Let n = size of the hash table hashTable = array of size n, w.pdf

Jun 30, 20230 likes2 views

aravlitraders2012

//Let n = size of the hash table // hashTable = array of size n, which represents the hash table // hashIndex = represents the index of array // If the list is EMPTY, we add the element into the list /*If the list is NOT EMPTY then we compare the newNode to the node already in the list in such a way to get sorted.*/ struct node { int key; T value; struct node *next; }; struct node *hashTable[n] = {NULL}; struct node * createNode(int key, char *name, int age) { struct node *newnode; newnode = (struct node *)malloc(sizeof(struct node)); newnode->key = key; newnode->value = value; newnode->next = NULL; return newnode; } add(int key, T value) { int hashIndex = key % n; struct node *newNode = creteNode(key,value); /*Adding new node to the list with No element */ if (hashTable[hashIndex] == NULL) { hashTable[hashIndex] = newnode; return; } /*Adding new node to the sorted list */ struct node *temp1; struct node *temp2 = hashTable[hashIndex]; temp1 = temp2; while(temp2 -> key < newNode -> key) { temp1 = temp2; temp2 = temp2 ->next; } newnode->next = temp2; temp1->next = newNode; return; } Solution //Let n = size of the hash table // hashTable = array of size n, which represents the hash table // hashIndex = represents the index of array // If the list is EMPTY, we add the element into the list /*If the list is NOT EMPTY then we compare the newNode to the node already in the list in such a way to get sorted.*/ struct node { int key; T value; struct node *next; }; struct node *hashTable[n] = {NULL}; struct node * createNode(int key, char *name, int age) { struct node *newnode; newnode = (struct node *)malloc(sizeof(struct node)); newnode->key = key; newnode->value = value; newnode->next = NULL; return newnode; } add(int key, T value) { int hashIndex = key % n; struct node *newNode = creteNode(key,value); /*Adding new node to the list with No element */ if (hashTable[hashIndex] == NULL) { hashTable[hashIndex] = newnode; return; } /*Adding new node to the sorted list */ struct node *temp1; struct node *temp2 = hashTable[hashIndex]; temp1 = temp2; while(temp2 -> key < newNode -> key) { temp1 = temp2; temp2 = temp2 ->next; } newnode->next = temp2; temp1->next = newNode; return; }.

//Let n = size of the hash table
// hashTable = array of size n, which represents the hash table
// hashIndex = represents the index of array
// If the list is EMPTY, we add the element into the list
/*If the list is NOT EMPTY then we compare the newNode to the node already in the list in such
a way to get sorted.*/
struct node
{
int key;
T value;
struct node *next;
};
struct node *hashTable[n] = {NULL};
struct node * createNode(int key, char *name, int age)
{
struct node *newnode;
newnode = (struct node *)malloc(sizeof(struct node));
newnode->key = key;
newnode->value = value;
newnode->next = NULL;
return newnode;
}
add(int key, T value)
{
int hashIndex = key % n;
struct node *newNode = creteNode(key,value);
/*Adding new node to the list with No element */
if (hashTable[hashIndex] == NULL)
{
hashTable[hashIndex] = newnode;
return;

}
/*Adding new node to the sorted list */
struct node *temp1;
struct node *temp2 = hashTable[hashIndex];
temp1 = temp2;
while(temp2 -> key < newNode -> key)
{
temp1 = temp2;
temp2 = temp2 ->next;
}
newnode->next = temp2;
temp1->next = newNode;
return;
}
Solution
//Let n = size of the hash table
// hashTable = array of size n, which represents the hash table
// hashIndex = represents the index of array
// If the list is EMPTY, we add the element into the list
/*If the list is NOT EMPTY then we compare the newNode to the node already in the list in such
a way to get sorted.*/
struct node
{
int key;
T value;
struct node *next;
};
struct node *hashTable[n] = {NULL};
struct node * createNode(int key, char *name, int age)
{
struct node *newnode;

newnode = (struct node *)malloc(sizeof(struct node));
newnode->key = key;
newnode->value = value;
newnode->next = NULL;
return newnode;
}
add(int key, T value)
{
int hashIndex = key % n;
struct node *newNode = creteNode(key,value);
/*Adding new node to the list with No element */
if (hashTable[hashIndex] == NULL)
{
hashTable[hashIndex] = newnode;
return;
}
/*Adding new node to the sorted list */
struct node *temp1;
struct node *temp2 = hashTable[hashIndex];
temp1 = temp2;
while(temp2 -> key < newNode -> key)
{
temp1 = temp2;
temp2 = temp2 ->next;
}
newnode->next = temp2;
temp1->next = newNode;
return;
}

The document discusses implementing a sorted linked list data structure in C++. It describes how to insert a new node into the linked list in the correct sorted position by finding the previous and current nodes between which the new node should be inserted. The key steps are to create a new node, find the previous and current nodes using a while loop, and connect the new node by updating the next pointers of the previous and current nodes.

Linked listA. S. M. Shafi

Linked lists are data structures that store data in scattered locations in memory. Each data item or node contains a data part that holds the actual data and a link part that points to the next node. The nodes are linked together using these links. A linked list requires an external pointer to point to the first node. Operations like traversing, searching, inserting and deleting nodes can be performed on linked lists by manipulating these node links.

Data Structures in C++I am really new to C++, so links are really .pdfrohit219406

Data Structures in C++ I am really new to C++, so links are really hard topic for me. It would be nice if you can provide explanations of what doubly linked lists are and of some of you steps... Thank you In this assignment, you will implement a doubly-linked list class, together with some list operations. To make things easier, you’ll implement a list of int, rather than a template class. Solution A variable helps us to identify the data. For ex: int a = 5; Here 5 is identified through variable a. Now, if we have collection of integers, we need some representation to identify them. We call it array. For ex: int arr[5] This array is nothing but a Data Structure. So, a Data Structure is a way to group the data. There are many Data Structures available like Arrays, Linked List, Doubly-Linked list, Stack, Queue, etc. Doubly-Linked list are the ones where you can traverse from the current node both in left and right directions. Why so many different types of Data Structures are required ? Answer is very simple, grouping of data, storage of data and accessing the data is different. For example, in case of Arrays we store all the data in contiguous locations. What if we are not able to store the data in contiguous locations because we have huge data. Answer is go for Linked List/Doubly-Linked list. Here we can store the data anywhere and link the data through pointers. I will try to provide comments for the code you have given. May be this can help you. #pragma once /* dlist.h Doubly-linked lists of ints */ #include class dlist { public: dlist() { } // Here we are creating a NODE, it has a integer value and two pointers. // One pointer is to move to next node and other to go back to previous node. struct node { int value; node* next; node* prev; }; // To return head pointer, i.e. start of the Doubly-Linked list. node* head() const { return _head; } // To return Tail pointer, i.e. end of the Doubly-Linked list. node* tail() const { return _tail; } // **** Implement ALL the following methods **** // Returns the node at a particular index (0 is the head). node* at(int index){ int cnt = 0; struct node* tmp = head(); while(tmp!=NULL) { if (cnt+1 == index) return tmp; tmp = tmp->next; } } // Insert a new value, after an existing one void insert(node *previous, int value){ // check if the given previous is NULL if (previous == NULL) { printf(\"the given previous node cannot be NULL\"); return; } // allocate new node struct node* new_node =(struct node*) malloc(sizeof(struct node)); // put in the data new_node->data = new_data; // Make next of new node as next of previous new_node->next = previous->next; // Make the next of previous as new_node previous->next = new_node; // Make previous as previous of new_node new_node->prev = previous; // Change previous of new_node\'s next node if (new_node->next != NULL) new_node->next->prev = new_node; } // Delete the given node void del(node* which){ struct node* head_ref = head(); /* base case */ if(*head_ref == NUL.

Inspect the class declaration for a doubly-linked list node in Node-h-.pdfvishalateen

Inspect the class declaration for a doubly-linked list node in Node.h. Access Node.h by clicking on the orange arrow next to main.cpp at the top of the coding window. The Node class has three member variables: a double data value, a pointer to the next node, and a pointer to the previous node. Each member variable is protected. So code outside of the class must use the provided getter and setter member functions to get or set a member variable. Node.h is read only, since no changes are required. Step 2: Implement the Insert() member function A class for a sorted, doubly-linked list is declared in SortedNumberList.h. Implement the SortedNumberList class's Insert() member function. The function must create a new node with the parameter value, then insert the node into the proper sorted position in the linked list. Ex: Suppose a SortedNumberList's current list is 23 47.25 86, then Insert(33.5) is called. A new node with data value 33.5 is created and inserted between 23 and 47.25, thus preserving the list's sorted order and yielding: 23 35.5 47.25 86 Step 3: Test in develop mode Code in main() takes a space- separated list of numbers and inserts each into a SortedNumberList. The list is displayed after each insertion. Ex: If input is 77 15 -42 63.5 then output is: List after inserting 77: 77 List after inserting 15: 15 77 List after inserting -42: -42 15 77 List after inserting 63.5: -42 15 63.5 77 Try various program inputs, ensuring that each outputs a sorted list. Step 4: Implement the Remove() member function Implement the SortedNumberList class's Remove() member function. The function takes a parameter for the number to be removed from the list. If the number does not exist in the list, the list is not changed and false is returned. Otherwise, the first instance of the number is removed from the list and true is returned. Uncomment the commented-out part in main() that reads a second input line and removes numbers from the list. Test in develop mode to ensure that insertion and removal both work properly, then submit code for grading. Ex: If input is 84 72 19 61 19 84 then output is: List after inserting 84: 84 List after inserting 72: 72 84 List after inserting 19: 19 72 84 List after inserting 61: 19 61 72 84 List after removing 19: 61 72 84 List after removing 84: 61 72 main.cpp #include #include #include #include "Node.h" #include "SortedNumberList.h" using namespace std; void PrintList(SortedNumberList& list); vector SpaceSplit(string source); int main(int argc, char *argv[]) { // Read the line of input numbers string inputLine; getline(cin, inputLine); // Split on space character vector terms = SpaceSplit(inputLine); // Insert each value and show the sorted list's contents after each insertion SortedNumberList list; for (auto term : terms) { double number = stod(term); cout << "List after inserting " << number << ": " << endl; list.Insert(number); PrintList(list); } /* // Read the input line with numbers to remove getline(cin, inputLi.

This assignment and the next (#5) involve design and development of a.pdfEricvtJFraserr

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.

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

1 #include #include #include /* Link list node */ struct node { int data; struct node* next; }; /* Given a reference (pointer to pointer) to the head of a list and an int, push a new node on the front of the list. */ void push(struct node** head_ref, int new_data) { /* allocate node */ struct node* new_node = (struct node*) malloc(sizeof(struct node)); /* put in the data */ new_node->data = new_data; /* link the old list off the new node */ new_node->next = (*head_ref); /* move the head to point to the new node */ (*head_ref) = new_node; } /* Takes head pointer of the linked list and index as arguments and return data at index*/ int GetNth(struct node* head, int index) { struct node* current = head; int count = 0; /* the index of the node we\'re currently looking at */ while (current != NULL) { if (count == index) return(current->data); count++; current = current->next; } /* if we get to this line, the caller was asking for a non-existent element so we assert fail */ assert(0); } /* Drier program to test above function*/ int main() { /* Start with the empty list */ struct node* head = NULL; /* Use push() to construct below list 1->12->1->4->1 */ push(&head, 1); push(&head, 4); push(&head, 1); push(&head, 12); push(&head, 1); /* Check the count function */ printf(\"Element at index 3 is %d\", GetNth(head, 3)); getchar(); } 2 #include #include #include /* Link list node */ struct node { int data; struct node* next; }; /* Given a reference (pointer to pointer) to the head of a list and an int, push a new node on the front of the list. */ void push(struct node** head_ref, int new_data) { /* allocate node */ struct node* new_node = (struct node*) malloc(sizeof(struct node)); /* put in the data */ new_node->data = new_data; /* link the old list off the new node */ new_node->next = (*head_ref); /* move the head to point to the new node */ (*head_ref) = new_node; } /* Takes head pointer of the linked list and index as arguments and return data at index*/ int GetNth(struct node* head, int index) { struct node* current = head; int count = 0; /* the index of the node we\'re currently looking at */ while (current != NULL) { if (count == index) return(current->data); count++; current = current->next; } /* if we get to this line, the caller was asking for a non-existent element so we assert fail */ assert(0); } /* Drier program to test above function*/ int main() { /* Start with the empty list */ struct node* head = NULL; /* Use push() to construct below list 1->12->1->4->1 */ push(&head, 1); push(&head, 4); push(&head, 1); push(&head, 12); push(&head, 1); /* Check the count function */ printf(\"Element at index 3 is %d\", GetNth(head, 3)); getchar(); } Solution 1 #include #include #include /* Link list node */ struct node { int data; struct node* next; }; /* Given a reference (pointer to pointer) to the head of a list and an int, push a new node on the front of the list. */ void push(struct node** head_ref, int new_data) { /* allocate node */ struct node*.

linked List.docx vhjgvjhvgjhjhbbjkhkjhkjhvasavim9

Data Structures & Algorithms - Lecture 3Faculty of Computers and Informatics, Suez Canal University, Ismailia, Egypt

linked-list.pptDikkySuryadiSKomMKom

Linked lists are a data structure made up of connected nodes, where each node contains data and a pointer to the next node. The three basic operations on linked lists are insertion, deletion, and traversal of nodes. There are variations like circular linked lists where the last node points to the first, and doubly linked lists where each node points to both the next and previous nodes. Linked lists have advantages over arrays in that they are dynamically sized and nodes can be easily inserted and deleted without moving other elements.

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

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.

week-13xKITE www.kitecolleges.com

LinkedlistMasud Parvaze

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 Homework HelpProgramming Homework Help

Describe a data structure to represent sets of elements (each element.pdfrajeshjain2109

Describe a data structure to represent sets of elements (each element with a unique key) that would support the following operations: insert(x, A) - insert an element x into A (assume the key of x is not in .4 yet) count (A, k) - return the number of elements in A with key at least k. Solution we can use Hashset for this.a simple code in hashset is given below public class MyNode { ... } public class MyDataStructure { private HashSet nodes = new HashSet(); /// /// Inserts an element to this data structure. /// If the element already exists, returns false. /// Complexity is averaged O(1). /// public bool Add(MyNode node) { return node != null && this.nodes.Add(node); } /// /// Removes a random element from the data structure. /// Returns the element if an element was found. /// Returns null if the data structure is empty. /// Complexity is averaged O(1). /// public MyNode Pop() { // This loop can execute 1 or 0 times. foreach (MyNode node in nodes) { this.nodes.Remove(node); return node; } return null; } } the unique key can also be achieved using dictioneries hence code for this is given below: public class FixedIntDictionary { // Our internal node structure. // We use structs instead of objects to not add pressure to the garbage collector. // We mantains our own way to manage garbage through the use of a free list. private struct Entry { // The key of the node internal int Key; // Next index in pEntries array. // This field is both used in the free list, if node was removed // or in the table, if node was inserted. // -1 means null. internal int Next; // The value of the node. internal T Value; } // The actual hash table. Contains indices to pEntries array. // The hash table can be seen as an array of singlt linked list. // We store indices to pEntries array instead of objects for performance // and to avoid pressure to the garbage collector. // An index -1 means null. private int[] pBuckets; // This array contains the memory for the nodes of the dictionary. private Entry[] pEntries; // This is the first node of a singly linked list of free nodes. // This data structure is called the FreeList and we use it to // reuse removed nodes instead of allocating new ones. private int pFirstFreeEntry; // Contains simply the number of items in this dictionary. private int pCount; // Contains the number of used entries (both in the dictionary or in the free list) in pEntries array. // This field is going only to grow with insertions. private int pEntriesCount; /// /// Creates a new FixedIntDictionary. /// tableBucketsCount should be a prime number /// greater than the number of items that this /// dictionary should store. /// The performance of this hash table will be very bad /// if you don\'t follow this rule! /// public FixedIntDictionary(int tableBucketsCount) { // Our free list is initially empty. this.pFirstFreeEntry = -1; // Initializes the entries array with a minimal amount of items. this.pEntries = new Entry[8]; // Allocate buckets and initial.

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.

This will need to be in a header file called LinkedList.hInser.pdfcleanhome88

This will need to be in a header file called "LinkedList.h" //Insertion void AddHead(const T& data); //(Create a new Node at the front of the list to store the passed in parameter.) void AddTail(const T& data); //(Create a new Node at the end of the list to store the passed in parameter.) void AddNodesHead(const T* data, unsigned int count); //Given an array of values, insert a node for each of those at the beginning of the list, maintaining the original order. void AddNodesTail(const T* data, unsigned int count); //except adding to the end of the list //Accessors unsigned int NodeCount() const; //Behaviors void PrintForward() const; //Iterator through all of the nodes and print out their values, one a time. void PrintReverse() const; //Exactly the same as PrintForward, except completely the opposite //Default Constructor LinkedList(); // //Destructor ~LinkedList(); //(Delete all the nodes created by the list.) main.cpp: #include #include #include #include "LinkedList.h" using namespace std; void TestAddHead(); void TestAddTail(); void TestAddingArrays(); int main() { int testNum; cin >> testNum; if (testNum == 1) TestAddHead(); else if (testNum == 2) TestAddTail(); else if (testNum == 3) TestAddingArrays(); return 0; } void TestAddHead() { cout << "=====Testing AddHead() functionality====" << endl; LinkedList data; for (int i = 0; i < 12; i += 2) data.AddHead(i); cout << "Node count: " << data.NodeCount() << endl; cout << "Print list forward:" << endl; data.PrintForward(); cout << "Print list in reverse:" << endl; data.PrintReverse(); } void TestAddTail() { cout << "=====Testing AddTail() functionality====" << endl; LinkedList data; for (int i = 0; i <= 21; i += 3) data.AddTail(i); cout << "Node count: " << data.NodeCount() << endl; cout << "Print list forward:" << endl; data.PrintForward(); cout << "Print list in reverse:" << endl; data.PrintReverse(); } void TestAddingArrays() { cout << "=====Testing AddNodesHead() and AddNodesTail() =====" << endl; string values[5]; values[0] = "*"; values[1] = "**"; values[2] = "***"; values[3] = "****"; values[4] = "*****"; LinkedList list; list.AddHead("**"); list.AddHead("***"); list.AddHead("****"); list.AddHead("*****"); list.AddHead("******"); list.AddHead("*******"); list.AddHead("********"); list.AddHead("*********"); list.AddHead("********"); list.AddHead("*******"); list.AddHead("******"); list.AddNodesHead(values, 5); list.AddNodesTail(values, 5); list.PrintForward(); }.

In this assignment you will implement insert() method for a singly l.pdffantasiatheoutofthef

In this assignment you will implement insert() method for a singly linked Link list. The code MyLinkedlist.java is given , it uses generic type as the linked list can be created with any object type. Some methods like add() and toString() are already implemented. The MyLinkedList class has a private nested class called Node. The objective of this assignment is to implement the insert() method that takes two inputs an element of generic type to be inserted and an index (value of integer type) at which it should be inserted. For this assignment we will assume the first element has the index 0. if the given index is less than zero do nothing but if it is greater than the size of the list, insert the element at the end of the list. if the index is between 0 and size , insert it in the appropriate location. You insert code should correctly update the theSize data member of the linked list. Return type for insert() method is boolean (true or false) depending on whether the insertion was successful or not. Use the following method header. } Main.java //add necessary libraries public class Main { //add your function here public static void main (String [] args) { } } MyLinkedList.java public class MyLinkedList { // A private nested Node class private static class Node{ public AnyType data; // Data members are public, this is fine as the Node class itself is private public Node next; public Node(AnyType d, Node n) { data = d; next=n; } @Override public String toString() { return data.toString(); } } // Data Members private Node head; // A node pointer to the first item in the List private int theSize; // a integer variable to hold the size of the List public MyLinkedList() { doClear(); } public void clear() { doClear(); } private void doClear() { head = null; theSize =0 ; } // An add method that will always add at the beginning of the List public boolean add(AnyType x) { if(head == null) // List is empty { head = new Node(x, null); theSize++; return true; } else // List has at least one item lets get to the end of it { Node newnode = new Node(x, head); head = newnode; theSize++; return true; } } @Override public String toString() { // This to String method will print the List from head to the end. String theList = ""; if(head==null) return theList; Node cur = head; while(cur.next!=null) { theList = theList + cur.toString() + " "; cur= cur.next; } return theList + cur.toString(); } public int getTheSize() { return theSize; } } /***************************************************************************** * * * SOLUTION CODE BELOW THIS HEADER * *****************************************************************************/ /***************************************************************************** * * * SOLUTION CODE ABOVE THIS HEADER * *****************************************************************************/ } Current file: Main.java Load default template... //add necessary Libraries public class Main \{ //add your function here public static vo.

in C++ , Design a linked list class named IntegerList to hold a seri.pdfeyewaregallery

in C++ , Design a linked list class named IntegerList to hold a series of integers. The class must have public member functions for appending , inserting, deleting and displaying nodes. The append function must append a node to the end of the list. The insert function must insert node in ascending order. The delete function must delete the node with the given value. The display function should display the list from head to tail. The IntegerList class should also have a destructor that destroys the list and release memory. The list node must be declared as a structure using keyword struct. This structure is a private member of IntegerList. You should also store the head of the list as a private member. Write a test program to test the implementation your IntegerList class works properly. Solution Here is the code for the above scenario: #include using namespace std; struct ListNode { float value; ListNode *next; }; ListNode *head; class LinkedList { public: int insertNode(float num); int appendNode(float num); void deleteNode(float num); void destroyList(); void displayList(); LinkedList(void) {head = NULL;} ~LinkedList(void) {destroyList();} }; int LinkedList::appendNode(float num) { ListNode *newNode, *nodePtr = head; newNode = new ListNode; if(newNode == NULL) { cout << \"Error allocating memory for new list member!\ \"; return 1; } newNode->value = num; newNode->next = NULL; if(head == NULL) { cout << \"List was empty - \" << newNode->value; cout << \" is part of list\'s first node.\ \"; head = newNode; } else { while(nodePtr->next != NULL) nodePtr = nodePtr->next; nodePtr->next = newNode; } return 0; } int LinkedList::insertNode(float num) { struct ListNode *newNode, *nodePtr = head, *prevNodePtr = NULL; newNode = new ListNode; if(newNode == NULL) { cout << \"Error allocating memory for new list member!\ \"; return 1; } newNode->value = num; newNode->next = NULL; if(head==NULL) { cout << \"List was empty - \" << newNode->value; cout << \" is part of list\'s first node.\ \"; head = newNode; } else { while((nodePtr != NULL) && (nodePtr->value < num)) { prevNodePtr = nodePtr; nodePtr = nodePtr->next; } if(prevNodePtr==NULL) newNode->next = head; else newNode->next = nodePtr; prevNodePtr->next = newNode; } return 0; } void LinkedList::deleteNode(float num) { struct ListNode *nodePtr = head, *prevNodePtr = NULL; if(head==NULL) { cout << \"The list was empty!\ \"; return; } if(head->value == num) { head = nodePtr->next; delete [] nodePtr; } else { while((nodePtr!= NULL)&&(nodePtr->value != num)) { prevNodePtr = nodePtr; nodePtr = nodePtr->next; } if(nodePtr==NULL) cout << \"The value \" << num << \" is not in this list!\ \"; else { prevNodePtr->next = nodePtr->next; delete [] nodePtr; } } } void LinkedList::destroyList() { struct ListNode *nodePtr = head, *nextNode = nodePtr; if(head==NULL) { cout << \"The list is empty!\ \"; return; } while (nodePtr != NULL) { nextNode = nodePtr->next; delete [] nodePtr; nodePtr = nextNode; } } void LinkedList.

