Homework 08:

Due by 11:59pm on Thursday, 11/1


Download hw08.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:

Grading: Homework is graded based on effort, not correctness. However, there is no partial credit; you must show substantial effort on every problem to receive any points.

Q1: Reverse

Write the procedure reverse, which takes in a list lst and outputs a reversed list. Hint: you may find the built-in function append useful.

(define (reverse lst)

Use Ok to test your code:

python3 ok -q reverse-simple

Q2: Longest increasing subsequence

Write the procedure longest-increasing-subsequence, which takes in a list lst and returns the longest subsequence in which all the terms are increasing. Note: the elements do not have to appear consecutively in the original list. For example, the longest increasing subsequence of (1 2 3 4 9 3 4 1 10 5) is (1 2 3 4 9 10). Assume that the longest increasing subsequence is unique.

Hint: The built-in procedures length and filter might be helpful to solving this problem.

(define (longest-increasing-subsequence lst)

Use Ok to test your code:

python3 ok -q longest-increasing-subsequence


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.

Q3: 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

Q4: 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

Q5: 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

Q6: 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!