# Homework 8

*Due by 11:59pm on Wednesday, 11/4*

## Instructions

Download hw08.zip. Inside the archive, you will find a file called hw08.scm, along with a copy of the OK autograder.

**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. See Lab 1 for 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:

### Question 1

Write a procedure `substitute`

that takes three arguments: a list `s`

, an `old`

word, and a `new`

word. It returns a list with the elements of `s`

, but with
every occurrence of `old`

replaced by `new`

, even within sub-lists.

```
(define (substitute s old new)
'YOUR-CODE-HERE
)
```

Use OK to unlock and test your code:

```
python3 ok -q substitute -u
python3 ok -q substitute
```

### Question 2

Write `sub-all`

, which takes a list `s`

, a list of `old`

words, and a
list of `new`

words; the last two lists must be the same length. It returns a
list with the elements of `s`

, but with each word that occurs in the second
argument replaced by the corresponding word of the third argument.

```
(define (sub-all s olds news)
'YOUR-CODE-HERE
)
```

Use OK to unlock and test your code:

```
python3 ok -q sub-all -u
python3 ok -q sub-all
```

### 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 exp 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))
```

### Question 3

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
```

### Question 4

Implement `derive-product`

, which applies the product
rule to differentiate
products:

```
(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
```

### Question 5

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)`

.

*Hint*: The built-in procedure `expt`

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
```

### Question 6

Implement `derive-exp`

, which uses the power
rule to derive
exps:

```
(define (derive-exp exp var)
'YOUR-CODE-HERE
)
```

Use OK to test your code:

`python3 ok -q derive-exp`