# CS 61A: Homework 9

Due by 11:59pm on Wednesday, 11/26

Submission: See Lab 1 for submission instructions. We have provided a hw9.py starter file for the questions below.

Readings: You might find the following references useful:

This extra-large homework is worth 6 points! While graded on effort, you should make progress on all problems to earn credit. Start early!

The Brackulator language shares an evaluator with the Calculator language, but uses a more concise syntax. Instead of using operator names or symbols, Brackulator indicates operations using different kinds of brackets:

• `[]`: add
• `()`: subtract
• `<>`: multiply
• `{}`: divide

Operand expressions are separated by spaces within these brackets. The following Brackulator expressions are followed by their Calculator equivalents.

``````<1 2 3>              (* 1 2 3)
(5)                  (- 5)
[2{4 8}]             (+ 2 (/ 4 8))
<[2{12 6}](3 4 5)>   (* (+ 2 (/ 12 6)) (- 3 4 5))``````

By solving the following problems, you will implement a parser, `brack_read`, that returns an expression for the `calc_eval` evaluator implemented in the Calculator example from lecture.

All of your solutions should be defined in terms of the following dictionaries of bracket types, which configure the parser to supply the correct operator for each bracket:

``````# A dictionary from pairs of matching brackets to the operators they indicate.
brackets = {('[', ']'): '+',
('(', ')'): '-',
('<', '>'): '*',
('{', '}'): '/'}

# A dictionary with left-bracket keys and corresponding right-bracket values.
left_to_right = {left: right for left, right in brackets}

# The set of all left and right brackets.
all_brackets = set(left_to_right.keys()).union(set(left_to_right.values()))``````

### Question 1

Complete `tokenize`, which splits a Brackulator expression into tokens, and raise a `ValueError` if any token is not a number or a known bracket. Hint: You can first surround each bracket with spaces using `line.replace`, then split on spaces. Afterward, check each token to ensure that it is legal. The provided `coerce_to_number` function may prove useful:

``````def tokenize(line):
"""Convert a string into a list of tokens.

>>> tokenize('2.3')
[2.3]
>>> tokenize('(2 3)')
['(', 2, 3, ')']
>>> tokenize('<2 3)')
['<', 2, 3, ')']
>>> tokenize('<[2{12.5 6.0}](3 -4 5)>')
['<', '[', 2, '{', 12.5, 6.0, '}', ']', '(', 3, -4, 5, ')', '>']

>>> tokenize('2.3.4')
Traceback (most recent call last):
...
ValueError: invalid token 2.3.4

>>> tokenize('?')
Traceback (most recent call last):
...
ValueError: invalid token ?

>>> tokenize('hello')
Traceback (most recent call last):
...
ValueError: invalid token hello

>>> tokenize('<(GO BEARS)>')
Traceback (most recent call last):
...
ValueError: invalid token GO
"""
# Surround all brackets by spaces so that they are separated by split.
for b in all_brackets:
line = line.replace(b, ' ' + b + ' ')

# Convert numerals to numbers and raise ValueErrors for invalid tokens.
tokens = []
for t in line.split():

def coerce_to_number(token):
"""Coerce a string to a number or return None.

>>> coerce_to_number('-2.3')
-2.3
>>> print(coerce_to_number('('))
None
"""
try:
return int(token)
except (TypeError, ValueError):
try:
return float(token)
except (TypeError, ValueError):
return None``````

### Question 2

Implement `brack_read`, which returns an expression tree for the first valid Brackulator expression in a list of tokens. The expression tree should contain Calculator operators that correspond to the bracket types. Raise a `SyntaxError` for any malformed expression. The `Pair` class and `nil` object from lecture appear at the bottom of this file. This function is similar to `scheme_read` from Calculator's Scheme reader file.

Hint: Introduce another function `read_tail` that reads the elements in a combination until reaching a closing bracket. In `brack_read` make sure that the closing bracket of an expression matches the opening bracket. The `left_to_right` dictionary defined above gives you the matching right bracket for each type of left bracket. The `brackets` dictionary gives you the corresponding operator (e.g. '+' for '[' and ']').

Once you complete this problem, you can place your homework file in the same directory as `scalc.py` (and its supporting files), then run `read_eval_print_loop` to interact with the Brackulator language:

``````def brack_read(tokens):
"""Return an expression tree for the first well-formed Brackulator
expression in tokens. Tokens in that expression are removed from tokens as
a side effect.

