Due at 11:59:59 pm on 11/21/2019.

Starter Files

Download lab09.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. See this article for more instructions on okpy and submitting assignments.

• To receive credit for this lab, you must complete Questions 1, 2, 3, 4, 5, and 6 in lab09.py and submit through OK.
• Questions 7 and 8 are extra practice. They can be found in the lab09_extra.py file. It is recommended that you complete these problems on your own time for extra practice.

A linked list is either an empty linked list (`Link.empty`) or a first value and the rest of the linked list.

``````class Link:
"""
>>> s
"""
empty = ()

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

def __repr__(self):
rest_str = ', ' + repr(self.rest)
else:
rest_str = ''

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 Print?" questions:

``python3 ok -q link -u``

If you get stuck, try loading lab09.py into an interpreter or drawing out the diagram for the linked list on a piece of paper.

``````>>> from lab09 import *
______1
______2
______True
______9001
______3
______1
______1
______2
______<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``

Question 3: Insert

Implement a function `insert` that takes a `Link`, a `value`, and an `index`, and inserts the `value` into the `Link` at the given `index`. You can assume the linked list already has at least one element. Do not return anything — `insert` should mutate the linked list.

Note: If the index is out of bounds, you can raise an `IndexError` with:

``raise IndexError``
``````def insert(link, value, index):
"""Insert a value into a Link at the given index.

<1 2 3>
<9001 1 2 3>
<9001 1 100 2 3>
IndexError
"""
if index == 0:
raise IndexError
else:

# iterative solution
index -= 1
if index == 0:
else:
raise IndexError``````

Use OK to test your code:

``python3 ok -q insert``

Trees

As we saw in lecture, we can also represent trees as objects.

``````class Tree:
def __init__(self, entry, branches=()):
self.entry = entry
for branch in branches:
assert isinstance(branch, Tree)
self.branches = list(branches)

def __repr__(self):
if self.branches:
branches_str = ', ' + repr(self.branches)
else:
branches_str = ''
return 'Tree({0}{1})'.format(self.entry, branches_str)

def is_leaf(self):
return not self.branches``````

Question 4: WWPP: Trees

Use OK to test your knowledge with the following "What Would Python Print?" questions:

``python3 ok -q trees -u``

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

``````>>> from lab09 import *
>>> t = Tree(1, Tree(2))
______Error
>>> t = Tree(1, [Tree(2)])
>>> t.entry
______1
>>> t.branches[0]
______Tree(2)
>>> t.branches[0].entry
______2
>>> t.entry = t.branches[0].entry
>>> 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)])``````

Question 5: Same Shape

Write a function `same_shape` that returns `True` if two `Tree`s have the same shape. Two trees have the same shape if they have the same number of children and each of their children have the same shape.

``````def same_shape(t1, t2):
"""Returns whether two Trees t1, t2 have the same shape. Two trees have the
same shape if they have the same number of branches and each of their
children have the same shape.

>>> t, s = Tree(1), Tree(3)
>>> same_shape(t, t)
True
>>> same_shape(t, s)
True
>>> t = Tree(1, [Tree(2), Tree(3)])
>>> same_shape(t, s)
False
>>> s = Tree(4, [Tree(7)])
>>> same_shape(t, s)
False
"""
return len(t1.branches) == len(t2.branches) and \
all(same_shape(st1, st2) for st1, st2 in zip(t1.branches, t2.branches))``````

Use OK to test your code:

``python3 ok -q same_shape``

Question 6: 23andTree

Write a function `computeAncesTree` that fills in the ethnic breakdown of everyone in your `FamilyTree`. `FamilyTree`s have a name and an ethnicity. An `ethnicity` is a dictionary mapping ethnicities to their percentages for that person.
A `FamilyTree` contains a list of parents. Parents are also `FamilyTree`s. A child's ethnic breakdown should be half of each of the ethnicities in each of their parents. A leaf `FamilyTree` is a tree with no parents. To begin, you can only assume that the `ethnicity` of the leafs of your `FamilyTree` are filled in.

``````class FamilyTree:
def __init__(self, name, ethnicity, parents=[]):
# Leaf FamilyTrees have no parents
self.name = name
self.ethnicity = ethnicity
self.parents = parents
if not self.is_leaf():
for p in parents:
assert isinstance(p, FamilyTree)

def __repr__(self):
if self.is_leaf():
return self.name
return '{} child of {}'.format(self.name, repr(self.parents))

def is_leaf(self):
return len(self.parents) == 0

def computeAncesTree(t):
"""
Fill in the ethnicities of all parents.
>>> gma1 = FamilyTree("Farah", {"Moroccan": 100.0})
>>> gpa1 = FamilyTree("Lorenzo", {"Italian" : 100.0})
>>> gpa2 = FamilyTree("Hai", {"Chinese":100.0})
>>> gma2 = FamilyTree("Gazala", {"Indian":100.0})
>>> papa1 = FamilyTree("Amjad", {}, [gma1, gpa1]) #  Son of Farah and Lorenzo
>>> papa2 = FamilyTree("Arjun", {}, [gma2, gpa2]) # Son of Hai and Gazala
>>> mama1 = FamilyTree("Anabella", {}, [gma1, gpa1]) # Daughter of Farah and Lorenzo
>>> mama2 = FamilyTree("Biyu", {}, [gma2, gpa2]) # Daughter of Hai and Gazala
>>> c1 = FamilyTree("Dipika", {}, [mama1, papa2]) # Daughter of Arjun and Anabella
>>> c2 = FamilyTree("Cosimo", {}, [mama1, papa2]) # Son of Arjun and Anabella
>>> c3 = FamilyTree("Jin", {}, [mama2, papa1]) # Son of Amjad and Biyu
>>> c4 = FamilyTree("Malika", {}, [mama2, papa1]) # Daughter of Amjad and Biyu
>>> eth = {a:25.0 for a in ["Moroccan", "Italian", "Chinese", "Indian"]}
>>> computeAncesTree(c1) == eth
True
>>> mama1.ethnicity["Moroccan"] == 50.0 and mama1.ethnicity["Italian"] == 50.0
True
>>> papa2.ethnicity["Indian"] == 50.0 and papa2.ethnicity["Chinese"] == 50.0
True
>>> mama2.ethnicity == papa1.ethnicity == c2.ethnicity == c3.ethnicity == c4.ethnicity
True
>>> computeAncesTree(c2) == computeAncesTree(c3) == computeAncesTree(c4) == eth
True
>>> papa1.ethnicity == mama1.ethnicity
True
>>> papa2.ethnicity == mama2.ethnicity
True
>>> sidepa = FamilyTree("Kahlil Gibran", {"Lebanese":75.0, "Moroccan":25.0})
>>> secret = FamilyTree("The Prophet", {}, [gma1, sidepa])
>>> eth2 = {"Lebanese":37.5, "Moroccan":62.5}
>>> computeAncesTree(secret) == eth2
True
"""
if t.is_leaf():
return t.ethnicity
else:
parents_ethnicities = [computeAncesTree(p) for p in t.parents]
new_ethnicity = {}
for ethnicity in parents_ethnicities:
for eth, val in ethnicity.items():
if eth in new_ethnicity:
new_ethnicity[eth] += val / 2
else:
new_ethnicity[eth] = val / 2
t.ethnicity = new_ethnicity
return t.ethnicity``````

Use OK to test your code:

``python3 ok -q computeAncesTree``

Extra Questions

The following questions are for extra practice — they can be found in the the lab09_extra.py file. It is recommended that you complete these problems on your own time.

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``

Question 8: 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
"""
lists = set()
return True
return False``````

Use OK to test your code:

``python3 ok -q has_cycle``

Extra for experts: Implement `has_cycle` 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
return False
elif fast == slow or fast.rest == 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``

Motivation

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. But let's not worry about that for now. The real reason, is that certain operations are faster with linked lists.

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).

You'll learn more about orders of growth this week, which will provide mathematical rigor when comparing the runtime of the same operations with different data structures.