Lab 8: Midterm Review
Due at 11:59pm on Monday, 3/20/2017.
Starter Files
Download lab08.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.
- To receive credit for this lab, you must complete Questions 1, 2, 3, 4 in lab08.py and submit through OK.
- The remaining questions are extra practice. They can be found in the lab08_extra.py file. It is recommended that you complete these problems on your own time.
Required Questions
Linked Lists
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:
"""A linked list.
>>> s = Link(1, Link(2, Link(3)))
>>> s.first
1
>>> s.rest
Link(2, Link(3))
"""
empty = ()
def __init__(self, first, rest=empty):
assert rest is Link.empty or isinstance(rest, Link)
self.first = first
self.rest = rest
def __repr__(self):
if self.rest is Link.empty:
return 'Link({})'.format(self.first)
else:
return 'Link({}, {})'.format(self.first, repr(self.rest))
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):
if link is Link.empty:
print('This linked list is empty!')
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.
Question 1: Linked Lists as Strings
Stan and Michelle like different ways of displaying the linked list
structure in Python. While Stan likes box and pointer diagrams,
Michelle prefers a more futuristic way. Write a function
make_to_string
that returns a function that converts the
linked list to a string in their preferred style.
Hint: You can convert numbers to strings using the str
function,
and you can combine strings together using +
.
>>> str(4)
'4'
>>> 'cs ' + str(61) + 'a'
'cs 61a'
def make_to_string(front, mid, back, empty_repr):
""" Returns a function that turns linked lists to strings.
>>> stans_to_string = make_to_string("[", "|-]-->", "", "[]")
>>> michelles_to_string = make_to_string("(", " . ", ")", "()")
>>> lst = Link(1, Link(2, Link(3, Link(4))))
>>> stans_to_string(lst)
'[1|-]-->[2|-]-->[3|-]-->[4|-]-->[]'
>>> stans_to_string(Link.empty)
'[]'
>>> michelles_to_string(lst)
'(1 . (2 . (3 . (4 . ()))))'
>>> michelles_to_string(Link.empty)
'()'
"""
"*** YOUR CODE HERE ***"
def printer(lnk):
if lnk is Link.empty:
return empty_repr
else:
return front + str(lnk.first) + mid + printer(lnk.rest) + back
return printer
Use OK to test your code:
python3 ok -q make_to_string
Question 2: Mutate Reverse
Implement mutate_reverse
, which takes a linked list instance and mutates it
so that its values appear in reverse order. For an extra challenge, find an
implementation that requires only linear time in the length of the list
(big-Theta n).
def mutate_reverse(link):
"""Mutates the Link so that its elements are reversed.
>>> link = Link(1)
>>> mutate_reverse(link)
>>> link
Link(1)
>>> link = Link(1, Link(2, Link(3)))
>>> mutate_reverse(link)
>>> link
Link(3, Link(2, Link(1)))
"""
"*** YOUR CODE HERE ***"
if link is Link.empty or link.rest is Link.empty:
return
mutate_reverse(link.rest)
while link.rest is not Link.empty:
link.first, link.rest.first = link.rest.first, link.first
link = link.rest
# O(n) solution
def mutate_reverse(link):
"""Mutates the Link so that its elements are reversed.
>>> link = Link(1)
>>> mutate_reverse(link)
>>> link
Link(1)
>>> link = Link(1, Link(2, Link(3)))
>>> mutate_reverse(link)
>>> link
Link(3, Link(2, Link(1)))
"""
link.rest = reverse_in_place(link.rest)
while link.rest is not Link.empty:
link.first, link.rest.first = link.rest.first, link.first
link = link.rest
def reverse_in_place(link):
result = Link.empty
while link is not Link.empty:
rest = link.rest
link.rest = result
result = link
link = rest
return result
Use OK to test your code:
python3 ok -q mutate_reverse
Trees
Question 3: Tree Map
Define the function tree_map
, which takes in a tree and a
one-argument function as arguments and returns a new tree which is the
result of mapping the function over the entries of the input tree.
def tree_map(fn, t):
"""Maps the function fn over the entries of tree and returns the
result in a new tree.
