Program to print all combination of size r in an array of size n.pdf
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// Program to print all combination of size r in an array of size n #include void combinationUtil(int arr[], int data[], int start, int end, int index, int r); // The main function that prints all combinations of size r // in arr[] of size n. This function mainly uses combinationUtil() void printCombination(int arr[], int n, int r) { // A temporary array to store all combination one by one int data[r]; // Print all combination using temprary array \'data[]\' combinationUtil(arr, data, 0, n-1, 0, r); } /* arr[] ---> Input Array data[] ---> Temporary array to store current combination start & end ---> Staring and Ending indexes in arr[] index ---> Current index in data[] r ---> Size of a combination to be printed */ void combinationUtil(int arr[], int data[], int start, int end, int index, int r) { // Current combination is ready to be printed, print it if (index == r) { if (data[0]+data[1]+data[2]+data[3]==24) { for (int j=0; j= r-index\" makes sure that including one element // at index will make a combination with remaining elements // at remaining positions for (int i=start; i<=end && end-i+1 >= r-index; i++) { data[index] = arr[i]; combinationUtil(arr, data, i+1, end, index+1, r); } } // Driver program to test above functions int main() { int arr[] = {1, 2, 3, 4, 5, 6, 7, 8, 9}; int r = 4; int n = sizeof(arr)/sizeof(arr[0]); printCombination(arr, n, r); } Solution // Program to print all combination of size r in an array of size n #include void combinationUtil(int arr[], int data[], int start, int end, int index, int r); // The main function that prints all combinations of size r // in arr[] of size n. This function mainly uses combinationUtil() void printCombination(int arr[], int n, int r) { // A temporary array to store all combination one by one int data[r]; // Print all combination using temprary array \'data[]\' combinationUtil(arr, data, 0, n-1, 0, r); } /* arr[] ---> Input Array data[] ---> Temporary array to store current combination start & end ---> Staring and Ending indexes in arr[] index ---> Current index in data[] r ---> Size of a combination to be printed */ void combinationUtil(int arr[], int data[], int start, int end, int index, int r) { // Current combination is ready to be printed, print it if (index == r) { if (data[0]+data[1]+data[2]+data[3]==24) { for (int j=0; j= r-index\" makes sure that including one element // at index will make a combination with remaining elements // at remaining positions for (int i=start; i<=end && end-i+1 >= r-index; i++) { data[index] = arr[i]; combinationUtil(arr, data, i+1, end, index+1, r); } } // Driver program to test above functions int main() { int arr[] = {1, 2, 3, 4, 5, 6, 7, 8, 9}; int r = 4; int n = sizeof(arr)/sizeof(arr[0]); printCombination(arr, n, r); }.
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![// Program to print all combination of size r in an array of size n
#include
void combinationUtil(int arr[], int data[], int start, int end,
int index, int r);
// The main function that prints all combinations of size r
// in arr[] of size n. This function mainly uses combinationUtil()
void printCombination(int arr[], int n, int r)
{
// A temporary array to store all combination one by one
int data[r];
// Print all combination using temprary array 'data[]'
combinationUtil(arr, data, 0, n-1, 0, r);
}
/* arr[] ---> Input Array
data[] ---> Temporary array to store current combination
start & end ---> Staring and Ending indexes in arr[]
index ---> Current index in data[]
r ---> Size of a combination to be printed */
void combinationUtil(int arr[], int data[], int start, int end,
int index, int r)
{
// Current combination is ready to be printed, print it
if (index == r)
{
if (data[0]+data[1]+data[2]+data[3]==24)
{
for (int j=0; j= r-index" makes sure that including one element
// at index will make a combination with remaining elements
// at remaining positions
for (int i=start; i<=end && end-i+1 >= r-index; i++)
{
data[index] = arr[i];
combinationUtil(arr, data, i+1, end, index+1, r);
}
}](https://image.slidesharecdn.com/programtoprintallcombinationofsizerinanarrayofsizen-230414005658-d3394e2d/85/Program-to-print-all-combination-of-size-r-in-an-array-of-size-n-pdf-1-320.jpg)
![// Driver program to test above functions
int main()
{
int arr[] = {1, 2, 3, 4, 5, 6, 7, 8, 9};
int r = 4;
int n = sizeof(arr)/sizeof(arr[0]);
printCombination(arr, n, r);
}
Solution
// Program to print all combination of size r in an array of size n
#include
void combinationUtil(int arr[], int data[], int start, int end,
int index, int r);
// The main function that prints all combinations of size r
// in arr[] of size n. This function mainly uses combinationUtil()
void printCombination(int arr[], int n, int r)
{
// A temporary array to store all combination one by one
int data[r];
// Print all combination using temprary array 'data[]'
combinationUtil(arr, data, 0, n-1, 0, r);
}
/* arr[] ---> Input Array
data[] ---> Temporary array to store current combination
start & end ---> Staring and Ending indexes in arr[]
index ---> Current index in data[]
r ---> Size of a combination to be printed */
void combinationUtil(int arr[], int data[], int start, int end,
int index, int r)
{
// Current combination is ready to be printed, print it
if (index == r)
{
if (data[0]+data[1]+data[2]+data[3]==24)](data:image/gif;base64,R0lGODlhAQABAIAAAAAAAP///yH5BAEAAAAALAAAAAABAAEAAAIBRAA7)
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Create an array of size 10 Assign values to the array Print the arra.docx
Create an array of size 10 Assign values to the array Print the array Swap the first and last elements in the array Print the array Shift all elements by one to the right Print the array Replace even elements with 0 Replace each element except the first and last by the larger of its two neighbors Print the array
Solution
/*Program to replace EVEN elements by 0 and ODD elements by 1.*/
#include <stdio.h>
/** funtion : readArray()
input : arr ( array of integer ), size
to read ONE-D integer array from standard input device (keyboard).
**/
void readArray(int arr[], int size)
{
int i =0;
printf(\"\ Enter elements : \ \");
for(i=0; i < size; i++)
{
printf(\"Enter arr[%d] : \",i);
scanf(\"%d\",&arr[i]);
}
}
/** funtion : printArray()
input : arr ( array of integer ), size
to display ONE-D integer array on standard output device (moniter).
**/
void printArray(int arr[], int size)
{
int i =0;
printf(\"\ Elements are : \");
for(i=0; i < size; i++)
{
printf(\"\ \\tarr[%d] : %d\",i,arr[i]);
}
printf(\"\ \");
}
/** funtion : replaceEvenOdd()
input : arr ( array of integer ), size
to replace EVEN elements by 0 and ODD elements by 1.
**/
void replaceEvenOdd(int arr[], int size)
{
int i=0;
for(i=0; i < size; i++)
{
if( arr[i] % 2 == 0 )
arr[i] = 0 ;
else
arr[i] = i;
}
}
int main()
{
int arr[10];
readArray(arr,10);
printf(\"\ Before replacement : \");
printArray(arr,10);
replaceEvenOdd(arr,10);
printf(\"\ After replacement : \");
printArray(arr,10);
return 0;
}
.
Write a program that obtains the execution time of selection sort, bu.pdf
Write a java program called Question39 that does the following: Gets input for temperature
Utilizing a branching statement: If temperature is 76-100, call method outputHot passing the
temperature input as an argument. If temperature is 0-39, call method outputCold passing the
temperature input as an argument. If temperature is 40 to 75, call method outputJustRight
passing the temperature input as an argument. If temperature is outside these ranges, output
“Temperature outside range” to the screen. Be precise, import modules, include comments,
prologue, etc. as needed.
Solution
Question39.java
import java.util.Scanner;
public class Question39 {
public static void main(String[] args) {
Scanner scan = new Scanner(System.in);
System.out.println(\"Enter temperature: \");
int temp = scan.nextInt();
if(temp >=76 && temp <=100){
outputHot(temp);
}
else if(temp >=0 && temp <=39){
outputCold(temp);
}
else if(temp >=40 && temp <=75){
outputJustRight(temp);
}
else{
System.out.println(\"Temperature outside range\");
}
}
public static void outputHot(int temp){
System.out.println(\"Temperature \"+temp+\" Hot \");
}
public static void outputCold(int temp){
System.out.println(\"Temperature \"+temp+\" Cold \");
}
public static void outputJustRight(int temp){
System.out.println(\"Temperature \"+temp+\" Just Right \");
}
}
Output:
Enter temperature:
45
Temperature 45 Just Right.
Merge Sort implementation in C++ The implementation for Mergesort gi.pdf
Merge Sort implementation in C++ The implementation for Mergesort given in Section 7.4 takes
an array as input and soils that array. At the beginning of Section 7.4 there is a simple
pseudocode implementation for sorting a linked list using Mergesort. Implement both a linked
list-based version of Mergesort and the array-based version of Mergesort. and compare and
analyze their running times.
Solution
/* Merger sort array-based version */
/* C program for Merge Sort */
#include
#include
// Merges two subarrays of arr[].
// First subarray is arr[l..m]
// Second subarray is arr[m+1..r]
void merge(int arr[], int l, int m, int r)
{
int i, j, k;
int n1 = m - l + 1;
int n2 = r - m;
/* create temp arrays */
int L[n1], R[n2];
/* Copy data to temp arrays L[] and R[] */
for (i = 0; i < n1; i++)
L[i] = arr[l + i];
for (j = 0; j < n2; j++)
R[j] = arr[m + 1+ j];
/* Merge the temp arrays back into arr[l..r]*/
i = 0; // Initial index of first subarray
j = 0; // Initial index of second subarray
k = l; // Initial index of merged subarray
while (i < n1 && j < n2)
{
if (L[i] <= R[j])
{
arr[k] = L[i];
i++;
}
else
{
arr[k] = R[j];
j++;
}
k++;
}
/* Copy the remaining elements of L[], if there
are any */
while (i < n1)
{
arr[k] = L[i];
i++;
k++;
}
/* Copy the remaining elements of R[], if there
are any */
while (j < n2)
{
arr[k] = R[j];
j++;
k++;
}
}
/* l is for left index and r is right index of the
sub-array of arr to be sorted */
void mergeSort(int arr[], int l, int r)
{
if (l < r)
{
// Same as (l+r)/2, but avoids overflow for
// large l and h
int m = l+(r-l)/2;
// Sort first and second halves
mergeSort(arr, l, m);
mergeSort(arr, m+1, r);
merge(arr, l, m, r);
}
}
/* UTILITY FUNCTIONS */
/* Function to print an array */
void printArray(int A[], int size)
{
int i;
for (i=0; i < size; i++)
printf(\"%d \", A[i]);
printf(\"\ \");
}
/* Driver program to test above functions */
int main()
{
int arr[] = {12, 11, 13, 5, 6, 7};
int arr_size = sizeof(arr)/sizeof(arr[0]);
printf(\"Given array is \ \");
printArray(arr, arr_size);
mergeSort(arr, 0, arr_size - 1);
printf(\"\ Sorted array is \ \");
printArray(arr, arr_size);
return 0;
}
--------------------------------------------------------------------------------------------------------------------
/* merge sort link-list version
#include
#include
/* Link list node */
struct node
{
int data;
struct node* next;
};
/* function prototypes */
struct node* SortedMerge(struct node* a, struct node* b);
void FrontBackSplit(struct node* source,
struct node** frontRef, struct node** backRef);
/* sorts the linked list by changing next pointers (not data) */
void MergeSort(struct node** headRef)
{
struct node* head = *headRef;
struct node* a;
struct node* b;
/* Base case -- length 0 or 1 */
if ((head == NULL) || (head->next == NULL))
{
return;
}
/* Split head into \'a\' and \'b\' sublists */
FrontBackSplit(head, &a, &b);
/* Recursively sort the sublists */
MergeSort(&a);
MergeSort(&b);
/* answer = merge the two sorted lists together */
*.
Write a C code to perform the following tasks In main Define an arr.pdf
Write a C code to perform the following tasks: In main: Define an array of SIZE = 7 and
initialize them with 3, 5, 7, 9, 11, 13, 15 Call function sum with the array and size, both passed in
pointer representations Print each element of the array Print the sum In function sum: Take the
parameters in pointer representations,
Solution
// C code
#include
#include
#include
#include
// sum function using pointers to add elements of array
int sumArray(int *array, int *SIZE)
{
int i;
int sum = 0;
for (i = 0; i < *SIZE; ++i)
{
sum = sum + *(array+i);
}
return sum;
}
int main()
{
int SIZE = 7;
int array[] = {3,5,7,9,11,13,15};
// ddeclare and initialize pointer
int *ptr;
ptr = array;
// pass array pointer and size of array to function
int sum = sumArray(ptr,&SIZE);
printf(\"Array: \");
int i;
for (i = 0; i < SIZE; ++i)
{
printf(\"%d \",array[i]);
}
printf(\"\ \");
printf(\"Sum of array: %d\ \",sum);
return 0;
}
/*
output:
Array: 3 5 7 9 11 13 15
Sum of array: 63
*/.
Write a program to implement and test the following sorting algorithm.docx
The document describes implementing and testing three sorting algorithms - insertion sort, heapsort, and shellsort - on arrays of 1,000 and 10,000 elements. It provides code samples to perform each algorithm and print the number of comparisons and item movements for analysis. The sorting algorithms are tested on sample arrays and the sorted output is displayed.
Program 1 (Practicing an example of function using call by referenc.pdf
Program 1: (Practicing an example of function using call by reference)
Write a program that reads a set of information related to students in C++ class and prints them
in a table format. The information is stored in a file called data.txt. Each row of the file contains
student number, grade for assignment 1, grade for assignment 2, grade for assignment 3, and
grade for assignment 4. Your main program should read each row, pass the grades for the three
assignments to a function called ProcessRow to calculate the average of the grades, minimum of
the four assignments, and the maximum of the four assignments. The results (average,
maximum, and minimum) should be returned to the main program and the main program prints
them on the screen in a table format. For example, if the file includes
126534 7 8 10 7
321345 5 6 4 9
324341 8 3 8 5
your program should print
Std-Id A1 A2 A3 A4 Min Max Average
----------------------------------------------------------------------------------
126534 7 8 10 7 7 10 8
321345 5 6 4 9 4 9 6
324341 8 3 8 5 3 8 6
You must use call by reference to do the above question.
_________________________________________________________________
Program 2: (Practicing an example of function using call by value)
Repeat the above question using call by value. This means you need to have three different
functions: one to calculate the average, another to calculate the minimum, and the third one to
calculate the maximum. This is how to call these functions:
Max = CalculateMax(A1,A2,A3, A4);
Min = CalculateMin(A1,A2,A3, A4);
Average = CalculateAvg(A1,A2,A3, A4);
_________________________________________________________________
Program 3:
Write a program with several functions that performs the following tasks. :
Create a function that reads the following 5 float numbers from the file data.txt into array Called
Arr1.
12.0 15.0 29.3 25.0 93.2
Copy array Arr1 into array Arr2 in reverse order
Print array Arr1.
Print array Arr2.
Find the number of elements in array Arr1 that are >= 80 and <=100.
Find the number of the elements in array Arr1 in which their contents are divisible by 5
Find the index of the elements in array Arr1 in which their contents are divisible by 3.
Find mean (average) in array Arr1.
Find the maximum number in array Arr1.
Ask the user to input a key. Then search for the key in array Arr1 and inform the user about the
existence (true / false) of the key in array
Your program should include several functions.
A function for filling up the information from file into an array (part a should call this function)
A function that does the copying of one array into another in reverse order. Arrays must have the
same size (part b should call this function)
Printing any array with any size (part c and d should call this function)
finding and returning the number of elements between 80 and 100 in any array with any size
(part e calls this function)
finding and returning the number of elements in an array that are divisible by 5 (par.
#include iostream using namespace std; void InsertionSort(int .pdf
#include
using namespace std;
void InsertionSort(int arr[],int size){
int temp,j;
for(int i=0; i=0){
arr[j+1] = arr[j];
j--;
}
arr[j+1] = temp;
}
}
// Merges two subarrays of arr[].
// First subarray is arr[l..m]
// Second subarray is arr[m+1..r]
void merge(int arr[], int l, int m, int r)
{
int i, j, k;
int n1 = m - l + 1;
int n2 = r - m;
/* create temp arrays */
int L[n1], R[n2];
/* Copy data to temp arrays L[] and R[] */
for (i = 0; i < n1; i++)
L[i] = arr[l + i];
for (j = 0; j < n2; j++)
R[j] = arr[m + 1+ j];
/* Merge arrays back into arr[l..r]*/
i = 0; // first subarray index
j = 0; // second subarray index
k = l; // merged subarray index
while (i < n1 && j < n2)
{
if (L[i] <= R[j])
{
arr[k] = L[i];
i++;
}
else
{
arr[k] = R[j];
j++;
}
k++;
}
/* Copy the remaining elements of L[], if there
are any */
while (i < n1)
{
arr[k] = L[i];
i++;
k++;
}
/* Copy the remaining elements of R[], if there
are any */
while (j < n2)
{
arr[k] = R[j];
j++;
k++;
}
}
/* l is for left index and r is right index of the
sub-array of arr to be sorted */
void mergeSort(int arr[], int l, int r)
{
if (l < r)
{
// Same as (l+r)/2, but avoids overflow for
// large l and h
int m = l+(r-l)/2;
// Sort first and second halves
mergeSort(arr, l, m);
mergeSort(arr, m+1, r);
merge(arr, l, m, r);
}
}
void printArray(int arr[], int size){//Function to print array
int i;
for (i=0; i < size; i++)
cout << arr[i] <<\" \";
cout << endl;
}
int main(){
int size = 20;
int array1[] = {17,33,88,54,96,82,12,36,69,74,79,66,32,11,1,98,76,55,3,4};
int array2[] = {20,1,19,2,18,3,17,4,16,5,15,6,14,7,13,8,12,9,11,10};
int array3[] = {220,209,198,187,176,165,154,143,132,121,110,99,88,77,66,55,44,33,22,11};
int array4[] = {25,27,23,2,7,99,96,1996,85,3,24,11,17,4,1,0,8,9,98,45};
int array5[] = {15,35,85,100,40,90,70,65,25,5,10,30,55,75,80,95,20,45,50,60};
//Output for array1
cout << \"Original array 1: \ \";
printArray(array1, size);
InsertionSort(array1, size);
cout << \"Array 1 sorted by insertion sort: \ \";
printArray(array1, size);
//CALL THE MERGE SORT HERE
//PRINT THE MERGE SORT HERE
cout << \"\ \";
//Output for array2
cout << \"Original array 2: \ \";
printArray(array2, size);
mergeSort(array2,0,size);
cout << \"Array 2 sorted by Merge sort: \ \";
printArray(array2, size);
cout << \"\ \";
//Output for array3
cout << \"Original array 3: \ \";
printArray(array3, size);
InsertionSort(array3, size);
cout << \"Array 3 sorted by insertion sort: \ \";
printArray(array3, size);
cout << \"\ \";
//Output for array4
cout << \"Original array 4: \ \";
printArray(array4, size);
mergeSort(array4,0,size);
cout << \"Array 4 sorted by Merge sort: \ \";
printArray(array4, size);
cout << \"\ \";
//Output for array5
cout << \"Original array 5: \ \";
printArray(array5, size);
InsertionSort(array5, size);
cout << \"Array 5 sorted by insertion sort: \ \";
printArray(array5, size);
cout << \"\ \";
return 0;
}
/* sample output
Note: As sorting an already sorted array does not result good. I have changed the array2,a.
