13 KiB
- Example: Arithmetic Operations for Rational Numbers
- Abstraction Barriers
- What is Meant by Data
- Extended Exercise: Interval Arithmetic
— title: 2.1 - Introduction to Data Abstraction layout: org —
Example: Arithmetic Operations for Rational Numbers
;; ===================================================================
;; 2.1.1: Example: Arithmetic Operators for Rational Numbers
;; ===================================================================
(define (add-rat x y)
(make-rat (+ (* (numer x) (denom y))
(* (numer y) (denom x)))
(* (denom x) (denom y))))
(define (sub-rat x y)
(make-rat (- (* (numer x) (denom y))
(* (numer y) (denom x)))
(* (denom x) (denom y))))
(define (mul-rat x y)
(make-rat (* (numer x) (numer y))
(* (denom x) (denom y))))
(define (div-rat x y)
(make-rat (* (numer x) (denom y))
(* (denom x) (numer y))))
(define (equal-rat? x y)
(= (* (numer x) (denom y))
(* (numer y) (denom x))))
(define (make-rat n d) (cons n d))
(define (numer x) (car x))
(define (denom x) (cdr x))
(define (print-rat x)
(newline)
(display (numer x))
(display "/")
(display (denom x)))
(define (gcd a b)
(if (= b 0)
a
(gcd b (remainder a b))))
(define (make-rat n d)
(let ((g (gcd n d)))
(cons (/ n g) (/ d g))))Exercise 2.1:
Define a better version of `make-rat' that handles both positive and negative arguments. `Make-rat' should normalize the sign so that if the rational number is positive, both the numerator and denominator are positive, and if the rational number is negative, only the numerator is negative.
;; -------------------------------------------------------------------
;; Exercise 2.1
;; -------------------------------------------------------------------
(define (make-rat n d)
(cond ((and (negative? n) (negative? d)) (make-rat (abs n) (abs d)))
((negative? d) (make-rat (- n) (- d)))
(else (let ((g (gcd n d)))
(cons (/ n g) (/ d g))))))Abstraction Barriers
Exercise 2.2
Consider the problem of representing line segments in a plane. Each segment is represented as a pair of points: a starting point and an ending point. Define a constructor `make-segment' and selectors `start-segment' and `end-segment' that define the representation of segments in terms of points. Furthermore, a point can be represented as a pair of numbers: the x coordinate and the y coordinate. Accordingly, specify a constructor `make-point' and selectors `x-point' and `y-point' that define this representation. Finally, using your selectors and constructors, define a procedure `midpoint-segment' that takes a line segment as argument and returns its midpoint (the point whose coordinates are the average of the coordinates of the endpoints). To try your procedures, you'll need a way to print points:
;; -------------------------------------------------------------------
;; Excercise 2.2
;; -------------------------------------------------------------------
(define (print-point p)
(newline)
(display "(")
(display (x-point p))
(display ",")
(display (y-point p))
(display ")")) (define make-point cons)
(define x-point car)
(define y-point cdr)
(define (midpoint-segment p1 p2)
(let ((average (lambda (x y) (/ (+ x y) 2))))
(make-point
(average (x-point p1) (x-point p2))
(average (y-point p1) (y-point p2)))))Exercise 2.3:
Implement a representation for rectangles in a plane. (Hint: You may want to make use of *Note Exercise 2-2::.) In terms of your constructors and selectors, create procedures that compute the perimeter and the area of a given rectangle. Now implement a different representation for rectangles. Can you design your system with suitable abstraction barriers, so that the same perimeter and area procedures will work using either representation?
;; -------------------------------------------------------------------
;; Exercise 2.3
;; -------------------------------------------------------------------
(define (perimeter-rectangle r)
(+ (* 2 (width-rectangle r))
(* 2 (height-rectangle r))))
(define (area-rectangle r)
(* (width-rectangle r)
(height-rectangle r)))
;; ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
;; Hard mode - Expose the 4 points of the rectangle
;; Width and Height have their own abstraction layer
;;~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
(define (width-rectangle r)
(abs (- (x2-rectangle r)
(x1-rectangle r))))
(define (height-rectangle r)
(abs (- (y2-rectangle r)
(y1-rectangle r))))
(define (x1-rectangle r) (x-point (top-left-point-rectangle r)))
(define (x2-rectangle r) (x-point (bottom-right-point-rectangle r)))
(define (y1-rectangle r) (y-point (top-left-point-rectangle r)))
(define (y2-rectangle r) (y-point (bottom-right-point-rectangle r)))
;; -------------------------------------------------------------------
;; Rectangle implementation using two points on a plane
(define make-rectangle cons)
(define top-left-point-rectangle car)
(define bottom-right-point-rectangle cdr)
(define (top-right-point-rectangle r)
(make-point (x-point (top-left-point-rectangle r))
(y-point (bottom-right-point-rectangle r))))
(define (bottom-left-point-rectangle r)
(make-point (x-point (top-left-point-rectangle r))
(y-point (bottom-right-point-rectangle r))))
;; -------------------------------------------------------------------
