Write an algorithm that reads a list of integers from the keyboard, .pdf

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Write an algorithm that reads a list of integers from the keyboard, creates a list of them using linked list implementation, and prints the result Solution /* Algorithm: Implementation of linked list. Create a node type class which will hold an integer and next pointer to point to another node. Create a constructor which will construct an object of type node with the integer type data as the only parameter and initializing next pointer to NULL. Create LinkedList class which will only have head pointer. Create a constructor which will initialize this head pointer to NULL. Create a append method which will recieve an input of type integer from the user and append it to LinkedList using below steps: 1.Create a Node type object. 1.IF head is NULL assign a Node type object created in 1st step. 2.IF head is not NULL then traverse to the end of the linked list till you found a Node which has next pointer set to NULL. 3.Add Node type object created in first step to the next pointer of last node. Create a print method which will traverse the list from the head of the linked list to the last node(Node with next pointer NULL) and while traversing it will keep printing the node data. Get the userinput: Ask user for integer continously till user input -1 and append all the integers to the linked list. */ PROGRAM: #include using namespace std; class Node { public: int data; Node *next; Node(int data) { this->data = data; this->next = NULL; } }; class LinkedList { private: Node *head; public: LinkedList() { head = NULL; } void addToFront(int data) { Node *temp = new Node(data); if(head == NULL) head = temp; else { temp->next = head; head = temp; } } void append(int data) { Node *temp = new Node(data); Node *current = head; if(current == NULL) { head = temp; return; } while(current->next != NULL) { current = current->next; } current->next = temp; } void print() { Node *current = head; if(current == NULL) cout << \"No element in list. \ \"; while(current != NULL) { cout << current->data << \" \"; current = current->next; } cout << endl; } void deleteLinkedList() { Node *prev; Node *current = head; while(current!=NULL) { prev = current; current = current->next; delete prev; } } void deleteLastNode() { Node *current = head; Node *prev = NULL; if(current == NULL) return; while(current->next != NULL) { prev = current; current = current->next; } if(prev == NULL) { delete current; head = NULL; } else { prev->next = NULL; delete current; } } ~LinkedList() { deleteLinkedList(); } }; int main() { LinkedList list; int data; while(1) { cout << \"Enter the integer(-1 to Quit) : \"; cin >> data; if(data==-1) break; list.append(data); } cout << \"Content of LinkedList is : \"; list.print(); return 0; } OUTPUT: Enter the integer(-1 to Quit) : 1 Enter the integer(-1 to Quit) : 2 Enter the integer(-1 to Quit) : 3 Enter the integer(-1 to Quit) : 4 Enter the integer(-1 to Quit) : -1 Content of LinkedList is : 1 2 3 4.

{
public:
int data;
Node *next;
Node(int data)
{
this->data = data;
this->next = NULL;
}
};
class LinkedList
{
private:
Node *head;
public:
LinkedList()
{
head = NULL;
}
void addToFront(int data)
{
Node *temp = new Node(data);
if(head == NULL)
head = temp;
else
{
temp->next = head;
head = temp;
}
}
void append(int data)
{
Node *temp = new Node(data);

Node *current = head;
if(current == NULL)
{
head = temp;
return;
}
while(current->next != NULL)
{
current = current->next;
}
current->next = temp;
}
void print()
{
Node *current = head;
if(current == NULL)
cout << "No element in list. ";
while(current != NULL)
{
cout << current->data << " ";
current = current->next;
}
cout << endl;
}
void deleteLinkedList()
{
Node *prev;
Node *current = head;

while(current!=NULL)
{
prev = current;
current = current->next;
delete prev;
}
}
void deleteLastNode()
{
Node *current = head;
Node *prev = NULL;
if(current == NULL)
return;
while(current->next != NULL)
{
prev = current;
current = current->next;
}
if(prev == NULL)
{
delete current;
head = NULL;
}
else
{
prev->next = NULL;
delete current;
}
}
~LinkedList()
{

deleteLinkedList();
}
};
int main()
{
LinkedList list;
int data;
while(1)
{
cout << "Enter the integer(-1 to Quit) : ";
cin >> data;
if(data==-1)
break;
list.append(data);
}
cout << "Content of LinkedList is : ";
list.print();
return 0;
}
OUTPUT:
Enter the integer(-1 to Quit) : 1
Enter the integer(-1 to Quit) : 2
Enter the integer(-1 to Quit) : 3
Enter the integer(-1 to Quit) : 4
Enter the integer(-1 to Quit) : -1
Content of LinkedList is : 1 2 3 4

