Homework 7 Solutions

Solution Files

Q0: Survey

Before you get started writing code, please fill out the midterm survey.

Important Submission Note

You're not done yet! Add the passphrase you receive at the end of the survey to passphrase at the top of the homework. For example, if the passphrase was CS61A (it isn't 🙂), then the first line of your file should read:

passphrase = 'CS61A'

Instead of:

passphrase = '*** PASSPHRASE HERE ***'

Use Ok to test your code:

python3 ok -q survey

Q1: Cadr and Caddr

Define the procedures cadr and caddr, which return the second and third elements of a list, respectively:

(define (cddr s)
  (cdr (cdr s)))

(define (cadr s)
(car (cdr s))
) (define (caddr s)
(car (cddr s))
)

Following the given example of cddr, defining cadr and caddr should be fairly straightforward.

Just for fun, if this were a Python linked list question instead, the solution might look something like this:

cadr = lambda l: l.rest.first
caddr = lambda l: l.rest.rest.first

Use Ok to unlock and test your code:

python3 ok -q cadr-caddr -u
python3 ok -q cadr-caddr

Conditional expressions

The cond special form is a general conditional expression, similar to a multi-clause conditional statement in Python. The general form of a conditional expression is:

(cond
    (<p1> <el1>)
    (<p2> <el2>)
    ...
    (<pn> <eln>)
    (else <else-expressions>))

This consists of the symbol cond followed by sequences of expressions (<p> <el>) called clauses.

The first expression in each pair is a predicate: an expression whose value is interpreted as either being true or false.

In Scheme, all values except the special boolean value False (historically called #f) are interpreted as true values. There is also a boolean value True (historically called #t). In the 61A Scheme interpreter, False and #f can be used interchangeably.

Conditional expressions are evaluated as follows:

  • The predicates <p1>, <p2>, ..., <pn> are evaluated in that order until one of them evaluates to a true value (anything but False).
  • If some predicate, such as <p2>, evaluates to a true value, then the following sequence of consequent expressions, such as <el2>, is evaluated, and its final expression provides the value of the whole cond expression.
  • Otherwise, if an else clause is present, then the sequence of else-expressions is evaluated, and its final expression provides the value of the whole cond expression.
  • If no clause has a true predicate and there is no else clause, then the value of the cond is unspecified and should not be used.

Q2: Sign

Using a cond expression, define a procedure sign that takes in one parameter x and returns -1 if x is negative, 0 if x is zero, and 1 if x is positive.

(define (sign x)
(cond ((> x 0) 1) ((= x 0) 0) ((< x 0) -1))
)

The order of the cases doesn't really matter, and we could also use an else clause in the place of one of the conditions.

The condition looks something like this in Python:

if x > 0:
    return 1
elif x == 0:
    return 0
elif x < 0:
    return -1

Opinions differ on which is better, but hopefully you can see that the Scheme cond is quite readable and compact once you get used to it.

Use Ok to unlock and test your code:

python3 ok -q sign -u
python3 ok -q sign

Q3: Pow

Implement a procedure pow for raising the number b to the power of a nonnegative integer n that runs in Θ(log n) time.

Hint: Consider the following observations:

  1. b2k = (bk)2
  2. b2k+1 = b(bk)2

You may use the built-in predicates even? and odd?.

(define (square x) (* x x))

(define (pow b n)
(cond ((= n 0) 1) ((even? n) (square (pow b (/ n 2)))) (else (* b (pow b (- n 1)))))
)

Use Ok to unlock and test your code:

python3 ok -q pow -u
python3 ok -q pow

The else clause shows the basic recursive version of pow that we've seen before in class.

We save time in computation by avoiding an extra n/2 multiplications of the base. Instead, we just square the result of b^(n/2). This behaviour is what gives us Θ(log n) time performance.

