# 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`

### 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.

### 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

### 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!