This assignment and the next (#5) involve design and development of a sequential non contiguous and dynamic datastructure called LinkedList. A linked list object is a container consisting of connected ListNode objects. As before, we are not going to use pre-fabricated classes from the c++ library, but construct the LinkedList ADT from scratch. The first step is construction and testing of the ListNode class. A ListNode object contains a data field and a pointer to the next ListNode object (note the recursive definition). #This assignment requires you to 1. Read the Assignment 4 Notes 2. Watch the Assignment 4 Support video 3. Implement the following methods of the ListNode class -custom constructor -setters for next pointer and data 4. Implement the insert and remove method in the main program 5. Scan the given template to find the above //TODO and implement the code needed //TODO in ListNodecpp.h file public: ListNode(T idata, ListNode<T> * newNext); public: void setNext(ListNode<T> * newNext); public: void setData(T newData); // TODO in main program void remove(ListNode<int> * &front,int value) void insert(ListNode<int> * &front,int value) # The driver is given ListNodeMain.cpp is given to you that does the following tests 1. Declares a pointer called front to point to a ListNode of datatype integer 2. Constructs four ListNodes with data 1,2,4 and adds them to form a linked list. 3. Inserts ListNode with data 3 to the list 4. Removes node 1 and adds it back to test removing and adding the first element 5. Removes node 3 to test removing a middle node 6. Removes node 4 to test removing the last node 7. Attempt to remove a non existent node 8. Remove all existing nodes to empty the list 9. Insert node 4 and then node 1 to test if insertions preserve order 10.Print the list Main.cpp #include <iostream> #include "ListNodecpp.h" // REMEMBER each ListNode has two parts : a data field // and an address field. The address is either null or points to the next node //in the chain //Requires: integer value for searching, address of front //Effects: traverses the list node chain starting from front until the end comparing search value with listnode getData. Returns the original search value if found, if not adds +1 to indicate not found //Modifies: Nothing int search(ListNode<int> * front, int value); //Requires: integer value for inserting, address of front //Effects: creates a new ListNode with value and inserts in proper position (increasing order)in the chain. If chain is empty, adds to the beginning //Modifies: front, if node is added at the beginning. //Also changes the next pointer of the previous node to point to the //newly inserted list node. the next pointer of the newly inserted pointer //points to what was the next of the previous node. //This way both previous and current links are adjusted //******** NOTE the use of & in passing pointer to front as parameter - // Why do you think this is needed ?********** void insert(ListNode<int> * &fr.

C++ Please write the whole code that is needed for this assignment- wr.docxBrianGHiNewmanv

In the class we extensively discussed a node class called IntNode in.pdfarjunstores123

In the class we extensively discussed a node class called IntNode in which each linked list node contains an integer and each node is linked forward to the next element in the list. Implement a node class called DoublyIntNode in which each node contains an integer number and each node is linked not only forward to the next element in the list but also backward to the previous element in the list. Your implementation of DoublyIntNode should adapt and implement all the methods of IntNode including the following methods: Constructor for DoublyIntNode addNodeAfter, removeNodeAfter getData, setData getForeLink, setForeLink getBackLink, setBackLink listCopy listCopyWithTail listLength listPart listPosition listSearch IntNode and Node classes: public class IntNode{ private int data; private IntNode link; // methods here } public class Node { private E data; private Node link; // methods here } Solution program #include using namespace std; struct list { struct list *prev; int data; struct list *next; } *node = NULL, *first = NULL, *last = NULL, *node1 = NULL, *node2 = NULL; class Dlist { public: void insert_beg() { 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; } } void insert_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; } } void display() { node = first; cout << \"List of data in Linked list in Ascending order!\" << endl; while(node != NULL) { cout << node->data << endl; node = node->next; } node = last; cout << \"List of data in Linked list in Descending order!\" << endl; while(node != NULL) { cout << node->data << endl; node = node->prev; } } void del() { int count = 0, number, i; node = node1 = node2 = first; for(node = first; node != NULL; node = node->next) cout << \"Enter value for the node:\" << endl; count++; display(); cout << count << \" nodes available here!\" << endl; cout << \"Enter the node number which you want to delete:\" << endl; cin >> number; if(number != 1) { if(number < count && number > 0) { for(i = 2; i <= number; i++) node = node->next; for(i = 2; i <= number-1; i++) node1 = node1->next; for(i = 2; i <= number+1; i++) node2 = node2->next; node2->prev = node1; node1->next = node2; node->prev = NULL; node->next = NULL; node = NULL; } else if(number == count) { node = last; last = node->prev; last->next = NULL; node = NULL; } else cout << \"Invalid node.

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10. Consider the following data structure used for implementing a linked list: struct Node int data. Node *next; Implement the function that takes a pointer to the first node of a linked list and an integer value as arguments. It returns -1 if the input value is not equal to the data value of any item in the list. Otherwise, it returns the location of an item in the list whose data value is equal to the input value. Assume that the location of the first item in the list is 0. int search List (Node *head, int value) write the function body Solution #include <iostream> using namespace std; struct Node{ int data; Node *next; }; //inserting element at front side in list void insert(Node **head,int data){ struct Node *temp; if(*head==NULL){ (*head)=(struct Node*)malloc(sizeof(struct Node)); (*head)->data=data; (*head)->next=NULL; }else{ temp=(struct Node*)malloc(sizeof(struct Node)); temp->data=data; temp->next=(*head); (*head)=temp; } } // searching element in list int searchList(Node *head, int value){ int i=0;int flag=0; while(head!=NULL){ if(head->data==value){ flag=1; return i; } i++; head=head->next; } if(flag==0) return -1; } void printElement(Node *head){ while(head!=NULL){ cout<<head->data; head=head->next; } } int main() { struct Node *head=NULL; insert(&head,10); insert(&head,20); insert(&head,30); insert(&head,40); insert(&head,50); //printElement(head); int rValue=searchList(head,30); if(rValue==-1){ cout<<\"Value do not exist in list\"<<endl; } else{ cout<<\"Value exist at position \"<<rValue<<endl; } return 0; } /*sample output*/ searchList(head,30); Value exist at 2 searchList(head,60); Value do not exist in list .

