# Homework 09

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

## Instructions

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)
'YOUR-CODE-HERE
)
```

Use Ok to unlock and test your code:

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

### Differentiation

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)
'YOUR-CODE-HERE
)
```

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)
'YOUR-CODE-HERE
)
```

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)
'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))
```

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)
'YOUR-CODE-HERE
)
```

Use Ok to unlock and test your code:

```
python3 ok -q derive-exp -u
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!