CS 61A Week 5 solutions
LAB EXERCISES:
Skipping the ones that just say "try this"...
5.
(define (+rat a b)
(make-rational (+ (* (numerator a) (denominator b))
(* (denominator a) (numerator b)))
(* (denominator a) (denominator b))))
2.2
;; Constructor and selector for segments
;; Note that we don't have to know what's inside a "point" for these
;; to work!
(define (make-segment point1 point2)
(cons point1 point2))
(define (start-segment seg)
(car seg))
(define (end-segment seg)
(cdr seg))
;; Constructor and selector for points
(define (make-point xcor ycor)
(cons xcor ycor))
(define (x-point point)
(car point))
(define (y-point point)
(cdr point))
;; midpoint of segment from P1(x1,y1) to P2(x2,y2) is
;; ( (x1+x2)/2 , (y1+y2)/2 )
(define (midpoint-segment seg)
(make-point (/ (+ (x-point (start-segment seg))
(x-point (end-segment seg)))
2)
(/ (+ (y-point (start-segment seg))
(y-point (end-segment seg)))
2)))
;; Alternate definition, maybe fewer keystrokes,
;; maybe even easier to read:
(define (midpoint-segment seg)
(let ((p1 (start-segment seg))
(p2 (end-segment seg)))
(make-point (/ (+ (x-point p1) (x-point p2)) 2)
(/ (+ (y-point p1) (y-point p2)) 2)))
Note: If you wanted to write (define (make-segment x1 y1 x2 y2) ...) or
something similar, it means you don't yet believe that abstract data
types are legitimate things, valid arguments to procedures and so on.
Just as in section 1.3 you had to learn to accept that procedures are
just as good as numbers, now you have to accept that points (or any
other abstract type) are just as good as numbers!
2.3
Since the hint suggests using exercise 2.2, let's represent a rectangle
as two adjacent sides (adjacent so that we get one length and one width).
(define (make-rectangle side1 side2)
(cons side1 side2))
(define (first-leg rect)
(car rect))
(define (second-leg rect)
(cdr rect))
Perimeter and area:
(define (perimeter rect)
(* 2 (+ (length-segment (first-leg rect))
(length-segment (second-leg rect)))))
(define (area rect)
(* (length-segment (first-leg rect))
(length-segment (second-leg rect))))
(define (length-segment seg)
(sqrt (+ (square (- (x-point (end-segment seg))
(x-point (start-segment seg))))
(square (- (y-point (end-segment seg))
(y-point (start-segment seg)))))))
Different representation for rectangles: Note that the representation
above really contains more information than necessary; it includes the
common point twice, and it doesn't take into account that the angle
between the two legs must be 90 degrees. So instead we could represent
a rectangle using a BASE, which is a segment, and a HEIGHT, which is
just a number (the length of the other segment).
(define (make-rectangle base height)
(cons base height))
(define (base-rectangle rect)
(car rect))
(define (height-rectangle rect)
(cdr rect))
Making the same perimeter and area procedures work: To do this
we have to redefine first-leg and second-leg in terms of the
new representation. The first leg can be just the base, but for
the second leg we need some analytic geometry. Specifically, we need
to know that if the slope of the base segment is Dy/Dx (using D for delta)
then the slope of a perpendicular height should be -Dx/Dy.
If we want the same perimeter and area procedures to work for either kind of
rectangle, we have to have first-leg and second-leg check which kind we
have, like this:
(define (first-leg rect)
(if (pair? (cdr rect))
(car rect)
(base-rectangle rect)))
(define (second-leg rect)
(if (pair? (cdr rect))
(cdr rect)
(let ((origin (start-segment (base-rectangle rect)))
(endpoint (end-segment (base-rectangle rect)))
(scale-factor (/ (height-rectangle rect)
(length-segment (base-rectangle rect)))))
(make-segment origin
(make-point (+ (x-point origin)
(* scale-factor
(- (y-point origin)
(y-point endpoint))))
(+ (y-point origin)
(* scale-factor
(- (x-point endpoint)
(x-point origin)))))))))
Alternatively, you might find it easier to redefine perimeter and area
in terms of the new representation, and then to make them work for the
old representation you'll have to define base-rectangle and height-rectangle
in terms of first-leg and second-leg:
(define (perimeter rect)
(* 2 (+ (length-segment (base-rectangle rect))
(height-rectangle rect))))
(define (area rect)
(* (length-segment (base-rectangle rect))
(height-rectangle rect)))
(define (base-rectangle rect)
(if (pair? (cdr rect))
(first-leg rect)
(car rect)))
(define (height-rectangle rect)
(if (pair? (cdr rect))
(length-segment (second-leg rect))
(cdr rect)))
Note that we don't want to check (pair? (cdr rect)) in the perimeter or
area procedure, because those procedures are above the abstraction
barrier -- they shouldn't have to know about the internal representation.