C Exam Help Programming Exam Help

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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 elements: 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.

C Homework HelpProgramming Homework Help

1#include stdio.h#include stdlib.h#include assert.h .pdfsudhinjv

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Linked lists are a data structure where each node contains a pointer to the next node in the list. Each node contains data and a pointer to the next node. The last node's pointer is set to NULL. Linked lists can dynamically grow and shrink during program execution. Nodes are implemented as structs containing data and a pointer. The head pointer points to the first node. Linked lists allow traversing the list by following node pointers until NULL is reached.

C++ problemPart 1 Recursive Print (40 pts)Please write the recu.pdfcallawaycorb73779

C++ problem Part 1: Recursive Print (40 pts) Please write the recursive List reverse print function, whose iterative version we wrote in class. Below are the function signatures for the functions you are going to need: public: /**Additional Operations*/ void print_reverse(); //Wrapper function that calls the reverse helper function to print a list in reverse //prints nothing if the List is empty private: void reverse(Nodeptr node); //Helper function for the public printReverse() function. //Recursively prints the data in a List in reverse. Why do we need the private helper function here? Since we are going to be reversing our list node by node, in a recursive fashion, we want to pass a one node at a time to our reverse function. However, since our nodes are private, we cannot access them if we call the function inside of main. Add these function signatures to your List.h file along with your other function prototypes inside the class definition. Make sure that you place the reverse function inside the private portion of your List class definition and the print_reverse function prototype to the public portion of your List class definition. Now, implement these two functions inside of List.h, under your section for additional operations. Important: Test each function carefully inside of your ListTest.cpp to make sure that it is working properly. Part 2: Adding an Index to Your List Nodes (20 pts) Next, you will add the following functions to your List.h /**Accessor Functions*/ int get_index(); //Indicates the index of the Node where the iterator is currently pointing //Nodes are numbered from 1 to length of the list //Pre: length != 0 //Pre: !off_end() ... int List::get_index() { //Implement the function here } /**Manipulation Procedures*/ void scroll_to_index(int index); //Moves the iterator to the node whose index is specified by the user //Pre: length != 0 ... void scroll_to_index(int index) { //Implement function here } Part 3: Implementing Search as Part of Your List (40 pts) Now, we are going to add two search functions to our List so that we can search for elements in our List. The first of these functions is going to be a simple linear search function. You will need to add the following function prototype and function definition to your List.h: /**Additional Operations*/ int linear_search(listitem item); //Searchs the list, element by element, from the start of the List to the end of the List //Returns the index of the element, if it is found in the List //Returns -1 if the element is not in the List //Pre: length != 0 ... int List::linear_search(listitem item) { //Implement the function here } You are also going to add a function to perform recursive binary search on your List. You will need to add the following function prototype and function definition to your List.h: int binary_search(int low, int high, listitem item); //Recursively searchs the list by dividing the search space in half //Returns the index of the element, if it is fo.

Unit7 Carnold 7490

- Data structures allow for efficient handling of large volumes of data through logical organization and relationships between data elements. Common linear data structures include arrays, lists, stacks, and queues, while trees and graphs are examples of non-linear data structures. - Abstract data types (ADTs) define a set of complex data objects and operations that can be performed on those objects without specifying their implementation. Data structures provide a way to implement the logical relationships and operations defined in an ADT. - Linked lists provide an alternative to arrays for storing data by linking each data element to the next using pointers, rather than requiring contiguous memory locations. This allows for more flexible insertion and deletion compared to arrays.

Data Structures and Agorithm: DS 04 Linked List.pptxRashidFaridChishti

C++ Please test your program before you submit the answer.pdfaashisha5

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.

Assignment isPage 349-350 #4 and #5 Use the Linked List lab.pdffortmdu

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) 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 #4 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 deleteLast(nodeType*& last, nodeType* first); int main() { nodeType *firs.