I have been tasked to write a code for a Singly Linked list that inc.pdfarkmuzikllc

I have been tasked to write a code for a Singly Linked list that includes the following: Implement a singly linked list class with the following methods: - `void insertAtBeginning(int value)`: Inserts a new node with the given value at the beginning of the list. - `void insertAtEnd(int value)`: Inserts a new node with the given value at the end of the list. - `void deleteNode(int value)`: Deletes the first occurrence of a node with the given value. - `void display()`: Displays the elements of the linked list. I have everything EXCEPT for the deleteNode. I have it included, but it is throwing an error....I just need a little extra help to fix that part. My Code: # #include using namespace std; class Node { //user defined type. public: // public class modifier to access the class node. int value; Node* next; }; void display(Node *head) { if(head == NULL){ cout << "NULL" << endl; } else { cout << head->value << endl; display(head->next); } } void insertAtBeginning(Node**head, int newValue){ // Steps to add a Node to the beginning of the linked list //1. Prepare a newNode Node * newNode = new Node(); newNode ->value = newValue; //2. Put it in front of the current head newNode->next = *head; //3. Move head of the list to point to the newNode *head= newNode; } void insertAtEnd(Node**head, int newValue) { //1.Prepare a newNode Node* newNode= new Node(); newNode ->value = newValue; newNode->next = NULL; //2. If Linked list is empty, newNode will be a head node if(*head ==NULL){ *head = newNode; return; } //3. Find last node Node* last= *head; while(last->next!=NULL){ last=last->next; } //4. Insert newNode after last node (at the end of List) last->next =newNode; } void deleteNode(Node**before_del){ Node* temp; temp = before_del->next; before_del->next = temp->next; delete temp; } int main() { Node* head=new Node(); Node* second=new Node(); Node* third=new Node(); Node* fourth=new Node(); head->value = 7; head->next = second; second->value = 12; second->next = third; third->value = 15; third->next = fourth; fourth->value = 20; fourth->next = NULL; insertAtBeginning(&head, 4); insertAtBeginning(&head, 6); insertAtEnd(&head, 23); del(&head(15)->next); display(head); }.

C Exam Help Programming Exam Help

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 [email protected]. You can also call on +1 678 648 4277 for any assistance with the C Exam.

Program to insert in a sorted list #includestdio.h#include.pdfsudhirchourasia86

* Program to insert in a sorted list */ #include #include /* Link list node */ struct node { int data; struct node* next; }; /* function to insert a new_node in a list. Note that this function expects a pointer to head_ref as this can modify the head of the input linked list (similar to push())*/ void sortedInsert(struct node** head_ref, struct node* new_node) { struct node* current; /* Special case for the head end */ if (*head_ref == NULL || (*head_ref)->data >= new_node->data) { new_node->next = *head_ref; *head_ref = new_node; } else { /* Locate the node before the point of insertion */ current = *head_ref; while (current->next!=NULL && current->next->data < new_node->data) { current = current->next; } new_node->next = current->next; current->next = new_node; } } /* BELOW FUNCTIONS ARE JUST UTILITY TO TEST sortedInsert */ /* A utility function to create a new node */ struct node *newNode(int new_data) { /* allocate node */ struct node* new_node = (struct node*) malloc(sizeof(struct node)); /* put in the data */ new_node->data = new_data; new_node->next = NULL; return new_node; } /* Function to print linked list */ void printList(struct node *head) { struct node *temp = head; while(temp != NULL) { printf(\"%d \", temp->data); temp = temp->next; } } /* Drier program to test count function*/ int main() { /* Start with the empty list */ struct node* head = NULL; struct node *new_node = newNode(5); sortedInsert(&head, new_node); new_node = newNode(10); sortedInsert(&head, new_node); new_node = newNode(7); sortedInsert(&head, new_node); new_node = newNode(3); sortedInsert(&head, new_node); new_node = newNode(1); sortedInsert(&head, new_node); new_node = newNode(9); sortedInsert(&head, new_node); printf(\"\ Created Linked List\ \"); printList(head); return 0; } public class Listnode { //*** fields *** private Object data; private Listnode next; //*** methods *** // 2 constructors public Listnode(Object d) { this(d, null); } public Listnode(Object d, Listnode n) { data = d; next = n; } // access to fields public Object getData() { return data; } public Listnode getNext() { return next; } // modify fields public void setData(Object ob) { data = ob; } public void setNext(Listnode n) { next = n; } } Thsi below program print out all the varibles in SLL #include #include #include /* Link list node */ struct node { int data; struct node* next; }; /* Given a reference (pointer to pointer) to the head of a list and an int, push a new node on the front of the list. */ void push(struct node** head_ref, int new_data) { /* allocate node */ struct node* new_node = (struct node*) malloc(sizeof(struct node)); /* put in the data */ new_node->data = new_data; /* link the old list off the new node */ new_node->next = (*head_ref); /* move the head to point to the new node */ (*head_ref) = new_node; } /* Takes head pointer of the linked list and index as arguments and return data at index*/ int GetNth(struct node* head, int index) { struct node* current = head; int .

Write the following using javaGiven a class ‘Node’ and ‘NodeList’,.pdffathimalinks

Write the following using java Given a class ‘Node’ and ‘NodeList’, that contains the below information diagram. -id: int -name: String -next: Node +Node(id: int, name: String) +setId(id: int): void +getId(): int +setName(name: String) : void +getName(): String +setNext(node: Node): void +getNext(): Node NodeList -size: int -root: Node +add(node: Node): void +size(): int +findNode(node: Node): boolean Implement add(Node), findNode(Node) and size methods in the NodeList class which is provided to you. The methods should work as: Using the following main method: public static void main(String[] args) { NodeList list = new NodeList(); Node node = new Node(1, \"Book\"); Node node2 = new Node(2, \"Lappy\"); list.add(node); list.add(node2); System.out.println(\"Length : \"+list.size()); Node node3 = new Node(3, \"Glass\"); Node node4 = new Node(4, \"Pen\"); list.add(node3); System.out.println(\"Length : \"+list.size()); if(list.findNode(node3)) System.out.println(\"Node found: \"+ node3.getName()); else System.out.println(\"Node not found: \"+ node3.getName()); if(list.findNode(node4)) System.out.println(\"Node found: \"+ node4.getName()); else System.out.println(\"Node not found: \"+ node4.getName()); } Then it should return the following output: Length : 2 Length : 3 Node found: Glass Node not found: Pen The given Node class is: public class Node { private int id = 0; private String name = \"\"; private Node next; public Node(int id, String name) { this.id = id; this.name = name; this.next = null; } public Node getNext() { return next; } public void setNext(Node node) { this.next = node } public int getId() { return id; { public void setId(int id) { this.id = id; } public String getName() { return name; } public void setName(String name) { this.name = name; } public String toString() { return \"ID : \"+this.id+\" Name : \"+this.name; } } The given NodeList class is: public class NodeList { private int size = 0; private Node root = null; /* * It has to take a new Node and add that to the next address of previous Node. * If the list is empty, assign it as the \"root\" * @Param - Node */ public void add(Node node) { // Implement this method!!! } /* * It has to return the size of the NodeList * * @return size */ public int size() { // Implement this method!!! } /* * It has to take a Node and checks if the node is in the list. * If it finds the node, it returns true, otherwise false * * @param - Node * @return boolean true/false */ public boolean findNode(Node node) { // Implement this method!!! } }Node -id: int -name: String -next: Node +Node(id: int, name: String) +setId(id: int): void +getId(): int +setName(name: String) : void +getName(): String +setNext(node: Node): void +getNext(): Node Solution Hi, Please find my implementation. Please let me know in case of any issue. ############ Node.java ############# public class Node { private int id = 0; private String name = \"\"; private Node next; public Node(int id, String name) { this.id = id; this..

