Description
1. [4 marks] Binary representation and algorithm analysis. Consider the following algorithm, which
manually counts up to a given number n, using an array of 0’s and 1’s to mimic binary notation.1
from math import floor, log2
def count(n: int) -> None:
# Precondition: n > 0.
p = floor(log2(n)) + 1 # The number of bits required to represent n.
bits = [0] * p # Initialize an array of length p with all 0’s.
for i in range(n): # i = 0, 1, …, n-1
# Increment the current count in the bits array. This adds 1 to
# the current number, basically using the loop to act as a “carry” operation.
j = p – 1
while bits[j] == 1:
bits[j] = 0
j -= 1
bits[j] = 1
For this question, assume each individual line of code in the above algorithm takes constant time, i.e.,
counts as a single step. (This includes the [0] * p line.)
(a) [3 marks] Prove that the running time of this algorithm is O(n log n).
(b) [1 mark] Prove that the running time of this algorithm is Ω(n).
1This is an extremely inefficient way of storing binary, and is certainly not how modern hardware does it. But it’s useful
as an interesting algorithm on which to perform runtime analysis.
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CSC165H1, Problem Set 4
2. [10 marks] Worst-case and best-case algorithm analysis. Consider the following function, which
takes in a list of integers.
def myprogram(L: List[int]) -> None:
n = len(L)
i = n – 1
x = 1
while i > 0:
if L[i] % 2 == 0:
i = i // 2 # integer division, rounds down
x += 1
else:
i -= x
Let W C(n) and BC(n) be the worst-case and best-case runtime functions of myprogram, respectively,
where n represents the length of the input list L. You may take the runtime of myprogram on a given list
L to be equal to the number of executions of the while loop.
(a) [3 marks] Prove that W C(n) ∈ O(n).
(b) [2 marks] Prove that W C(n) ∈ Ω(n).
(c) [2 marks] Prove that BC(n) ∈ O(log n).
(d) [3 marks] Prove that BC(n) ∈ Ω(log n).
Note: this is actually the hardest question of this problem set. A correct proof here needs to argue
that the variable x cannot be too big, so that the line i -= x doesn’t cause i to decrease too quickly!
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CSC165H1, Problem Set 4
3. [14 marks] Graph algorithm. Let G = (V, E) be a graph, and let V = {0, 1, . . . , n − 1} be the vertices
of the graph. One common way to represent graphs in a computer program is with an adjacency matrix,
a two-dimensional n-by-n array2 M containing 0’s and 1’s. The entry M[i][j] equals 1 if {i, j} ∈ E, and
0 otherwise; that is, the entries of the adjacency matrix represent the edges of the graph.
Keep in mind that graphs in our course are symmetric (an edge {i, j} is equivalent to an edge {j, i}), and
that no vertex can ever be adjacent to itself. This means that for all i, j ∈ {0, 1, . . . , n − 1}, M[i][j] ==
M[j][i], and that M[i][i] = 0.
The following algorithm takes as input an adjacency matrix M and determines whether the graph contains
at least one isolated vertex, which is a vertex that has no neighbours. If such a vertex is found, it then
does a very large amount of printing!
def has_isolated(M):
n = len(M) # n is the number of vertices of the graph
found_isolated = False
for i in range(n): # i = 0, 1, …, n-1
count = 0
for j in range(n): # j = 0, 1, …, n-1
count = count + M[i][j]
if count == 0:
found_isolated = True
break
if found_isolated:
for k in range(2 ** n):
print(’Degree too small’)
(a) [3 marks] Prove that the worst-case running time of this algorithm is Θ(2n
).
(b) [3 marks] Prove that the best-case running time of this algorithm is Θ(n
2
).
(c) [1 mark] Let n ∈ N. Find a formula for the number of adjacency matrices of size n-by-n that
represent valid graphs. For example, a graph G = (V, E) with |V | = 4 has 64 possible adjacency
matrices.
Note: a graph with the single edge (1, 2) is considered different from a graph with the single edge
(2, 3), and should be counted separately. (Even though these graphs have the same “shape”, the
vertices that are adjacent to each other are different for the two graphs.)
(d) [2 marks] Prove the formula that you derived in Part (c).
(e) [2 marks] Let n ∈ N. Prove that the number of n-by-n adjacency matrices that represent a graph
with at least one isolated vertex is at most n · 2
(n−1)(n−2)/2
.
(f) [3 marks] Finally, let AC(n) be the average-case runtime of the above algorithm, where the set of
inputs is simply all valid adjacency matrices (same as what you counted in part (c)).
Prove that AC(n) ∈ Θ(n
2
).
2
In Python, this would be a list of length n, each of whose elements is itself a list of length n.
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