Lec3-Linked list.pptxFaheemMahmood2

ccvcvbcbvcbvcbvcbvcb vngfg hgjhg gj jgjhgjhgjhg jhgh jhgjhgj gjhgjhgjhgjhgjhgjhgjhg jghgjhgjhgjhgjhg jjghjhgjhgjhgjhgjh hgjhgjhgjhgjhgjhgjhgjgjhgjhgjhg hgfgfhdfhtftfj jgyjyyyugyugyug yjgghgjhgjhf drdfhgfgfgfgj jgjhgjhgjhgjhgjhg jhgghjfgf fxdx gfd hdhfd tftyf yytfy ytftyf ytf ytfty fytftykgh kghjg jhghgh fgfghfgh fgfghfgf gfgfgfgfytf hgfggf ghfgfg gjhghjg cfffgh jgfgfhgffhg hghh yuy ygyyuyutyutyu jgfghfhgfghfh fghfhgfhgf fthfhfhf hfhfhgf hfhgfghf fghfghfghfhhgfghf jgjhgjhgjhgjhgjhgj jgjhgjh jgjhg jgjhgjhgjhg jhgjhgjhg jhgjhg jhgjhgollowing the announcement of Google+ API deprecation scheduled for March 2019, a number of changes will be made to Blogger’s Google+ integration on 4 February 2019. *Google+ widgets:* Support for the “+1 Button”, “Google+ Followers” and “Google+ Badge” widgets in Layout will no longer be available. All instances of these widgets will be removed from all blogs. *+1 buttons:* The +1/G+ buttons and Google+ share links below blog posts and in the navigation bar will be removed. Please note that if you have a custom template that includes Google+ features, you may need to update ... read more ollowing the announcement of Google+ API deprecation scheduled for March 2019, a number of changes will be made to Blogger’s Google+ integration on 4 February 2019. *Google+ widgets:* Support for the “+1 Button”, “Google+ Followers” and “Google+ Badge” widgets in Layout will no longer be available. All instances of these widgets will be removed from all blogs. *+1 buttons:* The +1/G+ buttons and Google+ share links below blog posts and in the navigation bar will be removed. Please note that if you have a custom template that includes Google+ features, you may need to update ... read moreollowing the announcement of Google+ API deprecation scheduled for March 2019, a number of changes will be made to Blogger’s Google+ integration on 4 February 2019. *Google+ widgets:* Support for the “+1 Button”, “Google+ Followers” and “Google+ Badge” widgets in Layout will no longer be available. All instances of these widgets will be removed from all blogs. *+1 buttons:* The +1/G+ buttons and Google+ share links below blog posts and in the navigation bar will be removed. Please note that if you have a custom template that includes Google+ features, you may need to update ... read moreollowing the announcement of Google+ API deprecation scheduled for March 2019, a number of changes will be made to Blogger’s Google+ integration on 4 February 2019. *Google+ widgets:* Support for the “+1 Button”, “Google+ Followers” and “Google+ Badge” widgets in Layout will no longer be available. All instances of these widgets will be removed from all blogs. *+1 buttons:* The +1/G+ buttons and Google+ share links below blog posts and in the navigation bar will be removed. Please note that if you have a custom template that includes Google+ features, you may need to update ... read moreollowing the announcement of Google+ API deprecation scheduled for March 2019, a nu

please help me in C++Objective Create a singly linked list of num.pdfaminbijal86

please help me in C++ Objective: Create a singly linked list of numbers based upon user input. Program logic: Ask for a number, add that number to the front of the list, print the list. Repeat until they enter -1 for the number. . Sample Input: 10, 15, 5, 2, 4, -1 Output: 4, 2, 5, 15, 10 Solution #include #include using namespace std; // It contains the value and the pointer to the next element in the list. struct element { int value; struct element *next; }; struct element *head = NULL; // The list is stored in the object of this class. class linkedlist { public: // To display list, we first see if head is NULL, // If head is NULL it means the list is empty. // Then we traverse along the list and display each item, until the next pointer is NULL. void display() { if(head == NULL){ cout << \"The list is empty\ \"; } else { struct element *temp; temp = head; while(temp != NULL){ cout << temp->value << \" \" ; temp = temp->next; } cout << \"\ \"; } } // To add an element to the list,create a newElement and store the value. // If head is NULL, add the newElement to the head. // To add an element at the starting of the list. // The next pointer of the newElement must point to the old head. // The new head must be pointed to the newElement. void add(int a){ struct element *newElement; newElement = (struct element *)malloc(sizeof(struct element)); newElement->value = a; newElement->next = NULL; if(head == NULL) { head = newElement; } else { newElement->next = head; head = newElement; } } }; int main() { // Create an object of the class linkedlist. linkedlist l = linkedlist(); int a; do { cout << \"Enter a number : \"; cin >> a; if(a!=-1) { // If the number is -1 do not add it to the list. l.add(a); // Add the number to the list. } l.display(); // Display the list. } while(a!=-1); // Keep asking for a number until -1 is pressed. return 0; }.

17 linkedlist (1)Himadri Sen Gupta

This document discusses linked lists and their implementation in C. It begins by describing the differences between array-based and linked list implementations. Specifically, it notes that linked lists use dynamic memory allocation and pointers to connect nodes rather than contiguous memory blocks. The document then covers creating and traversing linked lists using nodes with data fields and next pointers. It provides code examples for inserting, finding, deleting and destroying nodes in a singly linked list. Finally, it briefly lists some other common linked list operations like sorting, reversing and finding min/max elements.

File Type cppAdd the following to your linked list of strings pro.pdffootworld1

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\"<.

