# Lab 7: Recursive Objects lab07.zip

Due at 11:59pm on Friday, 10/13/2017.

## 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.

A linked list is either an empty linked list (`Link.empty`) or a Link object containing a first value and the rest of the linked list. Here is a part of the `Link` class. The entire class can be found in the assignment's files:

``````class Link:

>>> s.first
1
>>> s.rest
"""
empty = ()

def __init__(self, first, rest=empty):
self.first = first
self.rest = rest

def __repr__(self):
else:

def __str__(self):

>>> str(s)
'<1 2 3 4>'
'<1>'
'()'
"""
string = '<'
string += str(self.first) + ' '
self = self.rest
return string + str(self.first) + '>'``````

To check if a `Link` is empty, compare it against the class attribute `Link.empty`. For example, the below function prints out whether or not the link it is handed is empty:

``````def test_empty(link):
else:
print('This linked list is not empty!')``````

Note: Linked lists are recursive data structures! A linked list contains the first element of the list (`first`) and a reference to another linked list (`rest`) which contains the rest of the values in the list.

Use Ok to test your knowledge with the following "What Would Python Display?" questions:

``python3 ok -q link -u``

If you get stuck, try drawing out the diagram for the linked list on a piece of paper, or loading the `Link` class into the interpreter with `python3 -i` and the Python file you'd like to load.

``````>>> from link import *
______1
______2
______True
______9001
______3
______1
______1
______2
>>> print(link2) # Look at the __str__ method of Link in lab07.py
______<1 2 3 4>``````

Write a function `list_to_link` that converts a Python list to a `Link`.

``````def list_to_link(lst):
"""Takes a Python list and returns a Link with the same elements.

<1 2 3>
"""
if not lst:
else:

Use Ok to test your code:

``python3 ok -q list_to_link``

Write a function `link_to_list` that converts a given `Link` to a Python list.

``````def link_to_list(link):
"""Takes a Link and returns a Python list with the same elements.

[1, 2, 3, 4]
[]
"""
# Recursive solution
return []

# Iterative solution
result = []
return result``````

Use Ok to test your code:

``python3 ok -q link_to_list``

### Q4: 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.

>>> 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>
"""
return
else:

# alternate solution

Use Ok to test your code:

``python3 ok -q remove_all``

## Trees (Again)

Just like linked lists, we can also represent trees as objects.

``````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])``````

### Q5: WWPD: Trees

Use Ok to test your knowledge with the following "What Would Python Display?" questions:

``python3 ok -q trees -u``

Hint: Remember for all WWPD questions, enter `Function` if you believe the answer is `<function ...>` and `Error` if it errors.

``````>>> from lab07 import *
>>> t = Tree(1, Tree(2))
______Error
>>> t = Tree(1, [Tree(2)])
>>> t.label
______1
>>> t.branches
______Tree(2)
>>> t.branches.label
______2
>>> t.label = t.branches.label
>>> t
______Tree(2, [Tree(2)])
>>> t.branches.append(Tree(4, [Tree(8)]))
>>> len(t.branches)
______2
>>> t.branches
______Tree(2)
>>> t.branches
______Tree(4, [Tree(8)])``````

## Extra 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.

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 term, you'll be programming in Scheme, which is a programming language that uses linked lists for almost everything. During lecture, you learned that certain operations are faster with linked lists comparing to python-lists. Here you will test out those theories in practice.

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. To test this, in your terminal, enter the following command: ```python3 timing.py insert 100000```, which inserts 100,000 items into the beginning of both a linked list and a Python built-in list to compare the speed.

Now, say we want the item at index 3.

• In the built-in list, you can simply grab the item from the container with 3 labeled on it;
• In the linked list, you need to start at the first item, and go to its neighbor's neighbor's neighbor to finally reach the item at index 3.

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).

### Q6: 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)])
"""
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``

### Q7: Reverse Other

Write a function `reverse_other` that mutates the tree such that every other (odd_indexed) level of the tree's entries are all reversed. For example `Tree(1,[Tree(2), Tree(3)])` becomes `Tree(1,[Tree(3), Tree(2)])`

``````def reverse_other(t):
"""Reverse the entries of every other level of the tree using mutation.

>>> 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(5, [Tree(7), Tree(8)]), Tree(6)]), Tree(3)])
>>> reverse_other(t)
>>> t
Tree(1, [Tree(3, [Tree(5, [Tree(8), Tree(7)]), Tree(6)]), Tree(2)])
"""
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``

### Q8: 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.

Hint: The built-in `isinstance` function may be useful.

``````>>> s = Link(1, Link(2, Link(3, Link(4))))
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

Does not return the modified Link object.

>>> deep_map_mut(lambda x: x * x, link1)
<9 <16> 25 36>
"""
return
else:

Use Ok to test your code:

``python3 ok -q deep_map_mut``

### Q9: 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.rest.rest.rest = s
>>> has_cycle(s)
True
>>> has_cycle(t)
False
>>> has_cycle(u)
False
"""
return True
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.rest.rest.rest = s
>>> has_cycle_constant(s)
True
>>> has_cycle_constant(t)
False
"""
return False
``python3 ok -q has_cycle_constant``