SlideSet_4_Arraysnew.pdf
The document discusses arrays in C programming. Some key points include:
- An array is a collection of variables of the same type referred to by a common name. Each element has an index and arrays use contiguous memory locations.
- Arrays are declared with the type, name, and size. The first element is at index 0.
- One-dimensional arrays can be initialized, accessed, input from and output to the user. Multidimensional arrays like 2D arrays represent tables with rows and columns.
- Arrays can be passed to functions by passing the entire array or individual elements. Operations like searching, sorting and merging can be performed on arrays.
Recommended
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Solution
/*Program to replace EVEN elements by 0 and ODD elements by 1.*/
#include <stdio.h>
/** funtion : readArray()
input : arr ( array of integer ), size
to read ONE-D integer array from standard input device (keyboard).
**/
void readArray(int arr[], int size)
{
int i =0;
printf(\"\ Enter elements : \ \");
for(i=0; i < size; i++)
{
printf(\"Enter arr[%d] : \",i);
scanf(\"%d\",&arr[i]);
}
}
/** funtion : printArray()
input : arr ( array of integer ), size
to display ONE-D integer array on standard output device (moniter).
**/
void printArray(int arr[], int size)
{
int i =0;
printf(\"\ Elements are : \");
for(i=0; i < size; i++)
{
printf(\"\ \\tarr[%d] : %d\",i,arr[i]);
}
printf(\"\ \");
}
/** funtion : replaceEvenOdd()
input : arr ( array of integer ), size
to replace EVEN elements by 0 and ODD elements by 1.
**/
void replaceEvenOdd(int arr[], int size)
{
int i=0;
for(i=0; i < size; i++)
{
if( arr[i] % 2 == 0 )
arr[i] = 0 ;
else
arr[i] = i;
}
}
int main()
{
int arr[10];
readArray(arr,10);
printf(\"\ Before replacement : \");
printArray(arr,10);
replaceEvenOdd(arr,10);
printf(\"\ After replacement : \");
printArray(arr,10);
return 0;
}
.
Write a program that obtains the execution time of selection sort, bu.pdf
Write a java program called Question39 that does the following: Gets input for temperature
Utilizing a branching statement: If temperature is 76-100, call method outputHot passing the
temperature input as an argument. If temperature is 0-39, call method outputCold passing the
temperature input as an argument. If temperature is 40 to 75, call method outputJustRight
passing the temperature input as an argument. If temperature is outside these ranges, output
“Temperature outside range” to the screen. Be precise, import modules, include comments,
prologue, etc. as needed.
Solution
Question39.java
import java.util.Scanner;
public class Question39 {
public static void main(String[] args) {
Scanner scan = new Scanner(System.in);
System.out.println(\"Enter temperature: \");
int temp = scan.nextInt();
if(temp >=76 && temp <=100){
outputHot(temp);
}
else if(temp >=0 && temp <=39){
outputCold(temp);
}
else if(temp >=40 && temp <=75){
outputJustRight(temp);
}
else{
System.out.println(\"Temperature outside range\");
}
}
public static void outputHot(int temp){
System.out.println(\"Temperature \"+temp+\" Hot \");
}
public static void outputCold(int temp){
System.out.println(\"Temperature \"+temp+\" Cold \");
}
public static void outputJustRight(int temp){
System.out.println(\"Temperature \"+temp+\" Just Right \");
}
}
Output:
Enter temperature:
45
Temperature 45 Just Right.
Merge Sort implementation in C++ The implementation for Mergesort gi.pdf
Merge Sort implementation in C++ The implementation for Mergesort given in Section 7.4 takes
an array as input and soils that array. At the beginning of Section 7.4 there is a simple
pseudocode implementation for sorting a linked list using Mergesort. Implement both a linked
list-based version of Mergesort and the array-based version of Mergesort. and compare and
analyze their running times.
Solution
/* Merger sort array-based version */
/* C program for Merge Sort */
#include
#include
// Merges two subarrays of arr[].
// First subarray is arr[l..m]
// Second subarray is arr[m+1..r]
void merge(int arr[], int l, int m, int r)
{
int i, j, k;
int n1 = m - l + 1;
int n2 = r - m;
/* create temp arrays */
int L[n1], R[n2];
/* Copy data to temp arrays L[] and R[] */
for (i = 0; i < n1; i++)
L[i] = arr[l + i];
for (j = 0; j < n2; j++)
R[j] = arr[m + 1+ j];
/* Merge the temp arrays back into arr[l..r]*/
i = 0; // Initial index of first subarray
j = 0; // Initial index of second subarray
k = l; // Initial index of merged subarray
while (i < n1 && j < n2)
{
if (L[i] <= R[j])
{
arr[k] = L[i];
i++;
}
else
{
arr[k] = R[j];
j++;
}
k++;
}
/* Copy the remaining elements of L[], if there
are any */
while (i < n1)
{
arr[k] = L[i];
i++;
k++;
}
/* Copy the remaining elements of R[], if there
are any */
while (j < n2)
{
arr[k] = R[j];
j++;
k++;
}
}
/* l is for left index and r is right index of the
sub-array of arr to be sorted */
void mergeSort(int arr[], int l, int r)
{
if (l < r)
{
// Same as (l+r)/2, but avoids overflow for
// large l and h
int m = l+(r-l)/2;
// Sort first and second halves
mergeSort(arr, l, m);
mergeSort(arr, m+1, r);
merge(arr, l, m, r);
}
}
/* UTILITY FUNCTIONS */
/* Function to print an array */
void printArray(int A[], int size)
{
int i;
for (i=0; i < size; i++)
printf(\"%d \", A[i]);
printf(\"\ \");
}
/* Driver program to test above functions */
int main()
{
int arr[] = {12, 11, 13, 5, 6, 7};
int arr_size = sizeof(arr)/sizeof(arr[0]);
printf(\"Given array is \ \");
printArray(arr, arr_size);
mergeSort(arr, 0, arr_size - 1);
printf(\"\ Sorted array is \ \");
printArray(arr, arr_size);
return 0;
}
--------------------------------------------------------------------------------------------------------------------
/* merge sort link-list version
#include
#include
/* Link list node */
struct node
{
int data;
struct node* next;
};
/* function prototypes */
struct node* SortedMerge(struct node* a, struct node* b);
void FrontBackSplit(struct node* source,
struct node** frontRef, struct node** backRef);
/* sorts the linked list by changing next pointers (not data) */
void MergeSort(struct node** headRef)
{
struct node* head = *headRef;
struct node* a;
struct node* b;
/* Base case -- length 0 or 1 */
if ((head == NULL) || (head->next == NULL))
{
return;
}
/* Split head into \'a\' and \'b\' sublists */
FrontBackSplit(head, &a, &b);
/* Recursively sort the sublists */
MergeSort(&a);
MergeSort(&b);
/* answer = merge the two sorted lists together */
*.
Write a C code to perform the following tasks In main Define an arr.pdf
Write a C code to perform the following tasks: In main: Define an array of SIZE = 7 and
initialize them with 3, 5, 7, 9, 11, 13, 15 Call function sum with the array and size, both passed in
pointer representations Print each element of the array Print the sum In function sum: Take the
parameters in pointer representations,
Solution
// C code
#include
#include
#include
#include
// sum function using pointers to add elements of array
int sumArray(int *array, int *SIZE)
{
int i;
int sum = 0;
for (i = 0; i < *SIZE; ++i)
{
sum = sum + *(array+i);
}
return sum;
}
int main()
{
int SIZE = 7;
int array[] = {3,5,7,9,11,13,15};
// ddeclare and initialize pointer
int *ptr;
ptr = array;
// pass array pointer and size of array to function
int sum = sumArray(ptr,&SIZE);
printf(\"Array: \");
int i;
for (i = 0; i < SIZE; ++i)
{
printf(\"%d \",array[i]);
}
printf(\"\ \");
printf(\"Sum of array: %d\ \",sum);
return 0;
}
/*
output:
Array: 3 5 7 9 11 13 15
Sum of array: 63
*/.
Write a program to implement and test the following sorting algorithm.docx
The document describes implementing and testing three sorting algorithms - insertion sort, heapsort, and shellsort - on arrays of 1,000 and 10,000 elements. It provides code samples to perform each algorithm and print the number of comparisons and item movements for analysis. The sorting algorithms are tested on sample arrays and the sorted output is displayed.
Program 1 (Practicing an example of function using call by referenc.pdf
Program 1: (Practicing an example of function using call by reference)
Write a program that reads a set of information related to students in C++ class and prints them
in a table format. The information is stored in a file called data.txt. Each row of the file contains
student number, grade for assignment 1, grade for assignment 2, grade for assignment 3, and
grade for assignment 4. Your main program should read each row, pass the grades for the three
assignments to a function called ProcessRow to calculate the average of the grades, minimum of
the four assignments, and the maximum of the four assignments. The results (average,
maximum, and minimum) should be returned to the main program and the main program prints
them on the screen in a table format. For example, if the file includes
126534 7 8 10 7
321345 5 6 4 9
324341 8 3 8 5
your program should print
Std-Id A1 A2 A3 A4 Min Max Average
----------------------------------------------------------------------------------
126534 7 8 10 7 7 10 8
321345 5 6 4 9 4 9 6
324341 8 3 8 5 3 8 6
You must use call by reference to do the above question.
_________________________________________________________________
Program 2: (Practicing an example of function using call by value)
Repeat the above question using call by value. This means you need to have three different
functions: one to calculate the average, another to calculate the minimum, and the third one to
calculate the maximum. This is how to call these functions:
Max = CalculateMax(A1,A2,A3, A4);
Min = CalculateMin(A1,A2,A3, A4);
Average = CalculateAvg(A1,A2,A3, A4);
_________________________________________________________________
Program 3:
Write a program with several functions that performs the following tasks. :
Create a function that reads the following 5 float numbers from the file data.txt into array Called
Arr1.
12.0 15.0 29.3 25.0 93.2
Copy array Arr1 into array Arr2 in reverse order
Print array Arr1.
Print array Arr2.
Find the number of elements in array Arr1 that are >= 80 and <=100.
Find the number of the elements in array Arr1 in which their contents are divisible by 5
Find the index of the elements in array Arr1 in which their contents are divisible by 3.
Find mean (average) in array Arr1.
Find the maximum number in array Arr1.
Ask the user to input a key. Then search for the key in array Arr1 and inform the user about the
existence (true / false) of the key in array
Your program should include several functions.
A function for filling up the information from file into an array (part a should call this function)
A function that does the copying of one array into another in reverse order. Arrays must have the
same size (part b should call this function)
Printing any array with any size (part c and d should call this function)
finding and returning the number of elements between 80 and 100 in any array with any size
(part e calls this function)
finding and returning the number of elements in an array that are divisible by 5 (par.
#include iostream using namespace std; void InsertionSort(int .pdf
#include
using namespace std;
void InsertionSort(int arr[],int size){
int temp,j;
for(int i=0; i=0){
arr[j+1] = arr[j];
j--;
}
arr[j+1] = temp;
}
}
// Merges two subarrays of arr[].
// First subarray is arr[l..m]
// Second subarray is arr[m+1..r]
void merge(int arr[], int l, int m, int r)
{
int i, j, k;
int n1 = m - l + 1;
int n2 = r - m;
/* create temp arrays */
int L[n1], R[n2];
/* Copy data to temp arrays L[] and R[] */
for (i = 0; i < n1; i++)
L[i] = arr[l + i];
for (j = 0; j < n2; j++)
R[j] = arr[m + 1+ j];
/* Merge arrays back into arr[l..r]*/
i = 0; // first subarray index
j = 0; // second subarray index
k = l; // merged subarray index
while (i < n1 && j < n2)
{
if (L[i] <= R[j])
{
arr[k] = L[i];
i++;
}
else
{
arr[k] = R[j];
j++;
}
k++;
}
/* Copy the remaining elements of L[], if there
are any */
while (i < n1)
{
arr[k] = L[i];
i++;
k++;
}
/* Copy the remaining elements of R[], if there
are any */
while (j < n2)
{
arr[k] = R[j];
j++;
k++;
}
}
/* l is for left index and r is right index of the
sub-array of arr to be sorted */
void mergeSort(int arr[], int l, int r)
{
if (l < r)
{
// Same as (l+r)/2, but avoids overflow for
// large l and h
int m = l+(r-l)/2;
// Sort first and second halves
mergeSort(arr, l, m);
mergeSort(arr, m+1, r);
merge(arr, l, m, r);
}
}
void printArray(int arr[], int size){//Function to print array
int i;
for (i=0; i < size; i++)
cout << arr[i] <<\" \";
cout << endl;
}
int main(){
int size = 20;
int array1[] = {17,33,88,54,96,82,12,36,69,74,79,66,32,11,1,98,76,55,3,4};
int array2[] = {20,1,19,2,18,3,17,4,16,5,15,6,14,7,13,8,12,9,11,10};
int array3[] = {220,209,198,187,176,165,154,143,132,121,110,99,88,77,66,55,44,33,22,11};
int array4[] = {25,27,23,2,7,99,96,1996,85,3,24,11,17,4,1,0,8,9,98,45};
int array5[] = {15,35,85,100,40,90,70,65,25,5,10,30,55,75,80,95,20,45,50,60};
//Output for array1
cout << \"Original array 1: \ \";
printArray(array1, size);
InsertionSort(array1, size);
cout << \"Array 1 sorted by insertion sort: \ \";
printArray(array1, size);
//CALL THE MERGE SORT HERE
//PRINT THE MERGE SORT HERE
cout << \"\ \";
//Output for array2
cout << \"Original array 2: \ \";
printArray(array2, size);
mergeSort(array2,0,size);
cout << \"Array 2 sorted by Merge sort: \ \";
printArray(array2, size);
cout << \"\ \";
//Output for array3
cout << \"Original array 3: \ \";
printArray(array3, size);
InsertionSort(array3, size);
cout << \"Array 3 sorted by insertion sort: \ \";
printArray(array3, size);
cout << \"\ \";
//Output for array4
cout << \"Original array 4: \ \";
printArray(array4, size);
mergeSort(array4,0,size);
cout << \"Array 4 sorted by Merge sort: \ \";
printArray(array4, size);
cout << \"\ \";
//Output for array5
cout << \"Original array 5: \ \";
printArray(array5, size);
InsertionSort(array5, size);
cout << \"Array 5 sorted by insertion sort: \ \";
printArray(array5, size);
cout << \"\ \";
return 0;
}
/* sample output
Note: As sorting an already sorted array does not result good. I have changed the array2,a.
SlideSet_4_Arraysnew.pdf
The document discusses arrays in C programming. Some key points include:
- An array is a collection of variables of the same type referred to by a common name. Each element has an index and arrays use contiguous memory locations.
- Arrays are declared with the type, name, and size. The first element is at index 0.
- One-dimensional arrays can be initialized, accessed, input from and output to the user. Multidimensional arrays like 2D arrays represent tables with rows and columns.
- Arrays can be passed to functions by passing the entire array or individual elements. Operations like searching, sorting and merging can be performed on arrays.