Java AssignmentUsing the ListNode.java file below Write method.pdfambersushil

Java Assignment Using the ListNode.java file below Write methods called min and max that return the smallest and largest values in the linked list. These methods will be added to your ListNode class. For example if a variable called list stores {11, -7, 3, 42, 0, 14], the call of list.min() should return -7 and the call of list.max() should return 42. If the list is empty, return -1. Print the returned value. Write a method called insertNode that inserts a new node anywhere in your linked list. Display your linked list with new value. Write a program to test these methods to ensure accuracy and successful implementation. Upload your modified version of the ListNode java file only. Note: Original code should still be included. You will make additions and slight modifications where necessary. Throw exceptions where you think is necessary. // Listlode is a class for storing a single node of a linked // list. This node class is for a list of integer values. public class Listlode \{ public int data; //data stored in this node public Listliode front;//points to head node public Listliode next; /f link to next node in the list // post: constructs a node with data and null link public Listlode() \{ 3. // post: constructs a node with given data and null link: public Listlode(int data) \{ this(data, null); 3 // post: constructs a node with given data and given link. public Listlode(int data, Listliode next) \{ this.data = data; this. next = next; 3 public String toString() \{ if (front = null) : \} else \{ String result = " ["+ front data; Listliode current = front. next; while (current != null) \{ result +=m,m+ current data, current = current. next; 3 result += " ]n - return result; 3. }// post: appends the given value to the end of the list public void add(int value) \& puolic void aod if (front = nult) \{. front = new Listlode (value); 3 else \{ Listlode current = front; whistiode current = front; while (current next I= null) \{ current = current. next; 3 current. next = new Listliode (value ); 3. 3 J// post : returns the position of the first occurrence of the given /l value (-1 if not found ) public void remove (int index) \{ if ( index =){ \} else \{ Listiode current = nodeAt ( index - 1); current. next = current. next. next, 3 3 3// pre : <=i< size() // post: returns a reference to the node at the given index public Listlode nodeAt (int index) : List Node current = front; for (int i=0;i< index; i++) \{ current = current. next. \}. return current; 3.

Mi 103 linked listAmit Vats

A linked list is a data structure where each node contains a data field and a pointer to the next node. Nodes can be dynamically added or removed, allowing the list to grow and shrink. The key operations on a linked list are appending a node to the end, inserting a node in order, and deleting a node. Each node is allocated dynamically and linked together via next pointers, allowing flexible and efficient insertion and removal of nodes anywhere in the list.

Linked listMd. Afif Al Mamun

Linked lists are linear data structures where each node points to the next. Each node contains a data field and a pointer to the next node. There are three types: singly, doubly, and circular linked lists. Linked lists allow for constant-time insertions and deletions and do not require fixed size allocation. Common operations on linked lists include insertion, deletion, searching, and traversal. Linked lists are useful for implementations like stacks, queues, and dynamic data structures.

C++ please put everthing after you answer it- thanks Complete the stub.docxMatthPYNashd

C++ please put everthing after you answer it, thanks Complete the stub file for the DoublyLinkedList implementation. Please note: Any changes to the NodeType struct, not using the NodeType struct, or changing from template Type to using a specific data type will all be cause for deductions. You are required to implement the methods: insertFirst insertLast insertNode - (insert an item at a given index) deleteNode - (remove a specific item) The stub file contains the basic class format to implement the virtual methods from ListType. You should only need to complete the code for these four functions. Your code must maintain (or add) the node connections both forward and backward. In the double link list class these are called next and prev. Your code must also maintain the count variable. \} template Type ListType : : back ( ) const \{ assert(tail != NULL); //confirm tail exists return tail->data; //return the data of the tail node \} template void ListType >: :copyList (const ListType > \& otherList) \{ NodeType *newNode; //pointer to create a node NodeType current; //pointer to traverse the list if (head ! = NULL) //if the list is nonempty, make it empty destroyList(); if (otherList. head == NULL) //otherList is empty \{ head = NULL; tail = NULL; count = 0 ; \} else \{ current = otherList. head; // current points to the count = otherList . count; // list to be copied //copy the head node head = new NodeType < Type > ; / / create the node head->data = current > data; // copy the data head->next = NULL; // set the link field of tail = head; // the node to NULL // the next node //copy the remaining list while (current != NULL) \{ newNode = new NodeType < Type > ; // create a node newNode > data = current > data; // copy the data newNode->next = NULL; // set the link of // newNode to NULL tail->next = newNode; //attach newNode after tail tail = newNode; // make tail point to //the actual tail node current = current > next; 1/ make current point //to the next node \}/ /end while \}/lend else \}//end copylist .

Firefly luciferase is a commonly used reporter gene. The gene origina.pdfArrowdeepak

Firefly luciferase is a commonly used reporter gene. The gene originates from fireflies (a type of beetle) and when it combines with its substrate, luciferin, light is produced. Luciferin is very expensive to purchase. You would therefore like to figure out the biosynthetic pathway for making luciferin so that you can get bacteria to make luciferin. This will enable you pay off all your college loans! You have obtained 5 mutant firefly stocks that each fail to glow at night. The flies in each stock have the wild type luciferase gene but ultimately fail to make the luciferin substrate. Each stock has a unique mutation in one of the genes required to make luciferin. You know that the biosynthesis of luciferin involves 5 intermediate compounds: A, B, C, D, and E. However, you do not know the order that these compounds appear in the pathway. When you feed these intermediate compounds to the mutant fireflies, they sometimes are able glow (+) and sometimes still cannot glow (-). However, when you feed the mutant fireflies luciferin, they always glow. Based on the results in the table above, write the biosynthetic pathway for luciferin, showing the order of the intermediates (A-E) and the genes (M1-M5) required to convert each intermediate to the next in the pathway. Solution This is a probem, where in you can get the answer by the means of trial and error. Now that we have all pieces to this puzzle, lets join them. E>B>D>C>A this is the pathway in the production of Luciferase..

