Homework 08 Solutions

Solution Files

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)
(if (null? lst) nil (append (reverse (cdr lst)) (list (car 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)
(if (null? lst) lst (begin (define with-car (cons (car lst) (longest-increasing-subsequence (filter (lambda (x) (> x (car lst))) (cdr lst))))) (define without-car (longest-increasing-subsequence (cdr lst))) (if (> (length with-car) (length without-car)) with-car without-car)))

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)
(make-sum (derive (addend expr) var) (derive (augend expr) var))

Use Ok to unlock and test your code:

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

This question is deceptively simple; make sure you understand what the problem is asking!

To derive a sum of values, we simply derive the addend (the thing before the + in normal math) and the augend (the thing after the +).

Then, we have to combine the values again using a sum. In some cases, using the sum operator will work; in fact, since we've only implemented derivatives of sums and single variables, we can't do anything too complicated here!

But the correct solution requires the use of make-sum which will helpfully simplify arithmetic operations and handle symbols. This necessary is for cases where you have to derive something like the following (after implementing derive-exp):

  scm> (derive (make-sum x^3 x^2) 'x)
  (+ (* 3 (^ x 2)) (* 2 x))

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

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)
(make-sum (make-product (derive (multiplier expr) var) (multiplicand expr)) (make-product (multiplier expr) (derive (multiplicand expr) var)))

Use Ok to unlock and test your code:

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

Much like the derive-sum, make sure you understand what the problem is asking.

The main hiccup this time is that the derivative rules are a bit more complicated for products, and will require using products and sums. Just make sure to use make-sum and make-product, otherwise you may run into issues further on.

Video walkthrough: https://youtu.be/biE0VS5dFJA

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)
(cond ((= exponent 0) 1) ((= exponent 1) base) ((and (number? base) (integer? exponent)) (expt base exponent)) (else (list '^ base exponent)))
) (define (base exp)
(cadr exp)
) (define (exponent exp)
(caddr exp)
) (define (exp? exp)
(and (list? exp) (eq? (car 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

It may be helpful to refer to code for make-sum and make-prod to see how they handle some special cases.

  • For exponent of 0 or 1, we can return a simplified result.
  • If we're doing the exponent of two numbers, we can compute that right away using expt instead of creating an exponent abstraction. This is much like make-sum and make-prod where we calculate the result using + or *.
  • Otherwise, we create the exponent abstraction -- a list of the caret ^, base, and exponent.

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

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)
(let ((b (base exp)) (n (exponent exp))) (if (number? n) (let ((first (make-product n (make-exp b (- n 1))))) (make-product first (derive b var))) ;; Note: Chain rule isn't tested by ok. 'unknown))

Use Ok to unlock and test your code:

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

Applying the power rule here is fairly straightforward -- it's arguably not much different from derive-sum or derive-prod. The only real differences are that we're using the power rule and the result will require using make-exp and make-prod.

Our solution here also implements the chain rule, but it's not necessary to pass the tests. An implementation that doesn't use the chain rule could just return:

(make-product n (make-exp b (- n 1)))

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


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!