5 array
This document discusses arrays in C/C++. It defines an array as a collection of identical data objects stored in consecutive memory locations under a common name. It describes array declaration syntax, initialization, single and multi-dimensional arrays, and provides examples of using arrays to find the largest number, sort numbers, and calculate a sum. It also covers pointers, pointer arithmetic, and passing arrays to functions by value and reference.
write a program that given a list of 20 integers, sorts them accordi.pdf
write a program that given a list of 20 integers, sorts them according to insert sort. java
Solution
public class InsertionSort {
/**
* method to sort the array elements using insertion sorts *
*
* @param array
* @return
*/
public static int[] insertionSort(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;
}
return array;
}
/**
* method to print the array elements
*
* @param input
*/
private static void printNumbers(int[] input) {
for (int i = 0; i < input.length; i++) {
System.out.print(input[i] + \", \");
}
System.out.println(\"\ \");
}
/**
* @param args
*/
public static void main(String[] args) {
int[] input = { 3, 26, 67, 35, 9, -6, 43, 82, 10, 54, 14, 22, 21, 32,
121, 231, 12, 98, 77, 65 };
System.out.println(\"Before Sorting\");
printNumbers(input);
input = insertionSort(input);
System.out.println(\"Before Sorting\");
printNumbers(input);
}
}
OUTPUT:
Before Sorting
3, 26, 67, 35, 9, -6, 43, 82, 10, 54, 14, 22, 21, 32, 121, 231, 12, 98, 77, 65,
Before Sorting
-6, 3, 9, 10, 12, 14, 21, 22, 26, 32, 35, 43, 54, 65, 67, 77, 82, 98, 121, 231,.
c++ program I need to sort arrays using an insertion sort and a mer.pdf
c++ program: I need to sort arrays using an insertion sort and a merge sort (I already have). I
need to use these sorting methods to sort input arrays in the form of randomly generated numbers
in the size of 20, 100 and 200. So I need 3 randomily generated arrays and sorted and outputed
by both sorting method. I currently have it as given arrays and need help changing them to
randomily generated arrays
Here is my code so far:
#include
using namespace std;
void InsertionSort(int arr[],int size){
int temp,j;
for(int i=0; i=0){
arr[j+1] = arr[j];
j--;
}
arr[j+1] = temp;
}
}
//Merges two sub-arrays. First sub-array is arr[l..m],second is arr[m+1..r]
void merge(int arr[], int l, int m, int r){
int i, j, k;
int n1 = m - l + 1;
int n2 = r - m;
//Create temp arrays
int L[n1], R[n2];
//Copy data to temp arrays L[] and R[]
for (i = 0; i < n1; i++)
L[i] = arr[l + i];
for (j = 0; j < n2; j++)
R[j] = arr[m + 1+ j];
/* Merge arrays back into arr[l..r]*/
i = 0;
j = 0;
k = l;
while (i < n1 && j < n2){
if (L[i] <= R[j]){
arr[k] = L[i];
i++;
}
else{
arr[k] = R[j];
j++;
}
k++;
}
//Copy the remaining elements of L[], if there are any
while (i < n1){
arr[k] = L[i];
i++;
k++;
}
//Copy the remaining elements of R[], if there are any
while (j < n2){
arr[k] = R[j];
j++;
k++;
}
}
//l is left index and r is right index of the sub-array of array to be sorted
void MergeSort(int arr[], int l, int r){
if (l < r){
// Same as (l+r)/2, but avoids overflow for large l and h
int m = l+(r-l)/2;
// Sort first and second halves
MergeSort(arr, l, m);
MergeSort(arr, m+1, r);
merge(arr, l, m, r);
}
}
void printArray(int arr[], int size){//Function to print array
int i;
for (i=0; i < size; i++)
cout << arr[i] <<\" \";
cout << endl;
}
int main(){
int size = 20;
int array1[] = {17,33,88,54,96,82,12,36,69,74,79,66,32,11,1,98,76,55,3,4};
int array2[] = {20,1,19,2,18,3,17,4,16,5,15,6,14,7,13,8,12,9,11,10};
int array3[] = {220,209,198,187,176,165,154,143,132,121,110,99,88,77,66,55,44,33,22,11};
int array4[] = {15,35,85,100,40,90,70,65,25,5,10,30,55,75,80,95,20,45,50,60};
int array5[] = {100,99,98,97,96,95,94,93,92,91,90,89,88,87,86,85,84,83,82,81};
//Output for array1
cout << \"Original array 1: \ \";
printArray(array1, size);
InsertionSort(array1, size);
cout << \"Array 1 sorted by INSERTION SORT: \ \";
printArray(array1, size);
cout << \"\ \";
//Output for array2
cout << \"Original array 2: \ \";
printArray(array2, size);
MergeSort(array2,0,size);
cout << \"Array 2 sorted by MERGE SORT: \ \";
printArray(array2, size);
cout << \"\ \";
//Output for array3
cout << \"Original array 3: \ \";
printArray(array3, size);
InsertionSort(array3, size);
cout << \"Array 3 sorted by INSERTION SORT: \ \";
printArray(array3, size);
cout << \"\ \";
//Output for array4
cout << \"Original array 4: \ \";
printArray(array4, size);
MergeSort(array4,0,size);
cout << \"Array 4 sorted by MERGE SORT: \ \";
printArray(array4, size);
cout << \"\ \";
//Output for array5
cout << \"Original array 5: \ \";
printArray(arr.
For this homework, you will write a program to create and manipulate.pdf
For this homework, you will write a program to create and manipulate a simple linked list. For
each node in the linked list you will generate a random number, create a node that holds that
number, and insert that node into its sorted position in the linked list. Once the nodes are
inserted, you will traverse the list, printing the value held in each node. Then you will clean up
the list (deleting all the nodes) and exit the program. You will also learn about a tool that you can
use to help you check for memory errors in your code.
Specifications:
Define a struct appropriate for holding a random integer. This struct will also contain a \"next\"
pointer to reference a separate instance of the struct.
You may use the typedef keyword to give your struct a separate typename if you wish, but this is
not a requirement.
The program will read command line arguments. The first argument is the program name (as it
will always be) followed by a random number seed, the number of random numbers to generate
and ending with the Maximum Possible Value of the Random numbers generated (i.e. argc
should be 4).
Your program will use a \"dummy node\" for the linked list. In other words, the head of the list
will be an actual node, but the data field in that node will not be used for sorting purposes.
Your program must contain the following functions:
main() (for obvious reasons)
insertNodeSorted() - You may implement this either of two ways.
One way is to use the head of the list as one parameter and the integer value as the second. The
function will allocate memory for the node, initialize it and then insert it in the proper location in
the list.
The other way is to pass the head of the list as one parameter and a previously allocated and
initialized node as the other. The existing node is inserted in the proper location in the list.
printList() - Takes the head of the list as its only parameter, traverses the list, printing out the
data in sorted order.
deleteList() - Takes the head of the list as its only parameter, traverses the list, deleting all nodes.
The basic algorithm of your program is
Use a loop (based upon command line input values) that:
Generates a random number
Prints the number to stdout
Creates a node that contains that random number
Inserts the new node in sorted order into the existing list
Print the sorted linked list
Delete the linked list
You will use a Makefile to compile your program
Solution
#include
#include
#include
struct test_struct
{
int val;
struct test_struct *next;
};
struct test_struct *head = NULL;
struct test_struct *curr = NULL;
struct test_struct* create_list(int val)
{
printf(\"\ creating list with headnode as [%d]\ \",val);
struct test_struct *ptr = (struct test_struct*)malloc(sizeof(struct test_struct));
if(NULL == ptr)
{
printf(\"\ Node creation failed \ \");
return NULL;
}
ptr->val = val;
ptr->next = NULL;
head = curr = ptr;
return ptr;
}
struct test_struct* add_to_list(int val, bool add_to_end)
{
if(NULL == head)
{
retur.
The language is C. Thanks. Write a function that computes the intege.pdf
The language is C. Thanks. Write a function that computes the integer mean of an array of
integers. For example, the integer mean of - 1, 4, 2 is (- 1 + 4 + 2)/3 = 5/3 = 1, where/denotes
integer division. The function should implement the following specification;/* Computes the
integer mean of an array and stores it * where mn references. Returns -J for erroneous input *
(len
Solution
// C code
#include
int mean(int *a, int len, int *mn)
{
if(len <= 0)
return -1;
*mn = 0;
int i = 0;
for (i = 0; i < len; ++i)
{
*mn = *mn + *(a+i);
}
*mn = *mn/len;
return 0;
}
int main()
{
int size;
printf(\"Enter size of array: \");
scanf(\"%d\",&size);
int i;
int array[size];
for (i = 0; i < size; ++i)
{
printf(\"Enter array element at index %d: \",(i+1));
scanf(\"%d\",&array[i]);
}
int mn;
int result = mean(array, size, &mn);
if(result == 0)
printf(\"Mean: %d\ \", mn);
else
printf(\"Erroneous input\ \");
return 0;
}
/*
output:
Enter size of array: 3
Enter array element at index 1: -1
Enter array element at index 2: 4
Enter array element at index 3: 2
Mean: 1
*/.
Can you give an example of a binary heap programCan you give an .pdf
Can you give an example of a binary heap program
Can you give an example of a binary heap program
Solution
/*
* C Program to Implement a Heap & provide Insertion & Deletion Operation
*/
#include
int array[100], n;
main()
{
int choice, num;
n = 0;/*Represents number of nodes in the heap*/
while(1)
{
printf(\"1.Insert the element \ \");
printf(\"2.Delete the element \ \");
printf(\"3.Display all elements \ \");
printf(\"4.Quit \ \");
printf(\"Enter your choice : \");
scanf(\"%d\", &choice);
switch(choice)
{
case 1:
printf(\"Enter the element to be inserted to the list : \");
scanf(\"%d\", &num);
insert(num, n);
n = n + 1;
break;
case 2:
printf(\"Enter the elements to be deleted from the list: \");
scanf(\"%d\", &num);
delete(num);
break;
case 3:
display();
break;
case 4:
exit(0);
default:
printf(\"Invalid choice \ \");
}/*End of switch */
}/*End of while */
}/*End of main()*/
display()
{
int i;
if (n == 0)
{
printf(\"Heap is empty \ \");
return;
}
for (i = 0; i < n; i++)
printf(\"%d \", array[i]);
printf(\"\ \");
}/*End of display()*/
insert(int num, int location)
{
int parentnode;
while (location > 0)
{
parentnode =(location - 1)/2;
if (num <= array[parentnode])
{
array[location] = num;
return;
}
array[location] = array[parentnode];
location = parentnode;
}/*End of while*/
array[0] = num; /*assign number to the root node */
}/*End of insert()*/
delete(int num)
{
int left, right, i, temp, parentnode;
for (i = 0; i < num; i++) {
if (num == array[i])
break;
}
if (num != array[i])
{
printf(\"%d not found in heap list\ \", num);
return;
}
array[i] = array[n - 1];
n = n - 1;
parentnode =(i - 1) / 2; /*find parentnode of node i */
if (array[i] > array[parentnode])
{
insert(array[i], i);
return;
}
left = 2 * i + 1; /*left child of i*/
right = 2 * i + 2; /* right child of i*/
while (right < n)
{
if (array[i] >= array[left] && array[i] >= array[right])
return;
if (array[right] <= array[left])
{
temp = array[i];
array[i] = array[left];
array[left] = temp;
i = left;
}
else
{
temp = array[i];
array[i] = array[right];
array[right] = temp;
i = right;
}
left = 2 * i + 1;
right = 2 * i + 2;
}/*End of while*/
if (left == n - 1 && array[i]) {
temp = array[i];
array[i] = array[left];
array[left] = temp;
}
}
Output:.
Need this in JAVA We have N numbers as an array, you need to find a .pdf
Need this in JAVA We have N numbers as an array, you need to find a prefix array and a suffix
array, which we can get the maximum x or value with all elements in them. Notice that for
prefix[0, 1] and suffix[r, n - 1], do not intersect (I
Solution
// C++ program for a Trie based O(n) solution to find max
// subarray XOR
#include
using namespace std;
// Assumed int size
#define INT_SIZE 32
// A Trie Node
struct TrieNode
{
int value; // Only used in leaf nodes
TrieNode *arr[2];
};
// Utility function tp create a Trie node
TrieNode *newNode()
{
TrieNode *temp = new TrieNode;
temp->value = 0;
temp->arr[0] = temp->arr[1] = NULL;
return temp;
}
// Inserts pre_xor to trie with given root
void insert(TrieNode *root, int pre_xor)
{
TrieNode *temp = root;
// Start from the msb, insert all bits of
// pre_xor into Trie
for (int i=INT_SIZE-1; i>=0; i--)
{
// Find current bit in given prefix
bool val = pre_xor & (1<arr[val] == NULL)
temp->arr[val] = newNode();
temp = temp->arr[val];
}
// Store value at leaf node
temp->value = pre_xor;
}
// Finds the maximum XOR ending with last number in
// prefix XOR \'pre_xor\' and returns the XOR of this maximum
// with pre_xor which is maximum XOR ending with last element
// of pre_xor.
int query(TrieNode *root, int pre_xor)
{
TrieNode *temp = root;
for (int i=INT_SIZE-1; i>=0; i--)
{
// Find current bit in given prefix
bool val = pre_xor & (1<arr[1-val]!=NULL)
temp = temp->arr[1-val];
// If there is no prefix with opposite
// bit, then look for same bit.
else if (temp->arr[val] != NULL)
temp = temp->arr[val];
}
return pre_xor^(temp->value);
}
// Returns maximum XOR value of a subarray in arr[0..n-1]
int maxSubarrayXOR(int arr[], int n)
{
// Create a Trie and insert 0 into it
TrieNode *root = newNode();
insert(root, 0);
// Initialize answer and xor of current prefix
int result = INT_MIN, pre_xor =0;
// Traverse all input array element
for (int i=0; i
using namespace std;
// Assumed int size
#define INT_SIZE 32
// A Trie Node
struct TrieNode
{
int value; // Only used in leaf nodes
TrieNode *arr[2];
};
// Utility function tp create a Trie node
TrieNode *newNode()
{
TrieNode *temp = new TrieNode;
temp->value = 0;
temp->arr[0] = temp->arr[1] = NULL;
return temp;
}
// Inserts pre_xor to trie with given root
void insert(TrieNode *root, int pre_xor)
{
TrieNode *temp = root;
// Start from the msb, insert all bits of
// pre_xor into Trie
for (int i=INT_SIZE-1; i>=0; i--)
{
// Find current bit in given prefix
bool val = pre_xor & (1<arr[val] == NULL)
temp->arr[val] = newNode();
temp = temp->arr[val];
}
// Store value at leaf node
temp->value = pre_xor;
}
// Finds the maximum XOR ending with last number in
// prefix XOR \'pre_xor\' and returns the XOR of this maximum
// with pre_xor which is maximum XOR ending with last element
// of pre_xor.
int query(TrieNode *root, int pre_xor)
{
TrieNode *temp = root;
for (int i=INT_SIZE-1; i>=0; i--)
{
// Find current bit in given prefix
bool val = pre_xor & (1<arr[1-val]!=NULL)
temp = temp-.
There are a number of errors in the following program- All errors are.docx
There are a number of errors in the following program.
All errors are located in main() and structure definitions.
Function declarations and definitions are correct!
Locate all errors, fix them (as shown below), run the program and save its output
as a comment at the end of the source file.
Example:
int num = 10;
int *ptr;
num = &ptr; // <== Error: Comment the line and write the correct line below
// Write a short justification where appropriate
// num = &ptr; // Error #1
ptr = #
Name:
*/
#include <stdio.h>
#include <string.h>
#include <stdlib.h>
#include <time.h>
#define DUMMY_TRAILER '\177'
// octal ASCII code of the
// last character in the ASCII table
#define NUM_CITIES 10
typedef struct
{
char name[12];
int temperature[5];
} CITY;
// Stack and Queue Node
typedef struct node NODE;
struct node
{
CITY city;
node *next;
};
// Doubly Linked List Node
typedef struct d_node D_NODE;
struct d_node
{
CITY city;
NODE *forw;
NODE *back;
};
// Stack Functions
NODE *push(NODE *stack, const CITY *pStu);
NODE *pop(NODE **stack);
// Queue Functions
void enqueue(NODE **queue, NODE **rear, const CITY *pStu);
NODE *dequeue(NODE **queue, NODE **rear);
// Doubly Linked List Functions
D_NODE *init_list(void);
int insert(D_NODE *list, const CITY *pStu);
void traverse_forw(D_NODE *list);
void traverse_back(D_NODE *list);
// Other Functions
void printCity(const CITY *pCity);
int main (void)
{
CITY cList[NUM_CITIES] =
{
{"Cupertino", {88, 89, 87, 85, 89}},
{"Flagstaff", {81, 80, 88, 89, 89}},
{"Los Angeles", {87, 88, 89, 89, 90}},
{"Philadelphia", {96, 99, 99, 90, 95}},
{"Phoenix", {106, 109, 109, 100, 105}},
{"Portland", {89, 90, 85, 89, 90}},
{"Reno", {108, 105, 109, 100, 108}},
{"Salem", {85, 90, 85, 89, 90}},
{"Tucson", {107, 100, 109, 100, 108}},
{"Yreka", {101, 109, 100, 108, 109}}
};
NODE *stack = NULL;
NODE *top = NULL;
NODE *queue = NULL, *rear = NULL;
NODE *front;
D_NODE *list;
list = init_list();
// build stack and queue with data from an array of CITY structures
srand((unsigned int)time(NULL));
int count = rand() % 10;
for ( int n = 0; n < count; n++)
{
int i = rand() % NUM_CITIES;
int duplicate = insert(list, &cList[i]);
if(duplicate)
{
// already in the list!
push(stack, &cList[i]);
enqueue(&queue, &rear, cList[i]);
}
}
// display list
printf("\nLIST contents (forwards):\n");
traverse_forw(list);
printf("\nLIST contents (backwards):\n");
traverse_back(list);
// display stack
if (top)
{
printf("\nSTACK contents from top to bottom:\n");
while ((top = pop(stack)))
{
printCity(top->city);
}
}
else
printf ("Empty Stack!\n");
// display queue
if (front)
{
printf("\nQUEUE contents from front to rear:\n");
while ((front = dequeue( queue, rear)))
{
printCity(front->city);
}
}
else
printf ("Empty Queue!\n");
return 0;
}
/***************************************************
Displays the fileds of a CIS_CLASS structure
Pre pCls - a pointer to a CIS_CLASS structure
Post
*/
void printCity(const CITY *pCity)
{
printf("%-20s %3d\n", pCity->name, pCity->temperatur.
Write a program in C that generates a list of random numbers greater.pdf
Write a program in C that generates a list of random numbers greater than or equal to 0 and less
than 32, stores them in an array then have the program run through the numbers using a linked
lists then print out the numbers in the array.
Solution
#include
#include
#include
typedef struct node{ // create a struct for linked list
int data;
struct node *next;
}ll;
void main(){
srand(time(NULL)); // set the seed to random numbers
int n=10,i=0,num; // initilize the random numbers
ll *start = NULL; // set start = NULL
while(idata = rand()%32; // set a random number for a node
temp->next = start; // link node to start
start = temp; // move start to beginning
i++;
}
display(start);
}
void display(ll *start){
ll *temp = start;
while(temp!=NULL){ // while ll has elements
printf(\"%d->\",temp->data); // print data
temp = temp->next;
}
}
/* sample output
*/.
