*Due by 11:59pm on Friday, 5/2*

Homework 10 is in two parts, of which this is the first. The solutions to this part are in Python. Those for 10b are in logic notation.

**Submission.** See the online submission instructions.
Submit the files for hw10a and hw10b together as hw10.
We have provided a hw10a starter file for the questions below.

**Readings.** Sections
`4.2
4.3, and
4.4
of Composing Programs.

**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
leftandrightattributes of aMobileshould both beBranchinstances. Check that the types of constructor arguments forMobileareBranchinstances, and raise an appropriateTypeErrorfor incorrect argument types. See the doctest forMobilefor exception details.- The
lengthof aBranchand theweightof aWeightshould be positive numbers. Implementcheck_positiveto check if an objectxis a positive number.- Add a property
weightthat 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. Implement the method
is_balancedthat returnsTrueif and only if theMobileis balanced.

When you are finished, all doctests below should pass:

class Mobile: """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.is_balanced() True >>> m.left.contents = Mobile(Branch(1, Weight(1)), Branch(2, Weight(1))) >>> m.weight 3 >>> m.left.contents.is_balanced() False >>> m.is_balanced() # All submobiles must be balanced for m to be balanced False >>> m.left.contents.right.contents.weight = 0.5 >>> m.left.contents.is_balanced() True >>> m.is_balanced() False >>> m.right.length = 1.5 >>> m.is_balanced() True """ def __init__(self, left, right): "*** YOUR CODE HERE ***" self.left = left self.right = right @property def weight(self): """The total weight of the mobile.""" "*** YOUR CODE HERE ***" def is_balanced(self): """True if and only if the mobile is balanced.""" "*** YOUR CODE HERE ***" 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 """ "*** YOUR CODE HERE ***" class Branch: """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: """A weight.""" def __init__(self, weight): check_positive(weight) self.weight = weight def is_balanced(self): return 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).is_balanced() 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.is_balanced() 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'): "*** YOUR CODE HERE ***" return None

**Q3.** An enhancement of the `Stream` class from lecture appears below,
along with a function
that returns an infinite stream of integers. To extend it, implement an
`__iter__` method using a `yield` statement that returns a generator over
the elements of the stream. Also add a `__getitem__` method to support item
selection.

The `zip` function in the doctest for `__iter__` combines the corresponding
elements of two iterables into pairs until one iterator raises
`StopIteration`:

class Stream: """A lazily computed recursive 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)) def __iter__(self): """Return an iterator over the elements in the stream. >>> list(zip(range(6), ints)) [(0, 1), (1, 2), (2, 3), (3, 4), (4, 5), (5, 6)] """ "*** YOUR CODE HERE ***" def __getitem__(self, k): """Return the k-th element of the stream. >>> ints[5] 6 >>> increment_stream(ints)[7] 9 """ "*** YOUR CODE HERE ***" def increment_stream(s): """Increment all elements of a stream.""" return Stream(s.first+1, lambda: increment_stream(s.rest)) # The stream of consecutive integers starting at 1. ints = Stream(1, lambda: increment_stream(ints))

**Q4.** 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. >>> s = scale_stream(ints, 5) >>> s.first 5 >>> s.rest Stream(10, <...>) >>> scale_stream(s.rest, 10)[2] 200 """ "*** YOUR CODE HERE ***"

**Q5.** 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.

sbegins with1.- The elements of
scale_stream(s, 2)are also elements ofs.- The same is true for
scale_stream(s, 3)andscale-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.

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 increasing s0 and s1, removing repeats. >>> 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 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() >>> [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(): "*** YOUR CODE HERE ***" s = Stream(1, rest) return s