The following code fragment implements Horner's rule for evaluating a polynomial
$$ \begin{align} P(x) &= \sum_{k=0}^n a_kx^k = \\ &= a_0 + x(a_1 + x(a_2 + \cdots + x(a_{n-1} + xa_n) \cdots)) \end{align} $$
given the coefficients $a_0, a_1, \ldots ,a_n$ and a value for $x$:
y = 0 for i = n downto 0 y = aᵢ + x·y
In terms of $\Theta$-notation, what is the running time of this code fragment for Horner's rule?
Write pseudocode to implement the naive polynomial-evaluation algorithm that computes each term of the polynomial from scratch. What is the running time of this algorithm? How does it compare to Horner's rule?
Consider the following loop invariant:
At the start of each iteration of the for loop of lines 2-3,
$$ y = \sum_{k=0}^{n-(i+1)}a_{k+i+1}x^k $$
Interpret a summation with no terms as equaling 0. Following the structure of the loop invariant proof presented in this chapter, use this loop invariant to show that, at termination, $y = \sum_{k=0}^na_kx^k$.
Conclude by arguing that the given code fragment correctly evaluates a polynomial characterized by the coefficients $a_0,a_1,\ldots,a_n$.
Obviously $\Theta(n)$.
We assume that we have no exponentiation in the language. Thus:
y = 0
for i = 0 to n
m = 1
for k = 1 to n
m = m·x
y = y + aᵢ·m
The running time is $\Theta(n^2)$, because of the nested loop. It is, obviosly, slower.
Initialization: It is pretty trivial, since the summation has no terms, which implies $y = 0$.
Maintenance: By using the loop invariant, in the end of the $i$-th iteration, we have:
$$ \begin{align} y &= a_i + x\sum_{k=0}^{n-(i+1)}a_{k+i+1}x^k = a_ix^0 + \sum_{k=0}^{n-i-1}a_{k+i+1}x^{k+1} = \\ &= a_ix^0 \sum_{k=1}^{n-i}a_{k+i}x^k = \sum_{k=0}^{n-i}a_{k+i}x^k \end{align} $$
Termination: The loop terminates at $i = -1$. If we substitute, we get:
$$ y = \sum_{k=0}^{n-i-1}a_{k+i+1}x^k = \sum_{k=0}^na_kx^k $$
It should be fairly obvious, but the invariant of the loop is a sum that equals a polynomial with the given coefficients.