100
Pair('-', Pair(Pair('+', nil), nil))
(* (+ 2 (/ 12 6)) (- 3 4 5))
>>> brack_read(tokenize('(1)(1)')) # More than one expression is ok
Pair('-', Pair(1, nil))
>>> brack_read(tokenize('[])')) # Junk after a valid expression is ok
Pair('+', nil)

>>> brack_read(tokenize('([]')) # Missing right bracket
Traceback (most recent call last):
...
SyntaxError: unexpected end of line

>>> brack_read(tokenize('[)]')) # Extra right bracket
Traceback (most recent call last):
...
SyntaxError: unexpected )

Traceback (most recent call last):
...
SyntaxError: unexpected )

Traceback (most recent call last):
...
SyntaxError: unexpected end of line
"""
if not tokens:
raise SyntaxError('unexpected end of line')
token = tokens.pop(0)
n = coerce_to_number(token)
if n != None:
return n
elif token in left_to_right:

Support code for Brackulator (from the Calculator example):

``````###################################
# Support classes for Brackulator #
###################################

class Pair:
"""A pair has two instance attributes: first and second.  For a Pair to be
a well-formed list, second is either a well-formed list or nil.  Some
methods only apply to well-formed lists.

>>> s = Pair(1, Pair(2, nil))
>>> s
Pair(1, Pair(2, nil))
>>> print(s)
(1 2)
>>> len(s)
2
>>> s
2
>>> print(s.map(lambda x: x+4))
(5 6)
"""
def __init__(self, first, second):
self.first = first
self.second = second

def __repr__(self):
return "Pair({0}, {1})".format(repr(self.first), repr(self.second))

def __str__(self):
s = "(" + str(self.first)
second = self.second
while isinstance(second, Pair):
s += " " + str(second.first)
second = second.second
if second is not nil:
s += " . " + str(second)
return s + ")"

def __len__(self):
n, second = 1, self.second
while isinstance(second, Pair):
n += 1
second = second.second
if second is not nil:
raise TypeError("length attempted on improper list")
return n

def __getitem__(self, k):
if k < 0:
raise IndexError("negative index into list")
y = self
for _ in range(k):
if y.second is nil:
raise IndexError("list index out of bounds")
elif not isinstance(y.second, Pair):
raise TypeError("ill-formed list")
y = y.second
return y.first

def map(self, fn):
"""Return a Scheme list after mapping Python function FN to SELF."""
mapped = fn(self.first)
if self.second is nil or isinstance(self.second, Pair):
return Pair(mapped, self.second.map(fn))
else:
raise TypeError("ill-formed list")

class nil:
"""The empty list"""

def __repr__(self):
return "nil"

def __str__(self):
return "()"

def __len__(self):
return 0

def __getitem__(self, k):
if k < 0:
raise IndexError("negative index into list")
raise IndexError("list index out of bounds")

def map(self, fn):
return self

nil = nil() # Assignment hides the nil class; there is only one instance``````

To use the following function, you will need to place your homework solution in the same directory as the files from the Calculator Example:

``````def read_eval_print_loop():
"""Run a read-eval-print loop for the Brackulator language."""
global Pair, nil
from scalc import calc_eval

while True:
try:
src = tokenize(input('brack> '))
while len(src) > 0:
print(calc_eval(expression))
except (SyntaxError, ValueError, TypeError, ZeroDivisionError) as err:
print(type(err).__name__ + ':', err)
except (KeyboardInterrupt, EOFError):  # <Control>-D, etc.
return``````

### Question 3

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. Implement the method `is_balanced` that returns `True` if and only if the `Mobile` is 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):
self.left = left
self.right = right

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

def is_balanced(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(2)
>>> 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:
"""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``````

### Question 4

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'):
return None``````

### Question 5

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.

``````class Stream:

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.

>>> it = iter(ints)
>>> [next(it) for _ in range(6)]
[1, 2, 3, 4, 5, 6]
"""

def __getitem__(self, k):
"""Return the k-th element of the stream.

>>> ints
6
>>> increment_stream(ints)
9
>>> s = Stream(1, lambda: Stream(2))
>>> [s[i] for i in range(2)]
[1, 2]
>>> s
Traceback (most recent call last):
...
IndexError: Stream index out of range
"""

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))``````

### Question 6

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)
200
"""

### Question 7

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 the `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.

>>> 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

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``````

### Question 8

Implement a function `unique` that takes a stream and returns a stream in which duplicates of any value are filtered out. This should work for infinite streams as well as finite ones.

