CS 61A Structure and Interpretation of Computer Programs
Spring 2016
INSTRUCTIONS
Test 2 (corrected)
• You have 2 hours to complete the exam.
• The exam is open book, open notes, closed computer, closed calculator. The official CS 61A midterm 1 and 2
study guides will be provided.
• Mark your answers on the exam itself. We will not grade answers written on scratch paper.
Last name
First name
Student ID number
BearFacts email
Room in which you are taking this exam
TA
Name of the person to your left
Name of the person to your right
I pledge my honor that during this examination I have neither given nor received assistance. (please sign)
2
Reference. Some questions make use of the following class definitions from labs and homework:
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 is not Link.empty:
rest_str = ’, ’ + repr(self.rest) else:
rest_str = ’’
return ’Link({0}{1})’.format(repr(self.first), rest_str)
def __len__(self):
return 1 + len(self.rest)
def __getitem__(self, i): if i == 0:
return self.first else:
return self.rest[i-1]
def __str__(self): string = ’<’
while self.rest is not Link.empty: string += str(self.first) + ’, ’ self = self.rest
return string + str(self.first) + ’>’
class Tree:
def __init__(self, label, children=()):
self.label = label self.children = list(children)
def __repr__(self): if self.children:
children_str = ’, ’ + repr(self.children) else:
children_str = ’’
return ’Tree({0}{1})’.format(self.label, children_str)
def is_leaf(self):
return not self.children
Name: 3 1. (12 points) Pointers
For each of the following code fragments, add arrows and values to the object skeletons to the right to show the final state of the program. Single boxes are variables that contain pointers. Double boxes are Links. Not all boxes need be used.
(a) (3 pt)
L = Link(1, Link(2)) P=L
Q = Link(L, Link(P)) P.rest.rest = Q
L:
P:
Q:
(b) (3 pt)
L = Link.empty
for i in range(3): L:
L = Link(i, L)
4
(c) (3 pt) For the next two problems, show the result of executing the code on the left on the initial conditions displayed on the right. We’ve done the first statement for you in each case, so that the diagrams on the right show the state at the point marked # START. Use the empty object skeletons only for newly created Link objects. If any pointer is modified, neatly cross out the original pointer and draw in the replacement. Show only the final state, not any intermediate states.
if L is Link.empty: Q: return (Link.empty, Link.empty)
L1, L2 = crack1(L.rest)
return (Link(L.first, L2), L1) Q, R = crack1(P)
R:
(d) (3 pt)
P = Link(0, Link(1, Link(2)))
# START
def crack2(L):
012
012
P = Link(0, Link(1, Link(2)))
# START
def crack1(L):
P:
P:
Q:
L1, L2 = crack2(L.rest)
L.rest = L2
return (L, L1) R:
Q, R = crack2(P)
if L is Link.empty:
return (Link.empty, Link.empty)
Name: 5 2. (6 points) Complexity
As indicated in lecture, an assertion such as Θ(f(n)) ⊆ Θ(g(n)) means “any function that is in Θ(f(n)) is also in Θ(g(n)).”
(a) (1.5 pt) Circle each of the following that is true.
A. Θ(f(n)) ⊆ O(f(n))
B. Θ(2×2 + 1000x) ⊆ Θ(x2) C. Θ(x2) ̸= Θ(2×2 + 1000x) D. O(1/n) ⊆ O(1)
E. Θ(1/n) ⊆ Θ(1)
(b) (1.5 pt) Assume that M is an N × N array (an N -long Python list of N -long lists). Consider the following program:
def search(M, x): N = len(M)
Li, Uj = 0, N-1
while Li < N and Uj >= 0:
if M[Li][Uj] < x: Li += 1
elif M[Li][Uj] > x: Uj -= 1
else:
return True
return False
Circle the order of growth that best describes the worst-case execution time of a call to search as a function
of N.
A. Θ(N)
B. Θ(N2) C. Θ(logN) D. Θ(2N2) E. Θ(2N)
6
(c) (1.5 pt) Consider the following implementation of count, which takes in a linked list of numbers lst and an unordered Python list of numbers nums, and returns a count of the number of values in lst that appear in nums:
def count(lst, nums):
“””The number of elements in linked list LST that appear appear in the unordered Python list NUMS.
>>> L = Link(2, Link(4, Link(2, Link(3, Link(1)))))
>>> count(L, [2, 1, 5])
3″””
curr = lst
count = 0
while curr != Link.empty:
if curr.first in nums: count += 1
curr = curr.rest
return count
Circle the order of growth that best describes the worst-case execution time of count, as a function of n, the length of nums, and m, the length of lst. Since nums is a Python list, the in operator uses simple linear search.
