C++ Program to Implement Doubly Linked List #includei.pdf
•
0 likes•30 views
This C++ program implements a doubly linked list with the following functions: - create() adds a node to the empty list - append() adds a node to the end of the list - insert() inserts a node at a given index position - del() deletes a node with a given value - display() prints out the contents of the list It uses a node struct to define each list element and a DLL class to implement the linked list functions.
Report
Share
Report
Share
Download to read offline
Recommended
#includeiostream#includecstdio#includecstdlib Node .pdf
#include
#include
#include
/*
* Node Declaration
*/
using namespace std;
struct node
{
int info;
struct node *next;
struct node *prev;
}*start;
/*
Class Declaration
*/
class double_llist
{
public:
void create_list(int value);
void add_begin(int value);
void add_after(int value, int position);
void delete_element(int value);
void search_element(int value);
void display_dlist();
void count();
void reverse();
double_llist()
{
start = NULL;
}
};
/*
* Main: Conatins Menu
*/
int main()
{
int choice, element, position;
double_llist dl;
while (1)
{
cout<>choice;
switch ( choice )
{
case 1:
cout<<\"Enter the element: \";
cin>>element;
dl.create_list(element);
cout<>element;
dl.add_begin(element);
cout<>element;
cout<<\"Insert Element after postion: \";
cin>>position;
dl.add_after(element, position);
cout<>element;
dl.delete_element(element);
cout<info = value;
temp->next = NULL;
if (start == NULL)
{
temp->prev = NULL;
start = temp;
}
else
{
s = start;
while (s->next != NULL)
s = s->next;
s->next = temp;
temp->prev = s;
}
}
/*
* Insertion at the beginning
*/
void double_llist::add_begin(int value)
{
if (start == NULL)
{
cout<<\"First Create the list.\"<prev = NULL;
temp->info = value;
temp->next = start;
start->prev = temp;
start = temp;
cout<<\"Element Inserted\"<next;
if (q == NULL)
{
cout<<\"There are less than \";
cout<info = value;
if (q->next == NULL)
{
q->next = tmp;
tmp->next = NULL;
tmp->prev = q;
}
else
{
tmp->next = q->next;
tmp->next->prev = tmp;
q->next = tmp;
tmp->prev = q;
}
cout<<\"Element Inserted\"<info == value)
{
tmp = start;
start = start->next;
start->prev = NULL;
cout<<\"Element Deleted\"<next->next != NULL)
{
/*Element deleted in between*/
if (q->next->info == value)
{
tmp = q->next;
q->next = tmp->next;
tmp->next->prev = q;
cout<<\"Element Deleted\"<next;
}
/*last element deleted*/
if (q->next->info == value)
{
tmp = q->next;
free(tmp);
q->next = NULL;
cout<<\"Element Deleted\"<info<<\" <-> \";
q = q->next;
}
cout<<\"NULL\"<next;
cnt++;
}
cout<<\"Number of elements are: \"<next;
p1->next = NULL;
p1->prev = p2;
while (p2 != NULL)
{
p2->prev = p2->next;
p2->next = p1;
p1 = p2;
p2 = p2->prev;
}
start = p1;
cout<<\"List Reversed\"<
Solution
#include
#include
#include
/*
* Node Declaration
*/
using namespace std;
struct node
{
int info;
struct node *next;
struct node *prev;
}*start;
/*
Class Declaration
*/
class double_llist
{
public:
void create_list(int value);
void add_begin(int value);
void add_after(int value, int position);
void delete_element(int value);
void search_element(int value);
void display_dlist();
void count();
void reverse();
double_llist()
{
start = NULL;
}
};
/*
* Main: Conatins Menu
*/
int main()
{
int choice, element, position;
double_llist dl;
while (1)
{
cout<>choice;
switch ( choice )
{
case 1:
cout<<\"Enter the element: \";
cin>>element;
dl.create_list(element);
cout<>element;
dl.add_begin(element);
cout<>element;
cout<<\"Insert Element after postion: \";
cin>>position;
dl.add_after(element, position);
cout.
#includeiostream#includecstdio#includecstdlibusing names.pdf
#include
#include
#include
using namespace std;
/*
* Node Declaration
*/
struct node
{
int info;
struct node *next;
}*last;
/*
* Class Declaration
*/
class circular_llist
{
public:
void create_node(int value);
void add_begin(int value);
void add_after(int value, int position);
void delete_element(int value);
void search_element(int value);
void display_list();
void update();
void sort();
circular_llist()
{
last = NULL;
}
};
/*
* Create Circular Link List
*/
void circular_llist::create_node(int value)
{
struct node *temp;
temp = new(struct node);
temp->info = value;
if (last == NULL)
{
last = temp;
temp->next = last;
}
else
{
temp->next = last->next;
last->next = temp;
last = temp;
}
}
/*
* Insertion of element at beginning
*/
void circular_llist::add_begin(int value)
{
if (last == NULL)
{
cout<<\"First Create the list.\"<info = value;
temp->next = last->next;
last->next = temp;
}
/*
* Deletion of element from the list
*/
void circular_llist::delete_element(int value)
{
struct node *temp, *s;
s = last->next;
/* If List has only one element*/
if (last->next == last && last->info == value)
{
temp = last;
last = NULL;
free(temp);
return;
}
if (s->info == value) /*First Element Deletion*/
{
temp = s;
last->next = s->next;
free(temp);
return;
}
while (s->next != last)
{
/*Deletion of Element in between*/
if (s->next->info == value)
{
temp = s->next;
s->next = temp->next;
free(temp);
cout<<\"Element \"<next;
}
/*Deletion of last element*/
if (s->next->info == value)
{
temp = s->next;
s->next = last->next;
free(temp);
last = s;
return;
}
/*
* Display Circular Link List
*/
void circular_llist::display_list()
{
struct node *s;
if (last == NULL)
{
cout<<\"List is empty, nothing to display\"<
Solution
#include
#include
#include
using namespace std;
/*
* Node Declaration
*/
struct node
{
int info;
struct node *next;
}*last;
/*
* Class Declaration
*/
class circular_llist
{
public:
void create_node(int value);
void add_begin(int value);
void add_after(int value, int position);
void delete_element(int value);
void search_element(int value);
void display_list();
void update();
void sort();
circular_llist()
{
last = NULL;
}
};
/*
* Create Circular Link List
*/
void circular_llist::create_node(int value)
{
struct node *temp;
temp = new(struct node);
temp->info = value;
if (last == NULL)
{
last = temp;
temp->next = last;
}
else
{
temp->next = last->next;
last->next = temp;
last = temp;
}
}
/*
* Insertion of element at beginning
*/
void circular_llist::add_begin(int value)
{
if (last == NULL)
{
cout<<\"First Create the list.\"<info = value;
temp->next = last->next;
last->next = temp;
}
/*
* Deletion of element from the list
*/
void circular_llist::delete_element(int value)
{
struct node *temp, *s;
s = last->next;
/* If List has only one element*/
if (last->next == last && last->info == value)
{
temp = last;
last = NULL;
free(temp);
return;
}
if (s->info == value) /*First Element Deletion*/
{
temp = s;
last->next = s->next;
free(temp);
return;
}
while (s->next != .
I need help completing this C++ code with these requirements.instr.pdf
I need help completing this C++ code with these requirements.
instructions: IN C++ LANGUAGE PLEASE
Update the comments for each prototype by filling in the pre and post conditions.
Remember that pre conditions indicate what conditions must be met BEFORE the user calls the
function. The function will not work properly unless the pre conditions are met.
Post conditions indicate what will be the result of the function call. In other words, what
conditions will exist after the function is called.
See class notes from Thursday on pre and post conditions for some examples and further
explanation.
Code begins here:
#include //for NULL
class List
{
private:
struct Node
{
int data;
Node* next;
Node(int data): data(data), next(NULL){}
};
typedef struct Node* Nodeptr;
Nodeptr first;
Nodeptr last;
int size;
public:
/**Constructors and Destructors*/
List();
//Default constructor; initializes and empty list
//Postcondition:
~List();
//Destructor. Frees memory allocated to the list
//Postcondition:
/**Accessors*/
int getFirst();
//Returns the first element in the list
//Precondition:
int getLast();
//Returns the last element in the list
//Precondition:
bool isEmpty();
//Determines whether a list is empty.
int getSize();
//Returns the size of the list
/**Manipulation Procedures*/
void removeLast();
//Removes the value of the last element in the list
//Precondition:
//Postcondition:
void removeFirst();
//Removes the value of the first element in the list
//Precondition:
//Postcondition:
void insertLast(int data);
//Inserts a new element at the end of the list
//If the list is empty, the new element becomes both first and last
//Postcondition:
void insertFirst(int data);
//Inserts a new element at the start of the list
//If the list is empty, the new element becomes both first and last
//Postcondition:
/**Additional List Operations*/
void printList();
//Prints to the console the value of each element in the list sequentially
//and separated by a blank space
//Prints nothing if the list is empty
};
Solution
PROGRAM CODE:
#include
#include //for NULL
using namespace std;
class List
{
private:
struct Node
{
int data;
Node* next;
Node(int data): data(data), next(NULL){}
};
typedef struct Node* Nodeptr;
Nodeptr first;
Nodeptr last;
int size;
public:
/**Constructors and Destructors*/
List()
{
first = (Node*) malloc(sizeof(Node));
last = (Node*) malloc(sizeof(Node));
size = 0;
}
//Default constructor; initializes and empty list
//Postcondition:
~List()
{
delete first;
delete last;
}
//Destructor. Frees memory allocated to the list
//Postcondition:
/**Accessors*/
int getFirst()
{
return first->data;
}
//Returns the first element in the list
//Precondition:
int getLast()
{
return last->data;
}
//Returns the last element in the list
//Precondition:
bool isEmpty()
{
if(first == NULL)
return true;
else return false;
}
//Determines whether a list is empty.
int getSize()
{
return size;
}
//Returns the size of the list
/**Manipulation Procedures*/
void removeLast()
{
last = .
Write a program to implement below operations with both singly and d.pdf
Write a program to implement below operations with both singly and doubly linked list. Use
inheritance to define singly linked list and doubly linked list. Add Node() Delete Node() Update
Node() Delete All() Size()
Solution
#include
using namespace std;
class List;
class Node
{
friend class List;
protected:
int Data;
Node *Link;
};
class List:public Node
{
friend class Node;
private:
Node *Start;
public:
List()
{
Start=NULL;
}
Node Insert(int x);
Node Delete(int &x);
void Display();
};
Node List::Insert(int x)
{
Node *New=new Node;
New->Data=x;
New->Link=Start;
Start=New;
return *this;
}
Node List::Delete(int &x)//deletes first node
{
Node *Del=Start;
Start=Start->Link;
delete Del;
return *this;
}
void List::Display()
{
Node *P=Start;
while(P!=NULL)
{
cout<Data<<\" \";
P=P->Link;
}
}
int main()
{
List L;
L.Insert(4);
L.Insert(64);
L.Insert(3);
L.Insert(90);
L.Insert(66);
L.Insert(13);
L.Display();
}
Double Linked List.
Please need help on following program using c++ language. Please inc.pdf
Please need help on following program using c++ language. Please include all header files and
main file with their own title.
Extend the class linkedListType by adding the following operations:
Write a function that returns the info of the kth element of the linked list. If no such element
exists, terminate the program.
Write a function that deletes the kth element of the linked list. If no such element exists,
terminate the program.
Provide the definitions of these functions in the class linkedListType. Also, write a program to
test these functions. (Use either the class unorderedLinkedList or the
class orderedLinkedList to test your function.)
Solution
C++ Code for the given problem is below.....
.....................................................
/*
* C++ Program to Implement Singly Linked List
*/
#include
#include
#include
//using namespace std;
/*
* Node Declaration
*/
typedef struct nodetype
{
int info;
nodetype *next ;
} node;
/*
* Class Declaration
*/
class linkedListType
{
public:
void insfrombeg( node **head, int);
void delfromkth(node **head,int);
void findthekthnode(node *head,int);
void traverse(node *head);
int item,ch,kth;
node *ptr,*head,*temp;
};
/*
* Main :contains menu
*/
void main()
{
int item,ch,kth;
node *ptr,*head,*temp;
linkedListType s1;
clrscr();
while (1)
{
while(1)
{
cout<<\"PRESS (1) FOR INSERT INTO LIST\"<>ch;
switch(ch)
{
case 1:
cout<<\"ENTER THE VALUE \"<>item;
s1.insfrombeg(&head,item);
break;
case 2:
s1.traverse(head);
break;
case 3:
cout<<\"ENTER THE Kth Element which you want\'s to delete\"<>kth;
s1.delfromkth(&head,kth);
break;
case 4:
cout<<\"ENTER THE Kth Element which you want\'s to search \ \"<>kth;
s1.findthekthnode(head,kth);
break;
case 5:
exit(1);
}
}
// getch();
}
}
/*
* Inserting element in beginning
*/
void linkedListType::insfrombeg(node **head,int item)
{
temp= ( node * ) malloc (sizeof( node ));
temp->info= item;
temp->next= NULL;
if(*head == NULL)
{
temp->next= *head;
*head= temp;
}
else
{
temp->next= *head;
*head= temp;
}
}
/*
* Delete element at a given position
*/
void linkedListType::delfromkth(node **head,int kth)
{ node* temp1=*head,*temp2;
if(*head ==NULL)
{
cout<<\"LINK LIST IS EMPTY\ \"<next;
free(temp);
}
else
for(int j=0;jnext;
temp2= temp1->next;
temp1->next=temp2->next;
free(temp2);
}
/*
* Searching an element
*/
void linkedListType::findthekthnode(node *head,int kth)
{
ptr=head;
int i=1;
while((ptr!=NULL))
{
if(i==kth)
{
cout<<\"\ The Element at\"<< kth<< \"position with value is \\t
\"<info<info<next;
i++;
}
}
}
/*
* Traverse Link List
*/
void linkedListType::traverse(node *head)
{
ptr = head;
while((ptr->next)->next!=NULL)
{
cout<info<<\"\\t\";
ptr=ptr->next;
}
cout<
#include
#include
typedef struct nodetype
{
int info;
nodetype *next ;
} node;
int item,ch,kth,total=1;
node *ptr,*head,*temp;
void insfrombeg( node **head, int);
void delfromkth(node **head,int);
void findthekthnode(node *head,int);
void traverse(node *head);
void main()
{
clrscr();
while(1)
{
printf(\"PRESS .
C++ Doubly-Linked ListsThe goal of the exercise is to implement a.pdf
C++: Doubly-Linked Lists
The goal of the exercise is to implement a class List of doubly- linked lists.
My goal is to implement the class in a file doubly-linked.cpp.
The class List implements lists using the following structures for list ele- ments:
struct Node { int val ;
Node next;
Node prev; };
Each Node contains a value (an integer) and two pointers: one meant to point to the next element
of the list and one meant to point to the previous element in the list. The class provides the
following methods that you need to implement:
• A constructor and a destructor, with no arguments;
• void insert(int n): insert an integer at the end of the list;
• void reverse(): reverse the list;
• void print(): print the list to cout in the same format as the input (i.e. integers separated by
spaces);
doubly-linked.h
#ifndef __dll__
#define __dll__
#include
using namespace std;
// Basic structure to store elements of a list
struct Node {
int val; // contains the value
Node * next; // pointer to the next element in the list
Node * prev; // pointer to the previous element in the list
};
// Class List
class List {
public:
List(void); // Constructor
~List(void); // Destructor
void insert(int n); // This should insert n in the list
void reverse(void); // This should reverse the list
void print(void); // This shoiuld print the list
private:
Node * first; // Pointer to the first (if any) element in the list
};
#endif
main.cpp
#include
#include \"doubly-linked.h\"
using namespace std;
int main(void){
List l;
int n;
while(cin >> n){
l.insert(n);
}
// Print list as read from cin
l.print();
// Reverse the list and print it
l.reverse();
l.print();
// Reverse again and print it
l.reverse();
l.print();
return 0;
}
Solution
#include
using namespace std;
/* Linked list structure */
struct list {
struct list *prev;
int data;
struct list *next;
} *node = NULL, *first = NULL, *last = NULL, *node1 = NULL, *node2 = NULL;
class linkedlist {
public:
/* Function for create/insert node at the beginning of Linked list */
void insrt_frnt() {
list *addBeg = new list;
cout << \"Enter value for the node:\" << endl;
cin >> addBeg->data;
if(first == NULL) {
addBeg->prev = NULL;
addBeg->next = NULL;
first = addBeg;
last = addBeg;
cout << \"Linked list Created!\" << endl;
}
else {
addBeg->prev = NULL;
first->prev = addBeg;
addBeg->next = first;
first = addBeg;
cout << \"Data Inserted at the beginning of the Linked list!\" << endl;
}
}
/* Function for create/insert node at the end of Linked list */
void insrt_end() {
list *addEnd = new list;
cout << \"Enter value for the node:\" << endl;
cin >> addEnd->data;
if(first == NULL) {
addEnd->prev = NULL;
addEnd->next = NULL;
first = addEnd;
last = addEnd;
cout << \"Linked list Created!\" << endl;
}
else {
addEnd->next = NULL;
last->next = addEnd;
addEnd->prev = last;
last = addEnd;
cout << \"Data Inserted at the end of the Linked list!\" << endl;
}
}
/* Function for Display Linked list */
void display() {
node = first;
cout << \"List of data in L.
hi i have to write a java program involving link lists. i have a pro.pdf
hi i have to write a java program involving link lists. i have a problem with the nodes, as it is
posting errors with the nodes.
please help me with this problem. thank you.
public class LinkLists {
/* only need to store a single pointer to the node at the head
* of the list.
* The pointer is null if the list is empty.
* Also record the size of the list.
*/
protected Node head;
/* invariant: size is the number of nodes in the list pointed to by head */
protected int size;
/* no-arguments default constructor creates an empty list */
public LinkLists() {
head = null; // start with an empty list
size = 0;
}
/* accessor method */
public int size() {
return size;
}
/* value to add to the end of the list
*/
public void add(T value) {
head = addAtEnd(head, value);
size++;
}
/* node of the list to which the value should be added
* value to add to the end of the list
*/
private Node addAtEnd(Node node, T value) {
if (node == null) { // special case
return new Node(value, null);
} else if (node.getNext() == null) { // other special case
node.setNext(new Node(value, null));
} else {
addAtEnd(node.getNext(), value);
}
return node;
}
/* iterative implementation of the same method
* value to add to the end of the list
*/
public void add2(T value) {
if (head == null) {
head = new Node(value, null);
} else {
Node node = head; // guaranteed not to be null initially
while (node.getNext() != null) {
node = node.getNext(); // guaranteed not to be null here
}
// now, node.getNext() is guaranteed to be null
// similar to the second special case in addAtEnd
node.setNext(new Node(value, null));
}
size++;
}
public void remove(int position) throws BadItemCountException {
if ((position < 1) || (position > size)) {
throw new
BadItemCountException(\"invalid position \" + position +
\", only 1..\" + size + \" available\");
}
if (position == 1) {
head = head.getNext();
} else {
Node node = head;
for (int i = 2; i < position; i++) {
node = node.getNext();
}
node.setNext(node.getNext().getNext());
}
size--; // one less item
}
public String toString() {
return toString(head);
}
private String toString(Node node) {
if (node == null) {
return \"\";
} else {
return node.getValue() + \"\ \" + toString(node.getNext());
}
}
public static void main(String[] args) {
/* create two empty lists, make sure they print out correctly */
LinkLists list1 = new LinkLists();
LinkLists list2 = new LinkLists();
System.out.println(\"list1 = \'\" + list1 + \"\', list2 = \'\" + list2 + \"\'\");
System.out.println(\"list1.size() = \" + list1.size() +
\", list2.size() = \" + list2.size());
/* insert some items, keep checking */
list1.add(\"hello\");
list1.add(\"world\");
list2.add(\"foo\");
list2.add(\"bar\");
list2.add(\"baz\");
System.out.println(\"list1 = \'\" + list1 + \"\', list2 = \'\" + list2 + \"\'\");
System.out.println(\"list1.size() = \" + list1.size() +
\", list2.size() = \" + list2.size());
/* remove an item at an invalid position */
boolean caught = false;
try {
list2.remove(4);
} catch .
