# Homework 5: Data Abstraction, Mutable Functions & Generators hw05.zip

Due by 11:59pm on Thursday, 10/4

## Instructions

Note that this homework is longer than average - there will be no homework due 9/27 so there is as much content in this homework as two regular homework assignments. Therefore, it is recommended that you start as early as possible.

Additionally, some of the material on this homework (Mutable Functions and Generators) will not be covered until Week 6. Don't worry if you haven't seen them yet - you can work on the first two sections in the meantime!

• Questions 1-9 are required and can be found in `hw05.py`
• Questions 10-15 are optional and can be found in `hw05.py`

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.

Readings: You might find the following references useful:

Grading: Homework is graded based on effort, not correctness. However, there is no partial credit; you must show substantial effort on every problem to receive any points.

# Required questions

## Trees

### Q1: Replace Leaf

Define `replace_leaf`, which takes a tree `t`, a value `old`, and a value `new`. `replace_leaf` returns a new tree that's the same as `t` except that every leaf value equal to `old` has been replaced with `new`.

``````def replace_leaf(t, old, new):
"""Returns a new tree where every leaf value equal to old has
been replaced with new.

>>> yggdrasil = tree('odin',
...                  [tree('balder',
...                        [tree('thor'),
...                         tree('loki')]),
...                   tree('frigg',
...                        [tree('thor')]),
...                   tree('thor',
...                        [tree('sif'),
...                         tree('thor')]),
...                   tree('thor')])
>>> laerad = copy_tree(yggdrasil) # copy yggdrasil for testing purposes
>>> print_tree(replace_leaf(yggdrasil, 'thor', 'freya'))
odin
balder
freya
loki
frigg
freya
thor
sif
freya
freya
>>> laerad == yggdrasil # Make sure original tree is unmodified
True
"""
``````

Use Ok to test your code:

``python3 ok -q replace_leaf``

## Mobiles

Acknowledgements. This mobile example is based on a classic problem from Structure and Interpretation of Computer Programs, Section 2.2.2.

A mobile is a type of hanging sculpture. A binary mobile consists of two sides. Each side is a rod of a certain length, from which hangs either a weight or another mobile.

We will represent a binary mobile using the data abstractions below.

• A `mobile` has a left `side` and a right `side`.
• A `side` has a positive length and something hanging at the end, either a `mobile` or `weight`.
• A `weight` has a positive size.

### Q2: Weights

Implement the `weight` data abstraction by completing the `weight` constructor and the `size` selector so that a weight is represented using a two-element list where the first element is the string `'weight'`. The `total_weight` example is provided to demonstrate use of the mobile, side, and weight abstractions.

``````def mobile(left, right):
"""Construct a mobile from a left side and a right side."""
assert is_side(left), "left must be a side"
assert is_side(right), "right must be a side"
return ['mobile', left, right]

def is_mobile(m):
"""Return whether m is a mobile."""
return type(m) == list and len(m) == 3 and m[0] == 'mobile'

def left(m):
"""Select the left side of a mobile."""
assert is_mobile(m), "must call left on a mobile"
return m[1]

def right(m):
"""Select the right side of a mobile."""
assert is_mobile(m), "must call right on a mobile"
return m[2]``````
``````def side(length, mobile_or_weight):
"""Construct a side: a length of rod with a mobile or weight at the end."""
assert is_mobile(mobile_or_weight) or is_weight(mobile_or_weight)
return ['side', length, mobile_or_weight]

def is_side(s):
"""Return whether s is a side."""
return type(s) == list and len(s) == 3 and s[0] == 'side'

def length(s):
"""Select the length of a side."""
assert is_side(s), "must call length on a side"
return s[1]

def end(s):
"""Select the mobile or weight hanging at the end of a side."""
assert is_side(s), "must call end on a side"
return s[2]``````
``````def weight(size):
"""Construct a weight of some size."""
assert size > 0

def size(w):
"""Select the size of a weight."""
assert is_weight(w), 'must call size on a weight'

def is_weight(w):
"""Whether w is a weight."""
return type(w) == list and len(w) == 2 and w[0] == 'weight'``````

Use Ok to test your code:

``python3 ok -q total_weight``

### Q3: Balanced

Implement the `balanced` function, which returns whether `m` is a balanced mobile. A mobile is balanced if two conditions are met:

1. The torque applied by its left side is equal to that applied by its right side. Torque of the left side is the length of the left rod multiplied by the total weight hanging from that rod. Likewise for the right.
2. Each of the mobiles hanging at the end of its sides is balanced.

Hint: You may find it helpful to assume that weights themselves are balanced.

