Question

Question 2 (8 marks) (From the DPU textbook, Exercise 2.5) Using the Master theorem or recursion tree methods, solve the following recurence relations and give Θ bound for each of them a) T(n) 9T(n/3) +3n2 12n - 4, where n 21 b) T(n)-T(n-1) + n", where n > 1 and c > 0 c) T(n) = T(Vn) + 1, where n > b and T(b) = 2. Question 3: 10 marks) Suppose we have a subroutine merge2 to merge two sorted arrays in linear time (kn The purpose is to design a divide and conquer algorithm (Alg2) to merge k sorted arrays using merge2 recursively. a) Write a pseudocode for Alg2. (Hint: Assume that the input is given in a (k x n) rray with the rows and columns sorted in ascending order) b) Write the running time of Alg2 as a recurence relation. T(k)-? c) Construct a recursion tree with log k levels and (kn) work per level to represent the d) Since k is a constant, is Alg2 asymptotically faster than an algorithm with running time recurance relation obtained in part (b) Θ(n log n)? why?

Answer 1

Question 2:

(a)

T(n) = 9T(n/3) + 3n^2 + 12n - 4, where n greaterthanorequalto 1 Here, f(n) = 3n^2 + 12n - 4 = O (n^2) Applying Master's second case method, for a = 9 and b = 3 T(n) = Theta (n^2 log n) (b) T(n) = T(n - 1) + n^c, where n greaterthanorequalto 1 and c greaterthanorequalto 0 From above recursion tree, T(n) = Theta (n^c) (c) T(n) = T(squareroot (n)) + 1, wheren > b and T (b) = 2. substituting n = 2^m implies log n = m above recursion T(2^m) = T(2^m/2) + 1 Now, substituting , T(2^m) = S(m) S(m) = S(m/2) + 1. Now, applying cas 1 of Master's Method , (a = 1, b = 2) solution is: S (m) = Theta (log m) = Theta (log log n)

Applying Master's second case method, for a=9 and b=3.

(b)

From above recursion tree,

   $T(n)=Theta(n^c)$

(c)

Substituting, n=2^m=>log n=m in above recursion.

T(2^m)=T(2^m/2)+1.

Now, substituting, T(2^m)=S(m).

S(m)=S(m/2)+1.

Now, applying case 1 of Master's Method,(a=1, b=2) solution is: