# Homework 4: Sequences, Trees

*Due by 11:59pm on Thursday, October 6*

## 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

## Getting Started Videos

These videos may provide some helpful direction for tackling the coding problems on this assignment.

To see these videos, you should be logged into your berkeley.edu email.

### Q1: Mid-Semester Feedback

As part of this week's homework, please fill out the Mid-Semester Feedback form.

This survey is designed to help us make short term adjustments to the course so that it works better for you. We appreciate your feedback. We may not be able to make every change that you request, but we will read all the feedback and consider it.

Confidentiality: Your responses to the survey are confidential, and only the instructors and head TAs will be able to see this data unanonymized. More specifics on confidentiality can be found on the survey itself.

Once you finish the survey, you will be presented with a passphrase (if you miss
it, it should also be at the bottom of the confirmation email you receive). Put
this passphrase, as a string, on the line that says
`passphrase = '*** PASSPHRASE HERE ***'`

in the Python file for this assignment.

Use Ok to test your code:

`python3 ok -q midsem_survey`

## Arms-length recursion

Before we get started, a quick comment on recursion with tree data structures. Consider the following function.

```
def min_depth(t):
"""A simple function to return the distance between t's root and its closest leaf"""
if is_leaf(t):
return 0 # Base case---the distance between a node and itself is zero
h = float('inf') # Python's version of infinity
for b in branches(t):
if is_leaf(b): return 1 # !!!
h = min(h, 1 + min_depth(b))
return h
```

The line flagged with `!!!`

is an "arms-length" recursion violation. Although our code works correctly when it is present, by performing this check we are doing work that should be done by the next level of recursion—we already have an if-statement that handles any inputs to `min_depth`

that are leaves, so we should not include this line to eliminate redundancy in our code.

```
def min_depth(t):
"""A simple function to return the distance between t's root and its closest leaf"""
if is_leaf(t):
return 0
h = float('inf')
for b in branches(t):
# Still works fine!
h = min(h, 1 + min_depth(b))
return h
```

Arms-length recursion is not only redundant but often complicates our code and obscures the functionality of recursive functions, making writing recursive functions much more difficult. We always want our recursive case to be handling one and only one recursive level. We may or may not be checking your code periodically for things like this.

## Mobiles

**Acknowledgements**
This mobile example is based on
a classic problem from MIT 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. For example, the below diagram shows the left and right arms of Mobile A, and what hangs at the ends of each of those arms.

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

- A
`mobile`

must have both a left`arm`

and a right`arm`

. - An
`arm`

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

or`planet`

. - A
`planet`

has a positive mass, and nothing hanging from it.

### Q2: Weights

Below are the implementations of the various data abstractions used in mobiles.
The `mobile`

and `arm`

data abstractions have been completed for you.

Your job is to implement the `planet`

data abstraction by completing the `planet`

constructor
and the `mass`

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 mass.

*Implementation of the Mobile Data Abstraction* (**for your reference, no need to do anything here**):

```
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]
```

*Implementation of the Arm Data Abstraction*(

**for your reference, no need to do anything here**):

```
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]
```

(**Your turn!**)

*Fill out the implementation of the Planet Data Abstraction!*

```
def planet(mass):
"""Construct a planet of some mass."""
assert mass > 0
"*** YOUR CODE HERE ***"
def mass(w):
"""Select the mass of a planet."""
assert is_planet(w), 'must call mass 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'
```

*Total Weight implementation* (**for your reference and understanding, no need to do anything here**):

The `total_weight`

example is provided to demonstrate use of the mobile, arm,
and planet abstractions for your reference. You do not need to implement anything here. **You may use
the total_weight function in questions 3 and 4.**

The `examples`

function builds and returns three mobiles. These mobiles are used in the doctest examples
for `total_weight`

, a function that takes in either a mobile or a planet and returns its total weight. The total weight
of a planet is just its mass. The total weight of a mobile is the sum of the masses of all the planets hanging from the mobile.

```
def examples():
t = mobile(arm(1, planet(2)),
arm(2, planet(1)))
u = mobile(arm(5, planet(1)),
arm(1, mobile(arm(2, planet(3)),
arm(3, planet(2)))))
v = mobile(arm(4, t), arm(2, u))
return (t, u, v)
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
>>> from construct_check import check
>>> # checking for abstraction barrier violations by banning indexing
>>> check(HW_SOURCE_FILE, 'total_weight', ['Index'])
True
"""
if is_planet(m):
return mass(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`

### Q3: Balanced

Implement the `balanced`

function, which returns whether `m`

is a balanced
mobile. A mobile is balanced if both of the following conditions are met:

- The torque applied by its left arm is equal to that applied by its right
arm. The 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. For example,
if the left arm has a length of
`5`

, and there is a`mobile`

hanging at the end of the left arm of total weight`10`

, the torque on the left side of our mobile is`50`

. - Each of the mobiles hanging at the end of its arms is itself balanced.

Planets themselves are balanced, as there is nothing hanging off of them.

Reminder:You may use the`total_weight`

function defined for you in Question 2 (as well as any of the other functions defined in Question 2).