``````def balanced(m):
"""Return whether m is balanced.

>>> t, u, v = examples()
>>> balanced(t)
True
>>> balanced(v)
True
>>> w = mobile(side(3, t), side(2, u))
>>> balanced(w)
False
>>> balanced(mobile(side(1, v), side(1, w)))
False
>>> balanced(mobile(side(1, w), side(1, v)))
False
"""
``````

Use Ok to test your code:

``python3 ok -q balanced``

### Q4: Totals

Implement `totals_tree`, which takes a `mobile` (or `weight`) and returns a `tree` whose root is its total weight and whose branches are trees for the ends of the sides.

``````def totals_tree(m):
"""Return a tree representing the mobile with its total weight at the root.

>>> t, u, v = examples()
>>> print_tree(totals_tree(t))
3
2
1
>>> print_tree(totals_tree(u))
6
1
5
3
2
>>> print_tree(totals_tree(v))
9
3
2
1
6
1
5
3
2
"""
``````

Use Ok to test your code:

``python3 ok -q totals_tree``

## Mutable functions

### Q5: Counter

Define a function `make_counter` that returns a `counter` function, which takes a string and returns the number of times that the function has been called on that string.

``````def make_counter():
"""Return a counter function.

>>> c = make_counter()
>>> c('a')
1
>>> c('a')
2
>>> c('b')
1
>>> c('a')
3
>>> c2 = make_counter()
>>> c2('b')
1
>>> c2('b')
2
>>> c('b') + c2('b')
5
"""
``````

Use Ok to test your code:

``python3 ok -q make_counter``

### Q6: Next Fibonacci

Write a function `make_fib` that returns a function that returns the next Fibonacci number each time it is called. (The Fibonacci sequence begins with 0 and then 1, after which each element is the sum of the preceding two.) Use a `nonlocal` statement!

``````def make_fib():
"""Returns a function that returns the next Fibonacci number
every time it is called.

>>> fib = make_fib()
>>> fib()
0
>>> fib()
1
>>> fib()
1
>>> fib()
2
>>> fib()
3
>>> fib2 = make_fib()
>>> fib() + sum([fib2() for _ in range(5)])
12
"""
``````

Use Ok to test your code:

``python3 ok -q make_fib``

In lecture, we saw how to use functions to create mutable objects. Here, for example, is the function `make_withdraw` which produces a function that can withdraw money from an account:

``````def make_withdraw(balance):
"""Return a withdraw function with BALANCE as its starting balance.
>>> withdraw = make_withdraw(1000)
>>> withdraw(100)
900
>>> withdraw(100)
800
>>> withdraw(900)
'Insufficient funds'
"""
def withdraw(amount):
nonlocal balance
if amount > balance:
return 'Insufficient funds'
balance = balance - amount
return balance
return withdraw``````

Write a version of the `make_withdraw` function that returns password-protected withdraw functions. That is, `make_withdraw` should take a password argument (a string) in addition to an initial balance. The returned function should take two arguments: an amount to withdraw and a password.

A password-protected `withdraw` function should only process withdrawals that include a password that matches the original. Upon receiving an incorrect password, the function should:

1. Store that incorrect password in a list, and
2. Return the string 'Incorrect password'.

If a withdraw function has been called three times with incorrect passwords `p1`, `p2`, and `p3`, then it is locked. All subsequent calls to the function should return:

``"Your account is locked. Attempts: [<p1>, <p2>, <p3>]"``

The incorrect passwords may be the same or different:

``````def make_withdraw(balance, password):

>>> w = make_withdraw(100, 'hax0r')
>>> w(25, 'hax0r')
75
>>> error = w(90, 'hax0r')
>>> error
'Insufficient funds'
>>> error = w(25, 'hwat')
>>> error
>>> new_bal = w(25, 'hax0r')
>>> new_bal
50
>>> w(75, 'a')
>>> w(10, 'hax0r')
40
>>> w(20, 'n00b')
>>> w(10, 'hax0r')
"Your account is locked. Attempts: ['hwat', 'a', 'n00b']"
>>> w(10, 'l33t')
"Your account is locked. Attempts: ['hwat', 'a', 'n00b']"
>>> type(w(10, 'l33t')) == str
True
"""
``````

Use Ok to test your code:

``python3 ok -q make_withdraw``

### Q8: Joint Account

Suppose that our banking system requires the ability to make joint accounts. Define a function `make_joint` that takes three arguments.

