You can list anything, it doesnt matter. I just want to see code f.pdf

•

0 likes•2 views

F

fashionbigchennai

You can list anything, it doesn\'t matter. I just want to see code for this problem. Thank you Solution Include header files ....... Source code: public class SinglyLinkedListImpl { private Node head; private Node tail; public void add(T element){ Node nd = new Node(); nd.setValue(element); System.out.println(\"Adding: \"+element); if(head == null){ head = nd; tail = nd; } else { tail.setNextRef(nd); } tail = nd; } } public void addAfter(T element, T after){ Node tmp = head; Node refNode = null; System.out.println(\"Traversing to all nodes..\"); while(true){ if(tmp == null){ break; } if(tmp.compareTo(after) == 0){ refNode = tmp; break; } tmp = tmp.getNextRef(); } if(refNode != null){ Node nd = new Node(); nd.setValue(element); nd.setNextRef(tmp.getNextRef()); if(tmp == tail){ tail = nd; } tmp.setNextRef(nd); } else { System.out.println(\"Unable to find the given element...\"); } } public void deleteFront(){ if(head == null){ System.out.println(\"Underflow...\"); } Node tmp = head; head = tmp.getNextRef(); if(head == null){ tail = null; } System.out.println(\"Deleted: \"+tmp.getValue()); } public void deleteAfter(T after){ Node tmp = head; Node refNode = null; System.out.println(\"Traversing to all nodes..\"); while(true){ if(tmp == null){ break; } if(tmp.compareTo(after) == 0){ refNode = tmp; break; } tmp = tmp.getNextRef(); } if(refNode != null){ tmp = refNode.getNextRef(); refNode.setNextRef(tmp.getNextRef()); if(refNode.getNextRef() == null){ tail = refNode; } System.out.println(\"Deleted: \"+tmp.getValue()); } else { System.out.println(\"Unable to find the given element...\"); } } public void traverse(){ Node tmp = head; while(true){ if(tmp == null){ break; } System.out.println(tmp.getValue()); tmp = tmp.getNextRef(); } } public static void main(String a[]){ SinglyLinkedListImpl sl = new SinglyLinkedListImpl(); sl.add(3); sl.add(32); sl.add(54); sl.add(89); sl.addAfter(76, 54); sl.deleteFront(); sl.deleteAfter(76); sl.traverse(); } } class Node implements Comparable { private T value; private Node nextRef; public T getValue() { return value; } public void setValue(T value) { this.value = value; } public Node getNextRef() { return nextRef; } public void setNextRef(Node ref) { this.nextRef = ref; } public int compareTo(T arg) { if(arg == this.value){ return 0; } else { return 1; } } } This program implement the concept of single linked list..

You can list anything, it doesn't matter. I just want to see code for this problem. Thank you
Solution
Include header files .......
Source code:
public class SinglyLinkedListImpl {
private Node head;
private Node tail;
public void add(T element){
Node nd = new Node();
nd.setValue(element);
System.out.println("Adding: "+element);
if(head == null){
head = nd;
tail = nd;
} else {
tail.setNextRef(nd);
}
tail = nd;
}
}
public void addAfter(T element, T after){
Node tmp = head;
Node refNode = null;
System.out.println("Traversing to all nodes..");
while(true){
if(tmp == null){
break;
}
if(tmp.compareTo(after) == 0){
refNode = tmp;

break;
}
tmp = tmp.getNextRef();
}
if(refNode != null){
Node nd = new Node();
nd.setValue(element);
nd.setNextRef(tmp.getNextRef());
if(tmp == tail){
tail = nd;
}
tmp.setNextRef(nd);
} else {
System.out.println("Unable to find the given element...");
}
}
public void deleteFront(){
if(head == null){
System.out.println("Underflow...");
}
Node tmp = head;
head = tmp.getNextRef();
if(head == null){
tail = null;
}
System.out.println("Deleted: "+tmp.getValue());
}
public void deleteAfter(T after){
Node tmp = head;
Node refNode = null;
System.out.println("Traversing to all nodes..");
while(true){
if(tmp == null){
break;

}
if(tmp.compareTo(after) == 0){
refNode = tmp;
break;
}
tmp = tmp.getNextRef();
}
if(refNode != null){
tmp = refNode.getNextRef();
refNode.setNextRef(tmp.getNextRef());
if(refNode.getNextRef() == null){
tail = refNode;
}
System.out.println("Deleted: "+tmp.getValue());
} else {
System.out.println("Unable to find the given element...");
}
}
public void traverse(){
Node tmp = head;
while(true){
if(tmp == null){
break;
}
System.out.println(tmp.getValue());
tmp = tmp.getNextRef();
}
}
public static void main(String a[]){
SinglyLinkedListImpl sl = new SinglyLinkedListImpl();
sl.add(3);
sl.add(32);
sl.add(54);
sl.add(89);
sl.addAfter(76, 54);

sl.deleteFront();
sl.deleteAfter(76);
sl.traverse();
}
}
class Node implements Comparable {
private T value;
private Node nextRef;
public T getValue() {
return value;
}
public void setValue(T value) {
this.value = value;
}
public Node getNextRef() {
return nextRef;
}
public void setNextRef(Node ref) {
this.nextRef = ref;
}
public int compareTo(T arg) {
if(arg == this.value){
return 0;
} else {
return 1;
}
}
}
This program implement the concept of single linked list.

More Related Content

Similar to You can list anything, it doesnt matter. I just want to see code f.pdf

In this lab, you will be given a simple code for a min Heap, and youre going to complete some of the methods to handle some edge cases. Tasks for today: There are mainly two tasks, and the logic of both of them is provided in their bodies: 1) Fully implement AddItem(). 2) Fully implement HeapUp() Use the following start-up code. The main code and a demo class are provided as .cs files. Use them to test your code. Once you complete the above two tasks, upload the source code for your MinHeap class to Blackboard. class BinaryHeap : IEnumerable where T : IComparable { private T[] array; private int count; // Number of elements (nodes) public BinaryHeap(int size) { array = new T[size]; count = 0; } public T GetItem(int index) { return array[index]; } private void SetItem(int index, T value) { while (index >= array.Length) Grow(array.Length * 2); array[index] = value; } private void Grow(int newsize) { Array.Resize(ref array, newsize); } // Indices of left and right children private int LeftChildIndex(int pos) { return 2 * pos + 1; } private int RightChildIndex(int pos) { return 2 * pos + 2; } private int GetParentIndex(int pos) => (pos - 1) / 2; private T GetRightChild(int pos) => array[RightChildIndex(pos)]; private T GetLeftChild(int pos) => array[LeftChildIndex(pos)]; private T GetParent(int pos) => array[GetParentIndex(pos)]; // "Has" methods to determine if the indices exist private bool HasLeftChild(int pos) { if (LeftChildIndex(pos) < count) return true; else return false; } private bool HasRightChild(int pos) { if (RightChildIndex(pos) < count) return true; else return false; } private bool IsRoot(int pos) => pos == 0; // (true if element is root) // Swap, uh, swaps two values given two indicies. This should be private but I originally had it public for some reason. private void Swap(int position1, int position2) { T first = array[position1]; array[position1] = array[position2]; array[position2] = first; } public void AddItem(T value) { // This is part of la //Insert Logic // If the tree is empty, insert at the bottom (it does that already) // if not, insert at the end, // From the end you either need to swap to the root, and keep minheapify (Heapdown) // or, you should probably implement move up (for lab 6, you need to implement // for that you run a while loop, check if the current position both isn't the root, and is higher priority than a parent (in this case that probably means it's a lower value) // if it is, swap with parent, and keep doing that until it's its in the array[count] = value; count++; ReCalculateUp(); } private void HeapDown() { //CompareTo //this.CompareTo(value) returns < 0 if this < value //this.CompareTo(value) returns >0 if this > value int index = 0; while (HasLeftChild(index)) { var smallerIndex = LeftChildIndex(index); if (HasRightChild(index) && (GetRightChild(index).CompareTo(GetLeftChild(index))<0) ) //there's a set of ( ) around the right expression that are redundant but hopefully easier to read { smallerI.

