Driver.java import java.util.Scanner; import java.text.Decimal.pdf
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The Driver class contains methods to print numeric sequences, calculate triangular numbers, print arrays, and find the range between minimum and maximum values in an array. The main method calls each method, printing outputs including the sequences 0-10, -10 to 0, the sum of triangular numbers to 10, an array of values, and the range of the array.
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![}
else
return -1;
}
public static void printArray(int list [])
{
for (int i = 0; i < list.length ;i++ )
{
System.out.print(list[i] + " ");
}
System.out.println();
}
public static int range(int list [])
{
int max,min;
max = list[0];
min = list[0];
for (int i = 1; i < list.length ;i++ )
{
if(list[i] > max)
max = list[i];
if(list[i] < min)
min = list[i];
}
System.out.println("difference between the maximum and minimum values in the array: "+
(max-min));
return (max-min);
}
public static void main(String[] args)
{
printSequence(10);
printSequence(-10);
System.out.println();
if(trianglarNum(10) > 0)
System.out.println("Sum: "+ trianglarNum(10));
else](data:image/gif;base64,R0lGODlhAQABAIAAAAAAAP///yH5BAEAAAAALAAAAAABAAEAAAIBRAA7)
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The document contains 17 code snippets that demonstrate various Java programming concepts like loops, arrays, methods, classes, input/output, etc. Specifically, the code snippets show:
1. Using for loops to print numeric series
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3. Using StringTokenizer and Scanner classes
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The document is a 4 page sample code document containing examples of Java programming concepts. Page 1 contains code demonstrating if/else statements and printing output based on integer values. Page 2 imports a Scanner class and gets integer, float, and string user input. Page 3 briefly introduces arrays as a way to hold multiple values like columns in a spreadsheet. Page 4 indicates it will provide examples of arrays and loops.
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Java AssignmentWrite a program using sortingsorting bubble,sele.pdf
Java Assignment
Write a program using sorting
sorting: bubble,selection,insertion,quick,merge
Solution
Bubble Sort:
import java.util.Scanner;
class SortBubble {
public static void main(String []args) {
int number, c, d, swap;
Scanner in = new Scanner(System.in);
System.out.println(\"Please enter the number of elements to be sorted\");
number = in.nextInt();
int ar[] = new int[number];
System.out.println(\"Enter \" + number + \" integers\");
for (int i = 0; i < number; i++)
ar[i] = in.nextInt();
for (int i = 0; i < ( number - 1 ); i++) {
for (int j = 0; j < number - c - 1; j++) {
if (ar[j] > ar[j+1])
{
swap = ar[j]];
ar[j] = ar[j+1];
ar[j+1] = swap;
}
}
}
System.out.println(\"Sorted list of numbers\");
for (int c = 0; c < number; c++)
System.out.println(ar[c]);
}
}
b) Selection Sort:
public class SortSelection {
public static int[] sort(int[] arr){
for (int i = 0; i < arr.length - 1; i++)
{
int index = i;
for (int j = i + 1; j < arr.length; j++)
if (arr[j] < arr[index])
index = j;
int temp= arr[index];
arr[index] = arr[i];
arr[i] = temp;
}
return arr;
}
public static void main(String a[]){
int[] arr1 = {10,34,2,56,7,67,88,42};
int[] arr2 = sort(arr1);
for(int i:arr2){
System.out.print(i);
System.out.print(\", \");
}
}
}
c) Insertion
public class SortInsertion {
public static void main(String[] args) {
int[] array= { 4, 2, 9, 6, 23, 12, 34, 0, 1 };
sort(array);
}
static void printNumbers(int[] input) {
for (int i = 0; i < array.length; i++) {
System.out.print(array[i] + \", \");
}
System.out.println(\"\ \");
}
public static void sort(int array[]) {
int n = array.length;
for (int j = 1; j < n; j++) {
int key = array[j];
int i = j-1;
while ( (i > -1) && ( array [i] > key ) ) {
array [i+1] = array [i];
i--;
}
array[i+1] = key;
printNumbers(array);
}
}
}
d) Quick Sort:
public class SortQuick {
int array[];
int length;
public void sort(int[] inputArr) {
if (inputArr == null || inputArr.length == 0) {
return;
}
this.array = inputArr;
length = inputArr.length;
quickSort(0, length - 1);
}
private void quickSort(int lowerIndex, int higherIndex) {
int i = lowerIndex;
int j = higherIndex;
int pivot = array[lowerIndex+(higherIndex-lowerIndex)/2];
while (i <= j) {
while (array[i] < pivot) {
i++;
}
while (array[j] > pivot) {
j--;
}
if (i <= j) {
exchangeNumbers(i, j);
i++;
j--;
}
}
if (lowerIndex < j)
quickSort(lowerIndex, j);
if (i < higherIndex)
quickSort(i, higherIndex);
}
private void exchangeNumbers(int i, int j) {
int temp = array[i];
array[i] = array[j];
array[j] = temp;
}
public static void main(String a[]){
MyQuickSort sorter = new MyQuickSort();
int[] input = {24,2,45,20,56,75,2,56,99,53,12};
sorter.sort(input);
for(int i:input){
System.out.print(i);
System.out.print(\" \");
}
}
}.
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Refer to my progress on this assignment belowIn this problem you w.pdf
Refer to my progress on this assignment below
In this problem you will make it “more” object-oriented in the following ways:
-You will change its name to SortedList
-You will change the constructor that takes no arguments to be more traditional and initialize the
member fields to dummy values.
-You will add a constructor that takes in an initialized array and a size
-You will add an insert function that adds a value to the list and maintains its sorted-ness
-You will add a quicksort function check the below code.
-You will make updates as necessary to the main function so that it still runs and tests your code.
The code below seems to sort correctly but it skips over some items in the array and I am not
sure what is happening. Please help! Thanks!
import java.util.Scanner;
class SortedList
{
private static int array[];
private static int n;
public SortedList()
{
Scanner in = new Scanner(System.in);
System.out.println(\"Enter number of elements\");
n = in.nextInt();
array = new int[n];
if(n == 0) {
System.out.print(\"Since no arguments array set to: \");
} else {
System.out.print(\"Creating array size \" + n + \": \");
}
for(int i = 0; i < n; i++)
{
array[i] = 0;
}
//System.out.println(\"Enter \" + n + \" integers in ascending order\");
/*for (c = 0; c < n; c++)
array[c] = in.nextInt();*/
for(int i = 0; i < n; i++)
{
System.out.print(array[i] + \" \");
}
System.out.print(\"\ \");
}
public SortedList(int a[], int size)
{
array = a;
n = size;
}
public int binsearch(int search)
{
int first, last, middle;
first = 0;
last = n - 1;
middle = (first + last)/2;
while( first <= last )
{
if ( array[middle] < search )
first = middle + 1;
else if ( array[middle] == search )
{
//System.out.println(search + \" found at location \" + (middle + 1) + \".\");
return middle+1;//+1 for the non-CS people who don\'t start counting at zero.
}
else
last = middle - 1;
middle = (first + last)/2;
}
return -1;
//System.out.println(search + \" is not present in the list.\ \");
}
public static int partition(int input[], int p, int r)
{
int pivot = input[r];
while(p < r)
{
while(input[p] < pivot)
{
p++;
}
while(input[r] > pivot)
{
r--;
}
if(input[p] == input[r])
{
p++;
}
else if(p < r)
{
int tmp = input[p];
input[p] = input[r];
input[r] = tmp;
}
}
return r;
}
public static void quicksort(int input[], int p, int r)
{
if(p < r)
{
int j = partition(input, p, r);
quicksort(input, p, j-1);
quicksort(input, j+1, r);
}
}
public static void insert(int value, int cell)
{
array[cell] = value;
quicksort(array, 0, n-1);
}
public static void main(String args[])
{
int c;
Scanner in = new Scanner(System.in);
SortedList b = new SortedList();
System.out.println(\"Input numbers\");
for(int i = 0; i < n; i++)
{
c = in.nextInt();
in.nextLine();
insert(c, i);
}
in.close();
for(int j = 0; j < n; j++)
{
System.out.print(array[j] + \", \");
}
}
}
Solution
//the whole program that you wrote is correct except one line , when you call the quicksort
everytime you insert a //value dont pass n pass the i.
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JAVA PRACTICE QUESTIONS v1.4.pdf
This document contains 15 Java code examples demonstrating various Java concepts like:
1) Calculating total and average marks of a student
2) Using if-else statements to check voting eligibility and day of the week
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Java AssignmentWrite a program using sortingsorting bubble,sele.pdf
Java Assignment
Write a program using sorting
sorting: bubble,selection,insertion,quick,merge
Solution
Bubble Sort:
import java.util.Scanner;
class SortBubble {
public static void main(String []args) {
int number, c, d, swap;
Scanner in = new Scanner(System.in);
System.out.println(\"Please enter the number of elements to be sorted\");
number = in.nextInt();
int ar[] = new int[number];
System.out.println(\"Enter \" + number + \" integers\");
for (int i = 0; i < number; i++)
ar[i] = in.nextInt();
for (int i = 0; i < ( number - 1 ); i++) {
for (int j = 0; j < number - c - 1; j++) {
if (ar[j] > ar[j+1])
{
swap = ar[j]];
ar[j] = ar[j+1];
ar[j+1] = swap;
}
}
}
System.out.println(\"Sorted list of numbers\");
for (int c = 0; c < number; c++)
System.out.println(ar[c]);
}
}
b) Selection Sort:
public class SortSelection {
public static int[] sort(int[] arr){
for (int i = 0; i < arr.length - 1; i++)
{
int index = i;
for (int j = i + 1; j < arr.length; j++)
if (arr[j] < arr[index])
index = j;
int temp= arr[index];
arr[index] = arr[i];
arr[i] = temp;
}
return arr;
}
public static void main(String a[]){
int[] arr1 = {10,34,2,56,7,67,88,42};
int[] arr2 = sort(arr1);
for(int i:arr2){
System.out.print(i);
System.out.print(\", \");
}
}
}
c) Insertion
public class SortInsertion {
public static void main(String[] args) {
int[] array= { 4, 2, 9, 6, 23, 12, 34, 0, 1 };
sort(array);
}
static void printNumbers(int[] input) {
for (int i = 0; i < array.length; i++) {
System.out.print(array[i] + \", \");
}
System.out.println(\"\ \");
}
public static void sort(int array[]) {
int n = array.length;
for (int j = 1; j < n; j++) {
int key = array[j];
int i = j-1;
while ( (i > -1) && ( array [i] > key ) ) {
array [i+1] = array [i];
i--;
}
array[i+1] = key;
printNumbers(array);
}
}
}
d) Quick Sort:
public class SortQuick {
int array[];
int length;
public void sort(int[] inputArr) {
if (inputArr == null || inputArr.length == 0) {
return;
}
this.array = inputArr;
length = inputArr.length;
quickSort(0, length - 1);
}
private void quickSort(int lowerIndex, int higherIndex) {
int i = lowerIndex;
int j = higherIndex;
int pivot = array[lowerIndex+(higherIndex-lowerIndex)/2];
while (i <= j) {
while (array[i] < pivot) {
i++;
}
while (array[j] > pivot) {
j--;
}
if (i <= j) {
exchangeNumbers(i, j);
i++;
j--;
}
}
if (lowerIndex < j)
quickSort(lowerIndex, j);
if (i < higherIndex)
quickSort(i, higherIndex);
}
private void exchangeNumbers(int i, int j) {
int temp = array[i];
array[i] = array[j];
array[j] = temp;
}
public static void main(String a[]){
MyQuickSort sorter = new MyQuickSort();
int[] input = {24,2,45,20,56,75,2,56,99,53,12};
sorter.sort(input);
for(int i:input){
System.out.print(i);
System.out.print(\" \");
}
}
}.
Lab01.pptx
The document contains instructions for 14 programming exercises involving basic Java concepts like user input, arithmetic operations, loops, strings, and prime number identification. The exercises cover printing greetings, calculating triangle areas, generating even numbers, Fibonacci sequences, factorials, and determining parity, in addition to other foundational programming tasks. Students are guided to write Java code to prompt for input, perform calculations, and output results for each problem.
Refer to my progress on this assignment belowIn this problem you w.pdf
Refer to my progress on this assignment below
In this problem you will make it “more” object-oriented in the following ways:
-You will change its name to SortedList
-You will change the constructor that takes no arguments to be more traditional and initialize the
member fields to dummy values.
-You will add a constructor that takes in an initialized array and a size
-You will add an insert function that adds a value to the list and maintains its sorted-ness
-You will add a quicksort function check the below code.