3.linked listChandan Singh

This document discusses linked lists as a solution to problems with arrays. It provides definitions and examples of linked lists, including: - Each node contains a data element and a pointer to the next node - A linked list is a series of connected nodes with a head pointer pointing to the first node - Operations like insertion, deletion, and searching have linear time complexity of O(n) in the worst case - Variations include circular and doubly linked lists

could you implement this function please, im having issues with it..pdfferoz544

could you implement this function please, im having issues with it. void makeList (const ListNode::value_type [],const size_t& count) class ListNode { public: typedef int value_type; ListNode (value_type d = value_type(), ListNode* n = NULL) { datum = d; next = n; } //Assessor value_type getDatum () const { return datum; } ListNode const* getNext () const { return next; } //Mutator void setDatum (const value_type& d) {datum = d; } ListNode* getNext () { return next; } void setNext (ListNode* new_link) {next = new_link; } private: value_type datum; ListNode* next; }; class LinkedList { public: LinkedList (); virtual ~LinkedList (); void insertItem (ListNode::value_type); void makeList (const ListNode::value_type [],const size_t& count); void deleteList (); //The following friend function is implemented in lablinklist.cpp friend std::ostream& operator<<(std::ostream&, const LinkedList&); private: ListNode* head; }; This is the pseudocode, but i still have a hard time undertanding it. Creating a List (makeList(const ListNode::value_type [],const size_t& count)) This function receives an array in the order that we want to add it to the linkedlist. Index 0 will be the head node, index n will be the last node. First, create a node initialized with a data value and a NULL pointer. Set the \"head\" to point to the first node. Set up a current-pointer to the first node (or \"head\"). Get a data value for the next node. While more nodes to add { Create a new node initialized with the data value and a NULL pointer. Set the current-pointer link member (\"next\") to the new node. Set the current-pointer to point to the new node. Get a data value for the next node. } Thanks. Solution //linkedList.h #ifndef LINKEDLIST_H #define LINKEDLIST_H #include #include class ListNode { public: typedef int value_type; ListNode(value_type d = value_type(), ListNode* n = NULL) { datum = d; next = n; } //Assessor value_type getDatum() const { return datum; } ListNode const* getNext() const { return next; } //Mutator void setDatum(const value_type& d) { datum = d; } ListNode* getNext() { return next; } void setNext(ListNode* new_link) { next = new_link; } private: value_type datum; ListNode* next; }; class LinkedList { public: LinkedList(); virtual ~LinkedList(); void insertItem(ListNode::value_type); void makeList(const ListNode::value_type[], const size_t& count); void deleteList(); //The following friend function is implemented in lablinklist.cpp friend std::ostream& operator<<(std::ostream&, const LinkedList&); private: ListNode* head; }; #endif ------------------------------------------------------------- //linkedList.cpp #include \"linkedlist.h\" LinkedList::LinkedList() { head = NULL; } LinkedList::~LinkedList() { ListNode *tmp = head; while (tmp != NULL) { head = head->getNext(); delete tmp; tmp = head; } } void LinkedList::insertItem(ListNode::value_type item) { ListNode *newNode,*cur = head; newNode = new ListNode; newNode->setDatum(item); newNode->setNext(NULL);.

The hashtable youll be making will use Strings as the keys and Obje.pdfvicky309441

The hashtable you'll be making will use Strings as the keys and Object as the values. Similar to linked lists, by storing Object as values, you can store any kind of object in the hashtable. package Dictionary; import java.util.ArrayList; import java.util.Hashtable; import List.ListInterface; import List.MyLinkedList; public class MyHashtable implements DictionaryInterface { protected int tableSize; protected int size; // The LinkedList is of type Entry protected MyLinkedList[] table; public MyHashtable(int tableSize) { table = new MyLinkedList[tableSize]; this.tableSize = tableSize; } public int biggestBucket() { int biggestBucket = 0; for(int i = 0; i < table.length; i++) { // Loop through the table looking for non-null locations. if (table[i] != null) { // If you find a non-null location, compare the bucket size against the largest // bucket size found so far. If the current bucket size is bigger, set biggestBucket // to this new size. MyLinkedList bucket = table[i]; if (biggestBucket < bucket.size()) biggestBucket = bucket.size(); } } return biggestBucket; // Return the size of the biggest bucket found. } // Returns the average bucket length. Gives a sense of how many collisions are happening overall. public float averageBucket() { float bucketCount = 0; // Number of buckets (non-null table locations) float bucketSizeSum = 0; // Sum of the size of all buckets for(int i = 0; i < table.length; i++) { // Loop through the table if (table[i] != null) { // For a non-null location, increment the bucketCount and add to the bucketSizeSum MyLinkedList bucket = table[i]; bucketSizeSum += bucket.size(); bucketCount++; } } // Divide bucketSizeSum by the number of buckets to get an average bucket length. return bucketSizeSum/bucketCount; } public String toString() { String s = ""; for(int tableIndex = 0; tableIndex < tableSize; tableIndex++) { if (table[tableIndex] != null) { MyLinkedList bucket = table[tableIndex]; for(int listIndex = 0; listIndex < bucket.size(); listIndex++) { Entry e = (Entry)bucket.get(listIndex); s = s + "key: " + e.key + ", value: " + e.value + "\n"; } } } return s; } protected class Entry { String key; Object value; Entry(String key, Object value) { this.key = key; this.value = value; } } // Implement these methods public boolean isEmpty() {return true;} // returns true if the Dictionary is empty, false otherwise. public int size(){return -1;} //Returns the number of key/value pairs stored in the dictionary. // Adds a value stored under the given key. If the key has already been stored in the Dictionary, // replaces the value associated with the key and returns the old value. If the key isn't in the dictionary // returns null. public Object put(String key, Object value){ // 1. Compute an array index given the key // 2. If that location in the table is null, // that means nothing has been previously stored using a key with this hash code. // a. Create a new MyLinkedList to be the bucket. // b. Add the new Entry for the key/value.

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; }.

#include stdafx.h #include iostream using namespace std;vo.docxajoy21

This document contains C++ code to implement a linked list program that performs the following tasks: 1. Inputs 10 integers into a linked list while sorting them as they are inserted 2. Prints the sorted list and computes the average of the values 3. Deletes nodes from the list with values less than the average 4. Creates a second list of 10 sorted integers 5. Merges the two lists together, eliminating duplicates, and prints the final list

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.

Exception to indicate that Singly LinkedList is empty. .pdfaravlitraders2012

/** * Exception to indicate that Singly LinkedList is empty. */ class LinkedListEmptyException extends RuntimeException { public LinkedListEmptyException() { super(); } public LinkedListEmptyException(String message) { super(message); } } /** * Node class, which holds data and contains next which points to next Node. */ class Node { public int data; // data in Node. public Node next; // points to next Node in list. /** * Constructor */ public Node(int data) { this.data = data; } /** * Display Node\'s data */ public void displayNode() { System.out.print(data + \" \"); } } /** * Singly LinkedList class */ class LinkedList { private Node first; // ref to first link on list /** * LinkedList constructor */ public LinkedList() { first = null; } /** * Insert New Node at first position */ public void insertFirst(int data) { Node newNode = new Node(data); // Creation of New Node. newNode.next = first; // newLink ---> old first first = newNode; // first ---> newNode } /** * Method deletes specific Node from Singly LinkedList in java. */ public Node deleteSpecificNode(int deleteKey) { // Case1: when there is no element in LinkedList if (first == null) { // means LinkedList in empty, throw exception. throw new LinkedListEmptyException( \"LinkedList doesn\'t contain any Nodes.\"); } // Case2: when there is only one element in LinkedList- check whether we // have to delete that Node or not. if (first.data == deleteKey) { // means LinkedList consists of only one // element, delete that. Node tempNode = first; // save reference to first Node in tempNode- // so that we could return saved reference. first = first.next; System.out.println(\"Node with data=\" + tempNode.data + \" was found on first and has been deleted.\"); return tempNode; // return deleted Node. } // Case3: when there are atLeast two elements in LinkedList Node previous = null; Node current = first; while (current != null) { if (current.data == deleteKey) { System.out.println(\"Node with data=\" + current.data + \" has been deleted.\"); previous.next = current.next; // make previous node\'s next point // to current node\'s next. return current; // return deleted Node. } else { if (current.next == null) { // Means Node wasn\'t found. System.out.println(\"Node with data=\" + deleteKey + \" wasn\'t found for deletion.\"); return null; } previous = current; current = current.next; // move to next node. } } return null; } /** * Display Singly LinkedList */ public void displayLinkedList() { System.out.print(\"Displaying LinkedList [first--->last]: \"); Node tempDisplay = first; // start at the beginning of linkedList while (tempDisplay != null) { // Executes until we don\'t find end of // list. tempDisplay.displayNode(); tempDisplay = tempDisplay.next; // move to next Node System.out.print(\"-->\"); } System.out.println(); } } /** * Main class - To test LinkedList. */ public class SinglyLinkedListDeleteNodeExample { public static void main(String[] args) { LinkedList linkedList = new LinkedList(); /.