2.4
I hope you are awed by this problem. Isn't it beautiful that you can
use procedures to capture what you used to consider "data" like this?
Anyway, suppose we have two objects A and B, and we've made a pair out
of them with this version of cons. Just to make it easier to talk about,
let's give that pair a name:
(define mypair (cons A B))
The value of the symbol "mypair" is now the procedure created by
(lambda (m) (m A B))
Now what happens when we evaluate (car mypair)? By the substitution
rule, we must substitute the value of "mypair" for "z" in the body of car:
((lambda (m) (m A B)) (lambda (p q) p))
This expression has two elements, each of which is a lambda-expression.
As usual, we evaluate an expression by taking its first subexpression as
a function to be applied to the second subexpression as argument. The
body of the first subexpression is (m A B). Into that body we substitute
the argument for the formal parameter m:
((lambda (p q) p) A B)
To evaluate THAT expression, we substitute A for p and B for q in body of
the function. That body is just "p" so we get A as the value of the
original expression, as desired.
The corresponding definition of cdr is, of course,
(define (cdr z)
(z (lambda (p q) q)))
2.18. Reverse a list.
(define (reverse lst)
(define (iter old new)
(if (null? old)
new
(iter (cdr old) (cons (car old) new))))
(iter lst nil))
It's worth thinking about the fact that this function is MUCH easier to
write as an iterative procedure than as a recursive one. There's something
inherently reversing about the iterative structure, when applied to a list.
Here's the reason. The basic accumulator for lists is cons; this is analogous
to the role played by + in most of the earlier examples of recursive/iterative
problems. However, + is commutative; it doesn't matter whether you add a
series front-to-back or back-to-front. Cons is NOT commutative, and the
pair-diagram representation of a list is NOT left-right symmetric. The only
natural way to build up a list is right-to-left. If you want the list
(A B C) first you cons C onto nil, then you cons B onto that pair, then you
cons A onto that one.
The difference between the usual iterative control structure and the usual
recursive one is that the former accumulates its result "on the way down"
the chain of recursive invocations; the first invocation adds the first term
to the result and then does the second invocation. A recursive structure,
on the other hand, accumulates terms "on the way up" the chain; it keeps
doing recursive invocations, keeping pending the addition of the partial
result into the larger result. When you reach the end condition, the
invocations start terminating, and that's when the accumulation (using +
or whatever) happens.
In a list processing situation, the trace for recursion looks like this:
(recursive-proc '(a b c))
(cons (some-fn 'a) (recursive-proc '(b c)))
-----------------------
V
(cons (some-fn 'b) (recursive-proc '(c)))
---------------------
V
(cons (some-fn 'c) nil)
The trace for iteration is
(iterative-proc '(a b c))
(iterative-helper '(a b c) nil)
(iterative-helper '(b c) (cons (some-fn 'a) nil))
(iterative-helper '(c) (cons (some-fn 'b) (cons (some-fn 'a) nil)))
and so on. You can see that the reversal we wanted in this problem
"comes free" in the iterative solution.
If you want to write reverse recursively, you need left-right symmetrical
list operations. Using the word/sentence data abstraction you could say
(define (reverse lst)
(if (empty? lst)
nil
(sentence (reverse (butfirst lst)) (first lst))))
but this will be really slow, O(n^2). Also, it's a data abstraction
violation unless the argument is really a sentence, i.e., a list of
words, not a list of lists. You can avoid the DAV by saying
(define (reverse lst)
(if (null? lst)
nil
(append (reverse (cdr lst)) (list (car lst)))))
Compared to the SENTENCE solution, this one has an extra call to LIST
because APPEND requires lists as its arguments. This solution is
still O(n^2) time.
HOMEWORK:
2.7
(define (upper-bound interval)
(cdr interval))
(define (lower-bound interval)
(car interval))
2.8
;; The smallest possible value for A-B is (smallest A)-(largest B).
;; Likewise, the largest value is (largest A)-(smallest B).