```
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
>>> from construct_check import check
>>> # checking for abstraction barrier violations by banning indexing
>>> check(HW_SOURCE_FILE, 'balanced', ['Index'])
True
"""
"*** YOUR CODE HERE ***"
```

Use Ok to test your code:

`python3 ok -q balanced`

### Q4: Totals

Implement `totals_tree`

, which takes in a `mobile`

or `planet`

and returns a
`tree`

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

, `totals_tree`

should return a leaf. For a `mobile`

, `totals_tree`

should return a tree whose
label is that `mobile`

's total weight, and whose branches are `totals_tree`

s for the
ends of its arms. As a reminder, the description of the tree data abstraction can be found here.

```
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
>>> from construct_check import check
>>> # checking for abstraction barrier violations by banning indexing
>>> check(HW_SOURCE_FILE, 'totals_tree', ['Index'])
True
"""
"*** YOUR CODE HERE ***"
```

Use Ok to test your code:

`python3 ok -q totals_tree`

## More Trees

### Q5: Replace Loki at Leaf

Define `replace_loki_at_leaf`

, which takes a tree `t`

and a value `lokis_replacement`

. `replace_loki_at_leaf`

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

except
that every leaf label equal to `"loki"`

has been replaced with `lokis_replacement`

.

If you want to learn about the Norse mythology referenced in this problem, you can read about it here.

```
def replace_loki_at_leaf(t, lokis_replacement):
"""Returns a new tree where every leaf value equal to "loki" has
been replaced with lokis_replacement.
>>> yggdrasil = tree('odin',
... [tree('balder',
... [tree('loki'),
... tree('freya')]),
... tree('frigg',
... [tree('loki')]),
... tree('loki',
... [tree('sif'),
... tree('loki')]),
... tree('loki')])
>>> laerad = copy_tree(yggdrasil) # copy yggdrasil for testing purposes
>>> print_tree(replace_loki_at_leaf(yggdrasil, 'freya'))
odin
balder
freya
freya
frigg
freya
loki
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_loki_at_leaf`

### Q6: Has Path

Write a function `has_path`

that takes in a tree `t`

and a string `word`

. It
returns `True`

if there is a path that starts from the root where the entries
along the path spell out the `word`

, and `False`

otherwise. (This data structure
is called a trie, and it has a lot of cool applications, such as autocomplete).
You may assume that every node's `label`

is exactly one character.

```
def has_path(t, word):
"""Return whether there is a path in a tree where the entries along the path
spell out a particular word.
>>> greetings = tree('h', [tree('i'),
... tree('e', [tree('l', [tree('l', [tree('o')])]),
... tree('y')])])
>>> print_tree(greetings)
h
i
e
l
l
o
y
>>> has_path(greetings, 'h')
True
>>> has_path(greetings, 'i')
False
>>> has_path(greetings, 'hi')
True
>>> has_path(greetings, 'hello')
True
>>> has_path(greetings, 'hey')
True
>>> has_path(greetings, 'bye')
False
>>> has_path(greetings, 'hint')
False
"""
assert len(word) > 0, 'no path for empty word.'
"*** YOUR CODE HERE ***"
```

Use Ok to test your code:

`python3 ok -q has_path`

# Optional Questions

## Data Abstraction

Feel free to reference Section 2.2 for more information on data abstraction.

**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
measurements from physical devices) with known precision, so that
when computations are done with such approximate quantities the results
will be numbers of known precision. For example, if a measured quantity
x lies between two numbers a and b, Alyssa would like her system to
use this range in computations involving x.

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 also an interval, one that represents the range of the result.

Alyssa suggests the existence of an abstraction 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 create the interval using data abstraction. Using this constructor and the appropriate 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))
def add_interval(x, y):
"""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)
```

### 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."""
assert a <= b, 'Lower bound cannot be greater than upper bound'
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
```

### Q8: Interval Arithmetic

After implementing the abstraction, Alyssa decided to implement a few interval arithmetic functions.

This is her current implementation for **interval multiplication**.
Unfortunately there are some data abstraction violations,
so your task is to fix this 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
```

**Interval Subtraction**

Using a similar approach as `mul_interval`

and `add_interval`

, define a subtraction function for
intervals. If you find yourself repeating code, see if you can 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."""
"*** YOUR CODE HERE ***"
```

Use Ok to unlock and test your code:

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

**Interval Division**

Alyssa implements division below by multiplying by the reciprocal of
`y`

. A 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
```

### Q9: 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):
return div_interval(mul_interval(r1, r2), add_interval(r1, r2))
def par2(r1, r2):
one = interval(1, 1)
rep_r1 = div_interval(one, r1)
rep_r2 = div_interval(one, r2)
return div_interval(one, add_interval(rep_r1, rep_r2))
```

Lem points out that Alyssa's program gives different answers for the two ways of computing. Find two intervals `r1`

and `r2`

that demonstrate the difference in behavior between `par1`

and `par2`

when passed into each of the two functions.

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`

# Exam Practice

Homework assignments will also contain prior exam questions for you to try. These questions have no submission component; feel free to attempt them if you'd like some practice!

- Summer 2021 MT Q4: Maximum Exponen-tree-ation
- Summer 2019 MT Q8: Leaf It To Me
- Summer 2017 MT Q9: Temmie Flakes