5. Find the order of growth for solutions of the following recurrences. a. T(n)=4T (n/2) +n, T(1) = 1 b. T(n)= 4T (n/2) + n², T (1) = 1 c. T(n)= 4T (n/2) + n³, T (1) = 1 %3D %3D %3D

Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
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#5

5.1 Mergesort
1.
c. How does this algorithm compare with the brute-force algorithm for
problem?
4. As mentioned in Chapter 2, logarithm bases are irrelevant in most cont
arising in analyzing an algorithm's efficiency class. Is this true for both a
tions of the Master Theorem that include logarithms?
5. Find the order of growth for solutions of the following recurrences.
a. T(n)=4T(n/2) + n, T (1) =1
b. T(n)=4T (n/2) +n², T (1) = 1
c. T(n)=4T (n/2) + n³, T (1) = 1
%3D
%3D
%3D
%3D
6. Apply mergesort to sort the list E, X, A, M, P, L, E in alphabetical ord
7. Is mergesort a stable sorting algorithm?
8. a. Solve the recurrence relation for the number of key comparisons m
mergesort in the worst case. You may assume that n =
= 2k.
b. Set up a recurrence relation for the number of key comparisons
Transcribed Image Text:5.1 Mergesort 1. c. How does this algorithm compare with the brute-force algorithm for problem? 4. As mentioned in Chapter 2, logarithm bases are irrelevant in most cont arising in analyzing an algorithm's efficiency class. Is this true for both a tions of the Master Theorem that include logarithms? 5. Find the order of growth for solutions of the following recurrences. a. T(n)=4T(n/2) + n, T (1) =1 b. T(n)=4T (n/2) +n², T (1) = 1 c. T(n)=4T (n/2) + n³, T (1) = 1 %3D %3D %3D %3D 6. Apply mergesort to sort the list E, X, A, M, P, L, E in alphabetical ord 7. Is mergesort a stable sorting algorithm? 8. a. Solve the recurrence relation for the number of key comparisons m mergesort in the worst case. You may assume that n = = 2k. b. Set up a recurrence relation for the number of key comparisons
Expert Solution
Step 1

5(a) We have T(n) = 4T(n/2)+n

Given T(1) = 1

1 = 4T(1/2)+1

4T(1/2) = 0

T(1/2) = 0

Here n has the largest growth rate in T(n)

The order of growth for T(n) = 4T(n/2) +n is O(n)

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