# Homework 4: Data Abstraction, Trees, Nonlocal

*Due by 11:59pm on Thursday, March 5*

## Instructions

Download hw04.zip. Inside the archive, you will find a file called
hw04.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 2 points.**

# Required questions

## Abstraction

## Mobiles

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

We are making a planetarium mobile. A mobile is a type of hanging sculpture. A binary mobile consists of two arms. Each arm is a rod of a certain length, from which hangs either a planet or another mobile.

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

- A
`mobile`

has a left`arm`

and a right`arm`

. - An
`arm`

has a positive length and something hanging at the end, either a`mobile`

or`planet`

. - A
`planet`

has a positive size.

### Q1: Weights

Implement the `planet`

data abstraction by completing the `planet`

constructor
and the `size`

selector so that a planet is represented using a two-element list
where the first element is the string `'planet'`

and the second element is its size.
The `total_weight`

example is provided to demonstrate use of the mobile, arm, and planet abstractions.

```
def mobile(left, right):
"""Construct a mobile from a left arm and a right arm."""
assert is_arm(left), "left must be a arm"
assert is_arm(right), "right must be a arm"
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 arm of a mobile."""
assert is_mobile(m), "must call left on a mobile"
return m[1]
def right(m):
"""Select the right arm of a mobile."""
assert is_mobile(m), "must call right on a mobile"
return m[2]
```

```
def arm(length, mobile_or_planet):
"""Construct a arm: a length of rod with a mobile or planet at the end."""
assert is_mobile(mobile_or_planet) or is_planet(mobile_or_planet)
return ['arm', length, mobile_or_planet]
def is_arm(s):
"""Return whether s is a arm."""
return type(s) == list and len(s) == 3 and s[0] == 'arm'
def length(s):
"""Select the length of a arm."""
assert is_arm(s), "must call length on a arm"
return s[1]
def end(s):
"""Select the mobile or planet hanging at the end of a arm."""
assert is_arm(s), "must call end on a arm"
return s[2]
```

```
def planet(size):
"""Construct a planet of some size."""
assert size > 0
"*** YOUR CODE HERE ***"
def size(w):
"""Select the size of a planet."""
assert is_planet(w), 'must call size on a planet'
"*** YOUR CODE HERE ***"
def is_planet(w):
"""Whether w is a planet."""
return type(w) == list and len(w) == 2 and w[0] == 'planet'
```

```
def total_weight(m):
"""Return the total weight of m, a planet or mobile.
>>> t, u, v = examples()
>>> total_weight(t)
3
>>> total_weight(u)
6
>>> total_weight(v)
9
"""
if is_planet(m):
return size(m)
else:
assert is_mobile(m), "must get total weight of a mobile or a planet"
return total_weight(end(left(m))) + total_weight(end(right(m)))
```

Use Ok to test your code:

`python3 ok -q total_weight`

### Q2: Balanced

Hint:for more information on this problem (with more pictures!) please refer to this document.

Implement the `balanced`

function, which returns whether `m`

is a balanced
mobile. A mobile is balanced if two conditions are met:

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

Planets themselves are balanced.

```
def balanced(m):
"""Return whether m is balanced.
>>> t, u, v = examples()
>>> balanced(t)
True
>>> balanced(v)
True
>>> w = mobile(arm(3, t), arm(2, u))
>>> balanced(w)
False
>>> balanced(mobile(arm(1, v), arm(1, w)))
False
>>> balanced(mobile(arm(1, w), arm(1, v)))
False
"""
"*** YOUR CODE HERE ***"
```

Use Ok to test your code:

`python3 ok -q balanced`

### Q3: Totals

Implement `totals_tree`

, which takes a `mobile`

(or `planet`

) and returns a
`tree`

whose root is the total weight of the input. For a `planet`

, the result
should be a leaf. For a `mobile`

, the result's branches should be `totals_tree`

s
for the ends of its arms.

```
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
"""
"*** YOUR CODE HERE ***"
```

Use Ok to test your code:

`python3 ok -q totals_tree`

## Trees

### Q4: Replace Leaf

Define `replace_leaf`

, which takes a tree `t`

, a value `old`

, and a value
`replacement`

. `replace_leaf`

returns a new tree that's the same as `t`

except
that every leaf label equal to `old`

has been replaced with `replacement`

.

```
def replace_leaf(t, old, replacement):
"""Returns a new tree where every leaf value equal to old has
been replaced with replacement.
>>> yggdrasil = tree('odin',
... [tree('balder',
... [tree('thor'),
... tree('freya')]),
... 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
freya
frigg
freya
thor
sif
freya
freya
>>> laerad == yggdrasil # Make sure original tree is unmodified
True
"""
"*** YOUR CODE HERE ***"
```

Use Ok to test your code:

`python3 ok -q replace_leaf`

## Nonlocal

### Q5: Password Protected Account

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:

- Store that incorrect password in a list, and
- 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):
"""Return a password-protected withdraw function.
>>> w = make_withdraw(100, 'hax0r')
>>> w(25, 'hax0r')
75
>>> error = w(90, 'hax0r')
>>> error
'Insufficient funds'
>>> error = w(25, 'hwat')
>>> error
'Incorrect password'
>>> new_bal = w(25, 'hax0r')
>>> new_bal
50
>>> w(75, 'a')
'Incorrect password'
>>> w(10, 'hax0r')
40
>>> w(20, 'n00b')
'Incorrect password'
>>> 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
"""
"*** YOUR CODE HERE ***"
```

Use Ok to test your code:

`python3 ok -q make_withdraw`

### Q6: Joint Account

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

that takes three arguments.

- A password-protected
`withdraw`

function, - The password with which that
`withdraw`

function was defined, and - 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):
"""Return a password-protected withdraw function that has joint access to
the balance of withdraw.
>>> w = make_withdraw(100, 'hax0r')
>>> w(25, 'hax0r')
75
>>> make_joint(w, 'my', 'secret')
'Incorrect password'
>>> j = make_joint(w, 'hax0r', 'secret')
>>> w(25, 'secret')
'Incorrect password'
>>> 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(25, 'password')
'Incorrect password'
>>> j2(5, 'secret')
"Your account is locked. Attempts: ['my', 'secret', 'password']"
>>> j(5, 'secret')
"Your account is locked. Attempts: ['my', 'secret', 'password']"
>>> w(5, 'hax0r')
"Your account is locked. Attempts: ['my', 'secret', 'password']"
>>> make_joint(w, 'hax0r', 'hello')
"Your account is locked. Attempts: ['my', 'secret', 'password']"
"""
"*** YOUR CODE HERE ***"
```

Use Ok to test your code:

`python3 ok -q make_joint`

## Submit

Make sure to submit this assignment by running:

`python3 ok --submit`

# Extra Questions

### Q7: 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."""
"*** YOUR CODE HERE ***"
def upper_bound(x):
"""Return the upper bound of interval x."""
"*** YOUR CODE HERE ***"
```

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
```

### Q8: 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."""
"*** YOUR CODE HERE ***"
```

Use Ok to unlock and test your code:

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

### Q9: 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."""
"*** YOUR CODE HERE ***"
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
```

### Q10: 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:

Note: To make a multi-line string, you must use triple quotes `""" like this """`

.

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

### Q11: Quadratic

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'
"""
"*** YOUR CODE HERE ***"
```

Use Ok to test your code:

`python3 ok -q quadratic`