``````def unique(s):
"""Return a stream of the unique elements in s in the order that they
first appear.

>>> s = unique(to_stream([1, 2, 2, 1, 0, 4, 2, 3, 1, 9, 0]))
>>> [s[i] for i in range(6)]
[1, 2, 0, 4, 3, 9]
"""

def to_stream(lst):
if not lst:
return Stream.empty
return Stream(lst, lambda: to_stream(lst[1:]))``````

### Question 9

Run-length encoding is a very simple data compression technique, whereby runs of data are compressed and stored as a single value. A run is defined to be a contiguous sequence of the same number. For example, in the (finite) sequence

``1, 1, 1, 1, 1, 6, 6, 6, 6, 2, 5, 5, 5``

there are four runs: one each of 1, 6, 2, and 5. We can represent the same sequence as a sequence of tuples:

``(5, 1), (4, 6), (1, 2), (3, 5)``

Notice that the first element of each tuple is the number of times a particular number appears in a run, and the second element is the number in the run.

We will extend this idea to (possibly infinite) streams. Write a function called `rle` that takes in a stream of data, and returns a corresponding stream of tuples, which represents the run-length encoded version of the stream. It will also take in the maximum size of any given run (default 10) to prevent having to compress infinite runs.

``````def rle(s, max_run_length=10):
"""
>>> example_stream = to_stream([1, 1, 1, 2, 3, 3])
>>> encoded_example = rle(example_stream)
>>> [encoded_example[i] for i in range(3)]
[(3, 1), (1, 2), (2, 3)]
>>> shorter_encoded_example = rle(example_stream, 2)
>>> [shorter_encoded_example[i] for i in range(4)]
[(2, 1), (1, 1), (1, 2), (2, 3)]
>>> encoded_naturals = rle(ints)
>>> [encoded_naturals[i] for i in range(3)]
[(1, 1), (1, 2), (1, 3)]
>>> ones = Stream(1, lambda: ones)
>>> encoded_ones = rle(ones, max_run_length=3)
>>> [encoded_ones[i] for i in range(3)]
[(3, 1), (3, 1), (3, 1)]
"""

### Question 10: Challenge Problem (optional)

The Python Challenge is a website designed to teach people the many features of the Python Library. Each page of the site is a puzzle that can be solved simply in Python. The solution to each puzzle gives the URL of the next.

There is a function stub below to include your solution to puzzle 4 (the one with the picture of a wood carving). You will have to complete puzzles 0, 1, 2, and 3 to reach 4. You can start on puzzle 0 here.

Some hints:

• Puzzle 1. Try `str.maketrans` to make a dictionary and `str.translate` to generate a new string. Letters are listed in the `string` module.

``````>>> 'Borkozoy'.translate(str.maketrans('oz', 'el'))
'Berkeley'
>>> import string
>>> string.ascii_lowercase
'abcdefghijklmnopqrstuvwxyz'``````
• Puzzles 2 and 3. To view the source code of a web page in a browser, use

• Chrome: View > Developer > View Page Source
• Firefox: Tools > Web Developer > Page Source
• Safari: View > View Source

Uppercase letters are also in the `string` module.

``````>>> string.ascii_uppercase
'ABCDEFGHIJKLMNOPQRSTUVWXYZ'``````
• Puzzle 4. Here's an example of fetching the source of a web page in Python. The address below links to an archive of the first WWW site.

``````>>> base = 'http://www.w3.org/History/19921103-hypertext/hypertext'
>>> addr = base + '/WWW/TheProject.html'
>>> from urllib.request import urlopen
>>> contents.split('\n')
'<TITLE>The World Wide Web project</TITLE>'``````

As you work on this puzzle, make sure to print the result of each step.

The comments on the puzzle page say: "urllib may help. DON'T TRY ALL NOTHINGS, since it will never end. 400 times is more than enough."

You can include your solution to puzzle 4 below:

``````from urllib.request import urlopen

def puzzle_4():
"""Return the soluton to puzzle 4."""