;; Rectangle implementation using an origin point, width and height
(define (make-rectangle origin width height)
(cons origin (cons width height)))
(define (top-left-point-rectangle r) (car r))
(define (top-right-point-rectangle r)
(let ((x (x-point (car r)))
(y (y-point (car r)))
(width (car (cdr r))))
(make-point (+ x width) y)))
(define (bottom-left-point-rectangle r)
(let ((x (x-point (car r)))
(y (y-point (car r)))
(height (cdr (cdr r))))
(make-point x (+ y height))))
(define (bottom-right-point-rectangle r)
(let ((x (x-point (car r)))
(y (y-point (car r)))
(width (car (cdr r)))
(height (cdr (cdr r))))
(make-point (+ x width) (+ y height))))
;; ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
;; Simpler solution - Expose only width + height
;; ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
;; -------------------------------------------------------------------
;; Rectangle implementation using two points on a plane
(define make-rectangle cons)
(define (width-rectangle r)
(let ((p1 (car r))
(p2 (cdr r)))
(abs (- (x-point p1)
(x-point p2)))))
(define (height-rectangle r)
(let ((p1 (car r))
(p2 (cdr r)))
(abs (- (y-point p1)
(y-point p2)))))
;; -------------------------------------------------------------------
;; Rectangle implementation using an origin point, width and height
(define (make-rectangle origin width height)
(cons origin (cons width height)))
(define (width-rectangle r) (car (cdr r)))
(define (height-rectangle r) (cdr (cdr r)))What is Meant by Data
Exercise 2.4
Here is an alternative procedural representation of pairs. For this representation, verify that `(car (cons x y))' yields `x' for any objects `x' and `y'.
;; -------------------------------------------------------------------
;; Exercise 2.4
;; -------------------------------------------------------------------
(define (cons x y)
(lambda (m) (m x y)))
(define (car z)
(z (lambda (p q) p)))What is the corresponding definition of `cdr'? (Hint: To verify that this works, make use of the substitution model of section *Note 1-1-5::.)
(define (cdr z)
(z (lambda (p q) q)))Exercise 2.5
Show that we can represent pairs of nonnegative integers using only numbers and arithmetic operations if we represent the pair a and b as the integer that is the product 2^a 3^b. Give the corresponding definitions of the procedures `cons', `car', and `cdr'.
;; -------------------------------------------------------------------
;; Exercise 2.5
;; -------------------------------------------------------------------
(define (cons a b)
(* (expt 2 a) (expt 3 b)))
(define (factor-count n x count)
(if (= 0 (remainder x n))
(factor-count n (/ x n) (+ 1 count))
count))
(define (car p)
(factor-count 2 p 0))
(define (cdr p)
(factor-count 3 p 0))Exercise 2.6
In case representing pairs as procedures wasn't mind-boggling enough, consider that, in a language that can manipulate procedures, we can get by without numbers (at least insofar as nonnegative integers are concerned) by implementing 0 and the operation of adding 1 as
(define zero (lambda (f) (lambda (x) x)))
(define (add-1 n)
(lambda (f) (lambda (x) (f ((n f) x)))))This representation is known as "Church numerals", after its inventor, Alonzo Church, the logician who invented the [lambda] calculus.
Define `one' and `two' directly (not in terms of `zero' and `add-1'). (Hint: Use substitution to evaluate `(add-1 zero)'). Give a direct definition of the addition procedure `+' (not in terms of repeated application of `add-1').
(define one (lambda (f) (lambda (x) (f x))))
(define two (lambda (f) (lambda (x) (f (f x)))))
(define (add a b)
(lambda (f)
(lambda (x)
((a f) ((b f) x)))))Extended Exercise: Interval Arithmetic
;; ===================================================================
;; 2.1.4: Extended Exercise: Interval Arithmetic
;; ===================================================================
(define (add-interval x y)
(make-interval (+ (lower-bound x) (lower-bound y))
(+ (upper-bound x) (upper-bound y))))
(define (mul-interval x y)
(let ((p1 (* (lower-bound x) (lower-bound y)))
(p2 (* (lower-bound x) (upper-bound y)))
(p3 (* (upper-bound x) (lower-bound y)))
(p4 (* (upper-bound x) (upper-bound y))))
(make-interval (min p1 p2 p3 p4)
(max p1 p2 p3 p4))))
(define (div-interval x y)
(mul-interval x
(make-interval (/ 1.0 (upper-bound y))
(/ 1.0 (lower-bound y)))))Exercise 2.7
Alyssa's program is incomplete because she has not specified the implementation of the interval abstraction. Here is a definition of the interval constructor:
;; -------------------------------------------------------------------
;; Exercise 2.7
;; -------------------------------------------------------------------
(define (make-interval a b) (cons a b))Define selectors `upper-bound' and `lower-bound' to complete the implementation.
(define (upper-bound p)
(max (car p) (cdr p)))
(define (lower-bound p)
(min (car p) (cdr p)))Exercise 2.8:
Using reasoning analogous to Alyssa's, describe how the difference of two intervals may be computed. Define a corresponding subtraction procedure, called `sub-interval'.
;; -------------------------------------------------------------------
;; Exercise 2.8
;; -------------------------------------------------------------------