-You will make updates as necessary to the main function so that it still runs and tests your code.
The code below seems to sort correctly but it skips over some items in the array and I am not
sure what is happening. Please help! Thanks!
import java.util.Scanner;
class SortedList
{
private static int array[];
private static int n;
public SortedList()
{
Scanner in = new Scanner(System.in);
System.out.println(\"Enter number of elements\");
n = in.nextInt();
array = new int[n];
if(n == 0) {
System.out.print(\"Since no arguments array set to: \");
} else {
System.out.print(\"Creating array size \" + n + \": \");
}
for(int i = 0; i < n; i++)
{
array[i] = 0;
}
//System.out.println(\"Enter \" + n + \" integers in ascending order\");
/*for (c = 0; c < n; c++)
array[c] = in.nextInt();*/
for(int i = 0; i < n; i++)
{
System.out.print(array[i] + \" \");
}
System.out.print(\"\ \");
}
public SortedList(int a[], int size)
{
array = a;
n = size;
}
public int binsearch(int search)
{
int first, last, middle;
first = 0;
last = n - 1;
middle = (first + last)/2;
while( first <= last )
{
if ( array[middle] < search )
first = middle + 1;
else if ( array[middle] == search )
{
//System.out.println(search + \" found at location \" + (middle + 1) + \".\");
return middle+1;//+1 for the non-CS people who don\'t start counting at zero.
}
else
last = middle - 1;
middle = (first + last)/2;
}
return -1;
//System.out.println(search + \" is not present in the list.\ \");
}
public static int partition(int input[], int p, int r)
{
int pivot = input[r];
while(p < r)
{
while(input[p] < pivot)
{
p++;
}
while(input[r] > pivot)
{
r--;
}
if(input[p] == input[r])
{
p++;
}
else if(p < r)
{
int tmp = input[p];
input[p] = input[r];
input[r] = tmp;
}
}
return r;
}
public static void quicksort(int input[], int p, int r)
{
if(p < r)
{
int j = partition(input, p, r);
quicksort(input, p, j-1);
quicksort(input, j+1, r);
}
}
public static void insert(int value, int cell)
{
array[cell] = value;
quicksort(array, 0, n-1);
}
public static void main(String args[])
{
int c;
Scanner in = new Scanner(System.in);
SortedList b = new SortedList();
System.out.println(\"Input numbers\");
for(int i = 0; i < n; i++)
{
c = in.nextInt();
in.nextLine();
insert(c, i);
}
in.close();
for(int j = 0; j < n; j++)
{
System.out.print(array[j] + \", \");
}
}
}
Solution
//the whole program that you wrote is correct except one line , when you call the quicksort
everytime you insert a //value dont pass n pass the i.
1. import java.util.Scanner; public class Alphabetical_Order {.pdf
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14 thread
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Hi, Please find my codeimport java.util.Random;public class Pro.pdf
Hi, Please find my code:
import java.util.Random;
public class ProcessArray {
private int rows; //The attribute for number of rows in matrix
private int columns; //The attribute for number of columns in matrix
private int[][] firstArray; //The attribute for the first array
private int[][] secondArray; //The attribute for the second array
public int[][] getFirstArray() { return firstArray;}
public int[][] getSecondArray() { return secondArray;}
public ProcessArray(int rows, int columns){ //Constructor of object ProcessArray
this.rows = rows;
this.columns = columns;
int[][] array = new int[rows][columns];
initializeArray(array);
randomlyFillArray();
computeArrayValues();
printArray(secondArray);
}
public void initializeArray(int[][] array){ //Initializes first and second arrays and sets each value
to 0
firstArray = new int[rows][columns];
secondArray = new int[rows][columns];
}
public void randomlyFillArray(){ //Fills first array with random numbers
for(int i = 0; i < firstArray.length;i++){
for(int j = 0; j < firstArray[0].length;j++){
Random r = new Random();
int num = r.nextInt(16);
firstArray[i][j] = num;
}
}
}
public void computeArrayValues(){
int col = firstArray[0].length;
int row = firstArray.length;
for(int i = 0; i < row; i++){
for(int j = 0; j< col; j++){
secondArray[i][j] = 0;
if((i - 1) >=0){
secondArray[i][j] += firstArray[i-1][j];
if((j+1) < col)
secondArray[i][j] += firstArray[i-1][j+1];
if(j > 0)
secondArray[i][j] += firstArray[i-1][j-1];
}
if((j+1) < col)
secondArray[i][j] += firstArray[i][j+1];
if(j > 0)
secondArray[i][j] += firstArray[i][j-1];
if((i+1) < row){
secondArray[i][j] += firstArray[i+1][j];
if((j+1) < col)
secondArray[i][j] += firstArray[i+1][j+1];
if(j > 0)
secondArray[i][j] += firstArray[i+1][j-1];
}
}
}
}
public void printArray(int[][] Array){
System.out.println(\"\ Initial Array Filled With Random Numbers: \ \");
for(int a = 0; a < firstArray.length; a++){
for(int b = 0; b < firstArray[0].length; b++){
if(b == 0)
System.out.printf(\"%d \", firstArray[a][b]);
else
System.out.printf(\"%d \", firstArray[a][b]);
}System.out.println();
}System.out.println();
System.out.println(\"Computed Array: \ \");
for(int a = 0; a < secondArray.length; a++){
for(int b = 0; b < secondArray[0].length; b++){
if(b == 0)
System.out.printf(\"%d \", secondArray[a][b]);
else
System.out.printf(\"%d \", secondArray[a][b]);
}
System.out.println();
}System.out.println();
}
public static void main(String[] args) {
ProcessArray pr = new ProcessArray(3, 4);
}
}
/*
Sample Output:
Initial Array Filled With Random Numbers:
7 3 6 12
10 2 11 10
12 14 1 7
Computed Array:
15 36 38 27
38 64 55 37
26 36 44 22
*/
Solution
Hi, Please find my code:
import java.util.Random;
public class ProcessArray {
private int rows; //The attribute for number of rows in matrix
private int columns; //The attribute for number of columns in matrix
private int[][] firstArray; //The attribute for the first array
private int[][] secondArray; //The attribute for the second arr.
ArrayTest.java import java.util.Arrays; import java.util.Scann.pdf
ArrayTest.java
import java.util.Arrays;
import java.util.Scanner;
public class ArrayTest {
/**
* @param args
*/
public static void main(String[] args) {
// TODO Auto-generated method stub
int table[] = new int[10];
int count = 0;
Scanner scan = new Scanner(System.in);
for(int i=0; i<10; i++){
System.out.println(\"Enter Array Element:\");
int n = scan.nextInt();
if(n < 0){
System.out.println(\"Invalid Number.\");
i--;
if(count >= 5)
break;
}else{
table[i] = n;
count++;
}
}
int sum = 0;
System.out.println(\"Array Elements are: \"+Arrays.toString(table));
for(int i=0; i avg){
high++;
}
else{
equal++;
}
}
System.out.println(\"Number of elements less than average \"+less);
System.out.println(\"Number of elements greater than average \"+high);
System.out.println(\"Number of elements equal to average \"+equal);
for(;;){
System.out.println(\"Enter Search Element (-1 to quit): \");
int searchNum = scan.nextInt();
if(searchNum == -1){
break;
}
int index = find(searchNum, table, count);
if(index == -1){
System.out.println(\"Element not found\");
}
else{
System.out.println(\"Element found at \"+index);
}
}
smallest(table, count);
}
public static int find (int valueToSearchFor, int intArray[ ], int numberOfValidEnties){
for(int i=0; i a[i]){
min = a[i];
index = i;
}
}
System.out.println(\"The smallest element is \"+min+\" index is \"+index);
int temp = a[0];
a[0] = a[index];
a[index] = temp;
System.out.println(\"After swap array elements are :\"+Arrays.toString(a));
}
}
Output:
Enter Array Element:
3
Enter Array Element:
4
Enter Array Element:
5
Enter Array Element:
2
Enter Array Element:
1
Enter Array Element:
6
Enter Array Element:
7
Enter Array Element:
8
Enter Array Element:
-1
Invalid Number.
Array Elements are: [3, 4, 5, 2, 1, 6, 7, 8, 0, 0]
The Sum is 36
The Average is : 4.5
Number of elements less than average 4
Number of elements greater than average 4
Number of elements equal to average 0
Enter Search Element (-1 to quit):
2
Element found at 3
Enter Search Element (-1 to quit):
3
Element found at 0
Enter Search Element (-1 to quit):
1
Element found at 4
Enter Search Element (-1 to quit):
6
Element found at 5
Enter Search Element (-1 to quit):
-1
The smallest element is 1 index is 4
After swap array elements are :[1, 4, 5, 2, 3, 6, 7, 8, 0, 0]
Solution
ArrayTest.java
import java.util.Arrays;
import java.util.Scanner;
public class ArrayTest {
/**
* @param args
*/
public static void main(String[] args) {
// TODO Auto-generated method stub
int table[] = new int[10];
int count = 0;
Scanner scan = new Scanner(System.in);
for(int i=0; i<10; i++){
System.out.println(\"Enter Array Element:\");
int n = scan.nextInt();
if(n < 0){
System.out.println(\"Invalid Number.\");
i--;
if(count >= 5)
break;
}else{
table[i] = n;
count++;
}
}
int sum = 0;
System.out.println(\"Array Elements are: \"+Arrays.toString(table));
for(int i=0; i avg){
high++;
}
else{
equal++;
}
}
System.out.println(\"Number of elements less than average \"+less);
System.out.println(\"Number of.
import java.uti-WPS Office.docx
The document contains code for several Java programs. The first program takes a number input and checks if it is evenly divisible by 9. The second program calculates the areas of different shapes like rectangle, circle and triangle by taking length, width, radius or base and height as user input. The third program is a menu to order drinks from a pub by selecting different drinks and entering the quantity. It calculates and displays the total cost. The last program calculates distance traveled by taking speed and time as input.
Huraira_waris_Assgnment_4.docx
The document contains 5 questions related to matrix operations in Java. It includes code to find the sum of each row and product of each column in a matrix, use binary search to find a target value in a 2D array, print a specified row and column from a matrix, calculate frequency of numbers in a matrix using a HashMap, and find unique numbers and their frequencies in a matrix.
Sam wd programs
The document provides an index and descriptions of various topics related to web development including:
1. The modulus operator and examples of using it to check for divisibility.
2. Relational and logical operators like greater than, less than, equal to and examples of using them in code.
3. Descriptions of do-while and for loops with examples.
4. An example using a parameterized constructor to initialize cube dimensions.
5. Examples of string methods like startsWith, length, and trim.
6. Descriptions and examples of overloading methods and constructors.
7. An example of inheritance with overriding methods.
8. An interface example with animal classes
Java programs - bubble sort, iterator, linked list, hash set, reverse string,...
Java programs - bubble sort, iterator, linked list, hash set, reverse string,...Sunil Kumar Gunasekaran
The document contains 11 Java programming assignments involving arrays and strings:
1. Sorting an integer array using bubble sort
2. Printing only odd numbers from an integer array
3. Moving all even numbers to the beginning of an integer array
4. Finding unique numbers and occurrences of duplicates in an array
5. Checking if an integer array matches the Fibonacci series
6. Checking if an integer array is a palindrome
7. Finding unique words in a string
8. Reversing a string
9. Reversing each word in a string separately
10. Reversing the lines when writing a file
11. Creating a Java program with APIs for database SELECT and UPDATE operationsExample of JAVA Program
This Java program allows a user to perform arithmetic operations and generate random numbers on arrays of double values. The user is presented with a menu to choose between adding, subtracting, multiplying, dividing, dot product, or generating a random array. For each operation, the user can enter size of the arrays and values. The program then performs the selected operation and displays the results. It includes error handling for cases like dividing by zero. Methods are defined to get user input, perform the operations, and generate random numbers within a given range.
SumNumbers.java import java.util.Scanner;public class SumNumbe.pdf
SumNumbers.java
import java.util.Scanner;
public class SumNumbers {
/**
* @param args
*/
public static void main(String[] args) {
// TODO Auto-generated method stub
Scanner scan = new Scanner(System.in);
int sum = 0;
while(true){
System.out.println(\"Enter integer:\");
int n = scan.nextInt();
if(n == -1){
break;
}
sum = sum + n;
}
System.out.println(\"The sum is \"+sum);
}
}
Output:
Enter integer:
1
Enter integer:
2
Enter integer:
3
Enter integer:
4
Enter integer:
5
Enter integer:
6
Enter integer:
7
Enter integer:
8
Enter integer:
9
Enter integer:
-1
The sum is 45
Solution
SumNumbers.java
import java.util.Scanner;
public class SumNumbers {
/**
* @param args
*/
public static void main(String[] args) {
// TODO Auto-generated method stub
Scanner scan = new Scanner(System.in);
int sum = 0;
while(true){
System.out.println(\"Enter integer:\");
int n = scan.nextInt();
if(n == -1){
break;
}
sum = sum + n;
}
System.out.println(\"The sum is \"+sum);
}
}
Output:
Enter integer:
1
Enter integer:
2
Enter integer:
3
Enter integer:
4
Enter integer:
5
Enter integer:
6
Enter integer:
7
Enter integer:
8
Enter integer:
9
Enter integer:
-1
The sum is 45.