(Parent reference for BST) Redefine TreeNode by adding a reference to.pdf
(Parent reference for BST) Redefine TreeNode by adding a reference to a node\'s parent, as
shown below\': Reimplement the insert and delete methods in the BST class to update the parent
for each node in the tree. Add the following new method in BST:/** Returns the node for the
specified element. * Returns null if the element is not in the tree. */private TreeNode getNode(E
element)/** Returns true if the node for the element is a leaf */private boolean isLeaf(E
element)/** Returns the path of elements from the specified element * to the root in an array
list. */public ArrayList getPath(E e) Write a test program that prompts the user to enter 10
integers, adds them to the tree, deletes the first integer from the tree, and displays the paths for
all leaf nodes. Here is a sample run:
Solution
#include
#include
void swap(char *x, char *y)
{
char temp;
temp = *x;
*x = *y;
*y = temp;
void permute(char *a, int l, int r)
{
int i;
if (l == r)
printf(\"%s\ \", a);
else
{
for (i = l; i <= r; i++)
{
swap((a+l), (a+i));
permute(a, l+1, r);
swap((a+l), (a+i)); //backtrack
}
}
}
int main()
{
char str[] = \"ABC\";
int n = strlen(str);
permute(str, 0, n-1);
return 0;
}
part ii radix sort:
c++
#include
using namespace std;
int getMax(int arr[], int n)
{
int mx = arr[0];
for (int i = 1; i < n; i++)
if (arr[i] > mx)
mx = arr[i];
return main()
void countSort(int arr[], int n, int exp
for (i = 0; i < n; i++)
count[ (arr[i]/exp)%10 ]++;
for (i = 1; i < 10; i++)
count[i] += count[i - 1];
for (i = n - 1; i >= 0; i--)
{
output[count[ (arr[i]/exp)%10 ] - 1] = arr[i];
count[ (arr[i]/exp)%10 ]--;
}
for (i = 0; i < n; i++)
arr[i] = output[i];
}
adixsort(int arr[], int n)
{
int m = getMax(arr, n);
for (int exp = 1; m/exp > 0; exp *= 10)
countSort(arr, n, exp);
}
void print(int arr[], int n)
{
for (int i = 0; i < n; i++)
cout << arr[i] << \" \";
}
int main()
{
int arr[] = {170, 45, 75, 90, 802, 24, 2, 66};
int n = sizeof(arr)/sizeof(arr[0]);
radixsort(arr, n);
print(arr, n);
return 0;
}
java:
import java.io.*;
import java.util.*;
class Radix {
static int getMax(int arr[], int n)
{
int mx = arr[0];
for (int i = 1; i < n; i++)
if (arr[i] > mx)
mx = arr[i];
return mx;
}
static void countSort(int arr[], int n, int exp)
{
int output[] = new int[n]; // output array
int i;
int count[] = new int[10];
Arrays.fill(count,0);
for (i = 0; i < n; i++)
count[ (arr[i]/exp)%10 ]++;
for (i = 1; i < 10; i++)
count[i] += count[i - 1];
for (i = n - 1; i >= 0; i--)
{
output[count[ (arr[i]/exp)%10 ] - 1] = arr[i];
count[ (arr[i]/exp)%10 ]--;
}
for (i = 0; i < n; i++)
arr[i] = output[i];
}
static void radixsort(int arr[], int n)
{
int m = getMax(arr, n);
for (int exp = 1; m/exp > 0; exp *= 10)
countSort(arr, n, exp);
}
static void print(int arr[], int n)
{
for (int i=0; i
#include
void swap(char *x, char *y)
{
char temp;
temp = *x;
*x = *y;
*y = temp;
void permute(char *a, int l, int r)
{
int i;
if (l == r)
printf(\"%s\ \", a);
else
{
for (i = l; i <= r; i++)
{
swap((a+l), (a+i));
permute(a, l+1, .
Part 1)#include stdio.h #include stdlib.h #include pthrea.pdf
Part 1)
#include
#include
#include
#include
/*Error handling for pthread_create and pthread_join*/
/*from the pthread_create man page*/
#define handle_error_en(en, msg) \\
do { errno = en; perror(msg); exit(EXIT_FAILURE); } while (0)
/* # of running threads */
volatile int running_threads = 0;
pthread_t thread[3]; /*Descriptors for our 3 threads*/
int numOfElements;/*Total # of elements from the user*/
struct Results{ /*Struct to hold the statistical results*/
int min;
int max;
int average;
}Results;
/*This function finds the minimum element of an array*/
void *findMin(void *array_ptr){
int i; /*counter*/
int *elements = (int*)array_ptr; /*re reference void array pointer*/
Results.min = elements[0]; /*set minimum to first element */
for(i = 0; i < numOfElements; i++){ /*iterate through array*/
if(elements[i] < Results.min){ /*if the current element is less than the current min*/
Results.min = elements[i]; /*store the new min*/
}
}
running_threads -= 1; /*Decrement thread count*/
return NULL;
}
/*This function finds the maximum element of an array*/
void *findMax(void *array_ptr){
int i; /*counter*/
int *elements = (int*)array_ptr; /*re reference void array pointer*/
for(i = 0; i < numOfElements; i++){ /*iterate through array*/
if(elements[i] > Results.max){ /*store the new max*/
Results.max = elements[i];
}
}
running_threads -= 1; /*Decrement thread count*/
return NULL;
}
/*This function finds the average of an array*/
void *findAverage(void *array_ptr){
int i; /*counter*/
int *elements = (int*)array_ptr; /*re reference void array pointer*/
for(i = 0; i < numOfElements; i++){ /*iterate through array*/
Results.average += elements[i]; /*add element @ i to average*/
}
Results.average = Results.average/numOfElements; /*Divide the sum by the number of
elements*/
running_threads -= 1; /*Decrement running threads counter*/
return NULL;
}
/* This method accepts a int n(initial size of array) and
pointer to an array and returns # of elements in the array
*/
int getArrayInput(int n, int *array_ptr){
int input;/*Store user input */
int numberOfElements = 0;/*Number of Integers inputed*/
printf(\"Creating Dynamic Array...\ -\ \");
for(;;){ /*infinite loop*/
printf(\"Enter a positive value:\ Negative Number to Stop\ -\ \");
//Get Int from console, store at address of input
if (scanf(\"%d\",&input) != 1){
printf(\"\ Oops that wasn\'t an Integer\ lets try filling the array again\ Remember
INTEGERS only!\ \");
exit(EXIT_FAILURE);
}
if (input >= 0){
if (numberOfElements == n){
n += 1; //Make room for the current input
array_ptr = realloc(array_ptr, n * sizeof(int));//realloc array and set pointer
}
array_ptr[numberOfElements++] = input;//Store input at next empty element
} else {
printf(\"\ Number of Integers: %d\ \", numberOfElements);
break;
}
}
return numberOfElements;
}
/*This function joins our n number of threads */
void joinThreads(int numberOfThreads){
int i; /*count*/
int s; /*error #*/
while(numberOfThreads >= 0){ /*Join our threads*/
s = pthread_jo.
Digits.javapackage week04;import java.util.Scanner; public cla.pdf
Digits.java
package week04;
import java.util.Scanner;
public class Digits {
public static void main(String[] args)
{
//Declaring variables
int sum=0,number,rem;
//Scanner class Object is sued to read the inputs entered by the user
Scanner input = new Scanner(System.in);
//Getting the number entered by the user
System.out.print(\"Enter an integer : \");
number=input.nextInt();
//this loop continue to execute until the number is graeter than zero
while (number>0 )
{
//Getting the last digit of the number every time
rem=number%10;
/*Getting the remaining part of
* the number except last digit every time
*/
number=number/10;
//Calculating the sum of the digits
sum+=rem;
}
//Displaying the sum of the digits of the number
System.out.print(\"The sum of digits is: \" +sum);
}
}
______________________________________________
Output:
Enter an integer : 123456789
The sum of digits is: 45
__________________________________________Thank You
Solution
Digits.java
package week04;
import java.util.Scanner;
public class Digits {
public static void main(String[] args)
{
//Declaring variables
int sum=0,number,rem;
//Scanner class Object is sued to read the inputs entered by the user
Scanner input = new Scanner(System.in);
//Getting the number entered by the user
System.out.print(\"Enter an integer : \");
number=input.nextInt();
//this loop continue to execute until the number is graeter than zero
while (number>0 )
{
//Getting the last digit of the number every time
rem=number%10;
/*Getting the remaining part of
* the number except last digit every time
*/
number=number/10;
//Calculating the sum of the digits
sum+=rem;
}
//Displaying the sum of the digits of the number
System.out.print(\"The sum of digits is: \" +sum);
}
}
______________________________________________
Output:
Enter an integer : 123456789
The sum of digits is: 45
__________________________________________Thank You.
IN C CODEGiven an IntNode struct and the operating functions for a.pdf
In C ++
Complete ReplaceLevel() to output "levelPointer is null." if levelPointer is null. Otherwise,
assign levelValue to the variable pointed to by levelPointer. End with a newline.
Ex: If the input is Y A, then the output is:
Level is A.
#include
using namespace std;
void ReplaceLevel(char* levelPointer, char levelValue) {
/* Your code goes here */
}
int main() {
char level;
char* levelPointer;
char action;
char value;
level = 'Z';
cin >> action;
cin >> value;
if (action == 'Y') {
levelPointer = &level;
}
else {
levelPointer = nullptr;
}
ReplaceLevel(levelPointer, value);
cout << "Level is " << level << "." << endl;
return 0;
}.
I have written the code but cannot complete the assignment please help.pdf
I have written the code but cannot complete the assignment please help me to complete.
Please don't just copy other s answers as your own.
// Insertion sort
/*
#include <ctime>
#include <iostream>
using namespace std;
void insertionSort(int arr[], int n) {
int i, key, j;
for (i = 1; i < n; i++) {
key = arr[i];
j = i - 1;
// Move elements of arr[0..i-1], that are greater than key, to one position ahead of their current
position
while (j >= 0 && arr[j] > key) {
arr[j + 1] = arr[j];
j = j - 1;
}
arr[j + 1] = key;
}
}
int main() {
int n;
cout << "Enter the size of the array: ";
cin >> n;
int arr[n];
cout << "Enter the elements of the array: ";
for (int i = 0; i < n; i++) {
cin >> arr[i];
}
int num = sizeof(arr) / sizeof(arr[0]);
clock_t start, end;
double timetaken;
start = clock();
insertionSort(arr, num);
end = clock();
cout << "Sorted array: \n";
for (int i = 0; i < n; i++)
cout << arr[i] << " ";
cout << endl;
timetaken = ((double) (end - start)) / CLOCKS_PER_SEC;
cout << "Time taken : " << fixed << timetaken << "s" << endl;
return 0;
}
*/
// Shell Sort
/*
#include <ctime>
#include <iostream>
using namespace std;
int shellSort(int arr[], int n) {
for (int gap = n/2; gap > 0; gap /= 2) {
for (int i = gap; i < n; i += 1) {
int temp = arr[i];
int j;
for (j = i; j >= gap && arr[j - gap] > temp; j -= gap)
arr[j] = arr[j - gap];
arr[j] = temp;
}
}
return 0;
}
void printArray(int arr[], int n) {
for (int i=0; i<n; i++)
cout << arr[i] << " ";
}
int main() {
int n;
cout << "Enter the size of the array: ";
cin >> n;
int arr[n];
cout << "Enter the elements of the array: ";
for (int i = 0; i < n; i++) {
cin >> arr[i];
}
int num = sizeof(arr) / sizeof(arr[0]);
clock_t start, end;
double timetaken;
start = clock();
shellSort(arr, num);
end = clock();
cout << "\nArray after sorting: \n";
printArray(arr, n);
cout << endl;
timetaken = (double)(end - start) / CLOCKS_PER_SEC;
cout << "Time taken : " << fixed << timetaken << "s" << endl;
return 0;
}
*/
// MergeSort
/*
#include <ctime>
#include <iostream>
using namespace std;
void merge(int arr[], int l, int m, int r) {
int i, j, k;
int n1 = m - l + 1;
int n2 = r - m;
int L[n1], R[n2];
for (i = 0; i < n1; i++)
L[i] = arr[l + i];
for (j = 0; j < n2; j++)
R[j] = arr[m + 1 + j];
i = 0;
j = 0;
k = l;
while (i < n1 && j < n2) {
if (L[i] <= R[j]) {
arr[k] = L[i];
i++;
}
else {
arr[k] = R[j];
j++;
}
k++;
}
while (i < n1) {
arr[k] = L[i];
i++;
k++;
}
while (j < n2) {
arr[k] = R[j];
j++;
k++;
}
}
void mergeSort(int arr[], int l, int r) {
if (l < r) {
int m = l + (r - l) / 2;
mergeSort(arr, l, m);
mergeSort(arr, m + 1, r);
merge(arr, l, m, r);
}
}
int main() {
int n;
cout << "Enter the size of the array: ";
cin >> n;
int arr[n];
cout << "Enter the elements of the array: ";
for (int i = 0; i < n; i++) {
cin >> arr[i];
}
int arr_size = sizeof(arr) / sizeof(arr[0]);
clock_t start;
clock_t end;
double timetaken;
start = clock();
mergeSort(arr, 0, arr_size - 1);
end = clock();
timetaken = (double)(end - start) / CLOCKS_PER_SEC;
.
Embedded C - Lecture 2
* Preprocessor directives (continue).
* Data types (Primitive, derived, user defined).
* Qualifiers (Size, Sign and volatile modifiers and Storage classes).
* Scope and lifetime.
* Makefile introduction.
Write a program that reads in integer as many as the user enters from.docx
Write a program that reads in integer as many as the user enters from standard input and print out the Maximum and minimum values.
Solution
Answer
#include<stdio.h>
#include<conio.h>
//------main function-----//
main()
{
//------Variable Declaration--------//
int a[10],i,n,max,min;
//-----Enter the input value using printf() and scanf() function------//
printf(\"Enter the Value of N:\");
scanf(\"%d\",&n);
printf(\"\ \");
//----Input N Element using For Loop----//
for(i=0;i<n;i++)
{
printf(\"Enter the %d Element\",i+1);
scanf(\"%d\",&a[i]);
}
min=a[0]; //----Minimum Number Initialization-----//
max=a[0]; //----Maximum Number Initialization-----//
//----Minimum and Maximum Number Calculation using For Loop-----//
for(i=0;i<n;i++)
{
if(a[i]>max)
max=a[i];
if(a[i]<min)
min=a[i];
}
//---------- Dispaly Result-------//
printf(\"\ The Maximum Number is %d\ \",max);
printf(\"the Minimum Number is %d\ \",min);
getch(); //---Function for Hold Output Screen--//
return 0;
}
.
VIT351 Software Development VI Unit2
The document provides information about arrays, strings, and character handling functions in C language. It discusses:
1. Definitions and properties of arrays, including declaring, initializing, and accessing single and multi-dimensional arrays.
2. Built-in functions for testing and mapping characters from the ctype.h library, including isalnum(), isalpha(), iscntrl(), isdigit(), ispunct(), and isspace().
3. Strings in C being arrays of characters terminated by a null character. It discusses common string handling functions from string.h like strlen(), strrev(), strlwr(), strupr(), strcpy(), strcat(), and strcmp().
Given an IntNode struct and the operating functions for a linked list-.pdf
Given an IntNode struct and the operating functions for a linked list, complete the following
functions to extend the functionality of the linked list:
IntNode* IntNode_GetNth(IntNode* firstNode, int n)- Return a pointer to the nth node of the list
starting at firstNode.
void IntNode_PrintList(IntNode* firstNode) - Call IntNode_PrintNodeData() to output values of
the list starting at firstNode. Do not add extra space characters in between values.
int IntNode_SumList(IntNode* firstNode) - Return the sum of the values of all nodes starting at
firstNode.
Note: The code for IntNode_Create() provided here differs from the code shown in the book.
The given main() performs various actions to test IntNode_GetNth(), IntNode_PrintList(), and
IntNode_SumList().
main() reads 5 integers from a user:
The number of nodes to be added to a new list
The value of the first node of the list
An increment between the values of two consecutive nodes.
A value of a new node
The position of a node after which the new node will be added, with 1 indicating the first node
Ex: If the input is:
the output is:
Here's the code
#include <stdio.h>
#include <stdlib.h>
typedef struct IntNode_struct {
int dataVal;
struct IntNode_struct* nextNodePtr;
} IntNode;
// Allocate a node for initData
IntNode* IntNode_Create (int initData) {
IntNode* newNode = (IntNode*)malloc(sizeof(IntNode));
newNode->dataVal = initData;
newNode->nextNodePtr = NULL;
return newNode;
}
/* Insert newNode after node.
Before: thisNode -- next
After: thisNode -- newNode -- next
*/
void IntNode_InsertAfter (IntNode* thisNode, IntNode* newNode) {
IntNode* tmpNext = NULL;
tmpNext = thisNode->nextNodePtr; // Remember next
thisNode->nextNodePtr = newNode; // this -- new -- ?
newNode->nextNodePtr = tmpNext; // this -- new -- next
}
// Print dataVal
void IntNode_PrintNodeData(IntNode* thisNode) {
printf("%d ", thisNode->dataVal);
}
// Grab location pointed by nextNodePtr
IntNode* IntNode_GetNext(IntNode* thisNode) {
return thisNode->nextNodePtr;
}
/* ******** New functions ********/
// Return the length of the list
int IntNode_Length(IntNode* firstNode) {
int length = 0;
IntNode* currentNode = firstNode;
while(currentNode != NULL) {
++length;
currentNode = currentNode->nextNodePtr;
}
return length;
}
// Return the Nth element of the list. First node is 1.
