Exercise 2: Suppose we represent univariate polynomials as lists of monomials. Each monomial of the form $ax^n$ is represented as a list (a n) consisting of the coefficient a and the exponent n. Thus the polynomial $2-3x+x^2$ is represented by ((2 0) (-3 1) (1 2)). We assume that each exponent can occur in the polynomial representation at most once. E.g. ((1 0) (2 0)) is not a valid representation. Devise functions (add-pol p1 p2) and (mult-pol p1 p2) taking as arguments two polynomials p1,p2 and returning their sum and product respectively. For example, let p1 be ((1 0) (1 1)) (i.e., $p_1(x)=1+x$) and p2 ((-1 0) (1 1) (3 2)) (i.e., $p_2(x)=-1+x+3x^2$). Then (add-pol p1 p2) => ((2 1) (3 2)) (mult-pol p1 p2) => ((-1 0) (4 2) (3 3))Even though it might look like a tedious task, it is not so terrible because we will call higher-order functions to rescue us. Thanks to the folding function foldl, we will reduce our problem just to monomials. Let's start by defining simple operations on monomials. To make our code more comprehensible, we define the following functions extracting the coefficient and the exponent from a monomial.(define (get-coef m) (car m)) ; first component(define (get-exp m) (cadr m)) ; second componentNext it is easy to define addition of two monomials of the same exponents, namely $ax^n + bx^n = (a+b)x^n$. Similarly, multiplication of monomials is defined by $(ax^n)(bx^k)=abx^n+k$.
Now we come to the main trick. Suppose we have two polynomials $p_1(x)=a_0+a_1x$ and $p_2(x)=b_0+b_1x+b_2x^2$. We can express their sum as$p_1(x) + p_2(x) = ((p_1(x) + b_0) + b_1x) + b_2x^2$. Thus we need only to add a polynomial and a monomial at a single steps. Then the repetitive sum can be done by foldl function. Similarly for the multiplication, we can first compute products of all monomials by means of the function f-all-pairs from Exercise 1 and then express the results as $p_1(x)p_2(x) = (((a_0b_0 + a_0b_1x) + a_0b_2x^2) + a_1b_0x) + \cdots$.
Homework 5 Monomials All Operations
Thus we need a function adding a monomial mon and a polynomial pol. The function has to distinguish two cases: 1) we add a monomial whose exponent does not occur in pol, 2) or whose exponent occurs in pol. So we first filter monomials in pol according to their exponentsto obtain the monomial of the same exponent as mon and the remaining monomials. If there is no monomial of the same exponent, we just cons mon to the result, otherwise we add monomials of the same exponent and cons it to the result.
(define (add-mon-pol mon pol) (define (same-exp? m) (= (get-exp mon) (get-exp m))) ; #t if m has the same exponent as mon (define same-mon (filter same-exp? pol)) ; list containing the monomial of the same exponent or empty list (define rest (filter (compose not same-exp?) pol)) ; remaining monomials of different exponents (if (null? same-mon) (cons mon rest) (cons (add-mon mon (car same-mon)) rest)))
Finally, we can apply the folding function foldl to sum all monomials as was shown above. However, there are still two problems we have to deal with. 1) It may happen that the result contains monomials of the form $0x^n$. Such monomials can be clearly filtered out of the result. 2) It is common to sort monomials according to their exponents. Thus we define a function normalize solving these two problems.
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