A. Θ(n)
B. Θ(m)
C. Θ(n2)
D. Θ(n+m) E. Θ(nm)
F. Θ(mn2)
(d) (1.5 pt) Consider the following function for computing powers of a polynomial:
def polypow(P, k):
“””P ** k, where P is a polynomial and K is a non-negative integer.”””
result = Poly(1)
while k != 0:
if k % 2 == 1:
result = result.mult(P)
P = P.mult(P)
k = k // 2
Circle the order of growth that best describes the worst-case execution time of polypow, as a function of k, where execution time is measured in the number of times that the .mult method is called.
A. Θ(k)
B. Θ(k2) C. Θ(√k) D. Θ(logk) E. Θ(2k)
Name: 7 3. (8 points) Seeing Double
Fill in the functions below to produce linked lists in which each item of the original list is repeated immediately after that item. Your solutions should be iterative, not recursive.
(a) (4 pt) The function double1 is non-destructive, and produces a new list without disturbing the old.
def double1(L):
“””Returns a list in which each item in L appears twice in sequence. It is non-destructive.
>>> Q = Link(3, Link(4, Link(1)))
>>> double1(Q)
Link(3, Link(3, Link(4, Link(4, Link(1, Link(1))))))
>>> Q
Link(3, Link(4, Link(1)))
>>> double1(Link.empty)
()
“””
result = _______________________ last = None
while L is not Link.empty: if last is None:
_________________________________________
_________________________________________ else:
_________________________________________
_________________________________________
_______________________________________ return result
(b) (4 pt) The function double2 is destructive, and reuses Link objects in the original list wherever possible.
def double2(L):
“””Destructively modifies L to insert duplicates of each item immediately following the item, returning the result.
>>> Q = Link(3, Link(4, Link(1)))
>>> double2(Q)
Link(3, Link(3, Link(4, Link(4, Link(1, Link(1))))))
>>> Q
Link(3, Link(3, Link(4, Link(4, Link(1, Link(1))))))
“””
result = ___________________________________ while L is not Link.empty:
____________________________________________________
____________________________________________________ return result
8
4. (1 points) Extra
Last September, twin LIGO detectors observed gravitational waves that emanated from the merger of two black holes. In the process of this merger, three solar masses (roughly 6 × 1030 kg) were converted into gravitational energy. How many planets the size of earth (roughly 6 × 1024 kg) could this much energy accelerate to 1% of lightspeed (about 3000 km/sec)?
5. (8 points) Heaps of Trouble
A (min-)heap is a tree with the special property (the heap property) that every node has a label that is less than the labels of all its child nodes. This means that the minimum element of the heap is at the root, so it can be found in constant time. For example:
2
4 30 20
90 9 5 70 25
Suppose we have a heap containing at least two values. To remove and return its smallest element, while maintaining the heap property, we use the following function:
def remove_smallest(H):
“””Destructively remove and return the smallest value from heap H, restoring the heap property. Assumes H has at least two elements.”””
result = H.label
H.label = remove_leaf(H) # Step 1 reheapify(H) # Step 2 return result
The function remove_leaf removes one of the leaves from the heap, returning its label. The diagram on the left below shows the state of the heap above after executing Step 1 of remove_smallest. In general (as shown), this will cause the root to violate the heap property. To restore it, we use the function reheapify, which first swaps the root’s label with that of its smallest child (giving the tree in the middle below). If as a result, the heap property is still violated (as in the example), reheapify repeats the process on down the tree until the value inserted at the top reaches a point where it is smaller than all its children, which will always be true if it reaches a leaf, as happens in the example below (shown on the right), but can also happen before that.
90 4 4
4 30 20 90 30 20 5 30 20
9 5 70 25 9 5 70 25 9 90 70 25
Name: 9
(a) (4 pt) Write the function remove_leaf to remove a leaf from a heap destructively and return its label. Any leaf will do, but to be specifc, have it remove the leftmost leaf of the leftmost child of the leftmost child. . . of the root. Again, we assume that there are at least two values in the heap.
def remove_leaf(H):
“””Destructively remove far leftmost leaf of H, returning its label””” child = H.children[0]
if _______________________:
v = child.label
H.children = __________________________________
return v else:
return ____________________________________
(b) (4 pt) Write the function reheapify to restore the heap property of a heap destructively, assuming that
initially it is violated (if at all) only at the root.
def reheapify(H):
“””Destructively restore the heap property of H, assuming it is violated only at H itself, if at all.”””