Recommended
#includeiostream#includecstdio#includecstdlib Node .pdf
#include
#include
#include
/*
* Node Declaration
*/
using namespace std;
struct node
{
int info;
struct node *next;
struct node *prev;
}*start;
/*
Class Declaration
*/
class double_llist
{
public:
void create_list(int value);
void add_begin(int value);
void add_after(int value, int position);
void delete_element(int value);
void search_element(int value);
void display_dlist();
void count();
void reverse();
double_llist()
{
start = NULL;
}
};
/*
* Main: Conatins Menu
*/
int main()
{
int choice, element, position;
double_llist dl;
while (1)
{
cout<>choice;
switch ( choice )
{
case 1:
cout<<\"Enter the element: \";
cin>>element;
dl.create_list(element);
cout<>element;
dl.add_begin(element);
cout<>element;
cout<<\"Insert Element after postion: \";
cin>>position;
dl.add_after(element, position);
cout<>element;
dl.delete_element(element);
cout<info = value;
temp->next = NULL;
if (start == NULL)
{
temp->prev = NULL;
start = temp;
}
else
{
s = start;
while (s->next != NULL)
s = s->next;
s->next = temp;
temp->prev = s;
}
}
/*
* Insertion at the beginning
*/
void double_llist::add_begin(int value)
{
if (start == NULL)
{
cout<<\"First Create the list.\"<prev = NULL;
temp->info = value;
temp->next = start;
start->prev = temp;
start = temp;
cout<<\"Element Inserted\"<next;
if (q == NULL)
{
cout<<\"There are less than \";
cout<info = value;
if (q->next == NULL)
{
q->next = tmp;
tmp->next = NULL;
tmp->prev = q;
}
else
{
tmp->next = q->next;
tmp->next->prev = tmp;
q->next = tmp;
tmp->prev = q;
}
cout<<\"Element Inserted\"<info == value)
{
tmp = start;
start = start->next;
start->prev = NULL;
cout<<\"Element Deleted\"<next->next != NULL)
{
/*Element deleted in between*/
if (q->next->info == value)
{
tmp = q->next;
q->next = tmp->next;
tmp->next->prev = q;
cout<<\"Element Deleted\"<next;
}
/*last element deleted*/
if (q->next->info == value)
{
tmp = q->next;
free(tmp);
q->next = NULL;
cout<<\"Element Deleted\"<info<<\" <-> \";
q = q->next;
}
cout<<\"NULL\"<next;
cnt++;
}
cout<<\"Number of elements are: \"<next;
p1->next = NULL;
p1->prev = p2;
while (p2 != NULL)
{
p2->prev = p2->next;
p2->next = p1;
p1 = p2;
p2 = p2->prev;
}
start = p1;
cout<<\"List Reversed\"<
Solution
#include
#include
#include
/*
* Node Declaration
*/
using namespace std;
struct node
{
int info;
struct node *next;
struct node *prev;
}*start;
/*
Class Declaration
*/
class double_llist
{
public:
void create_list(int value);
void add_begin(int value);
void add_after(int value, int position);
void delete_element(int value);
void search_element(int value);
void display_dlist();
void count();
void reverse();
double_llist()
{
start = NULL;
}
};
/*
* Main: Conatins Menu
*/
int main()
{
int choice, element, position;
double_llist dl;
while (1)
{
cout<>choice;
switch ( choice )
{
case 1:
cout<<\"Enter the element: \";
cin>>element;
dl.create_list(element);
cout<>element;
dl.add_begin(element);
cout<>element;
cout<<\"Insert Element after postion: \";
cin>>position;
dl.add_after(element, position);
cout.
#includeiostream#includecstdio#includecstdlibusing names.pdf
#include
#include
#include
using namespace std;
/*
* Node Declaration
*/
struct node
{
int info;
struct node *next;
}*last;
/*
* Class Declaration
*/
class circular_llist
{
public:
void create_node(int value);
void add_begin(int value);
void add_after(int value, int position);
void delete_element(int value);
void search_element(int value);
void display_list();
void update();
void sort();
circular_llist()
{
last = NULL;
}
};
/*
* Create Circular Link List
*/
void circular_llist::create_node(int value)
{
struct node *temp;
temp = new(struct node);
temp->info = value;
if (last == NULL)
{
last = temp;
temp->next = last;
}
else
{
temp->next = last->next;
last->next = temp;
last = temp;
}
}
/*
* Insertion of element at beginning
*/
void circular_llist::add_begin(int value)
{
if (last == NULL)
{
cout<<\"First Create the list.\"<info = value;
temp->next = last->next;
last->next = temp;
}
/*
* Deletion of element from the list
*/
void circular_llist::delete_element(int value)
{
struct node *temp, *s;
s = last->next;
/* If List has only one element*/
if (last->next == last && last->info == value)
{
temp = last;
last = NULL;
free(temp);
return;
}
if (s->info == value) /*First Element Deletion*/
{
temp = s;
last->next = s->next;
free(temp);
return;
}
while (s->next != last)
{
/*Deletion of Element in between*/
if (s->next->info == value)
{
temp = s->next;
s->next = temp->next;
free(temp);
cout<<\"Element \"<next;
}
/*Deletion of last element*/
if (s->next->info == value)
{
temp = s->next;
s->next = last->next;
free(temp);
last = s;
return;
}
/*
* Display Circular Link List
*/
void circular_llist::display_list()
{
struct node *s;
if (last == NULL)
{
cout<<\"List is empty, nothing to display\"<
Solution
#include
#include
#include
using namespace std;
/*
* Node Declaration
*/
struct node
{
int info;
struct node *next;
}*last;
/*
* Class Declaration
*/
class circular_llist
{
public:
void create_node(int value);
void add_begin(int value);
void add_after(int value, int position);
void delete_element(int value);
void search_element(int value);
void display_list();
void update();
void sort();
circular_llist()
{
last = NULL;
}
};
/*
* Create Circular Link List
*/
void circular_llist::create_node(int value)
{
struct node *temp;
temp = new(struct node);
temp->info = value;
if (last == NULL)
{
last = temp;
temp->next = last;
}
else
{
temp->next = last->next;
last->next = temp;
last = temp;
}
}
/*
* Insertion of element at beginning
*/
void circular_llist::add_begin(int value)
{
if (last == NULL)
{
cout<<\"First Create the list.\"<info = value;
temp->next = last->next;
last->next = temp;
}
/*
* Deletion of element from the list
*/
void circular_llist::delete_element(int value)
{
struct node *temp, *s;
s = last->next;
/* If List has only one element*/
if (last->next == last && last->info == value)
{
temp = last;
last = NULL;
free(temp);
return;
}
if (s->info == value) /*First Element Deletion*/
{
temp = s;
last->next = s->next;
free(temp);
return;
}
while (s->next != .
I need help completing this C++ code with these requirements.instr.pdf
I need help completing this C++ code with these requirements.
instructions: IN C++ LANGUAGE PLEASE
Update the comments for each prototype by filling in the pre and post conditions.
Remember that pre conditions indicate what conditions must be met BEFORE the user calls the
function. The function will not work properly unless the pre conditions are met.
Post conditions indicate what will be the result of the function call. In other words, what
conditions will exist after the function is called.
See class notes from Thursday on pre and post conditions for some examples and further
explanation.
Code begins here:
#include //for NULL
class List
{
private:
struct Node
{
int data;
Node* next;
Node(int data): data(data), next(NULL){}
};
typedef struct Node* Nodeptr;
Nodeptr first;
Nodeptr last;
int size;
public:
/**Constructors and Destructors*/
List();
//Default constructor; initializes and empty list
//Postcondition:
~List();
//Destructor. Frees memory allocated to the list
//Postcondition:
/**Accessors*/
int getFirst();
//Returns the first element in the list
//Precondition:
int getLast();
//Returns the last element in the list
//Precondition:
bool isEmpty();
//Determines whether a list is empty.
int getSize();
//Returns the size of the list
/**Manipulation Procedures*/
void removeLast();
//Removes the value of the last element in the list
//Precondition:
//Postcondition:
void removeFirst();
//Removes the value of the first element in the list
//Precondition:
//Postcondition:
void insertLast(int data);
//Inserts a new element at the end of the list
//If the list is empty, the new element becomes both first and last
//Postcondition:
void insertFirst(int data);
//Inserts a new element at the start of the list
//If the list is empty, the new element becomes both first and last
//Postcondition:
/**Additional List Operations*/
void printList();
//Prints to the console the value of each element in the list sequentially
//and separated by a blank space
//Prints nothing if the list is empty
};
Solution
PROGRAM CODE:
#include
#include //for NULL
using namespace std;
class List
{
private:
struct Node
{
int data;
Node* next;
Node(int data): data(data), next(NULL){}
};
typedef struct Node* Nodeptr;
Nodeptr first;
Nodeptr last;
int size;
public:
/**Constructors and Destructors*/
List()
{
first = (Node*) malloc(sizeof(Node));
last = (Node*) malloc(sizeof(Node));
size = 0;
}
//Default constructor; initializes and empty list
//Postcondition:
~List()
{
delete first;
delete last;
}
//Destructor. Frees memory allocated to the list
//Postcondition:
/**Accessors*/
int getFirst()
{
return first->data;
}
//Returns the first element in the list
//Precondition:
int getLast()
{
return last->data;
}
//Returns the last element in the list
//Precondition:
bool isEmpty()
{
if(first == NULL)
return true;
else return false;
}
//Determines whether a list is empty.
int getSize()
{
return size;
}
//Returns the size of the list
/**Manipulation Procedures*/
void removeLast()
{
last = .
Write a program to implement below operations with both singly and d.pdf
Write a program to implement below operations with both singly and doubly linked list. Use
inheritance to define singly linked list and doubly linked list. Add Node() Delete Node() Update
Node() Delete All() Size()
Solution
#include
using namespace std;
class List;
class Node
{
friend class List;
protected:
int Data;
Node *Link;
};
class List:public Node
{
friend class Node;
private:
Node *Start;
public:
List()
{
Start=NULL;
}
Node Insert(int x);
Node Delete(int &x);
void Display();
};
Node List::Insert(int x)
{
Node *New=new Node;
New->Data=x;
New->Link=Start;
Start=New;
return *this;
}
Node List::Delete(int &x)//deletes first node
{
Node *Del=Start;
Start=Start->Link;
delete Del;
return *this;
}
void List::Display()
{
Node *P=Start;
while(P!=NULL)
{
cout<Data<<\" \";
P=P->Link;
}
}
int main()
{
List L;
L.Insert(4);
L.Insert(64);
L.Insert(3);
L.Insert(90);
L.Insert(66);
L.Insert(13);
L.Display();
}
Double Linked List.
Please need help on following program using c++ language. Please inc.pdf
Please need help on following program using c++ language. Please include all header files and
main file with their own title.
Extend the class linkedListType by adding the following operations:
Write a function that returns the info of the kth element of the linked list. If no such element
exists, terminate the program.
Write a function that deletes the kth element of the linked list. If no such element exists,
terminate the program.
Provide the definitions of these functions in the class linkedListType. Also, write a program to
test these functions. (Use either the class unorderedLinkedList or the
class orderedLinkedList to test your function.)
Solution
C++ Code for the given problem is below.....
.....................................................
/*
* C++ Program to Implement Singly Linked List
*/
#include
#include
#include
//using namespace std;
/*
* Node Declaration
*/
typedef struct nodetype
{
int info;
nodetype *next ;
} node;
/*
* Class Declaration
*/
class linkedListType
{
public:
void insfrombeg( node **head, int);
void delfromkth(node **head,int);
void findthekthnode(node *head,int);
void traverse(node *head);
int item,ch,kth;
node *ptr,*head,*temp;
};
/*
* Main :contains menu
*/
void main()
{
int item,ch,kth;
node *ptr,*head,*temp;
linkedListType s1;
clrscr();
while (1)
{
while(1)
{
cout<<\"PRESS (1) FOR INSERT INTO LIST\"<>ch;
switch(ch)
{
case 1:
cout<<\"ENTER THE VALUE \"<>item;
s1.insfrombeg(&head,item);
break;
case 2:
s1.traverse(head);
break;
case 3:
cout<<\"ENTER THE Kth Element which you want\'s to delete\"<>kth;
s1.delfromkth(&head,kth);
break;
case 4:
cout<<\"ENTER THE Kth Element which you want\'s to search \ \"<>kth;
s1.findthekthnode(head,kth);
break;
case 5:
exit(1);
}
}
// getch();
}
}
/*
* Inserting element in beginning
*/
void linkedListType::insfrombeg(node **head,int item)
{
temp= ( node * ) malloc (sizeof( node ));
temp->info= item;
temp->next= NULL;
if(*head == NULL)
{
temp->next= *head;
*head= temp;
}
else
{
temp->next= *head;
*head= temp;
}
}
/*
* Delete element at a given position
*/
void linkedListType::delfromkth(node **head,int kth)
{ node* temp1=*head,*temp2;
if(*head ==NULL)
{
cout<<\"LINK LIST IS EMPTY\ \"<next;
free(temp);
}
else
for(int j=0;jnext;
temp2= temp1->next;
temp1->next=temp2->next;
free(temp2);
}
/*
* Searching an element
*/
void linkedListType::findthekthnode(node *head,int kth)
{
ptr=head;
int i=1;
while((ptr!=NULL))
{
if(i==kth)
{
cout<<\"\ The Element at\"<< kth<< \"position with value is \\t
\"<info<info<next;
i++;
}
}
}
/*
* Traverse Link List
*/
void linkedListType::traverse(node *head)
{
ptr = head;
while((ptr->next)->next!=NULL)
{
cout<info<<\"\\t\";
ptr=ptr->next;
}
cout<
#include
#include
typedef struct nodetype
{
int info;
nodetype *next ;
} node;
int item,ch,kth,total=1;
node *ptr,*head,*temp;
void insfrombeg( node **head, int);
void delfromkth(node **head,int);
void findthekthnode(node *head,int);
void traverse(node *head);
void main()
{
clrscr();
while(1)
{
printf(\"PRESS .
C++ Doubly-Linked ListsThe goal of the exercise is to implement a.pdf
C++: Doubly-Linked Lists
The goal of the exercise is to implement a class List of doubly- linked lists.
My goal is to implement the class in a file doubly-linked.cpp.
The class List implements lists using the following structures for list ele- ments:
struct Node { int val ;
Node next;
Node prev; };
Each Node contains a value (an integer) and two pointers: one meant to point to the next element
of the list and one meant to point to the previous element in the list. The class provides the
following methods that you need to implement:
• A constructor and a destructor, with no arguments;
• void insert(int n): insert an integer at the end of the list;
• void reverse(): reverse the list;
• void print(): print the list to cout in the same format as the input (i.e. integers separated by
spaces);
doubly-linked.h
#ifndef __dll__
#define __dll__
#include
using namespace std;
// Basic structure to store elements of a list
struct Node {
int val; // contains the value
Node * next; // pointer to the next element in the list
Node * prev; // pointer to the previous element in the list
};
// Class List
class List {
public:
List(void); // Constructor
~List(void); // Destructor
void insert(int n); // This should insert n in the list
void reverse(void); // This should reverse the list
void print(void); // This shoiuld print the list
private:
Node * first; // Pointer to the first (if any) element in the list
};
#endif
main.cpp
#include
#include \"doubly-linked.h\"
using namespace std;
int main(void){
List l;
int n;
while(cin >> n){
l.insert(n);
}
// Print list as read from cin
l.print();
// Reverse the list and print it
l.reverse();
l.print();
// Reverse again and print it
l.reverse();
l.print();
return 0;
}
Solution
#include
using namespace std;
/* Linked list structure */
struct list {
struct list *prev;
int data;
struct list *next;
} *node = NULL, *first = NULL, *last = NULL, *node1 = NULL, *node2 = NULL;
class linkedlist {
public:
/* Function for create/insert node at the beginning of Linked list */
void insrt_frnt() {
list *addBeg = new list;
cout << \"Enter value for the node:\" << endl;
cin >> addBeg->data;
if(first == NULL) {
addBeg->prev = NULL;
addBeg->next = NULL;
first = addBeg;
last = addBeg;
cout << \"Linked list Created!\" << endl;
}
else {
addBeg->prev = NULL;
first->prev = addBeg;
addBeg->next = first;
first = addBeg;
cout << \"Data Inserted at the beginning of the Linked list!\" << endl;
}
}
/* Function for create/insert node at the end of Linked list */
void insrt_end() {
list *addEnd = new list;
cout << \"Enter value for the node:\" << endl;
cin >> addEnd->data;
if(first == NULL) {
addEnd->prev = NULL;
addEnd->next = NULL;
first = addEnd;
last = addEnd;
cout << \"Linked list Created!\" << endl;
}
else {
addEnd->next = NULL;
last->next = addEnd;
addEnd->prev = last;
last = addEnd;
cout << \"Data Inserted at the end of the Linked list!\" << endl;
}
}
/* Function for Display Linked list */
void display() {
node = first;
cout << \"List of data in L.
hi i have to write a java program involving link lists. i have a pro.pdf
hi i have to write a java program involving link lists. i have a problem with the nodes, as it is
posting errors with the nodes.
please help me with this problem. thank you.
public class LinkLists {
/* only need to store a single pointer to the node at the head
* of the list.
* The pointer is null if the list is empty.
* Also record the size of the list.