>>> numbers = Tree(1,
... [Tree(2,
... [Tree(3),
... Tree(4)]),
... Tree(5,
... [Tree(6,
... [Tree(7)]),
... Tree(8)])])
>>> print_tree(tree_map(lambda x: 2**x, numbers))
2
4
8
16
32
64
128
256
"""
"*** YOUR CODE HERE ***"
if t.is_leaf():
return Tree(fn(t.label), [])
mapped_subtrees = [tree_map(fn, b) for b in t.branches]
return Tree(fn(t.label), mapped_subtrees)
# Alternate solution
def tree_map(fn, t):
return Tree(fn(t.label), [tree_map(fn, b) for b in t.branches])
Use OK to test your code:
python3 ok -q tree_map
Question 4: Add trees
Define the function add_trees
, which takes in two trees and returns a new
tree where each corresponding node from the first tree is added with the node
from the second tree. If a node at any particular position is present in one
tree but not the other, it should be present in the new tree as well.
Hint: You may want to use the built-in zip function to iterate over multiple sequences at once.
def add_trees(t1, t2):
"""
>>> numbers = Tree(1,
... [Tree(2,
... [Tree(3),
... Tree(4)]),
... Tree(5,
... [Tree(6,
... [Tree(7)]),
... Tree(8)])])
>>> print_tree(add_trees(numbers, numbers))
2
4
6
8
10
12
14
16
>>> print_tree(add_trees(Tree(2), Tree(3, [Tree(4), Tree(5)])))
5
4
5
>>> print_tree(add_trees(Tree(2, [Tree(3)]), Tree(2, [Tree(3), Tree(4)])))
4
6
4
>>> print_tree(add_trees(Tree(2, [Tree(3, [Tree(4), Tree(5)])]), \
Tree(2, [Tree(3, [Tree(4)]), Tree(5)])))
4
6
8
5
5
"""
"*** YOUR CODE HERE ***"
if not t1:
return t2 # Could replace with copy_tree(t2)
if not t2:
return t1 # Could replace with copy_tree(t1)
new_label = t1.label + t2.label
t1_branches, t2_branches = list(t1.branches), list(t2.branches)
length_t1, length_t2 = len(t1_branches), len(t2_branches)
if length_t1 < length_t2:
t1_branches += [None for _ in range(length_t1, length_t2)]
elif length_t1 > length_t2:
t2_branches += [None for _ in range(length_t2, length_t1)]
return Tree(new_label, [add_trees(branch1, branch2) for branch1, branch2 in zip(t1_branches, t2_branches)])
Use OK to test your code:
python3 ok -q add_trees
Optional Questions
Objects
Question 5: WWPP: Methods
Use OK to test your knowledge with the following "What Would Python Print?" questions:
python3 ok -q foobar -u
Hint: Remember for all WWPP questions, enter
Function
if you believe the answer is<function ...>
andError
if it errors.
>>> class Foo:
... def print_one(self):
... print('foo')
... def print_two():
... print('foofoo')
>>> f = Foo()
>>> f.print_one()
______foo
>>> f.print_two()
______Error
>>> Foo.print_two()
______foofoo
>>> class Bar(Foo):
... def print_one(self):
... print('bar')
>>> b = Bar()
>>> b.print_one()
______bar
>>> Bar.print_two()
______foofoo
>>> Bar.print_one = lambda x: print('new bar')
>>> b.print_one()
______new bar
Question 6: WWPP: Attributes
Use OK to test your knowledge with the following "What Would Python Print?" questions:
python3 ok -q attributes -u
Hint: Remember for all WWPP questions, enter
Function
if you believe the answer is<function ...>
andError
if it errors.
