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---
title: 2.1 - Introduction to Data Abstraction
layout: org
---
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* Example: Arithmetic Operations for Rational Numbers
#+begin_src scheme :tangle yes
;; ===================================================================
;; 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))))
#+end_src
** 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.
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#+begin_src scheme :tangle yes
;; -------------------------------------------------------------------
;; 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))))))
#+end_src
* 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:
#+begin_src scheme :tangle yes
;; -------------------------------------------------------------------
;; Excercise 2.2
;; -------------------------------------------------------------------
(define (print-point p)
(newline)
(display "(")
(display (x-point p))
(display ",")
(display (y-point p))
(display ")"))
#+end_src
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#+begin_src scheme :tangle yes
(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)))))
#+end_src
** 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?
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#+begin_src scheme :tangle yes
;; -------------------------------------------------------------------
;; 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)))
#+end_src
* 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'.
#+begin_src scheme :tangle yes
;; -------------------------------------------------------------------
;; Exercise 2.4
;; -------------------------------------------------------------------
(define (cons x y)
(lambda (m) (m x y)))
(define (car z)
(z (lambda (p q) p)))
#+end_src
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::.)
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#+begin_src scheme :tangle yes
(define (cdr z)
(z (lambda (p q) q)))
#+end_src
** 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'.
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#+begin_src scheme :tangle yes
;; -------------------------------------------------------------------
;; 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))
#+end_src
** 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
#+begin_src scheme
(define zero (lambda (f) (lambda (x) x)))
(define (add-1 n)
(lambda (f) (lambda (x) (f ((n f) x)))))
#+end_src
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').
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#+begin_src scheme :tangle yes
(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)))))
#+end_src
* Extended Exercise: Interval Arithmetic
#+begin_src scheme :tangle yes
;; ===================================================================
;; 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)))))
#+end_src
** 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:
#+begin_src scheme :tangle yes
;; -------------------------------------------------------------------
;; Exercise 2.7
;; -------------------------------------------------------------------
(define (make-interval a b) (cons a b))
#+end_src
Define selectors `upper-bound' and `lower-bound' to complete the
implementation.
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#+begin_src scheme :tangle yes
(define (upper-bound p)
(max (car p) (cdr p)))
(define (lower-bound p)
(min (car p) (cdr p)))
#+end_src
** 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'.
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#+begin_src scheme :tangle yes
;; -------------------------------------------------------------------
;; Exercise 2.8
;; -------------------------------------------------------------------
#+end_src