# 61A Homework 12

Due by 11:59pm on Wednesday, 4/24

Submission. See the online submission instructions. We have provided a starter file for the questions below.

Readings. Section 4.2 of the online lecture notes.

Q1. (Objects and recursion review) A mobile is a type of hanging sculpture. A simple binary mobile consists of two branches, left and right. Each branch is a rod of a certain length, from which hangs either a weight or another mobile.

Improve the classes for Branch, Weight, and Mobile below in the following ways:

• The left and right attributes of a Mobile should both be Branch instances. Check that the types of constructor arguments for Mobile are Branch instances, and raise an appropriate TypeError for incorrect argument types. See the doctest for Mobile for exception details.
• The length of a Branch and the weight of a Weight should be positive numbers. Implement check_positive to check if an object x is a positive number.
• Add a property weight that gives the total weight of the mobile.
• A mobile is said to be balanced if the torque applied by its left branch is equal to that applied by its right branch (that is, if the length of the left rod multiplied by the weight hanging from that rod is equal to the corresponding product for the right side) and if each of the submobiles hanging off its branches is balanced. Add a property method isbalanced that returns True if and only if the Mobile is balanced.

When you are finished, all doctests below should pass:

```class Mobile(object):
"""A simple binary mobile that has branches of weights or other mobiles.

>>> Mobile(1, 2)
Traceback (most recent call last):
...
TypeError: 1 is not a Branch
>>> m = Mobile(Branch(1, Weight(2)), Branch(2, Weight(1)))
>>> m.weight
3
>>> m.isbalanced
True
>>> m.left.contents = Mobile(Branch(1, Weight(1)), Branch(2, Weight(1)))
>>> m.weight
3
>>> m.left.contents.isbalanced
False
>>> m.isbalanced # All submobiles must be balanced for m to be balanced
False
>>> m.left.contents.right.contents.weight = 0.5
>>> m.left.contents.isbalanced
True
>>> m.isbalanced
False
>>> m.right.length = 1.5
>>> m.isbalanced
True
"""

def __init__(self, left, right):
self.left = left
self.right = right

@property
def weight(self):
"""The total weight of the mobile."""

@property
def isbalanced(self):
"""True if and only if the mobile is balanced."""

def check_positive(x):
"""Check that x is a positive number, and raise an exception otherwise.

>>> check_positive('hello')
Traceback (most recent call last):
...
TypeError: hello is not a number
>>> check_positive('1')
Traceback (most recent call last):
...
TypeError: 1 is not a number
>>> check_positive(-2)
Traceback (most recent call last):
...
ValueError: -2 <= 0
"""

class Branch(object):
"""A branch of a simple binary mobile."""

def __init__(self, length, contents):
if type(contents) not in (Weight, Mobile):
raise TypeError(str(contents) + ' is not a Weight or Mobile')
check_positive(length)
self.length = length
self.contents = contents

@property
def torque(self):
"""The torque on the branch"""
return self.length * self.contents.weight

class Weight(object):
"""A weight."""
def __init__(self, weight):
check_positive(weight)
self.weight = weight
self.isbalanced = True
```

Q2. (Python evaluation and recursion review) Your partner designed a beautiful balanced Mobile, but forgot to fill in the classes of each part, instead just writing T.

T(T(4,T(T(4,T(1)),T(1,T(4)))),T(2,T(10)))

The built-in Python funciton eval takes a string argument, evaluates it as a Python expression, and returns its value.

Complete the definition of interpret_mobile so that it returns a well-formed mobile by guessing the class for each T. The function should exhaustively test all possible combinations of types, then attempt to eval the resulting string when no T remains, handling TypeErrors until a correct series of types is found.

Warning: Interpreting a large mobile is quite slow (can you say why?). You will want to remove the doctest for the large mobile during development:

```def interpret_mobile(s):
"""Return a Mobile described by string s by substituting one of the classes
Branch, Weight, or Mobile for each occurrenct of the letter T.

>>> simple = 'Mobile(T(2,T(1)), T(1,T(2)))'
>>> interpret_mobile(simple).weight
3
>>> interpret_mobile(simple).isbalanced
True
>>> s = 'T(T(4,T(T(4,T(1)),T(1,T(4)))),T(2,T(10)))'
>>> m = interpret_mobile(s)
>>> m.weight
15
>>> m.isbalanced
True
"""
next_T = s.find('T')        # The index of the first 'T' in s.
if next_T == -1:            # The string 'T' was not found in s
try:
return eval(s)      # Interpret s
except TypeError as e:
return None         # Return None if s is not a valid mobile
for t in ('Branch', 'Weight', 'Mobile'):
return None
```

Q3. Implement the function scale_iterator that returns an iterator over each element of an input iterator, scaled by k.

