Lab 7: Recursive Objects
Due at 11:59pm on Friday, 03/09/2018.
Starter Files
Download lab07.zip. Inside the archive, you will find starter files for the questions in this lab, along with a copy of the Ok autograder.
Submission
By the end of this lab, you should have submitted the lab with
python3 ok --submit
. You may submit more than once before the
deadline; only the final submission will be graded.
Check that you have successfully submitted your code on
okpy.org.
- To receive credit for this lab, you must complete Questions 1, 2, 3, 4, and 5 in lab07.py and submit through OK.
- Questions 6 through 9 are extra practice. They can be found in the lab07_extra.py file. It is recommended that you complete these problems on your own time.
Topics
Linked Lists
We've learned that a Python list is one way to store sequential values. Another type of list is a linked list. A Python list stores all of its elements in a single object, and each element can be accessed by using its index. A linked list, on the other hand, is a recursive object that only stores two things: its first value and a reference to the rest of the list, which is another linked list.
We can implement a class, Link
, that represents a linked list object. Each
instance of Link
has two instance attributes, first
and rest
.
class Link:
"""A linked list.
>>> s = Link(1)
>>> s.first
1
>>> s.rest is Link.empty
True
>>> s = Link(2, Link(3, Link(4)))
>>> s.second
3
>>> s.first = 5
>>> s.second = 6
>>> s.rest.rest = Link.empty
>>> s # Displays the contents of repr(s)
Link(5, Link(6))
>>> s.rest = Link(7, Link(Link(8, Link(9))))
>>> s
Link(5, Link(7, Link(Link(8, Link(9)))))
>>> print(s) # Prints str(s)
<5 7 <8 9>>
"""
empty = ()
def __init__(self, first, rest=empty):
assert rest is Link.empty or isinstance(rest, Link)
self.first = first
self.rest = rest
@property
def second(self):
return self.rest.first
@second.setter
def second(self, value):
self.rest.first = value
def __repr__(self):
if self.rest is not Link.empty:
rest_repr = ', ' + repr(self.rest)
else:
rest_repr = ''
return 'Link(' + repr(self.first) + rest_repr + ')'
def __str__(self):
string = '<'
while self.rest is not Link.empty:
string += str(self.first) + ' '
self = self.rest
return string + str(self.first) + '>'
A valid linked list can be one of the following:
- An empty linked list (
Link.empty
) - A
Link
object containing the first value of the linked list and a reference to the rest of the linked list
What makes a linked list recursive is that the rest
attribute of a single
Link
instance is another linked list! In the big picture, each Link
instance stores a single value of the list. When multiple Link
s are linked
together through each instance's rest
attribute, an entire sequence is
formed.
Note: This definition means that the
rest
attribute of anyLink
instance must be eitherLink.empty
or anotherLink
instance! This is enforced inLink.__init__
, which raises anAssertionError
if the value passed in forrest
is neither of these things.
We've also defined a pseudo-attribute second
with the @property
decorator
that will return the second element in the linked list as well as a
corresponding setter. Note that the second element of a linked list is really
just the first
attribute of the Link
instance stored in rest
. Don't worry
too much about the syntax of the setter function for now. See the docstring for
a closer look at how to use this property.
To check if a linked list is empty, compare it against the class attribute
Link.empty
. For example, the function below prints out whether or not the
link it is handed is empty:
def test_empty(link):
if link is Link.empty:
print('This linked list is empty!')
else:
print('This linked list is not empty!')
Motivation: Why linked lists
Since you are already familiar with Python's built-in lists, you might be wondering why we are teaching you another list representation. There are historical reasons, along with practical reasons. Later in the course, you'll be programming in Scheme, which is a programming language that uses linked lists for almost everything.
For now, let's compare linked lists and Python lists by looking at two common sequence operations: inserting an item and indexing.
Python's built-in list is like a sequence of containers with indices on them:
Linked lists are a list of items pointing to their neighbors. Notice that there's no explicit index for each item.
Suppose we want to add an item at the head of the list.
- With Python's built-in list, if you want to put an item into the container labeled with index 0, you must move all the items in the list into its neighbor containers to make room for the first item;
- With a linked list, you tell Python that the neighbor of the new item is the old beginning of the list.
We can compare the speed of this operation by timing how long it takes to insert a large number of items to the beginning of both types of lists. Enter the following command in your terminal to test this:
python3 timing.py insert 100000
Now, let's take a look at indexing. Say we want the item at index 3 from a list.