Electronegativity - Electronegativity is an atom.pdfaravlitraders2012

Electronegativity - Electronegativity is an atom\'s \'desire\' to grab another atom\'s electrons. In terms of ionization energy \"noble gas status\" is the golden standard. Elements in group one only need to give up one electron to reach that status. For this reason they are highly reactive and want go give up an electron to some other atom. Group 1 elements are the most reactive followed by group 2. In order of increasing reactivity: Calcium, Magnesium, Sodium, Potassium Solution Electronegativity - Electronegativity is an atom\'s \'desire\' to grab another atom\'s electrons. In terms of ionization energy \"noble gas status\" is the golden standard. Elements in group one only need to give up one electron to reach that status. For this reason they are highly reactive and want go give up an electron to some other atom. Group 1 elements are the most reactive followed by group 2. In order of increasing reactivity: Calcium, Magnesium, Sodium, Potassium.

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Program to insert in a sorted list #includestdio.h#include.pdfsudhirchourasia86

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could you implement this function please, im having issues with it..pdfferoz544

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Exception to indicate that Singly LinkedList is empty. .pdfaravlitraders2012

Electronegativity - Electronegativity is an atom.pdfaravlitraders2012

Sequence.h#ifndef MAIN #define MAIN #include cstdlibclass .pdfaravlitraders2012

Sequence.h #ifndef MAIN #define MAIN #include class Sequence { public: //TYPEDEFS and MEMBER CONSTANTS typedef double value; typedef std::size_t size_type; static const size_type DEFAULT_CAPACITY = 30; // CONSTRUCTORS and DESTRUCTOR Sequence(size_type initial_capacity = DEFAULT_CAPACITY); //constructor Sequence(const Sequence& start); // COPY CONSTRUCTOR ~Sequence( ); // DESTRUCTOR // MODIFICATION MEMBER FUNCTIONS void resize(size_type new_capacity); void start( ); void advance( ); void insert(const value& entry); void attach(const value& entry); void remove_current( ); void operator =(const Sequence& start); // CONSTANT MEMBER FUNCTIONS size_type size( ) const; bool is_item( ) const; value current( ) const; bool operator ==(const Sequence& other); bool operator !=(const Sequence& other); private: void shift_down(size_type index); void shift_up(size_type index); value* data; size_type used; size_type current_index; size_type capacity; }; #endif Sequence.cpp #include #include #include \"Sequence.h\" using namespace std; Sequence::Sequence(size_type initial_capacity) { data= new value [initial_capacity]; current_index=0; capacity=initial_capacity; used=0; } Sequence::Sequence(const Sequence& start) { data=new value[start.capacity]; capacity=start.capacity; current_index=start.current_index; used=start.used; copy(start.data,start.data+used,data); } Sequence::~Sequence( ) { delete [] data;//it releases dynamic memory afterwards. } void Sequence::resize(size_type new_capacity) { assert(new_capacity > size ()); // capacity is less than the number of items data=new double [size()]; delete [] data; double *tempdata;// temporary array tempdata=new double[size()]; for (unsigned int i=0; icurrent_index;i--) data[i]=data[i-1]; data[current_index]=entry; used++; } void Sequence::shift_down(size_type index) { } void Sequence::attach(const value& entry) { if(used == capacity) resize(used+1); if(current_index >= used) { start(); advance(); advance(); } if(used!=0) { for(size_type i = used; i > current_index+1; i--) { data[i] = data[i-1]; } data[current_index+1] = entry; current_index++; } data[current_index] = entry; ++used; } void Sequence::shift_up(size_type index) { } void Sequence::remove_current( ) { assert(is_item()==true); for (unsigned int i=0; i #include \"Sequence.h\" using namespace std; int main ( ) { Sequence s; s.insert(1); s.insert(2); s.insert(3); s.insert(4); s.insert(5); s.start( ); cout << s.current( ) << endl; system (\"pause\"); return 0; } Solution Sequence.h #ifndef MAIN #define MAIN #include class Sequence { public: //TYPEDEFS and MEMBER CONSTANTS typedef double value; typedef std::size_t size_type; static const size_type DEFAULT_CAPACITY = 30; // CONSTRUCTORS and DESTRUCTOR Sequence(size_type initial_capacity = DEFAULT_CAPACITY); //constructor Sequence(const Sequence& start); // COPY CONSTRUCTOR ~Sequence( ); // DESTRUCTOR // MODIFICATION MEMBER FUNCTIONS void resize(size_type new_capacity); void start( ); void advance( ); void inser.

In the 1st Question, As the polymorphisms occur in the non-coding se.pdfaravlitraders2012

In the 1st Question, As the polymorphisms occur in the non-coding sequence, thus, they does not contribute to phenotypic variations. As the polymorphisms are also neutral, thus, they are selectively neutral because during the course of evolution, the non-coding sequences are conserved and any polymorphisms in them become selectively neutral as they do not contribution to phenotypic variations. Hence, the answer is b. In the 2nd Question, according to the given data, at all loci the same polymorphism was found in nearly all of the tested species from lake victoria both lacustrine and riverine but different polymorphims at these loci were found in cichlids at other african lakes. This suggests that all african lakes cichlids are related by recent ancestry because the genetic loci where these polymorphism occur remains same and also that lake victoria was repopulated by a relatively small sample of cichlids because lake victoria cichlids have the same polymorphism at all the loci in the tested species. Hence, the answer is a and d (both apply). Solution In the 1st Question, As the polymorphisms occur in the non-coding sequence, thus, they does not contribute to phenotypic variations. As the polymorphisms are also neutral, thus, they are selectively neutral because during the course of evolution, the non-coding sequences are conserved and any polymorphisms in them become selectively neutral as they do not contribution to phenotypic variations. Hence, the answer is b. In the 2nd Question, according to the given data, at all loci the same polymorphism was found in nearly all of the tested species from lake victoria both lacustrine and riverine but different polymorphims at these loci were found in cichlids at other african lakes. This suggests that all african lakes cichlids are related by recent ancestry because the genetic loci where these polymorphism occur remains same and also that lake victoria was repopulated by a relatively small sample of cichlids because lake victoria cichlids have the same polymorphism at all the loci in the tested species. Hence, the answer is a and d (both apply)..

I noticed that temperature is not mentioned. A) .pdfaravlitraders2012

I noticed that temperature is not mentioned. A) Entropy of fusion = heat of fusion / melting temperature = (10.59kJ/mol) / melting temperature B) Sicne, fusion is a reversible process, delS (surroundings) + delS (system) = 0 so delS (surroundings) = -delS (system) = - (10.59kJ/mol) / melting temperature Solution I noticed that temperature is not mentioned. A) Entropy of fusion = heat of fusion / melting temperature = (10.59kJ/mol) / melting temperature B) Sicne, fusion is a reversible process, delS (surroundings) + delS (system) = 0 so delS (surroundings) = -delS (system) = - (10.59kJ/mol) / melting temperature.

1. Briefly describe the structured design approach and the object-or.pdfaravlitraders2012

1. Briefly describe the structured design approach and the object-oriented design. structured desing: 1)here it will divide the problem into small subparts or subproblems 2)it is also known as top-down approach 3)it will do step by step refinement and do basic programming 4)here the problems are divided in to smaller pieces and find the solution for each problem and combine them together to solve entire problem object oriented programming: 1)here each component is detrmined as an object 2)the main task is to fiding objects and doing required operations on them 3)each objects which change it state while doing operations 4)by combining and communicating with all required objects get an overall solution to the problem 2. How are these approaches similar? How do they differ? sol: one of the similar thing is both will divide problem in to small thing whiel structured desing will divided into sub problems and oop is divided in to objects 1)sturtured programming will save time instead of writing complex function in oop 2)as they are small program they are very easy to maintain by oop are not 3)structured programming are easy to understand 3. Which approach do you feel is most beneficial? Why? sol:i feel obejct oriented desing is more comfartable and benificial than structed desing ,because in object oriented desing every real world problem in to model of objects and by doing it will become easier to solve the problem in natural way Solution 1. Briefly describe the structured design approach and the object-oriented design. structured desing: 1)here it will divide the problem into small subparts or subproblems 2)it is also known as top-down approach 3)it will do step by step refinement and do basic programming 4)here the problems are divided in to smaller pieces and find the solution for each problem and combine them together to solve entire problem object oriented programming: 1)here each component is detrmined as an object 2)the main task is to fiding objects and doing required operations on them 3)each objects which change it state while doing operations 4)by combining and communicating with all required objects get an overall solution to the problem 2. How are these approaches similar? How do they differ? sol: one of the similar thing is both will divide problem in to small thing whiel structured desing will divided into sub problems and oop is divided in to objects 1)sturtured programming will save time instead of writing complex function in oop 2)as they are small program they are very easy to maintain by oop are not 3)structured programming are easy to understand 3. Which approach do you feel is most beneficial? Why? sol:i feel obejct oriented desing is more comfartable and benificial than structed desing ,because in object oriented desing every real world problem in to model of objects and by doing it will become easier to solve the problem in natural way.