Video walkthrough: https://youtu.be/-yt2EOv9ZLU

Q4: Ordered

Implement a procedure called ordered?, which takes a list of numbers and returns True if the numbers are in nondescending order, and False otherwise. Numbers are considered nondescending if each subsequent number is either larger or equal to the previous, that is:

1 2 3 3 4

Is nondescending, but:

1 2 3 3 2

Is not.

Hint: The built-in null? function returns whether its argument is nil.

(define (ordered? s)
(if (or (null? s) (null? (cdr s))) true (and (<= (car s) (cadr s)) (ordered? (cdr s))))
)

We approach this much like a standard Python linked list problem.

  • The base case is where s has zero or one items. Trivially, this is ordered.
  • Otherwise we check if the first element is in order -- that is, if it's smaller than the second element. If it's not, then the list is out of order. Otherwise, we check if the rest of s is in order.

You should verify for yourself that a Python implementation of this for linked lists is similar.

Use Ok to unlock and test your code:

python3 ok -q ordered -u
python3 ok -q ordered

Sets as Ordered Lists

A set is a type of collection that stores unique elements. The main operation associated with sets is checking whether a given value is in a set.

There is no such built-in set data type in Scheme, but one way to represent a set is by using an ordered list, where the ordering is used to make union/intersection functions more convenient. The following few questions explore this idea. Specifically, we will represent sets using Scheme lists ordered from least to greatest with no repeated elements.

Q5: Add

First, define add, which takes a set s and a value v as arguments. It returns a representation of a set containing the values in s and the value v. There should be no repeated elements in the return value.

We've provided a function empty? which returns True if the given set s has no elements.

(define (empty? s) (null? s))

(define (add s v)
(cond ((empty? s) (list v)) ((= (car s) v) s) ((> (car s) v) (cons v s)) ((< (car s) v) (cons (car s) (add (cdr s) v))))
)

Use Ok to unlock and test your code:

python3 ok -q add -u
python3 ok -q add

Inserting v into a sorted list is analogous to finding v in a sorted list. When we know that everything in s is larger than v, then we know that's the appropriate location to include v.

Video walkthrough: https://youtu.be/H49bWAcT_pk

Q6: Contains

Next, define contains?, which returns whether a set s contains value v.

; Sets as sorted lists
(define (contains? s v)
(cond ((empty? s) #f) ((> (car s) v) #f) ((= (car s) v) #t) ((< (car s) v) (contains? (cdr s) v)))
)

Use Ok to unlock and test your code:

python3 ok -q contains -u
python3 ok -q contains

Main ideas:

  • If the beginning of s is equal to v, then we're done.
  • If the beginning of s is smaller than v, then v could appear in the rest of s.
  • If the beginning of s is larger than v, then all the items of s are larger than v. So, v cannot be in s.

Video walkthrough: https://youtu.be/p3N-FhLjspg

Q7: Intersect and Union

Finally, define intersect, which returns a set containing only values that appear in both sets s and t, and union, which returns a set containing all values that appear in either set s or t.

Your implementation for both functions should run in linear time in the length of the input sets.

(define (intersect s t)
(cond ((or (empty? s) (empty? t)) nil) ((= (car s) (car t)) (cons (car s) (intersect (cdr s) (cdr t)))) ((< (car s) (car t)) (intersect (cdr s) t)) ((> (car s) (car t)) (intersect s (cdr t))))
) (define (union s t)
(cond ((empty? s) t) ((empty? t) s) ((= (car s) (car t)) (cons (car s) (union (cdr s) (cdr t)))) ((< (car s) (car t)) (cons (car s) (union (cdr s) t))) ((> (car s) (car t)) (cons (car t) (union s (cdr t)))))
)

Use Ok to unlock and test your code:

python3 ok -q intersect -u
python3 ok -q intersect
python3 ok -q union -u
python3 ok -q union

union and intersect are quite similar. The main difference is that we include items in union that get skipped in intersect (i.e. when (car s) and (car t) differ).

Video walkthrough: https://youtu.be/EWxT_Id9UlM