IntNode* IntNode_GetNth(IntNode* firstNode, int pos) {
/* Type your code here. */
}
// Print the entire list starting at firstNode
void IntNode_PrintList(IntNode* firstNode) {
/* Type your code here. */
}
// Sum the numbers in the list
int IntNode_SumList(IntNode* firstNode) {
/* Type your code here. */
}
int main(void) {
IntNode* firstNode = NULL;
IntNode* lastNode = NULL;
IntNode* newNode = NULL;
IntNode* curNode = NULL;
int listSize; // how many integers to read
int increment; // increment between consecutive node values
int firstNum; // first node value
// New node for newNum will be inserted after position newPos (first node = 1)
int newPos;
int newNum;
int i;
.
Arrays
The document outlines arrays and provides examples of declaring, initializing, and using arrays in C++ programs. It discusses declaring arrays with a specified size and type, initializing arrays using for loops or initializer lists, passing arrays to functions, and searching and sorting array elements. Examples demonstrate initializing arrays, outputting array contents, computing operations on array elements, and using arrays to count outcomes from rolling dice simulations.
Here is the code- I can't get it to work- I need a function that finds.pdf
Here is the code: I can't get it to work. I need a function that finds the maximum of two integers.
#include <stdio.h>
#include <stdlib.h>
#include <limits.h>
/* declare global variables including a table structure to hold scheduling
information */
/* optional: define a function that finds the maximum of two integers */
int total_processes = 0;
struct node{
int id;
int arrival;
int total_cycles;
int total_remaining;
int done;
int start;
int already_started;
int end;
int turnaround;
}* table = NULL;
typedef struct node table_type;
/***************************************************************/
void print_table() {
/*
1.declare local variables
2. print table header
3. for each process
4. print the contents (id, arrival time, total_cycles) of
each field of the table's index
5. if process has been scheduled ("done" field is 1,
6. print other contents (start time, end time, turnaround time)
*/
printf("ID/tArrival/tTotal/tStart/tEnd/tTurnaround/n");
printf("--------------------------------------------\n");
int i;
for(i=0; i < total_processes; i++){
printf("%d\t", table[i].id);
printf("%d\t", table[i].arrival);
printf("%d\t", table[i].total_cycles);
printf("%d\t", table[i].start);
printf("%d\t", table[i].end);
printf("%d\t", table[i].turnaround);
}
return;
}
/***************************************************************/
void procedure_1() {
/*"PROCEDURE FOR OPTION #1"
/* declare local variables */
/* prompt for total number of processes */
/* allocate memory for table to hold process parameters */
/* for each process */
/* prompt for process id, arrival time, and total cycle
time */
/* print contents of table */
int i;
printf("Enter total number of processes: ");
scanf("%d", &table[i].arrival);
table = (table_type *)malloc(total_processes * sizeof(table_type));
for(i = 0; i < total_processes; i++){
int arrival = -1;
int total_cycles = -1;
printf("Enter process id: ");
scanf("%d", &table[i].id);
}
printf("Enter arrival cycle for process P[%d]: ", i);
scanf("%d", &table[i].total_cycles);
/*
table[i].id = i;
table[i].arrival = -1;
table[i].total_cycles = -1;
table[i].start = -1;
table[i].end = -1;
table[i].turnaround = -1;
*/
print_table();
return;
}
/***************************************************************/
void procedure_2() {
/* "PROCEDURE FOR OPTION #2"
/* declare (and initilize when appropriate) local variables */
/* for each process, reset "done" field to 0 */
/* while there are still processes to schedule */
/* initilize the earliest arrival time to INT_MAX
(largest integer value) */
/* for each process not yet scheduled */
/* check if process has earlier arrival time
than current earliest and update */
/* set start time, end time, turnaround time, done fields
for unscheduled process with earliest arrival time */
/* update current cycle time and increment number of
processes scheduled */
/* print contents of table */
int i;
int done;
int min_value;
int current_cycle = 0;
int max;
int min_index;
table[i].done = 0;
while(done < 1){
min.
HeadPhone.java All the outputs were same because the variabl.pdf
// HeadPhone.java
/* All the outputs were same because the variables declared and defined in HeadPhone class
were static,
they will be updated only once during the program execution.
removing the static keyword from the variables which will be updated during the program
execution will
give the desired results
*/
public class HeadPhone {
private static final int LOW = 1;
private static final int MEDIUM = 2;
private static final int HIGH = 3;
private int volume = 0;
private boolean pluggedIn = false;
private String none = null;
private String manufacturer = none;
private String headPhoneColor = none;
private String volPos = none;
private String switchPos= none;
private String onOrOff= none;
// Default Headphone Constructor
public HeadPhone() {
volume = MEDIUM;
pluggedIn = false;
manufacturer = \"Sony\";
headPhoneColor = \"Gold\";
}
// Headphone Constructor
public HeadPhone (int volume2, boolean pluggedIn2, String manufacturer2, String
headPhoneColor2) {
volume = volume2;
pluggedIn = pluggedIn2;
manufacturer = manufacturer2;
headPhoneColor = headPhoneColor2;
}
// Setter Methods
// setVolume
public void setVolume(int volume2) {
volume = volume2;
}
// setPluggedIn
public void setPluggedIn(boolean pluggedIn2) {
pluggedIn = pluggedIn2;
}
// setManufacturer
public void setManufacturer(String manufacturer2) {
manufacturer = manufacturer2;
}
// setColor
public void setColor(String headPhoneColor2) {
headPhoneColor = headPhoneColor2;
}
// Getter Methods
// getVolume
public int getVolume() {
if(volume == 3) {
volPos = \"HIGH\";
}
else if (volume == 2) {
volPos = \"MEDIUM\";
}
else if(volume == 1) {
volPos = \"LOW\";
}
else {
System.out.println(\"Your entry is invalid, please try again.\");
}
return volume;
}
//getPluggedIn
public boolean getPluggedIn() {
if(pluggedIn == false) {
onOrOff = \"Off\";
}
else {
onOrOff = \"On\";
}
return pluggedIn;
}
//getManufacturer
public String getManufacturer() {
return manufacturer;
}
//getColor
public String getColor() {
return headPhoneColor;
}
// changeVolume method
public void changeVolume(int volume2) {
setVolume(volume2);
}
// toString method
public String toString() {
String str = \"toString() results: (volume=\" + volume + \", pluggedIn=\" + pluggedIn + \",
Manufacturer=\" + manufacturer + \", HeadPhone Color=\" + headPhoneColor +\")\";
return str;
}
}
// TestHeadPhone.java
import java.util.Scanner;
public class TestHeadPhone {
public static void main(String[] args) {
int row = 4;
HeadPhone [] hfArray = new HeadPhone[row];
HeadPhone hpObj1 = new HeadPhone();
hfArray[0] = hpObj1;
Scanner input = new Scanner(System.in);
for(int i=1; i<4; i++) {
HeadPhone hpObj2 = new HeadPhone();
System.out.println(\"Please enter the data for new headphones:\");
System.out.println(\"Would you like to turn the headphones on? (Type \'true\' for on, Type
\'false\' for off: \");
hpObj2.setPluggedIn(input.nextBoolean());
System.out.println(\"Please the the volume (Options are by integer (1 = Low, 2 = Medium, 3 =
High): \");
hpObj2.setV.
Methanol is soluble in water. methanol is a one carbon organic alco.pdf
Methanol is soluble in water. methanol is a one carbon organic alcohol. The -OH group forms
favorable hydrogen bonds with water, and the hydrophobic portion is small enough that it can
readily dissolve.
Solution
Methanol is soluble in water. methanol is a one carbon organic alcohol. The -OH group forms
favorable hydrogen bonds with water, and the hydrophobic portion is small enough that it can
readily dissolve..
More Related Content
Similar to Program to print all combination of size r in an array of size n.pdf
5 array
This document discusses arrays in C/C++. It defines an array as a collection of identical data objects stored in consecutive memory locations under a common name. It describes array declaration syntax, initialization, single and multi-dimensional arrays, and provides examples of using arrays to find the largest number, sort numbers, and calculate a sum. It also covers pointers, pointer arithmetic, and passing arrays to functions by value and reference.
write a program that given a list of 20 integers, sorts them accordi.pdf
write a program that given a list of 20 integers, sorts them according to insert sort. java
Solution
public class InsertionSort {
/**
* method to sort the array elements using insertion sorts *
*
* @param array
* @return
*/
public static int[] insertionSort(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;
}
return array;
}
/**
* method to print the array elements
*
* @param input
*/
private static void printNumbers(int[] input) {
for (int i = 0; i < input.length; i++) {
System.out.print(input[i] + \", \");
}
System.out.println(\"\ \");
}
/**
* @param args
*/
public static void main(String[] args) {
int[] input = { 3, 26, 67, 35, 9, -6, 43, 82, 10, 54, 14, 22, 21, 32,
121, 231, 12, 98, 77, 65 };
System.out.println(\"Before Sorting\");
printNumbers(input);
input = insertionSort(input);
System.out.println(\"Before Sorting\");
printNumbers(input);
}
}
OUTPUT:
Before Sorting
3, 26, 67, 35, 9, -6, 43, 82, 10, 54, 14, 22, 21, 32, 121, 231, 12, 98, 77, 65,
Before Sorting
-6, 3, 9, 10, 12, 14, 21, 22, 26, 32, 35, 43, 54, 65, 67, 77, 82, 98, 121, 231,.
c++ program I need to sort arrays using an insertion sort and a mer.pdf
c++ program: I need to sort arrays using an insertion sort and a merge sort (I already have). I
need to use these sorting methods to sort input arrays in the form of randomly generated numbers
in the size of 20, 100 and 200. So I need 3 randomily generated arrays and sorted and outputed
by both sorting method. I currently have it as given arrays and need help changing them to
randomily generated arrays
Here is my code so far:
#include
using namespace std;
void InsertionSort(int arr[],int size){
int temp,j;
for(int i=0; i=0){
arr[j+1] = arr[j];
j--;
}
arr[j+1] = temp;
}
}
//Merges two sub-arrays. First sub-array is arr[l..m],second is arr[m+1..r]
void merge(int arr[], int l, int m, int r){
int i, j, k;
int n1 = m - l + 1;
int n2 = r - m;
//Create temp arrays
int L[n1], R[n2];
//Copy data to temp arrays L[] and R[]
for (i = 0; i < n1; i++)
L[i] = arr[l + i];
for (j = 0; j < n2; j++)
R[j] = arr[m + 1+ j];
/* Merge arrays back into arr[l..r]*/
i = 0;
j = 0;
k = l;
while (i < n1 && j < n2){
if (L[i] <= R[j]){
arr[k] = L[i];
i++;
}
else{
arr[k] = R[j];
j++;
}
k++;
}
//Copy the remaining elements of L[], if there are any
while (i < n1){
arr[k] = L[i];
i++;
k++;
}
//Copy the remaining elements of R[], if there are any
while (j < n2){
arr[k] = R[j];
j++;
k++;
}
}
//l is left index and r is right index of the sub-array of array to be sorted
void MergeSort(int arr[], int l, int r){
if (l < r){
// Same as (l+r)/2, but avoids overflow for large l and h
int m = l+(r-l)/2;
// Sort first and second halves
MergeSort(arr, l, m);
MergeSort(arr, m+1, r);
merge(arr, l, m, r);
}
}
void printArray(int arr[], int size){//Function to print array
int i;
for (i=0; i < size; i++)
cout << arr[i] <<\" \";
cout << endl;
}
int main(){
int size = 20;
int array1[] = {17,33,88,54,96,82,12,36,69,74,79,66,32,11,1,98,76,55,3,4};
int array2[] = {20,1,19,2,18,3,17,4,16,5,15,6,14,7,13,8,12,9,11,10};
int array3[] = {220,209,198,187,176,165,154,143,132,121,110,99,88,77,66,55,44,33,22,11};
int array4[] = {15,35,85,100,40,90,70,65,25,5,10,30,55,75,80,95,20,45,50,60};
int array5[] = {100,99,98,97,96,95,94,93,92,91,90,89,88,87,86,85,84,83,82,81};
//Output for array1
cout << \"Original array 1: \ \";
printArray(array1, size);
InsertionSort(array1, size);
cout << \"Array 1 sorted by INSERTION SORT: \ \";
printArray(array1, size);
cout << \"\ \";
//Output for array2
cout << \"Original array 2: \ \";
printArray(array2, size);
MergeSort(array2,0,size);
cout << \"Array 2 sorted by MERGE SORT: \ \";
printArray(array2, size);
cout << \"\ \";
//Output for array3
cout << \"Original array 3: \ \";
printArray(array3, size);
InsertionSort(array3, size);
cout << \"Array 3 sorted by INSERTION SORT: \ \";
printArray(array3, size);
cout << \"\ \";
//Output for array4
cout << \"Original array 4: \ \";
printArray(array4, size);
MergeSort(array4,0,size);
cout << \"Array 4 sorted by MERGE SORT: \ \";
printArray(array4, size);
cout << \"\ \";
//Output for array5
cout << \"Original array 5: \ \";
printArray(arr.
For this homework, you will write a program to create and manipulate.pdf
For this homework, you will write a program to create and manipulate a simple linked list. For
each node in the linked list you will generate a random number, create a node that holds that
number, and insert that node into its sorted position in the linked list. Once the nodes are
inserted, you will traverse the list, printing the value held in each node. Then you will clean up
the list (deleting all the nodes) and exit the program. You will also learn about a tool that you can
use to help you check for memory errors in your code.
Specifications:
Define a struct appropriate for holding a random integer. This struct will also contain a \"next\"
pointer to reference a separate instance of the struct.
You may use the typedef keyword to give your struct a separate typename if you wish, but this is
not a requirement.
The program will read command line arguments. The first argument is the program name (as it
will always be) followed by a random number seed, the number of random numbers to generate
and ending with the Maximum Possible Value of the Random numbers generated (i.e. argc
should be 4).
Your program will use a \"dummy node\" for the linked list. In other words, the head of the list
will be an actual node, but the data field in that node will not be used for sorting purposes.
Your program must contain the following functions:
main() (for obvious reasons)
insertNodeSorted() - You may implement this either of two ways.
One way is to use the head of the list as one parameter and the integer value as the second. The
function will allocate memory for the node, initialize it and then insert it in the proper location in
the list.
The other way is to pass the head of the list as one parameter and a previously allocated and
initialized node as the other. The existing node is inserted in the proper location in the list.
printList() - Takes the head of the list as its only parameter, traverses the list, printing out the
data in sorted order.
deleteList() - Takes the head of the list as its only parameter, traverses the list, deleting all nodes.
The basic algorithm of your program is
Use a loop (based upon command line input values) that:
Generates a random number
Prints the number to stdout
Creates a node that contains that random number
Inserts the new node in sorted order into the existing list
Print the sorted linked list
Delete the linked list
You will use a Makefile to compile your program
Solution
#include
#include
#include
struct test_struct
{
int val;
struct test_struct *next;
};
struct test_struct *head = NULL;
struct test_struct *curr = NULL;
struct test_struct* create_list(int val)
{
printf(\"\ creating list with headnode as [%d]\ \",val);
struct test_struct *ptr = (struct test_struct*)malloc(sizeof(struct test_struct));
if(NULL == ptr)
{
printf(\"\ Node creation failed \ \");
return NULL;
}
ptr->val = val;
ptr->next = NULL;
head = curr = ptr;
return ptr;
}
struct test_struct* add_to_list(int val, bool add_to_end)
{
if(NULL == head)
{
retur.
The language is C. Thanks. Write a function that computes the intege.pdf
The language is C. Thanks. Write a function that computes the integer mean of an array of
integers. For example, the integer mean of - 1, 4, 2 is (- 1 + 4 + 2)/3 = 5/3 = 1, where/denotes
integer division. The function should implement the following specification;/* Computes the
integer mean of an array and stores it * where mn references. Returns -J for erroneous input *
(len
Solution
// C code
#include
int mean(int *a, int len, int *mn)
{
if(len <= 0)
return -1;
*mn = 0;
int i = 0;
for (i = 0; i < len; ++i)
{
*mn = *mn + *(a+i);
}
*mn = *mn/len;
return 0;
}
int main()
{
int size;
printf(\"Enter size of array: \");
scanf(\"%d\",&size);
int i;
int array[size];
for (i = 0; i < size; ++i)
{
printf(\"Enter array element at index %d: \",(i+1));
scanf(\"%d\",&array[i]);
}
int mn;
int result = mean(array, size, &mn);
if(result == 0)
printf(\"Mean: %d\ \", mn);
else
printf(\"Erroneous input\ \");
return 0;
}
/*
output:
Enter size of array: 3
Enter array element at index 1: -1
Enter array element at index 2: 4
Enter array element at index 3: 2
Mean: 1
*/.
Can you give an example of a binary heap programCan you give an .pdf
Can you give an example of a binary heap program
Can you give an example of a binary heap program
Solution
/*
* C Program to Implement a Heap & provide Insertion & Deletion Operation
*/
#include
int array[100], n;
main()
{
int choice, num;
n = 0;/*Represents number of nodes in the heap*/
while(1)
{
printf(\"1.Insert the element \ \");
printf(\"2.Delete the element \ \");
printf(\"3.Display all elements \ \");
printf(\"4.Quit \ \");
printf(\"Enter your choice : \");
scanf(\"%d\", &choice);
switch(choice)
{
case 1:
printf(\"Enter the element to be inserted to the list : \");
scanf(\"%d\", &num);
insert(num, n);
n = n + 1;
break;
case 2:
printf(\"Enter the elements to be deleted from the list: \");
scanf(\"%d\", &num);
delete(num);
break;
case 3:
display();
break;
case 4:
exit(0);
default:
printf(\"Invalid choice \ \");
}/*End of switch */
}/*End of while */
}/*End of main()*/
display()
{
int i;
if (n == 0)
{
printf(\"Heap is empty \ \");
return;
}
for (i = 0; i < n; i++)
printf(\"%d \", array[i]);
printf(\"\ \");
}/*End of display()*/
insert(int num, int location)
{
int parentnode;
while (location > 0)
{
parentnode =(location - 1)/2;
if (num <= array[parentnode])
{
array[location] = num;
return;
}
array[location] = array[parentnode];
location = parentnode;
}/*End of while*/
array[0] = num; /*assign number to the root node */
}/*End of insert()*/
delete(int num)
{
int left, right, i, temp, parentnode;
for (i = 0; i < num; i++) {
if (num == array[i])
break;
}
if (num != array[i])
{
printf(\"%d not found in heap list\ \", num);
return;
}
array[i] = array[n - 1];
n = n - 1;
parentnode =(i - 1) / 2; /*find parentnode of node i */
if (array[i] > array[parentnode])
{
insert(array[i], i);
return;
}
left = 2 * i + 1; /*left child of i*/
right = 2 * i + 2; /* right child of i*/
while (right < n)
{
if (array[i] >= array[left] && array[i] >= array[right])
return;
if (array[right] <= array[left])
{
temp = array[i];
array[i] = array[left];
array[left] = temp;
i = left;
}
else
{
temp = array[i];
array[i] = array[right];
array[right] = temp;
i = right;
}
left = 2 * i + 1;
right = 2 * i + 2;
}/*End of while*/
if (left == n - 1 && array[i]) {
temp = array[i];
array[i] = array[left];
array[left] = temp;
}
}
Output:.
