundamental proof technique used to establish the truth of an infinite number of cases. It is often employed to prove statements involving sequences or algorithms that recur over numbers. In our case, mathematical induction helps demonstrate the correctness of Horner's rule.
The inductive proof comprises two parts:
  • **Base Case:** We check the simplest case. For Horner's rule, when \( i = k-1 \), \( y = a_{k-1} \times x^0 = a_{k-1} \).
  • **Induction Step:** Assuming the rule works for \( i = j \), we show it also holds for the next step (\( i = j-1 \)). If at step \( j \), \( y \) represents a valid sum of terms, multiplying by \( x \) and adding \( a_{j-1} \) extends the correctness to the next step.
By successfully demonstrating both parts, we ensure that the method holds for any polynomial degree, confirming that every step in Horner's rule correctly computes required polynomial terms.
Algorithm Efficiency
Efficiency is a key feature of any algorithm, measuring how effectively it uses time and resources. In polynomial evaluation, two critical operations are multiplications and additions.
Horner's rule is prized for its efficiency. Generally, evaluating a polynomial through traditional methods would involve \( k(k+1)/2 \) multiplications and additions. With Horner's method, we reduce this to just \( k \) multiplications and \( k \) additions, as each iteration processes one term of the polynomial.
  • **Multiplications:** Only \( k \), as each step involves one multiplication of the current result by \( x \).
  • **Additions:** Also \( k \), since each iteration adds another coefficient \( a_i \).
This reduction is significant for applications requiring real-time processing or handling of high-degree polynomials. Efficient algorithms like Horner's rule allow complex calculations to be executed quickly, conserving computational resources. This explains why it's favored in programming and engineering scenarios.

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Most popular questions from this chapter

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 \(m \approx k^{1 / 2}\)

This exercise uses the FFT to develop a linear-time algorithm for integer multiplication; however, a rigorous analysis depends on an unproven conjecture (which follows from a generalization of the Riemann hypothesis). Suppose we want to multiply two positive integers \(a\) and \(b\), each of length at most \(\ell\) (represented internally using the data structure described in \(\S 3.3\) ). Throughout this exercise, assume that all computations are done on a RAM, and that arithmetic on integers of length \(O(\operatorname{len}(\ell))\) takes time \(O(1) .\) Let \(k\) be an integer parameter with \(k=\Theta(\operatorname{len}(\ell)),\) and let \(m:=\lceil\ell / k\rceil .\) We may write \(a=\sum_{i=0}^{m-1} a_{i} 2^{k i}\) and \(b=\sum_{i=0}^{m-1} b_{i} 2^{k i},\) where \(0 \leq a_{i}<2^{k}\) and \(0 \leq b_{i}<2^{k} .\) Let \(n\) be the integer determined by \(2 m \leq 2^{n}<4 m\).

State and re-work the analog of Exercise 3.52 for \(R[X],\) assuming \(2_{R} \in R^{*}\)

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

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 $$ \begin{array}{l} \left(\alpha_{0}, \ldots, \alpha_{2^{n-1}-1}\right) \leftarrow \mathrm{FFT}\left(n-1, \omega^{2} ; a_{0}, a_{2}, \ldots, a_{2^{n}-2}\right) \\\ \left(\beta_{0}, \ldots, \beta_{2^{n-1}-1}\right) \leftarrow \mathrm{FFT}\left(n-1, \omega^{2} ; a_{1}, a_{3}, \ldots, a_{2^{n}-1}\right) \\\ \text { for } i \leftarrow 0 \text { to } 2^{n-1}-1 \mathrm{do} \\ \quad \gamma_{i} \leftarrow \alpha_{i}+\beta_{i} \omega^{i}, \gamma_{i+2^{n-1}} \leftarrow \alpha_{i}-\beta_{i} \omega^{i} \\ \text { return }\left(\gamma_{0}, \ldots, \gamma_{2^{n}-1}\right) \end{array} $$

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