Question

Problem 5: Recurrence relations and detailed analysis of recursive algorithm efficiency

                g(n: non-negative integer)

                1. if n ≤ 1 then return n

                2. else return (5 * g(n─1) ─ 6 * g(n─2))

MergeSort divides the array to be sorted into two equal halves, calls itself recursively on each half to sort that subarray, and then calls the Merge algorithm to merge the two sorted halves in linear time. This leads to its two recurrence relations T(n)=2T(n/2)+cn, n>1; T(1)=c. We solved these recurrences using the Recursion Tree method in class to show that T(n)=cnlog₂n+cn and therefore its complexity order is Θ(nlog₂n). Now consider the question of whether Merge Sort will be faster if we split the array into three equal one-thirds, made a recursive call on each one-third to sort it and used a modified Merge algorithm to merge the three sorted subarrays into one wholly sorted array. Assuming that we can merge three sorted one-thirds in linear time, i.e., with the same cn work, the corresponding recurrences are:

                T(n)=3T(n/3)+cn, n>1; T(1)=c.

What is the complexity order of this new MergeSort algorithm, that you obtain by applying the Master Theorem?

A. Θ(n)                 B. Θ(nlgn)            C. Θ(cn)                D. Θ(nlog₃n)       E. Master Theorem cannot be applied

Solve the recurrences of this modified MergeSort using the more precise Recursion Tree method. It can be seen from the table that total work done by all recursive executions = T(n) = cn * (______________). What is the correct expression to fill in this blank?

A. lgn     B. log₃n                 C. log₃n+1            D. lgn+1                E. none of these are correct

level	Level number	Total # of recursive executions at this level	Input size to each recursive execution	Work done by each recursive execution, excluding the recursive calls	Total work done by the algorithm at this level = column3*column5
0	0	30	n/30	c(n/30)	cn
1	1	31	n/31	c(n/31)	cn
2	2	32	n/32	c(n/32)	cn
The level just above the base case level	(log3n-1)	3(logntobase3-1) = (nlog3tobase3)/3 = n/3	n/3(logntobase3-1)=3	c(3)	cn
Base case level	log3n	3logntobase3 = nlog3tobase3 = n	n/3logntobase3=1	c(1)	cn

Answer 1

For question 1:-

By applying Master Theorem:-

If f(n) = Θ(nc) where c = Logba then T(n) = Θ(ncLog n)

here c=1 , b=3, a=3

we get T(n) = Θ(nlogn) (Option B)

For question 2:-

By applying Recursion Tree method ,

we get log3n(Option B)