(define (sub-interval x y)
(add-interval x (make-interval (- (upper-bound y))
(- (lower-bound y)))))
;; It would also be okay to replicate the structure of add-interval instead of
;; using add-interval as a subprocedure, although this isn't exactly following
;; Alyssa's method as the problem suggests:
(define (sub-interval x y)
(make-interval (- (lower-bound x) (upper-bound y))
(- (upper-bound x) (lower-bound y)) ))
2.10
;; An interval spans zero if its lower bound is negative (or zero) and
;; its upper bound is positive (or zero).
;; It's only the divisor that we have to worry about.
(define (div-interval x y)
(if (and (<= (lower-bound y) 0)
(>= (upper-bound y) 0))
(error "Can't divide by an interval that spans zero.")
(mul-interval x
(make-interval (/ 1 (upper-bound y))
(/ 1 (lower-bound y))))))
2.12
(define (make-center-percent c p)
(let ((w (* c p 0.01)))
(make-interval (- c w) (+ c w))))
;; We multiply by 0.01 because p percent is a factor of p/100.
;; If you forgot about that part, it's a relatively minor error
;; for computer science purposes, although rather serious in a
;; practical engineering situation!
(define (percent i)
(* 100 (/ (/ (- (upper-bound i) (lower-bound i)) 2)
(/ (+ (lower-bound i) (upper-bound i)) 2))))
;; Slightly more efficient percent:
(define (percent i)
(* 100 (/ (- (upper-bound i) (lower-bound i))
(+ (lower-bound i) (upper-bound i)))))
;; Alternate versions, using the center-width procedures we already have:
(define (make-center-percent c p)
(make-center-width c (* c p 0.01)))
(define (percent i)
(* 100 (/ (width i) (center i))))
2.17
(define (last-pair lst)
(if (null? (cdr lst))
lst
(last-pair (cdr lst))))
2.20
The difficulty in writing recursive procedures that take any number of
arguments is that you're going to be tempted to make a recursive call
with only one argument, namely a list containing some of the original
arguments, sort of like this:
(define (same-parity . numbers)
(cond ((null? (cdr numbers)) numbers)
((equal? (even? (car numbers))
(even? (cadr numbers)))
(cons (car numbers)
(same-parity (cdr numbers)))) ; WRONG!
(else (same-parity (cons (car numbers) (cddr numbers)))))) ; WRONG!
(define (even? num)
(= (remainder num 2) 0))
Instead, the easiest thing to do is to define a helper procedure that
*does* expect a list of numbers as its one argument. An advantage is
that we can then separate out the first number, which is always
accepted, as a special case:
(define (same-parity tester . others)
(define (helper numlist)
(cond ((null? numlist) nil)
((equal? (even? tester) (even? (car numlist)))
(cons (car numlist) (helper (cdr numlist))))
(else (helper (cdr numlist)))))
(cons tester (helper others)))
Once we know about higher-order functions, there's an even easier
solution:
(define (same-parity tester . others)
(cons tester
(filter (lambda (num) (equal? (even? tester) (even? num)))
others)))
2.22. What's wrong with iterative square-list?
I've sort of answered this already, in talking about 2.18 (iterative
reverse). A list (as opposed to an arbitrary list structure) MUST be made
with
(cons new-member rest-of-list)
and not the other way around. That's why Louis's second try fails; he's
making something that isn't a list, even though he does have the members
in correct left-to-right order. He gets
((((() . 1) . 4) . 9) . 16)
instead of (1 4 9 16), which in dotted representation would be
(1 . (4 . (9 . (16 . ()))))
Louis's first try fails because he starts by consing the square of 1 onto
the empty list, etc.
Here's an iterative square-list:
(define (square-list x)
(define (iter list answer)
(if (null? list)
answer
(iter (cdr list)
(cons (square (car list))
answer))))
(reverse (iter x nil)))
In this version, iter is O(n) time and O(1) space, as desired. If we
use the iterative definition of reverse, it too is O(n) time and O(1) space.
So the whole thing is reasonably efficient in both time and space, even
though it has what seems to be a redundant pass through the list to reverse
it. If you're dealing with very large lists on very small computers, this
is sometimes the only way to get an answer at all.
2.23
(define (for-each proc lst)
(if (null? lst)
#t
(let ((ignored-result (proc (car lst))))
(for-each proc (cdr lst)))))
Several people said, "Why not just use map?" There are two answers.
First, it's inefficient because map will construct a list of all the
results, and we don't care about the results. Second, map does not
guarantee to apply the procedure argument to the list elements in
left-to-right order.