*/
protected Node head;
/* invariant: size is the number of nodes in the list pointed to by head */
protected int size;
/* no-arguments default constructor creates an empty list */
public LinkLists() {
head = null; // start with an empty list
size = 0;
}
/* accessor method */
public int size() {
return size;
}
/* value to add to the end of the list
*/
public void add(T value) {
head = addAtEnd(head, value);
size++;
}
/* node of the list to which the value should be added
* value to add to the end of the list
*/
private Node addAtEnd(Node node, T value) {
if (node == null) { // special case
return new Node(value, null);
} else if (node.getNext() == null) { // other special case
node.setNext(new Node(value, null));
} else {
addAtEnd(node.getNext(), value);
}
return node;
}
/* iterative implementation of the same method
* value to add to the end of the list
*/
public void add2(T value) {
if (head == null) {
head = new Node(value, null);
} else {
Node node = head; // guaranteed not to be null initially
while (node.getNext() != null) {
node = node.getNext(); // guaranteed not to be null here
}
// now, node.getNext() is guaranteed to be null
// similar to the second special case in addAtEnd
node.setNext(new Node(value, null));
}
size++;
}
public void remove(int position) throws BadItemCountException {
if ((position < 1) || (position > size)) {
throw new
BadItemCountException(\"invalid position \" + position +
\", only 1..\" + size + \" available\");
}
if (position == 1) {
head = head.getNext();
} else {
Node node = head;
for (int i = 2; i < position; i++) {
node = node.getNext();
}
node.setNext(node.getNext().getNext());
}
size--; // one less item
}
public String toString() {
return toString(head);
}
private String toString(Node node) {
if (node == null) {
return \"\";
} else {
return node.getValue() + \"\ \" + toString(node.getNext());
}
}
public static void main(String[] args) {
/* create two empty lists, make sure they print out correctly */
LinkLists list1 = new LinkLists();
LinkLists list2 = new LinkLists();
System.out.println(\"list1 = \'\" + list1 + \"\', list2 = \'\" + list2 + \"\'\");
System.out.println(\"list1.size() = \" + list1.size() +
\", list2.size() = \" + list2.size());
/* insert some items, keep checking */
list1.add(\"hello\");
list1.add(\"world\");
list2.add(\"foo\");
list2.add(\"bar\");
list2.add(\"baz\");
System.out.println(\"list1 = \'\" + list1 + \"\', list2 = \'\" + list2 + \"\'\");
System.out.println(\"list1.size() = \" + list1.size() +
\", list2.size() = \" + list2.size());
/* remove an item at an invalid position */
boolean caught = false;
try {
list2.remove(4);
} catch .
take the following code and give details of what each line of code i.pdf
take the following code and give details of what each line of code is doing
#include
#include
#include
using namespace std;
/*
* Node Declaration
*/
struct node
{
int info;
struct node *next;
}*start;
/*
* Class Declaration
*/
class single_llist
{
public:
node* create_node(int);
void insert_begin();
void insert_pos();
void insert_last();
void delete_pos();
void sort();
void search();
void update();
void reverse();
void display();
single_llist()
{
start = NULL;
}
};
/*
* Main :contains menu
*/
main()
{
int choice, nodes, element, position, i;
single_llist sl;
start = NULL;
while (1)
{
cout<>choice;
switch(choice)
{
case 1:
cout<<\"Inserting Node at Beginning: \"<info = value;
temp->next = NULL;
return temp;
}
}
/*
* Inserting element in beginning
*/
void single_llist::insert_begin()
{
int value;
cout<<\"Enter the value to be inserted: \";
cin>>value;
struct node *temp, *p;
temp = create_node(value);
if (start == NULL)
{
start = temp;
start->next = NULL;
}
else
{
p = start;
start = temp;
start->next = p;
}
cout<<\"Element Inserted at beginning\"<>value;
struct node *temp, *s;
temp = create_node(value);
s = start;
while (s->next != NULL)
{
s = s->next;
}
temp->next = NULL;
s->next = temp;
cout<<\"Element Inserted at last\"<info<<\"->\";
temp = temp->next;
}
cout<<\"NULL\"<
Solution
ANSWER ::
ASSMUING THE TOTAL LINE IN CODE ARE 158*(INCLUDING COMMENTS)
SO...
PROGRAM STARTS WITH SPECIFING HEADER FILES AND GLOBAL VARIABLE
DECLARATION
THEN Node Declaration USING STRUCTURES,
THEN WE WANT TO DECLARE THE CLASS USING SINGLE LINKED LIST AND THE
PROTOTYPE IS PUBLIC
IN THE PUBLIC WE WANT TO CREATE THE NODE AS INTEGER DATATYPE
THE NODE CONSIST OF insert_begin();
insert_pos();-> INDICATES THE CURRENT POSSITION
insert_last();->INSERT AT LAST
delete_pos();->TO DELETE THE NUMBER AT CURRENT POSITION
sort();->START THE SORTING
search();->USED TO SEARCH
update();->UPDATE THE NODE
reverse();->REVERSE NUMBERS
display();->TO DISPLAY THE NODE
single_llist()->FIFO
WE ARE STARTING TAKING NULL VALUE
AND THEN TAKE USER DEFIND FUNCTION MAIN()
CREATE THE DATATYPES choice, nodes, element, position, i
OBJECT CREATION OF SLL IS SL
USING WHILE LOOP THE NODE INSERT AT THE BEGINING OR ENDING STAGE
WE ARE USING SWITCH CASES TO PERFORM SLL ON NODE
NODE CREATION AS
node *single_llist::create_node(int value)
ALSO DECLARE THE TEMP IS TAKEN AS NULL VALUE
INSERT THE ELEMENT AT THE BEGINING
USING IF CONDITION WE CAN INSERT THE NODE AT THE BEGINING
struct node *temp, *p;
temp = create_node(value);
if (start == NULL)
{
start = temp;
start->next = NULL;
}
else
{
p = start;
start = temp;
start->next = p;
AND THEN INSERT THE NODE AT THE LAST
struct node *temp, *s;
temp = create_node(value);
s = start;
while (s->next != NULL)
{
s = s->next;
}
temp->next = NULL;
s->next = temp;
USING WHILE LOOP WE ARE INSERTING THE NODE AT LAST
USING THIS BELLOW CODE WE CAN SEARCH THE ELEMENTS IN THE NODE
struct node *temp;
if (start == NULL)
{
cout<<\"The List is Empty\"<info<<\"->\";
temp = temp->next;
}
cout.
Lecture 3, c++(complete reference,herbet sheidt)chapter-13
Slide of C++(Complete Reference,Herbet Sheildt book)
CSE Department
Rajshahi Engineering Science and Technology College
using set identitiesSolutionimport java.util.Scanner; c.pdf
using set identities
Solution
import java.util.Scanner;
/* category Node */
class Node
creator */
public Node()
/* builder */
public Node(int d,Node n)
/* perform to line link to next Node */
public void setLink(Node n)
/* perform to line knowledge to current Node */
public void setData(int d)
/* perform to urge link to next node */
public Node getLink()
come back link;
}
/* perform to urge knowledge from current Node */
public int getData()
come back data;
}
}
/* category linkedList */
class linkedList
begin ;
protected Node finish ;
public int size ;
/* builder */
public linkedList()
begin = null;
finish = null;
size = 0;
}
/* perform to envision if list is empty */
public mathematician isEmpty()
come back begin == null;
}
/* perform to urge size of the list */
public int getSize()
come back size;
}
/* perform to insert part at the begining */
public void insertAtStart(int val)
{
Node nptr = new Node(val,null);
nptr.setLink(start);
if(start == null)
{
begin = nptr;
nptr.setLink(start);
finish = start;
}
else
{
end.setLink(nptr);
begin = nptr;
}
size++ ;
}
/* perform to insert part at finish */
public void insertAtEnd(int val)
{
Node nptr = new Node(val,null);
nptr.setLink(start);
if(start == null)
{
begin = nptr;
nptr.setLink(start);
finish = start;
}
else
{
end.setLink(nptr);
finish = nptr;
}
size++ ;
}
/* perform to insert part at position */
public void insertAtPos(int val , int pos)
one ;
for (int i = 1; i < size - 1; i++)
ptr = ptr.getLink();
}
size++ ;
}
/* perform to delete part at position */
public void deleteAtPos(int pos)
one && pos == 1)
begin = null;
finish = null;
size = 0;
return ;
}
if (pos == 1)
begin = begin.getLink();
end.setLink(start);
size--;
return ;
}
if (pos == size)
whereas (s != end)
end = t;
end.setLink(start);
size --;
return;
}
Node ptr = start;
pos = pos - one ;
for (int i = 1; i < size - 1; i++)
ptr = ptr.getLink();
}
size-- ;
}
/* perform to show contents */
public void display()
making object of linkedList */
linkedList list = new linkedList();
System.out.println(\"Circular individually coupled List Test\ \");
char ch;
/* Perform list operations */
do
{
System.out.println(\"\ Circular individually coupled List Operations\ \");
System.out.println(\"1. insert at begining\");
System.out.println(\"2. insert at end\");
System.out.println(\"3. insert at position\");
System.out.println(\"4. delete at position\");
System.out.println(\"5. check empty\");
System.out.println(\"6. get size\");
int selection = scan.nextInt();
switch (choice).
Consider L = {a^nb^2nc^P p 0}. Prove L is not a context-free langu.pdf
Capital and Revenue Expenditures On February 13, Scanlon Co. paid $3,400 to repair the
transmission on one of its delivery vans.
In addition, Scanlon paid $20 to install a GPS system in its van.
Journalize the entries for the transmission. February 13? It should be two answers, Cr/Dr
Journalize the entry for GPS system expenditures. February 13? It should be two answers, Cr/Dr
Solution
Account Debit Credit Accumulated Depreciation- Delivery Van $3400 Cash $3400 To
record the repair of transmission on delivery vans.(Revenue Expenditure Delivery Van $20
Cash $20 (To record capital expenditure on Van (installing GPS).
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.
Using the provided table interface table.h and the sample linked lis.pdf
Using the provided table interface table.h and the sample linked list code linkedList.c, complete
an implementation of the Table ADT. Make sure that you apply the concepts of design by
contract (DbC) to your implementation.
Once you have fully implemented the table, create a main.c file that implements a testing
framework for your table.
Your table implementation must ensure that values inserted are unique, and internally sorted
within a linked list.
table.h
linkedList.c
Solution
khsdg
#include
#include
typedef enum BOOL { false, true } bool;
// Linked list node definition
typedef struct Node node;
struct Node
{
int number;
node *next;
};
static node *top = NULL;
// used to track where we are for the list traversal methods
static node *traverseNode = NULL;
// \"destroy\" will deallocate all nodes in a linked list object
// and will set \"top\" to NULL.
void destroy()
{
node *curr = top;
node *temp = NULL;
while ( curr != NULL )
{
// flip order to see it blow up...
temp = curr;
curr = curr->next;
free( temp );
}
top = NULL;
}
// \"build\" will create an ordered linked list consisting
// of the first \"size\" even integers.
void build( int size )
{
node *newNode = NULL;
int i = 0;
// make sure we don\'t have a list yet
destroy();
for ( i=size ; i>0 ; i-- )
{
newNode = malloc( sizeof( node ) );
newNode->number = i*2;
newNode->next = top;
top = newNode;
}
}
// starts a list traversal by getting the data at top.
// returns false if top == NULL.
bool firstNode( int *item )
{
bool result = false;
if ( top )
{
*item = top->number;
traverseNode = top->next;
result = true;
}
return result;
}
// gets the data at the current traversal node and increments the traversal.
// returns false if we\'re at the end of the list.
bool nextNode( int *item )
{
bool result = false;
if ( traverseNode )
{
*item = traverseNode->number;
traverseNode = traverseNode->next;
result = true;
}
return result;
}
// \"print\" will output an object\'s entire linked list
// to the standard output device -- one \"number\" per line.
void print()
{
int value;
if ( firstNode( &value ) )
{
do
{
printf( \"%d\ \", value );
} while ( nextNode( &value ) );
}
}
int main( int argc, char *argv[] )
{
build( 10 );
print();
destroy();
build( 5 );
build( 20 );
print();
destroy();
return 0;
}.
Assignment is Page 349-350 #4 and #5 Use the Linked Lis.pdf
Assignment is :
\"Page 349-350 #4 and #5 Use the \"Linked List lab\" you have been working on in class and add
the two functions the questions are asking you to develop: divideMid and divideAt. Be sure to
include comments Use meaningful identifier names (constants where appropriate) Do not work
together; no two people should have identical work!?!? Turn in .cpp file AND Turn in a \"print-
screen\' of your output (press \"print-screen\' on keyboard, then \'paste\' in MS-Word)\"
How do you solve QUESTION #5 in the book data structures using c++ by D.S. Malik in Visiual
Studios using the linked list below with what is being asked? Please need help
Linked list :
#include
#include
using namespace std;
struct nodeType
{
int info;
nodeType *link;
};
void createList(nodeType*& first, nodeType*& last);
void printList(nodeType*& first);
void insertFront(nodeType*& first);
void insertBack(nodeType*& last);
void deleteFirst(nodeType*& first);
void deleteLast(nodeType*& last, nodeType* first);
int main()
{
nodeType *first, *last;
int num;
createList(first, last);
int choice;
while(true)
{
cout<<\"1. Insert Front.\ 2. Insert Last.\ 3. Delete Front.\ 4. Delete Last.\ 5. Print List.\ 6. Exit.\
\";
cout<<\"Enter your choice: \";
cin>>choice;
switch(choice)
{
case 1: insertFront(first); break;
case 2: insertBack(last); break;
case 3: deleteFirst(first); break;
case 4: deleteLast(last, first); break;
case 5: printList(first); break;
case 6: return 0;
default: cout<<\"Invalid menu option. Try again.\"<>number;
while (number != -999)
{
newNode = new nodeType; // create new node
newNode->info = number;
newNode->link = NULL;
if (first == NULL)
{
first = newNode;
last = newNode;
}
else
{
last->link = newNode;
last = newNode;
}
cout<<\"Enter an integer (-999 to stop): \";
cin>>number;
} // end of while-loop
} // end of build list function
void deleteFirst(nodeType*& first)
{
nodeType *temp;
temp= first;
first= temp->link;
delete temp;
return;
}
void deleteLast(nodeType*& last, nodeType* current)
{
nodeType *temp;
while(current->link != NULL)
{
temp=current;
current=current->link;
}
temp=last;
current->link=NULL;
delete temp;
last = current;
return;
}
void insertFront(nodeType*& front)
{
int num;
cout<<\"\ Enter the number to insert: \";
cin>>num;
nodeType *newNode = new nodeType;
newNode->info=num;
newNode->link= front;
front= newNode;
return;
}
void insertBack(nodeType*& last)
{
int num;
cout<<\"\ Enter the number to insert: \";
cin>>num;
nodeType *newNode = new nodeType;
newNode->info=num;
newNode->link= NULL;
last->link= newNode;
last = newNode;
return;
}
void printList(nodeType*& first)
{
cout<<\"Inside printList...printing linked list...\ \"<info << \" \";
current = current->link;
}
cout<
#include
using namespace std;
struct nodeType
{
int info;
nodeType *link;
};
void createList(nodeType*& first, nodeType*& last);
void printList(nodeType*& first);
void insertFront(nodeType*& first);
void insertBack(nodeType*& last);
void deleteFirst(nodeType*& first);
void dele.
File Type cppAdd the following to your linked list of strings pro.pdf
File Type: cpp
Add the following to your linked list of strings program:
1. allow the user to search for an item (indicate the result of the search)
2. allow the user to insert an item (print the linked list after insertion)
3. allow the user to delete an item (print the linked list after deletion)
Here\'s the program I need to have remodified for this assignment as shown below.
#include
#include
#include
using namespace std;
int main()
{
struct nodeType
{
string info;
nodeType *link;
};
nodeType *first, *newNode, *last;
string x; //string input variable
cout << \"Enter a string of some kind. Enter \'Done\' when finished.\" << endl;
getline(cin, x);
first = NULL;
last = NULL;
while (x != \"Done\")
{
newNode = new nodeType;
newNode->info = x;
newNode->link = NULL;
if (first == NULL)
{
first = newNode;
last = newNode;
}
else
{
last->link = newNode;
last = newNode;
}
cout << newNode->info << \" \";
cout << \"Enter a string of some kind. Enter \'Done\' when finished.\" << endl;
getline(cin, x);
}
return 0;
}
Solution
In the main function, you need to enter the following:
main()
{
int ch, nodes, ele, posi;
sin_llist sl;
start = NULL;
while(1)
{
cout<<\"1.Search for an item\"<>ch;
switch(ch)
{
case 1:
cout<<\"Searching an item:\"<>val;
struct node *z;
z = start;
while (z != NULL)
{
posi++;
if (z->info == val)
{
flag = true;
cout<<\"Element \"<next;
}
if (!flag)
cout<<\"Element \"<>y;
struct node *tem, *k;
temp = create_node(y);
if (start == NULL)
{
start = tem;
start->next = NULL;
}
else
{
k = start;
start = tem;
start->next = k;
}
cout<<\"Element Inserted at beginning\"<>posi;
struct node *z, *ptr;
z = start;
if (posi == 1)
{
start = z->next;
}
else
{
while (z != NULL)
{
z = z->next;
count++;
}
if (posi > 0 && posi <= count)
{
z = start;
for (i = 1;i < pos;i++)
{
ptr = z;
z = z->next;
}
ptr->next = z->next;
}
else
{
cout<<\"Position out of range\"<.
C++ Please write the whole code that is needed for this assignment- wr.docx
Brianca plans to save $5,000, $1,000, and $42,000 a year over the next three years, respectively. How much would you need to deposit in one lump sum today to have the same amount as Brianca three years from now if you both earn 10.9 percent, compounded annually? $36,115 $35,192 $43,282 $41,635 $35,372
.
#includeiostream #includecstdio #includecstdlib using na.pdf
#include
#include
#include
using namespace std;
struct node
{
int val;
struct node *next;
}*startNode;
// declaring class
class LinkedList
{
public:
node* create_node(int);
void insert_last();
void sort();
void search();
void reverse();
void display();
LinkedList()
{
startNode = NULL;
}
};
// main
main()
{
int userChoice, linkedListNodes, i;
LinkedList LinkList;
startNode = NULL;
while (1)
{
// displaying menu
cout<<\" 1)Insert\";
cout<<\"2) Remove\";
cout<<\"3) Search\";
cout<<\"4) Display\";
cout<<\"5) Reverse\";
cout<<\"6) Exit \";
// reading user choice
cout<<\"Enter your userChoice : \";
cin>>userChoice;
switch(userChoice)
{
case 1:
cout<<\"Inserting node to linked list: \";
LinkList.insert_begin();
cout;
break;
case 2:
cout<<\"Delete element from linked list: \";
LinkList.delete();
break;
case 3:
cout<<\"Searching data in linked list: \";
LinkList.search();
cout;
break;
case 4:
cout<<\"Displaying linked list data: \";
LinkList.display();
cout;
break;
case 5:
cout<<\"Linked list reversing: \";
LinkList.reverse();
cout;
break;
case 6:
cout<<\"Exiting...\";
exit(1);
break;
default:
cout<<\"Enter correct userChoice\";
}
}
}
// node creating to store into list
node *LinkedList::create_node(int val)
{
struct node *tempNode, *tempNode;
tempNode = new(struct node);
if (tempNode == NULL)
{
cout<<\"No memory allocated \";
return 0;
}
else
{
tempNode->val = val;
tempNode->next = NULL;
return tempNode;
}
}
/*
* Inserting Node at last
*/
void LinkedList::insert_last()
{
int val;
cout<<\"Enter the val to be inserted: \";
cin>>val;
struct node *tempNode, *tempNode;
tempNode = create_node(val);
tempNode = startNode;
while (tempNode->next != NULL)
{
tempNode = tempNode->next;
}
tempNode->next = NULL;
tempNode->next = tempNode;
cout<<\"Element Inserted\";
}
// deleting element
void LinkedList::delete()
{
int position, i, count = 0;
if (startNode == NULL)
{
cout<<\"Linkedlist empty\"<>position;
struct node *tempNode, *ptr;
tempNode = startNode;
if (position == 1)
{
startNode = tempNode->next;
}
else
{
while (tempNode != NULL)
{
tempNode = tempNode->next;
count++;
}
if (position > 0 && position <= count)
{
tempNode = startNode;
for (i = 1;i < position;i++)
{
ptr = tempNode;
tempNode = tempNode->next;
}
ptr->next = tempNode->next;
}
else
{
cout<<\"Enter correct position\";
}
}
}
// search data in linked list
void LinkedList::search()
{
int val, position = 0;
bool flagValue = false;
if (startNode == NULL)
{
cout<<\"List is empty\";
return;
}
cout<<\"Enter the val to search: \";
cin>>val;
struct node *tempNode;
tempNode = startNode;
while (tempNode != NULL)
{
position++;
if (tempNode->val == val)
{
flagValue = true;
cout<<\"Data \"<next;
}
if (!flagValue)
cout<<\"data not found\";
}
// linkedlist reverese
void LinkedList::reverse()
{
struct node *pointer1, *pointer2, *pointer3;
if (startNode == NULL)
{
cout<<\"List is empty\"<next == NULL)
{
return;
}
pointer1 = startNode;
pointer2 = pointer1->next;
pointer3 = pointer2->next;
pointer1->next = NULL;
pointer2->.