Java programs
The document provides examples of Java code demonstrating various programming concepts like loops, conditional statements, methods, arrays and string manipulation. It includes code snippets to print "Hello World", check if a number is even or odd, print multiplication tables, convert temperatures between Fahrenheit and Celsius scales, take input from the user and perform basic operations on strings. The examples illustrate the use of various loops like for, while and do-while loops, if-else statements, methods and built-in string/array methods in Java.
Lab101.pptx
The document contains 12 code examples for Java programs that perform various tasks like greeting a user, calculating triangle area, printing even numbers, generating Fibonacci sequences, calculating factorials, determining if a number is even/odd, comparing two numbers, determining if a number is positive/negative, generating a multiplication table, and reversing a string. Each code example demonstrates the use of basic Java concepts like variables, conditionals, loops, methods, and classes to solve common programming problems.
import java.util.Scanner;public class Main {private static i.pdf
import java.util.Scanner;
public class Main {
private static int n;
private static int[] referenceString;
private static int[] physicalFrames;
private static int[] victimPages;
private static int pageFaults;
public static void main(String[] args) {
Scanner input = new Scanner(System.in);
int choice;
do {
System.out.println("0 Exit");
System.out.println("1 Input N");
System.out.println("2 Input the reference string");
System.out.println("3 Simulate the OPT algorithm");
System.out.println("4 Simulate the NEW algorithm");
System.out.print("Select option: ");
choice = input.nextInt();
switch (choice) {
case 0:
System.out.println("Exiting the program...");
break;
case 1:
System.out.print("Enter the value of N (2-8): ");
n = input.nextInt();
if (n < 2 || n > 8) {
System.out.println("Invalid value of N!");
n = 0;
} else {
physicalFrames = new int[n];
victimPages = new int[n];
System.out.println("N set to " + n + ".");
}
break;
case 2:
if (n == 0) {
System.out.println("Please enter the value of N first!");
} else {
System.out.print("Enter the reference string (length at least " + n + ", max length 20): ");
String[] strArr = input.next().split("");
int len = strArr.length;
if (len < n || len > 20) {
System.out.println("Invalid length of reference string!");
} else {
referenceString = new int[len];
for (int i = 0; i < len; i++) {
int page = Integer.parseInt(strArr[i]);
if (page < 0 || page > 9) {
System.out.println("Invalid page number!");
referenceString = null;
break;
}
referenceString[i] = page;
}
if (referenceString != null) {
System.out.println("Reference string set to " + input.next() + ".");
}
}
}
break;
case 3:
if (n == 0) {
System.out.println("Please enter the value of N first!");
} else if (referenceString == null) {
System.out.println("Please enter the reference string first!");
} else {
System.out.println("Simulating the OPT algorithm...");
simulateOPT();
}
break;
case 4:
if (n == 0) {
System.out.println("Please enter the value of N first!");
} else if (referenceString == null) {
System.out.println("Please enter the reference string first!");
} else {
System.out.println("Simulating the NEW algorithm...");
simulateNEW();
}
break;
default:
System.out.println("Invalid choice!");
break;
}
System.out.println();
} while (choice != 0);
}
private static void simulateOPT() {
pageFaults = 0;
int len = referenceString.length;
System.out.println("Reference String\tPhysical Frames\tPage Faults\tVictim Pages");
for (int i = 0; i < len; i++) {
int page = referenceString[i];
boolean found = false;
for (int j = 0; j < n; j++) {
if (physicalFrames[j] == page) {
found = true;
break;
}
}
if (!found) {
pageFaults++;
int maxDist = 0;
int victimPage = -1;
for (int j = 0; j < n; j++) {
int dist = 0;
for (int k = i + 1; k < len; k++) {
dist++;
if (physicalFrames[j] == referenceString[k]) {
break;
}
}
if (dist > maxDist) {
maxDist = dist;
victimPage = j;
}
}
physicalFrames[victimPage] = page;
victimPages[i] = physicalFrames[victimPage];
} else {
victimPages[i] = -1;.
FileName EX06_1java Programmer import ja.pdf
//FileName: EX06_1.java
//Programmer: ............
import java.util.Scanner;
import java.util.Arrays;
public class EX06_1
{
public static void main(String[] args)
{
Scanner input = new Scanner(System.in);
System.out.print("Enter a list of SAT scores: ");
String str = input.nextLine();
String[] items = str.split(" ");
double[] values = new double[items.length];
for (int i = 0; i<items.length; i++)
{
values[i] = Double.parseDouble(items[i]);
}
//count
System.out.print("Count: ");
int count = 0;
for (int i=0; i<values.length; i++)
{
count++;
}
System.out.println((count == values.length) ? values.length : count);
//sum
double sum = 0;
for (int i = 0; i<values.length; i++)
{
sum += values[i];
}
System.out.println("Sum: " + sum);
//mean
double mean = (sum/count);
System.out.println("Mean: " + mean);
//min and max
double min = values[0];
double max = values[0];
for (int i=1; i<values.length; i++)
{
if (values[i] < min)
{
min = values[i];
}
if (values[i] > max)
{
max = values[i];
}
}
System.out.println("Minimum: " + min);
System.out.println("Maximum: " + max);
System.out.println("Range: " + (max-min));
//sorting
Arrays.sort(values);
//median
int position = (count+1) / 2;
if (count%2 != 0)
{
System.out.println("Median: " + values[position-1]);
}
else
{
System.out.println("Median: " + (( values[position-1] + values[position]) / 2));
}
//Standard deviation
double sigmap = 0; //for population
double sigmas = 0; //for sample
for (int i=0; i<values.length; i++)
{
sigmap += (values[i] - mean)*(values[i] - mean);
}
sigmas = sigmap;
//sd for population
sigmap = sigmap / count;
System.out.println("Variance of population: " + sigmap);
double sdp = Math.sqrt(sigmap);
System.out.println("Standard Deviation of population: " + sdp);
//sd for sample
sigmas = sigmas / (count - 1);
System.out.println("Variance of sample: " + sigmas);
double sds = Math.sqrt(sigmas);
System.out.println("Standard Deviation of sample: " + sds);
//Skewness
double sigma = 0;
for (int i=0; i<values.length; i++)
{
sigma += (values[i] - mean)*(values[i] - mean)*(values[i] - mean);
}
double N = (double) count;
double sk = (Math.sqrt(N*(N-1)) / (N-2)) * (sigma / (sdp*sdp*sdp*N));
System.out.println("Skewness: " + sk);
//mode
int repeated = -1;
int[] frq = new int[values.length];
for (int i=0; i<values.length; i++)
{
int occ = 1;
for (int j=i+1; j<values.length; j++)
{
if (values[i] == values[j])
{
frq[j] = repeated;
occ++;
}
}
if (frq[i] != repeated)
{
frq[i] = occ;
}
}
int maxFrq = frq[0];
for (int i=0; i<frq.length; i++)
{
if (frq[i] > maxFrq)
{
maxFrq = frq[i];
}
}
System.out.print("Mode: ");
if (maxFrq == 1)
{
System.out.println("N/A");
}
else
{
for (int i=0; i<frq.length; i++)
{
if (frq[i] == maxFrq)
{
System.out.print(values[i] + " ");
}
}
System.out.print("n");
}
double standardError=sds/Math.sqrt(count);
System.out.println("Standard Error: "+standardError);
}
}
Programming Exercise #06_2:
Instruction:
1. Make a copy of your EX06_01.java file and rename it to EX06_2.java.
2. Open the EX06.
I need to create a data type that implement the following interface -.docx
I need to create a data type that implement the following interface ;
myNumber()
myNumber(int valueIn)
void printDecimal()
void printBinary()
String toString() //returns a string stating both the decimal and binary value
Solution
import java.util.*;
interface Myinterface
{
public void mynumber();
public void mynumber(int value);
public void printDecimal();
public void printbinary();
public void tostring();
}
class Sample implements Myinterface
{
public void mynumber()
{
int value=0;
System.out.println(\"implementation of mynumber()\");
}
public void mynumber(int value)
{
int a=value;
System.out.println(\"implementation of mynumber(int value):\"+a);
}
public void printDecimal()
{
int x=10;
System.out.println(\"Decimal number is:\"+x);
}
public void printbinary()
{
int x=10;
int arr[] = new int[100];
int i=0;
while(x>0)
{
arr[i]=x%2;
i++;
x=x/2;
}
int count=i-1;
System.out.println(\"Binary value :\");
for(i=count;i>=0;i--)
{
System.out.println(arr[i]);
}
}
public void tostring()
{
//toSting is used to return the string object value for the integer
int x=10;
int p=x;
int arr[] = new int[100];
int i=0;
while(x>0)
{
arr[i]=x%2;
i++;
x=x/2;
}
int count=i-1;
System.out.println(\"deimal is: \"+Integer.toString(p)); //implementing toSting object for decimal
System.out.println(\"Binary value :\");
for(i=count;i>=0;i--)
{
System.out.println(Integer.toString(arr[i]));
}
}
public static void main(String args[])
{
//defining objects for the interface and are called in the void main
Myinterface obj = new Sample();
obj.mynumber();
obj.mynumber(3);
obj.printDecimal();
obj.printbinary();
obj.tostring();
}
}
.
Write the program in MIPS that declares an array of positive integer.pdf
Write the program in MIPS that declares an array of positive integers. The size of the array is not
fixed. The final element of the array should have the value of 0xF, which is not used in
calculations, and should be used to calculate the size of the array. The program can prompt the
user for an integer input. Compare the input with the array. Display the index of the input in the
array if found. If the number does not match any element in the array, insert the number in the
array in sorted order. The program is terminated when input a negative integer. Print the new
array with appropriate message on the screen. For example assume an array as follows:
(array: .word 4, 5, 23, 5, 8, 3, 15, 67, 8, 9, 0xF)
Sorted array: 3, 4, 5, 5, 8, 8, 9, 15, 23, 67
Enter a number to search: 20
Number not found. Added to array.
Sorted array: 3, 4, 5, 5, 8, 8, 9, 15, 20, 23, 67
Enter a number to search: 5
Number at index 2
Solution
import java.util.Scanner;
public category Ascending _Order
for (int i = 0; i < n; i++)
temporary worker = a[i];
a[i] = a[j];
a[j] = temp;
}
}
}
System.out.print(\"Ascending Order:\");
for (int i = 0; i < n - 1; i++)
import java.util.Scanner;
public category Ascending _Order
for (int i = 0; i < n; i++)
temporary worker = a[i];
a[i] = a[j];
a[j] = temp;
}
}
}
System.out.print(\"Ascending Order:\");
for (int i = 0; i < n - 1; i++)
import java.util.Scanner;
public category Ascending _Order
for (int i = 0; i < n; i++)
temporary worker = a[i];
a[i] = a[j];
a[j] = temp;
}
}
}
System.out.print(\"Ascending Order:\");
for (int i = 0; i < n - 1; i++)
import java.util.Scanner;
public category Ascending _Order
for (int i = 0; i < n; i++)
temporary worker = a[i];
a[i] = a[j];
a[j] = temp;
}
}
}
System.out.print(\"Ascending Order:\");
for (int i = 0; i < n - 1; i++)
import java.util.Scanner;
public category Ascending _Order
for (int i = 0; i < n; i++)
temporary worker = a[i];
a[i] = a[j];
a[j] = temp;
}
}
}
System.out.print(\"Ascending Order:\");
for (int i = 0; i < n - 1; i++)
import java.util.Scanner;
public category Ascending _Order
for (int i = 0; i < n; i++)
temporary worker = a[i];
a[i] = a[j];
a[j] = temp;
}
}
}
System.out.print(\"Ascending Order:\");
for (int i = 0; i < n - 1; i++)
import java.util.Scanner;
public category Ascending _Order
for (int i = 0; i < n; i++)
temporary worker = a[i];
a[i] = a[j];
a[j] = temp;
}
}
}
System.out.print(\"Ascending Order:\");
for (int i = 0; i < n - 1; i++)
import java.util.Scanner;
public category Ascending _Order
for (int i = 0; i < n; i++)
temporary worker = a[i];
a[i] = a[j];
a[j] = temp;
}
}
}
System.out.print(\"Ascending Order:\");
for (int i = 0; i < n - 1; i++)
import java.util.Scanner;
public category Ascending _Order
for (int i = 0; i < n; i++)
temporary worker = a[i];
a[i] = a[j];
a[j] = temp;
}
}
}
System.out.print(\"Ascending Order:\");
for (int i = 0; i < n - 1; i++)
import java.util.Scanner;
public category Ascending _Order
for (int i = 0; i < n; i++)
temporary worker = a[i];
a[i] = a[j];
a.