Need this in JAVA We have N numbers as an array, you need to find a .pdf
Need this in JAVA We have N numbers as an array, you need to find a prefix array and a suffix
array, which we can get the maximum x or value with all elements in them. Notice that for
prefix[0, 1] and suffix[r, n - 1], do not intersect (I
Solution
// C++ program for a Trie based O(n) solution to find max
// subarray XOR
#include
using namespace std;
// Assumed int size
#define INT_SIZE 32
// A Trie Node
struct TrieNode
{
int value; // Only used in leaf nodes
TrieNode *arr[2];
};
// Utility function tp create a Trie node
TrieNode *newNode()
{
TrieNode *temp = new TrieNode;
temp->value = 0;
temp->arr[0] = temp->arr[1] = NULL;
return temp;
}
// Inserts pre_xor to trie with given root
void insert(TrieNode *root, int pre_xor)
{
TrieNode *temp = root;
// Start from the msb, insert all bits of
// pre_xor into Trie
for (int i=INT_SIZE-1; i>=0; i--)
{
// Find current bit in given prefix
bool val = pre_xor & (1<arr[val] == NULL)
temp->arr[val] = newNode();
temp = temp->arr[val];
}
// Store value at leaf node
temp->value = pre_xor;
}
// Finds the maximum XOR ending with last number in
// prefix XOR \'pre_xor\' and returns the XOR of this maximum
// with pre_xor which is maximum XOR ending with last element
// of pre_xor.
int query(TrieNode *root, int pre_xor)
{
TrieNode *temp = root;
for (int i=INT_SIZE-1; i>=0; i--)
{
// Find current bit in given prefix
bool val = pre_xor & (1<arr[1-val]!=NULL)
temp = temp->arr[1-val];
// If there is no prefix with opposite
// bit, then look for same bit.
else if (temp->arr[val] != NULL)
temp = temp->arr[val];
}
return pre_xor^(temp->value);
}
// Returns maximum XOR value of a subarray in arr[0..n-1]
int maxSubarrayXOR(int arr[], int n)
{
// Create a Trie and insert 0 into it
TrieNode *root = newNode();
insert(root, 0);
// Initialize answer and xor of current prefix
int result = INT_MIN, pre_xor =0;
// Traverse all input array element
for (int i=0; i
using namespace std;
// Assumed int size
#define INT_SIZE 32
// A Trie Node
struct TrieNode
{
int value; // Only used in leaf nodes
TrieNode *arr[2];
};
// Utility function tp create a Trie node
TrieNode *newNode()
{
TrieNode *temp = new TrieNode;
temp->value = 0;
temp->arr[0] = temp->arr[1] = NULL;
return temp;
}
// Inserts pre_xor to trie with given root
void insert(TrieNode *root, int pre_xor)
{
TrieNode *temp = root;
// Start from the msb, insert all bits of
// pre_xor into Trie
for (int i=INT_SIZE-1; i>=0; i--)
{
// Find current bit in given prefix
bool val = pre_xor & (1<arr[val] == NULL)
temp->arr[val] = newNode();
temp = temp->arr[val];
}
// Store value at leaf node
temp->value = pre_xor;
}
// Finds the maximum XOR ending with last number in
// prefix XOR \'pre_xor\' and returns the XOR of this maximum
// with pre_xor which is maximum XOR ending with last element
// of pre_xor.
int query(TrieNode *root, int pre_xor)
{
TrieNode *temp = root;
for (int i=INT_SIZE-1; i>=0; i--)
{
// Find current bit in given prefix
bool val = pre_xor & (1<arr[1-val]!=NULL)
temp = temp-.
There are a number of errors in the following program- All errors are.docx
There are a number of errors in the following program.
All errors are located in main() and structure definitions.
Function declarations and definitions are correct!
Locate all errors, fix them (as shown below), run the program and save its output
as a comment at the end of the source file.
Example:
int num = 10;
int *ptr;
num = &ptr; // <== Error: Comment the line and write the correct line below
// Write a short justification where appropriate
// num = &ptr; // Error #1
ptr = #
Name:
*/
#include <stdio.h>
#include <string.h>
#include <stdlib.h>
#include <time.h>
#define DUMMY_TRAILER '\177'
// octal ASCII code of the
// last character in the ASCII table
#define NUM_CITIES 10
typedef struct
{
char name[12];
int temperature[5];
} CITY;
// Stack and Queue Node
typedef struct node NODE;
struct node
{
CITY city;
node *next;
};
// Doubly Linked List Node
typedef struct d_node D_NODE;
struct d_node
{
CITY city;
NODE *forw;
NODE *back;
};
// Stack Functions
NODE *push(NODE *stack, const CITY *pStu);
NODE *pop(NODE **stack);
// Queue Functions
void enqueue(NODE **queue, NODE **rear, const CITY *pStu);
NODE *dequeue(NODE **queue, NODE **rear);
// Doubly Linked List Functions
D_NODE *init_list(void);
int insert(D_NODE *list, const CITY *pStu);
void traverse_forw(D_NODE *list);
void traverse_back(D_NODE *list);
// Other Functions
void printCity(const CITY *pCity);
int main (void)
{
CITY cList[NUM_CITIES] =
{
{"Cupertino", {88, 89, 87, 85, 89}},
{"Flagstaff", {81, 80, 88, 89, 89}},
{"Los Angeles", {87, 88, 89, 89, 90}},
{"Philadelphia", {96, 99, 99, 90, 95}},
{"Phoenix", {106, 109, 109, 100, 105}},
{"Portland", {89, 90, 85, 89, 90}},
{"Reno", {108, 105, 109, 100, 108}},
{"Salem", {85, 90, 85, 89, 90}},
{"Tucson", {107, 100, 109, 100, 108}},
{"Yreka", {101, 109, 100, 108, 109}}
};
NODE *stack = NULL;
NODE *top = NULL;
NODE *queue = NULL, *rear = NULL;
NODE *front;
D_NODE *list;
list = init_list();
// build stack and queue with data from an array of CITY structures
srand((unsigned int)time(NULL));
int count = rand() % 10;
for ( int n = 0; n < count; n++)
{
int i = rand() % NUM_CITIES;
int duplicate = insert(list, &cList[i]);
if(duplicate)
{
// already in the list!
push(stack, &cList[i]);
enqueue(&queue, &rear, cList[i]);
}
}
// display list
printf("\nLIST contents (forwards):\n");
traverse_forw(list);
printf("\nLIST contents (backwards):\n");
traverse_back(list);
// display stack
if (top)
{
printf("\nSTACK contents from top to bottom:\n");
while ((top = pop(stack)))
{
printCity(top->city);
}
}
else
printf ("Empty Stack!\n");
// display queue
if (front)
{
printf("\nQUEUE contents from front to rear:\n");
while ((front = dequeue( queue, rear)))
{
printCity(front->city);
}
}
else
printf ("Empty Queue!\n");
return 0;
}
/***************************************************
Displays the fileds of a CIS_CLASS structure
Pre pCls - a pointer to a CIS_CLASS structure
Post
*/
void printCity(const CITY *pCity)
{
printf("%-20s %3d\n", pCity->name, pCity->temperatur.
Write a program in C that generates a list of random numbers greater.pdf
Write a program in C that generates a list of random numbers greater than or equal to 0 and less
than 32, stores them in an array then have the program run through the numbers using a linked
lists then print out the numbers in the array.
Solution
#include
#include
#include
typedef struct node{ // create a struct for linked list
int data;
struct node *next;
}ll;
void main(){
srand(time(NULL)); // set the seed to random numbers
int n=10,i=0,num; // initilize the random numbers
ll *start = NULL; // set start = NULL
while(idata = rand()%32; // set a random number for a node
temp->next = start; // link node to start
start = temp; // move start to beginning
i++;
}
display(start);
}
void display(ll *start){
ll *temp = start;
while(temp!=NULL){ // while ll has elements
printf(\"%d->\",temp->data); // print data
temp = temp->next;
}
}
/* sample output
*/.
(Parent reference for BST) Redefine TreeNode by adding a reference to.pdf
(Parent reference for BST) Redefine TreeNode by adding a reference to a node\'s parent, as
shown below\': Reimplement the insert and delete methods in the BST class to update the parent
for each node in the tree. Add the following new method in BST:/** Returns the node for the
specified element. * Returns null if the element is not in the tree. */private TreeNode getNode(E
element)/** Returns true if the node for the element is a leaf */private boolean isLeaf(E
element)/** Returns the path of elements from the specified element * to the root in an array
list. */public ArrayList getPath(E e) Write a test program that prompts the user to enter 10
integers, adds them to the tree, deletes the first integer from the tree, and displays the paths for
all leaf nodes. Here is a sample run:
Solution
#include
#include
void swap(char *x, char *y)
{
char temp;
temp = *x;
*x = *y;
*y = temp;
void permute(char *a, int l, int r)
{
int i;
if (l == r)
printf(\"%s\ \", a);
else
{
for (i = l; i <= r; i++)
{
swap((a+l), (a+i));
permute(a, l+1, r);
swap((a+l), (a+i)); //backtrack
}
}
}
int main()
{
char str[] = \"ABC\";
int n = strlen(str);
permute(str, 0, n-1);
return 0;
}
part ii radix sort:
c++
#include
using namespace std;
int getMax(int arr[], int n)
{
int mx = arr[0];
for (int i = 1; i < n; i++)
if (arr[i] > mx)
mx = arr[i];
return main()
void countSort(int arr[], int n, int exp
for (i = 0; i < n; i++)
count[ (arr[i]/exp)%10 ]++;
for (i = 1; i < 10; i++)
count[i] += count[i - 1];
for (i = n - 1; i >= 0; i--)
{
output[count[ (arr[i]/exp)%10 ] - 1] = arr[i];
count[ (arr[i]/exp)%10 ]--;
}
for (i = 0; i < n; i++)
arr[i] = output[i];
}
adixsort(int arr[], int n)
{
int m = getMax(arr, n);
for (int exp = 1; m/exp > 0; exp *= 10)
countSort(arr, n, exp);
}
void print(int arr[], int n)
{
for (int i = 0; i < n; i++)
cout << arr[i] << \" \";
}
int main()
{
int arr[] = {170, 45, 75, 90, 802, 24, 2, 66};
int n = sizeof(arr)/sizeof(arr[0]);
radixsort(arr, n);
print(arr, n);
return 0;
}
java:
import java.io.*;
import java.util.*;
class Radix {
static int getMax(int arr[], int n)
{
int mx = arr[0];
for (int i = 1; i < n; i++)
if (arr[i] > mx)
mx = arr[i];
return mx;
}
static void countSort(int arr[], int n, int exp)
{
int output[] = new int[n]; // output array
int i;
int count[] = new int[10];
Arrays.fill(count,0);
for (i = 0; i < n; i++)
count[ (arr[i]/exp)%10 ]++;
for (i = 1; i < 10; i++)
count[i] += count[i - 1];
for (i = n - 1; i >= 0; i--)
{
output[count[ (arr[i]/exp)%10 ] - 1] = arr[i];
count[ (arr[i]/exp)%10 ]--;
}
for (i = 0; i < n; i++)
arr[i] = output[i];
}
static void radixsort(int arr[], int n)
{
int m = getMax(arr, n);
for (int exp = 1; m/exp > 0; exp *= 10)
countSort(arr, n, exp);
}
static void print(int arr[], int n)
{
for (int i=0; i
#include
void swap(char *x, char *y)
{
char temp;
temp = *x;
*x = *y;
*y = temp;
void permute(char *a, int l, int r)
{
int i;
if (l == r)
printf(\"%s\ \", a);
else
{
for (i = l; i <= r; i++)
{
swap((a+l), (a+i));
permute(a, l+1, .
Part 1)#include stdio.h #include stdlib.h #include pthrea.pdf
Part 1)
#include
#include
#include
#include
/*Error handling for pthread_create and pthread_join*/
/*from the pthread_create man page*/
#define handle_error_en(en, msg) \\
do { errno = en; perror(msg); exit(EXIT_FAILURE); } while (0)
/* # of running threads */
volatile int running_threads = 0;
pthread_t thread[3]; /*Descriptors for our 3 threads*/
int numOfElements;/*Total # of elements from the user*/
struct Results{ /*Struct to hold the statistical results*/
int min;
int max;
int average;
}Results;
/*This function finds the minimum element of an array*/
void *findMin(void *array_ptr){
int i; /*counter*/
int *elements = (int*)array_ptr; /*re reference void array pointer*/
Results.min = elements[0]; /*set minimum to first element */
for(i = 0; i < numOfElements; i++){ /*iterate through array*/
if(elements[i] < Results.min){ /*if the current element is less than the current min*/
Results.min = elements[i]; /*store the new min*/
}
}
running_threads -= 1; /*Decrement thread count*/
return NULL;
}
/*This function finds the maximum element of an array*/
void *findMax(void *array_ptr){
int i; /*counter*/
int *elements = (int*)array_ptr; /*re reference void array pointer*/
for(i = 0; i < numOfElements; i++){ /*iterate through array*/
if(elements[i] > Results.max){ /*store the new max*/
Results.max = elements[i];
}
}
running_threads -= 1; /*Decrement thread count*/
return NULL;
}
/*This function finds the average of an array*/
void *findAverage(void *array_ptr){
int i; /*counter*/
int *elements = (int*)array_ptr; /*re reference void array pointer*/
for(i = 0; i < numOfElements; i++){ /*iterate through array*/
Results.average += elements[i]; /*add element @ i to average*/
}
Results.average = Results.average/numOfElements; /*Divide the sum by the number of
elements*/
running_threads -= 1; /*Decrement running threads counter*/
return NULL;
}
/* This method accepts a int n(initial size of array) and
pointer to an array and returns # of elements in the array
*/
int getArrayInput(int n, int *array_ptr){
int input;/*Store user input */
int numberOfElements = 0;/*Number of Integers inputed*/
printf(\"Creating Dynamic Array...\ -\ \");
for(;;){ /*infinite loop*/
printf(\"Enter a positive value:\ Negative Number to Stop\ -\ \");
//Get Int from console, store at address of input
if (scanf(\"%d\",&input) != 1){
printf(\"\ Oops that wasn\'t an Integer\ lets try filling the array again\ Remember
INTEGERS only!\ \");
exit(EXIT_FAILURE);
}
if (input >= 0){
if (numberOfElements == n){
n += 1; //Make room for the current input
array_ptr = realloc(array_ptr, n * sizeof(int));//realloc array and set pointer
}
array_ptr[numberOfElements++] = input;//Store input at next empty element
} else {
printf(\"\ Number of Integers: %d\ \", numberOfElements);
break;
}
}
return numberOfElements;
}
/*This function joins our n number of threads */
void joinThreads(int numberOfThreads){
int i; /*count*/
int s; /*error #*/
while(numberOfThreads >= 0){ /*Join our threads*/
s = pthread_jo.
Digits.javapackage week04;import java.util.Scanner; public cla.pdf
Digits.java
package week04;
import java.util.Scanner;
public class Digits {
public static void main(String[] args)
{
//Declaring variables
int sum=0,number,rem;
//Scanner class Object is sued to read the inputs entered by the user
Scanner input = new Scanner(System.in);
//Getting the number entered by the user
System.out.print(\"Enter an integer : \");
number=input.nextInt();
//this loop continue to execute until the number is graeter than zero
while (number>0 )
{
//Getting the last digit of the number every time
rem=number%10;
/*Getting the remaining part of
* the number except last digit every time
*/
number=number/10;
//Calculating the sum of the digits
sum+=rem;
}
//Displaying the sum of the digits of the number
System.out.print(\"The sum of digits is: \" +sum);
}
}
______________________________________________
Output:
Enter an integer : 123456789
The sum of digits is: 45
__________________________________________Thank You
Solution
Digits.java
package week04;
import java.util.Scanner;
public class Digits {
public static void main(String[] args)
{
//Declaring variables
int sum=0,number,rem;
//Scanner class Object is sued to read the inputs entered by the user
Scanner input = new Scanner(System.in);
//Getting the number entered by the user
System.out.print(\"Enter an integer : \");
number=input.nextInt();
//this loop continue to execute until the number is graeter than zero
while (number>0 )
{
//Getting the last digit of the number every time
rem=number%10;
/*Getting the remaining part of
* the number except last digit every time
*/
number=number/10;
//Calculating the sum of the digits
sum+=rem;
}
//Displaying the sum of the digits of the number
System.out.print(\"The sum of digits is: \" +sum);
}
}
______________________________________________
Output:
Enter an integer : 123456789
The sum of digits is: 45
__________________________________________Thank You.
IN C CODEGiven an IntNode struct and the operating functions for a.pdf
In C ++
Complete ReplaceLevel() to output "levelPointer is null." if levelPointer is null. Otherwise,
assign levelValue to the variable pointed to by levelPointer. End with a newline.
