Homework 5 Solutions
Solution Files
You can find the solutions in hw05.py.
Required Questions
Note: This homework is due on Tuesday, March 15 instead of the standard deadline of Thursday, March 10.
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Mid-Semester Feedback
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 instructor (Pamela) and head TA (Vanshaj) 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_surveyParsons Problems
Q2: Chain
For this question, we will define a chain as a path from the root of a tree t to any leaf
such that all nodes on the path share the same label. Implement the function chain, which,
given a tree t, returns True if there exists any chain in the tree, and False otherwise.
def chain(t):
"""
Returns whether there exists a path in t where all nodes
share the same label.
>>> all_fives = Tree(5, [Tree(5), Tree(5, [Tree(5)])])
>>> chain(all_fives)
True
>>> t1 = Tree(1, [Tree(3, [Tree(4)]), Tree(1)])
>>> chain(t1)
True
>>> t2 = Tree(1, [Tree(3, [Tree(4)]), Tree(5)])
>>> chain(t2)
False
"""
if t.is_leaf():
return True
for b in t.branches:
if t.label == b.label and chain(b):
return True
return FalseQ3: Flatten Link
Write a function flatten_link that takes in a linked list lnk and returns the
sequence as a Python list. If lnk has nested linked lists, flatten_link should
flatten lnk.
def flatten_link(lnk):
"""Takes a linked list and returns a flattened Python list with the same elements.
>>> link = Link(1, Link(2, Link(3, Link(4))))
>>> flatten_link(link)
[1, 2, 3, 4]
>>> flatten_link(Link.empty)
[]
>>> deep_link = Link(Link(1, Link(2, Link(3, Link(4)))), Link(Link(5), Link(6)))
>>> flatten_link(deep_link)
[1, 2, 3, 4, 5, 6]
"""
if lnk is Link.empty:
return []
if isinstance(lnk.first, Link):
flattened = flatten_link(lnk.first)
return flattened + flatten_link(lnk.rest)
return [lnk.first] + flatten_link(lnk.rest)Code Writing Questions
Q4: Has Path
Write a function has_path that takes in a Tree t and a string term. It
returns True if there is a path that starts from the root where the entries
along the path spell out the term, and False otherwise. You may assume
that every node's label is exactly one character.
This data structure is called a trie, and it has a lot of cool applications, such as autocomplete.
def has_path(t, term):
"""Return whether there is a path in a Tree where the entries along the path
spell out a particular term.
>>> greetings = Tree('h', [Tree('i'),
... Tree('e', [Tree('l', [Tree('l', [Tree('o')])]),
... Tree('y')])])
>>> print(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(term) > 0, 'no path for empty term.'
if t.label != term[0]:
return False
elif len(term) == 1:
return True
for b in t.branches:
if has_path(b, term[1:]):
return True
return FalseUse Ok to test your code:
python3 ok -q has_pathQ5: Duplicate Link
Write a function duplicate_link that takes in a linked list lnk and a value. duplicate_link will mutate lnk such that if there is a linked list node that has a first equal to value, that node will be duplicated. Note that you should be mutating the original link list lnk; you will need to create new Links, but you should not be returning a new linked list.
Note: in order to insert a link into a linked list, you need to modify the
.restof certain links. We encourage you to draw out a doctest to visualize!
def duplicate_link(lnk, val):
"""Mutates `lnk` such that if there is a linked list
node that has a first equal to value, that node will
be duplicated. Note that you should be mutating the
original link list.
>>> x = Link(5, Link(4, Link(3)))
>>> duplicate_link(x, 5)
>>> x
Link(5, Link(5, Link(4, Link(3))))
>>> y = Link(2, Link(4, Link(6, Link(8))))
>>> duplicate_link(y, 10)
>>> y
Link(2, Link(4, Link(6, Link(8))))
"""
if lnk is Link.empty:
return
elif lnk.first == val:
remaining = lnk.rest
lnk.rest = Link(val, remaining)
duplicate_link(remaining, val)
else:
duplicate_link(lnk.rest, val)Use Ok to test your code:
python3 ok -q duplicate_linkQ6: Mutable Mapping
Implement deep_map_mut(fn, link), which applies a function fn onto
all elements in the given linked list lnk. If an element is itself a
linked list, apply fn to each of its elements, and so on.
Your implementation should mutate the original linked list. Do not create any new linked lists.
Hint: The built-in
isinstancefunction may be useful.>>> s = Link(1, Link(2, Link(3, Link(4)))) >>> isinstance(s, Link) True >>> isinstance(s, int) False
Construct Check: The last doctest of this question ensures that you do not create new linked lists. If you are failing this doctest, ensure that you are not creating link lists by calling the constructor, i.e.
s = Link(1)
def deep_map_mut(fn, lnk):
"""Mutates a deep link lnk by replacing each item found with the
result of calling fn on the item. Does NOT create new Links (so
no use of Link's constructor).
Does not return the modified Link object.
>>> link1 = Link(3, Link(Link(4), Link(5, Link(6))))
>>> # Disallow the use of making new Links before calling deep_map_mut
>>> Link.__init__, hold = lambda *args: print("Do not create any new Links."), Link.__init__
>>> try:
... deep_map_mut(lambda x: x * x, link1)
... finally:
... Link.__init__ = hold
>>> print(link1)
<9 <16> 25 36>
"""
if lnk is Link.empty:
return
elif isinstance(lnk.first, Link):
deep_map_mut(fn, lnk.first)
else:
lnk.first = fn(lnk.first)
deep_map_mut(fn, lnk.rest)Use Ok to test your code:
python3 ok -q deep_map_mutSubmit
Make sure to submit this assignment by running:
python3 ok --submitOptional Questions
Homework assignments will also contain prior exam-level questions for you to take a look at. These questions have no submission component; feel free to attempt them if you'd like a challenge!