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30.1 Representing polynomials


Multiply the polynomials $A(x) = 7x^3 - x^2 + x - 10$ and $B(x) = 8x^3 - 6x + 3$ using equations $\text{(30.1)}$ and $\text{(30.2)}$.

$$ \begin{array}{rl} & 56x^6 - 8x^5 + (8 - 42)x^4 + (-80 + 6 + 21)x^3 + (-3 - 6)x^2 + (60 + 3)x - 30 \\ = & 56x^6 - 8x^5 - 34x^4 - 53x^3 - 9x^2 + 63x - 30. \end{array} $$


Another way to evaluate a polynomial $A(x)$ of degree-bound $n$ at a given point $x_0$ is to divide $A(x)$ by the polynomial $(x - x_0)$, obtaining a quotient polynomial $q(x)$ of degree-bound $n - 1$ and a remainder $r$, such that

$$A(x) = q(x)(x - x_0) + r.$$

Clearly, $A(x_0) = r$. Show how to compute the remainder $r$ and the coefficients of $q(x)$ in time $\Theta(n)$ from $x_0$ and the coefficients of $A$.

Let $A$ be the matrix with $1$'s on the diagonal, $−x_0$'s on the super diagonal, and $0$'s everywhere else. Let $q$ be the vector $(r, q_0, q_1, \dots, x_{n − 2})$. If $a = (a_0, a_1, \dots, a_{n − 1})$ then we need to solve the matrix equation $Aq = a$ to compute the remainder and coefficients. Since $A$ is tridiagonal, Problem 28-1 (e) tells us how to solve this equation in linear time.


Derive a point-value representation for $A^\text{rev}(x) = \sum_{j = 0}^{n - 1} a_{n - 1 - j}x^j$ from a point-value representation for $A(x) = \sum_{j = 0}^{n - 1} a_jx^j$, assuming that none of the points is $0$.

For each pair of points, $(p, A(p))$, we can compute the pair $(\frac{1}{p}, A^{rev}(\frac{1}{p}))$. To do this, we note that

$$ \begin{aligned} A^{rev}(\frac{1}{p}) & = \sum_{j = 0}^{n - 1} a_{n - 1 - j} (\frac{1}{p})^j \\ & = \sum_{j = 0}^{n - 1} a_j(\frac{1}{p})^{n - 1 - j} \\ & = p^{1 - n} \sum_{j = 0}^{n - 1} a_jp^j \\ & = p^{1 - n}A(p), \end{aligned} $$

since we know what $A(p)$ is, we can compute $A^{rev}(\frac{1}{p})$ of course, we are using the fact that $p \ne 0$ because we are dividing by it. Also, we know that each of these points are distinct, because $\frac{1}{p} = \frac{1}{p'}$ implies that $p = p'$ by cross multiplication. So, since all the $x$ values were distinct in the point value representation of $A$, they will be distinct in this point value representation of $A^{rev}$ that we have made.


Prove that $n$ distinct point-value pairs are necessary to uniquely specify a polynomial of degree-bound $n$, that is, if fewer than $n$ distinct point-value pairs are given, they fail to specify a unique polynomial of degree-bound $n$. ($\textit{Hint:}$ Using Theorem 30.1, what can you say about a set of $n - 1$ point-value pairs to which you add one more arbitrarily chosen point-value pair?)

Suppose that just $n − 1$ point-value pairs uniquely determine a polynomial $P$ which satisfies them. Append the point value pair $(x_{n - 1}, y_{n − 1})$ to them, and let $P'$ be the unique polynomial which agrees with the $n$ pairs, given by Theorem 30.1. Now append instead $(x_{n - 1}, y'_{n − 1})$ where $y_{n − 1} \ne y'_{n - 1}$, and let $P''$ be the polynomial obtained from these points via Theorem 30.1. Since polynomials coming from $n$ pairs are unique, $P' \ne P''$. However, $P'$ and $P''$ agree on the original $n − 1$ point-value pairs, contradicting the fact that $P$ was determined uniquely.