Question 1,2,4 ------------------------------------Please check.pdf
Question 1,2,4 :-
-----------------------------------
Please check comments in Programme
#include
/**
* Method for Linear Search
* Loop through given array and search for given key
* IF found return index and stop
* else -1 will be returned at the end
*
*/
int search(int array[],int key,int size){
int index = -1;
for (int c = 0; c < size; c++)
{
if (array[c] == key) /* if required element found */
{
printf(\"%d is present at location %d.\ \", key, c+1);
index = c;
break;
}
}
if(index == -1)
printf(\"%d is not found.\ \", key);
return index;
}
/**
* Function for Binary search
* Binary search is used only for sorted arrays
* Check key with element at middle position
* if key == then return middle positon
* if key > middle position , apply same for sub array range from [middle...last]
* if key < middle position , apply same for sub array range from [0..middle]
*
*/
int binary_search(int array[],int key,int start,int end){
int index = -1;
int first = start;
int last = end;
int middle = (first+last)/2;
while (first <= last) {
//Print array between first nad last for each step
printf(\"Start binary search step.\ \");
for(int i = first ; i < last ; i++){
printf(\"Binary search step element.%d\ \",array[i]);
}
printf(\"Stop binary search step.\ \");
if (array[middle] < key)
first = middle + 1;
else if (array[middle] == key) {
index = middle;
printf(\"%d found at location %d.\ \", key, middle+1);
break;
}
else
last = middle - 1;
middle = (first + last)/2;
}
return index;
}
/**
* Recursive linear search
* Instead of for loop call same method with next index
* first pass 0, check if key found at index 0, if not found call same method with index 1 .....
*/
int linear_search_recursive(int array[],int key,int size,int index){
if(array[index] == key){
printf(\"%d found at location %d.\ \",key, index);
return index;
}
else if (index >= size-1){
return -1;
}
else{
return linear_search_recursive(array,key,size,++index);
}
}
/**
* Recursive Binary Search
* Recursively call Binary Search , check key found at element middle
* if key == then return middle positon
* if key > middle position , call recursive method for sub array range from [middle...last]
* if key < middle position , call recursive method for sub array range from [0..middle]
*/
int binary_search_recursive(int array[],int key,int start,int end){
//Key not found
if(start> end){
return -1;
}
int index = -1;
int first = start;
int last = end;
int middle = (first+last)/2;
//Print array between first nad last for each step
printf(\"Start binary search step.\ \");
for(int i = first ; i < last ; i++){
printf(\"Binary search step element.%d\ \",array[i]);
}
printf(\"Stop binary search step.\ \");
if (array[middle] < key){
first = middle + 1;
return binary_search_recursive(array,key,first,last) ;
}else if (array[middle] == key) {
printf(\"%d found at location %d.\ \", key, middle+1);
return middle;
}
else{
last = middle - 1;
return binary_search_recursive(array,key,first,last) ;
}
return -1;
}
/**
*Function t.
The answer is C. securin.APC degrade securin to release separase .pdf
The answer is: C. securin.
APC degrade securin to release separase which degrades cohesin.
Solution
The answer is: C. securin.
APC degrade securin to release separase which degrades cohesin..
HiCan you please summarise the question. More information confuse .pdf
Hi
Can you please summarise the question. More information confuse to sole this question
Solution
Hi
Can you please summarise the question. More information confuse to sole this question.
What property of proteins us used to separate them in HICHydrophob.pdf
What property of proteins us used to separate them in HIC
Hydrophobic interaction chromatography (HIC) isolates Proteins in light of their hydrophobicity,
and is regularly utilized as a middle stride as a part of a cleaning plan. Proteins are bound to a
stationary stage in a high ionic quality cushion and, along these lines, HIC can commonly be
performed instantly after particle trade chromatography with no cradle trade or weakening
required. HIC is likewise usually performed after an ammonium sulfate precipitation, a strategy
that can be utilized to rapidly expel proteins by accelerating a few, yet not all, proteins with salt.
HIC is infrequently material in early strides of a purging plan or as a last stride in the expulsion
of follow polluting influences from the protein of intrigue.
====================================================
Described mobile and stationary phase in HIC
Hydrophobic association chromatography (HIC) is an other option to turned around stage
chromatography for abusing the hydrophobic properties of proteins.
The expansion of a salt to the mobile stage support and test arrangement advances protein-
medium connections. The proteins are adsorbed to the medium in a mobile stage containing a
high grouping of salt. The vast majority of the bound proteins are viably desorbed by essentially
washing with water or a weaken, close impartial support. Since HIC utilizes a more polar, less
denaturing environment than RPLC, it is getting to be well known for protein filtration, regularly
in blend with particle trade or gel filtration chromatography.
Solution
What property of proteins us used to separate them in HIC
Hydrophobic interaction chromatography (HIC) isolates Proteins in light of their hydrophobicity,
and is regularly utilized as a middle stride as a part of a cleaning plan. Proteins are bound to a
stationary stage in a high ionic quality cushion and, along these lines, HIC can commonly be
performed instantly after particle trade chromatography with no cradle trade or weakening
required. HIC is likewise usually performed after an ammonium sulfate precipitation, a strategy
that can be utilized to rapidly expel proteins by accelerating a few, yet not all, proteins with salt.
HIC is infrequently material in early strides of a purging plan or as a last stride in the expulsion
of follow polluting influences from the protein of intrigue.
====================================================
Described mobile and stationary phase in HIC
Hydrophobic association chromatography (HIC) is an other option to turned around stage
chromatography for abusing the hydrophobic properties of proteins.
The expansion of a salt to the mobile stage support and test arrangement advances protein-
medium connections. The proteins are adsorbed to the medium in a mobile stage containing a
high grouping of salt. The vast majority of the bound proteins are viably desorbed by essentially
washing with water or a weaken, close impartial support. Since HIC uti.
upload complete priblem somthing is missingSolutionupload comp.pdf
The document discusses an upload issue where the process completed but something was missing. No other details are provided about the upload, missing content, or possible solutions. The document repeats the same short statement twice without providing any useful information to understand or resolve the issue.
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1. import java.util.Scanner; public class Alphabetical_Order {.pdf
The document contains code for two Java programs. The first program takes user input of names and sorts them alphabetically. The second program takes a number from the user and checks which numbers between 1 and that number are prime numbers.
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Hi, Please find my codeimport java.util.Random;public class Pro.pdf
Hi, Please find my code:
import java.util.Random;
public class ProcessArray {
private int rows; //The attribute for number of rows in matrix
private int columns; //The attribute for number of columns in matrix
private int[][] firstArray; //The attribute for the first array
private int[][] secondArray; //The attribute for the second array
public int[][] getFirstArray() { return firstArray;}
public int[][] getSecondArray() { return secondArray;}
public ProcessArray(int rows, int columns){ //Constructor of object ProcessArray
this.rows = rows;
this.columns = columns;
int[][] array = new int[rows][columns];
initializeArray(array);
randomlyFillArray();
computeArrayValues();
printArray(secondArray);
}
public void initializeArray(int[][] array){ //Initializes first and second arrays and sets each value
to 0
firstArray = new int[rows][columns];
secondArray = new int[rows][columns];
}
public void randomlyFillArray(){ //Fills first array with random numbers
for(int i = 0; i < firstArray.length;i++){
for(int j = 0; j < firstArray[0].length;j++){
Random r = new Random();
int num = r.nextInt(16);
firstArray[i][j] = num;
}
}
}
public void computeArrayValues(){
int col = firstArray[0].length;
int row = firstArray.length;
for(int i = 0; i < row; i++){
for(int j = 0; j< col; j++){
secondArray[i][j] = 0;
if((i - 1) >=0){
secondArray[i][j] += firstArray[i-1][j];
if((j+1) < col)
secondArray[i][j] += firstArray[i-1][j+1];
if(j > 0)
secondArray[i][j] += firstArray[i-1][j-1];
}
if((j+1) < col)
secondArray[i][j] += firstArray[i][j+1];
if(j > 0)
secondArray[i][j] += firstArray[i][j-1];
if((i+1) < row){
secondArray[i][j] += firstArray[i+1][j];
if((j+1) < col)
secondArray[i][j] += firstArray[i+1][j+1];
if(j > 0)
secondArray[i][j] += firstArray[i+1][j-1];
}
}
}
}
public void printArray(int[][] Array){
System.out.println(\"\ Initial Array Filled With Random Numbers: \ \");
for(int a = 0; a < firstArray.length; a++){
for(int b = 0; b < firstArray[0].length; b++){
if(b == 0)
System.out.printf(\"%d \", firstArray[a][b]);
else
System.out.printf(\"%d \", firstArray[a][b]);
}System.out.println();
}System.out.println();
System.out.println(\"Computed Array: \ \");
for(int a = 0; a < secondArray.length; a++){
for(int b = 0; b < secondArray[0].length; b++){
if(b == 0)
System.out.printf(\"%d \", secondArray[a][b]);
else
System.out.printf(\"%d \", secondArray[a][b]);
}
System.out.println();
}System.out.println();
}
public static void main(String[] args) {
ProcessArray pr = new ProcessArray(3, 4);
}
}
/*
Sample Output:
Initial Array Filled With Random Numbers:
7 3 6 12
10 2 11 10
12 14 1 7
Computed Array:
15 36 38 27
38 64 55 37
26 36 44 22
*/
Solution
Hi, Please find my code:
import java.util.Random;
public class ProcessArray {
private int rows; //The attribute for number of rows in matrix
private int columns; //The attribute for number of columns in matrix
private int[][] firstArray; //The attribute for the first array
private int[][] secondArray; //The attribute for the second arr.
ArrayTest.java import java.util.Arrays; import java.util.Scann.pdf
ArrayTest.java
import java.util.Arrays;
import java.util.Scanner;
public class ArrayTest {
/**
* @param args
*/
public static void main(String[] args) {
// TODO Auto-generated method stub
int table[] = new int[10];
int count = 0;
Scanner scan = new Scanner(System.in);
for(int i=0; i<10; i++){
System.out.println(\"Enter Array Element:\");
int n = scan.nextInt();
if(n < 0){
System.out.println(\"Invalid Number.\");
i--;
if(count >= 5)
break;
}else{
table[i] = n;
count++;
}
}
int sum = 0;
System.out.println(\"Array Elements are: \"+Arrays.toString(table));
for(int i=0; i avg){
high++;
}
else{
equal++;
}
}
System.out.println(\"Number of elements less than average \"+less);
System.out.println(\"Number of elements greater than average \"+high);
System.out.println(\"Number of elements equal to average \"+equal);
for(;;){
System.out.println(\"Enter Search Element (-1 to quit): \");
int searchNum = scan.nextInt();
if(searchNum == -1){
break;
}
int index = find(searchNum, table, count);
if(index == -1){
System.out.println(\"Element not found\");
}
else{
System.out.println(\"Element found at \"+index);
}
}
smallest(table, count);
}
public static int find (int valueToSearchFor, int intArray[ ], int numberOfValidEnties){
for(int i=0; i a[i]){
min = a[i];
index = i;
}
}
System.out.println(\"The smallest element is \"+min+\" index is \"+index);
int temp = a[0];
a[0] = a[index];
a[index] = temp;
System.out.println(\"After swap array elements are :\"+Arrays.toString(a));
}
}
Output:
Enter Array Element:
3
Enter Array Element:
4
Enter Array Element:
5
Enter Array Element:
2
Enter Array Element:
1
Enter Array Element:
6
Enter Array Element:
7
Enter Array Element:
8
Enter Array Element:
-1
Invalid Number.