In C++ I need help with this method that Im trying to write fillLi.pdf
In C++ I need help with this method that I'm trying to write: fillListFromFile( ) : void -- this
method will add elements to the list from a file called linkedListData1.txt HINT: this is where
you will keep track of the number of elements in the list
I have all of the other methods, but I'm stuck on this one and I've added it as the last method in
the application. The code I have so far is below.
.cpp file:
.h file:
main:
#include
#include
#include
#include "Node.h"
using namespace std;
void displayList(Node*);
void insertFront(Node** head, int value);
void insertBack(Node** ptrToHead, int value);
void insertAfter(Node* previous, int value);
void deleteAtN(Node** ptrToHead, int n);
void deleteNode(Node** ptrToHead, int searchVal);
void clearList(Node**);
bool searchVal(Node** head, int value);
int countFrequencyOf(Node** ptrTohead, int value);
int size(Node** ptrTohead);
void displayFromFrontToN(Node** ptrToHead, int n);
void displayFromNToEnd(Node** ptrToHead, int n);
void insertInOrder(Node** head, Node* ptrToHead);
void displayInRows(Node** ptrToHead);
int main()
{
Node* node1 = new Node();
node1->setValue(10);
Node* node2 = new Node();
node2->setValue(20);
Node* node3 = new Node();
node3->setValue(30);
node1->next = node2;
node2->next = node3;
Node* head = node1;
displayList(head);
insertFront(&head, 5);
displayList(head);
insertBack(&head, 50);
displayList(head);
insertBack(&head, 60);
displayList(head);
insertAfter(node2, 25);
displayList(head);
deleteAtN(&head, 2);
displayList(head);
deleteNode(&head, 50);
displayList(head);
clearList(&head);
displayList(head);
insertFront(&head, 5);
insertFront(&head, 10);
insertFront(&head, 20);
insertBack(&head, 50);
searchVal(&head, 50);
displayList(head);
countFrequencyOf(&head, 25);
displayList(head);
insertBack(&head, 25);
insertFront(&head, 25);
countFrequencyOf(&head, 25);
displayList(head);
size(&head);
cout << "add one item to list" << endl;
insertBack(&head, 25);
size(&head);
displayList(head);
cout << "Display From Front To N" << endl;
displayFromFrontToN(&head, 4);
cout << "Display From N To End" << endl;
displayFromNToEnd(&head, 3);
cout << "Display In Rows" << endl;
displayInRows(&head);
insertFront(&head, 12);
insertFront(&head, 17);
insertFront(&head, 22);
insertBack(&head, 58);
cout << "Display In Rows" << endl;
displayInRows(&head);//call displayInRows()
return 0;
}//end main()
void displayList(Node* ptr) {
if (ptr == nullptr)
cout << "empty list" << endl;
else {
while (ptr != nullptr) {
cout << ptr->getValue() << endl;
ptr = ptr->next;
}
}
cout << "---------------------------" << endl;
}//end displayList()
void insertFront(Node** head, int value) {
Node* newNode = new Node();
newNode->setValue(value);
//cout<< newNode->getValue() << endl;
newNode->next = *head;
*head = newNode;
}
void insertBack(Node** ptrToHead, int value) {
Node* newNode = new Node();
newNode->setValue(value);
newNode->next = nullptr;
if (*ptrToHead == nullptr)
{
*ptrToHead = newNode;
}//end if
else
{
Node* last .
write recursive function that calculates and returns the length of a.pdf
write recursive function that calculates and returns the length of a linked list
Solution
// C code to determine length of linked list recursively
#include
#include
// node of linked list
struct node
{
int key;
struct node* next;
};
// push key to linked list
void push(struct node** root, int new_element)
{
struct node* nodeNew = (struct node*) malloc(sizeof(struct node));
nodeNew->key = new_element;
nodeNew->next = (*root);
(*root) = nodeNew;
}
// recursively get length of linked list
int lengthLinkedlist(struct node* root)
{
// base case
if (root == NULL)
return 0;
// recursively call the function
return 1 + lengthLinkedlist(root->next);
}
int main()
{
struct node* root = NULL;
push(&root, 11);
push(&root, 34);
push(&root, 4);
push(&root, 23);
push(&root, 14);
push(&root, 88);
push(&root, 56);
push(&root, 84);
push(&root, 6);
/* Check the count function */
printf(\"Length of Linked list: %d\ \", lengthLinkedlist(root));
return 0;
}
/*
output:
Length of Linked list: 9
*/.
C++ Please test your program before you submit the answer.pdf
C++. Please, test your program before you submit the answer.
Background:
Circular Linked List: A circular linked list is a linked list where a non-empty list has no null pointers.
Unless empty, each node has a pointer to the next node, and the last node points back to the first
node.
Assigment:
Write a cpp program that will serve the purpose of a test driver for previously written CLinkedList.h
file. Your program will:
1) Create a choice menu and prompt the user for input. Must be case insensitive!
2) Read the input and call for functions outside of main, in order to perform operations chosen by
the user. These functions involve: Inserting Interger, Removing Interger, Forward Print and
Backward Print.
Requirements and Notes:
1. Forwards Print does not require a subfunction in order to produce output. All the other
operations must call corresponding subfunctions.
2. Nodes of the circular list must be arranged in ascending order.
3. User Input must be processed case insensitive, and implemented using the loop.
4. Document your code completely.
5. CLinkedList.h file contains 3 classes.
- Circular linked lists class with following member functions: constructor, copy constructor,
destructor, assignment operator, insert and remove.
- Node class included in the file header.
- A list iterator class provides a public interface to the list. The pointer points at nodes in the linked
list. Its member functions are as follows:
Constructor: Assigns the linked list data member to the parameter, a reference to a circular linked
list.
begin() sets the iterator to the first node of the linked list (or NULL if the list is empty)
isEmpty() returns whether or not the wrapped linked list is empty
isFirstNode() returns whether the present node is the first node of the linked list
isLastNode() returns whether the present node is the last node of the linked list
operator*() returns the data of the node currently pointed at. (You need 2)
operator++() pre-increment operator advances the pointer to the next node in the list, if there is
one
operator++(int) post-increment operator advances the pointer to the next node in the list, if there
is one
6. Insert function must give feedback where it says the position of where the int was inserted for
example: "Inserting 25 in the middle."
7. Remove function must give either one of these types of feedback:
- "Removing 25" if the int given by the user was found in the list
-"Cannot find 25" if it was not. The program will then move the user back to the choice menu.
Files:
// File: Node.h
// Singly-linked list node definition/implementation (also good for circular)
#ifndef NODE_H
#define NODE_H
// Need to prototype template classes if they are to be friends
template <typename eltType> class CLinkedList;
template <typename eltType> class CListItr;
template <typename eltType> class Node
{ private:
Node(eltType info, Node* link = NULL ) :
data(info), next(link)
{};
eltType data;
Node *next;
friend class CLinkedList<eltType>;
friend c.
C++ Background Circular Linked List A circular linked list.pdf
C++
Background:
Circular Linked List: A circular linked list is a linked list where a non-empty list has no null pointers.
Unless empty, each node has a pointer to the next node, and the last node points back to the first
node.
Assigment:
Write a cpp program that will serve the purpose of a test driver for previously written CLinkedList.h
file. Your program will:
1) Create a choice menu and prompt the user for input. Must be case insensitive!
2) Read the input and call for functions outside of main, in order to perform operations chosen by
the user. These functions involve: Inserting Interger, Removing Interger, Forward Print and
Backward Print.
Requirements and Notes:
1. Forwards Print does not require a subfunction in order to produce output. All the other
operations must call corresponding subfunctions.
2. Nodes of the circular list must be arranged in ascending order.
3. User Input must be processed case insensitive, and implemented using the loop.
4. Document your code completely.
5. CLinkedList.h file contains 3 classes.
- Circular linked lists class with following member functions: constructor, copy constructor,
destructor, assignment operator, insert and remove.
- Node class included in the file header.
- A list iterator class provides a public interface to the list. The pointer points at nodes in the linked
list. Its member functions are as follows:
Constructor: Assigns the linked list data member to the parameter, a reference to a circular linked
list.
begin() sets the iterator to the first node of the linked list (or NULL if the list is empty)
isEmpty() returns whether or not the wrapped linked list is empty
isFirstNode() returns whether the present node is the first node of the linked list
isLastNode() returns whether the present node is the last node of the linked list
operator*() returns the data of the node currently pointed at. (You need 2)
operator++() pre-increment operator advances the pointer to the next node in the list, if there is
one
operator++(int) post-increment operator advances the pointer to the next node in the list, if there
is one
6. Insert function must give feedback where it says the position of where the int was inserted for
example: "Inserting 25 in the middle."
7. Remove function must give either one of these types of feedback:
- "Removing 25" if the int given by the user was found in the list
-"Cannot find 25" if it was not. The program will then move the user back to the choice menu.
Files:
// File: Node.h
// Singly-linked list node definition/implementation (also good for circular)
#ifndef NODE_H
#define NODE_H
// Need to prototype template classes if they are to be friends
template <typename eltType> class CLinkedList;
template <typename eltType> class CListItr;
template <typename eltType> class Node
{ private:
Node(eltType info, Node* link = NULL ) :
data(info), next(link)
{};
eltType data;
Node *next;
friend class CLinkedList<eltType>;
friend class CListItr<eltType>;
};
#endif
// File: CLinkedList.h
.
AnswerNote LinkedList.cpp is written and driver program main.cpp.pdf
Answer:
Note: LinkedList.cpp is written and driver program main.cpp is also given.
//ListInterface.h
#ifndef _LIST_INTERFACE
#define _LIST_INTERFACE
template
class ListInterface
{
public:
virtual bool isEmpty() const = 0;
virtual int getLength() const = 0;
virtual bool insert(int newPosition, const ItemType& newEntry) = 0;
virtual bool remove(int position) = 0;
virtual void clear() = 0;
virtual ItemType getEntry(int position) const = 0;
virtual void replace(int position, const ItemType& newEntry) = 0;
}; // end ListInterface
#endif
//Node.h
#ifndef NODE_
#define NODE_
template
class Node
{
private:
ItemType item; // A data item
Node* next; // Pointer to next node
public:
Node();
Node(const ItemType& anItem);
Node(const ItemType& anItem, Node* nextNodePtr);
void setItem(const ItemType& anItem);
void setNext(Node* nextNodePtr);
ItemType getItem() const ;
Node* getNext() const ;
}; // end Node
#endif
//Node.cpp
#include \"Node.h\"
template
Node::Node() : next(nullptr)
{
} // end default constructor
template
Node::Node(const ItemType& anItem) : item(anItem), next(nullptr)
{
} // end constructor
template
Node::Node(const ItemType& anItem, Node* nextNodePtr) :
item(anItem), next(nextNodePtr)
{
} // end constructor
template
void Node::setItem(const ItemType& anItem)
{
item = anItem;
} // end setItem
template
void Node::setNext(Node* nextNodePtr)
{
next = nextNodePtr;
} // end setNext
template
ItemType Node::getItem() const
{
return item;
} // end getItem
template
Node* Node::getNext() const
{
return next;
} // end getNext
//PrecondViolatedExcept.h
#ifndef PRECOND_VIOLATED_EXCEPT_
#define PRECOND_VIOLATED_EXCEPT_
#include
#include
class PrecondViolatedExcept : public std::logic_error
{
public:
PrecondViolatedExcept(const std::string& message = \"\");
}; // end PrecondViolatedExcept
#endif
//PrecondViolatedExcept.cpp
#include \"PrecondViolatedExcept.h\"
PrecondViolatedExcept::PrecondViolatedExcept(const std::string& message)
: std::logic_error(\"Precondition Violated Exception: \" + message)
{
} // end constructor
//LinkedList.h
#ifndef LINKED_LIST_
#define LINKED_LIST_
#include \"ListInterface.h\"
#include \"Node.h\"
#include
#include \"PrecondViolatedExcept.h\"
template
class LinkedList : public ListInterface
{
private:
Node* headPtr; // Pointer to first node in the chain;
// (contains the first entry in the list)
int itemCount; // Current count of list items
// Locates a specified node in this linked list.
Node* getNodeAt(int position) const;
public:
LinkedList();
LinkedList(const LinkedList& aList);
virtual ~LinkedList();
bool isEmpty() const;
int getLength() const;
bool insert(int newPosition, const ItemType& newEntry);
bool remove(int position);
void clear();
ItemType getEntry(int position) const;
void replace(int position, const ItemType& newEntry);
}; // end LinkedList
#endif
//LinkedList.cpp
#include \"LinkedList.h\"
#include
#include
#include
#include
//Default constructor
template
LinkedList::LinkedList() : headPtr(nullptr), itemCount(0)
{
.
C code on linked list #include stdio.h #include stdlib.h.pdf
For the ethanol water graph the simple distillation should be a steady smooth
increase, the fractional distillation should look like an s, that sort of takes a straight up jump in
the middle. I am unsure about the other mixture though sorry, hope that helps
Solution
For the ethanol water graph the simple distillation should be a steady smooth
increase, the fractional distillation should look like an s, that sort of takes a straight up jump in
the middle. I am unsure about the other mixture though sorry, hope that helps.
steps are as follows Dissolve the (pressumably.pdf
steps are as follows : Dissolve the (pressumably solid) mixture in an organic
solven,,,,,,, Then, in a seperating funnel, add an (Aqueous) solution of sodium hydroxide. The
Phenol, and benzoic acid, as acids, will react with the base to form phenolate and benzenoate
ions, which will dissolve into the water layer. Collect this layer, and re-precipitate the organic
molecules by adding concentrated Hydrochloric acid. The phenol and Benzoic acid can be
seperated via their varying solubilities, (8.3g/100ml vs 0.34g/100ml. respectively) by heating the
mixture in water to dissolve the phenol, and then filtering of the (practically) insoluble benzoic
acid. The solution can then be acidified (with HCl) to return the phenol..,,,,,,,,,,,,, Now add a
solution of hydrochloric acid to the seperating funnel, which (as before) will react with the basic
anniline dossilving it into the water. Collect the aqueous layer and re-precipitate the anniline
with excess sodium hydroxide. ,,,,,,,,,,,,,,, finally boil off the remaining organic solvent to obtain
the napthalene, which was left untouched by all the acids and bases.,,,,,,,,, Flow chart : Mixture
+ xs HCl ----> anilinium(aq) Mixture - aniline + weak base (Na2CO3) ---> benzoate(aq)
Mixture - aniline - benzoic acid + base (NaOH) ---> p-bromophenoxide(aq) Mixture - aniline -
benzoic acid - p-bromophenol = naphthalene
Solution
steps are as follows : Dissolve the (pressumably solid) mixture in an organic
solven,,,,,,, Then, in a seperating funnel, add an (Aqueous) solution of sodium hydroxide. The
Phenol, and benzoic acid, as acids, will react with the base to form phenolate and benzenoate
ions, which will dissolve into the water layer. Collect this layer, and re-precipitate the organic
molecules by adding concentrated Hydrochloric acid. The phenol and Benzoic acid can be
seperated via their varying solubilities, (8.3g/100ml vs 0.34g/100ml. respectively) by heating the
mixture in water to dissolve the phenol, and then filtering of the (practically) insoluble benzoic
acid. The solution can then be acidified (with HCl) to return the phenol..,,,,,,,,,,,,, Now add a
solution of hydrochloric acid to the seperating funnel, which (as before) will react with the basic
anniline dossilving it into the water. Collect the aqueous layer and re-precipitate the anniline
with excess sodium hydroxide. ,,,,,,,,,,,,,,, finally boil off the remaining organic solvent to obtain
the napthalene, which was left untouched by all the acids and bases.,,,,,,,,, Flow chart : Mixture
+ xs HCl ----> anilinium(aq) Mixture - aniline + weak base (Na2CO3) ---> benzoate(aq)
Mixture - aniline - benzoic acid + base (NaOH) ---> p-bromophenoxide(aq) Mixture - aniline -
benzoic acid - p-bromophenol = naphthalene.
the two features of the asynchronus connection that allow the receiv.pdf
the two features of the asynchronus connection that allow the receiver to stay in sync with the
incoming data stream are the start bit and short frame connections . asynchronus connection are
asynchronus serial communications in which the communicating end points interfaces are not
continously synchronized by common clock signal and the data stream contains synchronization
in the form of start and stop signals before and after each unit of transmission .asynchronus start
and stop signals was widley used for dial up modem sources to time sharing computers and bbs
systems asynchronus start stop is the physical layer is used to connect computers to modems for
many dial up internet access applications using data link frame protocol such as ppp to create
packets made upto out of asynchronous serial characters .the examples that would benefit from
asynchronous connection are modems and usb that are connectedd to interfaces
Solution
the two features of the asynchronus connection that allow the receiver to stay in sync with the
incoming data stream are the start bit and short frame connections . asynchronus connection are
asynchronus serial communications in which the communicating end points interfaces are not
continously synchronized by common clock signal and the data stream contains synchronization
in the form of start and stop signals before and after each unit of transmission .asynchronus start
and stop signals was widley used for dial up modem sources to time sharing computers and bbs
systems asynchronus start stop is the physical layer is used to connect computers to modems for
many dial up internet access applications using data link frame protocol such as ppp to create
packets made upto out of asynchronous serial characters .the examples that would benefit from
asynchronous connection are modems and usb that are connectedd to interfaces.
More Related Content
Similar to C++ Program to Implement Doubly Linked List #includei.pdf
take the following code and give details of what each line of code i.pdf
take the following code and give details of what each line of code is doing
#include
#include
#include
using namespace std;
/*
* Node Declaration
*/
struct node
{
int info;
struct node *next;
}*start;
/*
* Class Declaration
*/
class single_llist
{
public:
node* create_node(int);
void insert_begin();
void insert_pos();
void insert_last();
void delete_pos();
void sort();
void search();
void update();
void reverse();
void display();
single_llist()
{
start = NULL;
}
};
/*
* Main :contains menu
*/
main()
{
int choice, nodes, element, position, i;
single_llist sl;
start = NULL;
while (1)
{
cout<>choice;
switch(choice)
{
case 1:
cout<<\"Inserting Node at Beginning: \"<info = value;
temp->next = NULL;
return temp;
}
}
/*
* Inserting element in beginning
*/
void single_llist::insert_begin()
{
int value;
cout<<\"Enter the value to be inserted: \";
cin>>value;
struct node *temp, *p;
temp = create_node(value);
if (start == NULL)
{
start = temp;
start->next = NULL;
}
else
{
p = start;
start = temp;
start->next = p;
}
cout<<\"Element Inserted at beginning\"<>value;
struct node *temp, *s;
temp = create_node(value);
s = start;
while (s->next != NULL)
{
s = s->next;
}
temp->next = NULL;
s->next = temp;
cout<<\"Element Inserted at last\"<info<<\"->\";
temp = temp->next;
}
cout<<\"NULL\"<
Solution
ANSWER ::
ASSMUING THE TOTAL LINE IN CODE ARE 158*(INCLUDING COMMENTS)
SO...
