# Homework 7

*Due by 11:59pm on Friday, 4/15*

## Instructions

Download hw07.zip. Inside the archive, you will find a file called hw07.py, 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 0 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:

### Generators

### Question 1: 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.
>>> 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:** This
problem yields to recursion. If you had the permutations of a subsequence of
a that was missing one element, how could generate from it the permutations
of the full sequence?

Use OK to test your code:

`python3 ok -q permutations`

### Streams

The `Stream`

class defines a lazy sequence, a lazily
evaluated linked list. In other words, a Stream's elements (except for the
first element) are only evaluated as those values are needed.

```
class Stream:
"""A lazily computed linked list."""
class empty:
def __repr__(self):
return 'Stream.empty'
empty = empty()
def __init__(self, first, compute_rest=lambda: Stream.empty):
assert callable(compute_rest), 'compute_rest must be callable.'
self.first = first
self._compute_rest = compute_rest
@property
def rest(self):
"""Return the rest of the stream, computing it if necessary."""
if self._compute_rest is not None:
self._rest = self._compute_rest()
self._compute_rest = None
return self._rest
def __repr__(self):
return 'Stream({0}, <...>)'.format(repr(self.first))
```

Instead of specifying all of the elements in `__init__`

, we
provide a function, `compute_rest`

, that encapsulates the algorithm
used to calculate the remaining elements of the stream.
The code in the function body is not evaluated until it is called,
which lets us implement the desired evaluation behavior.

This implementation of streams also uses *memoization*. The first time
a program asks a `Stream`

for its `rest`

field, the `Stream`

code
computes the required value using `compute_rest`

, saves the resulting
value, and then returns it. After that, every time the `rest`

field is
referenced, the stored value is simply returned and it is not computed
again.

Here is an example (which you may use in solutions):

```
def make_integer_stream(first=1):
def compute_rest():
return make_integer_stream(first+1)
return Stream(first, compute_rest)
```

We start out with a stream whose first
element is 1, and whose `compute_rest`

function creates another stream.
So when we do compute the `rest`

, we get another stream whose first
element is one greater than the previous element, and whose
`compute_rest`

creates another stream. Hence, we effectively get an
infinite stream of integers, computed one at a time. This is almost
like an infinite recursion, but one which can be viewed one step at a
time, and so does not crash.

Another example:

```
def map_stream(fn, s):
if s is Stream.empty:
return s
return Stream(fn(s.first), lambda: map_stream(fn, s.rest))
```

### Question 2: 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 3: Efficient Map

The `map-stream`

function:

```
def map_stream(fn, s):
if s is Stream.empty:
return s
return Stream(fn(s.first), lambda: map_stream(fn, s.rest))
```

creates a new lambda function each time the tail of the list is
expanded via the `rest`

attribute, because it calls `map_stream`

, which
then executes `lambda: map_stream(fn, s.rest)`

. The value of a lambda
expression is a new Python object, requiring space. But each function
that is given as a second argument to `Stream`

is only called once, and then
thrown away (`self._compute_rest = None`

). See if you can find a way to
avoid creating a new function, and instead re-use the same function for each
`_compute_rest`

value. Obviously, the crux of the problem is to devise a way
to get this (parameterless) function to return a different value each time.

```
def lst_to_stream(lst):
"""A utility for converting iterator or iterable LST into a corresponding
Stream value. (Use for testing only)"""
if not lst:
return Stream.empty
return Stream(lst[0], lambda: lst_to_stream(lst[1:]))
def efficient_map_stream(fn, s):
"""Compute the Stream resulting from applying FN to each value of Stream
S in turn. [This implementation should only create a finite number of
functions in total, regardless of the length of the stream S.]
>>> str = lst_to_stream([7, -9, -3, 0])
>>> ab = efficient_map_stream(abs, str)
>>> stream_to_list(ab)
[7, 9, 3, 0]
>>> stream_to_list(efficient_map_stream(abs, Stream.empty))
[]
>>> stream_to_list(efficient_map_stream(abs, Stream(-3)))
[3]
>>> stream_to_list(efficient_map_stream(lambda x: x*x, make_integer_stream(1)))
[1, 4, 9, 16, 25, 36, 49, 64, 81, 100]
"""
"*** YOUR CODE HERE ***"
```

Use OK to test your code:

`python3 ok -q efficient_map_stream`

### Question 4: 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).

```
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, <...>)]
"""
"*** YOUR CODE HERE ***"
```

Use OK to test your code:

`python3 ok -q make_stream_of_streams`

### 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`