Array Elements are: [3, 4, 5, 2, 1, 6, 7, 8, 0, 0]
The Sum is 36
The Average is : 4.5
Number of elements less than average 4
Number of elements greater than average 4
Number of elements equal to average 0
Enter Search Element (-1 to quit):
2
Element found at 3
Enter Search Element (-1 to quit):
3
Element found at 0
Enter Search Element (-1 to quit):
1
Element found at 4
Enter Search Element (-1 to quit):
6
Element found at 5
Enter Search Element (-1 to quit):
-1
The smallest element is 1 index is 4
After swap array elements are :[1, 4, 5, 2, 3, 6, 7, 8, 0, 0]
Solution
ArrayTest.java
import java.util.Arrays;
import java.util.Scanner;
public class ArrayTest {
/**
* @param args
*/
public static void main(String[] args) {
// TODO Auto-generated method stub
int table[] = new int[10];
int count = 0;
Scanner scan = new Scanner(System.in);
for(int i=0; i<10; i++){
System.out.println(\"Enter Array Element:\");
int n = scan.nextInt();
if(n < 0){
System.out.println(\"Invalid Number.\");
i--;
if(count >= 5)
break;
}else{
table[i] = n;
count++;
}
}
int sum = 0;
System.out.println(\"Array Elements are: \"+Arrays.toString(table));
for(int i=0; i avg){
high++;
}
else{
equal++;
}
}
System.out.println(\"Number of elements less than average \"+less);
System.out.println(\"Number of.
import java.uti-WPS Office.docx
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The document contains 5 questions related to matrix operations in Java. It includes code to find the sum of each row and product of each column in a matrix, use binary search to find a target value in a 2D array, print a specified row and column from a matrix, calculate frequency of numbers in a matrix using a HashMap, and find unique numbers and their frequencies in a matrix.
Sam wd programs
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2. Relational and logical operators like greater than, less than, equal to and examples of using them in code.
3. Descriptions of do-while and for loops with examples.
4. An example using a parameterized constructor to initialize cube dimensions.
5. Examples of string methods like startsWith, length, and trim.
6. Descriptions and examples of overloading methods and constructors.
7. An example of inheritance with overriding methods.
8. An interface example with animal classes
Java programs - bubble sort, iterator, linked list, hash set, reverse string,...
Java programs - bubble sort, iterator, linked list, hash set, reverse string,...Sunil Kumar Gunasekaran
The document contains 11 Java programming assignments involving arrays and strings:
1. Sorting an integer array using bubble sort
2. Printing only odd numbers from an integer array
3. Moving all even numbers to the beginning of an integer array
4. Finding unique numbers and occurrences of duplicates in an array
5. Checking if an integer array matches the Fibonacci series
6. Checking if an integer array is a palindrome
7. Finding unique words in a string
8. Reversing a string
9. Reversing each word in a string separately
10. Reversing the lines when writing a file
11. Creating a Java program with APIs for database SELECT and UPDATE operationsExample of JAVA Program
This Java program allows a user to perform arithmetic operations and generate random numbers on arrays of double values. The user is presented with a menu to choose between adding, subtracting, multiplying, dividing, dot product, or generating a random array. For each operation, the user can enter size of the arrays and values. The program then performs the selected operation and displays the results. It includes error handling for cases like dividing by zero. Methods are defined to get user input, perform the operations, and generate random numbers within a given range.
SumNumbers.java import java.util.Scanner;public class SumNumbe.pdf
SumNumbers.java
import java.util.Scanner;
public class SumNumbers {
/**
* @param args
*/
public static void main(String[] args) {
// TODO Auto-generated method stub
Scanner scan = new Scanner(System.in);
int sum = 0;
while(true){
System.out.println(\"Enter integer:\");
int n = scan.nextInt();
if(n == -1){
break;
}
sum = sum + n;
}
System.out.println(\"The sum is \"+sum);
}
}
Output:
Enter integer:
1
Enter integer:
2
Enter integer:
3
Enter integer:
4
Enter integer:
5
Enter integer:
6
Enter integer:
7
Enter integer:
8
Enter integer:
9
Enter integer:
-1
The sum is 45
Solution
SumNumbers.java
import java.util.Scanner;
public class SumNumbers {
/**
* @param args
*/
public static void main(String[] args) {
// TODO Auto-generated method stub
Scanner scan = new Scanner(System.in);
int sum = 0;
while(true){
System.out.println(\"Enter integer:\");
int n = scan.nextInt();
if(n == -1){
break;
}
sum = sum + n;
}
System.out.println(\"The sum is \"+sum);
}
}
Output:
Enter integer:
1
Enter integer:
2
Enter integer:
3
Enter integer:
4
Enter integer:
5
Enter integer:
6
Enter integer:
7
Enter integer:
8
Enter integer:
9
Enter integer:
-1
The sum is 45.
Java programs
The document provides examples of Java code demonstrating various programming concepts like loops, conditional statements, methods, arrays and string manipulation. It includes code snippets to print "Hello World", check if a number is even or odd, print multiplication tables, convert temperatures between Fahrenheit and Celsius scales, take input from the user and perform basic operations on strings. The examples illustrate the use of various loops like for, while and do-while loops, if-else statements, methods and built-in string/array methods in Java.
Lab101.pptx
The document contains 12 code examples for Java programs that perform various tasks like greeting a user, calculating triangle area, printing even numbers, generating Fibonacci sequences, calculating factorials, determining if a number is even/odd, comparing two numbers, determining if a number is positive/negative, generating a multiplication table, and reversing a string. Each code example demonstrates the use of basic Java concepts like variables, conditionals, loops, methods, and classes to solve common programming problems.
import java.util.Scanner;public class Main {private static i.pdf
import java.util.Scanner;
public class Main {
private static int n;
private static int[] referenceString;
private static int[] physicalFrames;
private static int[] victimPages;
private static int pageFaults;
public static void main(String[] args) {
Scanner input = new Scanner(System.in);
int choice;
do {
System.out.println("0 Exit");
System.out.println("1 Input N");
System.out.println("2 Input the reference string");
System.out.println("3 Simulate the OPT algorithm");
System.out.println("4 Simulate the NEW algorithm");
System.out.print("Select option: ");
choice = input.nextInt();
switch (choice) {
case 0:
System.out.println("Exiting the program...");
break;
case 1:
System.out.print("Enter the value of N (2-8): ");
n = input.nextInt();
if (n < 2 || n > 8) {
System.out.println("Invalid value of N!");
n = 0;
} else {
physicalFrames = new int[n];
victimPages = new int[n];
System.out.println("N set to " + n + ".");
}
break;
case 2:
if (n == 0) {
System.out.println("Please enter the value of N first!");
} else {
System.out.print("Enter the reference string (length at least " + n + ", max length 20): ");
String[] strArr = input.next().split("");
int len = strArr.length;
if (len < n || len > 20) {
System.out.println("Invalid length of reference string!");
} else {
referenceString = new int[len];
for (int i = 0; i < len; i++) {
int page = Integer.parseInt(strArr[i]);
if (page < 0 || page > 9) {
System.out.println("Invalid page number!");
referenceString = null;
break;
}
referenceString[i] = page;
}
if (referenceString != null) {
System.out.println("Reference string set to " + input.next() + ".");
}
}
}
break;
case 3:
if (n == 0) {
System.out.println("Please enter the value of N first!");
} else if (referenceString == null) {
System.out.println("Please enter the reference string first!");
} else {
System.out.println("Simulating the OPT algorithm...");
simulateOPT();
}
break;
case 4:
if (n == 0) {
System.out.println("Please enter the value of N first!");
} else if (referenceString == null) {
System.out.println("Please enter the reference string first!");
} else {
System.out.println("Simulating the NEW algorithm...");
simulateNEW();
}
break;
default:
System.out.println("Invalid choice!");
break;
}
System.out.println();
} while (choice != 0);
}
private static void simulateOPT() {
pageFaults = 0;
int len = referenceString.length;
System.out.println("Reference String\tPhysical Frames\tPage Faults\tVictim Pages");
for (int i = 0; i < len; i++) {
int page = referenceString[i];
boolean found = false;
for (int j = 0; j < n; j++) {
if (physicalFrames[j] == page) {
found = true;
break;
}
}
if (!found) {
pageFaults++;
int maxDist = 0;
int victimPage = -1;
for (int j = 0; j < n; j++) {
int dist = 0;
for (int k = i + 1; k < len; k++) {
dist++;
if (physicalFrames[j] == referenceString[k]) {
break;
}
}
if (dist > maxDist) {
maxDist = dist;
victimPage = j;
}
}
physicalFrames[victimPage] = page;
victimPages[i] = physicalFrames[victimPage];
} else {
victimPages[i] = -1;.
FileName EX06_1java Programmer import ja.pdf
//FileName: EX06_1.java
//Programmer: ............
import java.util.Scanner;
import java.util.Arrays;
public class EX06_1
{
public static void main(String[] args)
{
Scanner input = new Scanner(System.in);
System.out.print("Enter a list of SAT scores: ");
String str = input.nextLine();
String[] items = str.split(" ");
double[] values = new double[items.length];
for (int i = 0; i<items.length; i++)
{
values[i] = Double.parseDouble(items[i]);
}
//count
System.out.print("Count: ");
int count = 0;
for (int i=0; i<values.length; i++)
{
count++;
}
System.out.println((count == values.length) ? values.length : count);
//sum
double sum = 0;
for (int i = 0; i<values.length; i++)
{
sum += values[i];
}
System.out.println("Sum: " + sum);
//mean
double mean = (sum/count);
System.out.println("Mean: " + mean);
//min and max
double min = values[0];
double max = values[0];
for (int i=1; i<values.length; i++)
{
if (values[i] < min)
{
min = values[i];
}
if (values[i] > max)
{
max = values[i];
}
}
System.out.println("Minimum: " + min);
System.out.println("Maximum: " + max);
System.out.println("Range: " + (max-min));
//sorting
Arrays.sort(values);
//median
int position = (count+1) / 2;
if (count%2 != 0)
{
System.out.println("Median: " + values[position-1]);
}
else
{
System.out.println("Median: " + (( values[position-1] + values[position]) / 2));
}
//Standard deviation
double sigmap = 0; //for population
double sigmas = 0; //for sample
for (int i=0; i<values.length; i++)
{
sigmap += (values[i] - mean)*(values[i] - mean);
}
sigmas = sigmap;
//sd for population
sigmap = sigmap / count;
System.out.println("Variance of population: " + sigmap);
double sdp = Math.sqrt(sigmap);
System.out.println("Standard Deviation of population: " + sdp);
//sd for sample
sigmas = sigmas / (count - 1);
System.out.println("Variance of sample: " + sigmas);
double sds = Math.sqrt(sigmas);
System.out.println("Standard Deviation of sample: " + sds);
//Skewness
double sigma = 0;
for (int i=0; i<values.length; i++)
{
sigma += (values[i] - mean)*(values[i] - mean)*(values[i] - mean);
}
double N = (double) count;
double sk = (Math.sqrt(N*(N-1)) / (N-2)) * (sigma / (sdp*sdp*sdp*N));
System.out.println("Skewness: " + sk);
//mode
int repeated = -1;
int[] frq = new int[values.length];
for (int i=0; i<values.length; i++)
{
int occ = 1;
for (int j=i+1; j<values.length; j++)
{
if (values[i] == values[j])
{
frq[j] = repeated;
occ++;
}
}
if (frq[i] != repeated)
{
frq[i] = occ;
}
}
int maxFrq = frq[0];
for (int i=0; i<frq.length; i++)
{
if (frq[i] > maxFrq)
{
maxFrq = frq[i];
}
}
System.out.print("Mode: ");
if (maxFrq == 1)
{
System.out.println("N/A");
}
else
{
for (int i=0; i<frq.length; i++)
{
if (frq[i] == maxFrq)
{
System.out.print(values[i] + " ");
}
}
System.out.print("n");
}
double standardError=sds/Math.sqrt(count);
System.out.println("Standard Error: "+standardError);
}
}
Programming Exercise #06_2:
Instruction:
1. Make a copy of your EX06_01.java file and rename it to EX06_2.java.
2. Open the EX06.
I need to create a data type that implement the following interface -.docx
I need to create a data type that implement the following interface ;
myNumber()
myNumber(int valueIn)
void printDecimal()
void printBinary()
String toString() //returns a string stating both the decimal and binary value
Solution
import java.util.*;
interface Myinterface
{
public void mynumber();
public void mynumber(int value);
public void printDecimal();
public void printbinary();
public void tostring();
}
class Sample implements Myinterface
{
public void mynumber()
{
int value=0;
System.out.println(\"implementation of mynumber()\");
}
public void mynumber(int value)
{
int a=value;
System.out.println(\"implementation of mynumber(int value):\"+a);
}
public void printDecimal()
{
int x=10;
System.out.println(\"Decimal number is:\"+x);
}
public void printbinary()
{
int x=10;
int arr[] = new int[100];
int i=0;
while(x>0)
{
arr[i]=x%2;
i++;
x=x/2;
}
int count=i-1;
System.out.println(\"Binary value :\");
for(i=count;i>=0;i--)
{
System.out.println(arr[i]);
}
}
public void tostring()
{
//toSting is used to return the string object value for the integer
int x=10;
int p=x;
int arr[] = new int[100];
int i=0;
while(x>0)
{
arr[i]=x%2;
i++;
x=x/2;
}
int count=i-1;
System.out.println(\"deimal is: \"+Integer.toString(p)); //implementing toSting object for decimal
System.out.println(\"Binary value :\");
for(i=count;i>=0;i--)
{
System.out.println(Integer.toString(arr[i]));
}
}
public static void main(String args[])
{
//defining objects for the interface and are called in the void main
Myinterface obj = new Sample();
obj.mynumber();
obj.mynumber(3);
obj.printDecimal();
obj.printbinary();
obj.tostring();
}
}
.
