Question

1. [5 marks Show the following hold using the definition of Big Oh: a) 2 mark 1729 is O(1) b) 3 marks 2n2-4n -3 is O(n2) 2. [3 marks] Using the definition of Big-Oh, prove that 2n2(n 1) is not O(n2) 3. 6 marks Let f(n),g(n), h(n) be complexity functions. Using the definition of Big-Oh, prove the following two claims a) 3 marks Let k be a positive real constant and f(n) is O(g(n)), then k f(n) is O(g(n)) b) 3 marks] If f(n) is O(g(n) and g(n) is O(h(n)), then f(n) is O(h(n)). 4. 11 marks Consider the element distinctness problem: Given an array A storing n inte- gers, determine whether all the elements in the array are distinct or not. That is, if all the elements in A are unique, return true; return false if there is at least one duplicate element in A. Below four different problem instances are provided; elements bolded are duplicate elements 0 1 2 3 4 5 6 12312 3 4 false 0 1 2 3 4 5 41 22 6 8 false 0 1 2 3 true 0 1 2 3 45 6 7 true Design an algorithm that solves the element distinctness problem. a) [4 marks Write pseudocode for the algorithm b) Prove your algorithm is correct, do this by proving the two following parts: i marks] Show that the algorithm terminates in finite time ii) 2 marks Show that the algorithm always produces correct output e) 1 mark Explain what the worst case for the algorithm is. d) 3 marks Perform worst-case analysis to compute the time complexity of the algorithm. You must give the order of your complexity function using Big-Oh notation, and you must explain how you computed the time complexity

Answer 1

O(g(n)) = { f(n): there exist positive constants c and 
                  n0 such that 0 <= f(n) <= c*g(n) for 
                  all n >= n0}

1.

a) Clearly, for c = 1730 and n >= 1, we have

0 <= 1729 <= c

Hence, 1729 = O(1)

b) For c = 2 and n >= 0, we have

0 <= 2n² - 4n - 3 <= c*n²

Hence, 2n² - 4n - 3 = O(n²)

2. f(n) = 2n²(n - 1) = 2n³ - 2n²

There does not exist any positive integer c such that

0 <= f(n) <= c*n² where n >= 0

Hecne, f(n) != O(n²)

3.

a) Given that f(n) = O(g(n))

=> 0 <= f(n) <= c*g(n) where c is a positive constant

=> 0 <= k*f(n) <= k*c*g(n)

=> 0 <= k*f(n) <= c1*g(n) where c1 = k*c

Clearly, k*f(n) = O(g(n))

b)

Given that f(n) = O(g(n))

=> 0 <= f(n) <= c1*g(n) where c1 is a positive constant ------> (1)

Given that g(n) = O(h(n))

=> 0 <= g(n) <= c2*h(n) where c2 is a positive constant ------> (2)

Combining 1 and 2, we get

0 <= f(n) <= c1*g(n) <= c1*c2*h(n)

or, 0 <= f(n) <= c1*c2*h(n)

=> 0 <= f(n) <= c3*h(n) where c3 = c1*C2

Hence, f(n) = O(h(n))

NOTE: As per HOMEWORKLIB POLICY, I am allowed to answer only 4 questions (including sub-parts) on a single post. Kindly post the remaining questions separately and I will try to answer them. Sorry for the inconvenience caused.