Warning

## Lab 3: Higher-order functions

Exercise 1: Write a function (mult-all-pairs lst1 lst2) taking two lists and returning a list of all possible binary products between elements from lst1 and elements from lst2. Mathematically, it could be written by a comprehension term as [ x*y | x in lst1, y in lst2 ]. E.g. (mult-all-pairs '(1 2 3) '(-2 0)) ⇒ (-2 0 -4 0 -6 0). Once you have it, generalize your function to (f-all-pairs f lst1 lst2) so that the multiplication is replaced by any binary function f. E.g. (f-all-pairs cons '(1 2 3) '(a b)) ⇒ ((1 . a) (1 . b) (2 . a) (2 . b) (3 . a) (3 . b)).

Hint: These functions are just applications of two nested map functions. For each element x in lst1 we multiply by x all elements in lst2. A function multiplying by x can be created from the multiplication function * by partial application of x, i.e., we take the curryfied version of * and apply it to x yielding ((curry *) x). Once we map ((curry *) x) along lst2, the result is a list. So doing it for each x in lst1 results in a list of lists. Thus we have to flatten the result and append all the lists. This can be done by apply-ing append.

Solution

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 component
Next 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 $p_1(x)p_2(x) = (((a_0b_0 + a_0b_1x) + a_0b_2x^2) + a_1b_0x) + \cdots$.

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 exponents to 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.

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.

Final solution

Task 1: Write a function linear-combination taking a list of vectors, a list of coefficients and returning the corresponding linear combination. The function should be created in the curried form (the list of vectors being the first argument) because it will be convenient for the next task. For example, consider a linear combination $2\cdot(1, 2, 3) - 1\cdot(1, 0, 1) + 3\cdot(0, 2, 0) = (1,10,5)$. Then your implementation should work as follow:

 ((linear-combination '((1 2 3) (1 0 1) (0 2 0))) '(2 -1 3)) => (1 10 5)

Hint: Create first a binary function computing scalar multiplication of a scalar and a vector using map. Then use the fact that map can apply the scalar multiplication to two lists simultaneously (in our case the list of coefficients and the list of vectors). This results in a list of vectors multiplied by respective coefficients. Then it suffices to sum them component by component.

Solution

Task 2: Use the function from the previous task to define a function (matrix-mult m1 m2) computing the matrix multiplication of m1 and m2. Then apply foldl function to define the power of a square matrix, i.e., a function (matrix-power k mat) computing k-fold product of mat. You can assume that $k\geq 1$ so that there is no need to define the identity matrix. E.g.

 (matrix-mult '((1 2 3)
(-1 0 2))
'((1 -1)
(2 0)
(0 3))) => ((5 8) (-1 7))

(matrix-power 3 '((2 3) (0 -1))) => ((8 9) (0 -1))

Hint: Use the fact that the matrix multiplication is just a repeated application of the linear-combination function. More precisely, consider $m_1\cdot m_2$. The $i$-th row of the result is just the linear combination of the rows of $m_2$ with the coefficients taken from the $i$-th row of $m_1$. So it suffices to apply the (linear-combination m_2) to each row of $m_1$.

To define the matrix power, use the foldl function applied to a list composed of the same matrix mat. To create such a list, use function (make-list n el) ⇒ (el el … el).

Solution