Write the program in MIPS that declares an array of positive integer.pdf
Write the program in MIPS that declares an array of positive integers. The size of the array is not
fixed. The final element of the array should have the value of 0xF, which is not used in
calculations, and should be used to calculate the size of the array. The program can prompt the
user for an integer input. Compare the input with the array. Display the index of the input in the
array if found. If the number does not match any element in the array, insert the number in the
array in sorted order. The program is terminated when input a negative integer. Print the new
array with appropriate message on the screen. For example assume an array as follows:
(array: .word 4, 5, 23, 5, 8, 3, 15, 67, 8, 9, 0xF)
Sorted array: 3, 4, 5, 5, 8, 8, 9, 15, 23, 67
Enter a number to search: 20
Number not found. Added to array.
Sorted array: 3, 4, 5, 5, 8, 8, 9, 15, 20, 23, 67
Enter a number to search: 5
Number at index 2
Solution
import java.util.Scanner;
public category Ascending _Order
for (int i = 0; i < n; i++)
temporary worker = a[i];
a[i] = a[j];
a[j] = temp;
}
}
}
System.out.print(\"Ascending Order:\");
for (int i = 0; i < n - 1; i++)
import java.util.Scanner;
public category Ascending _Order
for (int i = 0; i < n; i++)
temporary worker = a[i];
a[i] = a[j];
a[j] = temp;
}
}
}
System.out.print(\"Ascending Order:\");
for (int i = 0; i < n - 1; i++)
import java.util.Scanner;
public category Ascending _Order
for (int i = 0; i < n; i++)
temporary worker = a[i];
a[i] = a[j];
a[j] = temp;
}
}
}
System.out.print(\"Ascending Order:\");
for (int i = 0; i < n - 1; i++)
import java.util.Scanner;
public category Ascending _Order
for (int i = 0; i < n; i++)
temporary worker = a[i];
a[i] = a[j];
a[j] = temp;
}
}
}
System.out.print(\"Ascending Order:\");
for (int i = 0; i < n - 1; i++)
import java.util.Scanner;
public category Ascending _Order
for (int i = 0; i < n; i++)
temporary worker = a[i];
a[i] = a[j];
a[j] = temp;
}
}
}
System.out.print(\"Ascending Order:\");
for (int i = 0; i < n - 1; i++)
import java.util.Scanner;
public category Ascending _Order
for (int i = 0; i < n; i++)
temporary worker = a[i];
a[i] = a[j];
a[j] = temp;
}
}
}
System.out.print(\"Ascending Order:\");
for (int i = 0; i < n - 1; i++)
import java.util.Scanner;
public category Ascending _Order
for (int i = 0; i < n; i++)
temporary worker = a[i];
a[i] = a[j];
a[j] = temp;
}
}
}
System.out.print(\"Ascending Order:\");
for (int i = 0; i < n - 1; i++)
import java.util.Scanner;
public category Ascending _Order
for (int i = 0; i < n; i++)
temporary worker = a[i];
a[i] = a[j];
a[j] = temp;
}
}
}
System.out.print(\"Ascending Order:\");
for (int i = 0; i < n - 1; i++)
import java.util.Scanner;
public category Ascending _Order
for (int i = 0; i < n; i++)
temporary worker = a[i];
a[i] = a[j];
a[j] = temp;
}
}
}
System.out.print(\"Ascending Order:\");
for (int i = 0; i < n - 1; i++)
import java.util.Scanner;
public category Ascending _Order
for (int i = 0; i < n; i++)
temporary worker = a[i];
a[i] = a[j];
a.
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1. import java.util.Scanner; public class Alphabetical_Order {.pdf
Hi, Please find my codeimport java.util.Random;public class Pro.pdf
Hi, Please find my codeimport java.util.Random;public class Pro.pdf
ArrayTest.java import java.util.Arrays; import java.util.Scann.pdf
ArrayTest.java import java.util.Arrays; import java.util.Scann.pdf
Java programs - bubble sort, iterator, linked list, hash set, reverse string,...
Java programs - bubble sort, iterator, linked list, hash set, reverse string,...
SumNumbers.java import java.util.Scanner;public class SumNumbe.pdf
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I need to create a data type that implement the following interface -.docx
I need to create a data type that implement the following interface -.docx
Write the program in MIPS that declares an array of positive integer.pdf
Write the program in MIPS that declares an array of positive integer.pdf
More from anandhomeneeds
Question 1,2,4 ------------------------------------Please check.pdf
Question 1,2,4 :-
-----------------------------------
Please check comments in Programme
#include
/**
* Method for Linear Search
* Loop through given array and search for given key
* IF found return index and stop
* else -1 will be returned at the end
*
*/
int search(int array[],int key,int size){
int index = -1;
for (int c = 0; c < size; c++)
{
if (array[c] == key) /* if required element found */
{
printf(\"%d is present at location %d.\ \", key, c+1);
index = c;
break;
}
}
if(index == -1)
printf(\"%d is not found.\ \", key);
return index;
}
/**
* Function for Binary search
* Binary search is used only for sorted arrays
* Check key with element at middle position
* if key == then return middle positon
* if key > middle position , apply same for sub array range from [middle...last]
* if key < middle position , apply same for sub array range from [0..middle]
*
*/
int binary_search(int array[],int key,int start,int end){
int index = -1;
int first = start;
int last = end;
int middle = (first+last)/2;
while (first <= last) {
//Print array between first nad last for each step
printf(\"Start binary search step.\ \");
for(int i = first ; i < last ; i++){
printf(\"Binary search step element.%d\ \",array[i]);
}
printf(\"Stop binary search step.\ \");
if (array[middle] < key)
first = middle + 1;
else if (array[middle] == key) {
index = middle;
printf(\"%d found at location %d.\ \", key, middle+1);
break;
}
else
last = middle - 1;
middle = (first + last)/2;
}
return index;
}
/**
* Recursive linear search
* Instead of for loop call same method with next index
* first pass 0, check if key found at index 0, if not found call same method with index 1 .....
*/
int linear_search_recursive(int array[],int key,int size,int index){
if(array[index] == key){
printf(\"%d found at location %d.\ \",key, index);
return index;
}
else if (index >= size-1){
return -1;
}
else{
return linear_search_recursive(array,key,size,++index);
}
}
/**
* Recursive Binary Search
* Recursively call Binary Search , check key found at element middle
* if key == then return middle positon
* if key > middle position , call recursive method for sub array range from [middle...last]
* if key < middle position , call recursive method for sub array range from [0..middle]
*/
int binary_search_recursive(int array[],int key,int start,int end){
//Key not found
if(start> end){
return -1;
}
int index = -1;
int first = start;
int last = end;
int middle = (first+last)/2;
//Print array between first nad last for each step
printf(\"Start binary search step.\ \");
for(int i = first ; i < last ; i++){
printf(\"Binary search step element.%d\ \",array[i]);
}
printf(\"Stop binary search step.\ \");
if (array[middle] < key){
first = middle + 1;
return binary_search_recursive(array,key,first,last) ;
}else if (array[middle] == key) {
printf(\"%d found at location %d.\ \", key, middle+1);
return middle;
}
else{
last = middle - 1;
return binary_search_recursive(array,key,first,last) ;
}
return -1;
}
/**
*Function t.
The answer is C. securin.APC degrade securin to release separase .pdf
The answer is: C. securin.
APC degrade securin to release separase which degrades cohesin.
Solution
The answer is: C. securin.
APC degrade securin to release separase which degrades cohesin..
HiCan you please summarise the question. More information confuse .pdf
Hi
Can you please summarise the question. More information confuse to sole this question
Solution
Hi
Can you please summarise the question. More information confuse to sole this question.
What property of proteins us used to separate them in HICHydrophob.pdf
What property of proteins us used to separate them in HIC
Hydrophobic interaction chromatography (HIC) isolates Proteins in light of their hydrophobicity,
and is regularly utilized as a middle stride as a part of a cleaning plan. Proteins are bound to a
stationary stage in a high ionic quality cushion and, along these lines, HIC can commonly be
performed instantly after particle trade chromatography with no cradle trade or weakening
required. HIC is likewise usually performed after an ammonium sulfate precipitation, a strategy
that can be utilized to rapidly expel proteins by accelerating a few, yet not all, proteins with salt.
HIC is infrequently material in early strides of a purging plan or as a last stride in the expulsion
of follow polluting influences from the protein of intrigue.
====================================================
Described mobile and stationary phase in HIC
Hydrophobic association chromatography (HIC) is an other option to turned around stage
chromatography for abusing the hydrophobic properties of proteins.
The expansion of a salt to the mobile stage support and test arrangement advances protein-
medium connections. The proteins are adsorbed to the medium in a mobile stage containing a
high grouping of salt. The vast majority of the bound proteins are viably desorbed by essentially
washing with water or a weaken, close impartial support. Since HIC utilizes a more polar, less
denaturing environment than RPLC, it is getting to be well known for protein filtration, regularly
in blend with particle trade or gel filtration chromatography.
Solution
What property of proteins us used to separate them in HIC
Hydrophobic interaction chromatography (HIC) isolates Proteins in light of their hydrophobicity,
and is regularly utilized as a middle stride as a part of a cleaning plan. Proteins are bound to a
stationary stage in a high ionic quality cushion and, along these lines, HIC can commonly be
performed instantly after particle trade chromatography with no cradle trade or weakening
required. HIC is likewise usually performed after an ammonium sulfate precipitation, a strategy
that can be utilized to rapidly expel proteins by accelerating a few, yet not all, proteins with salt.
HIC is infrequently material in early strides of a purging plan or as a last stride in the expulsion
of follow polluting influences from the protein of intrigue.
====================================================
Described mobile and stationary phase in HIC
Hydrophobic association chromatography (HIC) is an other option to turned around stage
chromatography for abusing the hydrophobic properties of proteins.
The expansion of a salt to the mobile stage support and test arrangement advances protein-
medium connections. The proteins are adsorbed to the medium in a mobile stage containing a
high grouping of salt. The vast majority of the bound proteins are viably desorbed by essentially
washing with water or a weaken, close impartial support. Since HIC uti.
upload complete priblem somthing is missingSolutionupload comp.pdf
The document discusses an upload issue where the process completed but something was missing. No other details are provided about the upload, missing content, or possible solutions. The document repeats the same short statement twice without providing any useful information to understand or resolve the issue.
the phenomena responsible for this type of inheritancce is Epistasis.pdf
the phenomena responsible for this type of inheritancce is Epistasis. in this particular example
two recessive allele at one locus has the ability to affect the allele of other locus.say for example,
lets take first locus to be C and other to be P.
later on it was discovered that it is linked to the biochemistry of the pea plant.
Solution
the phenomena responsible for this type of inheritancce is Epistasis. in this particular example
two recessive allele at one locus has the ability to affect the allele of other locus.say for example,
lets take first locus to be C and other to be P.
later on it was discovered that it is linked to the biochemistry of the pea plant..
The substrate in urease test is urea.The end products in urease te.pdf
The substrate in urease test is urea.
The end products in urease test are urea, carbondioxide and water
(NH2)2 CO+2H2O CO2+ H2O + 2NH3
Urease test is used to determine the ability of an organism to hydrolyse urea into ammonia and
carbon dioxide. The ammonia formed during the reaction turns the media into alkaline, which is
indicated by phenol red, changing its color from light orange (at pH 6.8) to pink (at pH 8.1).
Rapid urease positive organisms like Helicobacter pylori, turns the medium pink within 24
hours. Weakly positive organisms take several days, whereas negative organisms produce no
colour.