Find all hazards in this circuit. Redesign the circuit as a three-le.pdfArrowdeepak

Find all hazards in this circuit. Redesign the circuit as a three-level NOR circuit that is free of all hazards Show how four 2-to-1 and one 4-to-1 multiplexers could be connected to form an 8- to-1 MUX with three control inputs Solution 1 University of Wisconsin - Madison ECE/Comp Sci 352 Digital Systems Fundamentals Charles R. Kime Section 2 – Fall 2001 Chapter 2 – Combinational Logic Circuits – Part 7 Charles Kime & Thomas Kaminski © 2001 Prentice Hall, Inc Logic and Computer Design Fundamentals Logic and Computer Design Fundamentals © 2001 Prentice Hall, Inc Chapter 2-Part 7 2 NAND and NOR Implementation We found that we could implement general Boolean equations with these three primitives: • AND • OR • NOT In this section we will find that either of two gates, the NAND gate or the NOR gate can be used to implement arbitrary logic functions. We use the Positive Logic Convention (where all signals are active high) and a small circle to on a symbol to represent NOT or invert. 2 Logic and Computer Design Fundamentals © 2001 Prentice Hall, Inc Chapter 2-Part 7 3 NAND Gates The basic positive logic NAND gate is denoted by the following symbol: • AND-Invert (NAND) NAND comes from NOT AND, I. e., the AND function with a NOT applied. We call this symbol for a NAND gate an AND-Invert. The small circle represents the invert function. If we apply DeMorgan\'s Law we get: X Y Z X YZ = X + Y + Z F(X, Y,Z) = X Y Z Logic and Computer Design Fundamentals © 2001 Prentice Hall, Inc Chapter 2-Part 7 4 NAND Gates (Cont.) Applying DeMorgan\'s Law gives: • Invert-OR (NAND) We call this symbol for a NAND gate the Invert - OR since all inputs are inverted, followed by the OR function. Both symbols represent the NAND gate - it is sometimes more logically descriptive to use one form over the other. A NAND gate with one input degenerates to an inverter. X Y Z F(X, Y,Z) = X + Y + Z 3 Logic and Computer Design Fundamentals © 2001 Prentice Hall, Inc Chapter 2-Part 7 5 NAND Function Implementation NAND gates can implement a simplified Sum-ofProducts form. Constructing two level NAND-NAND gate circuit: The first level is two 2-input NAND gates using ANDInvert. The second level is one 2- input NAND gate using Invert-OR. Using the NAND relationship, we have: G(A,B,C,D) = AB CD = AB + CD = A B + CD A B C D G(A,B,C,D) = AB + CD Logic and Computer Design Fundamentals © 2001 Prentice Hall, Inc Chapter 2-Part 7 6 NAND Implementation (Cont.) In the implementation, note that the bubbles are on opposite ends of the same line. Thus, they can be combined and deleted: A B C D G(A,B,C,D) This form of the implementation is the Sum-of- Products form. 4 Logic and Computer Design Fundamentals © 2001 Prentice Hall, Inc Chapter 2-Part 7 7 NAND Implementation (Cont.) In the implementation, the bubbles are on opposite ends of the same line. By , they can be combined and deleted: A sum-of-products (SOP) form results To implement an equation like: F(A,B,C) = A + BC, the NAND for A degenerates t.

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Explain the “life” of a secreted protein molecule - trace the pathwa.pdfArrowdeepak

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7. Using to Table 1 to assist you, explain the rational for the diver.pdfArrowdeepak

7. Using to Table 1 to assist you, explain the rational for the diversity of the mycobiome, as compared to the number of bacterial species identified in the lungs of CF patients. This should be a full paragraph, not one or two sentences. Solution From the tabulated data given, it can be clearly noticed that majority of the population suffers from ascomycete infections. Particularly, the most infectious species can be identified as Candida albicans followed by C. dubliniensis. Saccharomyces cerevisiae infections are also very high followed by C. parapsilosis and C. glabrata. Other types of infections are also evident in patients but these organisms remain the most frequent infective agents for patients with cystic fibrosis. Since all of these microbes spread the disease from one organism to another, they are highly communicable in nature. Also, since very high frequency of some of the species can be noticed from the data, it is also suggested that the infections are opportunistic in nature i.e. they will infect the individual as soon as his immunity against these microbes is minimized. Importantly, since a huge population is suffering form nearly similar kind of infections, it can be suggested that fixation of these microbial infections is possible in this population which can be clinically deleterious in advanced stages because eradication of these microbes using multiple antibiotic treatment would be difficult. Hence, it can be evaluated from the data that the statisitical figures are clinically alarming and demand immediate attention..

A. list two vertebrae that spinal nerve t12 travels between .pdfArrowdeepak

A. list two vertebrae that spinal nerve t12 travels between B. Define the function of ground substance in connective tissue proper. A. list two vertebrae that spinal nerve t12 travels between B. Define the function of ground substance in connective tissue proper. B. Define the function of ground substance in connective tissue proper. Solution A. Two vertebrae that spinal nerve t12 travels between are-cervical spine vertebrae and lumbar spine vertebrae B. Groud substance is an amorphous gel-like substance present in the connective tissue proper, which surrounds the cells. Glycoproteins and Hydrophilic muco-polysacharides present in the ground substance attract water to induce variation in viscosity. Groud substance of all connective tissue proper make intimate contact with the cells. Hence, these cells doesnot occupy distinct space within the matrix..