PROGRAM STARTS WITH SPECIFING HEADER FILES AND GLOBAL VARIABLE
DECLARATION
THEN Node Declaration USING STRUCTURES,
THEN WE WANT TO DECLARE THE CLASS USING SINGLE LINKED LIST AND THE
PROTOTYPE IS PUBLIC
IN THE PUBLIC WE WANT TO CREATE THE NODE AS INTEGER DATATYPE
THE NODE CONSIST OF insert_begin();
insert_pos();-> INDICATES THE CURRENT POSSITION
insert_last();->INSERT AT LAST
delete_pos();->TO DELETE THE NUMBER AT CURRENT POSITION
sort();->START THE SORTING
search();->USED TO SEARCH
update();->UPDATE THE NODE
reverse();->REVERSE NUMBERS
display();->TO DISPLAY THE NODE
single_llist()->FIFO
WE ARE STARTING TAKING NULL VALUE
AND THEN TAKE USER DEFIND FUNCTION MAIN()
CREATE THE DATATYPES choice, nodes, element, position, i
OBJECT CREATION OF SLL IS SL
USING WHILE LOOP THE NODE INSERT AT THE BEGINING OR ENDING STAGE
WE ARE USING SWITCH CASES TO PERFORM SLL ON NODE
NODE CREATION AS
node *single_llist::create_node(int value)
ALSO DECLARE THE TEMP IS TAKEN AS NULL VALUE
INSERT THE ELEMENT AT THE BEGINING
USING IF CONDITION WE CAN INSERT THE NODE AT THE BEGINING
struct node *temp, *p;
temp = create_node(value);
if (start == NULL)
{
start = temp;
start->next = NULL;
}
else
{
p = start;
start = temp;
start->next = p;
AND THEN INSERT THE NODE AT THE LAST
struct node *temp, *s;
temp = create_node(value);
s = start;
while (s->next != NULL)
{
s = s->next;
}
temp->next = NULL;
s->next = temp;
USING WHILE LOOP WE ARE INSERTING THE NODE AT LAST
USING THIS BELLOW CODE WE CAN SEARCH THE ELEMENTS IN THE NODE
struct node *temp;
if (start == NULL)
{
cout<<\"The List is Empty\"<info<<\"->\";
temp = temp->next;
}
cout.
Lecture 3, c++(complete reference,herbet sheidt)chapter-13
Slide of C++(Complete Reference,Herbet Sheildt book)
CSE Department
Rajshahi Engineering Science and Technology College
using set identitiesSolutionimport java.util.Scanner; c.pdf
using set identities
Solution
import java.util.Scanner;
/* category Node */
class Node
creator */
public Node()
/* builder */
public Node(int d,Node n)
/* perform to line link to next Node */
public void setLink(Node n)
/* perform to line knowledge to current Node */
public void setData(int d)
/* perform to urge link to next node */
public Node getLink()
come back link;
}
/* perform to urge knowledge from current Node */
public int getData()
come back data;
}
}
/* category linkedList */
class linkedList
begin ;
protected Node finish ;
public int size ;
/* builder */
public linkedList()
begin = null;
finish = null;
size = 0;
}
/* perform to envision if list is empty */
public mathematician isEmpty()
come back begin == null;
}
/* perform to urge size of the list */
public int getSize()
come back size;
}
/* perform to insert part at the begining */
public void insertAtStart(int val)
{
Node nptr = new Node(val,null);
nptr.setLink(start);
if(start == null)
{
begin = nptr;
nptr.setLink(start);
finish = start;
}
else
{
end.setLink(nptr);
begin = nptr;
}
size++ ;
}
/* perform to insert part at finish */
public void insertAtEnd(int val)
{
Node nptr = new Node(val,null);
nptr.setLink(start);
if(start == null)
{
begin = nptr;
nptr.setLink(start);
finish = start;
}
else
{
end.setLink(nptr);
finish = nptr;
}
size++ ;
}
/* perform to insert part at position */
public void insertAtPos(int val , int pos)
one ;
for (int i = 1; i < size - 1; i++)
ptr = ptr.getLink();
}
size++ ;
}
/* perform to delete part at position */
public void deleteAtPos(int pos)
one && pos == 1)
begin = null;
finish = null;
size = 0;
return ;
}
if (pos == 1)
begin = begin.getLink();
end.setLink(start);
size--;
return ;
}
if (pos == size)
whereas (s != end)
end = t;
end.setLink(start);
size --;
return;
}
Node ptr = start;
pos = pos - one ;
for (int i = 1; i < size - 1; i++)
ptr = ptr.getLink();
}
size-- ;
}
/* perform to show contents */
public void display()
making object of linkedList */
linkedList list = new linkedList();
System.out.println(\"Circular individually coupled List Test\ \");
char ch;
/* Perform list operations */
do
{
System.out.println(\"\ Circular individually coupled List Operations\ \");
System.out.println(\"1. insert at begining\");
System.out.println(\"2. insert at end\");
System.out.println(\"3. insert at position\");
System.out.println(\"4. delete at position\");
System.out.println(\"5. check empty\");
System.out.println(\"6. get size\");
int selection = scan.nextInt();
switch (choice).
Consider L = {a^nb^2nc^P p 0}. Prove L is not a context-free langu.pdf
Capital and Revenue Expenditures On February 13, Scanlon Co. paid $3,400 to repair the
transmission on one of its delivery vans.
In addition, Scanlon paid $20 to install a GPS system in its van.
Journalize the entries for the transmission. February 13? It should be two answers, Cr/Dr
Journalize the entry for GPS system expenditures. February 13? It should be two answers, Cr/Dr
Solution
Account Debit Credit Accumulated Depreciation- Delivery Van $3400 Cash $3400 To
record the repair of transmission on delivery vans.(Revenue Expenditure Delivery Van $20
Cash $20 (To record capital expenditure on Van (installing GPS).
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.
Using the provided table interface table.h and the sample linked lis.pdf
Using the provided table interface table.h and the sample linked list code linkedList.c, complete
an implementation of the Table ADT. Make sure that you apply the concepts of design by
contract (DbC) to your implementation.
Once you have fully implemented the table, create a main.c file that implements a testing
framework for your table.
Your table implementation must ensure that values inserted are unique, and internally sorted
within a linked list.
table.h
linkedList.c
Solution
khsdg
#include
#include
typedef enum BOOL { false, true } bool;
// Linked list node definition
typedef struct Node node;
struct Node
{
int number;
node *next;
};
static node *top = NULL;
// used to track where we are for the list traversal methods
static node *traverseNode = NULL;
// \"destroy\" will deallocate all nodes in a linked list object
// and will set \"top\" to NULL.
void destroy()
{
node *curr = top;
node *temp = NULL;
while ( curr != NULL )
{
// flip order to see it blow up...
temp = curr;
curr = curr->next;
free( temp );
}
top = NULL;
}
// \"build\" will create an ordered linked list consisting
// of the first \"size\" even integers.
void build( int size )
{
node *newNode = NULL;
int i = 0;
// make sure we don\'t have a list yet
destroy();
for ( i=size ; i>0 ; i-- )
{
newNode = malloc( sizeof( node ) );
newNode->number = i*2;
newNode->next = top;
top = newNode;
}
}
// starts a list traversal by getting the data at top.
// returns false if top == NULL.
bool firstNode( int *item )
{
bool result = false;
if ( top )
{
*item = top->number;
traverseNode = top->next;
result = true;
}
return result;
}
// gets the data at the current traversal node and increments the traversal.
// returns false if we\'re at the end of the list.
bool nextNode( int *item )
{
bool result = false;
if ( traverseNode )
{
*item = traverseNode->number;
traverseNode = traverseNode->next;
result = true;
}
return result;
}
// \"print\" will output an object\'s entire linked list
// to the standard output device -- one \"number\" per line.
void print()
{
int value;
if ( firstNode( &value ) )
{
do
{
printf( \"%d\ \", value );
} while ( nextNode( &value ) );
}
}
int main( int argc, char *argv[] )
{
build( 10 );
print();
destroy();
build( 5 );
build( 20 );
print();
destroy();
return 0;
}.
Assignment is Page 349-350 #4 and #5 Use the Linked Lis.pdf
Assignment is :
\"Page 349-350 #4 and #5 Use the \"Linked List lab\" you have been working on in class and add
the two functions the questions are asking you to develop: divideMid and divideAt. Be sure to
include comments Use meaningful identifier names (constants where appropriate) Do not work
together; no two people should have identical work!?!? Turn in .cpp file AND Turn in a \"print-
screen\' of your output (press \"print-screen\' on keyboard, then \'paste\' in MS-Word)\"
How do you solve QUESTION #5 in the book data structures using c++ by D.S. Malik in Visiual
Studios using the linked list below with what is being asked? Please need help
Linked list :
#include
#include
using namespace std;
struct nodeType
{
int info;
nodeType *link;
};
void createList(nodeType*& first, nodeType*& last);
void printList(nodeType*& first);
void insertFront(nodeType*& first);
void insertBack(nodeType*& last);
void deleteFirst(nodeType*& first);
void deleteLast(nodeType*& last, nodeType* first);
int main()
{
nodeType *first, *last;
int num;
createList(first, last);
int choice;
while(true)
{
cout<<\"1. Insert Front.\ 2. Insert Last.\ 3. Delete Front.\ 4. Delete Last.\ 5. Print List.\ 6. Exit.\
\";
cout<<\"Enter your choice: \";
cin>>choice;
switch(choice)
{
case 1: insertFront(first); break;
case 2: insertBack(last); break;
case 3: deleteFirst(first); break;
case 4: deleteLast(last, first); break;
case 5: printList(first); break;
case 6: return 0;
default: cout<<\"Invalid menu option. Try again.\"<>number;
while (number != -999)
{
newNode = new nodeType; // create new node
newNode->info = number;
newNode->link = NULL;
if (first == NULL)
{
first = newNode;
last = newNode;
}
else
{
last->link = newNode;
last = newNode;
}
cout<<\"Enter an integer (-999 to stop): \";
cin>>number;
} // end of while-loop
} // end of build list function
void deleteFirst(nodeType*& first)
{
nodeType *temp;
temp= first;
first= temp->link;
delete temp;
return;
}
void deleteLast(nodeType*& last, nodeType* current)
{
nodeType *temp;
while(current->link != NULL)
{
temp=current;
current=current->link;
}
temp=last;
current->link=NULL;
delete temp;
last = current;
return;
}
void insertFront(nodeType*& front)
{
int num;
cout<<\"\ Enter the number to insert: \";
cin>>num;
nodeType *newNode = new nodeType;
newNode->info=num;
newNode->link= front;
front= newNode;
return;
}
void insertBack(nodeType*& last)
{
int num;
cout<<\"\ Enter the number to insert: \";
cin>>num;
nodeType *newNode = new nodeType;
newNode->info=num;
newNode->link= NULL;
last->link= newNode;
last = newNode;
return;
}
void printList(nodeType*& first)
{
cout<<\"Inside printList...printing linked list...\ \"<info << \" \";
current = current->link;
}
cout<
#include
using namespace std;
struct nodeType
{
int info;
nodeType *link;
};
void createList(nodeType*& first, nodeType*& last);
void printList(nodeType*& first);
void insertFront(nodeType*& first);
void insertBack(nodeType*& last);
void deleteFirst(nodeType*& first);
void dele.
File Type cppAdd the following to your linked list of strings pro.pdf
File Type: cpp
Add the following to your linked list of strings program:
1. allow the user to search for an item (indicate the result of the search)
2. allow the user to insert an item (print the linked list after insertion)
3. allow the user to delete an item (print the linked list after deletion)
Here\'s the program I need to have remodified for this assignment as shown below.
#include
#include
#include
using namespace std;
int main()
{
struct nodeType
{
string info;
nodeType *link;
};
nodeType *first, *newNode, *last;
string x; //string input variable
cout << \"Enter a string of some kind. Enter \'Done\' when finished.\" << endl;
getline(cin, x);
first = NULL;
last = NULL;
while (x != \"Done\")
{
newNode = new nodeType;
newNode->info = x;
newNode->link = NULL;
if (first == NULL)
{
first = newNode;
last = newNode;
}
else
{
last->link = newNode;
last = newNode;
}
cout << newNode->info << \" \";
cout << \"Enter a string of some kind. Enter \'Done\' when finished.\" << endl;
getline(cin, x);
}
return 0;
}
Solution
In the main function, you need to enter the following:
main()
{
int ch, nodes, ele, posi;
sin_llist sl;
start = NULL;
while(1)
{
cout<<\"1.Search for an item\"<>ch;
switch(ch)
{
case 1:
cout<<\"Searching an item:\"<>val;
struct node *z;
z = start;
while (z != NULL)
{
posi++;
if (z->info == val)
{
flag = true;
cout<<\"Element \"<next;
}
if (!flag)
cout<<\"Element \"<>y;
struct node *tem, *k;
temp = create_node(y);
if (start == NULL)
{
start = tem;
start->next = NULL;
}
else
{
k = start;
start = tem;
start->next = k;
}
cout<<\"Element Inserted at beginning\"<>posi;
struct node *z, *ptr;
z = start;
if (posi == 1)
{
start = z->next;
}
else
{
while (z != NULL)
{
z = z->next;
count++;
}
if (posi > 0 && posi <= count)
{
z = start;
for (i = 1;i < pos;i++)
{
ptr = z;
z = z->next;
}
ptr->next = z->next;
}
else
{
cout<<\"Position out of range\"<.
C++ Please write the whole code that is needed for this assignment- wr.docx
Brianca plans to save $5,000, $1,000, and $42,000 a year over the next three years, respectively. How much would you need to deposit in one lump sum today to have the same amount as Brianca three years from now if you both earn 10.9 percent, compounded annually? $36,115 $35,192 $43,282 $41,635 $35,372
.
#includeiostream #includecstdio #includecstdlib using na.pdf
#include
#include
#include
using namespace std;
struct node
{
int val;
struct node *next;
}*startNode;
// declaring class
class LinkedList
{
public:
node* create_node(int);
void insert_last();
void sort();
void search();
void reverse();
void display();
LinkedList()
{
startNode = NULL;
}
};
// main
main()
{
int userChoice, linkedListNodes, i;
LinkedList LinkList;
startNode = NULL;
while (1)
{
// displaying menu
cout<<\" 1)Insert\";
cout<<\"2) Remove\";
cout<<\"3) Search\";
cout<<\"4) Display\";
cout<<\"5) Reverse\";
cout<<\"6) Exit \";
// reading user choice
cout<<\"Enter your userChoice : \";
cin>>userChoice;
switch(userChoice)
{
case 1:
cout<<\"Inserting node to linked list: \";
LinkList.insert_begin();
cout;
break;
case 2:
cout<<\"Delete element from linked list: \";
LinkList.delete();
break;
case 3:
cout<<\"Searching data in linked list: \";
LinkList.search();
cout;
break;
case 4:
cout<<\"Displaying linked list data: \";
LinkList.display();
cout;
break;
case 5:
cout<<\"Linked list reversing: \";
LinkList.reverse();
cout;
break;
case 6:
cout<<\"Exiting...\";
exit(1);
break;
default:
cout<<\"Enter correct userChoice\";
}
}
}
// node creating to store into list
node *LinkedList::create_node(int val)
{
struct node *tempNode, *tempNode;
tempNode = new(struct node);
if (tempNode == NULL)
{
cout<<\"No memory allocated \";
return 0;
}
else
{
tempNode->val = val;
tempNode->next = NULL;
return tempNode;
}
}
/*
* Inserting Node at last
*/
void LinkedList::insert_last()
{
int val;
cout<<\"Enter the val to be inserted: \";
cin>>val;
struct node *tempNode, *tempNode;
tempNode = create_node(val);
tempNode = startNode;
while (tempNode->next != NULL)
{
tempNode = tempNode->next;
}
tempNode->next = NULL;
tempNode->next = tempNode;
cout<<\"Element Inserted\";
}
// deleting element
void LinkedList::delete()
{
int position, i, count = 0;
if (startNode == NULL)
{
cout<<\"Linkedlist empty\"<>position;
struct node *tempNode, *ptr;
tempNode = startNode;
if (position == 1)
{
startNode = tempNode->next;
}
else
{
while (tempNode != NULL)
{
tempNode = tempNode->next;
count++;
}
if (position > 0 && position <= count)
{
tempNode = startNode;
for (i = 1;i < position;i++)
{
ptr = tempNode;
tempNode = tempNode->next;
}
ptr->next = tempNode->next;
}
else
{
cout<<\"Enter correct position\";
}
}
}
// search data in linked list
void LinkedList::search()
{
int val, position = 0;
bool flagValue = false;
if (startNode == NULL)
{
cout<<\"List is empty\";
return;
}
cout<<\"Enter the val to search: \";
cin>>val;
struct node *tempNode;
tempNode = startNode;
while (tempNode != NULL)
{
position++;
if (tempNode->val == val)
{
flagValue = true;
cout<<\"Data \"<next;
}
if (!flagValue)
cout<<\"data not found\";
}
// linkedlist reverese
void LinkedList::reverse()
{
struct node *pointer1, *pointer2, *pointer3;
if (startNode == NULL)
{
cout<<\"List is empty\"<next == NULL)
{
return;
}
pointer1 = startNode;
pointer2 = pointer1->next;
pointer3 = pointer2->next;
pointer1->next = NULL;
pointer2->.
In C++ I need help with this method that Im trying to write fillLi.pdf
In C++ I need help with this method that I'm trying to write: fillListFromFile( ) : void -- this
method will add elements to the list from a file called linkedListData1.txt HINT: this is where
you will keep track of the number of elements in the list
I have all of the other methods, but I'm stuck on this one and I've added it as the last method in
the application. The code I have so far is below.