In this lab, you will be given a simple code for a min Heap, and you.pdf

In this lab, you will be given a simple code for a min Heap, and you.pdf

In this lab, you will be given a simple code for a min Heap, and you.pdf

137 Lab-2.2.pdf

137 Lab-2.2.pdf

137 Lab-2.2.pdf

21E135MAHIESHWARJ

Solve using Java programming language. //************************** SLL.java ********************************* // a generic singly linked list class public class SLL<T> { private class SLLNode<T> { private T info; private SLLNode<T> next; public SLLNode() { this(null,null); } public SLLNode(T el) { this(el,null); } public SLLNode(T el, SLLNode<T> ptr) { info = el; next = ptr; } } protected SLLNode<T> head, tail; public SLL() { head = tail = null; } public boolean isEmpty() { return head == null; } public void addToHead(T el) { head = new SLLNode<T>(el,head); if (tail == null) tail = head; } public void addToTail(T el) { if (!isEmpty()) { tail.next = new SLLNode<T>(el); tail = tail.next; } else head = tail = new SLLNode<T>(el); } public T deleteFromHead() { // delete the head and return its info; if (isEmpty()) return null; T el = head.info; if (head == tail) // if only one node on the list; head = tail = null; else head = head.next; return el; } public T deleteFromTail() { // delete the tail and return its info; if (isEmpty()) return null; T el = tail.info; if (head == tail) // if only one node in the list; head = tail = null; else { // if more than one node in the list, SLLNode<T> tmp; // find the predecessor of tail; for (tmp = head; tmp.next != tail; tmp = tmp.next); tail = tmp; // the predecessor of tail becomes tail; tail.next = null; } return el; } public void delete(T el) { // delete the node with an element el; if (!isEmpty()) if (head == tail && el.equals(head.info)) // if only one head = tail = null; // node on the list; else if (el.equals(head.info)) // if more than one node on the list; head = head.next; // and el is in the head node; else { // if more than one node in the list SLLNode<T> pred, tmp;// and el is in a nonhead node; for (pred = head, tmp = head.next; tmp != null && !tmp.info.equals(el); pred = pred.next, tmp = tmp.next); if (tmp != null) { // if el was found; pred.next = tmp.next; if (tmp == tail) // if el is in the last node; tail = pred; } } } @Override public String toString() { if(head == null) return "[ ]"; String str = "[ "; SLLNode<T> tmp = head; while(tmp != null){ str += tmp.info + " "; tmp = tmp.next; } return str+"]"; } public boolean contains(T el) { if(head == null) return false; SLLNode<T> tmp = head; while(tmp != null){ if(tmp.info.equals(el)) return true; tmp = tmp.next; } return false; } public int size(){ if(head == null) return 0; int count = 0; SLLNode<T> p = head; while(p != null) { count++; p = p.next; } return count; } } Given the code above, solve the following tasks (please show the solution for every task clearly): For part (c) this is the code for the test program: public class SLL_Driver { public static void main(String[] args) { SLL<String> myList = new SLL<String>(); String[] cityNames = {"Jubail", "Riyadh", "Abha", "Dammam", "Taif"}; for(int i = 0; i < cityNames.length; i++) myList.addToHead(cityNames[i]); System.out.println("The list is: " + myList); System.out.println("It is " + myLis.

Solve using Java programming language- ----------------------------.pdf

Solve using Java programming language- ----------------------------.pdf

Solve using Java programming language- ----------------------------.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 .

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 pro.pdf

hi i have to write a java program involving link lists. i have a pro.pdf

archgeetsenterprises

Write the Java source code necessary to build a solution for the problem below: Create a MyLinkedList class. Create methods in the class to add an item to the head, tail, or middle of a linked list; remove an item from the head, tail, or middle of a linked list; check the size of the list; and search for an element in the list. Create a test class to use the newly created MyLinkedList class. Add the following names in to the list: James, John, Michael, Peter, Allison, Daniel, George, Simon, Jason, and Mark. Your program should allow the user to enter a name from the console, and then search to see if the name exists in the list. Solution Hi, Please find my implementation. Please let me know in case of any issue. public class MyLinkedList { static class Node{ String data; Node next; Node(String d){ data = d; next =null; } } private Node head; public MyLinkedList() { head = null; } public void append(String data){ if(head == null) head = new Node(data); else{ Node temp = head; while(temp.next != null) temp = temp.next; temp.next = new Node(data); } } public void addFront(String data){ Node newNode = new Node(data); newNode.next = head; head = newNode; } public void removeFromLast(){ if(head == null || head.next == null) head = null; else{ Node temp = head; while(temp.next.next != null) temp = temp.next; temp.next = null; // removing last element } } public void removeFirst(){ if(head == null) return; else head = head.next; } public boolean search(String data){ Node temp = head; while(temp != null){ if(temp.data.equals(data)) return true; temp = temp.next; } return false; } public void addAtMiddle(String data){ Node slow = head; Node fast = head; Node prev = null; Node newNode = new Node(data); if(head == null || head.next == null){ newNode.next = head; head = newNode; return; } while(fast != null && fast.next != null){ fast = fast.next.next; prev = slow; slow = slow.next; } if(prev != null) prev.next =newNode; newNode.next = slow; } public void removeAtMiddle(){ Node slow = head; Node fast = head; Node prev = null; if(head == null || head.next == null){ head = null; return; } while(fast != null && fast.next != null){ fast = fast.next.next; prev = slow; slow = slow.next; } prev.next = slow.next; } public void display(){ Node temp = head; while(temp != null){ System.out.print(temp.data+\" \"); temp =temp.next; } System.out.println(); } } public class MyLinkedListTest { public static void main(String[] args) { MyLinkedList list = new MyLinkedList(); list.addAtMiddle(\"James\"); list.display(); list.addAtMiddle(\"John\"); list.display(); list.addAtMiddle(\"Michael\"); list.display(); list.addAtMiddle(\"Peter\"); list.display(); list.addAtMiddle(\"Allison\"); list.display(); list.addFront(\"Daniel\"); list.display(); list.append(\"George\"); list.display(); list.addFront(\"Simon\"); list.display(); list.append(\"Jason\"); list.display(); list.addAtMiddle(\"Mark\"); list.display(); list.removeAtMiddle(); list.display(); list.removeFirst(); list.displ.

Write the Java source code necessary to build a solution for the pro.pdf

Write the Java source code necessary to build a solution for the pro.pdf

Write the Java source code necessary to build a solution for the pro.pdf

public class SLL { protected SLLNode head, tail; public SLL() { head = tail = null; } public void addToTail(T e){ SLLNode newNode = new SLLNode(e); if(head == null) head = tail = newNode; else{ tail.next = newNode; tail = tail.next; } } public void addToHead(T e) { SLLNode newNode = new SLLNode(e, head); if(head == null) tail = newNode; head = newNode; } public String toString(){ if (head == null) return "[]"; String out = "["; SLLNode p = head; while (p.next != null){ out += p.info + " "; p = p.next; } out += p.info + "]"; return out; } } public class SLLNode { protected T info; protected SLLNode next; public SLLNode() { this(null, null); } public SLLNode(T e) { this(e, null); } public SLLNode(T e, SLLNode ptr) { info = e; next = ptr; } } public class SortedSLL> extends SLL{ public void insert(T e){ } public void merge(SortedSLL list){ } } public class SortedSLL T extends Comparable T extends SLLT{ public void insert (T e) \{ \} public void merge(SortedSLL T list) \{ \} public class SortedSLL T extends Comparable T extends SLLT{ public void insert ( T e) \{ \} public void merge(SortedSLL T list) \{ \} \} ublic class SLLT{ protected SLLNodeT head, tail; public SLL() \{ head = tail = null; \} public void addToTail( T e) \{ SLLNode newNode = new SLLNodeT(e); if ( head == null) head = tail = newNode; else\{ tail. next = newNode; tail = tail. next; \} \} public void addToHead( T e) \{ SLLNode T newNode = new SLLNodeT(e, head ); if (head == null) tail = newNode; head = newNode; \} public string tostring()\{ if (head == null) return "[]"; String out = "["; SLLNode Tp= head; while (p.next != null) \{ out +=p.info +" "; p=p next; \} out += p.info + "]"; return out; \} ublic class SLLNode T{ protected T info; protected SLLNode T next; public SLLNode() \{ this(null, null); \} public SLLNode( T e) \{ this (e, null); \} public SLLNode( T e, SLLNode T ptr ){ info =e; next =ptr; \} Write a subclass of SLL named SortedSLL that represents a sorted SinglyLinkedList (sorted in nondecending order). The class should contain the following methods: 1- [40 Points] void insert ( T e) that inserts an SLLNode in its correct position so that the resulting list remains sorted. Make sure to consider special cases such as: - The list is empty. - The list has one element. - The inserted element is smaller than the head, or larger than the tail.. 2- [60 Points] void merge(SortedSLL T list) that merges the calling list with the argument list. Note that this can be done by calling the insert method for each element in list, but this is highly ineffiecient, answer the following question: Q2.1: What is the complexity of this approach? Explain. [10 Points] Instead, you should come up with a solution that traverse both lists just one time. Furthermore, the argument list should be empty after the merge is completed. Q2.2: What is the complexity of this approach? Explain. [10 Points] Hint: Make sure the class SortedSLL accepts types T that is comparable. You can do this by writing .

public class SLLT { protected SLLNodeT head, tail; pub.pdf

public class SLLT { protected SLLNodeT head, tail; pub.pdf

public class SLLT { protected SLLNodeT head, tail; pub.pdf

I am Gabriel C. I am a C Exam Expert at programmingexamhelp.com. I hold a PhD. in Business analyst of Information Technology, Montreal College of Information Technology, Canada. I have been helping students with their exams for the past 8 years. You can hire me to take your exam in C. Visit programmingexamhelp.com or email support@programmingexamhelp.com. You can also call on +1 678 648 4277 for any assistance with the C Exam.

C Exam Help

Programming Exam Help

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 .

Please need help on following program using c++ language. Please inc.pdf

Please need help on following program using c++ language. Please inc.pdf

Please need help on following program using c++ language. Please inc.pdf

Doublylinklist

DSA(1).pptx

Lab-2.2 717822E504.pdf

Lab-2.2 717822E504.pdf

Lab-2.2 717822E504.pdf

21E135MAHIESHWARJ

I am Andrew O. I am a C Homework Expert at programminghomeworkhelp.com. I hold a master’s in Programming from, the University of Southampton, UK. I have been helping students with their homework for the past 10 years. I solve homework related to C. Visit programminghomeworkhelp.com or email support@programminghomeworkhelp.com. You can also call on +1 678 648 4277 for any assistance with C Homework.

