# Homework 5: Nonlocal, Iterators & Generators hw05.zip

Due by 11:59pm on Tuesday, July 21

## Instructions

Download hw05.zip. Inside the archive, you will find a file called hw05.py, along with a copy of the `ok` autograder.

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 correctness. Each incorrect problem will decrease the total score by one point. There is a homework recovery policy as stated in the syllabus. This homework is out of 3 points.

# Required Questions

## Nonlocal

### Q1: Make Bank

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 new function `make_bank`, which should create a bank account with value `balance` and should also return another function. This new function should be able to withdraw and deposit money. The second function will take in two arguments: `message` and `amount`. When the message passed in is `'deposit'`, the bank will deposit `amount` into the account. When the message passed in is `'withdraw'`, the bank will attempt to withdraw `amount` from the account. If the account does not have enough money for a withdrawal, the string `'Insufficient funds'` will be returned. If the `message` passed in is neither of the two commands, the function should return `'Invalid message'` Examples are shown in the doctests.

``````def make_bank(balance):
"""Returns a bank function with a starting balance. Supports
withdrawals and deposits.

>>> bank = make_bank(100)
>>> bank('withdraw', 40)    # 100 - 40
60
>>> bank('hello', 500)      # Invalid message passed in
'Invalid message'
>>> bank('deposit', 20)     # 60 + 20
80
>>> bank('withdraw', 90)    # 80 - 90; not enough money
'Insufficient funds'
>>> bank('deposit', 100)    # 80 + 100
180
>>> bank('goodbye', 0)      # Invalid message passed in
'Invalid message'
>>> bank('withdraw', 60)    # 180 - 60
120
"""
def bank(message, amount):
return bank``````

Use Ok to test your code:

``python3 ok -q make_bank``

Write a version of the `make_withdraw` function shown in the previous question 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 frozen. All subsequent calls to the function should return:

``"Too many incorrect attempts. 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')
"Too many incorrect attempts. Attempts: ['hwat', 'a', 'n00b']"
>>> w(10, 'l33t')
"Too many incorrect attempts. Attempts: ['hwat', 'a', 'n00b']"
>>> type(w(10, 'l33t')) == str
True
"""
``````

Use Ok to test your code:

``python3 ok -q make_withdraw``

## Iterators and Generators

### Q3: Repeated

Implement a function (not a generator function) that returns the first value in the iterator `t` that appears `k` times in a row. As described in lecture, iterators can provide values using either the `next(t)` function or with a for-loop. Do not worry about cases where the function reaches the end of the iterator without finding a suitable value, all lists passed in for the tests will have a value that should be returned. If you are receiving an error where the iterator has completed then the program is not identifying the correct value. Iterate through the items such that if the same iterator is passed into `repeated` twice, it continues in the second call at the point it left off in the first. An example of this behavior is shown in the doctests.

``````def repeated(t, k):
"""Return the first value in iterator T that appears K times in a row. Iterate through the items such that
if the same iterator is passed into repeated twice, it continues in the second call at the point it left off
in the first.

>>> lst = iter([10, 9, 10, 9, 9, 10, 8, 8, 8, 7])
>>> repeated(lst, 2)
9
>>> lst2 = iter([10, 9, 10, 9, 9, 10, 8, 8, 8, 7])
>>> repeated(lst2, 3)
8
>>> s = iter([3, 2, 2, 2, 1, 2, 1, 4, 4, 5, 5, 5])
>>> repeated(s, 3)
2
>>> repeated(s, 3)
5
>>> s2 = iter([4, 1, 6, 6, 7, 7, 8, 8, 2, 2, 2, 5])
>>> repeated(s2, 3)
2
"""
assert k > 1
``````

Use Ok to test your code:

``python3 ok -q repeated``

### Q4: Merge

Implement `merge(incr_a, incr_b)`, which takes two iterables `incr_a` and `incr_b` whose elements are ordered. `merge` yields elements from `incr_a` and `incr_b` in sorted order, eliminating repetition. You may assume `incr_a` and `incr_b` themselves do not contain repeats, and that none of the elements of either are `None`. You may not assume that the iterables are finite; either may produce an infinite stream of results.

