Lab 3: Higher-order functions
Exercise 1
Implement 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 ]
For example:
(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
:
> (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
(define (mult-all-pairs lst1 lst2)
(apply ; flatten the result
append
(map
(lambda (x) (map ((curry *) x) lst2)) ; multiply all elements of lst2 by x
lst1))) ; do it for each element x in lst1
(define (f-all-pairs f lst1 lst2)
(apply
append
(map
(lambda (x) (map ((curry f) x) lst2))
lst1)))
Exercise 2
Suppose we represent univariate polynomials as lists of monomials. Each monomial of the form is represented as a list (a n)
consisting of the coefficient a
and the exponent n
. Thus the polynomial 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., ) and p2
((-1 0) (1 1) (3 2))
(i.e., ). Then
(add-pol p1 p2) => ((2 1) (3 2))
(mult-pol p1 p2) => ((-1 0) (4 2) (3 3))
Even though it might look tedious, 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
Similarly, multiplication of monomials is defined by
Addition and multiplication on monomials
(define (add-mon m1 m2)
(list (+ (get-coef m1) (get-coef m2)) ; sum coefficients
(get-exp m1))) ; keep exponent
(define (mult-mon m1 m2)
(list (* (get-coef m1) (get-coef m2)) ; multiply coefficients
(+ (get-exp m1) (get-exp m2)))) ; sum exponents
Now we come to the main trick. Suppose we have two polynomials and . We can express their sum as . 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 .
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.
Addition of polynomial and monomial
(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 . 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
(define (normalize p)
(define (non-zero-coef? m) (not (= 0 (get-coef m))))
(sort
(filter non-zero-coef? p)
(lambda (p1 p2) (< (get-exp p1) (get-exp p2)))))
(define (add-pol p1 p2)
(normalize (foldl add-mon-pol p1 p2)))
(define (mult-pol p1 p2)
(normalize (foldl add-mon-pol '() (f-all-pairs mult-mon p1 p2))))
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 . 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
(define (scalar-mult coef vec)
(map (curry * coef) vec))
(define (linear-combination vectors)
(lambda (coefs)
(apply map + (map scalar-mult coefs vectors))))
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 , so 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 that the matrix multiplication is just a repeated application of the linear-combination
function. More precisely, consider . The -th row of the result is just the linear combination of the rows of with the coefficients taken from the -th row of . So it suffices to apply the (linear-combination m_2)
to each row of .
To define the matrix power, use the foldl
function applied to a list composed of the same matrix mat
. To create such a list, you can use the function
> (make-list n el)
(el el ... el)
Solution
(define (matrix-mult m1 m2)
(map (linear-combination m2) m1))
(define (matrix-power k mat)
(foldl matrix-mult mat (make-list (- k 1) mat)))