C Homework Help

C Homework Help

C Homework Help

Programming Homework Help

Lab-2.4 101.pdf

Lab-2.4 101.pdf

Lab-2.4 101.pdf

21E135MAHIESHWARJ

Binary Tree

#include struct node { char value; struct node *next; }; class StringOfNode { struct node * head; public: StringOfNode() { head=NULL; } int size() { struct node *p=head; int count=1; while(p) { count++; p=p->next; } return count; } struct node * begin() { return head; } char &operator[](int index) { int i=0; struct node *p=head; while(index!=i) { p=p->next; i++; } return p->value; } void insertFront(char valuee) { struct node * temp=(struct node *)malloc(sizeof(struct node)); temp->value=valuee; temp->next=NULL; if(head==NULL) head=temp; else { temp->next=head; head=temp; } } void insert(char valuee, int pos) { struct node *p=head; struct node * temp=(struct node *)malloc(sizeof(struct node)); temp->value=valuee; temp->next=NULL; int i=0; if(p==NULL) { head=temp; } else { while(i!=pos-1) { p=p->next; i++; } temp->next=p->next; p->next=temp; } } void insertBack(char valuee) { struct node *p=head; struct node * temp=(struct node *)malloc(sizeof(struct node)); temp->value=valuee; temp->next=NULL; if(p==NULL) { head=temp; } else { while(p->next!=NULL) p=p->next; p->next=temp; } } void print() { struct node *p=head; while(p) { std::cout<value<<\"\ \"; p=p->next; } } StringOfNode operator+(const StringOfNode& b) { StringOfNode *S; struct node *p=head; while(p->next!=NULL) p=p->next; p->next=S->begin(); } }; int main() { StringOfNode *p1=new StringOfNode; p1->insertBack(\'b\'); p1->insertBack(\'c\'); p1->insertBack(\'e\'); p1->insert(\'d\',2); p1->insertFront(\'a\'); p1->print(); system(\"pause\"); return 0; } Solution #include struct node { char value; struct node *next; }; class StringOfNode { struct node * head; public: StringOfNode() { head=NULL; } int size() { struct node *p=head; int count=1; while(p) { count++; p=p->next; } return count; } struct node * begin() { return head; } char &operator[](int index) { int i=0; struct node *p=head; while(index!=i) { p=p->next; i++; } return p->value; } void insertFront(char valuee) { struct node * temp=(struct node *)malloc(sizeof(struct node)); temp->value=valuee; temp->next=NULL; if(head==NULL) head=temp; else { temp->next=head; head=temp; } } void insert(char valuee, int pos) { struct node *p=head; struct node * temp=(struct node *)malloc(sizeof(struct node)); temp->value=valuee; temp->next=NULL; int i=0; if(p==NULL) { head=temp; } else { while(i!=pos-1) { p=p->next; i++; } temp->next=p->next; p->next=temp; } } void insertBack(char valuee) { struct node *p=head; struct node * temp=(struct node *)malloc(sizeof(struct node)); temp->value=valuee; temp->next=NULL; if(p==NULL) { head=temp; } else { while(p->next!=NULL) p=p->next; p->next=temp; } } void print() { struct node *p=head; while(p) { std::cout<value<<\"\ \"; p=p->next; } } StringOfNode operator+(const StringOfNode& b) { StringOfNode *S; struct node *p=head; while(p->next!=NULL) p=p->next; p->next=S->begin(); } }; int main() { StringOfNode *p1=new StringOfNode; p1->insertBack(\'b\'); p1->insertBack(\'c\'); p1->insertBack(\'e\'); p1->insert(\'d\',2.

#includeiostream struct node { char value; struct no.pdf

#includeiostream struct node { char value; struct no.pdf

#includeiostream struct node { char value; struct no.pdf

ankitmobileshop235

C# code please Write a program that creates a link list object of 10 characters, then creates a second list object containing a copy of the first list but in reverse order. Solution using System; class LinkedList { private char data; private LinkedList next; public LinkedList() { data = \'c\'; next = null; } public LinkedList(char value) { data = value; next = null; } public LinkedList Append(char value) { LinkedList node = new LinkedList(value); if(this.next == null) { // Easy to handle node.next = null; // already set in constructor this.next = node; } else { // Insert in the middle LinkedList temp = this.next; node.next = temp; this.next = node; } return node; } public void Display(LinkedList node) { if(node == null) node = this; System.Console.WriteLine(\"\ \ LinkedList:\ \ \"); while(node != null) { System.Console.Write(node.data+\" -> \"); node = node.next; } System.Console.Write(\"\ \"); } public LinkedList Reverse_with_copy(LinkedList head) { LinkedList p = head; LinkedList newhead = null; while(true) { if(p == null) return newhead; LinkedList newnode = new LinkedList(p.data); newnode.next = newhead; newhead = newnode; p = p.next; } } } class Program { static void Main(string[] args) { LinkedList head = new LinkedList(\'a\'); LinkedList p = head.Append(\'c\'); p = p.Append(\'e\'); p = p.Append(\'h\'); p = p.Append(\'b\'); p = p.Append(\'d\'); p = p.Append(\'f\'); p = p.Append(\'g\'); head.Display(null); LinkedList newhead = head.Reverse_with_copy(head); newhead.Display(null); } } /* sample output */.

C# code pleaseWrite a program that creates a link list object of 1.pdf

C# code pleaseWrite a program that creates a link list object of 1.pdf

C# code pleaseWrite a program that creates a link list object of 1.pdf

THE CODE HAS A SEGMENTATION FAULT BUT I CANNOT FIND OUT WHERE. NEED HELP FIXING THIS CODE? THIS IS THE PROBLEM Write a program that allows the user to enter two positive integers and outputs their sum. Sounds simple? Here\'s the catch: The numbers may be any size! They may be way too large to store in a single variable. You may not assume any particular maximum number of digits. HERE IS THE CODE #include #include struct node /* Linked list node*/ { int data; struct node* next; }; struct node *newNode(int data) /* Function to create a new node with giving the data*/ { struct node *new_node = (struct node *) malloc(sizeof(struct node)); new_node->data = data; new_node->next = NULL; return new_node; } void push(struct node** head_ref, int new_data) /* Function to insert a node at the beginning of the Doubly Linked List*/ { struct node* new_node = newNode(new_data); /*allocate a node*/ new_node->next = (*head_ref); /* link the old list off to the new node*/ (*head_ref) = new_node; /*move the head to point to the new node*/ } /* Adding the contents of two linked lists and return the head node of resultant list*/ struct node* addTwoLists (struct node* first, struct node* second) { struct node* res = NULL; /* res is head node of the resultant list*/ struct node *temp, *prev = NULL; int carry = 0, sum; while (first != NULL || second != NULL) /*while both the lists exists*/ { /* Calculate value of next digit in resultant list. // The next digit is sum of following things // (i) Carry // (ii) Next digit of first list (if there is a next digit) // (ii) Next digit of second list (if there is a next digit)*/ sum = carry + (first? first->data: 0) + (second? second->data: 0); carry = (sum >= 10)? 1 : 0; /* update carry for next calulation*/ sum = sum % 10; /*update sum if it is greater than 10*/ temp = newNode(sum); /*Create a new node with sum as data*/ if(res == NULL) /*if this is the first node then set it as head of the resultant list*/ res = temp; else /* If this is not the first node then connect it to the rest.*/ prev->next = temp; prev = temp; /* Set prev for next insertion*/ if (first) first = first->next; /* Move first pointers to next node*/ if (second) second = second->next; /*Move second pointer to the next node*/ } if (carry > 0) temp->next = newNode(carry); return res; /* return head of the resultant list*/ } void printList(struct node *node) /* function to print a linked list*/ { while(node != NULL) { printf(\"%d \", node->data); node = node->next; } printf(\"\ \"); } int main(void) /* program to test above function*/ { struct node* res = NULL; struct node* first = NULL; struct node* second = NULL; printf(\"Enter first positive integer: \"); scanf(\"%d\",&first); printList(first); printf(\"Enter second positive integer: \"); scanf(\"%d\",&second); printList(second); /*Add the two lists and see result*/ res = addTwoLists(first, second); printf(\"Sum is: %d\"); printList(res); return 0; } Solution #include #include typedef struct Node { int da.

THE CODE HAS A SEGMENTATION FAULT BUT I CANNOT FIND OUT WHERE. NEED .pdf

THE CODE HAS A SEGMENTATION FAULT BUT I CANNOT FIND OUT WHERE. NEED .pdf

THE CODE HAS A SEGMENTATION FAULT BUT I CANNOT FIND OUT WHERE. NEED .pdf

fathimahardwareelect

I want help in the following C++ programming task. Please do coding in C++. Persistent Data Structures A persistent data structure is one in which we can make changes (such as inserting and removing elements) while still preserving the original data structure Of course, an easy way to do this is to create a new copy of the entire data structure every time we perform an operation that changes the data structure, but it is much more efficient to only make a copy of the smallest possible portion of the data structure. First, let\'s look at one of the easiest possible data structures — a stack, implemented using a linked list. Persistent Stack Let\'s say that we have a stack S1, and we want to create a new stack S2, which is the result of pushing a new element X onto S1, without changing S1 at all. This is easy — we just add a new element that points to S1: Starting with the stack S1: We push X on S1, to get S2, leaving S1 as it was. So after the operation we have two versions of the stack S1 the older version with elements A, B and C and S2 the newer version with elements X, A, B and C. What about popping? That works in the same way. Starting with our original S1 (elements A, B and C), we can pop off the first element (A), to get S2: What if we want to push something on to this new stack? It works in the same was as before – if we push an X onto S2 above to get S3, we have So for every operation you make a new version of the stack while reusing any existing stack elements. You will have to make a linked list of pointers that will point to the head of the different stack versions. Also each stack is also implemented as a linked list of elements SAB Solution #include #include #include #include #include #include #include #include #include #include #include namespace dts { template > class PersistentSet { public: PersistentSet(); PersistentSet(Func); bool add(const T&); bool add(T&&); bool remove(const T& key); bool empty() const; size_t history_size() const; class TreeIterator : public std::iterator, std::ptrdiff_t, const T*, const T&> { using node = typename dts::PersistentSet< std::remove_cv_t, Func>::Nodeptr; node itr; node nil; std::stack path; node find_successor(node n) { n = n->rigth; if (n != nil) { while (n->left != nil) { path.push(n); n = n->left; } } else { n = path.top(); path.pop(); } return n; } public: explicit TreeIterator(node n, node pnil) : nil(pnil) //begin { if (n == nil) itr = nil; else { path.push(nil); while (n->left != nil) { path.push(n); n = n->left; } itr = n; } } explicit TreeIterator(node pnil) // end : itr(pnil), nil(pnil) { } TreeIterator& operator++ () { itr = find_successor(itr); return *this; } TreeIterator operator++ (int) { TreeIterator tmp(*this); itr = find_successor(itr); return tmp; } bool operator == (const TreeIterator& rhs) const { return itr == rhs.itr; } bool operator != (const TreeIterator& rhs) const { return itr != rhs.itr; } const T& operator* () const { return itr- >key; } const T& operator-> () c.