Ex: If the input is Y A, then the output is:
Level is A.
#include
using namespace std;
void ReplaceLevel(char* levelPointer, char levelValue) {
/* Your code goes here */
}
int main() {
char level;
char* levelPointer;
char action;
char value;
level = 'Z';
cin >> action;
cin >> value;
if (action == 'Y') {
levelPointer = &level;
}
else {
levelPointer = nullptr;
}
ReplaceLevel(levelPointer, value);
cout << "Level is " << level << "." << endl;
return 0;
}.
I have written the code but cannot complete the assignment please help.pdf
I have written the code but cannot complete the assignment please help me to complete.
Please don't just copy other s answers as your own.
// Insertion sort
/*
#include <ctime>
#include <iostream>
using namespace std;
void insertionSort(int arr[], int n) {
int i, key, j;
for (i = 1; i < n; i++) {
key = arr[i];
j = i - 1;
// Move elements of arr[0..i-1], that are greater than key, to one position ahead of their current
position
while (j >= 0 && arr[j] > key) {
arr[j + 1] = arr[j];
j = j - 1;
}
arr[j + 1] = key;
}
}
int main() {
int n;
cout << "Enter the size of the array: ";
cin >> n;
int arr[n];
cout << "Enter the elements of the array: ";
for (int i = 0; i < n; i++) {
cin >> arr[i];
}
int num = sizeof(arr) / sizeof(arr[0]);
clock_t start, end;
double timetaken;
start = clock();
insertionSort(arr, num);
end = clock();
cout << "Sorted array: \n";
for (int i = 0; i < n; i++)
cout << arr[i] << " ";
cout << endl;
timetaken = ((double) (end - start)) / CLOCKS_PER_SEC;
cout << "Time taken : " << fixed << timetaken << "s" << endl;
return 0;
}
*/
// Shell Sort
/*
#include <ctime>
#include <iostream>
using namespace std;
int shellSort(int arr[], int n) {
for (int gap = n/2; gap > 0; gap /= 2) {
for (int i = gap; i < n; i += 1) {
int temp = arr[i];
int j;
for (j = i; j >= gap && arr[j - gap] > temp; j -= gap)
arr[j] = arr[j - gap];
arr[j] = temp;
}
}
return 0;
}
void printArray(int arr[], int n) {
for (int i=0; i<n; i++)
cout << arr[i] << " ";
}
int main() {
int n;
cout << "Enter the size of the array: ";
cin >> n;
int arr[n];
cout << "Enter the elements of the array: ";
for (int i = 0; i < n; i++) {
cin >> arr[i];
}
int num = sizeof(arr) / sizeof(arr[0]);
clock_t start, end;
double timetaken;
start = clock();
shellSort(arr, num);
end = clock();
cout << "\nArray after sorting: \n";
printArray(arr, n);
cout << endl;
timetaken = (double)(end - start) / CLOCKS_PER_SEC;
cout << "Time taken : " << fixed << timetaken << "s" << endl;
return 0;
}
*/
// MergeSort
/*
#include <ctime>
#include <iostream>
using namespace std;
void merge(int arr[], int l, int m, int r) {
int i, j, k;
int n1 = m - l + 1;
int n2 = r - m;
int L[n1], R[n2];
for (i = 0; i < n1; i++)
L[i] = arr[l + i];
for (j = 0; j < n2; j++)
R[j] = arr[m + 1 + j];
i = 0;
j = 0;
k = l;
while (i < n1 && j < n2) {
if (L[i] <= R[j]) {
arr[k] = L[i];
i++;
}
else {
arr[k] = R[j];
j++;
}
k++;
}
while (i < n1) {
arr[k] = L[i];
i++;
k++;
}
while (j < n2) {
arr[k] = R[j];
j++;
k++;
}
}
void mergeSort(int arr[], int l, int r) {
if (l < r) {
int m = l + (r - l) / 2;
mergeSort(arr, l, m);
mergeSort(arr, m + 1, r);
merge(arr, l, m, r);
}
}
int main() {
int n;
cout << "Enter the size of the array: ";
cin >> n;
int arr[n];
cout << "Enter the elements of the array: ";
for (int i = 0; i < n; i++) {
cin >> arr[i];
}
int arr_size = sizeof(arr) / sizeof(arr[0]);
clock_t start;
clock_t end;
double timetaken;
start = clock();
mergeSort(arr, 0, arr_size - 1);
end = clock();
timetaken = (double)(end - start) / CLOCKS_PER_SEC;
.
Embedded C - Lecture 2
* Preprocessor directives (continue).
* Data types (Primitive, derived, user defined).
* Qualifiers (Size, Sign and volatile modifiers and Storage classes).
* Scope and lifetime.
* Makefile introduction.
Write a program that reads in integer as many as the user enters from.docx
Write a program that reads in integer as many as the user enters from standard input and print out the Maximum and minimum values.
Solution
Answer
#include<stdio.h>
#include<conio.h>
//------main function-----//
main()
{
//------Variable Declaration--------//
int a[10],i,n,max,min;
//-----Enter the input value using printf() and scanf() function------//
printf(\"Enter the Value of N:\");
scanf(\"%d\",&n);
printf(\"\ \");
//----Input N Element using For Loop----//
for(i=0;i<n;i++)
{
printf(\"Enter the %d Element\",i+1);
scanf(\"%d\",&a[i]);
}
min=a[0]; //----Minimum Number Initialization-----//
max=a[0]; //----Maximum Number Initialization-----//
//----Minimum and Maximum Number Calculation using For Loop-----//
for(i=0;i<n;i++)
{
if(a[i]>max)
max=a[i];
if(a[i]<min)
min=a[i];
}
//---------- Dispaly Result-------//
printf(\"\ The Maximum Number is %d\ \",max);
printf(\"the Minimum Number is %d\ \",min);
getch(); //---Function for Hold Output Screen--//
return 0;
}
.
VIT351 Software Development VI Unit2
The document provides information about arrays, strings, and character handling functions in C language. It discusses:
1. Definitions and properties of arrays, including declaring, initializing, and accessing single and multi-dimensional arrays.
2. Built-in functions for testing and mapping characters from the ctype.h library, including isalnum(), isalpha(), iscntrl(), isdigit(), ispunct(), and isspace().
3. Strings in C being arrays of characters terminated by a null character. It discusses common string handling functions from string.h like strlen(), strrev(), strlwr(), strupr(), strcpy(), strcat(), and strcmp().
Given an IntNode struct and the operating functions for a linked list-.pdf
Given an IntNode struct and the operating functions for a linked list, complete the following
functions to extend the functionality of the linked list:
IntNode* IntNode_GetNth(IntNode* firstNode, int n)- Return a pointer to the nth node of the list
starting at firstNode.
void IntNode_PrintList(IntNode* firstNode) - Call IntNode_PrintNodeData() to output values of
the list starting at firstNode. Do not add extra space characters in between values.
int IntNode_SumList(IntNode* firstNode) - Return the sum of the values of all nodes starting at
firstNode.
Note: The code for IntNode_Create() provided here differs from the code shown in the book.
The given main() performs various actions to test IntNode_GetNth(), IntNode_PrintList(), and
IntNode_SumList().
main() reads 5 integers from a user:
The number of nodes to be added to a new list
The value of the first node of the list
An increment between the values of two consecutive nodes.
A value of a new node
The position of a node after which the new node will be added, with 1 indicating the first node
Ex: If the input is:
the output is:
Here's the code
#include <stdio.h>
#include <stdlib.h>
typedef struct IntNode_struct {
int dataVal;
struct IntNode_struct* nextNodePtr;
} IntNode;
// Allocate a node for initData
IntNode* IntNode_Create (int initData) {
IntNode* newNode = (IntNode*)malloc(sizeof(IntNode));
newNode->dataVal = initData;
newNode->nextNodePtr = NULL;
return newNode;
}
/* Insert newNode after node.
Before: thisNode -- next
After: thisNode -- newNode -- next
*/
void IntNode_InsertAfter (IntNode* thisNode, IntNode* newNode) {
IntNode* tmpNext = NULL;
tmpNext = thisNode->nextNodePtr; // Remember next
thisNode->nextNodePtr = newNode; // this -- new -- ?
newNode->nextNodePtr = tmpNext; // this -- new -- next
}
// Print dataVal
void IntNode_PrintNodeData(IntNode* thisNode) {
printf("%d ", thisNode->dataVal);
}
// Grab location pointed by nextNodePtr
IntNode* IntNode_GetNext(IntNode* thisNode) {
return thisNode->nextNodePtr;
}
/* ******** New functions ********/
// Return the length of the list
int IntNode_Length(IntNode* firstNode) {
int length = 0;
IntNode* currentNode = firstNode;
while(currentNode != NULL) {
++length;
currentNode = currentNode->nextNodePtr;
}
return length;
}
// Return the Nth element of the list. First node is 1.
IntNode* IntNode_GetNth(IntNode* firstNode, int pos) {
/* Type your code here. */
}
// Print the entire list starting at firstNode
void IntNode_PrintList(IntNode* firstNode) {
/* Type your code here. */
}
// Sum the numbers in the list
int IntNode_SumList(IntNode* firstNode) {
/* Type your code here. */
}
int main(void) {
IntNode* firstNode = NULL;
IntNode* lastNode = NULL;
IntNode* newNode = NULL;
IntNode* curNode = NULL;
int listSize; // how many integers to read
int increment; // increment between consecutive node values
int firstNum; // first node value
// New node for newNum will be inserted after position newPos (first node = 1)
int newPos;
int newNum;
int i;
.
Arrays
The document outlines arrays and provides examples of declaring, initializing, and using arrays in C++ programs. It discusses declaring arrays with a specified size and type, initializing arrays using for loops or initializer lists, passing arrays to functions, and searching and sorting array elements. Examples demonstrate initializing arrays, outputting array contents, computing operations on array elements, and using arrays to count outcomes from rolling dice simulations.
Here is the code- I can't get it to work- I need a function that finds.pdf
Here is the code: I can't get it to work. I need a function that finds the maximum of two integers.
#include <stdio.h>
#include <stdlib.h>
#include <limits.h>
/* declare global variables including a table structure to hold scheduling
information */
/* optional: define a function that finds the maximum of two integers */
int total_processes = 0;
struct node{
int id;
int arrival;
int total_cycles;
int total_remaining;
int done;
int start;
int already_started;
int end;
int turnaround;
}* table = NULL;
typedef struct node table_type;
/***************************************************************/
void print_table() {
/*
1.declare local variables
2. print table header
3. for each process
4. print the contents (id, arrival time, total_cycles) of
each field of the table's index
5. if process has been scheduled ("done" field is 1,
6. print other contents (start time, end time, turnaround time)
*/
printf("ID/tArrival/tTotal/tStart/tEnd/tTurnaround/n");
printf("--------------------------------------------\n");
int i;
for(i=0; i < total_processes; i++){
printf("%d\t", table[i].id);
printf("%d\t", table[i].arrival);
printf("%d\t", table[i].total_cycles);
printf("%d\t", table[i].start);
printf("%d\t", table[i].end);
printf("%d\t", table[i].turnaround);
}
return;
}
/***************************************************************/
void procedure_1() {
/*"PROCEDURE FOR OPTION #1"
/* declare local variables */
/* prompt for total number of processes */
/* allocate memory for table to hold process parameters */
/* for each process */
/* prompt for process id, arrival time, and total cycle
time */
/* print contents of table */
int i;
printf("Enter total number of processes: ");
scanf("%d", &table[i].arrival);
table = (table_type *)malloc(total_processes * sizeof(table_type));
for(i = 0; i < total_processes; i++){
int arrival = -1;
int total_cycles = -1;
printf("Enter process id: ");
scanf("%d", &table[i].id);
}
printf("Enter arrival cycle for process P[%d]: ", i);
scanf("%d", &table[i].total_cycles);
/*
table[i].id = i;
table[i].arrival = -1;
table[i].total_cycles = -1;
table[i].start = -1;
table[i].end = -1;
table[i].turnaround = -1;
*/
print_table();
return;
}
/***************************************************************/
void procedure_2() {
/* "PROCEDURE FOR OPTION #2"
/* declare (and initilize when appropriate) local variables */
/* for each process, reset "done" field to 0 */
/* while there are still processes to schedule */
/* initilize the earliest arrival time to INT_MAX
(largest integer value) */
/* for each process not yet scheduled */
/* check if process has earlier arrival time
than current earliest and update */
/* set start time, end time, turnaround time, done fields
for unscheduled process with earliest arrival time */
/* update current cycle time and increment number of
processes scheduled */
/* print contents of table */
int i;
int done;
int min_value;
int current_cycle = 0;
int max;
int min_index;
table[i].done = 0;
while(done < 1){
min.
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HeadPhone.java All the outputs were same because the variabl.pdf
// HeadPhone.java
/* All the outputs were same because the variables declared and defined in HeadPhone class
were static,
they will be updated only once during the program execution.
removing the static keyword from the variables which will be updated during the program
execution will
give the desired results
*/
public class HeadPhone {
private static final int LOW = 1;
private static final int MEDIUM = 2;
private static final int HIGH = 3;
private int volume = 0;
private boolean pluggedIn = false;
private String none = null;
private String manufacturer = none;
private String headPhoneColor = none;
private String volPos = none;
private String switchPos= none;
private String onOrOff= none;
// Default Headphone Constructor
public HeadPhone() {
volume = MEDIUM;
pluggedIn = false;
manufacturer = \"Sony\";
headPhoneColor = \"Gold\";
}
// Headphone Constructor
public HeadPhone (int volume2, boolean pluggedIn2, String manufacturer2, String
headPhoneColor2) {
volume = volume2;
pluggedIn = pluggedIn2;
manufacturer = manufacturer2;
headPhoneColor = headPhoneColor2;
}
// Setter Methods
// setVolume
public void setVolume(int volume2) {
volume = volume2;
}
// setPluggedIn
public void setPluggedIn(boolean pluggedIn2) {
pluggedIn = pluggedIn2;
}
// setManufacturer
public void setManufacturer(String manufacturer2) {
manufacturer = manufacturer2;
}
// setColor
public void setColor(String headPhoneColor2) {
headPhoneColor = headPhoneColor2;
}
// Getter Methods
// getVolume
public int getVolume() {
if(volume == 3) {
volPos = \"HIGH\";
}
else if (volume == 2) {
volPos = \"MEDIUM\";
}
else if(volume == 1) {
volPos = \"LOW\";
}
else {
System.out.println(\"Your entry is invalid, please try again.\");
}
return volume;
}
//getPluggedIn
public boolean getPluggedIn() {
if(pluggedIn == false) {
onOrOff = \"Off\";
}
else {
onOrOff = \"On\";
}
return pluggedIn;
}
//getManufacturer
public String getManufacturer() {
return manufacturer;
}
//getColor
public String getColor() {
return headPhoneColor;
}
// changeVolume method
public void changeVolume(int volume2) {
setVolume(volume2);
}
// toString method
public String toString() {
String str = \"toString() results: (volume=\" + volume + \", pluggedIn=\" + pluggedIn + \",
Manufacturer=\" + manufacturer + \", HeadPhone Color=\" + headPhoneColor +\")\";
return str;
}
}
// TestHeadPhone.java
import java.util.Scanner;
public class TestHeadPhone {
public static void main(String[] args) {
int row = 4;
HeadPhone [] hfArray = new HeadPhone[row];
HeadPhone hpObj1 = new HeadPhone();
hfArray[0] = hpObj1;
Scanner input = new Scanner(System.in);
for(int i=1; i<4; i++) {
HeadPhone hpObj2 = new HeadPhone();
System.out.println(\"Please enter the data for new headphones:\");
System.out.println(\"Would you like to turn the headphones on? (Type \'true\' for on, Type
\'false\' for off: \");
hpObj2.setPluggedIn(input.nextBoolean());
System.out.println(\"Please the the volume (Options are by integer (1 = Low, 2 = Medium, 3 =
High): \");
hpObj2.setV.
Methanol is soluble in water. methanol is a one carbon organic alco.pdf
Methanol is soluble in water. methanol is a one carbon organic alcohol. The -OH group forms
favorable hydrogen bonds with water, and the hydrophobic portion is small enough that it can
readily dissolve.
Solution
Methanol is soluble in water. methanol is a one carbon organic alcohol. The -OH group forms
favorable hydrogen bonds with water, and the hydrophobic portion is small enough that it can
readily dissolve..
Solubility in water happens when something is pol.pdf
Solubility in water happens when something is polar or charged. Charged is much
easier to understand--it\'s when there\'s an actual positive or negative charge on any given
molecule. Polarity is just like charge, but instead of the electrons being specifically on a given
atom and not on another (like in Na+ and Cl-), they\'re shared in a way such that they spend
more time on one atom vs another (like CH3OH--the electrons spend more time on the O than
they do on either the C or the Hs). How do you know which atoms like to hold onto electrons
more than other atoms? Electronegativity differences. Certain atoms like F, N, and O are really
electronegative, which means they like to hold onto electrons more than other atoms, like, say, C
and H. So, if F were bound to C, the electrons would be on the F more often than the C. On the
other hand, if you have to atoms, like C and H, which have very similar electronegativity values,
they will share the electrons more readily, and the electrons don\'t stay on one or the other atoms
more often. Such molecules are called non-polar, and are, in fact, not soluble in water. Instead,
they\'re fat-soluble, because fat is just a whole bunch of C-H bonds next to each other. So,
looking at the structures of those given vitamins, vitamin A consists mostly of a bunch of C-H
bonds; it\'s likely fat soluble. Vitamin B6 has a bunch of C-O-H bonds and an N, too. It\'s much
more polar, and likely water soluble. Vitamin C has tons of oxygens and is definitely water
soluble. Vitamin E is mostly C-H bonds and not too many O-H or O-C bonds, so it\'s likely fat
soluble. Hope that helps! :)
Solution
Solubility in water happens when something is polar or charged. Charged is much
easier to understand--it\'s when there\'s an actual positive or negative charge on any given
molecule. Polarity is just like charge, but instead of the electrons being specifically on a given
atom and not on another (like in Na+ and Cl-), they\'re shared in a way such that they spend
more time on one atom vs another (like CH3OH--the electrons spend more time on the O than
they do on either the C or the Hs). How do you know which atoms like to hold onto electrons
more than other atoms? Electronegativity differences. Certain atoms like F, N, and O are really
electronegative, which means they like to hold onto electrons more than other atoms, like, say, C
and H. So, if F were bound to C, the electrons would be on the F more often than the C. On the
other hand, if you have to atoms, like C and H, which have very similar electronegativity values,
they will share the electrons more readily, and the electrons don\'t stay on one or the other atoms
more often. Such molecules are called non-polar, and are, in fact, not soluble in water. Instead,
they\'re fat-soluble, because fat is just a whole bunch of C-H bonds next to each other. So,
looking at the structures of those given vitamins, vitamin A consists mostly of a bunch of C-H
bonds; it\'s likely fat soluble. Vi.
moles = 0.144 .pdf
The document provides the value of moles as 0.144. No other information or context is given about what this value represents.
