Description

General instructions
• Your problem sets are graded on both correctness and clarity of communication. Solutions that are technically correct but poorly written will not receive full marks. Please read over your solutions carefully before submitting them.
• Solutions must be typeset electronically, and submitted as a PDF with the correct filename. Handwritten submissions will receive a grade of ZERO.
The required filename for this problem set is problem set4.pdf.
• Problem sets must be submitted online through MarkUs. If you haven’t used MarkUs before, give yourself plenty of time to figure it out, and ask for help if you need it! If you are working with a partner, you must form a group on MarkUs, and make one submission per group. “I didn’t know how to use MarkUs” is not a valid excuse for submitting late work.
Additional instructions
• All final Big-Oh, Omega, and Theta expressions should be fully simplified according to three rules: don’t include constant factors (so O(n), not O(3n)), don’t include slower-growing terms (so O(n2), not O(n2 + n)), and don’t include floor or ceiling functions.
• When proving a bound on WC(n) or BC(n) you must start by writing the statement that you are proving. In addition, use the phrases ‘at most’ and ‘at least’ correctly.
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.
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(nlogn).
(b) [1 mark] Prove that the running time of this algorithm is Ω(n).
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
        

(a) [3 marks] Prove that WC(n) ∈ O(n).
(b) [2 marks] Prove that WC(n) ∈ Ω(n).
(c) [2 marks] Prove that BC(n) ∈ O(logn).
(d) [3 marks] Prove that BC(n) ∈ Ω(logn).
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!
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 array 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(n2).
(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(n2).