I want help in the following C++ programming task. Please do coding .pdf

I want help in the following C++ programming task. Please do coding .pdf

I want help in the following C++ programming task. Please do coding .pdf

bermanbeancolungak45

coding in C; Create a function called reverseList that takes the head of a linked list, reverses the order of all the nodes. For example, if the list contained 1, 2, 3, 4 in its nodes, the list will now contain 4, 3, 2, 1. please add a main test. Solution Please find the required program below: #include<stdio.h> #include<stdlib.h> struct node { int data; struct node* next; }; void push(struct node** head, int new_element) { struct node* new_node = (struct node*) malloc(sizeof(struct node)); new_node->data = new_element; new_node->next = (*head); (*head) = new_node; } void printList(struct node *head) { struct node *temp = head; while(temp != NULL) { printf(\"%d \", temp->data); temp = temp->next; } } void reverseList(struct node** head) { struct node* prev = NULL; struct node* current = *head; struct node* next; while (current != NULL) { next = current->next; current->next = prev; prev = current; current = next; } *head = prev; } int main() { struct node* head = NULL; push(&head, 12); push(&head, 14); push(&head, 5); push(&head, 8); printf(\"\ Linked list Before Reversing :\ \"); printList(head); reverseList(&head); printf(\"\ Linked list After Reversing :\ \"); printList(head); } -------------------------------------------------------------------------------- OUTPUT: Linked list Before Reversing : 8Â Â 5Â Â 14Â Â 12 Linked list After Reversing : 12Â Â 14Â Â 5Â Â 8 .

coding in C- Create a function called reverseList that takes the head.docx

coding in C- Create a function called reverseList that takes the head.docx

coding in C- Create a function called reverseList that takes the head.docx

#include #include using namespace std; template class Node{ private: T value; Node *next; public: Node(); Node (T data); T remove(); T getValue(); Node* getNext(); void setValue(T data); void setNext(Node *nextNode); }; template Node::Node(){ T value =NULL; } template Node::Node(T data){ T value=data; } template T Node::getValue(){ return value; } template Node* Node::getNext(){ return next; } template void Node::setValue(T data){ T value=data; } template void Node::setNext(Node *nextNode){ next=nextNode; } template class MyLinkedList{ Node start; //Define a constructor MyLinkedList() { start->setValue(0); start->setNext(NULL); } //Member Functions: // Adds a new element to the list void add (T val) { /* creating a new node to hold value*/ Nodetemp(val,NULL); /* If no start node till now, this node is made as start node*/ if (start->next == NULL) { start.setNext(temp); } /* if start node exists, insert new node at end of list*/ else { s = start->next; /* locating end of list*/ while (s->next != NULL) { s= s->next; } temp->next = NULL; /* attaching new node */ s->next = temp; cout<<\"Element Inserted\"<next->next != NULL) { s = s->next; } //After locating the end node ptr=s->next; s->next=NULL; //deletting end node return ptr; } } //Prints elements of the list void printContent() { struct node *ptr = start; cout<data; ptr=ptr->next; } } } //Prints total number of elements int getSize() { struct node *ptr, *s; int count=0; if (start == NULL) { cout<<\"The List is empty\"<next; count=1; while (ptr != NULL) { count++; ptr=ptr->next; } return count; } //Returns true if there is no element in the list. bool empty() { if(start==NULL) { return empty; } } }; Solution #include #include using namespace std; template class Node{ private: T value; Node *next; public: Node(); Node (T data); T remove(); T getValue(); Node* getNext(); void setValue(T data); void setNext(Node *nextNode); }; template Node::Node(){ T value =NULL; } template Node::Node(T data){ T value=data; } template T Node::getValue(){ return value; } template Node* Node::getNext(){ return next; } template void Node::setValue(T data){ T value=data; } template void Node::setNext(Node *nextNode){ next=nextNode; } template class MyLinkedList{ Node start; //Define a constructor MyLinkedList() { start->setValue(0); start->setNext(NULL); } //Member Functions: // Adds a new element to the list void add (T val) { /* creating a new node to hold value*/ Nodetemp(val,NULL); /* If no start node till now, this node is made as start node*/ if (start->next == NULL) { start.setNext(temp); } /* if start node exists, insert new node at end of list*/ else { s = start->next; /* locating end of list*/ while (s->next != NULL) { s= s->next; } temp->next = NULL; /* attaching new node */ s->next = temp; cout<<\"Element Inserted\"<next->next != NULL) { s = s->next; } //After locating the end node ptr=s->next; s->next=NULL; //deletting end node return ptr; } } //Prints elements of the list void printContent() { struct .

#include iostream #include cstddefusing namespace std;temp.pdf

#include iostream #include cstddefusing namespace std;temp.pdf

#include iostream #include cstddefusing namespace std;temp.pdf

Similar to You can list anything, it doesnt matter. I just want to see code f.pdf (20)

In this lab, you will be given a simple code for a min Heap, and you.pdf

In this lab, you will be given a simple code for a min Heap, and you.pdf

In this lab, you will be given a simple code for a min Heap, and you.pdf

137 Lab-2.2.pdf

137 Lab-2.2.pdf

137 Lab-2.2.pdf

Solve using Java programming language- ----------------------------.pdf

Solve using Java programming language- ----------------------------.pdf

Solve using Java programming language- ----------------------------.pdf

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 pro.pdf

hi i have to write a java program involving link lists. i have a pro.pdf

Write the Java source code necessary to build a solution for the pro.pdf

Write the Java source code necessary to build a solution for the pro.pdf

Write the Java source code necessary to build a solution for the pro.pdf

public class SLLT { protected SLLNodeT head, tail; pub.pdf

public class SLLT { protected SLLNodeT head, tail; pub.pdf

public class SLLT { protected SLLNodeT head, tail; pub.pdf

C Exam Help

Please need help on following program using c++ language. Please inc.pdf

Please need help on following program using c++ language. Please inc.pdf

Please need help on following program using c++ language. Please inc.pdf

Doublylinklist

DSA(1).pptx

Lab-2.2 717822E504.pdf

Lab-2.2 717822E504.pdf

Lab-2.2 717822E504.pdf

C Homework Help

C Homework Help

C Homework Help

Lab-2.4 101.pdf

Lab-2.4 101.pdf

Lab-2.4 101.pdf

Binary Tree

#includeiostream struct node { char value; struct no.pdf

#includeiostream struct node { char value; struct no.pdf

#includeiostream struct node { char value; struct no.pdf

C# code pleaseWrite a program that creates a link list object of 1.pdf

C# code pleaseWrite a program that creates a link list object of 1.pdf

C# code pleaseWrite a program that creates a link list object of 1.pdf

THE CODE HAS A SEGMENTATION FAULT BUT I CANNOT FIND OUT WHERE. NEED .pdf

THE CODE HAS A SEGMENTATION FAULT BUT I CANNOT FIND OUT WHERE. NEED .pdf

THE CODE HAS A SEGMENTATION FAULT BUT I CANNOT FIND OUT WHERE. NEED .pdf

I want help in the following C++ programming task. Please do coding .pdf

I want help in the following C++ programming task. Please do coding .pdf

I want help in the following C++ programming task. Please do coding .pdf

coding in C- Create a function called reverseList that takes the head.docx

coding in C- Create a function called reverseList that takes the head.docx

coding in C- Create a function called reverseList that takes the head.docx

#include iostream #include cstddefusing namespace std;temp.pdf

#include iostream #include cstddefusing namespace std;temp.pdf

#include iostream #include cstddefusing namespace std;temp.pdf

More from fashionbigchennai

Write a Java program that performs these operationsPrompt the use.pdf

Write a Java program that performs these operationsPrompt the use.pdf

Write a Java program that performs these operationsPrompt the use.pdf

fashionbigchennai

Write a complete PUP enabled website using WebMatrix. At the beginning, choose the template of empty PHP website or empty HTML website. In the major PUP file of this site, i.e. index.php, inside PHP script block, write PHP code to print out first 10 prime numbers inclusively starting with 2. Print these numbers in one line and separate them with two spaces. You MUST use PHP output statement, flow control statements for selection and loop control structures in PHP. Solution Need to create two files named template.php and index.php template.php PHP include Website Template index.php.