Do people from different cultures experience emotions differently O.pdfArrowdeepak

Do people from different cultures experience emotions differently? One study designed to examine this question collected data from 411 college students from five different cultures. The participants were asked to record, on a 1 (never) to 7 (always) scale, how much of the time they typically felt eight specific emotions. These were averaged to produce the global emotion score for each participant. Here is a summary of this measure: Note that the convention of giving the standard deviations in parentheses after the means saves a great deal of space in a table such as this. (a) The ANOVA F statistic was reported as 5.57. Give the degrees of freedom and either an approximate (from Table E) or an exact (from software) P- value. Sketch a picture of the F distribution that illustrates the P-value. (Round your P-value to four decimal places.) What do you conclude? We have strong evidence at the 5% level that the means are not all the same. We have no evidence at the 5% level that the means are different. (b) Without doing any additional formal analysis, describe the pattern in the means that appears to be responsible for your conclusion in part (a). Are there pairs of means that are quite similar? European Americans have the highest emotion scores, Indians are in the middle, and the other three cultures are lowest. Hispanic Americans have the highest emotion scores, Japanese are in the middle, and the other three cultures are lowest. Japanese have the highest emotion scores, Hispanic Americans are in the middle, and the other three cultures are lowest. Asian Americans have the highest emotion scores, Japanese are in the middle, and the other three cultures are lowest. European Americans have the highest emotion scores, Hispanic Americans are in the middle, and the other three cultures are lowest. Culture n Mean (s)European American Asian American Japanese Indian Hispanic American48 32 91 160 804.39 (1.01) 4.35 (1.18) 4.71 (1.13) 4.34 (1.26) 5.04 (1.16) Solution (a) numerator df= 5-1=4 denominator df = 48+32+91+160+80-5 =406 P-value = P(F with df=4 and df=406 >5.57 ) =0.0002 (from F table) We have strong evidence at the 5% level that the means are not all the same. --------------------------------------------------------------------------------------------------------- (b)Hispanic Americans have the highest emotion scores, Japanese are in the middle, and the other three cultures are lowest..

Describe the basic elements found in robot controllers. (Choose all t.pdfArrowdeepak

Consider the following probability distribution The Normal distribut.pdfArrowdeepak

1. Please explain 1. What is cosolidation of a soi 2. What are the .pdfArrowdeepak

1) Consolidation is the process by which a saturated soil reduces in volume by expelling water from its pores under an applied load over time. 2) The one-dimensional consolidation theory proposed by Terzaghi makes assumptions that the soil is fully saturated, uniformly distributed loading, drainage occurs only vertically, and soil is homogeneous and isotropic. 3) The main difference between compaction and consolidation is that compaction involves expelling air from soil pores, while consolidation involves expelling water.

What type of gametes did Sutton predict Mendel (parental, recombin.pdfArrowdeepak

What type of gametes did Sutton predict? Mendel? (parental, recombinant, both) Solution 1. According to Mendel’s law of segregation : 2 . The Chromosomal Theory of Inheritance (Theodor Boveri and Walter Sutton --1903) was consistent with Mendel\'s laws and the basic postulates were: -----------------------These lead to the conclusion that both Sutton and Mendel predicted parental types of gametes because in case of parental type , gametes are produced in equal proportion but recombinant type produces gametes in unequal proportions. Besides this , In 1911, after observing that linked traits could on occasion be inherited separately, Thomas Hunt Morgan suggested that \"crossovers\" can occur between linked genes and crossover is the prime reason for recombination.

With reference to Barth (2014), discuss how the author assessed the .pdfArrowdeepak

With reference to Barth (2014), discuss how the author assessed the extent to which unmodified historical cost, modified historical cost, and fair value measurement possess the qualitative characteristics specified in the IFRS Conceptual Framework. Solution The conceptual framework in IFRS states the qualitative characteristics of the financial statements to be useful which are as follows: 1. Fundamental : Fundamental qualitative characteristics are relevance and faithful representation. 2. Enhancing: Enhancing qualitative characteristics are comparability, completeness, verifiablity and understandability. According to Barth(2014), Historical cost does not possess the qualitative characteristics of the IFRS conceptual framework as it does not facilitates comparability and relevance but yes it is sometimes relevant. The historical cost should be modified to use in financial statements, it should be indexed for Inflation. Modified historical cost possess the qualitative characteristics of the IFRS. It is understandable, complete, accurate, but not considered as faithful representation. This is widely used but both modified historical cost and unmodified historical cost lack meaning. Whereas the Fair value presentation is more consistent with the new framework and hence it should be widely used. This paper concludes that assets and liabilities at large should be measured at fair value to provide the financial statements with the qualitative characteristics..