.cpp file:
.h file:
main:
#include
#include
#include
#include "Node.h"
using namespace std;
void displayList(Node*);
void insertFront(Node** head, int value);
void insertBack(Node** ptrToHead, int value);
void insertAfter(Node* previous, int value);
void deleteAtN(Node** ptrToHead, int n);
void deleteNode(Node** ptrToHead, int searchVal);
void clearList(Node**);
bool searchVal(Node** head, int value);
int countFrequencyOf(Node** ptrTohead, int value);
int size(Node** ptrTohead);
void displayFromFrontToN(Node** ptrToHead, int n);
void displayFromNToEnd(Node** ptrToHead, int n);
void insertInOrder(Node** head, Node* ptrToHead);
void displayInRows(Node** ptrToHead);
int main()
{
Node* node1 = new Node();
node1->setValue(10);
Node* node2 = new Node();
node2->setValue(20);
Node* node3 = new Node();
node3->setValue(30);
node1->next = node2;
node2->next = node3;
Node* head = node1;
displayList(head);
insertFront(&head, 5);
displayList(head);
insertBack(&head, 50);
displayList(head);
insertBack(&head, 60);
displayList(head);
insertAfter(node2, 25);
displayList(head);
deleteAtN(&head, 2);
displayList(head);
deleteNode(&head, 50);
displayList(head);
clearList(&head);
displayList(head);
insertFront(&head, 5);
insertFront(&head, 10);
insertFront(&head, 20);
insertBack(&head, 50);
searchVal(&head, 50);
displayList(head);
countFrequencyOf(&head, 25);
displayList(head);
insertBack(&head, 25);
insertFront(&head, 25);
countFrequencyOf(&head, 25);
displayList(head);
size(&head);
cout << "add one item to list" << endl;
insertBack(&head, 25);
size(&head);
displayList(head);
cout << "Display From Front To N" << endl;
displayFromFrontToN(&head, 4);
cout << "Display From N To End" << endl;
displayFromNToEnd(&head, 3);
cout << "Display In Rows" << endl;
displayInRows(&head);
insertFront(&head, 12);
insertFront(&head, 17);
insertFront(&head, 22);
insertBack(&head, 58);
cout << "Display In Rows" << endl;
displayInRows(&head);//call displayInRows()
return 0;
}//end main()
void displayList(Node* ptr) {
if (ptr == nullptr)
cout << "empty list" << endl;
else {
while (ptr != nullptr) {
cout << ptr->getValue() << endl;
ptr = ptr->next;
}
}
cout << "---------------------------" << endl;
}//end displayList()
void insertFront(Node** head, int value) {
Node* newNode = new Node();
newNode->setValue(value);
//cout<< newNode->getValue() << endl;
newNode->next = *head;
*head = newNode;
}
void insertBack(Node** ptrToHead, int value) {
Node* newNode = new Node();
newNode->setValue(value);
newNode->next = nullptr;
if (*ptrToHead == nullptr)
{
*ptrToHead = newNode;
}//end if
else
{
Node* last .
write recursive function that calculates and returns the length of a.pdf
write recursive function that calculates and returns the length of a linked list
Solution
// C code to determine length of linked list recursively
#include
#include
// node of linked list
struct node
{
int key;
struct node* next;
};
// push key to linked list
void push(struct node** root, int new_element)
{
struct node* nodeNew = (struct node*) malloc(sizeof(struct node));
nodeNew->key = new_element;
nodeNew->next = (*root);
(*root) = nodeNew;
}
// recursively get length of linked list
int lengthLinkedlist(struct node* root)
{
// base case
if (root == NULL)
return 0;
// recursively call the function
return 1 + lengthLinkedlist(root->next);
}
int main()
{
struct node* root = NULL;
push(&root, 11);
push(&root, 34);
push(&root, 4);
push(&root, 23);
push(&root, 14);
push(&root, 88);
push(&root, 56);
push(&root, 84);
push(&root, 6);
/* Check the count function */
printf(\"Length of Linked list: %d\ \", lengthLinkedlist(root));
return 0;
}
/*
output:
Length of Linked list: 9
*/.
C++ Please test your program before you submit the answer.pdf
C++. Please, test your program before you submit the answer.
Background:
Circular Linked List: A circular linked list is a linked list where a non-empty list has no null pointers.
Unless empty, each node has a pointer to the next node, and the last node points back to the first
node.
Assigment:
Write a cpp program that will serve the purpose of a test driver for previously written CLinkedList.h
file. Your program will:
1) Create a choice menu and prompt the user for input. Must be case insensitive!
2) Read the input and call for functions outside of main, in order to perform operations chosen by
the user. These functions involve: Inserting Interger, Removing Interger, Forward Print and
Backward Print.
Requirements and Notes:
1. Forwards Print does not require a subfunction in order to produce output. All the other
operations must call corresponding subfunctions.
2. Nodes of the circular list must be arranged in ascending order.
3. User Input must be processed case insensitive, and implemented using the loop.
4. Document your code completely.
5. CLinkedList.h file contains 3 classes.
- Circular linked lists class with following member functions: constructor, copy constructor,
destructor, assignment operator, insert and remove.
- Node class included in the file header.
- A list iterator class provides a public interface to the list. The pointer points at nodes in the linked
list. Its member functions are as follows:
Constructor: Assigns the linked list data member to the parameter, a reference to a circular linked
list.
begin() sets the iterator to the first node of the linked list (or NULL if the list is empty)
isEmpty() returns whether or not the wrapped linked list is empty
isFirstNode() returns whether the present node is the first node of the linked list
isLastNode() returns whether the present node is the last node of the linked list
operator*() returns the data of the node currently pointed at. (You need 2)
operator++() pre-increment operator advances the pointer to the next node in the list, if there is
one
operator++(int) post-increment operator advances the pointer to the next node in the list, if there
is one
6. Insert function must give feedback where it says the position of where the int was inserted for
example: "Inserting 25 in the middle."
7. Remove function must give either one of these types of feedback:
- "Removing 25" if the int given by the user was found in the list
-"Cannot find 25" if it was not. The program will then move the user back to the choice menu.
Files:
// File: Node.h
// Singly-linked list node definition/implementation (also good for circular)
#ifndef NODE_H
#define NODE_H
// Need to prototype template classes if they are to be friends
template <typename eltType> class CLinkedList;
template <typename eltType> class CListItr;
template <typename eltType> class Node
{ private:
Node(eltType info, Node* link = NULL ) :
data(info), next(link)
{};
eltType data;
Node *next;
friend class CLinkedList<eltType>;
friend c.
C++ Background Circular Linked List A circular linked list.pdf
C++
Background:
Circular Linked List: A circular linked list is a linked list where a non-empty list has no null pointers.
Unless empty, each node has a pointer to the next node, and the last node points back to the first
node.
Assigment:
Write a cpp program that will serve the purpose of a test driver for previously written CLinkedList.h
file. Your program will:
1) Create a choice menu and prompt the user for input. Must be case insensitive!
2) Read the input and call for functions outside of main, in order to perform operations chosen by
the user. These functions involve: Inserting Interger, Removing Interger, Forward Print and
Backward Print.
Requirements and Notes:
1. Forwards Print does not require a subfunction in order to produce output. All the other
operations must call corresponding subfunctions.
2. Nodes of the circular list must be arranged in ascending order.
3. User Input must be processed case insensitive, and implemented using the loop.
4. Document your code completely.
5. CLinkedList.h file contains 3 classes.
- Circular linked lists class with following member functions: constructor, copy constructor,
destructor, assignment operator, insert and remove.
- Node class included in the file header.
- A list iterator class provides a public interface to the list. The pointer points at nodes in the linked
list. Its member functions are as follows:
Constructor: Assigns the linked list data member to the parameter, a reference to a circular linked
list.
begin() sets the iterator to the first node of the linked list (or NULL if the list is empty)
isEmpty() returns whether or not the wrapped linked list is empty
isFirstNode() returns whether the present node is the first node of the linked list
isLastNode() returns whether the present node is the last node of the linked list
operator*() returns the data of the node currently pointed at. (You need 2)
operator++() pre-increment operator advances the pointer to the next node in the list, if there is
one
operator++(int) post-increment operator advances the pointer to the next node in the list, if there
is one
6. Insert function must give feedback where it says the position of where the int was inserted for
example: "Inserting 25 in the middle."
7. Remove function must give either one of these types of feedback:
- "Removing 25" if the int given by the user was found in the list
-"Cannot find 25" if it was not. The program will then move the user back to the choice menu.
Files:
// File: Node.h
// Singly-linked list node definition/implementation (also good for circular)
#ifndef NODE_H
#define NODE_H
// Need to prototype template classes if they are to be friends
template <typename eltType> class CLinkedList;
template <typename eltType> class CListItr;
template <typename eltType> class Node
{ private:
Node(eltType info, Node* link = NULL ) :
data(info), next(link)
{};
eltType data;
Node *next;
friend class CLinkedList<eltType>;
friend class CListItr<eltType>;
};
#endif
// File: CLinkedList.h
.
AnswerNote LinkedList.cpp is written and driver program main.cpp.pdf
Answer:
Note: LinkedList.cpp is written and driver program main.cpp is also given.
//ListInterface.h
#ifndef _LIST_INTERFACE
#define _LIST_INTERFACE
template
class ListInterface
{
public:
virtual bool isEmpty() const = 0;
virtual int getLength() const = 0;
virtual bool insert(int newPosition, const ItemType& newEntry) = 0;
virtual bool remove(int position) = 0;
virtual void clear() = 0;
virtual ItemType getEntry(int position) const = 0;
virtual void replace(int position, const ItemType& newEntry) = 0;
}; // end ListInterface
#endif
//Node.h
#ifndef NODE_
#define NODE_
template
class Node
{
private:
ItemType item; // A data item
Node* next; // Pointer to next node
public:
Node();
Node(const ItemType& anItem);
Node(const ItemType& anItem, Node* nextNodePtr);
void setItem(const ItemType& anItem);
void setNext(Node* nextNodePtr);
ItemType getItem() const ;
Node* getNext() const ;
}; // end Node
#endif
//Node.cpp
#include \"Node.h\"
template
Node::Node() : next(nullptr)
{
} // end default constructor
template
Node::Node(const ItemType& anItem) : item(anItem), next(nullptr)
{
} // end constructor
template
Node::Node(const ItemType& anItem, Node* nextNodePtr) :
item(anItem), next(nextNodePtr)
{
} // end constructor
template
void Node::setItem(const ItemType& anItem)
{
item = anItem;
} // end setItem
template
void Node::setNext(Node* nextNodePtr)
{
next = nextNodePtr;
} // end setNext
template
ItemType Node::getItem() const
{
return item;
} // end getItem
template
Node* Node::getNext() const
{
return next;
} // end getNext
//PrecondViolatedExcept.h
#ifndef PRECOND_VIOLATED_EXCEPT_
#define PRECOND_VIOLATED_EXCEPT_
#include
#include
class PrecondViolatedExcept : public std::logic_error
{
public:
PrecondViolatedExcept(const std::string& message = \"\");
}; // end PrecondViolatedExcept
#endif
//PrecondViolatedExcept.cpp
#include \"PrecondViolatedExcept.h\"
PrecondViolatedExcept::PrecondViolatedExcept(const std::string& message)
: std::logic_error(\"Precondition Violated Exception: \" + message)
{
} // end constructor
//LinkedList.h
#ifndef LINKED_LIST_
#define LINKED_LIST_
#include \"ListInterface.h\"
#include \"Node.h\"
#include
#include \"PrecondViolatedExcept.h\"
template
class LinkedList : public ListInterface
{
private:
Node* headPtr; // Pointer to first node in the chain;
// (contains the first entry in the list)
int itemCount; // Current count of list items
// Locates a specified node in this linked list.
Node* getNodeAt(int position) const;
public:
LinkedList();
LinkedList(const LinkedList& aList);
virtual ~LinkedList();
bool isEmpty() const;
int getLength() const;
bool insert(int newPosition, const ItemType& newEntry);
bool remove(int position);
void clear();
ItemType getEntry(int position) const;
void replace(int position, const ItemType& newEntry);
}; // end LinkedList
#endif
//LinkedList.cpp
#include \"LinkedList.h\"
#include
#include
#include
#include
//Default constructor
template
LinkedList::LinkedList() : headPtr(nullptr), itemCount(0)
{
.
C code on linked list #include stdio.h #include stdlib.h.pdf
For the ethanol water graph the simple distillation should be a steady smooth
increase, the fractional distillation should look like an s, that sort of takes a straight up jump in
the middle. I am unsure about the other mixture though sorry, hope that helps
Solution
For the ethanol water graph the simple distillation should be a steady smooth
increase, the fractional distillation should look like an s, that sort of takes a straight up jump in
the middle. I am unsure about the other mixture though sorry, hope that helps.
Similar to C++ Program to Implement Doubly Linked List #includei.pdf (20)
take the following code and give details of what each line of code i.pdf
take the following code and give details of what each line of code i.pdf
Lecture 3, c++(complete reference,herbet sheidt)chapter-13
Lecture 3, c++(complete reference,herbet sheidt)chapter-13
using set identitiesSolutionimport java.util.Scanner; c.pdf
using set identitiesSolutionimport java.util.Scanner; c.pdf
Consider L = {a^nb^2nc^P p 0}. Prove L is not a context-free langu.pdf
Consider L = {a^nb^2nc^P p 0}. Prove L is not a context-free langu.pdf
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.docx
Using the provided table interface table.h and the sample linked lis.pdf
Using the provided table interface table.h and the sample linked lis.pdf
Assignment is Page 349-350 #4 and #5 Use the Linked Lis.pdf
Assignment is Page 349-350 #4 and #5 Use the Linked Lis.pdf
File Type cppAdd the following to your linked list of strings pro.pdf
File Type cppAdd the following to your linked list of strings pro.pdf
C++ Please write the whole code that is needed for this assignment- wr.docx
C++ Please write the whole code that is needed for this assignment- wr.docx
#includeiostream #includecstdio #includecstdlib using na.pdf
#includeiostream #includecstdio #includecstdlib using na.pdf
In C++ I need help with this method that Im trying to write fillLi.pdf
In C++ I need help with this method that Im trying to write fillLi.pdf
write recursive function that calculates and returns the length of a.pdf
write recursive function that calculates and returns the length of a.pdf
C++ Please test your program before you submit the answer.pdf
C++ Please test your program before you submit the answer.pdf
C++ Background Circular Linked List A circular linked list.pdf
C++ Background Circular Linked List A circular linked list.pdf
AnswerNote LinkedList.cpp is written and driver program main.cpp.pdf
AnswerNote LinkedList.cpp is written and driver program main.cpp.pdf
C code on linked list #include stdio.h #include stdlib.h.pdf
C code on linked list #include stdio.h #include stdlib.h.pdf
More from Lalkamal2
steps are as follows Dissolve the (pressumably.pdf
steps are as follows : Dissolve the (pressumably solid) mixture in an organic
solven,,,,,,, Then, in a seperating funnel, add an (Aqueous) solution of sodium hydroxide. The
Phenol, and benzoic acid, as acids, will react with the base to form phenolate and benzenoate
ions, which will dissolve into the water layer. Collect this layer, and re-precipitate the organic
molecules by adding concentrated Hydrochloric acid. The phenol and Benzoic acid can be
seperated via their varying solubilities, (8.3g/100ml vs 0.34g/100ml. respectively) by heating the
mixture in water to dissolve the phenol, and then filtering of the (practically) insoluble benzoic
acid. The solution can then be acidified (with HCl) to return the phenol..,,,,,,,,,,,,, Now add a
solution of hydrochloric acid to the seperating funnel, which (as before) will react with the basic
anniline dossilving it into the water. Collect the aqueous layer and re-precipitate the anniline
with excess sodium hydroxide. ,,,,,,,,,,,,,,, finally boil off the remaining organic solvent to obtain
the napthalene, which was left untouched by all the acids and bases.,,,,,,,,, Flow chart : Mixture
+ xs HCl ----> anilinium(aq) Mixture - aniline + weak base (Na2CO3) ---> benzoate(aq)
Mixture - aniline - benzoic acid + base (NaOH) ---> p-bromophenoxide(aq) Mixture - aniline -
benzoic acid - p-bromophenol = naphthalene
Solution
steps are as follows : Dissolve the (pressumably solid) mixture in an organic
solven,,,,,,, Then, in a seperating funnel, add an (Aqueous) solution of sodium hydroxide. The
Phenol, and benzoic acid, as acids, will react with the base to form phenolate and benzenoate
ions, which will dissolve into the water layer. Collect this layer, and re-precipitate the organic
molecules by adding concentrated Hydrochloric acid. The phenol and Benzoic acid can be
seperated via their varying solubilities, (8.3g/100ml vs 0.34g/100ml. respectively) by heating the
mixture in water to dissolve the phenol, and then filtering of the (practically) insoluble benzoic
acid. The solution can then be acidified (with HCl) to return the phenol..,,,,,,,,,,,,, Now add a
solution of hydrochloric acid to the seperating funnel, which (as before) will react with the basic
anniline dossilving it into the water. Collect the aqueous layer and re-precipitate the anniline
with excess sodium hydroxide. ,,,,,,,,,,,,,,, finally boil off the remaining organic solvent to obtain
the napthalene, which was left untouched by all the acids and bases.,,,,,,,,, Flow chart : Mixture
+ xs HCl ----> anilinium(aq) Mixture - aniline + weak base (Na2CO3) ---> benzoate(aq)
Mixture - aniline - benzoic acid + base (NaOH) ---> p-bromophenoxide(aq) Mixture - aniline -
benzoic acid - p-bromophenol = naphthalene.
the two features of the asynchronus connection that allow the receiv.pdf
the two features of the asynchronus connection that allow the receiver to stay in sync with the
incoming data stream are the start bit and short frame connections . asynchronus connection are
asynchronus serial communications in which the communicating end points interfaces are not
continously synchronized by common clock signal and the data stream contains synchronization
in the form of start and stop signals before and after each unit of transmission .asynchronus start
and stop signals was widley used for dial up modem sources to time sharing computers and bbs
systems asynchronus start stop is the physical layer is used to connect computers to modems for
many dial up internet access applications using data link frame protocol such as ppp to create
packets made upto out of asynchronous serial characters .the examples that would benefit from
asynchronous connection are modems and usb that are connectedd to interfaces
Solution
the two features of the asynchronus connection that allow the receiver to stay in sync with the
incoming data stream are the start bit and short frame connections . asynchronus connection are
asynchronus serial communications in which the communicating end points interfaces are not
continously synchronized by common clock signal and the data stream contains synchronization
in the form of start and stop signals before and after each unit of transmission .asynchronus start
and stop signals was widley used for dial up modem sources to time sharing computers and bbs
systems asynchronus start stop is the physical layer is used to connect computers to modems for
many dial up internet access applications using data link frame protocol such as ppp to create
packets made upto out of asynchronous serial characters .the examples that would benefit from
asynchronous connection are modems and usb that are connectedd to interfaces.
Telomerase is the enzyme responsible for maintenance of the length o.pdf
Telomerase is the enzyme responsible for maintenance of the length of telomeres by addition of
guanine-rich repetitive sequences, (Zvereva, 2010). Thus Telomerase, is a Specialized
Polymerase That Carries Its Own RNA Template. Thus it accounts for a faithful replication of
chromosomal ends as theTelomerase contains an RNA molecule that acts as a template for
extending the leading strand.
Solution
Telomerase is the enzyme responsible for maintenance of the length of telomeres by addition of
guanine-rich repetitive sequences, (Zvereva, 2010). Thus Telomerase, is a Specialized
Polymerase That Carries Its Own RNA Template. Thus it accounts for a faithful replication of
chromosomal ends as theTelomerase contains an RNA molecule that acts as a template for
extending the leading strand..
Suppose Bob sends an encrypted document to Alice. To be verifiable, .pdf
Suppose Bob sends an encrypted document to Alice. To be verifiable, Alice must be able to
convince herself that Bob sent the encrypted document. To be non-forgeable, Alice must be able
to convince herself that only Bob could have sent the encrypted document Alice must be able to
convince someone else that only Bob could have sent the document. To illustrate the latter
distinction, suppose Bob and Alice share a secret key, and they are the only ones in the world
who know the key. If Alice receives a document that was encrypted with the key, and knows that
she did not encrypt the document herself, then the document is known to be verifiable and non-
forgeable (assuming a suitably strong encryption system was used). However, Alice can not
convince someone else that Bob must have sent the document, since in fact Alice knew the key
herself and could have encrypted/sent the document.