Write a complete PUP enabled website using WebMatrix. At the beginnin.pdf

Write a complete PUP enabled website using WebMatrix. At the beginnin.pdf

Write a complete PUP enabled website using WebMatrix. At the beginnin.pdf

fashionbigchennai

Which of the following is a flaw, loophole, oversight, or error that makes an organization susceptible to attack or damage? A. Risk B. Vulnerability C. Threat D. Exploit Solution B. Vulnerability is a flaw, loophole, oversight, or error that makes the organization susceptible to attack or damage. All other answers are incorrect: A risk can be defined as the potential harm that can arise from some present process or from some future event; a threat is an unwanted event that can result in harm to an asset or service; and an exploit takes advantage of a bug, glitch, or vulnerability..

Which of the following is a flaw, loophole, oversight, or error that.pdf

Which of the following is a flaw, loophole, oversight, or error that.pdf

Which of the following is a flaw, loophole, oversight, or error that.pdf

fashionbigchennai

what trade agreements were initiated by the United States and Canada.pdf

what trade agreements were initiated by the United States and Canada.pdf

what trade agreements were initiated by the United States and Canada.pdf

fashionbigchennai

Use the series and parallel commands in Matlab to find the transfer functions of 1. G(s) in series/parallel with H(s) 2. H(s) in series/parallel with T(s) G(s) = (s+1)/(s^2+4s+10) H(s) =(s ^2+5s+1)/(s^4+s^3+3s+10) T(s) = (s+1)/(s+1)(s+2)(s+3)(s+4) Solution MATLAB CODE G = tf([1,1],[1 3 0]) A.) % SERIES CONNECTION S = G * H % PARALLEL CONNECTION P = G + H B.) % SERIES CONNECTION S1= H * T % PARALLEL CONNECTION P1= H + T.

Use the series and parallel commands in Matlab to find the transfer .pdf

Use the series and parallel commands in Matlab to find the transfer .pdf

Use the series and parallel commands in Matlab to find the transfer .pdf

fashionbigchennai

True or False Even if there is a strong correlation between the ind.pdf

True or False Even if there is a strong correlation between the ind.pdf

True or False Even if there is a strong correlation between the ind.pdf

fashionbigchennai

This is my code but not complete, please complete my code related by lab#4 And i wans results for output in black screen. #include #include #define MAX 100; using namespace std; struct date { int day,month,year\' }; struct STUDENT_TYPE { string name; int id; double GPA; date DOB; }; int i=0; int main() { STUDENT_TYPE student[MAX]; int choice=-1; while(choice!=0) { cout<<\"\ 0-Exit\ 1-Insert\ 2-Delete_last\ 3-Display\ 4-Search\ 5-Deletion_particular_record\ 6- Insert_after_record\ 7-Sort(acending)\"; cout<<\"Enter your choice:\"; cin>>choice; switch(choice) { case 0exit(0); case 1:insert();break; case 2:delete();break; case 3:display();break; case 4:searcht();break; case 5:delete();break; case 6:insert();break; case 7:sort();break; default:cout<<\"Wrong selection\"; } } return 0; } void insert() { cout<<\"Enter the student details:\"; cout<<\"student name:\"; cin>>student[i].name; cout<<\"student id:\"; cin>>student[i].id; cout<<\"student GPA:\"; cin>>student[i].GPA; cout<<\"student DOB\"; cin>>student[i].DOB.day>>student[i].DOB.month>>student[i].DOB.year; i++; } void delete() { //assign 0 to last pointed record student[i].name=0; student[i].id=0; student[i].GPA=0; student[i].DOB.day=0;student[i].DOB.month=0;student[i].DOB.year=0; i--; } void display () { cout<<\"student details are\ \"; for(int n=0; n>id; for(int n=0;n>record; for(int n=0;n>id; for(int n=0;npos;p--) { // move all record towards right to make space for insertion of new record student[p].name=student[p-1].name; student[p].id=student[p-1].id; student[p].GPA=student[p-1].GPA; student[p].DOB.day=student[p-1].DOB.day; student[p].DOB.month=student[p-1].DOB.month; student[p].DOB.year=student[p-1].DOB.year; } cout<<\"enter the student details:\"; cout<<\"student name:\"; cin>>student[pos].name; cout<<\"student id:\"; cin>>student[pos].id; cout<<\"student GPA:\"; cin>>student[pos].GPA; cout<<\"student DOB:\"; cin>>student[pos].DOB.day>>student[pos].DOB.month>>student[pos].DOB.year; } void sort() { // bubble sorting technique is used struct STUDENT_TYPE TEMP[2]; for(intx=0;xstudent[y+1]id) { //move all the record information to temp and swap them temp[0].name=student[j].name; temp[j].name=student[j+1].name; temp[j+1].name=student[0].name; //the above three lines are repeated for all the student fields } } } } #include #include #define MAX 100; using namespace std; struct date { int day,month,year\' }; struct STUDENT_TYPE { string name; int id; double GPA; date DOB; }; int i=0; int main() { STUDENT_TYPE student[MAX]; int choice=-1; while(choice!=0) { cout<<\"\ 0-Exit\ 1-Insert\ 2-Delete_last\ 3-Display\ 4-Search\ 5-Deletion_particular_record\ 6- Insert_after_record\ 7-Sort(acending)\"; cout<<\"Enter your choice:\"; cin>>choice; switch(choice) { case 0exit(0); case 1:insert();break; case 2:delete();break; case 3:display();break; case 4:searcht();break; case 5:delete();break; case 6:insert();break; case 7:sort();break; default:cout<<\"Wrong selection\"; } } return 0; } void insert() { cout.

This is my code but not complete, please complete my.pdf

This is my code but not complete, please complete my.pdf

This is my code but not complete, please complete my.pdf

fashionbigchennai

The terminology employed in factory automation is often confusing in part because the names are so similar and in part because the sub areas do indeed overlap. Carefully distinguish among CIM , CAD , CAE , CAM and CAPP indicating any overlaps. Solution CIM(computer integrated manufacturing): It is the approach of using computers for controling the entire production process. This integration allows individual processes to exchange information with each other and initiate actions among them. It simply means the fully use of computer controlled machines and fully automation system in manufacturing products. CAD(computer aided design): It is a software and is used by architects, engineers or artists for the creation of precision drawing for technical overview. it is also helpful in the creation of 2D or 3D models. CAE(computer aided engineering): Is a process of solving problems related to engineering through the help of interactive graphical software.It can be graphical or mathematical or modeling. CAM(computer aided manufacturing): is the software for the controlling of maching tools or to assist all the process for the manufacturing of the machine. CIPP(computer aided process planning): Is the use of technology for the process and planning of a part or product in manufacturing. CAPP is the process in between CAD and CAM. It provides the planning for the process for producing a specific part. ///////////////////////////////////////////////////////////////// differences //////////////////////////////////////////////////////////////// The logic is that CAD/CAM provide the tools, CIM provides the internal logic and that is used while organizing the computers or the programs and the information that flows among them. Computer-aided manufacturing moto is to use the computer software to control machine tools and related machinery(or machine parts) in the manufacturing process. It is not technically or logically considered a system for engineering software programs, but rather than for hardware solution.But on the other side computer aided engineering involves the processes the software work, like the analysis or simulation part of the actual work done..

The terminology employed in factory automation is often confusing in.pdf

The terminology employed in factory automation is often confusing in.pdf

The terminology employed in factory automation is often confusing in.pdf

fashionbigchennai

The probability that A hits a target is 1/4 and the probability that B hits a target 1/3. (a) If each fires twice, what is the probability that the target will be hit at least once? (b) If each fire once and the target is hit only once, what is the probability that A hit the target? Solution probability that A hits a target = 1/4 probability that B hits a target = 1/3 a) after hitting twice, probability that the target will be hit at least once = 1 - probability that the target is not hit at all probability that the target is not hit at all = Probability that A does not hit the target the first time * Probability that B does not hit the target the first time * Probability that A does not hit the target the second time * Probability that B does not hit the target the second time = (1 - 1/4) * (1 - 1/3) * (1 - 1/4) * (1 - 1/3) = 3/4 * 2/3 * 3/4 * 2/3 = 1/4 after hitting twice, probability that the target will be hit at least once = 1 - 1/4 = 3/4 ------------------------------------------ b)If each fire once and the target is hit only once, the probability that A hit the target = probability that A hit the target * probability that B missed + probability that A hit the target * probability that B hit the target = 1/4 * (1 - 1/3) + 1/4 * 1/3 = 1/4 * 2/3 + 1/4 * 1/3 = 1/4.

The probability that A hits a target is 14 and the probability that.pdf

The probability that A hits a target is 14 and the probability that.pdf

The probability that A hits a target is 14 and the probability that.pdf

fashionbigchennai

Step 2: Formulate a Hypothesis Formulate a hypothesis by making an explicit statement about the cause of the common cold based on the observations and generalizations you made in Step 1. based on the 3 things below HypothesisObservation 1: Common cold is caused by bacterial or viral infection and transmitted from one person to other through droplets from infected person. The person infected with common cold show protect his nose or mouth with steril mask, which covers mouth and nose. If nose or mouth is not covered by mask, droplets can transmit common cold to many people in surrounding environment. Observation 2: Common cold is widely transmitted very easily, if a person infected with common cold and he sneezes in public places which is having high population, several thousands of droplets can be transported through air at onetime, when person sneezes, so avoid moving in public places when common cold is rampant in specific area. Observation 3: Doctors who examine the persons suffering with common cold should be careful, the physician should use antibiotics or vaccines to prevent common cold infection. Solution The hypothesis is this statement: Common cold is a viral infectious disease which can be transmistted from one person to another on sneezing.It can be prevented by taking antibiotics and vaccines by the physicians while check-up..

