# Homework 5 hw05.zip

Due by 11:59pm on Thursday, 3/1

## Instructions

We recommend that you begin this homework early. In the past, it used to be two separate one-week homeworks, so expect to spend twice the amount of time and effort compared to a regular homework.

Finally, the Object Oriented Programming and Mutation problems correspond to the Objects and Mutable Functions lectures in week 6. The remaining questions can be completed with knowledge of the coursework up to week 5.

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:

# Required Questions

### 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``

### Q2: Towers of Hanoi

A classic puzzle called the Towers of Hanoi is a game that consists of three rods, and a number of disks of different sizes which can slide onto any rod. The puzzle starts with `n` disks in a neat stack in ascending order of size on a `start` rod, the smallest at the top, forming a conical shape.

The objective of the puzzle is to move the entire stack to an `end` rod, obeying the following rules:

• Only one disk may be moved at a time.
• Each move consists of taking the top (smallest) disk from one of the rods and sliding it onto another rod, on top of the other disks that may already be present on that rod.
• No disk may be placed on top of a smaller disk.

Complete the definition of `move_stack`, which prints out the steps required to move `n` disks from the `start` rod to the `end` rod without violating the rules.

``````def print_move(origin, destination):
"""Print instructions to move a disk."""
print("Move the top disk from rod", origin, "to rod", destination)

def move_stack(n, start, end):
"""Print the moves required to move n disks on the start pole to the end
pole without violating the rules of Towers of Hanoi.

n -- number of disks
start -- a pole position, either 1, 2, or 3
end -- a pole position, either 1, 2, or 3

There are exactly three poles, and start and end must be different. Assume
that the start pole has at least n disks of increasing size, and the end
pole is either empty or has a top disk larger than the top n start disks.

>>> move_stack(1, 1, 3)
Move the top disk from rod 1 to rod 3
>>> move_stack(2, 1, 3)
Move the top disk from rod 1 to rod 2
Move the top disk from rod 1 to rod 3
Move the top disk from rod 2 to rod 3
>>> move_stack(3, 1, 3)
Move the top disk from rod 1 to rod 3
Move the top disk from rod 1 to rod 2
Move the top disk from rod 3 to rod 2
Move the top disk from rod 1 to rod 3
Move the top disk from rod 2 to rod 1
Move the top disk from rod 2 to rod 3
Move the top disk from rod 1 to rod 3
"""
assert 1 <= start <= 3 and 1 <= end <= 3 and start != end, "Bad start/end"
``````

Use Ok to test your code:

``python3 ok -q move_stack``

## 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, which use the `tree` data abstraction for their representation.

• A `mobile` has a left side (index 0) and a right side (index 1).
• A `side` has a length and a structure, which is either a `mobile` or `weight`.
• A `weight` has a size, which is a positive number.

### Q3: Weights

Implement the `weight` data abstraction by completing the `weight` constructor, the `size` selector, and the `is_weight` predicate so that a weight is represented using a tree. 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."""
return tree('mobile', [left, right])

def is_mobile(m):
return is_tree(m) and label(m) == 'mobile'

def sides(m):
"""Select the sides of a mobile."""
assert is_mobile(m), "must call sides on a mobile"
return branches(m)

def is_side(m):
return not is_mobile(m) and not is_weight(m) and type(label(m)) == int``````
``````def side(length, mobile_or_weight):
"""Construct a side: a length of rod with a mobile or weight at the end."""
return tree(length, [mobile_or_weight])

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

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 branches(s)[0]``````
``````def weight(size):
"""Construct a weight of some size."""
assert size > 0

def size(w):
"""Select the size of a weight."""

def is_weight(w):
"""Whether w is a weight, not a mobile."""
``````

Use Ok to test your code:

``python3 ok -q total_weight``

### Q4: 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 (a similar calculation is used for the right side).
2. Each of the submobiles hanging off its sides is balanced.

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

It is a DAV (data abstraction violation) to assume that a mobile itself is a list. Instead, we should use selectors to get the sides of mobile.

However, since we know that the `sides` of a mobile is a list of sides, we can index into the list returned by `sides(m)` for some mobile `m` without committing a DAV.

Take a look at `total_weight` to see this in action.