Solution
Suppose Bob sends an encrypted document to Alice. To be verifiable, Alice must be able to
convince herself that Bob sent the encrypted document. To be non-forgeable, Alice must be able
to convince herself that only Bob could have sent the encrypted document Alice must be able to
convince someone else that only Bob could have sent the document. To illustrate the latter
distinction, suppose Bob and Alice share a secret key, and they are the only ones in the world
who know the key. If Alice receives a document that was encrypted with the key, and knows that
she did not encrypt the document herself, then the document is known to be verifiable and non-
forgeable (assuming a suitably strong encryption system was used). However, Alice can not
convince someone else that Bob must have sent the document, since in fact Alice knew the key
herself and could have encrypted/sent the document..
probablity = 212 100 = 16.6 Solutionprobablity = 212 10.pdf
probablity = 2/12 * 100 = 16.6 %
Solution
probablity = 2/12 * 100 = 16.6 %.
Platinum complexes like Asplastin are powerful antitumor medications.pdf
Platinum complexes like Asplastin are powerful antitumor medications and they are having
ability to cross the cell membrane and then reaches the nucleus and will attach to DNA and stop
cell growth. The transfer of this Asplastin is facilitated by a copper transporter (Ctr1-membrane-
dwelling protein which main function is to bring copper into cells). The |Ctr1 structure has of
three helical segments which appears to sit in the membrane, and one end can be seen protruding
into the cell, while the other end will be seen on the outside and a channel like structure will be
formed when three such molecules assemble together.
When Asplastin is injected in the body of a cancer patient, the Asplastin which have a central
platinum ion and four ligands (which includes two neighboring chloride ions and two
neighboring amino groups). The Ctr1 will displaces these four ligands from Asplastin and will
bind to the platinum ion itself and it will transfer it through the channel and released in the
cytosol of the cell. In addition to this few amount of Asplastin molecules will interact with the
methionine-rich extracellular end of the copper transporter to form a trimeric channel structure to
get transported as a little interior bubble filled with the outer medium and will released into
various organelles including nucleus.
How it acts on cancer cells: The 4 steps how Asplastin will act is (i) Cellular uptake of drug, (ii)
aquation/activation of Asplastin, (iii) Drug binding to DNA and (iv) Lesions formed on DNA
will make the cell to opt for apoptosis.
N7 sites of deoxyguanosine residues on the DNA are the most nucleophilic positions and are
preferentially platinated. Asplastin will form a monofunctional adducts on DNA (You can see
only one covalent bond to the genomic polymer). Like in second reaction second guanine base
will get substituted by chloride ligand, thus forming a cross-link on the DNA (either on the same
strand or it may be on different strands). The most common lesion are formed from 1,2-d(GpG)
intrastrand cross-links and finally due to drug action we can see bending and unwinding of the
double helix and cells with damaged DNA will be arrested at the G2/M transition of the cell
cycle and undergoes apoptosis.
Solution
Platinum complexes like Asplastin are powerful antitumor medications and they are having
ability to cross the cell membrane and then reaches the nucleus and will attach to DNA and stop
cell growth. The transfer of this Asplastin is facilitated by a copper transporter (Ctr1-membrane-
dwelling protein which main function is to bring copper into cells). The |Ctr1 structure has of
three helical segments which appears to sit in the membrane, and one end can be seen protruding
into the cell, while the other end will be seen on the outside and a channel like structure will be
formed when three such molecules assemble together.
When Asplastin is injected in the body of a cancer patient, the Asplastin which have a central
platinum ion and f.
Part 1 1)#include stdio.hint testWhileLoop() ; int testFo.pdf
Part 1:
1)
#include
int testWhileLoop() ;
int testForLoop() ;
int main()
{
printf(\"While loop test \ \");
testWhileLoop();
printf(\"For loop test \ \");
testForLoop();
return 0;
}
Part 2:
1)
#include
int testWhileLoop() ;
int main()
{
printf(\"While loop test \ \");
testWhileLoop();
return 0;
}
int testWhileLoop()
{
int count=0;
int n;
while (count<10)
{
printf(\"Enter an integer: \");
scanf(\"%d\",&n);
printf(\"You Entered: %d\ \",n);
count++;
}
}
2)
#include
int testWhileLoop() ;
int main()
{
printf(\"While loop test \ \");
testWhileLoop();
return 0;
}
int testWhileLoop()
{
int count=0;
int n;
int sum=0;
while (count<10)
{
printf(\"Enter an integer: \");
scanf(\"%d\",&n);
printf(\"You Entered: %d\ \",n);
count++;
sum=sum+n;
}
printf(\"Total sum of integers is: %d\ \",sum);
}
3)
#include
int testWhileLoop() ;
int main()
{
printf(\"While loop test \ \");
testWhileLoop();
return 0;
}
int testWhileLoop()
{
int count=0;
int n;
int sum=0;
while (count<10)
{
printf(\"Enter an integer between 5 and 20: \");
scanf(\"%d\",&n);
while(n<5 || n>20)
{
printf(\"Error: You must enter a value between 5 and 20: please renter: \");
scanf(\"%d\",&n);
}
printf(\"You Entered: %d\ \",n);
count++;
sum=sum+n;
}
printf(\"Total sum of integers is: %d\ \",sum);
}
Part 3:
#include
int testForLoop() ;
int main()
{
printf(\"For loop test \ \");
testForLoop();
return 0;
}
int testForLoop()
{
int count=0;
int n;
int sum=0;
for (count=0;count<10;count++)
{
printf(\"Enter an integer: \");
scanf(\"%d\",&n);
printf(\"You Entered: %d\ \",n);
sum=sum+n;
}
printf(\"Total sum of integers is: %d\ \",sum);
}
Solution
Part 1:
1)
#include
int testWhileLoop() ;
int testForLoop() ;
int main()
{
printf(\"While loop test \ \");
testWhileLoop();
printf(\"For loop test \ \");
testForLoop();
return 0;
}
Part 2:
1)
#include
int testWhileLoop() ;
int main()
{
printf(\"While loop test \ \");
testWhileLoop();
return 0;
}
int testWhileLoop()
{
int count=0;
int n;
while (count<10)
{
printf(\"Enter an integer: \");
scanf(\"%d\",&n);
printf(\"You Entered: %d\ \",n);
count++;
}
}
2)
#include
int testWhileLoop() ;
int main()
{
printf(\"While loop test \ \");
testWhileLoop();
return 0;
}
int testWhileLoop()
{
int count=0;
int n;
int sum=0;
while (count<10)
{
printf(\"Enter an integer: \");
scanf(\"%d\",&n);
printf(\"You Entered: %d\ \",n);
count++;
sum=sum+n;
}
printf(\"Total sum of integers is: %d\ \",sum);
}
3)
#include
int testWhileLoop() ;
int main()
{
printf(\"While loop test \ \");
testWhileLoop();
return 0;
}
int testWhileLoop()
{
int count=0;
int n;
int sum=0;
while (count<10)
{
printf(\"Enter an integer between 5 and 20: \");
scanf(\"%d\",&n);
while(n<5 || n>20)
{
printf(\"Error: You must enter a value between 5 and 20: please renter: \");
scanf(\"%d\",&n);
}
printf(\"You Entered: %d\ \",n);
count++;
sum=sum+n;
}
printf(\"Total sum of integers is: %d\ \",sum);
}
Part 3:
#include
int testForLoop() ;
int main()
{
printf(\"For loop test \ \");
testForLoop();
return 0;
}
int testForLoop()
{
int c.
none of them are correct!Solutionnone of them are correct!.pdf
none of them are correct!
Solution
none of them are correct!.
Negative control plate with no DNA transforms competent cells withou.pdf
Negative control plate with no DNA transforms competent cells without any DNA on selection
media. , If more colonies are produced, shows the problem with antibiotic in the agar plate.
No ligase control plates provide information about digestion efficiency. it transform competent
cells withdigested plasmid or DNA. If several colonies are found, it shows that digestion did not
completed properly.
Solution
Negative control plate with no DNA transforms competent cells without any DNA on selection
media. , If more colonies are produced, shows the problem with antibiotic in the agar plate.
No ligase control plates provide information about digestion efficiency. it transform competent
cells withdigested plasmid or DNA. If several colonies are found, it shows that digestion did not
completed properly..
Modifications highlighted in bold lettersDropOutStack.javaim.pdf
//Modifications highlighted in bold letters
//DropOutStack.java
import java.util.Arrays;
public class DropOutStack {
/**
* Defines the interface to a stack collection.
*
* @author Java Foundations
* @version 4.0
*/
public interface StackADT {
/**
* Adds the specified element to the top of this stack.
*
* @param element
* element to be pushed onto the stack
*/
public void push(T element);
/**
* Removes and returns the top element from this stack.
*
* @return the element removed from the stack
*/
public T pop();
/**
* Returns without removing the top element of this stack.
*
* @return the element on top of the stack
*/
public T peek();
/**
* Returns true if this stack contains no elements.
*
* @return true if the stack is empty
*/
public boolean isEmpty();
/**
* Returns the number of elements in this stack.
*
* @return the number of elements in the stack
*/
public int size();
/**
* Returns a string representation of this stack.
*
* @return a string representation of the stack
*/
public String toString();
}
/**
* Represents the situation in which a collection is empty.
*
* @author Java Foundations
* @version 4.0
*/
public static class EmptyCollectionException extends RuntimeException {
/**
* Sets up this exception with an appropriate message.
*
* @param collection
* the name of the collection
*/
public EmptyCollectionException(String collection) {
super(\"The \" + collection + \" is empty.\");
}
}
/**
* Program entry point for stack testing.
*
* @param args
* Argument list.
*/
public static void main(String[] args) {
ArrayDropOutStack stack = new ArrayDropOutStack(4);
System.out.println(\"DROP-OUT STACK TESTING\");
stack.push(1);
stack.push(2);
stack.push(3);
stack.push(4);
System.out.println(\"Stack : \"+stack.toString());
System.out.println(\"pushing 5 element \");
stack.push(5);
System.out.println(\"Now stack elements\");
System.out.println(stack.toString());
System.out.println(\"The size of the stack is: \" + stack.size());
if (!stack.isEmpty())
System.out.println(\"The stack contains:\ \" + stack.toString());
stack.pop();
System.out.println(\"poping element 5\"+stack.toString());
System.out.println(\"The size of the stack is: \" + stack.size());
System.out.println(\"pushing 7 element \");
stack.push(7);
System.out.println(\"Now stack elements\");
System.out.println(stack.toString());
System.out.println(\"pushing 8 element \");
stack.push(8);
System.out.println(\"Now stack elements\");
System.out.println(stack.toString());
if (!stack.isEmpty())
System.out.println(\"The stack contains:\ \" + stack.toString());
}
/**
* An array implementation of a stack in which the bottom of the stack is
* fixed at index 0.
*
* @author Java Foundations
* @version 4.0
*/
public static class ArrayDropOutStack implements StackADT {
private final static int DEFAULT_CAPACITY = 100;
private int top;
private T[] stack;
/**
* Creates an empty stack using the default capacity.
*/
public ArrayDropOutStack() {
this(DEFAULT_CAPACITY);
}
/**
* Creates an empty stack using.
Message types used by the DHCP boot sequenceop Message type.h.pdf
Message types used by the DHCP boot sequence
op
: Message type.
htype
: Hardware address type defined at Internet Assigned Numbers Authority (IANA).
hlen
: Hardware address length, in octets.
hops
: Value set to zero by DHCP clients. Optionally used to count the number of relay agents that
forwarded the message.
xid
: A random number used to associate messages and responses between a client and a server.
secs
: Seconds elapsed since client began address acquisition or renewal process.
flags
: Flags set by client. The Broadcast flag is set if the client cannot receive unicast IP datagrams
(for example, before it is configured with an IP address).
ciaddr
: Used if the client has an IP address and can respond to Address Resolution Protocol (ARP)
requests.
yiaddr
: Address given to the DHCP client by the DHCP server.
siaddr
: IP address of the server that is offering a lease.
giaddr
: DHCP relay agent IP address.
chaddr
: Client hardware address.
sname
: Optional server host name. Not used in Windows Server 2008.
file
: The name of the file containing the boot image for a BOOTP client.
options
: Optional parameters field. In the DHCP protocol packet, each option begins with a single octet
tag, which holds the option code, and a second octet, which describes the option data length, in
bytes. For a complete list of the DHCP options available by default on a DHCP server running
on Windows Server 2008, see DHCP Tools and Options.
op
: Message type.
htype
: Hardware address type defined at Internet Assigned Numbers Authority (IANA).
hlen
: Hardware address length, in octets.
hops
: Value set to zero by DHCP clients. Optionally used to count the number of relay agents that
forwarded the message.
xid
: A random number used to associate messages and responses between a client and a server.
secs
: Seconds elapsed since client began address acquisition or renewal process.
flags
: Flags set by client. The Broadcast flag is set if the client cannot receive unicast IP datagrams
(for example, before it is configured with an IP address).
ciaddr
: Used if the client has an IP address and can respond to Address Resolution Protocol (ARP)
requests.
yiaddr
: Address given to the DHCP client by the DHCP server.
siaddr
: IP address of the server that is offering a lease.
giaddr
: DHCP relay agent IP address.
chaddr
: Client hardware address.
sname
: Optional server host name. Not used in Windows Server 2008.
file
: The name of the file containing the boot image for a BOOTP client.
options
: Optional parameters field. In the DHCP protocol packet, each option begins with a single octet
tag, which holds the option code, and a second octet, which describes the option data length, in
bytes. For a complete list of the DHCP options available by default on a DHCP server running
on Windows Server 2008, see DHCP Tools and Options.
Solution
Message types used by the DHCP boot sequence
op
: Message type.
htype
: Hardware address type defined at Internet Assigned Numbers Authority.
Law of conservation of matter(mass)In any closed system subjected .pdf
Law of conservation of matter(mass)
In any closed system subjected to no external forces,the mass is constant irrespective of its
changes in different form. Matter cannot be created or destroyed in an isolated system. It only
changes from one form into another.
Solution
Law of conservation of matter(mass)
In any closed system subjected to no external forces,the mass is constant irrespective of its
changes in different form. Matter cannot be created or destroyed in an isolated system. It only
changes from one form into another..
It is a unit of measurement for the amount of substance.The mole is .pdf
It is a unit of measurement for the amount of substance.The mole is defined as the amount of
substance that contains as many elementary entities (e.g., atoms, molecules, ions, electrons) as
there are atoms in 12 g of the isotope carbon-12. The experimentally determined value of a mole
as adopted 6.02 × 1023.
Solution
It is a unit of measurement for the amount of substance.The mole is defined as the amount of
substance that contains as many elementary entities (e.g., atoms, molecules, ions, electrons) as
there are atoms in 12 g of the isotope carbon-12. The experimentally determined value of a mole
as adopted 6.02 × 1023..
In insurance, the term risk pooling refers to the spreading of f.pdf
In insurance, the term \"risk pooling\" refers to the spreading of financial risks evenly among a
large number of contributors to the program. Insurance is the transference of risks from
individuals or corporations who cannot bear a possible unplanned financial catastrophe to the
capital markets, which can bear them easily -- at least in theory.
Pure risk is a category of risk in which loss is the only possible outcome, which is the opposite of
speculative risk. There are products that can be purchased to mitigate pure risk, such as home
insurance being used to protect homeowners against their homes being destroyed. Other
examples of pure risk events include premature death, identity theft and career-ending
disabilities.1) in cases of home insurance and health insurance risk pooling can be effectively
used as of all the individuals who invest in the insurance by the law of large number a miniscule
probability of the people actually suffer the real risk. 2) Risk pooling is an important concept in
supply chain management . It suggests that demand variability is reduced if one aggregates
demand across locations because, as demand is aggregated across different locations, it becomes
more likely that high demand from one customer will be offset by low demand from another.
However,consider the 1)\"high risk pools\", which simply means separating out the sickest, most
expensive people in the country and dumping them into a separate program.With the \"bad
apples\" (ie, human beings with terrible medical problems) safely tucked out of the way, the
average cost of treating everyone else supposedly suddenly becomes less pricey.This is the exact
opposite of the entire point of health insurance in the first place...spreading the risk and hence
risk pooling fails here.
2)Liability risk includes economic losses from being held responsible for harming others or their
property. This cannot be covered by risk pooling.
Solution
In insurance, the term \"risk pooling\" refers to the spreading of financial risks evenly among a
large number of contributors to the program. Insurance is the transference of risks from
individuals or corporations who cannot bear a possible unplanned financial catastrophe to the
capital markets, which can bear them easily -- at least in theory.
Pure risk is a category of risk in which loss is the only possible outcome, which is the opposite of
speculative risk. There are products that can be purchased to mitigate pure risk, such as home
insurance being used to protect homeowners against their homes being destroyed. Other
examples of pure risk events include premature death, identity theft and career-ending
disabilities.1) in cases of home insurance and health insurance risk pooling can be effectively
used as of all the individuals who invest in the insurance by the law of large number a miniscule
probability of the people actually suffer the real risk. 2) Risk pooling is an important concept in
supply chain management . It suggests that.
Some CO is Added The Pressure of CH4 and H2O will.pdf
Some CO is Added The Pressure of CH4 and H2O will Increase Equilibrium will
shift to the Left Some CH4 is Removed Pressure of H2O will Increase Pressure of CO will
Decrease Equilibrium will shift to the Left The Volume of Vessel is Reduced Pressure of CH4
will Increase Pressure of H2O will Increase Equilibrium will shift to the Left
Solution
Some CO is Added The Pressure of CH4 and H2O will Increase Equilibrium will
shift to the Left Some CH4 is Removed Pressure of H2O will Increase Pressure of CO will
Decrease Equilibrium will shift to the Left The Volume of Vessel is Reduced Pressure of CH4
will Increase Pressure of H2O will Increase Equilibrium will shift to the Left.
Comments Figures are not properly numbered. Further signal of figur.pdf
Comments: Figures are not properly numbered. Further signal of figure 1 is not shown and hence
the above figure (assumed to figure 2) can not be simulated.
Solution
Comments: Figures are not properly numbered. Further signal of figure 1 is not shown and hence
the above figure (assumed to figure 2) can not be simulated..