Step 2 Formulate a HypothesisFormulate a hypothesis by making an .pdf

Step 2 Formulate a HypothesisFormulate a hypothesis by making an .pdf

Step 2 Formulate a HypothesisFormulate a hypothesis by making an .pdf

fashionbigchennai

Since the directions of v and a are entirely independent of each other the derivative of the speed d|b|/dt can take any value between -|a| and + |a| including plusminus |a|. True False Solution False Though direction of change of velocity with respect to time is independent of velocity, The the magnitude of change of velocity with respect to time always represent accleration |a|, and its value can be +/-|a|.

Since the directions of v and a are entirely independent of each othe.pdf

Since the directions of v and a are entirely independent of each othe.pdf

Since the directions of v and a are entirely independent of each othe.pdf

fashionbigchennai

Reaction Statistics Background When collecting experimental data from chemical reactions, it’s often useful to generate statistics based on the data. One experimental measure is the reaction rate in moles per second, representing the amount of product formed per unit time. If we have a set of these reaction rates collected in a data file, we can calculate summary statistical information, such as the minimum and maximum values, the arithmetic mean, variance, and standard deviation. Finding the minimum and maximum are straightforward: we scan through all the data, and keep track of the smallest and largest values encountered. The arithmetic mean (or average) is defined as: m = (X1+X2+…+Xn)/n where n is the number of reaction rates, and xi represents one experimental reaction rate. Once you have the arithmetic mean, the variance can be calculated as the mean of the squares of the deviations from the mean: v= ((Xn-m)^2+(X2 – m^2) + …+(Xn-m)^2)/n where n is the number of reaction rates, xi represents one experimental reaction rate, and m is the arithmetic mean of the reaction rates. Once you have the variance, you can calculate the standarddeviation as: s = sqrt(v) Assignment You will develop a C program that reads data from an input text file containing chemical reaction rates (in moles per second), and computes the minimum, maximum, arithmetic mean, variance, and standard deviation for that set of data. Your instructor will provide input text files, which will each contain a series of double values, each on a line of its own within the file. Your program will read one of these input files into an array of doubles (i.e., it will populate the array using the data values from the file). Your program will then calculate statistics using that array of doubles, and will write the results out to a separate output text file. The goals of this assignment are to provide you with experience reading and writing text data files, provide you with experience passing an array into a function, and give you more experience organizing your program into separate C functions. When defining your C functions, you may either: Define the functions before they are used by any other functions, OR Place function prototypes near the top of your code (after all #include directives), and then define the functions in any order. Part 1 – Opening Files and Reading Data Create a new Visual Studio Win32 Console project named reactionstats. Create a new C source file named project4.c within that project. At the top of the source file, #define _CRT_SECURE_NO_WARNINGS, and then include stdio.h, math.h, stdlib.h, stdbool.h, and float.h. Inside your main function, define the following: A one-dimensional array of 600 doubles. They do not need to be initialized to anything at this stage. An integer variable to hold the number of elements in the array, initialized using the approach demonstrated in class, using sizeof. A FILE pointer variable, which will refer to the input data text file..

Reaction StatisticsBackgroundWhen collecting experimental data f.pdf

Reaction StatisticsBackgroundWhen collecting experimental data f.pdf

Reaction StatisticsBackgroundWhen collecting experimental data f.pdf

fashionbigchennai

Please explain in detaile the European debt crisis Solution In 2007, EU economies, on the surface, seemed to be doing relatively well – with positive economic growth and low inflation. Public debt was often high, but (apart from Greece) it appeared to be manageable assuming a positive trend in economic growth. However, the global credit crunch changed many things. EU Bond Yields Uncompetitiveness Eurozone countries with debt problems are also generally uncompetitive with a higher inflation rate and higher labour costs. This means there is less demand for their exports, higher current account deficit and lower economic growth. (The UK became uncompetitive, but being outside the Euro, the Pound could depreciate 20% restoring competitiveness. Poor Prospects for Growth People have been selling Greek and Italian bonds for two reasons. Firstly because of high structural debt, but also because of very poor prospects for growth. Countries facing debt crisis have to cut spending and implement austerity budgets. This causes lower growth, higher unemployment and lower tax revenues. However, they have nothing to stimulate economic growth. Individual Cases Greece’s Case In early 2010 these difficult developments reflected in rising spreads on sovereign bond yields between the affected peripheral member states of Greece, Ireland, Portugal and Spain, and most notably Germany. The Greek yield diverged in early 2010 with Greece needing eurozone assistance by May 2010. Greece received two bailouts from the EU over the following five years during which the country adopted EU-mandated austerity measures to cut costs while experiencing a further economic recession as well as social unrest. In June 2015 Greece, with divided political and fiscal leadership and a continued recession, was facing a sovereign default. However, on July 5, 2015 the Greek people voted against further EU austerity measures, with a possibility of Greece leaving the European Monetary Union entirely. The withdrawal of a nation from the EMU is unprecedented, and the speculated effects on Greece\'s economy if the currency is returned to the drachma range from total economic collapse to a surprise recovery. Ireland Case Ireland followed Greece in requiring a bailout in November 2010, with Portugal next in May 2011. Italy and Spain were also vulnerable, with Spain requiring official assistance in June 2012 along with Cyprus. By 2014, Ireland, Portugal and Spain, due to various fiscal reforms, domestic austerity measures and other unique economic factors, all successfully exited their bailout programs requiring no further assistance. The road to full economic recovery is still underway. Cyprus, too, reported a slow but steady ongoing recovery, averting further financial crisis thus far. Summary of Main Causes of Debt Crisis.

Please explain in detaile the European debt crisisSolutionIn 2.pdf

Please explain in detaile the European debt crisisSolutionIn 2.pdf

Please explain in detaile the European debt crisisSolutionIn 2.pdf

fashionbigchennai

Part 1: Many phone companies are replacing the wire in their phone systems with fiber optic cable. Name at least two advantages they expect to gain from doing so. In one paragraph each, describe why fiber optic cable gives each advantage. Part 2: Explain the difference between an amplifier and a repeater for transmitting data over distances. In your answer, tell why repeaters are preferred for digital data. Part 3: Explain why a hardware timer that can generate interrupts is necessary to support concurrent processing. Part 4: Explain why a graphical or window-based operating system interface is more readily amenable to running multiple tasks than is a command line interface. Part 5: What is the spectrum of a signal? What is the relationship between the spectrum of a signal and the bandwidth of a channel? (Use the textbook; you\'ll likely get a wrong answer from a web search!) Part 6: Name and briefly describe the three characteristics of a sine wave signal that can be modulated to allow the signal to carry information. Part 7: Explain how frequency division multiplexing can be used to carry two or more independent information streams on one communication medium. Part 8: A modern CPU has a bit that can place the CPU in the system state or the application state. Briefly describe the difference between \"system state\" and \"application state.\" Part 9: What does it mean to \"mount\" or \"unmount\" a storage device? Explain why unmounting a device is necessary in the Windows operating system. Part 10: Explain what it means to bootstrap a computer. In your answer, describe the steps executed by the hardware and software in bootstrapping a computer that is initially turned off. Solution Part 1: Due to much lower attenuation and interference, provides long distance, Fiber optic signals are stronger.These are much thinner and lighter than metal wires.Data can be transmitted digitally.It has less susceptible to interference. Part 2: An amplifier can not discriminate between the intended signal and noise,it amplifies equally everything fed into it.A repeater does not amplify the signal.It regenerates the signal.When it receives a weakened or corrupted signal,it creates a copy,bit for bit at the original strength. A repeater is a regenerator. Using the large number of amplifiers over a long, lossy link inevitablyincreases the noise and decreases the signal to noise levels.A repeater is usually a receiver followed by a regenerating logic and transmitter.In this case regeneration receives the digital signals and retimes it to remove jitter and then retransmits it with minimal baseband noise. Part 4: Graphical User Interface uses graphics along with a keyboard and mouse,to provide an easy to use interface to a program.It provides windows,pulldown menus,buttons,scrollbars,iconic images,wizards and other icons and the mouse to enable the users to interact with the operating system. users easy to control,manipulate and toggle through multiple programs and fo.

Part 1 Many phone companies are replacing the wire in their phone s.pdf

Part 1 Many phone companies are replacing the wire in their phone s.pdf

Part 1 Many phone companies are replacing the wire in their phone s.pdf

fashionbigchennai

Mary is an excellent waitress. In the restaurant, she can easily handle 8 tables at a time without getting overloaded/overworked. During a normal day the restaurant has an an average of 4 tables to serve at a time. What is the probability Mary will be overworked in a given day? (Hint: Find the probability there are more than 8 tables occupied) Solution This is a Poisson distribution with random variable X= no of tables Mary handles at a time having a Poisson distribution with mena value = 4 P(X>8) = 0.0214 Prob for Mary overworking is 0.0214.

Mary is an excellent waitress. In the restaurant, she can easily hand.pdf

Mary is an excellent waitress. In the restaurant, she can easily hand.pdf

Mary is an excellent waitress. In the restaurant, she can easily hand.pdf

fashionbigchennai

List three characteristics that influence the rate of simple diffusion of small molecules. Solution Answer= Three characteristics that influence the rate of simple diffusion of small molecules 1] Particle size = Rate of diffusion is inversaly proportional to particle size Since the heat of environment is the energy source for diffusion the smaller particles moves at a faster rate than larger paricles at a given temperature 2] temperature = Rate of diffusion is directly proportional to temperature As temperature increses rate of diffusion also increases At high temp. particles move faster because more energy is available to diffuse them 3] Concentration difference = rate of increses as conc. difference increases The substance diffusing between 2 areas exibits a conc. diffference as the particle diffuse from one side of the wall to other side 4] particle distance = rate of diffusion is directly proportional to particle distance it takes longer for a particle to diffuse when it is further away than it does for a particle that is closer to the point of diffusion.