``````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``

## Object Oriented Programming

### Q5: Retirement

Add a `time_to_retire` method to the `Account` class that takes an `amount`. It returns how many years the holder would need to wait in order for the current `balance` to grow to at least `amount`, assuming that the bank adds `balance` times the `interest` rate at the end of every year.

``````class Account:
"""An account has a balance and a holder.

>>> a = Account('John')
>>> a.deposit(10)
10
>>> a.balance
10
>>> a.interest
0.02

>>> a.time_to_retire(10.25) # 10 -> 10.2 -> 10.404
2
>>> a.balance               # balance should not change
10
>>> a.time_to_retire(11)    # 10 -> 10.2 -> ... -> 11.040808032
5
>>> a.time_to_retire(100)
117
"""

interest = 0.02  # A class attribute

def __init__(self, account_holder):
self.holder = account_holder
self.balance = 0

def deposit(self, amount):
self.balance = self.balance + amount
return self.balance

def withdraw(self, amount):
"""Subtract amount from balance if funds are available."""
if amount > self.balance:
return 'Insufficient funds'
self.balance = self.balance - amount
return self.balance

def time_to_retire(self, amount):
"""Return the number of years until balance would grow to amount."""
assert self.balance > 0 and amount > 0 and self.interest > 0
``````

Use Ok to test your code:

``python3 ok -q Account``

### Q6: Free Checking

Implement `FreeChecking`, which is like the `CheckingAccount` from lecture except that it only charges a withdraw fee after 2 free withdrawals. Such a deal! Even unsuccessful withdrawals count against the free quota, but only successful withdrawals actually incur a fee.

``````class FreeChecking(Account):
"""A bank account that charges for withdrawals, but the first two are free!

>>> ch = FreeChecking('Jack')
>>> ch.balance = 20
>>> ch.withdraw(100)  # First one's free
'Insufficient funds'
>>> ch.withdraw(3)    # And the second
17
>>> ch.balance
17
>>> ch.withdraw(3)    # Ok, two free withdrawals is enough
13
>>> ch.withdraw(3)
9
>>> ch2 = FreeChecking('John')
>>> ch2.balance = 10
>>> ch2.withdraw(3) # No fee
7
>>> ch.withdraw(3)  # ch still charges a fee
5
>>> ch.withdraw(5)  # Not enough to cover fee + withdraw
'Insufficient funds'
"""
withdraw_fee = 1
free_withdrawals = 2

``````

Use Ok to test your code:

``python3 ok -q FreeChecking``

## Mutation

### Q7: 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``

### Q8: 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
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``

### Q10: 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``

# 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)``````

### Q11: 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``````

### Q12: Sub Interval

Using reasoning analogous to Alyssa's, define a subtraction function for intervals. Try to reuse functions that have already been implemented.

``````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``````

### Q13: 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``````

### Q14: 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``

### Q15: 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'
"""
``````

Use Ok to test your code:

``python3 ok -q quadratic``

### Q17: Polynomial

Write a function `polynomial` that takes an interval `x` and a list of coefficients `c`, and returns the interval containing all values of `f(t)` for `t` in interval `x`, where:

``f(t) = c[k-1] * pow(t, k-1) + c[k-2] * pow(t, k-2) + ... + c[0] * 1``

Like `quadratic`, your `polynomial` function should return the smallest such interval, one that does not suffer from the multiple references problem.

Hint: You can approximate this result. Try using Newton's method.

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

>>> str_interval(polynomial(interval(0, 2), [-1, 3, -2]))
'-3 to 0.125'
>>> str_interval(polynomial(interval(1, 3), [1, -3, 2]))
'0 to 10'
>>> str_interval(polynomial(interval(0.5, 2.25), [10, 24, -6, -8, 3]))
'18.0 to 23.0'
"""
``python3 ok -q polynomial``