If you think it's inelegant to have a variable ignored-result whose
value is something we don't care about, you're right. In chapter 3
we'll learn about an alternative notation:
(define (for-each proc lst)
(if (null? lst)
#t
(begin (proc (car lst))
(for-each proc (cdr lst)))))
And it turns out that a COND clause can contain more than two expressions,
so this can also be written as
(define (for-each proc lst)
(cond ((null? lst) #t)
(else (proc (car lst))
(for-each proc (cdr lst)))))
SUBSTITUTE:
For reference, here's the SUBSTITUTE for sentences from week 2 lab:
(define (substitute sent old new)
(cond ((empty? sent) '())
((equal? (first sent) old)
(se new (substitute (butfirst sent) old new)))
(else (se (first sent) (substitute (butfirst sent) old new)))))
If we just change the sentence constructor/selector functions to the proper
ones for lists, and one formal parameter's name, we get this:
(define (substitute LST old new) ;; first try
(cond ((NULL? LST) '())
((equal? (CAR LST) old)
(CONS new (substitute (CDR LST) old new)))
(else (CONS (CAR LST) (substitute (CDR LST) old new)))))
Why CONS rather than LIST or APPEND? Because we're combining some function of
the CAR of a pair with some function (namely a recursive call) of the CDR,
trying to build a new structure with the same shape as the original. LIST or
APPEND would change the shape; if we used LIST, all the lists we made would
have exactly two elements, but if we used APPEND, we would be flattening
sublists into, essentially, a sentence.
But this isn't good enough, because it doesn't examine sublists. The most
obvious solution would be to add a clause to the COND to distinguish sublists
from atomic (non-list) elements:
(define (substitute lst old new)
(cond ((null? lst) '())
((equal? (car lst) old)
(cons new (substitute (cdr lst) old new)))
((PAIR? (CAR LST))
(CONS (SUBSTITUTE (CAR LST) OLD NEW)
(SUBSTITUTE (CDR LST) OLD NEW)))
(else (cons (car lst) (substitute (cdr lst) old new)))))
This works. But once we're familiar with the idioms of tree recursion
(week 6), we can see a slightly more elegant solution:
(define (substitute lst old new)
(cond ((null? lst) '())
((equal? LST old) NEW)
((pair? LST)
(cons (substitute (car lst) old new)
(substitute (cdr lst) old new)))
(else LST)))
The ELSE clause is reached when the LST argument is a word rather than
a list, and is not equal to OLD. (A list is either null or a pair.)
SUBSTITUTE2:
This will have the same program structure as SUBSTITUTE, but the EQUAL? test
is replaced by something a little more complicated.
The straightforward solution is to use MEMBER in place of EQUAL?, and then, if
a true answer is returned, look for the right thing to substitute:
(define (substitute2 lst olds news)
(cond ((null? lst) '())
((MEMBER lst olds)
(FIND-MATCH LST OLDS NEWS))
((pair? lst)
(cons (substitute2 (car lst) olds news)
(substitute2 (cdr lst) olds news)))
(else lst)))
(define (find-match element olds news)
(cond ((null? olds) #f) ; This shouldn't happen
((equal? element (car olds))
(car news))
(else (find-match element (cdr olds) (cdr news)))))
This works fine. But it's a little inelegant that we end up looking for the
element in OLDS twice, once with MEMBER and once again with FIND-MATCH. We
can photograph two birds with one stone if we replace the first COND clause
of find-match with one that returns the value we actually want if ELEMENT
isn't found in OLDS:
(define (substitute2 lst olds news)
(cond ((null? lst) '())
((pair? lst)
(cons (substitute2 (car lst) olds news)
(substitute2 (cdr lst) olds news)))
(else (FIND-MATCH LST OLDS NEWS))))
(define (find-match element olds news)
(cond ((null? olds) ELEMENT)
((equal? element (car olds))
(car news))
(else (find-match element (cdr olds) (cdr news)))))
This version of SUBSTITUTE2 has one fewer COND clause. The two clauses
for lists remain, and the ELSE clause now handles all words, whether or
not they're members of OLDS.
Extra for experts:
(define (cxr-function name)
(define (helper name)
(if (empty? name)
(lambda (x) x)
(compose (if (equal? (first name) 'a) car cdr)
(helper (bf name)))))
(helper (bf (bl name)))) ; eliminate C and R from name
(define (compose f g)
(lambda (x) (f (g x))))
If you didn't think to use COMPOSE, you could of course do the same thing
without it, a little less neatly:
(define (cxr-function name)
(define (helper name)
(if (empty? name)
(lambda (x) x)
(lambda (x) ((if (equal? (first name) 'a) car cdr)
((helper (bf name)) x)))))
(helper (bf (bl name)))) ; eliminate C and R from name