Exercise 3. For each of the following program fragments give a (.) estimation of therunning time as a function of n.(a) sum = 0;(b)(c)(d)(e)for (int i = 0; i< n * n; i++) {for(int j = 0; j < n/2; j++)sum++;}sum = 0;for (int isum++;+; }for (int j}sum =for=}=0; j < n/2; j++) {sum++; }0;(int i = 0; i< n * n; i++) {for (int j = 0; j < n * n; j++)sum++sum =0;for (int i =0; i

Here in this question we have given five code fragments and we have asked to find running time…

Answered: Exercise 3. For each of the following…

Exercise 3. For each of the following program fragments give a (.) estimation of the inning time as a function of n. (a) sum= 0; for (int i = 0; i< n * n; i++) { for (int j = 0; j

Database System Concepts

7th Edition

ISBN:9780078022159

Author:Abraham Silberschatz Professor, Henry F. Korth, S. Sudarshan

Publisher:Abraham Silberschatz Professor, Henry F. Korth, S. Sudarshan

Chapter1: Introduction

Section: Chapter Questions

Problem 1PE

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Concept explainers

Computational Systems

The computation systems can be defined as the systems that are capable of solving a problem that includes calculations either mathematical or logical, and are able to produce the result as an output. For example, a simple calculator that is given a set of numbers and operators as input and produces the result as an output can be defined as a computational system. The above example is a simple illustration only, the computational systems range from simple devices to complex devices such as the systems that control the SpaceX companies, Starlink satellites which are being quite recently seen in the skies at many places globally.

Question

Exercise 3. For each of the following program fragments give a (.) estimation of the
running time as a function of n.
(a) sum = 0;
(b)
(c)
(d)
(e)
for (int i = 0; i< n * n; i++) {
for(int j = 0; j < n/2; j++)
sum++;
}
sum = 0;
for (int i
sum++;
+; }
for (int j
}
sum =
for
=
}
=
0; j < n/2; j++) {
sum++; }
0;
(int i = 0; i< n * n; i++) {
for (int j = 0; j < n * n; j++)
sum++
sum =
0;
for (int i =
0; i<n; i++) {
sum = 0;
for (int i =
1; i<n; i = 2*i)
sum++
0; i<n; i++) {
for(int j = 1; j < n * n; j = 2*j)
sum++

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