Homework 4: Data Abstraction, Trees
Files: hw04.zip
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, zip up your assignment and submit it through Gradescope. 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. This homework is out of 4 points.
Required questions
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 hThe 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 hArms-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
mobilemust have both a leftarmand a rightarm. - An
armhas a positive length and must have something hanging at the end, either amobileorplanet. - A
planethas a positive size, and nothing hanging from it.
Q2: 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.
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 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)The total_weight example is provided to demonstrate use of the mobile,
arm, and planet abstractions.
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 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
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 amobilehanging at the end of the left arm of weight10, the torque on the left side of our mobile is50. - Each of the mobiles hanging at the end of its arms is balanced.
Planets themselves are balanced, as there is nothing hanging off of them.
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_trees 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
Submit
Submissions will be in Canvas.
It is highly recommended that you test your code before you submit it. To run all of the tests for the required questions, type:
python3 ok
Optional Questions
Trees
Q7: Preorder
Define the function preorder, which takes in a tree as an argument and
returns a list of all the entries in the tree in the order that
print_tree would print them.
The following diagram shows the order that the nodes would get printed, with the arrows representing function calls.
Note: This ordering of the nodes in a tree is called a preorder traversal.
def preorder(t):
"""Return a list of the entries in this tree in the order that they
would be visited by a preorder traversal (see problem description).
>>> numbers = tree(1, [tree(2), tree(3, [tree(4), tree(5)]), tree(6, [tree(7)])])
>>> preorder(numbers)
[1, 2, 3, 4, 5, 6, 7]
>>> preorder(tree(2, [tree(4, [tree(6)])]))
[2, 4, 6]
"""
"*** YOUR CODE HERE ***"Use Ok to test your code:
python3 ok -q preorder
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)Q8: 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
Q9: 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
Q10: 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, r2Use Ok to test your code:
python3 ok -q check_par