Rapid Urease Test (RUT) is a popular and simple test which uses the above principle , for
detecting the presence of Helicobacter pylori on gastric mucosa. The test is also known as CLO
test (Campylobacter like organism test). It is also useful in identifying several genera and species
of Enterobacteriaceae like Proteus spp, Klebsiella spp and Yersinia spp. Also for identifying
Cryptococcus spp and Brucella spp.
Solution
The substrate in urease test is urea.
The end products in urease test are urea, carbondioxide and water
(NH2)2 CO+2H2O CO2+ H2O + 2NH3
Urease test is used to determine the ability of an organism to hydrolyse urea into ammonia and
carbon dioxide. The ammonia formed during the reaction turns the media into alkaline, which is
indicated by phenol red, changing its color from light orange (at pH 6.8) to pink (at pH 8.1).
Rapid urease positive organisms like Helicobacter pylori, turns the medium pink within 24
hours. Weakly positive organisms take several days, whereas negative organisms produce no
colour.
Rapid Urease Test (RUT) is a popular and simple test which uses the above principle , for
detecting the presence of Helicobacter pylori on gastric mucosa. The test is also known as CLO
test (Campylobacter like organism test). It is also useful in identifying several genera and species
of Enterobacteriaceae like Proteus spp, Klebsiella spp and Yersinia spp. Also for identifying
Cryptococcus spp and Brucella spp..
later we create a specialstack class ,which inherits from stack cl.pdf
//later we create a specialstack class ,which inherits from stack class.there we return the smallest
integer
//implementation
#include
#include
using namespace std;
/* A simple stack class named IntStack with push and pop operations */
class IntStack
{
private:
static const int max = 100;
int arr[max];
int top;
public:
//constructor for initialising value of top
IntStack()
{
top = -1;
}
void push(int x);
int pop();
bool isEmpty();
bool isFull();
};
/* Stack\'s member method to check if the stack is iempty */
bool IntStack::isEmpty()
{
if(top == -1)
return true;
return false;
}
/* Stack\'s member method to check if the stack is full */
bool IntStack::isFull()
{
if(top == max - 1)
return true;
return false;
}
/* Stack\'s member method push() to insert an element */
void IntStack::push(int x)
{
if(isFull())
{
cout<<\"Stack Overflow\";
}
top++;
arr[top] = x;
}
/* Stack\'s member method pop() to remove an element from it */
int IntStack::pop()
{
if(isEmpty())
{
cout<<\"Stack Underflow\";
abort();
}
int x = arr[top];
top--;
return x;
}
/* A class that supports all the stack operations and one additional operation getMIN() that
returns the minimum element from stack at any time. This class inherits from the stack class and
uses an auxiliarry stack that holds minimum elements */
class SpecialStack: public IntStack
{
IntStack min;
public:
//function declarations
void push(int x);
int pop();
int getMIN();
};
/* SpecialStack\'s member method to insert an element to it. This method
makes sure that the min stack is also updated with appropriate minimum
values */
void SpecialStack::push(int x)
{
if(isEmpty()==true)
{
IntStack::push(x);
min.push(x);
}
else
{
IntStack::push(x);
int y = min.pop();
min.push(y);
/* push only when the incoming element of main stack is smaller
than or equal to top of auxiliary stack */
if( x <= y )
min.push(x);
}
}
/* SpecialStack\'s member method to remove an element from it. This method
removes top element from min stack also. */
int SpecialStack::pop()
{
int x = IntStack::pop();
int y = min.pop();
/* Push the popped element y back only if it is not equal to x */
if ( y != x )
min.push(y);
return x;
}
/* SpecialStack\'s member method to get minimum element from it. */
int SpecialStack::getMIN()
{
int x = min.pop();
min.push(x);
return x;
}
/* main program to test SpecialStack methods */
//execution starts here
int main()
{
SpecialStack s;
//push 3 elements and get the minimum
s.push(15);
s.push(25);
s.push(33);
cout<
#include
using namespace std;
/* A simple stack class named IntStack with push and pop operations */
class IntStack
{
private:
static const int max = 100;
int arr[max];
int top;
public:
//constructor for initialising value of top
IntStack()
{
top = -1;
}
void push(int x);
int pop();
bool isEmpty();
bool isFull();
};
/* Stack\'s member method to check if the stack is iempty */
bool IntStack::isEmpty()
{
if(top == -1)
return true;
return false;
}
/* Stack\'s member method to check if the stack is full */
bool IntStac.
I think that 34SolutionI think that 34.pdf
This very short document does not provide enough context or information to generate a meaningful 3 sentence summary. It contains a statement about the number 34 but does not elaborate on the topic or point being discussed.
D(ANSWER)Polycrystalline or multicrystalline materials, or polycry.pdf
D(ANSWER)
Polycrystalline or multicrystalline materials, or polycrystals are solids that are composed of
many crystallites of varying size and orientation. Crystallites are also referred to as grains. They
are small or even microscopic crystals and form during the cooling of many materials. Their
orientation can be random with no preferred direction, called random texture, or directed,
possibly due to growth and processing conditions. Fiber texture is an example of the latter. The
areas where crystallite grains meet are known as grain boundaries.
Grain boundaries are interfaces where crystals of different orientations meet. A grain boundary is
a single-phase interface, with crystals on each side of the boundary being identical except in
orientation. The term \"crystallite boundary\" is sometimes, though rarely, used. Grain boundary
areas contain those atoms that have been perturbed from their original lattice sites, dislocations,
and impurities that have migrated to the lower energy grain boundary.
Treating a grain boundary geometrically as an interface of a single crystal cut into two parts, one
of which is rotated, we see that there are five variables required to define a grain boundary. The
first two numbers come from the unit vector that specifies a rotation axis. The third number
designates the angle of rotation of the grain. The final two numbers specify the plane of the grain
boundary (or a unit vector that is normal to this plane).
Grain boundaries disrupt the motion of dislocations through a material. Dislocation propagation
is impeded because of the stress field of the grain boundary defect region and the lack of slip
planes and slip directions and overall alignment across the boundaries. Reducing grain size is
therefore a common way to improve strength, often without any sacrifice in toughness because
the smaller grains create more obstacles per unit area of slip plane. This crystallite size-strength
relationship is given by the Hall-Petch relationship. The high interfacial energy and relatively
weak bonding in grain boundaries makes them preferred sites for the onset of corrosion and for
the precipitation of new phases from the solid.
Grain boundary migration plays an important role in many of the mechanisms of creep. Grain
boundary migration occurs when a shear stress acts on the grain boundary plane and causes the
grains to slide. This means that fine-grained materials actually have a poor resistance to creep
relative to coarser grains, especially at high temperatures, because smaller grains contain more
atoms in grain boundary sites. Grain boundaries also cause deformation in that they are sources
and sinks of point defects. Voids in a material tend to gather in a grain boundary, and if this
happens to a critical extent, the material could fracture.
During grain boundary migration, the rate determining step depends on the angle between two
adjacent grains. In a small angle dislocation boundary, the migration rate depends on vac.
1.It is the process of directing the television assistances are tran.pdf
1.It is the process of directing the television assistances are transported with the help of intrenet
protocol insdicating a packed switched network like local area network.
2.This iptv provides a power to bourn the media in tiny batches.moreover it provides a com-pell
alternative as a commanding method to appart the content whe help of digital rights
management.
3.Selecting a iptv lattice is depending upon the multiples containing service portfilo and network
investment .
4.super head end : It is brain of iptv system.this is the place where broadcasting channels are
received and transmited through the sources like fibre feeds,satelites.the video demand storage
content is present at this stage.
5.Coustmer premise equipement:It is joined with home wiring to produce the home network.it
contains a residential gate which is used to handle the control request from set top box.originally
the cost of home wiring is costly,service providers extremely prefers to rewire the homes at the
time of begining installation.
6.Core network: It is the middle part of a communication system offering a interconnection shift
between the boarder networks.it is utilized to offer a soloitary connection.
7.headend end :It is a slice of television system.modulators demodulators processors and
antennna....etc apparatus are used at this headend.
protocols:
1.This iptv layers on multicast ans unicast.the protocols that are utilized in iptv are as below.
1.service based provider streaming
2.web based unicast only live and vod
3.web based multicast live and unicast vod streaming
4.joined tv\'s set top boxes
Devices
The devices utilized in internet protocol telivision are as below
1.android
2.mobiles
3.smart tvs
4.spark
5.royal
6.sun price ................etc
Features specific to television services
1.Time shifting :A combo of iptv and digital video recorder allows the time shifting of
scheduling theme.
2.Personalization:End-to-end iptv methodology braces the bidirectional communicationa and
permits the end users to indidualize their tv watching atitude by permiting the users to mark as
per their mood and what they want to see.
3.bandwidth requirements:This iptv system supports as per the channel requested by user.hence
the iptv system allows the cable operators in order to preserve the bandwidth.
4.Support to interact tv:This iptv system offers two -way proficiency to perit the network
operators to carry a complete output interactive of tv applications like games,internet browsing.
Solution
1.It is the process of directing the television assistances are transported with the help of intrenet
protocol insdicating a packed switched network like local area network.
2.This iptv provides a power to bourn the media in tiny batches.moreover it provides a com-pell
alternative as a commanding method to appart the content whe help of digital rights
management.
3.Selecting a iptv lattice is depending upon the multiples containing service portfilo and network
investment .
4.super head en.
Answer The major objective of implementing a health information man.pdf
Answer: The major objective of implementing a health information management system is to
automate and streamline the process of capturing, processing, maintaining/storing, and analysing
the information related to health and health case. This information is just not related to one thing,
but covers many aspects and involves many stakeholders. It is related to patient\'s profile
(personal as well as medical), treatment prescribed to him/her, doctor\'s and health officials\'
profile, other relevant health records like details and results of various tests (if performed),
accounting, health related human resources among many other things.
Out of these, the most important part is - information related to patient\'s profile, treatment
prescribed, and other relevant health records. Collectively, this is termed as Electronic Health
Records (EHRs) which lies at the core of any HIMS. The nature and kind of information that is
present in EHRs is very personal and sensitive. Hence, ensuring security, privacy and
confidentiality of EHRs in HMIS should be utmost concern because if this information is
breached or comromised, its after-effects can be desasterous.
Hence, while implementing any HMIS, it is very critical to think of required and relevant
security and data protection mechanisms and make provisions for them into the system.
Solution
Answer: The major objective of implementing a health information management system is to
automate and streamline the process of capturing, processing, maintaining/storing, and analysing
the information related to health and health case. This information is just not related to one thing,
but covers many aspects and involves many stakeholders. It is related to patient\'s profile
(personal as well as medical), treatment prescribed to him/her, doctor\'s and health officials\'
profile, other relevant health records like details and results of various tests (if performed),
accounting, health related human resources among many other things.
Out of these, the most important part is - information related to patient\'s profile, treatment
prescribed, and other relevant health records. Collectively, this is termed as Electronic Health
Records (EHRs) which lies at the core of any HIMS. The nature and kind of information that is
present in EHRs is very personal and sensitive. Hence, ensuring security, privacy and
confidentiality of EHRs in HMIS should be utmost concern because if this information is
breached or comromised, its after-effects can be desasterous.
Hence, while implementing any HMIS, it is very critical to think of required and relevant
security and data protection mechanisms and make provisions for them into the system..
The key assumptions made by the Hadoop Distributed File System(HDFS).pdf
The key assumptions made by the Hadoop Distributed File System(HDFS) approach are:
1. High Volume: The volume of data in HDFS is expected to be really large, some like being
measured in terabytes or even petabytes.
2. High Availability: The cluster can contain thousands of file servers, and some of them may be
down(unavailable) at some given instance.
3. Reliable Communication: Communication is reliable among working machines.
4. The filesystem is not designed for interactive operations, instead, we prefer high throughput
over low latency.
5. Servers are distributed among racks and data centers, to ensure redundancy, and therefore
availability.
Solution
The key assumptions made by the Hadoop Distributed File System(HDFS) approach are:
1. High Volume: The volume of data in HDFS is expected to be really large, some like being
measured in terabytes or even petabytes.
2. High Availability: The cluster can contain thousands of file servers, and some of them may be
down(unavailable) at some given instance.
3. Reliable Communication: Communication is reliable among working machines.
4. The filesystem is not designed for interactive operations, instead, we prefer high throughput
over low latency.
5. Servers are distributed among racks and data centers, to ensure redundancy, and therefore
availability..