1) Protocols are needed so your computer can interact with other com.pdfaravlitraders2012

1) Protocols are needed so your computer can interact with other computers for things such as using the internet and sending and receiving files and emails. Protocols are the \"language\" computers use to talk to each other. Standards are important so all networks have to follow the same rules. Standards are a set of guidelines on how different companies need to make their equipment. 2) a protocol is the special set of rules that end points in a telecommunication connection use when they communicate. Protocols specify interactions between the communicating entities. Protocols exist at several levels in a telecommunication connection. For example, there are protocols for the data interchange at the hardware device level and protocols for data interchange at the application program level. In the standard model known as Open Systems Interconnection (OSI), there are one or more protocols at each layer in the telecommunication exchange that both ends of the exchange must recognize and observe. Transmission Control Protocol (TCP), which uses a set of rules to exchange messages with other Internet points at the information packet level Internet Protocol (IP), which uses a set of rules to send and receive messages at the Internet address level Others are Hypertext Transfer Protocol (HTTP) and File Transfer Protocol (FTP), each with defined sets of rules to use with corresponding programs elsewhere on the Internet 3) Standards play an important role in networking. Without standards, manufacturers of networking products have no common ground on which to build their systems. Interconnecting products from various vendors would be difficult, if not impossible. Standardization can make or break networking products. These days many vendors are hesitant to support new technology unless there is a standardization base from which to work. Vendors want to know there will be some measure of interoperability for their hardware and software. Examples : IEEE 802.1 (LAN/MAN) IEEE 802.3 (Ethernet) IEEE 802.5 (Token Ring) IEEE 802.11 (Wireless LAN) 4) IEEE 802 group of standards https://en.m.wikipedia.org/wiki/IEEE_802 5) File transfer protocol ( ftp) https://en.m.wikipedia.org/wiki/File_Transfer_Protocol Solution 1) Protocols are needed so your computer can interact with other computers for things such as using the internet and sending and receiving files and emails. Protocols are the \"language\" computers use to talk to each other. Standards are important so all networks have to follow the same rules. Standards are a set of guidelines on how different companies need to make their equipment. 2) a protocol is the special set of rules that end points in a telecommunication connection use when they communicate. Protocols specify interactions between the communicating entities. Protocols exist at several levels in a telecommunication connection. For example, there are protocols for the data interchange at the hardware device level and protocols for data interchange at t.

argon is inert gas , doesnot chemically reacts wi.pdfaravlitraders2012

#include sstream #include linkylist.h #include iostream.pdfaravlitraders2012

#include #include \"linkylist.h\" #include #include #include #include using namespace std; int main(int argc, char *argv[]) { //node* head = new node; LinkedList list(1); ifstream infile1 (argv[1]); //make sure the file opens properly and read the contents of the file in to a list if(!infile1) { cout<<\"Your Testfile could not be opened!\"<>users; if (users.find(\"I\")!=string::npos) { //int spot = atoi((users[3]).c_str()); string pot = string(users.substr(2)); stringstream spot(pot); string data; cin>>data; insertAfter(list.getNode(spot),data); } /*else if (users.find(\"D\")!=string::npos) { int spoty = atoi((users[3]).c_str()); deleteNode(spoty); }*/ else if (users==\"L\") { list.traverse_and_print(); } } return 0; #include using namespace std; class Node { friend class LinkedList; private: int value; /* data, can be any data type, but use integer for easiness */ Node *Next; /* pointer to the next node */ public: /* Constructors with No Arguments */ Node(void) : Next(NULL) { } /* Constructors with a given value */ Node(int val) : value(val), Next(NULL) { } /* Constructors with a given value and a link of the next node */ Node(int val, Node* next) : value(val), Next(next) {} /* Getters */ int getValue(void) { return value; } Node* getNext(void) { return Next; } }; /* definition of the linked list class */ class LinkedList { private: /* pointer of head node */ Node *Head; /* pointer of tail node */ Node *Tail; public: /* Constructors with a given value of a list node */ LinkedList(int val); /* Destructor */ ~LinkedList(void); /* Function to append a node to the end of a linked list */ void tailAppend(string val); /* Remove a node with a specific value if it exists */ void remove(int val); void insertAfter(Node* afterMe, string val); Node* getNode(int pos); /* Traversing the list and printing the value of each node */ void traverse_and_print(); }; #include #include \"linkylist.h\" #include using namespace std; LinkedList::LinkedList(int val) { /* Create a new node, acting as both the head and tail node */ Head = new Node(val); Tail = Head; } Node* LinkedList::getNode(int pos) { Node* p = Head; /* Counter */ while (pos--) { if (p != NULL) p = p->Next; else return NULL; } return p; } void LinkedList::insertAfter(Node* afterMe, string val) { if(afterme != NULL) { afterMe->Next = new Node(val, afterMe->Next); } } LinkedList::~LinkedList() { } void LinkedList::tailAppend(string val) { /* The list is empty? */ if (Head == NULL) { /* the same to create a new list with a given value */ Tail = Head = new Node(val); } else { /* Append the new node to the tail */ Tail->Next = new Node(val); /* Update the tail pointer */ Tail = Tail->Next; } } void LinkedList::remove(int val) { Node *pPre = NULL, *pDel = NULL; /* Check whether it is the head node? if it is, delete and update the head node */ if (Head->value == val) { /* point to the node to be deleted */ pDel = Head; /* update */ Head = pDel->Next; delete pDel; return; } pPre = Head; pDel = Head->Next; /*.

Function header If a program contain a function definition, that .pdfaravlitraders2012

Function header : If a program contain a function definition, that function defnition consist of a Function header write by taking pointers or in the absense of pointers. Example : A function header look like that includes pointers. void (*current_menu1)(int) // here current_menu1 is a pointer void (*payment_details)(int); // here payment_details is a pointer void (*payments_receipts)(double ); / /payments_receipts is a pointer Solution Function header : If a program contain a function definition, that function defnition consist of a Function header write by taking pointers or in the absense of pointers. Example : A function header look like that includes pointers. void (*current_menu1)(int) // here current_menu1 is a pointer void (*payment_details)(int); // here payment_details is a pointer void (*payments_receipts)(double ); / /payments_receipts is a pointer.