- In the built-in list, you can use Python list indexing, e.g.
lst[3]
, to easily get the item at index 3. - In the linked list, you need to start at the first item and repeatedly follow
the
rest
attribute, e.g.link.rest.rest.first
. How does this scale if the index you were trying to access was very large?
To test this, enter the following command in your terminal
python3 timing.py index 10000
This program compares the speed of randomly accessing 10,000 items from both a linked list and a built-in Python list (each with length 10,000).
What conclusions can you draw from these tests? Can you think of situations where you would want to use one type of list over another? In this class, we aren't too worried about performance. However, in future computer science courses, you'll learn how to make performance tradeoffs in your programs by choosing your data structures carefully.
Trees (Again)
We've already seen trees as abstract data types. Recall that a tree is a
recursive data type that has a label
(the value stored in the root of the
tree) and branches
(a list of trees directly underneath the root).
The tree data type that we studied was simply an abstract representation of a tree structure. Behind the scenes, the tree ADT was implemented using Python lists.
Now, we'll be working with trees as actual objects with attributes and methods! Here is the class definition:
class Tree:
def __init__(self, label, branches=[]):
for c in branches:
assert isinstance(c, Tree)
self.label = label
self.branches = list(branches)
def __repr__(self):
if self.branches:
branches_str = ', ' + repr(self.branches)
else:
branches_str = ''
return 'Tree({0}{1})'.format(self.label, branches_str)
def is_leaf(self):
return not self.branches
def __eq__(self, other):
return type(other) is type(self) and self.label == other.label \
and self.branches == other.branches
def __str__(self):
def print_tree(t, indent=0):
tree_str = ' ' * indent + str(t.label) + "\n"
for b in t.branches:
tree_str += print_tree(b, indent + 1)
return tree_str
return print_tree(self).rstrip()
def copy_tree(self):
return Tree(self.label, [b.copy_tree() for b in self.branches])
You'll see that the Tree
class is pretty similar to the tree ADT, namely
because the class is simple a formalization of the abstract data type into a
real, user-defined data type. Here is a summary of the differences:
- | Tree ADT | Tree class |
---|---|---|
Constructing a tree | Calling the constructor function tree(...) returns a tree ADT |
Calling the class constructor Tree(...) (which calls Tree.__init__(...) ) returns a Tree object |
Label and branches | Returned by selector functions label(...) and branches(...) |
Stored in instance attributes label and branches |
Mutability | The tree ADT is immutable | The label and branches attributes of a Tree instance can be reassigned, mutating the tree |
Checking if a tree is a leaf | The convenience function is_leaf(...) returns whether or not a tree ADT is a leaf. |
The bound method t.is_leaf() returns whether or not a Tree object is a leaf. This method can only be called on Tree objects. |
Required Questions
What Would Python Display?
Q1: WWPD: Linked Lists
Read over the Link
class in lab07.py
. Make sure you understand the
doctests.
Use Ok to test your knowledge with the following "What Would Python Display?" questions:
python3 ok -q link -u
Enter
Function
if you believe the answer is<function ...>
,Error
if it errors, andNothing
if nothing is displayed.If you get stuck, try drawing out the box-and-pointer diagram for the linked list on a piece of paper or loading the
Link
class into the interpreter withpython3 -i lab07.py
.
>>> from lab07 import *
>>> link = Link(1000)
>>> link.first
______1000
>>> link.rest is Link.empty
______True
>>> link = Link(1000, 2000)
______AssertionError
>>> link = Link(1000, Link())
______TypeError
>>> from lab07 import *
>>> link = Link(1, Link(2, Link(3)))
>>> link.first
______1
>>> link.rest.first
______2
>>> link.rest.rest.rest is Link.empty
______True
>>> link.first = 9001
>>> link.first
______9001
>>> link.rest = link.rest.rest
>>> link.rest.first
______3
>>> link = Link(1)
>>> link.rest = link
>>> link.rest.rest.rest.rest.first
______1
>>> link = Link(2, Link(3, Link(4)))
>>> link2 = Link(1, link)
>>> link2.first
______1
>>> link2.rest.first
______2
>>> from lab07 import *
>>> link = Link(5, Link(6, Link(7)))
>>> link.second
______6
>>> link.rest.second
______7
>>> link.second = 10
>>> link # Look at the __repr__ method of Link
______Link(5, Link(10, Link(7)))
>>> link.second = Link(8, Link(9))
>>> link.second.first
______8
>>> print(link) # Look at the __str__ method of Link
______<5 <8 9> 7>
Q2: WWPD: Trees
Use Ok to test your knowledge with the following "What Would Python Display?" questions:
python3 ok -q trees -u
Enter
Function
if you believe the answer is<function ...>
,Error
if it errors, andNothing
if nothing is displayed.
