Homework 09

Due by 11:59pm on Thursday, 4/5


Download hw09.zip.

Our course uses a custom version of Scheme (which you will build for Project 4) included in the starter ZIP archive. To start the interpreter, type python3 scheme. To run a Scheme program interactively, type python3 scheme -i <file.scm>. To exit the Scheme interpreter, type (exit).

Submission: When you are done, submit with python3 ok --submit. You may submit more than once before the deadline; only the final submission will be scored. Check that you have successfully submitted your code on okpy.org. See Lab 0 for more instructions on submitting assignments.

Using Ok: If you have any questions about using Ok, please refer to this guide.

Readings: You might find the following references useful:

Q1: How Many Dots?

Implement how-many-dots, which takes in a nested scheme list s and returns the number of dots that appear when it is displayed by the interpreter (not counting decimal points). You may assume that s is a nested list that contains only numbers.

Hints: A dot appears when the second element of a pair is not a well formed list. The procedures pair?, null?, and number? test whether a value is a pair, nil, or a number, respectively.

(define (how-many-dots s)

Use Ok to unlock and test your code:

python3 ok -q how-many-dots -u
python3 ok -q how-many-dots


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 EXPR 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 (list? 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 (list? x) (eq? (car x) '*)))
(define (multiplier p) (cadr p))
(define (multiplicand p) (caddr p))

Note that we will not test whether your solutions to this question correctly apply the chain rule. For more info, check out the extensions section.

Q2: Derive Sum

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

(define (derive-sum expr var)

Use Ok to unlock and test your code:

python3 ok -q derive-sum -u
python3 ok -q derive-sum

Q3: Derive Product

Implement derive-product, which applies the product rule to differentiate products. This means taking the multiplier and multiplicand, and then summing the result of multiplying one by the derivative of the other.

The ok tests expect the terms of the result in a particular order. First, multiply the derivative of the multiplier by the multiplicand. Then, multiply the multiplier by the derivative of the multiplicand. Sum these two terms to form the derivative of the original product.

(define (derive-product expr var)

Use Ok to unlock and test your code:

python3 ok -q derive-product -u
python3 ok -q derive-product

Q4: Make Exp

Implement a data abstraction 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).

You may want to use the built-in procedure expt, which 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))

Use Ok to unlock and test your code:

python3 ok -q make-exp -u
python3 ok -q make-exp

Q5: Derive Exp

Implement derive-exp, which uses the power rule to derive exponents. Reduce the power of the exponent by one, and multiply the entire expression by the original exponent.

(define (derive-exp exp var)

Use Ok to unlock and test your code:

python3 ok -q derive-exp -u
python3 ok -q derive-exp


There are many ways to extend this symbolic differentiation system. For example, you could simplify nested exponentiation expression such as (^ (^ x 3) 2), products of exponents such as (* (^ x 2) (^ x 3)), and sums of products such as (+ (* 2 x) (* 3 x)). You could apply the chain rule when deriving exponents, so that expressions like (derive '(^ (^ x y) 3) 'x) are handled correctly. Enjoy!