List three characteristics that influence the rate of simple diffusi.pdf

List three characteristics that influence the rate of simple diffusi.pdf

List three characteristics that influence the rate of simple diffusi.pdf

fashionbigchennai

It\'s all worked out, but I\'m still very confused by the steps they took to get each part. Like why multiply everything by (x^2 + 9) and then in the next step, how did they end up with those answers? I\'m just very confused. Solution they just multiplide (x^2 + 9) for simpification LIKE EXAMPLE eq 2+3=5 if we multify both sides with 2 (2+3)2=5*2 10=10 equality not changed ,,,like we use this in mathematics after solving eq they got X parameters and they cheked in original eq.

Its all worked out, but Im still very confused by the steps .pdf

Its all worked out, but Im still very confused by the steps .pdf

Its all worked out, but Im still very confused by the steps .pdf

fashionbigchennai

Is the world overpopulated? What are the arguments for the notion that the world is overpopulated? What are some alternative approaches to this topic? Solution The human popualtion has been growing ever since the evolution of mankind. Overpopulation is a condition when the population exceeds the carrying capacity of the environment. There is continuous increase in the birth rate and a significant decrease in the mortality rate. This results in increase in population at an alarming rate. The population of the world is predicted to keep on rising 2050. The sources and resources are being exhausted. This would lead to environment\'s inability to support life, reduced quality of life..

Is the world overpopulated What are the arguments for the notion th.pdf

Is the world overpopulated What are the arguments for the notion th.pdf

Is the world overpopulated What are the arguments for the notion th.pdf

fashionbigchennai

In the ellipse shown below, each point marked with a dot is called a(.pdf

In the ellipse shown below, each point marked with a dot is called a(.pdf

In the ellipse shown below, each point marked with a dot is called a(.pdf

fashionbigchennai

Identify the following: Level of caudal end of subarachnoid space Epidural, subarachnoid spaces Level of caudal end of spinal cord Solution Answer:- Level of caudal end of Subarachnoid space is lower border S-2 Vertebra. Epidural- Basically epidural space is define as the outermost part of the spinal canal.it is also called as the epidural cavity. this space lying outside th edura meter. Dura meter enclsoed by the subarachnoid space. it contains a lumphatics spinal nerve roots and network of internal vertebral venous plexuses.level is L1 & L2. Level of caudal end of spinal cord. - The Spinal cord can be defined as the long , thin tubular bundles of nervous tissue and it supports cells and it run in vertebral column. the caudal end of spianl cord is generaly cone shaped it is also called as the conus medullaris. and its ends at S2 level..

Identify the following Level of caudal end of subarachnoid space E.pdf

Identify the following Level of caudal end of subarachnoid space E.pdf

Identify the following Level of caudal end of subarachnoid space E.pdf

fashionbigchennai

More from fashionbigchennai (20)

Write a Java program that performs these operationsPrompt the use.pdf

Write a Java program that performs these operationsPrompt the use.pdf

Write a Java program that performs these operationsPrompt the use.pdf

Write a complete PUP enabled website using WebMatrix. At the beginnin.pdf

Write a complete PUP enabled website using WebMatrix. At the beginnin.pdf

Write a complete PUP enabled website using WebMatrix. At the beginnin.pdf

Which of the following is a flaw, loophole, oversight, or error that.pdf

Which of the following is a flaw, loophole, oversight, or error that.pdf

Which of the following is a flaw, loophole, oversight, or error that.pdf

what trade agreements were initiated by the United States and Canada.pdf

what trade agreements were initiated by the United States and Canada.pdf

what trade agreements were initiated by the United States and Canada.pdf

Use the series and parallel commands in Matlab to find the transfer .pdf

Use the series and parallel commands in Matlab to find the transfer .pdf

Use the series and parallel commands in Matlab to find the transfer .pdf

True or False Even if there is a strong correlation between the ind.pdf

True or False Even if there is a strong correlation between the ind.pdf

True or False Even if there is a strong correlation between the ind.pdf

This is my code but not complete, please complete my.pdf

This is my code but not complete, please complete my.pdf

This is my code but not complete, please complete my.pdf

The terminology employed in factory automation is often confusing in.pdf

The terminology employed in factory automation is often confusing in.pdf

The terminology employed in factory automation is often confusing in.pdf

The probability that A hits a target is 14 and the probability that.pdf

The probability that A hits a target is 14 and the probability that.pdf

The probability that A hits a target is 14 and the probability that.pdf

Step 2 Formulate a HypothesisFormulate a hypothesis by making an .pdf

Step 2 Formulate a HypothesisFormulate a hypothesis by making an .pdf

Step 2 Formulate a HypothesisFormulate a hypothesis by making an .pdf

Since the directions of v and a are entirely independent of each othe.pdf

Since the directions of v and a are entirely independent of each othe.pdf

Since the directions of v and a are entirely independent of each othe.pdf

Reaction StatisticsBackgroundWhen collecting experimental data f.pdf

Reaction StatisticsBackgroundWhen collecting experimental data f.pdf

Reaction StatisticsBackgroundWhen collecting experimental data f.pdf

Please explain in detaile the European debt crisisSolutionIn 2.pdf

Please explain in detaile the European debt crisisSolutionIn 2.pdf

Please explain in detaile the European debt crisisSolutionIn 2.pdf

Part 1 Many phone companies are replacing the wire in their phone s.pdf

Part 1 Many phone companies are replacing the wire in their phone s.pdf

Part 1 Many phone companies are replacing the wire in their phone s.pdf

Mary is an excellent waitress. In the restaurant, she can easily hand.pdf

Mary is an excellent waitress. In the restaurant, she can easily hand.pdf

Mary is an excellent waitress. In the restaurant, she can easily hand.pdf

List three characteristics that influence the rate of simple diffusi.pdf

List three characteristics that influence the rate of simple diffusi.pdf

List three characteristics that influence the rate of simple diffusi.pdf

Its all worked out, but Im still very confused by the steps .pdf

Its all worked out, but Im still very confused by the steps .pdf

Its all worked out, but Im still very confused by the steps .pdf

Is the world overpopulated What are the arguments for the notion th.pdf

Is the world overpopulated What are the arguments for the notion th.pdf

Is the world overpopulated What are the arguments for the notion th.pdf

In the ellipse shown below, each point marked with a dot is called a(.pdf

In the ellipse shown below, each point marked with a dot is called a(.pdf

In the ellipse shown below, each point marked with a dot is called a(.pdf

Identify the following Level of caudal end of subarachnoid space E.pdf

Identify the following Level of caudal end of subarachnoid space E.pdf

Identify the following Level of caudal end of subarachnoid space E.pdf

Recently uploaded

Python Notes for mca i year students osmania university.docx

Python Notes for mca i year students osmania university.docx

Python Notes for mca i year students osmania university.docx

Ramakrishna Reddy Bijjam

FICTIONAL SALESMAN/SALESMAN SNSW 2024.pdf

FICTIONAL SALESMAN/SALESMAN SNSW 2024.pdf

FICTIONAL SALESMAN/SALESMAN SNSW 2024.pdf

Pondicherry University

Graduate Outcomes Presentation Slides - English

Graduate Outcomes Presentation Slides - English

Graduate Outcomes Presentation Slides - English

On_Translating_a_Tamil_Poem_by_A_K_Ramanujan.pptx

On_Translating_a_Tamil_Poem_by_A_K_Ramanujan.pptx

On_Translating_a_Tamil_Poem_by_A_K_Ramanujan.pptx

TỔNG ÔN TẬP THI VÀO LỚP 10 MÔN TIẾNG ANH NĂM HỌC 2023 - 2024 CÓ ĐÁP ÁN (NGỮ Â...

TỔNG ÔN TẬP THI VÀO LỚP 10 MÔN TIẾNG ANH NĂM HỌC 2023 - 2024 CÓ ĐÁP ÁN (NGỮ Â...

TỔNG ÔN TẬP THI VÀO LỚP 10 MÔN TIẾNG ANH NĂM HỌC 2023 - 2024 CÓ ĐÁP ÁN (NGỮ Â...

Nguyen Thanh Tu Collection

80 ĐỀ THI THỬ TUYỂN SINH TIẾNG ANH VÀO 10 SỞ GD – ĐT THÀNH PHỐ HỒ CHÍ MINH NĂ...

80 ĐỀ THI THỬ TUYỂN SINH TIẾNG ANH VÀO 10 SỞ GD – ĐT THÀNH PHỐ HỒ CHÍ MINH NĂ...

80 ĐỀ THI THỬ TUYỂN SINH TIẾNG ANH VÀO 10 SỞ GD – ĐT THÀNH PHỐ HỒ CHÍ MINH NĂ...

Nguyen Thanh Tu Collection

This PowerPoint helps students to consider the concept of infinity.

This PowerPoint helps students to consider the concept of infinity.

This PowerPoint helps students to consider the concept of infinity.

christianmathematics

Philosophy of china and it's charactistics

Philosophy of china and it's charactistics

Philosophy of china and it's charactistics

FSB Advising Checklist - Orientation 2024

FSB Advising Checklist - Orientation 2024

FSB Advising Checklist - Orientation 2024

Elizabeth Walsh

Mehran University Newsletter Vol-X, Issue-I, 2024

Mehran University Newsletter Vol-X, Issue-I, 2024

Mehran University Newsletter Vol-X, Issue-I, 2024

Mehran University of Engineering & Technology, Jamshoro

Wizards are very useful for creating a good user experience. In all businesses, interactive sessions are most beneficial. To improve the user experience, wizards in Odoo provide an interactive session. For creating wizards, we can use transient models or abstract models. This gives features of a model class except the data storing. Transient and abstract models have permanent database persistence. For them, database tables are made, and the records in such tables are kept until they are specifically erased.