>>> class Foo:
... a = 10
... def __init__(self, a):
... self.a = a
>>> class Bar(Foo):
... b = 1
>>> a = Foo(5)
>>> b = Bar(2)
>>> a.a
______5
>>> b.a
______2
>>> Foo.a
______10
>>> Bar.b
______1
>>> Bar.a
______10
>>> b.b
______1
>>> Foo.c = 15
>>> Foo.c
______15
>>> a.c
______15
>>> b.c
______15
>>> Bar.c
______15
>>> b.b = 3
>>> b.b
______3
>>> Bar.b
______1
Nonlocal
Question 7: Advanced Counter
Complete the definition of make_advanced_counter_maker
,
which creates a function that creates counters. These counters can not
only update their personal count, but also a shared count for all
counters. They can also reset either count.
def make_advanced_counter_maker():
"""Makes a function that makes counters that understands the
messages "count", "global-count", "reset", and "global-reset".
See the examples below:
>>> make_counter = make_advanced_counter_maker()
>>> tom_counter = make_counter()
>>> tom_counter('count')
1
>>> tom_counter('count')
2
>>> tom_counter('global-count')
1
>>> jon_counter = make_counter()
>>> jon_counter('global-count')
2
>>> jon_counter('count')
1
>>> jon_counter('reset')
>>> jon_counter('count')
1
>>> tom_counter('count')
3
>>> jon_counter('global-count')
3
>>> jon_counter('global-reset')
>>> tom_counter('global-count')
1
"""
"*** YOUR CODE HERE ***"
global_count = 0
def make_counter():
count = 0
def counter(msg):
nonlocal global_count, count
if msg == 'count':
count += 1
return count
elif msg == 'reset':
count = 0
elif msg == 'global-count':
global_count += 1
return global_count
elif msg == 'global-reset':
global_count = 0
return counter
return make_counter
Use OK to test your code:
python3 ok -q make_advanced_counter_maker
Lists
Question 8: Trade
In the integer market, each participant has a list of positive integers to trade. When two participants meet, they trade the smallest non-empty prefix of their list of integers. A prefix is a slice that starts at index 0.
Write a function trade
that exchanges the first m
elements of list first
with the first n
elements of list second
, such that the sums of those
elements are equal, and the sum is as small as possible. If no such prefix
exists, return the string 'No deal!'
and do not change either list. Otherwise
change both lists and return 'Deal!'
. A partial implementation is provided.
def trade(first, second):
"""Exchange the smallest prefixes of first and second that have equal sum.
>>> a = [1, 1, 3, 2, 1, 1, 4]
>>> b = [4, 3, 2, 7]
>>> trade(a, b) # Trades 1+1+3+2=7 for 4+3=7
'Deal!'
>>> a
[4, 3, 1, 1, 4]
>>> b
[1, 1, 3, 2, 2, 7]
>>> c = [3, 3, 2, 4, 1]
>>> trade(b, c)
'No deal!'
>>> b
[1, 1, 3, 2, 2, 7]
>>> c
[3, 3, 2, 4, 1]
>>> trade(a, c)
'Deal!'
>>> a
[3, 3, 2, 1, 4]
>>> b
[1, 1, 3, 2, 2, 7]
>>> c
[4, 3, 1, 4, 1]
"""
m, n = 1, 1
"*** YOUR CODE HERE ***"
equal_prefix = lambda: sum(first[:m]) == sum(second[:n])
while m < len(first) and n < len(second) and not equal_prefix():
if sum(first[:m]) < sum(second[:n]):
m += 1
else:
n += 1
if False: # change this line!
if equal_prefix(): first[:m], second[:n] = second[:n], first[:m]
return 'Deal!'
else:
return 'No deal!'
Use OK to test your code:
python3 ok -q trade
Orders of Growth
Question 9: Boom (Orders of Growth)
What is the order of growth in time for the following function boom? Use big-θ notation.
def boom(n):
sum = 0
a, b = 1, 1
while a <= n*n:
while b <= n*n:
sum += (a*b)
b += 1
b = 0
a += 1
return sum
θ(n4)
Use OK to test your code:
python3 ok -q boom -u