1. Peter Grant - the evolutionary biologist who has studied finches .pdfArrowdeepak

1. Peter Grant - the evolutionary biologist who has studied finches on Daphne Major (an island in the Galapagos) for several decades - collected data on birds\' beak sizes, weight, diet and mating behavior to test his hypotheses. This kind of data is an example of: 2. In the video \"How Does Evolution Really Work?\", biologists described an example where the length of individual hummingbird beaks grew longer in response to changes in their food source (flowers). True False 3 The observation that the individuals possessing favorable traits are more likely to survive and reproduce than those possessing less favorable traits is the basis for: 4. Which one of the following descriptions, pertaining to a scientific hypothesis or theory, is NOT true? 5. Which of the following is involved in the scientific method (MARK ALL THAT APPLY)? making observations 6. Due to ___________________, \"bad science” (e.g. fake data or fake results) is eventually discovered and revealed. powerful police forces 7. Darwin\'s Hypothesis of Evolution by Natural Selection included the observation that populations often grow faster than the food supply and therefore individuals need to compete for resources. Darwin borrowed this idea from the work of: 8. Lamarck\'s \"Inheritance of Acquired Characteristics\" can best be described as: disease and starvation are required to keep population size in check 9. Aristotle and other early philosophers believed that species were \"immutable\" or \"fixed\", which means: species are always changing species don\'t change over timeA.taxonomy B.artificial selectionC.empirical evidenceD.non-factual informationE.a belief Solution 1)B)Artificial Selection as it tells about the way birds survive and their mating. 2)True yes it is rightly stated. 3)D)Natural selection which says traits move from one generation to other. and the ones with more traits survive more than other here. 4)E)it must deal with evidence that is empirical (i.e. the kind that can be observed).

COMMUTER PASSES Five different types of monthly commuter passes are o.pdfArrowdeepak

If a researcher rejects a null hypothesis when that hypothesis is ac.pdfArrowdeepak

If a researcher rejects a null hypothesis when that hypothesis is actually true, he has committed: a type I error a type II error a type III error no error a type I error a type II error a type III error no error Solution If a researcher rejects a null hypothesis when that hypothesis is actually true, he has committed: a type I error a type II error a type III error no error A type I error is the incorrectly rejecting a true null hypothesis. a type I error a type II error a type III error no error.

Which of the following values of r allows a perfect prediction of sc.pdfArrowdeepak

Which of the following is a way in which meiosis differs from mitosi.pdfArrowdeepak

Which of the following is a way in which meiosis differs from mitosis? In meiosis, a spindle of microtubules does not form. Meiosis produces only diploid daughter cells. Meiosis is found in asexual reproduction. In meiosis, the parent cell divides twice. Meiosis never produces gametes (sex cells). Question Part Solution Answer is 4 In meiosis parent cell divides twice whereas in mitosis parent cell divides only once. Meiosis is a reductional division which occurs in diploid cell to produce four tabloid cells whereas mitosis is a educational division which either occur in haploid or diploid cell to produce haploid or diploid daughter cells. Meiosis produces balloon gametes. Also, meiosis is not found in asexual reproduction. Meiosis and mitosis both involves formation of spindle of microtubules..

What is the inheritance of the following pedigree X-linked recessiv.pdfArrowdeepak

What is the profitability index for an investment with the following.pdfArrowdeepak

What is the profitability index for an investment with the following cash flows given a 6 percent required return? 1.38 0.72 1.06 0.81 1.23 What is the profitability index for an investment with the following cash flows given a 6 percent required return? Solution Present value of inflows=cash inflow*Present value of discounting factor(rate%,time period) 15700/1.06+18000/1.06^2+17400/1.06^3 =$45440.6322 PI=Present value of inflows/Present value of outflows =$45440.6322/$37000 which is equal to =1.23(Approx)..

What are the characteristics of each microscope 1.compound2.s.pdfArrowdeepak

What are the characteristics of each microscope 1.compound 2.stereo ( dissecting) 3. Phase contrast 4. Scanning (SEM) 5. Transmission (TEM) What are the characteristics of each microscope 1.compound 2.stereo ( dissecting) 3. Phase contrast 4. Scanning (SEM) 5. Transmission (TEM) 1.compound 2.stereo ( dissecting) 3. Phase contrast 4. Scanning (SEM) 5. Transmission (TEM) Solution The characteristics of each microscope is as follows: 1.COMPOUND MICROSCOPE In compound microscope the objective lens forms a real, inverted and magnified image of the object. The eye piece acts like a magnifying glass and forms a virtual erect and magnified image. Object is placed outside the principal focus of the objective lens. It consists of two convex lens, ocular, close to the eye and s3cond objective lens. It also reduces chromatic abberrations. 2. STEREO (DISSECTING) It is an optical microscope which is designed for low magnification, it uses light reflected from the surfacevof an object rather than transmitted through it. It forms three dimensional erect image as the prism under the eye piece reverses the image. 3. PHASE CONTRAST It is used to examine biological tissues (live microorganisms). It is a type of light microscopy that increases contrasts of transparent and colourless objects. This is achieved by influencing the optical path of the light. Thus contrast of light and dark makes the specimen visible to the human eye. 4. SCANNING (SEM). It is used for producing high resolution images of a sample surface. The images have 3-D appearance and best used to judge the surface structure of the sample. It produces image using focused beam of electrons. In SEM, the picture is shown on the monitor. 5.TRANSMISSION (TEM). It is based on transmitted electrons. It provides details of internal composition (show morphology, crystallisation, stress or even magnetic domains of the sample). In TEM, picture is shown on fluorescent screens. It provides 2-D picture of the sample. It uses electromagnetic lenses to focus the electrons into very thin beams..