How to Create and Manage Wizard in Odoo 17

How to Create and Manage Wizard in Odoo 17

How to Create and Manage Wizard in Odoo 17

Spellings Wk 4 and Wk 5 for Grade 4 at CAPS

Spellings Wk 4 and Wk 5 for Grade 4 at CAPS

Spellings Wk 4 and Wk 5 for Grade 4 at CAPS

HMCS Max Bernays Pre-Deployment Brief (May 2024).pptx

HMCS Max Bernays Pre-Deployment Brief (May 2024).pptx

HMCS Max Bernays Pre-Deployment Brief (May 2024).pptx

The basics of sentences session 3pptx.pptx

The basics of sentences session 3pptx.pptx

The basics of sentences session 3pptx.pptx

NO1 Top Black Magic Specialist In Lahore Black magic In Pakistan Kala Ilam Ex...

NO1 Top Black Magic Specialist In Lahore Black magic In Pakistan Kala Ilam Ex...

NO1 Top Black Magic Specialist In Lahore Black magic In Pakistan Kala Ilam Ex...

On National Teacher Day, meet the 2024-25 Kenan Fellows

On National Teacher Day, meet the 2024-25 Kenan Fellows

On National Teacher Day, meet the 2024-25 Kenan Fellows

Simple, Complex, and Compound Sentences Exercises.pdf

Simple, Complex, and Compound Sentences Exercises.pdf

Simple, Complex, and Compound Sentences Exercises.pdf

stareducators107

Jamworks pilot and AI at Jisc (20/03/2024)

Jamworks pilot and AI at Jisc (20/03/2024)

Jamworks pilot and AI at Jisc (20/03/2024)

dusjagr & nano talk on open tools for agriculture research and learning

dusjagr & nano talk on open tools for agriculture research and learning

dusjagr & nano talk on open tools for agriculture research and learning

Marc Dusseiller Dusjagr

Beyond_Borders_Understanding_Anime_and_Manga_Fandom_A_Comprehensive_Audience_...

Beyond_Borders_Understanding_Anime_and_Manga_Fandom_A_Comprehensive_Audience_...

Beyond_Borders_Understanding_Anime_and_Manga_Fandom_A_Comprehensive_Audience_...

Recently uploaded (20)

Python Notes for mca i year students osmania university.docx

Python Notes for mca i year students osmania university.docx

Python Notes for mca i year students osmania university.docx

FICTIONAL SALESMAN/SALESMAN SNSW 2024.pdf

FICTIONAL SALESMAN/SALESMAN SNSW 2024.pdf

FICTIONAL SALESMAN/SALESMAN SNSW 2024.pdf

Graduate Outcomes Presentation Slides - English

Graduate Outcomes Presentation Slides - English

Graduate Outcomes Presentation Slides - English

On_Translating_a_Tamil_Poem_by_A_K_Ramanujan.pptx

On_Translating_a_Tamil_Poem_by_A_K_Ramanujan.pptx

On_Translating_a_Tamil_Poem_by_A_K_Ramanujan.pptx

TỔNG ÔN TẬP THI VÀO LỚP 10 MÔN TIẾNG ANH NĂM HỌC 2023 - 2024 CÓ ĐÁP ÁN (NGỮ Â...

TỔNG ÔN TẬP THI VÀO LỚP 10 MÔN TIẾNG ANH NĂM HỌC 2023 - 2024 CÓ ĐÁP ÁN (NGỮ Â...

TỔNG ÔN TẬP THI VÀO LỚP 10 MÔN TIẾNG ANH NĂM HỌC 2023 - 2024 CÓ ĐÁP ÁN (NGỮ Â...

80 ĐỀ THI THỬ TUYỂN SINH TIẾNG ANH VÀO 10 SỞ GD – ĐT THÀNH PHỐ HỒ CHÍ MINH NĂ...

80 ĐỀ THI THỬ TUYỂN SINH TIẾNG ANH VÀO 10 SỞ GD – ĐT THÀNH PHỐ HỒ CHÍ MINH NĂ...

80 ĐỀ THI THỬ TUYỂN SINH TIẾNG ANH VÀO 10 SỞ GD – ĐT THÀNH PHỐ HỒ CHÍ MINH NĂ...

This PowerPoint helps students to consider the concept of infinity.

This PowerPoint helps students to consider the concept of infinity.

This PowerPoint helps students to consider the concept of infinity.

Philosophy of china and it's charactistics

Philosophy of china and it's charactistics

Philosophy of china and it's charactistics

FSB Advising Checklist - Orientation 2024

FSB Advising Checklist - Orientation 2024

FSB Advising Checklist - Orientation 2024

Mehran University Newsletter Vol-X, Issue-I, 2024

Mehran University Newsletter Vol-X, Issue-I, 2024

Mehran University Newsletter Vol-X, Issue-I, 2024

How to Create and Manage Wizard in Odoo 17

How to Create and Manage Wizard in Odoo 17

How to Create and Manage Wizard in Odoo 17

Spellings Wk 4 and Wk 5 for Grade 4 at CAPS

Spellings Wk 4 and Wk 5 for Grade 4 at CAPS

Spellings Wk 4 and Wk 5 for Grade 4 at CAPS

HMCS Max Bernays Pre-Deployment Brief (May 2024).pptx

HMCS Max Bernays Pre-Deployment Brief (May 2024).pptx

HMCS Max Bernays Pre-Deployment Brief (May 2024).pptx

The basics of sentences session 3pptx.pptx

The basics of sentences session 3pptx.pptx

The basics of sentences session 3pptx.pptx

NO1 Top Black Magic Specialist In Lahore Black magic In Pakistan Kala Ilam Ex...

NO1 Top Black Magic Specialist In Lahore Black magic In Pakistan Kala Ilam Ex...

NO1 Top Black Magic Specialist In Lahore Black magic In Pakistan Kala Ilam Ex...

On National Teacher Day, meet the 2024-25 Kenan Fellows

On National Teacher Day, meet the 2024-25 Kenan Fellows

On National Teacher Day, meet the 2024-25 Kenan Fellows

Simple, Complex, and Compound Sentences Exercises.pdf

Simple, Complex, and Compound Sentences Exercises.pdf

Simple, Complex, and Compound Sentences Exercises.pdf

Jamworks pilot and AI at Jisc (20/03/2024)

Jamworks pilot and AI at Jisc (20/03/2024)

Jamworks pilot and AI at Jisc (20/03/2024)

dusjagr & nano talk on open tools for agriculture research and learning

dusjagr & nano talk on open tools for agriculture research and learning

dusjagr & nano talk on open tools for agriculture research and learning

Beyond_Borders_Understanding_Anime_and_Manga_Fandom_A_Comprehensive_Audience_...

Beyond_Borders_Understanding_Anime_and_Manga_Fandom_A_Comprehensive_Audience_...

Beyond_Borders_Understanding_Anime_and_Manga_Fandom_A_Comprehensive_Audience_...

You can list anything, it doesnt matter. I just want to see code f.pdf

1. You can list anything, it doesn't matter. I just want to see code for this problem. Thank you Solution Include header files ....... Source code: public class SinglyLinkedListImpl { private Node head; private Node tail; public void add(T element){ Node nd = new Node(); nd.setValue(element); System.out.println("Adding: "+element); if(head == null){ head = nd; tail = nd; } else { tail.setNextRef(nd); } tail = nd; } } public void addAfter(T element, T after){ Node tmp = head; Node refNode = null; System.out.println("Traversing to all nodes.."); while(true){ if(tmp == null){ break; } if(tmp.compareTo(after) == 0){ refNode = tmp;

2. break; } tmp = tmp.getNextRef(); } if(refNode != null){ Node nd = new Node(); nd.setValue(element); nd.setNextRef(tmp.getNextRef()); if(tmp == tail){ tail = nd; } tmp.setNextRef(nd); } else { System.out.println("Unable to find the given element..."); } } public void deleteFront(){ if(head == null){ System.out.println("Underflow..."); } Node tmp = head; head = tmp.getNextRef(); if(head == null){ tail = null; } System.out.println("Deleted: "+tmp.getValue()); } public void deleteAfter(T after){ Node tmp = head; Node refNode = null; System.out.println("Traversing to all nodes.."); while(true){ if(tmp == null){ break;

3. } if(tmp.compareTo(after) == 0){ refNode = tmp; break; } tmp = tmp.getNextRef(); } if(refNode != null){ tmp = refNode.getNextRef(); refNode.setNextRef(tmp.getNextRef()); if(refNode.getNextRef() == null){ tail = refNode; } System.out.println("Deleted: "+tmp.getValue()); } else { System.out.println("Unable to find the given element..."); } } public void traverse(){ Node tmp = head; while(true){ if(tmp == null){ break; } System.out.println(tmp.getValue()); tmp = tmp.getNextRef(); } } public static void main(String a[]){ SinglyLinkedListImpl sl = new SinglyLinkedListImpl(); sl.add(3); sl.add(32); sl.add(54); sl.add(89); sl.addAfter(76, 54);

4. sl.deleteFront(); sl.deleteAfter(76); sl.traverse(); } } class Node implements Comparable { private T value; private Node nextRef; public T getValue() { return value; } public void setValue(T value) { this.value = value; } public Node getNextRef() { return nextRef; } public void setNextRef(Node ref) { this.nextRef = ref; } public int compareTo(T arg) { if(arg == this.value){ return 0; } else { return 1; } } } This program implement the concept of single linked list.