1. A password-protected `withdraw` function,
2. The password with which that `withdraw` function was defined, and
3. A new password that can also access the original account.

The `make_joint` function returns a `withdraw` function that provides additional access to the original account using either the new or old password. Both functions draw from the same balance. Incorrect passwords provided to either function will be stored and cause the functions to be locked after three wrong attempts.

Hint: The solution is short (less than 10 lines) and contains no string literals! The key is to call `withdraw` with the right password and amount, then interpret the result. You may assume that all failed attempts to withdraw will return some string (for incorrect passwords, locked accounts, or insufficient funds), while successful withdrawals will return a number.

Use `type(value) == str` to test if some `value` is a string:

``````def make_joint(withdraw, old_password, new_password):
the balance of withdraw.

>>> w = make_withdraw(100, 'hax0r')
>>> w(25, 'hax0r')
75
>>> make_joint(w, 'my', 'secret')
>>> j = make_joint(w, 'hax0r', 'secret')
>>> w(25, 'secret')
>>> j(25, 'secret')
50
>>> j(25, 'hax0r')
25
>>> j(100, 'secret')
'Insufficient funds'

>>> j2 = make_joint(j, 'secret', 'code')
>>> j2(5, 'code')
20
>>> j2(5, 'secret')
15
>>> j2(5, 'hax0r')
10

>>> j2(5, 'secret')
>>> j(5, 'secret')
>>> w(5, 'hax0r')
>>> make_joint(w, 'hax0r', 'hello')
"""
``````

Use Ok to test your code:

``python3 ok -q make_joint``

## Generators

### Q9: Generate Paths

Define a generator function `generate_paths` which takes in a tree `t`, a value `x`, and yields each path from the root of `t` to a node that has label `x`. Each path should be represented as a list of the labels along that path in the tree. You may yield the paths in any order.

``````def generate_paths(t, x):
"""Yields all possible paths from the root of t to a node with the label x
as a list.

>>> t1 = tree(1, [tree(2, [tree(3), tree(4, [tree(6)]), tree(5)]), tree(5)])
>>> print_tree(t1)
1
2
3
4
6
5
5
>>> next(generate_paths(t1, 6))
[1, 2, 4, 6]
>>> path_to_5 = generate_paths(t1, 5)
>>> sorted(list(path_to_5))
[[1, 2, 5], [1, 5]]

>>> t2 = tree(0, [tree(2, [t1])])
>>> print_tree(t2)
0
2
1
2
3
4
6
5
5
>>> path_to_2 = generate_paths(t2, 2)
>>> sorted(list(path_to_2))
[[0, 2], [0, 2, 1, 2]]
"""
``````

Use Ok to test your code:

``python3 ok -q generate_paths``

# Extra Questions

Extra questions are not worth extra credit and are entirely optional. They are designed to challenge you to think creatively!

## Interval

Acknowledgements. This interval arithmetic example is based on a classic problem from Structure and Interpretation of Computer Programs, Section 2.1.4.

Introduction. Alyssa P. Hacker is designing a system to help people solve engineering problems. One feature she wants to provide in her system is the ability to manipulate inexact quantities (such as measured parameters of physical devices) with known precision, so that when computations are done with such approximate quantities the results will be numbers of known precision.

Alyssa's idea is to implement interval arithmetic as a set of arithmetic operations for combining "intervals" (objects that represent the range of possible values of an inexact quantity). The result of adding, subracting, multiplying, or dividing two intervals is itself an interval, representing the range of the result.

Alyssa postulates the existence of an abstract object called an "interval" that has two endpoints: a lower bound and an upper bound. She also presumes that, given the endpoints of an interval, she can construct the interval using the data constructor interval. Using the constructor and selectors, she defines the following operations:

``````def str_interval(x):
"""Return a string representation of interval x."""
return '{0} to {1}'.format(lower_bound(x), upper_bound(x))

"""Return an interval that contains the sum of any value in interval x and
any value in interval y."""
lower = lower_bound(x) + lower_bound(y)
upper = upper_bound(x) + upper_bound(y)
return interval(lower, upper)``````

### Q10: Interval Abstraction

Alyssa's program is incomplete because she has not specified the implementation of the interval abstraction. She has implemented the constructor for you; fill in the implementation of the selectors.

``````def interval(a, b):
"""Construct an interval from a to b."""
return [a, b]

def lower_bound(x):
"""Return the lower bound of interval x."""

def upper_bound(x):
"""Return the upper bound of interval x."""
``````

Use Ok to unlock and test your code:

``````python3 ok -q interval -u
python3 ok -q interval``````

Louis Reasoner has also provided an implementation of interval multiplication. Beware: there are some data abstraction violations, so help him fix his code before someone sets it on fire.

``````def mul_interval(x, y):
"""Return the interval that contains the product of any value in x and any
value in y."""
p1 = x[0] * y[0]
p2 = x[0] * y[1]
p3 = x[1] * y[0]
p4 = x[1] * y[1]
return [min(p1, p2, p3, p4), max(p1, p2, p3, p4)]``````

Use Ok to unlock and test your code:

``````python3 ok -q mul_interval -u
python3 ok -q mul_interval``````

### Q11: Sub Interval

Using reasoning analogous to Alyssa's, define a subtraction function for intervals. Try to reuse functions that have already been implemented if you find yourself repeating code.

``````def sub_interval(x, y):
"""Return the interval that contains the difference between any value in x
and any value in y."""
``````

Use Ok to unlock and test your code:

``````python3 ok -q sub_interval -u
python3 ok -q sub_interval``````

### Q12: Div Interval

Alyssa implements division below by multiplying by the reciprocal of `y`. Ben Bitdiddle, an expert systems programmer, looks over Alyssa's shoulder and comments that it is not clear what it means to divide by an interval that spans zero. Add an `assert` statement to Alyssa's code to ensure that no such interval is used as a divisor:

``````def div_interval(x, y):
"""Return the interval that contains the quotient of any value in x divided by
any value in y. Division is implemented as the multiplication of x by the
reciprocal of y."""
reciprocal_y = interval(1/upper_bound(y), 1/lower_bound(y))
return mul_interval(x, reciprocal_y)``````

Use Ok to unlock and test your code:

``````python3 ok -q div_interval -u
python3 ok -q div_interval``````

### Q13: Par Diff

After considerable work, Alyssa P. Hacker delivers her finished system. Several years later, after she has forgotten all about it, she gets a frenzied call from an irate user, Lem E. Tweakit. It seems that Lem has noticed that the formula for parallel resistors can be written in two algebraically equivalent ways:

``par1(r1, r2) = (r1 * r2) / (r1 + r2)``

or

``par2(r1, r2) = 1 / (1/r1 + 1/r2)``

He has written the following two programs, each of which computes the `parallel_resistors` formula differently::

``````def par1(r1, r2):

def par2(r1, r2):
one = interval(1, 1)
rep_r1 = div_interval(one, r1)
rep_r2 = div_interval(one, r2)

Lem complains that Alyssa's program gives different answers for the two ways of computing. This is a serious complaint.

Demonstrate that Lem is right. Investigate the behavior of the system on a variety of arithmetic expressions. Make some intervals `r1` and `r2`, and show that `par1` and `par2` can give different results.

``````def check_par():
"""Return two intervals that give different results for parallel resistors.

>>> r1, r2 = check_par()
>>> x = par1(r1, r2)
>>> y = par2(r1, r2)
>>> lower_bound(x) != lower_bound(y) or upper_bound(x) != upper_bound(y)
True
"""
r1 = interval(1, 1) # Replace this line!
r2 = interval(1, 1) # Replace this line!
return r1, r2``````

Use Ok to test your code:

``python3 ok -q check_par``

### Q14: Multiple References

Eva Lu Ator, another user, has also noticed the different intervals computed by different but algebraically equivalent expressions. She says that the problem is multiple references to the same interval.

The Multiple References Problem: a formula to compute with intervals using Alyssa's system will produce tighter error bounds if it can be written in such a form that no variable that represents an uncertain number is repeated.

Thus, she says, `par2` is a better program for parallel resistances than `par1`. Is she right? Why? Write a function that returns a string containing a written explanation of your answer:

``````def multiple_references_explanation():
return """The multiple reference problem..."""
``````

Write a function `quadratic` that returns the interval of all values `f(t)` such that `t` is in the argument interval `x` and `f(t)` is a quadratic function:

``f(t) = a*t*t + b*t + c``

Make sure that your implementation returns the smallest such interval, one that does not suffer from the multiple references problem.

Hint: the derivative `f'(t) = 2*a*t + b`, and so the extreme point of the quadratic is `-b/(2*a)`:

``````def quadratic(x, a, b, c):
"""Return the interval that is the range of the quadratic defined by
coefficients a, b, and c, for domain interval x.

>>> str_interval(quadratic(interval(0, 2), -2, 3, -1))
'-3 to 0.125'
>>> str_interval(quadratic(interval(1, 3), 2, -3, 1))
'0 to 10'
"""
``python3 ok -q quadratic``