CS 61A: Homework 8

Due by 11:59pm on Wednesday, 11/12

Submission: See Lab 1 for submission instructions. We have provided a hw8.scm starter file for the questions below.

Readings: You might find the following references useful:

Table of Contents

To complete this homework assignment on your own computer, you will either need to install Scheme or use our online Scheme interpreter.

You can load the starter file in STK using the command

stk -load hw8.scm

Scheme does not have a built-in testing framework. To verify behavior, we will use the following assert-equal procedure, which tests whether an expression evaluates to an expected value.

(define (assert-equal value expression)
  (if (equal? value (eval expression))
    (print 'ok)
    (print (list 'for expression ':
                 'expected value
                 'but 'got (eval expression)))))

When you evaluate your solution, a long sequence of ok should be displayed. Any test failures will describe the error instead.

Question 1

Write the function deep-map, which takes a function fn and a list s that may contain numbers or other lists. It returns a list with identical structure to s, but replacing each non-list element by the result of applying fn on it, even for elements within sub-lists.

(define (deep-map fn s)

(define (square x) (* x x))
(define (double x) (* 2 x))
(define (test-deep-map)
  (assert-equal '(4 9) '(deep-map square '(2 3)))
  (assert-equal '(4 (6 8)) '(deep-map double '(2 (3 4))))
  (assert-equal '(1 4 (9 16 (25 36) ((49)) 64) (81 100))
                '(deep-map square
                '(1 2 (3 4  (5  6)  ((7))  8)  (9  10)))))


Hint: You can use the predicate list? to check if a value is a list.

Question 2

Write a procedure substitute that takes three arguments: a list s, an old word, and a new word. It returns a list with the elements of s, but with every occurrence of old replaced by new, even within sub-lists.

(define (substitute-list s olds news)

(define (test-substitute-list)
  (assert-equal '((four calling birds) (three french hens) (two turtle doves))
                  '((4 calling birds) (3 french hens) (2 turtle doves))
                  '(1 2 3 4)
                  '(one two three four))))


Question 3

Write substitute-list, that takes a list s, a list of old words, and a list of new words; the last two lists must be the same length. It returns a list with the elements of s, but with each word that occurs in the second argument replaced by the corresponding word of the third argument.


The following problems develop a system for symbolic differentiation of algebraic expressions. The derive Scheme procedure takes an algebraic expression and a variable and returns the derivative of the expression with respect to the variable. Symbolic differentiation is of special historical significance in Lisp. It was one of the motivating examples behind the development of the language. Differentiating is a recursive process that applies different rules to different kinds of expressions:

; Derive returns the derivative of exp with respect to var.
(define (derive expr var)
  (cond ((number? expr) 0)
        ((variable? expr) (if (same-variable? expr var) 1 0))
        ((sum? expr) (derive-sum expr var))
        ((product? expr) (derive-product expr var))
        ((exp? expr) (derive-exp expr var))
        (else 'Error)))

To implement the system, we will use the following data abstraction. Sums and products are lists, and they are simplified on construction:

; Variables are represented as symbols
(define (variable? x) (symbol? x))
(define (same-variable? v1 v2)
  (and (variable? v1) (variable? v2) (eq? v1 v2)))

; Numbers are compared with =
(define (=number? expr num)
  (and (number? expr) (= expr num)))

; Sums are represented as lists that start with +.
(define (make-sum a1 a2)
  (cond ((=number? a1 0) a2)
        ((=number? a2 0) a1)
        ((and (number? a1) (number? a2)) (+ a1 a2))
        (else (list '+ a1 a2))))
(define (sum? x)
  (and (pair? x) (eq? (car x) '+)))
(define (addend s) (cadr s))
(define (augend s) (caddr s))

; Products are represented as lists that start with *.
(define (make-product m1 m2)
(cond ((or (=number? m1 0) (=number? m2 0)) 0)
      ((=number? m1 1) m2)
      ((=number? m2 1) m1)
      ((and (number? m1) (number? m2)) (* m1 m2))
      (else (list '* m1 m2))))
(define (product? x)
  (and (pair? x) (eq? (car x) '*)))
(define (multiplier p) (cadr p))
(define (multiplicand p) (caddr p))

(define (test-sum)
  (assert-equal '(+ a x)       '(make-sum 'a 'x))
  (assert-equal '(+ a (+ x 1)) '(make-sum 'a (make-sum 'x 1)))
  (assert-equal 'x             '(make-sum 'x 0))
  (assert-equal 'x             '(make-sum 0 'x))
  (assert-equal 4              '(make-sum 1 3)))

(define (test-product)
  (assert-equal '(* a x) '(make-product 'a 'x))
  (assert-equal 0        '(make-product 'x 0))
  (assert-equal 'x       '(make-product 1 'x))
  (assert-equal 6        '(make-product 2 3)))


Question 4

Implement derive-sum, a procedure that differentiates a sum by summing the derivatives of the addend and augend. Use the abstract data type for a sum:

(define (derive-sum expr var)

(define (test-derive-sum)
  (assert-equal 1 '(derive '(+ x 3) 'x)))


Question 5

Implement derive-product, which applies the product rule to differentiate products:

(define (derive-product expr var)

(define (test-derive-product)
  (assert-equal 'y '(derive '(* x y) 'x))
  (assert-equal '(+ (* x y) (* y (+ x 3)))
                '(derive '(* (* x y) (+ x 3)) 'x)))


Question 6

Implement an abstract data type for exponentiation: a base raised to the power of an exponent. The base can be any expression, but assume that the exponent is a non-negative integer. You can simplify the cases when exponent is 0 or 1, or when base is a number, by returning numbers from the constructor make-exp. In other cases, you can represent the exp as a triple (^ base exponent).

Hint: The built-in procedure expt takes two number arguments and raises the first to the power of the second.

; Exponentiations are represented as lists that start with ^.
(define (make-exp base exponent)

(define (base exp)

(define (exponent exp)

(define (exp? exp)

(define x^2 (make-exp 'x 2))
(define x^3 (make-exp 'x 3))

(define (test-exp)
  (assert-equal 'x '(make-exp 'x 1))
  (assert-equal 1  '(make-exp 'x 0))
  (assert-equal 16 '(make-exp 2 4))
  (assert-equal '(^ x 2) 'x^2)
  (assert-equal 'x    '(base x^2))
  (assert-equal 2     '(exponent x^2))
  (assert-equal true  '(exp? x^2))
  (assert-equal false '(exp? 1))
  (assert-equal false '(exp? 'x))


Question 7

Implement derive-exp, which uses the power rule to derive exps:

(define (derive-exp exp var)

(define (test-derive-exp)
  (assert-equal '(* 2 x)                   '(derive x^2 'x))
  (assert-equal '(* 3 (^ x 2))             '(derive x^3 'x))
  (assert-equal '(+ (* 3 (^ x 2)) (* 2 x)) '(derive (make-sum x^3 x^2) 'x))