It is used to remove water from the organic layer.pdf
It is used to remove water from the organic layer. During the experiment some
water gets into the organic layer because of the partial miscibility of the organic products in
water. Saturated sodium chloride will pull the water from the organic layer following simple
solute rules. Sat. NaCl wants to become more dilute and salts have a higher affinity for water
than for organic solvents. Therefore removing the bulk of the water in the organic layer.
Solution
It is used to remove water from the organic layer. During the experiment some
water gets into the organic layer because of the partial miscibility of the organic products in
water. Saturated sodium chloride will pull the water from the organic layer following simple
solute rules. Sat. NaCl wants to become more dilute and salts have a higher affinity for water
than for organic solvents. Therefore removing the bulk of the water in the organic layer..
nitrogen is more electronegative than phosphorus .pdf
nitrogen is more electronegative than phosphorus , so nitrogen tends to attracts
more of the electron cloud towards itself. hence nitrogen phosphorous bond in not fully covalent.
hence there is no equal sharing of electrons
Solution
nitrogen is more electronegative than phosphorus , so nitrogen tends to attracts
more of the electron cloud towards itself. hence nitrogen phosphorous bond in not fully covalent.
hence there is no equal sharing of electrons.
ionic equation is 2 Cu+2(aq) + 4Cl-(aq) + 4 K+(aq.pdf
ionic equation is 2 Cu+2(aq) + 4Cl-(aq) + 4 K+(aq)+ 4I-(aq) --> 2 CuI(s) + I2(aq) +
4 K+(aq)+ 4Cl-(aq)
Solution
ionic equation is 2 Cu+2(aq) + 4Cl-(aq) + 4 K+(aq)+ 4I-(aq) --> 2 CuI(s) + I2(aq) +
4 K+(aq)+ 4Cl-(aq).
hydration of alkene. .pdf
The document discusses hydration of alkene. Alkene undergoes addition reaction with water to form an alcohol. Water adds across the carbon-carbon double bond in alkene resulting in a saturated alcohol product.
H2O + NO + NO2 -- 2HNO2 Three molecules collide.pdf
H2O + NO + NO2 --> 2HNO2 Three molecules collide at an instant is rare, thus it
is not an elementary reaction that can occur at a significant rate.
Solution
H2O + NO + NO2 --> 2HNO2 Three molecules collide at an instant is rare, thus it
is not an elementary reaction that can occur at a significant rate..
he coordination compounds , metal complexes or si.pdf
he coordination compounds , metal complexes or simply complexes are compounds
which contain an ion or a central atom, usually a metal, surrounded by a collection of ions or
molecules. The chance discovery dell\'esamminocobalto (III) chloride, CoCl 3 6 NH 3 , due to
Tassaert in 1798 marks the beginning of the chemistry of the complexes. Many of the jobs
immediately after they were made ??with ammonia and complex that resulted were called
metallammine. Subsequently it was found that other amines and other anions such as CN - , NO
2 - , NCS - , Cl - can form complexes with metals. Many compounds were prepared with these
anions and, at first, were called by the name of the chemical that prepared them first.
Subsequently, because of the fact that many of these complexes was colored, was used a scheme
of nomenclature based on the color. Our current knowledge about the nature of the metal
complexes are due to Werner , in 1893 at the age of 26 years suggested that even today is called
coordination theory of Werner. His most important are three postulates: Most of the elements
have two types of meaning: a) primary valence and b) secondary valence. In modern terminology
a) corresponds to the oxidation state and b) the coordination number Each element tends to
satisfy the valence both primary and secondary The secondary valence is directed towards fixed
positions of the space Consider the complex cloroamminocobalto (III). According to the theory
of Werner the first member of the series has the formula Co [(NH 3 ) 6 ] Cl 3 and is represented
as shown in the figure: (i upload for u on this address: Structures
http://i45.tinypic.com/2l2654.gif http://i50.tinypic.com/jaykgy.jpg) The primary valence or
oxidation state of the cobalt (III) is 3. The three ions Cl - primary saturate the valence of cobalt
chloride ions and then the three primary saturate the valence of cobalt negative ions that
neutralize the positive charge of the metal ion affect the primary value.
Solution
he coordination compounds , metal complexes or simply complexes are compounds
which contain an ion or a central atom, usually a metal, surrounded by a collection of ions or
molecules. The chance discovery dell\'esamminocobalto (III) chloride, CoCl 3 6 NH 3 , due to
Tassaert in 1798 marks the beginning of the chemistry of the complexes. Many of the jobs
immediately after they were made ??with ammonia and complex that resulted were called
metallammine. Subsequently it was found that other amines and other anions such as CN - , NO
2 - , NCS - , Cl - can form complexes with metals. Many compounds were prepared with these
anions and, at first, were called by the name of the chemical that prepared them first.
Subsequently, because of the fact that many of these complexes was colored, was used a scheme
of nomenclature based on the color. Our current knowledge about the nature of the metal
complexes are due to Werner , in 1893 at the age of 26 years suggested that even today is called
coordination .
PdCl2 + 2DMSO PdCl22DMSO It forms a complexe..pdf
PdCl2 + 2DMSO ? PdCl2*2DMSO It forms a complexe.
Solution
PdCl2 + 2DMSO ? PdCl2*2DMSO It forms a complexe..
melting point of the mixture decreases when small.pdf
melting point of the mixture decreases when small amount of naphthalene is added
Solution
melting point of the mixture decreases when small amount of naphthalene is added.
Polymers are made by joining small molecules into large ones.But m.pdf
Polymers are made by joining small molecules into large ones.But most of these monomer
molecules are perfectly stable as they are, so they have devised two general methods to make
them react with each other, building up the backbone chain as the reaction proceeds.
This method (also known as step-growth) requires that the monomers possess two or more kinds
of functional groups that are able to react with each other in such a way that parts of these groups
combine to form a small molecule (often H2O) which is eliminated from the two pieces.
2 Additional Polymerisation:
Polymer play a major role in various industry sector and which are used in plastic bags ,water
pipes outer layer of
electric wires clothing and,woven fabric and also other materials.
PLASTIC BAGS:
WATER PIPES
OUTER LAYER OF ELECTRIC WIRES:
Outer layer of insulator is also called as electric insulator.
It is the most commonly used thermoplastic insulator for cables.
It is cheap, durable and widely available.
However, the chlorine in PVC (a halogen) causes the production of thick, toxic, black smoke
when burnt and can be a health hazard in areas where low smoke and toxicity are required
Normal operating temperatures are typically between 75C and 105C (depending on PVC type).
CLOTHING:
Solution
Polymers are made by joining small molecules into large ones.But most of these monomer
molecules are perfectly stable as they are, so they have devised two general methods to make
them react with each other, building up the backbone chain as the reaction proceeds.
This method (also known as step-growth) requires that the monomers possess two or more kinds
of functional groups that are able to react with each other in such a way that parts of these groups
combine to form a small molecule (often H2O) which is eliminated from the two pieces.
2 Additional Polymerisation:
Polymer play a major role in various industry sector and which are used in plastic bags ,water
pipes outer layer of
electric wires clothing and,woven fabric and also other materials.
PLASTIC BAGS:
WATER PIPES
OUTER LAYER OF ELECTRIC WIRES:
Outer layer of insulator is also called as electric insulator.
It is the most commonly used thermoplastic insulator for cables.
It is cheap, durable and widely available.
However, the chlorine in PVC (a halogen) causes the production of thick, toxic, black smoke
when burnt and can be a health hazard in areas where low smoke and toxicity are required
Normal operating temperatures are typically between 75C and 105C (depending on PVC type).
CLOTHING:.
What is Operational ExcellenceAchieving Operational Excellence requ.pdf
What is Operational Excellence?Achieving Operational Excellence requires the successful
implementation of a integrated Business Execution System that effectively and seamlessly
integrates the following four building blocks: 1)Strategy Deployment2)Performance
Management3)Process Excellence and4)High Performance Work Teams.Strategy or Policy
Deployment is the process that aligns and links business strategy and execution. Strategy
Deployment Performance Management High Performance Work Teams Process Excellence
Operational Excellence Business Execution System Strategy Deployment Process
Tools: 1)The Hoshin X-Matrix is a tool that visualizes an organization’sstrategic
objectives,strategic initiatives,key performance indicators,key projects &action items, andhuman
resources requirements in one simple matrix. 2)The Hoshin X-Matrixenables an organization to
easily review the alignment of its strategic objectives, strategicinitiatives, key performance
indicators, key action items and human resources.
Supply chain management:If you manage a small firm that makes a few products or sells a few
services, chances are you will have a small number of suppliers. Supply chain: 1)A firm\'s
supply chain is a network of organization and business processes for procuring raw materials,
transforming these materials into intermediate and finished products, and distributing the
finished products to customers.2)Supply chain software is classified as either software to help
businesses plan their supply chains (supply chain planning) or software to help them execute the
supply chain steps. 3)it manage the flow of products through distribution centers and warehouses
to ensure that products are delivered to the right locations in the most efficient manner.
Customer Relationship Management Systems 1) You were probably heard phrases such as the
customer is always right or the customer comes first. Today these words ring more true than
ever. 2)Because competitive advantage based on an innovative new product or service is often
very short lived, companies are realizing that their only enduring competitive strength may be
their relationships with their customers. 3) Some say that the basis of competition has switched
from who sells the most products and services to whom own the customer, and that customer
relationships represent a firm\'s most valuable asset. 4) Customer relationship management
systems typically provide software and online tools for sales, customer service, and marketing.
5)Partner relationship management (PRM) uses many of the same data, tools, and systems as
customer relationship management to enhance collaboration between a company and its selling
partners.
Enterprise Application: 1)Enterprise application require not only deep seated technological
changes but also fundamental changes in the way the business operates. 2)Companies must make
sweeping changes to their business processes to work with the software. 3)Companies adopting
enterprise applications can .
the name arsenic acid, like phosphoric acid (H3PO4), does notconvey .pdf
the name arsenic acid, like phosphoric acid (H3PO4), does notconvey the structure, but does fit
into a general patternwhereby the ending ‘ic’ denotes a higher or the
highest possible oxidation state (compare nitric acid, sulfuricacid).
HClO: Hypochlorous Acid
Solution
the name arsenic acid, like phosphoric acid (H3PO4), does notconvey the structure, but does fit
into a general patternwhereby the ending ‘ic’ denotes a higher or the
highest possible oxidation state (compare nitric acid, sulfuricacid).
HClO: Hypochlorous Acid.
Sucrose is a molecular compound and does not ionize in solution.KC.pdf
Sucrose is a molecular compound and does not ionize in solution.
KCl is a soluble ionic compound and fully ionizes in solution:
KCl(aq) => K+(aq) + Cl-(aq)
Sucrose and KCl do not interact in solution.
Mass balance equation: [KCl] = [K+]
Charge balance equation: [K+] = [Cl-]
Solution
Sucrose is a molecular compound and does not ionize in solution.
KCl is a soluble ionic compound and fully ionizes in solution:
KCl(aq) => K+(aq) + Cl-(aq)
Sucrose and KCl do not interact in solution.
Mass balance equation: [KCl] = [K+]
Charge balance equation: [K+] = [Cl-].
C. Increasing the temperature of the system .pdf
Increasing the temperature of the system is proposed as a solution. No other details are provided about the system, the current temperature, the proposed increased temperature, or the expected effects of a higher temperature. The summary is limited due to the brevity of the original document.
because Electron Affinity is the amount of energy.pdf
because Electron Affinity is the amount of energy needed to add an electron to an
atom or molecule.. Thus Decreasing its Stabillity and Whenever We have Decrease in Stability
We Have A Negative Energy Released thus energy required and hence negative electron
affinity..The Electron affinity of an atom or molecule is defined as the amount of energy released
when an electron is added to a neutral atom or molecule to form a negative ion. X + e- ? X-
(note: the correct equation in words is X + electron -> X + energy) This property is measured for
atoms and molecules in the gaseous state only, since in the solid or liquid states their energy
levels would be changed by contact with other atoms or molecules. A list of the electron
affinities was used by Robert S. Mulliken to develop an electronegativity scale for atoms, equal
to the average of the electron affinity and ionization potential.Other theoretical concepts that use
electron affinity include electronic chemical potential and chemical hardness. Another example,
a molecule or atom that has a more positive value of electron affinity than another is often called
an electron acceptor and the less positive an electron donor. Together they may undergo charge-
transfer reactions. In solids, the electron affinity is the energy difference between the vacuum
energy and the conduction band minimum. To use electron affinities properly, it is essential to
keep track of sign. For any reaction that releases energy, the change in energy, ?E, has a negative
value and the reaction is called an exothermic process. Electron capture for almost all non-noble
gas atoms involves the release of energy and thus are exothermic. The positive values that are
listed in tables of Eea are amounts or magnitudes. It is the word, released within the definition
energy released that supplies the negative sign. Confusion arises in mistaking Eea for a change in
energy, ?E, in which case the positive values listed in tables would be for an endo- not exo-
thermic process. The relation between the two is, Eea = - ?E(attach). However, if the value
assigned to Eea is negative, the negative sign implies a reversal of direction, and energy is
required to attach an electron. In this case, the electron capture is an endothermic process and the
relationship, Eea = - ?E(attach) is still valid. Negative values typically arise for the capture of a
second electron, but also for the nitrogen atom. The usual expression for calculating Eea when
an electron is attached is Eea = (Einitial - Efinal)attach = - ?E(attach) This expression does
follow the convention ?X = X(final) - X(initial) since - ?E = - (E(final) - E(initial)) = E(initial) -
E(final). Equivalently, electron affinity can also be defined as the amount of energy required to
detach an electron from a singly charged negative ion,i.e. the energy change for the process X- ?
X + e- If the same table is employed for the forward and reverse reactions, without switching
signs, care must be take.
Ques-1Clostredium botulinum, bacteria that cause food poisoning &.pdf
Ques-1:
Clostredium botulinum, bacteria that cause food poisoning & it is the etiological agent in this
outbreak of food poisoning
Ques-2:
This food intoxication because \"botulinum\" is the toxin generated by these bacteria by
putrefying food material and these bacteria is anaerobic. These toxins are made of exotoxins,
endotoxins (tetanus toxin, botulism toxin & cholera toxin) as these are composed of proteins,
lipoppolysaccharides produced through cell wall to act on the host cell immune system. These
are virulent and cause damage to host cells through inflammatory mediators released via
phagocytosis by immune system. I would declare the situation as a sortumme an outbreak, this is
because acute episodes of \"gastrointestinal illness is mainly due to food borne illness caused by
\"clostridium botulinum\". It has observed clear symptoms of disease characterized by nausea,
headache, blurred vision, bronchoconstriction due to stiffen throat & neck
Fever, headache, nausea, and severe respiratory symptoms followed by blurred vision; this is
because botulinum toxin is a neurotoxin that potentially inhibits release of acetylcholine from
nerve endings and leading to drooping eyelid and slurred speech.
Ques-3:
Potato salad was transported in a loose in a pickup truck & it is not covered or packed before
preparing meals therefore anaerobic fermentation of this carbohydrate is the source of
contamination of this bacteria & cause food borne illness
Solution
Ques-1:
Clostredium botulinum, bacteria that cause food poisoning & it is the etiological agent in this
outbreak of food poisoning
Ques-2:
This food intoxication because \"botulinum\" is the toxin generated by these bacteria by
putrefying food material and these bacteria is anaerobic. These toxins are made of exotoxins,
endotoxins (tetanus toxin, botulism toxin & cholera toxin) as these are composed of proteins,
lipoppolysaccharides produced through cell wall to act on the host cell immune system. These
are virulent and cause damage to host cells through inflammatory mediators released via
phagocytosis by immune system. I would declare the situation as a sortumme an outbreak, this is
because acute episodes of \"gastrointestinal illness is mainly due to food borne illness caused by
\"clostridium botulinum\". It has observed clear symptoms of disease characterized by nausea,
headache, blurred vision, bronchoconstriction due to stiffen throat & neck
Fever, headache, nausea, and severe respiratory symptoms followed by blurred vision; this is
because botulinum toxin is a neurotoxin that potentially inhibits release of acetylcholine from
nerve endings and leading to drooping eyelid and slurred speech.
Ques-3:
Potato salad was transported in a loose in a pickup truck & it is not covered or packed before
preparing meals therefore anaerobic fermentation of this carbohydrate is the source of
contamination of this bacteria & cause food borne illness.
Programprinting.c#include stdio.hint main() { cha.pdf
//Program
//printing.c
#include
int main()
{
char ch=\'A\';
int i,j;
//to print a)
for(i=6;i>=0;i--)
{
for(j=0;j
Solution
//Program
//printing.c
#include
int main()
{
char ch=\'A\';
int i,j;
//to print a)
for(i=6;i>=0;i--)
{
for(j=0;j.
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