# Homework 9

*Due by 11:59pm on Monday, 8/7*

## Instructions

Download hw09.zip.

**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.

# Homework Questions

## Iterators

### Question 1: Link Iterable

To make the`Link`

class iterable, implement the `LinkIterator`

class.
```
class LinkIterator:
"""
>>> lnk = Link(1, Link(2, Link(3)))
>>> lnk_iter = LinkIterator(lnk)
>>> next(lnk_iter)
1
>>> next(lnk_iter)
2
"""
def __init__(self, link):
"*** YOUR CODE HERE ***"
def __iter__(self):
"*** YOUR CODE HERE ***"
def __next__(self):
"*** YOUR CODE HERE ***"
```

Use OK to test your code:

`python3 ok -q LinkIterator`

## Generators

### Question 2: In Order

Write a function that yields the entries of a valid binary search tree in sorted order.

```
def in_order(t):
"""
Yields the entries of a valid binary search tree in sorted order.
>>> b = BTree(5, BTree(3, BTree(2), BTree(4)), BTree(6))
>>> list(in_order(b))
[2, 3, 4, 5, 6]
>>> list(in_order(bst([1, 3, 5, 7, 9, 11, 13])))
[1, 3, 5, 7, 9, 11, 13]
>>> list(in_order(BTree(1)))
[1]
"""
"*** YOUR CODE HERE ***"
```

Hint:The`yield from`

expression is helpful if you want to yield all the values from another sequence.

Use OK to test your code:

`python3 ok -q in_order`

### Question 3: Generate Permutations

Given a list of unique elements, a *permutation* of the list is a
reordering of the elements. For example, `[2, 1, 3]`

, `[1, 3, 2]`

, and
`[3, 2, 1]`

are all permutations of the list `[1, 2, 3]`

.

Implement `permutations`

, a generator function that takes in a `lst`

and outputs
all permutations of `lst`

, each as a list (see doctest for an example).

```
def permutations(lst):
"""Generates all permutations of sequence LST. Each permutation is a
list of the elements in LST in a different order.
The order of the permutations does not matter.
>>> sorted(permutations([1, 2, 3]))
[[1, 2, 3], [1, 3, 2], [2, 1, 3], [2, 3, 1], [3, 1, 2], [3, 2, 1]]
>>> type(permutations([1, 2, 3]))
<class 'generator'>
>>> sorted(permutations((10, 20, 30)))
[[10, 20, 30], [10, 30, 20], [20, 10, 30], [20, 30, 10], [30, 10, 20], [30, 20, 10]]
>>> sorted(permutations("ab"))
[['a', 'b'], ['b', 'a']]
"""
if not lst:
yield []
return
"*** YOUR CODE HERE ***"
```

The order in which you generate permutations is irrelevant.

Hint:If you had the permutations of`lst`

minus one element, how could you use that to generate the permutations of the full`lst`

?

Use OK to test your code:

`python3 ok -q permutations`

## Streams

### Question 4: 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))
[]
"""
"*** YOUR CODE HERE ***"
```

Use OK to test your code:

`python3 ok -q scale_stream`

### Question 5: 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
"*** YOUR CODE HERE ***"
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():
"*** YOUR CODE HERE ***"
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 6: 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.
>>> 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]
"""
"*** YOUR CODE HERE ***"
```

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`

### Question 7: 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_of_streams.rest
Stream(Stream(2, <...>), <...>)
>>> stream_of_streams.rest.rest
Stream(Stream(3, <...>), <...>)
>>> stream_of_streams
Stream(Stream(1, Stream(2, Stream(3, <...>))), Stream(Stream(2, Stream(3, <...>)), Stream(Stream(3, <...>), <...>)))
"""
"*** YOUR CODE HERE ***"
```

Use OK to test your code:

`python3 ok -q make_stream_of_streams`