CS 61A Structure and Interpretation of Computer Programs Fall 2020 Quiz 6 Solutions
INSTRUCTIONS
• Please review this worksheet before the exam prep session. Coming prepared will help greatly, as the TA will be live solving without allocating much time for individual work.
• Either Sean or Derek will be on video live solving these questions. The other TA will be answering questions in the chat. It is in your best interest to come prepared with specific questions.
• This is not graded, and you do not need to turn this in to anyone.
• Fall 2020 students: the boxes below are an artifact from more typical semesters to simulate exam environments. Obviously this doesn’t apply to this semester’s exams, but we just kept the fields to keep our materials looking professional 🙂 Feel free to ignore them.
• For multiple choice questions, fill in each option or choice completely.
– means mark all options that apply – means mark a single choice
Last name
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Discussion Section
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1. Quarantine Dieting
For each of the expressions in the table below, write the output displayed by the interactive Python interpreter when the expression is evaluated. The output may have multiple lines. If an error occurs, write “Error”, but include all output displayed before the error. If a function value is displayed, write “Function”.
Assume that you have started python3 and executed the following statements. Also assume that effects from a previous subpart persist to future subparts.
class VendingMachine: k=0
def __init__(self, k, v):
self.soda = JunkDrink(self)
self.k = k
if v:
print(isinstance(self.soda.machine, VendingMachine))
class JunkDrink:
def __init__(self, machine):
self.machine = machine
a = VendingMachine(1, False)
b = VendingMachine.__init__(a, 2, False)
VendingMachine.__init__(VendingMachine, 10, False)
Expression
Interactive Output
a.k
2
b.k
Error
VendingMachine.k
10
isinstance(b, VendingMachine)
False
a is a.soda.machine
True
VendingMachine is a.soda.machine
False
c = VendingMachine
c.__init__(c, 11, True)
False
c.soda.machine is VendingMachine
True
a.k == c.k
False
c.soda.machine.k
11
2. COVID Party
Fill in each of the blanks in the code such that the expressions and outputs in the table below are consistent. Assume that effects from a previous subpart persist to future subparts. The Link class is provided on the next page. Do not use the Link constructor unless it is already provided in the skeleton
class Party:
guests = Link.empty
def __init__(self, time):
Party.guests = Link(time + 1, Party.guests)
def attend(self):
self.guests.rest = Link(self.guests.rest)
return (self.guests is Party.guests) and Party.guests
class Costume(Party):
def __init__(self, bow, tie):
Party.guests.rest, self.ie = Link(bow), Link(self)
def attend(self):
print(repr(self.ie)) # A: Costume(5, 6).attend() would have been “whacky” with no quotes
Party.attend = lambda self: Party(9).guests
def __repr__(self):
print(“Nice”)
return “Costume”
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Expression
Interactive Output
Link(1, Link.empty)
Link(1)
Link(1, Link(2))
Link(1, Link(2))
Party(1).guests
Link(2)
Party(3).attend()
Link(4, Link(Link(2)))
Costume(5, 6).attend()
Nice
Link(Costume)
Party(7).attend()
Link(10, Link(8, Link(4, Link(5))))
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3. Link class
class Link:
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:
rest = “, ” + repr(self.rest)
else:
rest = “”
return “Link(” + repr(self.first) + rest + “)”
def __str__(self):
return “whacky”
4. Max Path
(a) Given a tree, find the maximum path sum. A path sum is the sum of a sequence of connected nodes. The path can start anywhere (not necessarily the root or leaf), end anywhere, and move from parent to child or child to parent. The only constraint is that the ith node in the sequence has to be either the parent or direct child of the (i – 1)th node in the sequence. As an example, the function should return 23 for the following tree. The path is 6 => 3 => 2 => 4 => 8. Assume each node has at most 2 branches for simplicity.
def maxPathSum(tree):
# Every node is either
# a) starting point of a path
# b) an intermediate node from a left child going upward
# c) an intermediate node from a right child going upward
# d) the connector node for a path between left and right
# e) not in the max path at all
tree.options = [tree.label, tree.label, tree.label] # encodes options b through d
if tree.is_leaf():
return tree.label
b = tree.branches # this just makes it easier to fit solution on the page
x = maxPathSum(b[0]) # (e) for `tree`
p = max(max(b[0].options[:-1]), b[0].label) # checks (a)-(c) in left subtree
tree.options[0] = tree.label + p # (b) for `tree`
if len(b) == 2:
x = max(maxPathSum(b[1]), x) # checks (e) for `tree`
q = max(max(b[1].options[:-1]), b[1].label) # checks (a)-(c) in right subtree
tree.options[1] = tree.label + q # (c) for `tree`
tree.options[2] = tree.label + p + q # (d) for `tree`
return max(x, tree.label, max(tree.options)) # checks (a)-(e) for `tree`
An alternate and easier to read solution is posted here: https://piazza.com/class/kdz4wzqnb6052o?cid=2558
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