Due by 11:59pm on Thursday, November 7

Instructions

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:

• Scheme Specification
• Scheme Built-in Procedure Reference

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. This homework is out of 2 points.

Scheme Editor

How to launch

In your hw07 folder you will find a new editor. To run this editor, run python3 editor. This should pop up a window in your browser; if it does not, please navigate to localhost:31415 and you should see it.

Make sure to run python3 ok in a separate tab or window so that the editor keeps running.

Features

The hw07.scm file should already be open. You can edit this file and then run Run to run the code and get an interactive terminal or Test to run the ok tests.

Environments will help you diagram your code, and Debug works with environments so you can see where you are in it. We encourage you to try out all these features.

Reformat is incredibly useful for determining whether you have parenthesis based bugs in your code. You should be able to see after formatting if your code looks weird where the issue is.

By default, the interpreter uses Lisp-style formatting, where the parens are all put on the end of the last line

(define (f x)
    (if (> x 0)
        x
        (- x)))

However, if you would prefer the close parens to be on their own lines as so

(define (f x)
    (if (> x 0)
        x
        (- x)
    )
)

you can go to Settings and select the second option.

Questions

Q1: 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 (documentation) and filter (documentation) might be helpful to solving this problem.

; helper function
; returns the values of lst that are bigger than x
; e.g., (larger-values 3 '(1 2 3 4 5 1 2 3 4 5)) --> (4 5 4 5)
(define (larger-values x lst)
    'YOUR-CODE-HERE)

(define (longest-increasing-subsequence lst)
    ; the following skeleton is optional, remove if you like
    (if (null? lst)
        nil
        (begin
            (define first (car lst))
            (define rest (cdr lst))
            (define large-values-rest
                (larger-values first rest))
            (define with-first
                'YOUR-CODE-HERE)
            (define without-first
                'YOUR-CODE-HERE)
            (if 'YOUR-CONDITION-HERE
                with-first
                without-first))))

To test your code, if you are in the local Scheme editor, hit Test. You can click on a case, press Run, and then use the Debug and Environments features to figure out why your code is not functioning correctly.

You can also test your code from the terminal by running

python3 ok -q longest-increasing-subsequence

Derivative

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.

Note: the formula for the derivative of a sum is (f(x) + g(x))' = f'(x) + g'(x)

(define (derive-sum expr var)
  'YOUR-CODE-HERE
)

The tests for this section aren't exhaustive, but tests for later parts will fully test it.

Before you start, check your understanding by running

python3 ok -q derive-sum -u

To test your code, if you are in the local Scheme editor, hit Test. You can click on a case, press Run, and then use the Debug and Environments features to figure out why your code is not functioning correctly.

You can also test your code from the terminal by running

python3 ok -q derive-sum

Q3: Derive Product

Note: the formula for the derivative of a product is (f(x) g(x))' = f'(x) g(x) + f(x) g'(x)

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. In other words, f' g + f g' not some other ordering

(define (derive-product expr var)
  'YOUR-CODE-HERE
)

Before you start, check your understanding by running

python3 ok -q derive-product -u

To test your code, if you are in the local Scheme editor, hit Test. You can click on a case, press Run, and then use the Debug and Environments features to figure out why your code is not functioning correctly.

You can also test your code from the terminal by running

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)
  'YOUR-CODE-HERE
)

(define (base exp)
  'YOUR-CODE-HERE
)

(define (exponent exp)
  'YOUR-CODE-HERE
)

(define (exp? exp)
  'YOUR-CODE-HERE
)

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

Before you start, check your understanding by running

python3 ok -q make-exp -u

To test your code, if you are in the local Scheme editor, hit Test. You can click on a case, press Run, and then use the Debug and Environments features to figure out why your code is not functioning correctly.

You can also test your code from the terminal by running

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)
  'YOUR-CODE-HERE
)

Before you start, check your understanding by running

python3 ok -q derive-exp -u

To test your code, if you are in the local Scheme editor, hit Test. You can click on a case, press Run, and then use the Debug and Environments features to figure out why your code is not functioning correctly.

You can also test your code from the terminal by running

python3 ok -q derive-exp

Extensions

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!