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UP | HOME

Higher-order Functions

Table of Contents

1 Introduction

We illustrate programming with higher-order functions. This is one of the main features of a functional programming language.

2 Compose

The compose function o in mathematics takes two functions \(f:A\rightarrow B\) assumed to be of type and \(g:B\rightarrow C\) and returns a function of type \(A\rightarrow C\) that takes an argument \(x\) and returns \(g\) applied to the result $f to \(x\).

;;; Signature
;;; o: [A -> B], [B -> C] -> [A -> C]

;;; usage
;;; ((o add1 *2) 3) => 8
(define o
  (lambda (f g) ; f: [A -> B]
		; g: [B -> C]
    (lambda (x) ; x: A
      (g (f x)))))

In Racket, we could also have written this definition as follows:

(define ((o f g) x)
   (g (f x)))

Note that this corresponds to the definition \((f\circ g)(x) = g(f(x))\).

To test this, consider the function *2:

(define *2
  (lambda (n)
    (* n 2)))

The expression (o add1 *2) returns a function, which may be applied to a number. Thus, ((o add1 *2) 3).

2.1 Tracing the execution of ((o add1 *2) 3)


;; ((o add1 *2) 3) =>

;; ((lambda (x) (g (f x))) {f: add1, g: *2} 3)

;; ((lambda (x) (*2 (add1 x))) 3) =>

;; (*2 (add1 x)) {x: 3} =>

;; (*2 (add1 3)) =>

;; (*2 4) =>

;; ((lambda (n) (* n 2)) 4) =>

;; (* n 2) {n: 4} =>

;; (* 4 2) =>

;; 8

3 Iteration

3.1 Definition

;;; (iter f n) returns the nth iterate of the function f.
;;; iter : [A -> A], nat? -> [A -> A]

;;; (iter f n) = identity, if n=0
;;; (iter f n) = (o f (iter f (sub1 n))), if n>0

;;; (iter f 0) = identity
;;; (iter f 1) = f
;;; (iter f 2) = (o f f)

(define identity
  (lambda (x)
    x))

(define iter
  (lambda (f n) ; f: A -> A
		; n: nat?
    (if (= n 0)
	identity
	(o f (iter f (sub1 n))))))

3.2 Theorem relating (iter f 2) and (o f f)

;;; Theorem:  (iter f 2) === (o f f)
;;; Proof:

(iter f 2) => beta

(if (= n 0)
    identity
    (o f (iter f (sub1 n)))) {f: f, n: 2}  => subst

(o f (iter f 1)) => beta 

(o f (if (= n 0)
       identity
       (o f (iter f (sub1 n)))){f: f, n:1}) => subst

(o f (o f (iter f 0))) => beta

(o f (o f (if (= n 0)
              identity
              (o f (iter f (sub1 n)))){f:f, n:0})) => subst

(o f (o f identity)) => ...

(o f (lambda (x) (identity (f x)))) =>
(o f (lambda (x) ((lambda (x) x) (f x)))) =>
(o f (lambda (x) (f x))) => eta
(o f f)

4 Derivative

;;; The Derivative operator

;;;
;;; ((D f) x) = (/ (- (f (+ x dx)) (f x)) dx)


(define dx 0.01)

;;; D : [num? -> num?] -> [num? -> num?]
(define D
  (lambda (f)  ; f: num? -> num?
    (lambda (x) ; x: num?
      (/ (- (f (+ x dx)) (f x))
	 dx))))



;;; D2 : [num? -> num?, num?] -> num?
(define D2
  (lambda (f x)  ; f: num? -> num?
    (/ (- (f (+ x dx)) (f x))
       dx)))

5 Lists

5.1 Map

;;; map: [A -> B, (list-of A)] -> (list-of B)
;;; usage:
;;; (map add1 '(2 5 9)) => '(3 6 10)
;;; (map add1 '()) => '()
;;; (map f '()) => '()
;;; (map f (cons x ls)) => (cons (f x) (map f ls))


(define map
  (lambda (f ls)   ; f: A -> B
    (if (null? ls) ; ls: (list-of A)
	'()
	(let ([x (first ls)]
	      [ls2 (rest ls)])
	  (cons (f x)
		(map f ls2))))))

5.2 Digression into let

(let ((x 3) (y 4)) (+ x y))

is syntactic shorthand for

((lambda (x y) (+ x y)) 3 4)

5.3 Filter


;;; filter

;;; (filter f ls) takes a predicate f and a list ls and returns a list
;;; all of whose elements satisfy f.

;;; Examples
;;; -------

;;; (filter number? '(a 3 8 "hello")) => '(3 8)

;;; (filter even? '( 2 4 9)) => '(2 4)

;;; (filter f? '()) => '()
;;; (filter f? (cons x ls)) =
;;;   = (filter f? ls),    if (f? x) is false
;;;   = (cons x (filter f? ls)), if (f? x) is true


(define filter
  (lambda (f? ls)
    (if (null? ls)
	'()
	(let ([x (first ls)]
	      [lsr (rest ls)])
	  (if (f? x)
	      (cons x (filter f? lsr))
	      (filter f? lsr))))))

Author: Venkatesh Choppella

Created: 2024-08-29 Thu 12:23

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