>>> from lab07 import *
>>> t = Tree(1, Tree(2))
______Error
>>> t = Tree(1, [Tree(2)])
>>> t.label
______1
>>> t.branches[0]
______Tree(2)
>>> t.branches[0].label
______2
>>> t.label = t.branches[0].label
>>> t
______Tree(2, [Tree(2)])
>>> t.branches.append(Tree(4, [Tree(8)]))
>>> len(t.branches)
______2
>>> t.branches[0]
______Tree(2)
>>> t.branches[1]
______Tree(4, [Tree(8)])
Coding Practice
Q3: Link to List
Write a function link_to_list
that takes in a linked list and returns the
sequence as a Python list. You may assume that the input list is shallow; none
of the elements is another linked list.
Try to find both an iterative and recursive solution for this problem!
def link_to_list(link):
"""Takes a linked list and returns a Python list with the same elements.
>>> link = Link(1, Link(2, Link(3, Link(4))))
>>> link_to_list(link)
[1, 2, 3, 4]
>>> link_to_list(Link.empty)
[]
"""
"*** YOUR CODE HERE ***"
# Recursive solution
if link is Link.empty:
return []
return [link.first] + link_to_list(link.rest)
# Iterative solution
def link_to_list(link):
result = []
while link is not Link.empty:
result.append(link.first)
link = link.rest
return result
Use Ok to test your code:
python3 ok -q link_to_list
Q4: Store Digits
Write a function store_digits
that takes in an integer n
and returns
a linked list where each element of the list is a digit of n
.
def store_digits(n):
"""Stores the digits of a positive number n in a linked list.
>>> s = store_digits(1)
>>> s
Link(1)
>>> store_digits(2345)
Link(2, Link(3, Link(4, Link(5))))
>>> store_digits(876)
Link(8, Link(7, Link(6)))
"""
"*** YOUR CODE HERE ***"
result = Link.empty
while n > 0:
result = Link(n % 10, result)
n //= 10
return result
Use Ok to test your code:
python3 ok -q store_digits
Q5: Cumulative Sum
Write a function cumulative_sum
that mutates the Tree t
, where each node's
label becomes the sum of all entries in the subtree rooted at the node.
def cumulative_sum(t):
"""Mutates t where each node's root becomes the sum of all entries in the
corresponding subtree rooted at t.
>>> t = Tree(1, [Tree(3, [Tree(5)]), Tree(7)])
>>> cumulative_sum(t)
>>> t
Tree(16, [Tree(8, [Tree(5)]), Tree(7)])
"""
"*** YOUR CODE HERE ***"
for st in t.branches:
cumulative_sum(st)
t.label = sum([st.label for st in t.branches]) + t.label
Use Ok to test your code:
python3 ok -q cumulative_sum
Optional Questions
The following questions are for extra practice -- they can be found in the the lab07_extra.py file. It is recommended that you complete these problems on your own time.
Linked List Practice
Q6: Remove All
Implement a function remove_all
that takes a Link
, and a value
,
and remove any linked list node containing that value. You can assume the
list already has at least one node containing value
and the first element is
never removed. Notice that you are not returning anything, so you should mutate the list.
def remove_all(link , value):
"""Remove all the nodes containing value. Assume there exists some
nodes to be removed and the first element is never removed.
