COMPUTER SCIENCE ILLUMIN.-TEXT
6th Edition
ISBN: 2810016866372
Author: Dale
Publisher: JONES+BART
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Chapter 1, Problem 48E
Explanation of Solution
Abstraction:
- Abstraction is the concept of showing only the essential details to the user.
- It hides the background complex data from user and the user can view only those details which is relevant to them.
- The data displaying to the user will be very simple.
Abstraction relating to the Figure 1.2:
The figure 1.2 contains two pictures. First one is a car engine and the second one is a user who is driving the car...
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Please answer JAVA OOP problem below:
You are working at a university that tracks students. Each student is identified by their name and faculty advisor. Each faculty advisor is identified by their name, department, and maximum number of students they can advise.
Using solid OO design principles, create a modular program that implements all the classes for the problem and also creates an implementation class that gathers user input for one student and then prints out the information gathered by creating the appropriate data definition and implementation classes. All data must be validated.
I have given the code so far:
Implementation:
import javax.swing.JOptionPane;
public class Implementation {
public static void main(String[] args) {
FacultyAdvisor facultyAdvisor = new FacultyAdvisor("Sharmin Sultana", "IT", 30);
Student student = new Student("John", facultyAdvisor);
JOptionPane.showMessageDialog(null, student.toString());
}
}
Student:
public…
Exercise 1 Function and Structure [30 pts]
Please debug the following program and answer the following questions. There is a cycle in a linked
list if some node in the list can be reached again by continuously following the next pointer.
#include
typedef struct node {
int value;
struct node *next;
} node;
int 11_has_cycle (node *first)
if (first
==
node *head
{
NULL) return 0;B
= first;
while (head->next != NULL) {
if (head == first) {
return 1; }
head
head->next;
}
return 0;
void test_11_has_cycle() {
int i;
node nodes [6];
for (i = 0; i < 6; i++)
nodes [i] .next = NULL;
nodes [i].value
i;
}
nodes [0] .next
=
&nodes [1];
nodes [1] .next = &nodes [2];
nodes [2] .next
=
&nodes [3];
nodes [3] .next = & nodes [4];
nodes [4] .next
=
NULL;
nodes [5] .next = &nodes [0];
printf("1. Checking first list for cycles. \n Function 11_has_cycle says it
hass cycle\n\n", 11_has_cycle (&nodes [0]) ?"a":"no");
printf("2. Checking length-zero list for cycles. \n Function 11_has_cycle
says it has %s…
checkpoint exercice for my students for Amortized Analysis
Chapter 1 Solutions
COMPUTER SCIENCE ILLUMIN.-TEXT
Ch. 1 - Prob. 1ECh. 1 - Prob. 2ECh. 1 - Prob. 3ECh. 1 - Prob. 4ECh. 1 - Prob. 5ECh. 1 - Prob. 6ECh. 1 - Prob. 7ECh. 1 - Prob. 8ECh. 1 - Prob. 9ECh. 1 - Prob. 10E
Ch. 1 - Prob. 11ECh. 1 - Prob. 12ECh. 1 - Prob. 13ECh. 1 - Prob. 14ECh. 1 - Prob. 15ECh. 1 - Prob. 16ECh. 1 - Prob. 17ECh. 1 - Prob. 18ECh. 1 - Prob. 19ECh. 1 - Prob. 20ECh. 1 - Prob. 21ECh. 1 - Prob. 22ECh. 1 - Prob. 23ECh. 1 - Prob. 24ECh. 1 - Prob. 25ECh. 1 - Prob. 26ECh. 1 - Prob. 27ECh. 1 - Prob. 28ECh. 1 - Prob. 29ECh. 1 - Prob. 30ECh. 1 - Prob. 31ECh. 1 - Prob. 32ECh. 1 - Prob. 33ECh. 1 - Prob. 34ECh. 1 - Prob. 35ECh. 1 - Prob. 36ECh. 1 - Prob. 37ECh. 1 - Prob. 38ECh. 1 - Prob. 39ECh. 1 - Prob. 40ECh. 1 - Prob. 41ECh. 1 - Prob. 42ECh. 1 - Prob. 43ECh. 1 - Prob. 44ECh. 1 - Prob. 45ECh. 1 - Prob. 46ECh. 1 - Prob. 47ECh. 1 - Prob. 48ECh. 1 - Prob. 50ECh. 1 - Prob. 51ECh. 1 - Prob. 52ECh. 1 - Prob. 53ECh. 1 - Prob. 54ECh. 1 - Prob. 55ECh. 1 - Prob. 56ECh. 1 - Prob. 57ECh. 1 - Prob. 58ECh. 1 - Prob. 59ECh. 1 - Prob. 1TQCh. 1 - Prob. 2TQCh. 1 - Prob. 3TQCh. 1 - Prob. 4TQ
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