# Homework 8 Solutions

## Solution Files

You can find the solutions in hw08.py and hw08.scm

# Homework Questions

## Some Review: Sets

One really convenient thing about Python sets is that many operations on sets (adding elements, removing elements, checking membership) run in θ(1) (constant) time (usually).

Some of the problems use a utility method called `timeit`

, which takes a
parameterless function as argument, executes it, and returns the time required
to do so. It's a variation on the function `timeit.timeit`

function in the
Python3 library.

### Question 1: Missing Value

Write the following function so it (usually) runs in θ(n) time, where n is
the length of `lst0`

.

```
def missing_val(lst0, lst1):
"""Assuming that lst0 contains all the values in lst1, but lst1 is missing
one value in lst0, return the missing value. The values need not be
numbers.
>>> from random import shuffle
>>> missing_val(range(10), [1, 0, 4, 5, 7, 9, 2, 6, 3])
8
>>> big0 = [str(k) for k in range(15000)]
>>> big1 = [str(k) for k in range(15000) if k != 293 ]
>>> shuffle(big0)
>>> shuffle(big1)
>>> missing_val(big0, big1)
'293'
>>> timeit(lambda: missing_val(big0, big1)) < 1.0
True
"""
return list(set(lst0) - set(lst1))[0]
```

Use OK to test your code:

`python3 ok -q missing_val`

### Question 2: Find duplicates k

Write the following function so it runs in O(n) time.

Hint:Sets have an`add`

and a`remove`

method.

```
def find_duplicates_k(k, lst):
"""Returns True if there are any duplicates in lst that are within k
indices apart.
>>> find_duplicates_k(3, [1, 2, 3, 4, 1])
False
>>> find_duplicates_k(4, [1, 2, 3, 4, 1])
True
>>> find_duplicates_k(4, [1, 1, 2, 3, 3])
True
"""
prev_set = set()
for i, elem in enumerate(lst):
if elem in prev_set:
return True
prev_set.add(elem)
if i - k >= 0:
prev_set.remove(lst[i - k])
return False
```

Use OK to test your code:

`python3 ok -q find_duplicates_k`

## More Review: Scheme

Note:Q3 and Q4 (Substitute, Sub All) were extra lab questions from Lab 9. You may check the solutions if you are stuck, but we highly recommend you work through the problem on your own for practice.

### Question 3: Substitute

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.

*Hint*: The built-in `pair?`

predicate returns True if its argument is a `cons`

pair.

```
(define (substitute s old new)
(cond ((null? s) nil)
((pair? (car s)) (cons (substitute (car s) old new)
(substitute (cdr s) old new)))
((equal? (car s) old) (cons new
(substitute (cdr s) old new)))
(else (cons (car s) (substitute (cdr s) old new)))))
```

Use OK to test your code:

`python3 ok -q substitute`

### Question 4: Sub All

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)
(if (null? olds)
s
(sub-all (substitute s (car olds) (car news))
(cdr olds)
(cdr news))))
```

Use OK to test your code:

`python3 ok -q sub-all`

## Streams

### Question 5: Scale Stream

Implement the function `scale_stream`

, which returns a stream over each
element of an input stream, scaled by `k`

:

```
def scale_stream(s, k):
"""Return a stream of the elements of S scaled by a number K.
>>> ints = make_integer_stream(1)
>>> s = scale_stream(ints, 5)
>>> stream_to_list(s, 5)
[5, 10, 15, 20, 25]
>>> s = scale_stream(Stream("x", lambda: Stream("y")), 3)
>>> stream_to_list(s)
['xxx', 'yyy']
>>> stream_to_list(scale_stream(Stream.empty, 10))
[]
"""
if s is Stream.empty:
return Stream.empty
return Stream(s.first * k, lambda: scale_stream(s.rest, k))
```

Use OK to test your code:

`python3 ok -q scale_stream`

### Question 6: Regular Numbers

**Acknowledgements.** This exercise is taken from
*Structure and Interpretation of Computer Programs*,
Section 3.5.2.

A famous problem, first raised by Richard Hamming, is to enumerate, in
ascending order with no repetitions, all positive integers with no
prime factors other than 2, 3, or 5. These are called
regular numbers.
One obvious way to do this is to simply test each integer in turn to
see whether it has any factors other than 2, 3, and 5. But this is very
inefficient, since, as the integers get larger, fewer and fewer of them
fit the requirement. As an alternative, we can build a stream of such
numbers. Let us call the required stream of numbers `s`

and notice the
following facts about it.

`s`

begins with`1`

.- The elements of
`scale_stream(s, 2)`

are also elements of`s`

. - The same is true for
`scale_stream(s, 3)`

and`scale_stream(s, 5)`

. - These are all of the elements of
`s`

.

Now all we have to do is combine elements from these sources. For this
we define a `merge`

function that combines two ordered streams into
one ordered result stream, eliminating repetitions.

Fill in the definition of `merge`

, then fill in the definition of
`make_s`

below:

```
def merge(s0, s1):
"""Return a stream over the elements of strictly increasing s0 and s1,
removing repeats. Assume that s0 and s1 have no repeats.
>>> ints = make_integer_stream(1)
>>> twos = scale_stream(ints, 2)
>>> threes = scale_stream(ints, 3)
>>> m = merge(twos, threes)
>>> stream_to_list(m, 10)
[2, 3, 4, 6, 8, 9, 10, 12, 14, 15]
"""
if s0 is Stream.empty:
return s1
elif s1 is Stream.empty:
return s0
e0, e1 = s0.first, s1.first
if e0 < e1:
return Stream(e0, lambda: merge(s0.rest, s1))
elif e1 < e0:
return Stream(e1, lambda: merge(s0, s1.rest))
else:
return Stream(e0, lambda: merge(s0.rest, s1.rest))
def make_s():
"""Return a stream over all positive integers with only factors 2, 3, & 5.
>>> s = make_s()
>>> stream_to_list(s, 20)
[1, 2, 3, 4, 5, 6, 8, 9, 10, 12, 15, 16, 18, 20, 24, 25, 27, 30, 32, 36]
"""
def rest():
two_three = merge(scale_stream(s, 2), scale_stream(s, 3))
return merge(two_three, scale_stream(s, 5)) s = Stream(1, rest)
return s
```

Use OK to test your code:

`python3 ok -q merge`

Use OK to test your code:

`python3 ok -q make_s`

### Question 7: Linear Congruential Generator

A common method of producing pseudo-random numbers is by means of the following recurrence relation:

*R*_{0}= seed value*R*) %_{i+1}= (a*R_{i}+ c*n*where*R*denotes the ith pseudo-random number in the stream;_{i}*a*,*c*, and*n*are constant integers, and*seed value*is some initial value provided by the user or chosen automatically by the system.

Define a function that returns a stream of random numbers that uses this linear-congruential formula.

```
from operator import add, mul, mod
def make_random_stream(seed, a, c, n):
"""The infinite stream of pseudo-random numbers generated by the
recurrence r[0] = SEED, r[i+1] = (r[i] * A + C) % N. Your solution
must not use any lambdas or def's that we have not supplied in the
skeleton.
>>> s = make_random_stream(25, 29, 5, 32)
>>> stream_to_list(s, 10)
[25, 26, 23, 0, 5, 22, 3, 28, 17, 18]
>>> s = make_random_stream(17, 299317, 13, 2**20)
>>> stream_to_list(s, 10)
[17, 894098, 115783, 383424, 775373, 994174, 941859, 558412, 238793, 718506]
"""
astr = constant_stream(a)
cstr = constant_stream(c)
nstr = constant_stream(n)
result = Stream(seed,
lambda: combine_streams(mod,
combine_streams(add,
combine_streams(mul, result, astr),
cstr),
nstr))
return result
```

Your solution must use *only* the functions defined in the skeleton, without
defining any additional ones.
Likewise, any lambda expressions should contain only calls to these
functions.

Use OK to test your code:

`python3 ok -q make_random_stream`

## Extra questions

Extra questions are not worth extra credit and are entirely optional.

### Question 8: Stream of Streams

Write the function`make_stream_of_streams`

, which returns an infinite
stream, where the element at position `i`

, counting from 1, is an
infinite stream of integers that start from `i`

. Your solution should
have the form
```
result = Stream(..., lambda: ...)
return result
```

and should *not* require any additional helper functions (i.e., just use
recursively defined streams, and any additional functions supplied in your
starter file). You may find the `map_stream`

function useful.

```
def make_stream_of_streams():
"""
>>> stream_of_streams = make_stream_of_streams()
>>> stream_of_streams
Stream(Stream(1, <...>), <...>)
>>> stream_to_list(stream_of_streams, 3)
[Stream(1, <...>), Stream(2, <...>), Stream(3, <...>)]
>>> stream_to_list(stream_of_streams, 4)
[Stream(1, <...>), Stream(2, <...>), Stream(3, <...>), Stream(4, <...>)]
"""
result = Stream(make_integer_stream(1),
lambda: map_stream(lambda x: x.rest, result))
return result
```

Use OK to test your code:

`python3 ok -q make_stream_of_streams`

### 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))
```

### Question 9: 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 test your code:

`python3 ok -q derive-sum`

### Question 10: Derive Product

Implement `derive-product`

, which applies the product
rule to differentiate
products:

```
(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 test your code:

`python3 ok -q derive-product`

### Question 11: 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)`

.

*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)
(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 test your code:

`python3 ok -q make-exp`

### Question 12: Derive Exp

Implement `derive-exp`

, which uses the power
rule to derive
exps:

```
(define (derive-exp exp var)
(let ((b (base exp))
(n (exponent exp)))
(if (number? n)
(let ((n-1 (make-sum n -1)))
(let ((first (make-product n (make-exp b n-1))))
(make-product first (derive b var))))
'unknown)))
```

Use OK to test your code:

`python3 ok -q derive-exp`