Show how to use equation $\text{(30.5)}$ to interpolate in time $\Theta(n^2)$. ($\textit{Hint:}$ First compute the coefficient representation of the polynomial $\prod_j (x - x_j)$ and then divide by $(x - x_k)$ as necessary for the numerator of each term; see Exercise 30.1-2. You can compute each of the $n$ denominators in time $O(n)$.)

First, we show that we can compute the coefficient representation of $\prod_j (x - x_j)$ in time $\Theta(n^2)$. We will do it by recursion, showing that multiplying $\prod_{j < k} (x − x_j)$ by $(x − x_k)$ only takes time $O(n)$, since this only needs to be done $n$ times, this gets is total runtime of $O(n)$. Suppose that $\sum_{i = 0}^{k - 1} k_ix^i$ is a coefficient representation of $\prod_{j < k} (x − x_j)$. To multiply this by $(x − x_k)$, we just set $(k + 1)_i = k_{i − 1} - x_kk_i$ for $i = 1, \dots, k$ and $(k + 1)_0 = −x_k \cdot k_0$. Each of these coefficients can be computed in constant time, since there are only linearly many coefficients, then, the time to compute the next partial product is just $O(n)$.

Now that we have a coefficient representation of $\prod_j (x − x_j)$, we need to compute, for each $k \prod_{j − k} (x − x_j)$, each of which can be computed in time $\Theta(n)$ by problem 30.1-2. Since the polynomial is defined as a product of things containing the thing we are dividing by, we have that the remainder in each case is equal to $0$. Lets call these polynomials $f_k$. Then, we need only compute the sum $\sum_k y_k \frac{f_k(x)}{f_k(x_k)}$. That is, we compute $f(x_k)$ each in time $\Theta(n)$, so all told, only $\Theta(n^2)$ time is spent computing all the $f(x_k)$ values. For each of the terms in the sum, dividing the polynomial $f_k(x)$ by the number $f_k(x_k)$ and multiplying by $y_k$ only takes time $\Theta(n)$, so total it takes time $\Theta(n^2)$. Lastly, we are adding up $n$ polynomials, each of degree bound $n − 1$, so the total time taken there is $\Theta(n^2)$.


Explain what is wrong with the "obvious" approach to polynomial division using a point-value representation, i.e., dividing the corresponding $y$ values. Discuss separately the case in which the division comes out exactly and the case in which it doesn't.

If we wish to compute $P / Q$ but $Q$ takes on the value zero at some of these points, then we can't carry out the "obvious" method. However, as long as all point value pairs $(x_i, y_i)$ we choose for $Q$ are such that $y_i \ne 0$, then the approach comes out exactly as we would like.


Consider two sets $A$ and $B$, each having $n$ integers in the range from $0$ to $10n$. We wish to compute the Cartesian sum of $A$ and $B$, defined by

$$C = \{x + y: x \in A \text{ and } y \in B\}.$$

Note that the integers in $C$ are in the range from $0$ to $20n$. We want to find the elements of $C$ and the number of times each element of $C$ is realized as a sum of elements in $A$ and $B$. Show how to solve the problem in $O(n\lg n)$ time. ($\textit{Hint:}$ Represent $A$ and $B$ as polynomials of degree at most $10n$.)

For the set $A$, we define the polynomial $f_A$ to have a coefficient representation that has $a_i$ equal zero if $i \notin A$ and equal to $1$ if $i \in A$. Similarly define $f_B$. Then, we claim that looking at $f_C := f_A \cdot f_B$ in coefficient form, we have that the $i$th coefficient, $c_i$ is exactly equal to the number of times that $i$ is realized as a sum of elements from $A$ and $B$. Since we can perform the polynomial multiplication in time $O(n \lg n)$ by the methods of this chapter, we can get the final answer in time $O(n \lg n)$. To see that $f_C$ has the nice property described, we'll look at the ways that we could end up having a term of $x^i$ appear. Each contribution to that coefficient must come from there being some $k$ so that $a_k \ne 0$ and $b_{i − k} \ne 0$, because the powers of $x$ attached to each are additive when we multiply. Since each of these contributions is only ever $1$, the final coefficient is counting the total number of such contributions, therefore counting the number of $k \in A$ such that $−k \in B$, which is exactly what we claimed $f_C$ was counting.