You will probably find it helpful to use the two-argument version of the built-in `next` function: `next(incr, v)` is the same as `next(incr)`, except that instead of raising `StopIteration` when `incr` runs out of elements, it returns `v`.

See the doctest for examples of behavior.

``````def merge(incr_a, incr_b):
"""Yield the elements of strictly increasing iterables incr_a and incr_b, removing
repeats. Assume that incr_a and incr_b have no repeats. incr_a or incr_b may be infinite
sequences.

>>> m = merge([0, 2, 4, 6, 8, 10, 12, 14], [0, 3, 6, 9, 12, 15])
>>> type(m)
<class 'generator'>
>>> list(m)
[0, 2, 3, 4, 6, 8, 9, 10, 12, 14, 15]
>>> def big(n):
...    k = 0
...    while True: yield k; k += n
>>> m = merge(big(2), big(3))
>>> [next(m) for _ in range(11)]
[0, 2, 3, 4, 6, 8, 9, 10, 12, 14, 15]
"""
iter_a, iter_b = iter(incr_a), iter(incr_b)
next_a, next_b = next(iter_a, None), next(iter_b, None)
``````

Watch the hints video below for somewhere to start:

Use Ok to test your code:

``python3 ok -q merge``

## Submit

Make sure to submit this assignment by running:

``python3 ok --submit``

# Extra Questions

### Q5: 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.

If the password is incorrect or cannot be verified because the underlying account is locked, the `make_joint` should propagate the error. Otherwise, it 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_pass, new_pass):
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')
"Too many incorrect attempts. Attempts: ['my', 'secret', 'password']"
>>> j(5, 'secret')
"Too many incorrect attempts. Attempts: ['my', 'secret', 'password']"
>>> w(5, 'hax0r')
"Too many incorrect attempts. Attempts: ['my', 'secret', 'password']"
>>> make_joint(w, 'hax0r', 'hello')
"Too many incorrect attempts. Attempts: ['my', 'secret', 'password']"
"""
``````

Use Ok to test your code:

``python3 ok -q make_joint``

### Q6: Remainder Generator

Like functions, generators can also be higher-order. For this problem, we will be writing `remainders_generator`, which yields a series of generator objects.

`remainders_generator` takes in an integer `m`, and yields `m` different generators. The first generator is a generator of multiples of `m`, i.e. numbers where the remainder is 0. The second is a generator of natural numbers with remainder 1 when divided by `m`. The last generator yields natural numbers with remainder `m - 1` when divided by `m`.

Hint: You can call the `naturals` function to create a generator of infinite natural numbers.

Hint: Consider defining an inner generator function. Each yielded generator varies only in that the elements of each generator have a particular remainder when divided by `m`. What does that tell you about the argument(s) that the inner function should take in?

``````def remainders_generator(m):
"""
Yields m generators. The ith yielded generator yields natural numbers whose
remainder is i when divided by m.

>>> import types
>>> [isinstance(gen, types.GeneratorType) for gen in remainders_generator(5)]
[True, True, True, True, True]
>>> remainders_four = remainders_generator(4)
>>> for i in range(4):
...     print("First 3 natural numbers with remainder {0} when divided by 4:".format(i))
...     gen = next(remainders_four)
...     for _ in range(3):
...         print(next(gen))
First 3 natural numbers with remainder 0 when divided by 4:
4
8
12
First 3 natural numbers with remainder 1 when divided by 4:
1
5
9
First 3 natural numbers with remainder 2 when divided by 4:
2
6
10
First 3 natural numbers with remainder 3 when divided by 4:
3
7
11
"""
Note that if you have implemented this correctly, each of the generators yielded by `remainder_generator` will be infinite - you can keep calling `next` on them forever without running into a `StopIteration` exception.
``python3 ok -q remainders_generator``