Hint: Define scale_iterator as a generator function:

```def scale_iterator(s, k):
"""Return an iterator over the elements of s scaled by a number k.

>>> s = scale_iterator(iter(make_integer_stream(3)), 5)
>>> next(s)
15
>>> next(s)
20
"""
```

Q4. Implement the function iterator_to_stream that converts an iterator into a stream, allowing the elements in an iterator to be indexed:

```def iterator_to_stream(iterator):
"""Convert an iterator into a stream.

>>> s = iterator_to_stream(iter([3, 2, 1]))
>>> s.first
3
>>> s.rest
Stream(2, <...>)
>>> s.rest.rest.rest
Rlist.empty
"""
```

Q5. A famous problem, first raised by R. Hamming, is to enumerate, in ascending order with no repetitions, all positive integers with no prime factors other than 2, 3, or 5. 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 procedure merge that combines two ordered streams into one ordered result stream, eliminating repetitions.

The following is a definition of scale_stream in terms of scale_iterator:

```def scale_stream(s, k):
"""Return a stream over the elements of s scaled by a number k.

>>> s = scale_stream(make_integer_stream(3), 5)
>>> s.first
15
>>> s.rest
Stream(20, <...>)
>>> scale_stream(s.rest, 10)
300
"""
return iterator_to_stream(scale_iterator(iter(s), k))
```

Fill in the definition of merge, then use merge and scale_stream to fill in the definition of make_s below:

```def merge(s0, s1):
"""Return a stream over the elements of increasing s0 and s1, removing
repeats.

>>> ints = make_integer_stream(1)
>>> twos = scale_stream(ints, 2)
>>> threes = scale_stream(ints, 3)
>>> m = merge(twos, threes)
>>> [m[i] for i in range(10)]
[2, 3, 4, 6, 8, 9, 10, 12, 14, 15]
"""
if s0 is Stream.empty:
return s1
if 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:

def make_s():
"""Return a stream over all positive integers with only factors 2, 3, & 5.

>>> s = make_s()
>>> [s[i] for i in range(20)]
[1, 2, 3, 4, 5, 6, 8, 9, 10, 12, 15, 16, 18, 20, 24, 25, 27, 30, 32, 36]
"""
def rest():
s = Stream(1, rest)
return s
```

Q6. (Extra for Experts) An iterator can enumerate the items in a tree as well as a sequence. Implement values, which interleaves the values of the branches of a tree.

Implement values in terms of the helper function interleave, described below. Both of these functions, values and interleave, should return generators that yield values incrementally rather than constructing a sequence.

Hint: The itertools module provides functions and recipes that may be helpful.

```class Tree:
"""An n-ary tree with internal values."""
def __init__(self, value, branches=[]):
self.value = value
self.branches = branches

def values(t):
"""Yield values of a tree by interleaving iterators of the branches.

>>> T = Tree
>>> t = T(1, [T(2, [T(4), T(6, [T(8)])]), T(3, [T(5), T(7)])])
>>> tuple(values(t))
(1, 2, 3, 4, 5, 6, 7, 8)
"""

def interleave(*iterables):
"""Interleave elements of iterables.

>>> tuple(interleave([1, 4], [2, 5, 7, 8], [3, 6]))
(1, 2, 3, 4, 5, 6, 7, 8)
"""
```

The Rlist and Stream classes from lecture, along with a function to create an infinite stream of integers:

```class Rlist(object):
"""A recursive list consisting of a first element and the rest.

>>> s = Rlist(1, Rlist(2, Rlist(3)))
>>> len(s)
3
>>> s
1
>>> s
2
>>> s
3
"""
class EmptyList(object):
def __repr__(self):
return 'Rlist.empty'
def __len__(self):
return 0
empty = EmptyList()

def __init__(self, first, rest=empty):
self.first = first
self.rest = rest

def __repr__(self):
f = repr(self.first)
if self.rest is Rlist.empty:
return 'Rlist({0})'.format(f)
else:
return 'Rlist({0}, {1})'.format(f, repr(self.rest))

def __len__(self):
return 1 + len(self.rest)

def __getitem__(self, k):
if k == 0:
return self.first
return self.rest[k - 1]

def __iter__(self):
current = self
while current is not Rlist.empty:
yield current.first
current = current.rest

class Stream(Rlist):
"""A lazily computed recursive list.

>>> s = Stream(1, lambda: Stream(2+3, lambda: Stream(9)))
>>> s.first
1
>>> s.rest.first
5
>>> s.rest
Stream(5, <...>)
>>> s.rest.rest.first
9
>>> s = Stream(1, lambda: Stream(1+2, lambda: Stream(9)))
>>> s
9
>>> s
1
>>> s
3
"""
def __init__(self, first, compute_rest=lambda: Stream.empty):
if not callable(compute_rest):
raise TypeError('compute_rest must be callable')
self.first = first
self._compute_rest = compute_rest
self._rest = None

@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 __len__(self):
raise NotImplementedError('length not supported on Streams')

def __repr__(self):
return 'Stream({0}, <...>)'.format(repr(self.first))

def make_integer_stream(first=1):
"""Return an infinite stream of increasing integers."""
def compute_rest():
return make_integer_stream(first+1)
return Stream(first, compute_rest)
```