Problem 2
Question
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 \(R\).
Step-by-Step Solution
Verified Answer
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.
1Step 1: 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.
2Step 2: 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.
3Step 3: Output the result
Upon completion of the iteration, the final value of y is the result of the polynomial evaluation at x:
$$
g(x) = y
$$
4Step 4: 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.
5Step 5: 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.
Key Concepts
Polynomial EvaluationMathematical InductionAlgorithm Efficiency
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:
With this form, Horner's rule simplifies the steps:
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:
- \( g(x) = (((a_{k-1}x + a_{k-2})x + a_{k-3})x + \cdots + a_{1})x + a_0 \)
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.
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:
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.
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.
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 \( a_i \).
Other exercises in this chapter
Problem 3
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 \(f,\) w
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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}\
View solution Problem 5
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) \operat
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