>>> l1 = Link(0, Link(2, Link(2, Link(3, Link(1, Link(2, Link(3)))))))
>>> print(l1)
<0 2 2 3 1 2 3>
>>> remove_all(l1, 2)
>>> print(l1)
<0 3 1 3>
>>> remove_all(l1, 3)
>>> print(l1)
<0 1>
"""
"*** YOUR CODE HERE ***"
if link is Link.empty or link.rest is Link.empty:
return
if link.rest.first == value:
link.rest = link.rest.rest
remove_all(link, value)
else:
remove_all(link.rest, value)
# alternate solution
if link is not Link.empty and link.rest is not Link.empty:
remove_all(link.rest, value)
if link.rest.first == value:
link.rest = link.rest.rest
Use Ok to test your code:
python3 ok -q remove_all
Q7: Mutable Mapping
Implement deep_map_mut(fn, link)
, which applies a function fn
onto
all elements in the given linked list link
. 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
isinstance
function may be useful.>>> s = Link(1, Link(2, Link(3, Link(4)))) >>> isinstance(s, Link) True >>> isinstance(s, int) False
def deep_map_mut(fn, link):
"""Mutates a deep link 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))))
>>> deep_map_mut(lambda x: x * x, link1)
>>> print(link1)
<9 <16> 25 36>
"""
"*** YOUR CODE HERE ***"
if link is Link.empty:
return
elif isinstance(link.first, Link):
deep_map_mut(fn, link.first)
else:
link.first = fn(link.first)
deep_map_mut(fn, link.rest)
Use Ok to test your code:
python3 ok -q deep_map_mut
Q8: Cycles
The Link
class can represent lists with cycles. That is, a list may
contain itself as a sublist.
>>> s = Link(1, Link(2, Link(3)))
>>> s.rest.rest.rest = s
>>> s.rest.rest.rest.rest.rest.first
3
Implement has_cycle
,that returns whether its argument, a Link
instance, contains a cycle.
Hint: Iterate through the linked list and try keeping track of which
Link
objects you've already seen.
def has_cycle(link):
"""Return whether link contains a cycle.
>>> s = Link(1, Link(2, Link(3)))
>>> s.rest.rest.rest = s
>>> has_cycle(s)
True
>>> t = Link(1, Link(2, Link(3)))
>>> has_cycle(t)
False
>>> u = Link(2, Link(2, Link(2)))
>>> has_cycle(u)
False
"""
"*** YOUR CODE HERE ***"
links = []
while link is not Link.empty:
if link in links:
return True
links.append(link)
link = link.rest
return False
Use Ok to test your code:
python3 ok -q has_cycle
As an extra challenge, implement has_cycle_constant
with only constant space. (If you followed
the hint above, you will use linear space.) The solution is short (less than 20
lines of code), but requires a clever idea. Try to discover the solution
yourself before asking around:
def has_cycle_constant(link):
"""Return whether link contains a cycle.
>>> s = Link(1, Link(2, Link(3)))
>>> s.rest.rest.rest = s
>>> has_cycle_constant(s)
True
>>> t = Link(1, Link(2, Link(3)))
>>> has_cycle_constant(t)
False
"""
"*** YOUR CODE HERE ***"
if link is Link.empty:
return False
slow, fast = link, link.rest
while fast is not Link.empty:
if fast.rest == Link.empty:
return False
elif fast is slow or fast.rest is slow:
return True
else:
slow, fast = slow.rest, fast.rest.rest
return False
Use Ok to test your code:
python3 ok -q has_cycle_constant
Tree Practice
Q9: Reverse Other
Write a function reverse_other
that mutates the tree such that nodes on
every other (even_indexed) level have the labels of their branches all reversed.
For example Tree(1,[Tree(2), Tree(3)])
becomes Tree(1,[Tree(3), Tree(2)])
def reverse_other(t):
"""Mutates the tree such that nodes on every other (even_indexed) level
have the labels of their branches all reversed.
>>> t = Tree(1, [Tree(2), Tree(3), Tree(4)])
>>> reverse_other(t)
>>> t
Tree(1, [Tree(4), Tree(3), Tree(2)])
>>> t = Tree(1, [Tree(2, [Tree(3, [Tree(4), Tree(5)]), Tree(6, [Tree(7)])]), Tree(8)])
>>> reverse_other(t)
>>> t
Tree(1, [Tree(8, [Tree(3, [Tree(5), Tree(4)]), Tree(6, [Tree(7)])]), Tree(2)])
"""
"*** YOUR CODE HERE ***"
def reverse_helper(t, need_reverse):
if t.is_leaf():
return
new_labs = [child.label for child in t.branches][::-1]
for i in range(len(t.branches)):
child = t.branches[i]
reverse_helper(child, not need_reverse)
if need_reverse:
child.label = new_labs[i]
reverse_helper(t, True)
Use Ok to test your code:
python3 ok -q reverse_other