if _____________________: return
else:
s = H.children[0] for c in H.children:
if __________________________: s=c
if ____________________:
s.label, H.label = _____________________________ ___________________________
10
6. (8 points) OOPs
Given the class definitions on the left, fill in the blanks to show what the Python interpreter would print. Print “ERROR” for cases that would cause an exception. Put “
class Person:
name = “Outis”
def get_name(self): return self.name
def response(self, question): v = self.cogitate(question) if v is None:
return “I do not know” else:
return v
def cogitate(self, question): return None
def set_name(self, new_name): self.name = new_name
def __str__(self): return self.name
class Learner(Person):
def __init__(self): self.facts = {}
def learn(self, question, answer): self.facts[question] = answer return ’Got it’
def cogitate(self, question): if question in self.facts:
return self.facts[question]
class Beginner(Learner):
def __init__(self, name):
Learner.__init__(self) self.set_name(name)
def response(self, question):
r = Person.response(self, question) return “I think ” + r
>>> odysseus = Learner()
>>> odysseus.learn(’god’, ’Athena’)
>>> hipp = Beginner(’Hippothales’)
>>> hipp.learn(’favorite person’, ’Lysis’)
>>> odysseus.get_name() >>> hipp.get_name()
>>> Person.name = “Nemo” >>> hipp.get_name()
>>> odysseus.get_name()
>>> odysseus.set_name(odysseus.get_name()) >>> Person.name = “Nobody”
>>> odysseus.get_name()
>>> someone = Person()
>>> someone.learn(’Earth mass’, ’5.972e24 kg’)
>>> someone.response(’Earth mass’) >>> hipp.response(’favorite person’) >>> odysseus.response(’god’)
Name: 11 7. (8 points) Evicted!
An LRU cache (stands for “least recently used”) is a kind of dictionary that can only hold a fixed, finite number of keys (its capacity) and corresponding values. When addition of a new key would exceed that capacity, the least recently accessed key in the cache is removed (“evicted”) and replaced with the new value. Such caches are used to speed up access to some relatively slow, but much larger dictionaries. For example, most computers have a large main memory and various caches for saving and retrieving recently accessed memory values; the latter can be 200 times faster than the former.
(a) (2 pt)
Consider the following “slow” dictionary implementation:
class SlowData:
“””
Simulates a basic read-only memory store of KEY => VALUE mappings >>> slow_data = SlowData(((0, ’a’), (1, ’b’), (2, ’c’)))
>>> slow_data[1]
’b’
>>> slow_data[2]
’c’
“””
def __init__(self, data):
self._data = data
# A sequence of (KEY, VALUE) tuples
def __getitem__(self, key):
“””Get the value associated with KEY, or None if there is none.””” for curr_key, curr_value in self._data:
if key == curr_key: return curr_value
return None
If mem is a SlowData containing N tuples, what is the worst-case execution time for the following code
fragment?
result = 0
for i in range(N): result += mem[i]
Circle the correct answer below.
A. Θ(N) B. Θ(N log N) C. Θ(N2) D. Θ(N3)
12
(b) (4 pt)
An LRUCache object is intended to provide access to values from a SlowData in such a way that the results of some recent accesses to the SlowData object are saved and subsequently accessed quickly. To do this, the cache keeps a list of key/value tuples whose size has a fixed upper limit. If a key that is in the cache is accessed, its corresponding value is fetched from this list without consulting the SlowData object. If a key is not in the cache, it is fetched from the SlowData object. Each time a value is referenced, it is placed at or moved to the end of the cache list, and if that makes the list too long (longer than the capacity), the first item in the list is removed (so that it will have to be retrieved from the SlowData object if accessed again).
Fill in the code below to have this behavior. (A convenient way to remove the item at index k from a list L is del L[k].)
class LRUCache:
def __init__(self, capacity, slow_data):
self._capacity = capacity self._slow_data = slow_data self._cache = []
def __getitem__(self, key):
for
v =
i in range(len(self._cache)): pair = self._cache[i]
if ______________________:
_____________________________
_____________________________
return pair[1] self._slow_data[key]
self._cache____________________________ if len(self._cache) > self._capacity:
del ________________________________ return v
(c) (1 pt) If mem is a SlowData containing N tuples, what is the worst-case execution time for the following code fragment?
cached_mem = LRUCache(4, mem)
result = 0
for i in range(N): result += cached_mem[i]
Circle the correct answer below.
A. Θ(N) B. Θ(N log N) C. Θ(N2) D. Θ(N3)
(d) (1 pt) If cached_mem is as above, what is the worst-case execution time for the following code fragment? result = 0
for i in range(N): result += cached_mem[i % 4]
A. Θ(N) B. Θ(N log N) C. Θ(N2) D. Θ(N3)