Change the value of jug1 and jug 2 accordingly. 1 File WaterJugBSF.j.pdf
Change the value of jug1 and jug 2 accordingly. 1 File WaterJugBSF.java
import java.util.*;
import java.util.LinkedList;
class Test {
HashSet uniqueStates;
void letsRoll() {
/*
* The capacity of jug1(jug1 capacity is greater than jug2 capacity)
*/
int jug1 = 3;
/*
* The capacity of jug two
*/
int jug2 = 2;
/*
* Final Amount needed in jug1
*/
int amtNeeded = 2;
State initialState = new State(0, 0);
State finalState = new State(amtNeeded, 0);
State finalPath = null;
uniqueStates = new HashSet();
/*
* Change queue to stack to convert to DFS
*/
LinkedList queue = new LinkedList();
queue.add(initialState);
while (!queue.isEmpty()) {
State currState = queue.poll();
if (currState.equals(finalState)) {
finalPath = currState;
break;
}
// if x is zero fill it
if (currState.x == 0) {
State nextState = new State(jug1, currState.y, currState);
//System.out.println(currState);
checkUniqueStates(uniqueStates, nextState, queue);
}
// if y is empty fill it
if (currState.y == 0) {
State nextState = new State(currState.x, jug2, currState);
checkUniqueStates(uniqueStates, nextState, queue);
}
// if x is not empty , empty it
if (currState.x > 0) {
State nextState = new State(0, currState.y, currState);
checkUniqueStates(uniqueStates, nextState, queue);
}
// if y is not empty, empty it
if (currState.y > 0) {
State nextState = new State(currState.x, 0, currState);
checkUniqueStates(uniqueStates, nextState, queue);
}
// transfer from x to y, when x non empty and y is not full
if (currState.x > 0 && currState.y < jug2) {
int amtToTransfer = Math.min(currState.x, jug2 - currState.y);
State nextState = new State(currState.x - amtToTransfer, currState.y
+ amtToTransfer,
currState);
checkUniqueStates(uniqueStates, nextState, queue);
}
// transfer from y to x, when y is not empty and x is not full
if (currState.y > 0 && currState.x < jug1) {
int amtToTransfer = Math.min(currState.y, jug1 - currState.x);
State nextState = new State(currState.x + amtToTransfer, currState.y
- amtToTransfer,
currState);
checkUniqueStates(uniqueStates, nextState, queue);
}
}
if (finalPath != null) {
System.out.println(\"Final Result\");
System.out.println(\"J1 J2\");
System.out.println(finalPath);
}
else{
System.out.println(\"Not Possible\");
}
}
/*
* Checks whether State toCheck has been generated before, if not it is
* added to queue and uniqueStates Set
*/
void checkUniqueStates(HashSet uniqueStates, State toCheck,
LinkedList queue) {
if (!uniqueStates.contains(toCheck)) {
uniqueStates.add(toCheck);
queue.add(toCheck);
}
}
public static void main(String[] args) {
new Test().letsRoll();
}
}
2 State.java
class State {
/*
* Amount in jug1 for current State
*/
int x;
/*
* Amount in jug2 for current State
*/
int y;
/*
* Parent of current State
*/
State pre;
public State(int x, int y) {
super();
this.x = x;
this.y = y;
}
public State(int x, int y, State pre) {
super();
this.x = x;
this.y = y;
this.pre = pre;
System.out.println(\"(\"+x+\",\"+y+\")\");
}
@Override
public int hashCode() {
final int .
Balance sheetBalance sheet is the statement of financial position .pdf
Balance sheet
Balance sheet is the statement of financial position which reports all the assets, liabilities and
equity held by the company at a specific date. It helps in telling about the investments, resources,
claims and owner’s equity of the company. It revolves around the equation i.e. Assets =
Liabilities + equity.
What is shown in balance sheet:
The balance sheet has three major categories which includes Assets, liabilities and equity. All
these are calssified on the basis of their nature.
Assets can be said as the resources of the company and are classified as: Current assets, long
term investment, property, plant and equipment, intangible assets and other assets
Assets includes:
•Current assets – Cash, accounts receivables, prepaid expenses, inventories
•Long term investments – investment in stock
•Plant property and equipment – Land, building, equipment
•Intangible assets – Goodwill, patents, trademarks
Liabilities are the claims and can be classified as current liabilities, long term liabilities
Liabilities includes:
•Current liabilities: Accounts payable, income tax payable. Salaries payable, accrued expenses,
unearned revenue, current portion of long term debt due
•Long term liabilities- bonds payable, notes payable
Owner;s equity can be classified as the Capital stock which includes common stock and
preferred stock
It also includes Additional paid in capital, Retained earnings
Retained earnings are the accumulated earnings of the company. It is calculated as Beginning
balance, Add: Net income less: Dividends
Importance:
Balance sheet is important in every business. Considering the hospitality firm, it is important to
analyze different aspects of teh firm to know its financial postion. As the hospitality firm
emphasize on services, it is important to know the financial position of the company through its
financial statements. the hospitality firms like hotel needs to report their assets (building) held by
the company along with other assets, liabilities and equity.
Balance sheet is the statement which helps in analysing the financial position of the company, it
can also help in predicting the cash flows and the financial position of the company. With the
help of the balance sheet we can help in analysing the ability of the firm to pay off its short term
and long term debts.
Balance sheet reports its assets and liabilities on historical basis and uses different estimates to
analyse the amounts like uncollectibles
Solution
Balance sheet
Balance sheet is the statement of financial position which reports all the assets, liabilities and
equity held by the company at a specific date. It helps in telling about the investments, resources,
claims and owner’s equity of the company. It revolves around the equation i.e. Assets =
Liabilities + equity.
What is shown in balance sheet:
The balance sheet has three major categories which includes Assets, liabilities and equity. All
these are calssified on the basis of their nature.
Assets can be .
More from Lalkamal2 (20)
steps are as follows Dissolve the (pressumably.pdf
steps are as follows Dissolve the (pressumably.pdf
the two features of the asynchronus connection that allow the receiv.pdf
the two features of the asynchronus connection that allow the receiv.pdf
Telomerase is the enzyme responsible for maintenance of the length o.pdf
Telomerase is the enzyme responsible for maintenance of the length o.pdf
Suppose Bob sends an encrypted document to Alice. To be verifiable, .pdf
Suppose Bob sends an encrypted document to Alice. To be verifiable, .pdf
probablity = 212 100 = 16.6 Solutionprobablity = 212 10.pdf
probablity = 212 100 = 16.6 Solutionprobablity = 212 10.pdf
Platinum complexes like Asplastin are powerful antitumor medications.pdf
Platinum complexes like Asplastin are powerful antitumor medications.pdf
Part 1 1)#include stdio.hint testWhileLoop() ; int testFo.pdf
Part 1 1)#include stdio.hint testWhileLoop() ; int testFo.pdf
none of them are correct!Solutionnone of them are correct!.pdf
none of them are correct!Solutionnone of them are correct!.pdf
Negative control plate with no DNA transforms competent cells withou.pdf
Negative control plate with no DNA transforms competent cells withou.pdf
Modifications highlighted in bold lettersDropOutStack.javaim.pdf
Modifications highlighted in bold lettersDropOutStack.javaim.pdf
Message types used by the DHCP boot sequenceop Message type.h.pdf
Message types used by the DHCP boot sequenceop Message type.h.pdf
Law of conservation of matter(mass)In any closed system subjected .pdf
Law of conservation of matter(mass)In any closed system subjected .pdf
It is a unit of measurement for the amount of substance.The mole is .pdf
It is a unit of measurement for the amount of substance.The mole is .pdf
In insurance, the term risk pooling refers to the spreading of f.pdf
In insurance, the term risk pooling refers to the spreading of f.pdf
Some CO is Added The Pressure of CH4 and H2O will.pdf
Some CO is Added The Pressure of CH4 and H2O will.pdf
Comments Figures are not properly numbered. Further signal of figur.pdf
Comments Figures are not properly numbered. Further signal of figur.pdf
Change the value of jug1 and jug 2 accordingly. 1 File WaterJugBSF.j.pdf
Change the value of jug1 and jug 2 accordingly. 1 File WaterJugBSF.j.pdf
Balance sheetBalance sheet is the statement of financial position .pdf
Balance sheetBalance sheet is the statement of financial position .pdf
Recently uploaded
June 3, 2024 Anti-Semitism Letter Sent to MIT President Kornbluth and MIT Cor...
Letter from the Congress of the United States regarding Anti-Semitism sent June 3rd to MIT President Sally Kornbluth, MIT Corp Chair, Mark Gorenberg
Dear Dr. Kornbluth and Mr. Gorenberg,
The US House of Representatives is deeply concerned by ongoing and pervasive acts of antisemitic
harassment and intimidation at the Massachusetts Institute of Technology (MIT). Failing to act decisively to ensure a safe learning environment for all students would be a grave dereliction of your responsibilities as President of MIT and Chair of the MIT Corporation.
This Congress will not stand idly by and allow an environment hostile to Jewish students to persist. The House believes that your institution is in violation of Title VI of the Civil Rights Act, and the inability or
unwillingness to rectify this violation through action requires accountability.
Postsecondary education is a unique opportunity for students to learn and have their ideas and beliefs challenged. However, universities receiving hundreds of millions of federal funds annually have denied
students that opportunity and have been hijacked to become venues for the promotion of terrorism, antisemitic harassment and intimidation, unlawful encampments, and in some cases, assaults and riots.
The House of Representatives will not countenance the use of federal funds to indoctrinate students into hateful, antisemitic, anti-American supporters of terrorism. Investigations into campus antisemitism by the Committee on Education and the Workforce and the Committee on Ways and Means have been expanded into a Congress-wide probe across all relevant jurisdictions to address this national crisis. The undersigned Committees will conduct oversight into the use of federal funds at MIT and its learning environment under authorities granted to each Committee.
• The Committee on Education and the Workforce has been investigating your institution since December 7, 2023. The Committee has broad jurisdiction over postsecondary education, including its compliance with Title VI of the Civil Rights Act, campus safety concerns over disruptions to the learning environment, and the awarding of federal student aid under the Higher Education Act.
• The Committee on Oversight and Accountability is investigating the sources of funding and other support flowing to groups espousing pro-Hamas propaganda and engaged in antisemitic harassment and intimidation of students. The Committee on Oversight and Accountability is the principal oversight committee of the US House of Representatives and has broad authority to investigate “any matter” at “any time” under House Rule X.
• The Committee on Ways and Means has been investigating several universities since November 15, 2023, when the Committee held a hearing entitled From Ivory Towers to Dark Corners: Investigating the Nexus Between Antisemitism, Tax-Exempt Universities, and Terror Financing. The Committee followed the hearing with letters to those institutions on January 10, 202
Chapter 3 - Islamic Banking Products and Services.pptx
Chapter 3 - Islamic Banking Products and Services.pptxMohd Adib Abd Muin, Senior Lecturer at Universiti Utara Malaysia
This slide is prepared for master's students (MIFB & MIBS) UUM. May it be useful to all.Best Digital Marketing Institute In NOIDA
Safalta Digital marketing institute in Noida, provide complete applications that encompass a huge range of virtual advertising and marketing additives, which includes search engine optimization, virtual communication advertising, pay-per-click on marketing, content material advertising, internet analytics, and greater. These university courses are designed for students who possess a comprehensive understanding of virtual marketing strategies and attributes.Safalta Digital Marketing Institute in Noida is a first choice for young individuals or students who are looking to start their careers in the field of digital advertising. The institute gives specialized courses designed and certification.
for beginners, providing thorough training in areas such as SEO, digital communication marketing, and PPC training in Noida. After finishing the program, students receive the certifications recognised by top different universitie, setting a strong foundation for a successful career in digital marketing.
Embracing GenAI - A Strategic Imperative
Artificial Intelligence (AI) technologies such as Generative AI, Image Generators and Large Language Models have had a dramatic impact on teaching, learning and assessment over the past 18 months. The most immediate threat AI posed was to Academic Integrity with Higher Education Institutes (HEIs) focusing their efforts on combating the use of GenAI in assessment. Guidelines were developed for staff and students, policies put in place too. Innovative educators have forged paths in the use of Generative AI for teaching, learning and assessments leading to pockets of transformation springing up across HEIs, often with little or no top-down guidance, support or direction.
This Gasta posits a strategic approach to integrating AI into HEIs to prepare staff, students and the curriculum for an evolving world and workplace. We will highlight the advantages of working with these technologies beyond the realm of teaching, learning and assessment by considering prompt engineering skills, industry impact, curriculum changes, and the need for staff upskilling. In contrast, not engaging strategically with Generative AI poses risks, including falling behind peers, missed opportunities and failing to ensure our graduates remain employable. The rapid evolution of AI technologies necessitates a proactive and strategic approach if we are to remain relevant.
Operation Blue Star - Saka Neela Tara
Operation “Blue Star” is the only event in the history of Independent India where the state went into war with its own people. Even after about 40 years it is not clear if it was culmination of states anger over people of the region, a political game of power or start of dictatorial chapter in the democratic setup.
The people of Punjab felt alienated from main stream due to denial of their just demands during a long democratic struggle since independence. As it happen all over the word, it led to militant struggle with great loss of lives of military, police and civilian personnel. Killing of Indira Gandhi and massacre of innocent Sikhs in Delhi and other India cities was also associated with this movement.
TESDA TM1 REVIEWER FOR NATIONAL ASSESSMENT WRITTEN AND ORAL QUESTIONS WITH A...
TESDA TM1 REVIEWER FOR NATIONAL ASSESSMENT WRITTEN AND ORAL QUESTIONS WITH ANSWERS.
Biological Screening of Herbal Drugs in detailed.
Biological screening of herbal drugs: Introduction and Need for
Phyto-Pharmacological Screening, New Strategies for evaluating
Natural Products, In vitro evaluation techniques for Antioxidants, Antimicrobial and Anticancer drugs. In vivo evaluation techniques
for Anti-inflammatory, Antiulcer, Anticancer, Wound healing, Antidiabetic, Hepatoprotective, Cardio protective, Diuretics and
Antifertility, Toxicity studies as per OECD guidelines
A Survey of Techniques for Maximizing LLM Performance.pptx
A Survey of Techniques for Maximizing LLM Performance
Introduction to AI for Nonprofits with Tapp Network
Dive into the world of AI! Experts Jon Hill and Tareq Monaur will guide you through AI's role in enhancing nonprofit websites and basic marketing strategies, making it easy to understand and apply.
Supporting (UKRI) OA monographs at Salford.pptx
How libraries can support authors with open access requirements for UKRI funded books
Wednesday 22 May 2024, 14:00-15:00.
How libraries can support authors with open access requirements for UKRI fund...
How libraries can support authors with open access requirements for UKRI funded books
Wednesday 22 May 2024, 14:00-15:00.
Home assignment II on Spectroscopy 2024 Answers.pdf
Answers to Home assignment on UV-Visible spectroscopy: Calculation of wavelength of UV-Visible absorption
1.4 modern child centered education - mahatma gandhi-2.pptx
Child centred education is an educational approach that priorities the interest, needs and abilities of the child in the learning process.
Thesis Statement for students diagnonsed withADHD.ppt
Presentation required for the master in Education.
Recently uploaded (20)
June 3, 2024 Anti-Semitism Letter Sent to MIT President Kornbluth and MIT Cor...
June 3, 2024 Anti-Semitism Letter Sent to MIT President Kornbluth and MIT Cor...
Chapter 3 - Islamic Banking Products and Services.pptx
Chapter 3 - Islamic Banking Products and Services.pptx
TESDA TM1 REVIEWER FOR NATIONAL ASSESSMENT WRITTEN AND ORAL QUESTIONS WITH A...
TESDA TM1 REVIEWER FOR NATIONAL ASSESSMENT WRITTEN AND ORAL QUESTIONS WITH A...
Group Presentation 2 Economics.Ariana Buscigliopptx
Group Presentation 2 Economics.Ariana Buscigliopptx
A Survey of Techniques for Maximizing LLM Performance.pptx
A Survey of Techniques for Maximizing LLM Performance.pptx
Introduction to AI for Nonprofits with Tapp Network
Introduction to AI for Nonprofits with Tapp Network
How libraries can support authors with open access requirements for UKRI fund...
How libraries can support authors with open access requirements for UKRI fund...
Home assignment II on Spectroscopy 2024 Answers.pdf
Home assignment II on Spectroscopy 2024 Answers.pdf
1.4 modern child centered education - mahatma gandhi-2.pptx
1.4 modern child centered education - mahatma gandhi-2.pptx
Thesis Statement for students diagnonsed withADHD.ppt
Thesis Statement for students diagnonsed withADHD.ppt
C++ Program to Implement Doubly Linked List #includei.pdf
- 1. /* * C++ Program to Implement Doubly Linked List */ #include #include #include /* * Node Declaration */ using namespace std; struct node { int info; struct node *next; struct node *prev; }*head; /* Class Declaration */ class dll { public: void create(int value); void append(int value); void insert(int value, int index); void del(int value); void display(); dll() { head = NULL; } /* destructor*/ ~dll() { node *ptr;
- 2. ptr = head; while(ptr!=0){ node *next = ptr->next; delete ptr; ptr = next; } } }; /* * Main: Conatins Menu */ int main() { int option, element, index; dll dl; while (1) { cout<>option; switch ( option ) { case 1: cout<<"Enter the element: "; cin>>element; dl.create(element); cout<>element; dl.append(element); cout<>element; cout<<"Insert Element after postion: "; cin>>index; dl.insert(element, index); cout<>element; dl.del(element); cout<info = value; temp->next = NULL; if (head == NULL) {
- 3. temp->prev = NULL; head = temp; } else { s = head; while (s->next != NULL) s = s->next; s->next = temp; temp->prev = s; } } /* * Insertion at the beginning */ void dll::append(int value) { if (head == NULL) { cout<<"First Create the list."<prev = NULL; temp->info = value; temp->next = head; head->prev = temp; head = temp; cout<<"Element Inserted"<next; if (q == NULL) { cout<<"There are less than "; cout<info = value; if (q->next == NULL) { q->next = tmp; tmp->next = NULL; tmp->prev = q; } else
- 4. { tmp->next = q->next; tmp->next->prev = tmp; q->next = tmp; tmp->prev = q; } cout<<"Element Inserted"<info == value) { tmp = head; head = head->next; head->prev = NULL; cout<<"Element Deleted"<next->next != NULL) { /*Element deleted in between*/ if (q->next->info == value) { tmp = q->next; q->next = tmp->next; tmp->next->prev = q; cout<<"Element Deleted"<next; } /*last element deleted*/ if (q->next->info == value) { tmp = q->next; free(tmp); q->next = NULL; cout<<"Element Deleted"<info<<" <-> "; q = q->next; } cout<<"NULL"< Solution /* * C++ Program to Implement Doubly Linked List */
- 5. #include #include #include /* * Node Declaration */ using namespace std; struct node { int info; struct node *next; struct node *prev; }*head; /* Class Declaration */ class dll { public: void create(int value); void append(int value); void insert(int value, int index); void del(int value); void display(); dll() { head = NULL; } /* destructor*/ ~dll() { node *ptr; ptr = head; while(ptr!=0){ node *next = ptr->next; delete ptr;
- 6. ptr = next; } } }; /* * Main: Conatins Menu */ int main() { int option, element, index; dll dl; while (1) { cout<>option; switch ( option ) { case 1: cout<<"Enter the element: "; cin>>element; dl.create(element); cout<>element; dl.append(element); cout<>element; cout<<"Insert Element after postion: "; cin>>index; dl.insert(element, index); cout<>element; dl.del(element); cout<info = value; temp->next = NULL; if (head == NULL) { temp->prev = NULL; head = temp; } else
- 7. { s = head; while (s->next != NULL) s = s->next; s->next = temp; temp->prev = s; } } /* * Insertion at the beginning */ void dll::append(int value) { if (head == NULL) { cout<<"First Create the list."<prev = NULL; temp->info = value; temp->next = head; head->prev = temp; head = temp; cout<<"Element Inserted"<next; if (q == NULL) { cout<<"There are less than "; cout<info = value; if (q->next == NULL) { q->next = tmp; tmp->next = NULL; tmp->prev = q; } else { tmp->next = q->next; tmp->next->prev = tmp; q->next = tmp;
- 8. tmp->prev = q; } cout<<"Element Inserted"<info == value) { tmp = head; head = head->next; head->prev = NULL; cout<<"Element Deleted"<next->next != NULL) { /*Element deleted in between*/ if (q->next->info == value) { tmp = q->next; q->next = tmp->next; tmp->next->prev = q; cout<<"Element Deleted"<next; } /*last element deleted*/ if (q->next->info == value) { tmp = q->next; free(tmp); q->next = NULL; cout<<"Element Deleted"<info<<" <-> "; q = q->next; } cout<<"NULL"<