Properties areMeanModeMedianQuartilesPercentilesRan.pdf
Properties are:
Mean
Mode
Median
Quartiles
Percentiles
Range
Mean = Sum of values / n
Standard Variation = Sqrt( variance )
Variance = Sum ( X - Mean )^2 / n
Median = Middle Most Observation
Mode = Most Repeated Observation
Range = Maximum - Minimum
Solution
Properties are:
Mean
Mode
Median
Quartiles
Percentiles
Range
Mean = Sum of values / n
Standard Variation = Sqrt( variance )
Variance = Sum ( X - Mean )^2 / n
Median = Middle Most Observation
Mode = Most Repeated Observation
Range = Maximum - Minimum.
MgCO3 Increasing acidity will cause the following reaction Mg.pdf
MgCO3:
Increasing acidity will cause the following reaction:
MgCO3(s) + 2H+ ==> Mg(2+) + H2O + CO2(g)
and thus the solubility will increase.
AgI:
Increasing acidity will NOT affect the solubility.
CuS:
Increasing acidity will cause the following reaction:
CuS(s) + 2H+ ==> Cu(2+) + H2S(g)
and thus the solubility will increase.
PbBr2:
Increasing acidity will NOT affect the solubility.
Ba(OH)2
The solubility of Ba(OH)2 in water is only 5.6 g/100 g. By acidifying the solution, the following
reaction take place:
H+ + OH- ==> H2O
and thus [OH-] would drop, and thus Ba(OH)2 solubility would be increased.
Solution
MgCO3:
Increasing acidity will cause the following reaction:
MgCO3(s) + 2H+ ==> Mg(2+) + H2O + CO2(g)
and thus the solubility will increase.
AgI:
Increasing acidity will NOT affect the solubility.
CuS:
Increasing acidity will cause the following reaction:
CuS(s) + 2H+ ==> Cu(2+) + H2S(g)
and thus the solubility will increase.
PbBr2:
Increasing acidity will NOT affect the solubility.
Ba(OH)2
The solubility of Ba(OH)2 in water is only 5.6 g/100 g. By acidifying the solution, the following
reaction take place:
H+ + OH- ==> H2O
and thus [OH-] would drop, and thus Ba(OH)2 solubility would be increased..
Environmental conditions play a key role in defining the function an.pdf
Environmental conditions play a key role in defining the function and distribution of plants, in
combination with other factors. Changes in long term environmental conditions that can be
collectively coined climate change are known to have had enormous impacts on plant diversity
patterns in the future and are seen as having significant current impacts. It is predicted that
climate change will remain one of the major drivers of biodiversity patterns in the future.
The Earth has experienced a constantly changing climate in the time since plants first evolved. In
comparison to the present day, this history has seen Earth as cooler, warmer, drier and wetter,
and CO2 (carbon dioxide) concentrations have been both higher and lower. These changes have
been reflected by constantly shifting vegetation, for example forest communities dominating
most areas in interglacial periods, and herbaceous communities dominating during glacial
periods. It has been shown that past climatic change has been a major driver of the processes of
speciation and extinction. The best known example of this is the Carboniferous Rainforest
Collapse which occurred 350 million years ago. This event decimated amphibian populations
and spurred on the evolution of reptiles
Greenhouse effect - the mechanism
The sun radiates solar energy on earth. The larger part of this energy (45%) is radiated back into
space. Greenhouse gases in the atmosphere contribute to global warming by adsorption and
reflection of atmospheric and solar energy. This natural phenomenon is what we call the
greenhouse effect. It is agreed that the greenhouse effect is correlated with global temperature
change. The primary greenhouse gases in the Earth’s atmosphere are water vapor, carbon
dioxide, methane, nitrous oxide, and ozone
Mechanism of global warming:
i. The incoming radiation from the Sun is mostly in the form of visible light and nearby
wavelengths, largely in the range 0.2 – 4 1m, corresponding to the Sun’s radioactive temperature
of 6,000 K. Almost half the radiation is in the form of “visible” light, which our eyes are adapted
to use.
ii. About 50% of the Sun’s energy is absorbed at the earth’s surface and the rest is reflected or
absorbed by the atmosphere. The reflection of light back into space – largely by clouds – does
not much affect the basic mechanism; this light, effectively, is lost to the system.
iii (a) Earth absorbs most of the sunlight it receives; The absorbed energy warms the surface; (b)
Earth then emits the absorbed light’s energy as infrared light; (c) greenhouse gases absorb a lot
of the infrared light before it can leave our atmosphere; (d) being absorbed slows the rate at
which energy escapes to space; and (e) the slower passage of energy heats up the atmosphere,
water, and ground. By increasing the amount of greenhouse gases in the atmosphere, humans are
increasing the atmosphere’s absorption of infrared light, thereby warming Earth and disrupting
global climate patterns.
Effec.
1. Autralopith is a known genus of hominids which is extinct today..pdf
1. Autralopith is a known genus of hominids which is extinct today.
This can be A. biosei species. This genus dates back to 2.3-1.2 million years. Teeth and bones
examination is the major way to know what exactly is the age of the skeleton.
The proof of them being bipedal is that their foot imprints match to that of the present day
humans. Pelvis, femur and foot bones should be examined to confirm that the organism used to
walk bipedally.
Long arms and sloping fave are the ape like characters.
It was an adult. This can be known by examining the layers deposited on the body which may
come from volcanic eruption. Topmost layer and lowermost layers can be examined.
Solution
1. Autralopith is a known genus of hominids which is extinct today.
This can be A. biosei species. This genus dates back to 2.3-1.2 million years. Teeth and bones
examination is the major way to know what exactly is the age of the skeleton.
The proof of them being bipedal is that their foot imprints match to that of the present day
humans. Pelvis, femur and foot bones should be examined to confirm that the organism used to
walk bipedally.
Long arms and sloping fave are the ape like characters.
It was an adult. This can be known by examining the layers deposited on the body which may
come from volcanic eruption. Topmost layer and lowermost layers can be examined..
Mobile IP1) Mobile IP(MIP) is an Internet Engineering Task Force(.pdf
Mobile IP:
1) Mobile IP(MIP) is an Internet Engineering Task Force(IETF) standard communications
protocol.
2) This is designed to allow mobile device users to move from one network to another while
maintaining a permanent IP address.
3) Mobile IP for IPv4 is described in IETF RFC 5944 and extensions are defined in IETF RFC
4721.
4) Mobile IPv6 the IP mobility implementation for the next generation of the Internet Protocol
IPv6 is described in RFC 6275.
Advantages of using Mobile IP
• It allows fast, continuous lowcost access to corporate networks in remote areas where there
is no public telephone system or cellular coverage.
• It supports a wide range of applications from Internet access and email to ecommerce.
• Users can be permanently connected to their Internet provider and charged only for the data
packets that are sent and received.
• Lower equipment and utilization costs for those requiring reliable highspeed data connections
in remote locations worldwide.
• A user can take a palmtop or laptop computer anywhere without losing the connection to the
home network.
• Mobile IP finds local IP routers and connects automatically. It is phonejack and wirefree.
• Other than mobile nodes/routers, the remaining routers and hosts will still use current IP.
Mobile IP leaves transport and higher protocols
unaffected.
• Authentication is performed to ensure that rights are being protected.
• Mobile IP can move from one type of medium to another without losing connectivity. It is
unique in its ability to accommodate heterogeneous
mobility in addition to homogenous mobility.
Disadvantage of Mobile IP
• There is a routing inefficiency problem caused by the “triangle routing” formed by the home
agent, correspondent host and the foreign agent.
• Security risks are the most important problem facing Mobile IP. Besides the traditional security
risks with
IP, one has to worry about faked careof addresses.
• Another issue related to the security is how to make Mobile IP coexist with the security
features coming in
use within the Internet.
Characteristics of Mobile IP (Baseline Requirements)
• Compatibility: A new standard cannot require changes for applications or network protocols
already in use.
• Transparency: Mobility should remain “invisible” for many higher layer protocols and
applications.
• Scalability and efficiency: Introducing a new mechanism into the Internet must not degrade the
efficiency of the network.
• Security: All messages used to transmit information to another node about the location of a
mobile node must be authenticated to protect against remote redirection attacks
The Mobile IP in Wireless Networks
• new proposed standard of the IETF designed to support mobile users .
• new Internet standard for the Web and private networks.
• currently two standards one to support the current IPv4 one for the upcoming IPv6.
• One of the big requirements on the new standard is that it should support both ordinary and
wireless network
• The big issue is.
Answer1. Cells can generate many different tissue types endoderm,.pdf
Answer
1. Cells can generate many different tissue types: endoderm, mesoderm or ectoderm
2. Cells can generate any type of mesoderm tissue.
3. Cells can generate any type of muscle.
4. cells can generate heart muscle.
Solution
Answer
1. Cells can generate many different tissue types: endoderm, mesoderm or ectoderm
2. Cells can generate any type of mesoderm tissue.
3. Cells can generate any type of muscle.
4. cells can generate heart muscle..
3 a) The trace of an nSolution3 a) The trace of an n.pdf
The document discusses the trace of a matrix. The trace of an n x n matrix is defined as the sum of its diagonal entries. To find the trace, we add the entries along the main diagonal of the matrix from top left to bottom right. The trace provides some useful information about the matrix.
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- 1. //Driver.java import java.util.Scanner; import java.text.DecimalFormat; public class Driver { public static void printSequence (int num) { if(num > 0) { for (int i = 0; i <= num ;i++ ) { System.out.print(i + " "); } System.out.println(); } else if(num < 0) { for (int i = 0; i >= num ;i-- ) { System.out.print(i + " "); } System.out.println(); } } public static int trianglarNum (int num) { int sum = 0; if(num > 0) { for (int i = 1; i <= num ;i++ ) { sum = sum + i; } return sum;
- 2. } else return -1; } public static void printArray(int list []) { for (int i = 0; i < list.length ;i++ ) { System.out.print(list[i] + " "); } System.out.println(); } public static int range(int list []) { int max,min; max = list[0]; min = list[0]; for (int i = 1; i < list.length ;i++ ) { if(list[i] > max) max = list[i]; if(list[i] < min) min = list[i]; } System.out.println("difference between the maximum and minimum values in the array: "+ (max-min)); return (max-min); } public static void main(String[] args) { printSequence(10); printSequence(-10); System.out.println(); if(trianglarNum(10) > 0) System.out.println("Sum: "+ trianglarNum(10)); else
- 3. System.out.println("Negative input "); if(trianglarNum(-10) > 0) System.out.println("Sum: "+ trianglarNum(-10)); else System.out.println("Negative input "); int array[] = {1,2,4,5,-4,7,8,11,5,7}; printArray(array); System.out.println(); System.out.println("difference between the maximum and minimum values in the array: "+ range(array)); } } /* output: 0 1 2 3 4 5 6 7 8 9 10 0 -1 -2 -3 -4 -5 -6 -7 -8 -9 -10 Sum: 55 Negative input 1 2 4 5 -4 7 8 11 5 7 difference between the maximum and minimum values in the array: 15 difference between the maximum and minimum values in the array: 15 */ Solution //Driver.java import java.util.Scanner; import java.text.DecimalFormat; public class Driver { public static void printSequence (int num) { if(num > 0) {
- 4. for (int i = 0; i <= num ;i++ ) { System.out.print(i + " "); } System.out.println(); } else if(num < 0) { for (int i = 0; i >= num ;i-- ) { System.out.print(i + " "); } System.out.println(); } } public static int trianglarNum (int num) { int sum = 0; if(num > 0) { for (int i = 1; i <= num ;i++ ) { sum = sum + i; } return sum; } else return -1; } public static void printArray(int list []) { for (int i = 0; i < list.length ;i++ ) { System.out.print(list[i] + " "); } System.out.println();
- 5. } public static int range(int list []) { int max,min; max = list[0]; min = list[0]; for (int i = 1; i < list.length ;i++ ) { if(list[i] > max) max = list[i]; if(list[i] < min) min = list[i]; } System.out.println("difference between the maximum and minimum values in the array: "+ (max-min)); return (max-min); } public static void main(String[] args) { printSequence(10); printSequence(-10); System.out.println(); if(trianglarNum(10) > 0) System.out.println("Sum: "+ trianglarNum(10)); else System.out.println("Negative input "); if(trianglarNum(-10) > 0) System.out.println("Sum: "+ trianglarNum(-10)); else System.out.println("Negative input "); int array[] = {1,2,4,5,-4,7,8,11,5,7}; printArray(array); System.out.println(); System.out.println("difference between the maximum and minimum values in the array: "+ range(array)); }
- 6. } /* output: 0 1 2 3 4 5 6 7 8 9 10 0 -1 -2 -3 -4 -5 -6 -7 -8 -9 -10 Sum: 55 Negative input 1 2 4 5 -4 7 8 11 5 7 difference between the maximum and minimum values in the array: 15 difference between the maximum and minimum values in the array: 15 */