GameOfLife.cs using System; using System.Collections.Generic;.pdfaravlitraders2012

GameOfLife.cs using System; using System.Collections.Generic; using System.Linq; using System.Text; using System.Threading; namespace A03LeeJoe { class GameOfLife { private int[, ,] grid; private int displayedGrid = 0; private int tempGrid = 1; private const int DEAD = 0; private const int ALIVE = 1; private const int TOP = 5; private const int LEFT = 20; //constructor public GameOfLife() { grid = CreateGrid(); } //methods private int[, ,] CreateGrid() { int[, ,] newGrid = new int[25, 40, 2]; for (int col = 14; col < 27; col++) { newGrid[13, col, 0] = ALIVE; } for (int row = 8; row < 11; row++) { newGrid[row, 12, 0] = ALIVE; newGrid[row, 20, 0] = ALIVE; newGrid[row, 28, 0] = ALIVE; } return newGrid; } public void Start() { //Display Grid Console.SetWindowSize(80, 40); Setup(); //loop until user presses a key while (!Console.KeyAvailable) { Thread.Sleep(1000); //perform next move NextMove(); DisplayGrid(); SetColor(1); Console.WriteLine(); Console.CursorLeft = LEFT; Console.WriteLine(\" Press any key to end \", displayedGrid); } } private void Setup() { CreateHeader(); DisplayGrid(); //prompt user to press enter Console.WriteLine(); Console.CursorLeft = LEFT; SetColor(1); Console.WriteLine(\" Please press enter to begin \"); Console.ReadLine(); } private void CreateHeader() { Console.SetCursorPosition(LEFT, TOP); SetColor(1); Console.WriteLine(\"{0,40}\", \" \"); Console.CursorLeft = LEFT; Console.WriteLine(\" G A M E O F L I F E \"); Console.CursorLeft = LEFT; Console.WriteLine(\"{0,40}\", \" \"); } private void NextMove() { //iterate through grid for (int row = 0; row <= grid.GetUpperBound(0); row++) { for (int col = 0; col <= grid.GetUpperBound(1); col++) { //each iteration check neighbor count int neighborCount = GetNeighborCount(row, col); //determine new life or death and update on temp grid\' grid[row, col, tempGrid] = UpdateCell(grid[row, col, displayedGrid], neighborCount); } } //switch the grid\'s 3rd dimension to display new grid SwitchGrid(); } private void SwitchGrid() { if (displayedGrid == 0) { displayedGrid = 1; tempGrid = 0; } else { displayedGrid = 0; tempGrid = 1; } } private int UpdateCell(int stateOfCell, int neighborCount) { if (stateOfCell == ALIVE) { switch (neighborCount) { case 0: case 1: return DEAD; case 2: case 3: return ALIVE; default: return DEAD; } } else if (neighborCount == 3) { return ALIVE; } return stateOfCell; } private void DisplayGrid() { Console.SetCursorPosition(LEFT, TOP + 4); for (int row = 0; row <= grid.GetUpperBound(0); row++) { Console.CursorLeft = LEFT; for (int col = 0; col <= grid.GetUpperBound(1); col++) { SetColor(grid[row, col, displayedGrid]); Console.Write(\".\"); } Console.WriteLine(); } } private void SetColor(int state) { if (state == 1) { Console.BackgroundColor = ConsoleColor.Red; Console.ForegroundColor = ConsoleColor.Yellow; } else { Console.BackgroundColor = ConsoleColor.Blue; Console.ForegroundColor = ConsoleColor.Black; } } private int GetNeighborCount(int row, int col) { int ne.

With the vast increase in technology, the number of ways that the us.pdfaravlitraders2012

With the vast increase in technology, the number of ways that the use of cybertechnology impact the personalities are also increasing. Many of the social scientists has proofs for this. One incident, in the 1980s, that quickly caught their attention involved a male psychologist who joined an online forum for disabled persons, where he identified himself as a woman who had become crippled as a result of an automobile accident. Mitch Parsell (2008) describes the Internet as a \"powerful new force\" for what he calls \"the manufacture of identity.\" He believes that this technological medium offers an unparalleled ability to create ourselves in our own image. It gives users an unprecedented capacity to determine their initial presentations to others.In short, it enables users to be masters of their identity. Now when we talk about Ethics and Technology: Controversies, Questions, and Strategies for Ethical Computing: Cybertechnology as a \"Medium of Self-Expression\" Sherry Turkle (2005) notes that by the mid-1980s, computers had become an \"evocative object\"—i.e., an object that \"provoked self- reflection.\" Turkle suggested that a computer could be viewed as a model for analyzing and constructing one\'s identity. She also noted that the computer has become a medium through which people could discover their personal identity; for example, computers provide a context in which individuals can try out different cognitive styles and different methods of problem solving to ultimately discover which style or method they prefer. Because people develop their own unique style of computing, Turkle argues that the computational environment becomes an extension of themselves, in much the same way that their manner of dress is an extension of their personality. So, in her view, a computer can function as a \"medium for self-expression\" as well as for \"self-discovery. Impact of cybertechnology on Sense of Self We have examined some effects that one\'s interactions in virtual or computer-mediated environments, including MOOs and MUDs, can have for one\'s personal identity. 2 main factors are: our relation to nature, our relation to (and sense of place in) the universe. Williams notes that in the aftermath of these milestones, we have a much humbler view of our place in the universe than our ancestors did. How, exactly, has cybertechnology also has influenced the way that we see ourselves in relation to both the factors—i.e., how has this relatively recent technology already begun to define us as human beings? To support Bolter\'s thesis that we have come to see ourselves more and more in computer-like ways, we have only to reflect for a moment on some of the expressions that we now use to describe ourselves. For example, Bolter points out that when psychologists speak of \"input and output states of the brain,\" or of the brain\'s hardware and software, they exemplify Turing\'s men. And when cognitive psychologists study the \"mind\'s algorithm for searchi.

What is the need of the t Distribution According to the central.pdfaravlitraders2012

What is the need of the t Distribution? According to the central limit theorem, the sampling distribution of a statistic (like a sample mean) will follow a normal distribution, as long as the sample size is sufficiently large. Therefore, when we know the standard deviation of the population, we can compute a z-score, and use the normal distribution to evaluate probabilities with the sample mean. But sample sizes are sometimes small, and often we do not know the standard deviation of the population. When either of these problems occur, statisticians rely on the distribution of the t statistic (also known as the t score), whose values are given by: t = [ x - ? ] / [ s / sqrt( n ) ] where x is the sample mean, ? is the population mean, s is the standard deviation of the sample, and n is the sample size. The distribution of the t statistic is called the t distribution or the Student t distribution. The t distribution allows us to conduct statistical analyses on certain data sets that are not appropriate for analysis, using the normal distribution. Degrees of Freedom There are actually many different t distributions. The particular form of the t distribution is determined by its degrees of freedom. The degrees of freedom refers to the number of independent observations in a set of data. When estimating a mean score or a proportion from a single sample, the number of independent observations is equal to the sample size minus one. Hence, the distribution of the t statistic from samples of size 8 would be described by a t distribution having 8 - 1 or 7 degrees of freedom. Similarly, a t distribution having 15 degrees of freedom would be used with a sample of size 16. Solution What is the need of the t Distribution? According to the central limit theorem, the sampling distribution of a statistic (like a sample mean) will follow a normal distribution, as long as the sample size is sufficiently large. Therefore, when we know the standard deviation of the population, we can compute a z-score, and use the normal distribution to evaluate probabilities with the sample mean. But sample sizes are sometimes small, and often we do not know the standard deviation of the population. When either of these problems occur, statisticians rely on the distribution of the t statistic (also known as the t score), whose values are given by: t = [ x - ? ] / [ s / sqrt( n ) ] where x is the sample mean, ? is the population mean, s is the standard deviation of the sample, and n is the sample size. The distribution of the t statistic is called the t distribution or the Student t distribution. The t distribution allows us to conduct statistical analyses on certain data sets that are not appropriate for analysis, using the normal distribution. Degrees of Freedom There are actually many different t distributions. The particular form of the t distribution is determined by its degrees of freedom. The degrees of freedom refers to the number of independent observations in a set of da.

We can see in course of evolution of plant there was a transition fr.pdfaravlitraders2012

We can see in course of evolution of plant there was a transition from aquatic to terrestrial life has taken place.They have adapted lots of features to servive on land. 1.Thay became so large comparative ao aquatic plants.This feature evolve to colonise the land. 2.They have developed well organised vascular system allowing plant to absorb water from soil. 3 .They have evolved strong roots to adhere by the substratum.They can go few meters inside the soil. Solution We can see in course of evolution of plant there was a transition from aquatic to terrestrial life has taken place.They have adapted lots of features to servive on land. 1.Thay became so large comparative ao aquatic plants.This feature evolve to colonise the land. 2.They have developed well organised vascular system allowing plant to absorb water from soil. 3 .They have evolved strong roots to adhere by the substratum.They can go few meters inside the soil..

Wave CharacterMany of the things that light does are only expla.pdfaravlitraders2012

>> Wave Character Many of the things that light does are only explained sufficiently by thinking of light as a wave. Refraction and diffraction are two examples. Light refracts when it travels from one medium to another, because waves travel at different speeds through different media. In a similar way, light diffracts when it travels between or around objects, because obstacles make the light waves bend. So, obviously we\'re not wrong about light behaving like a wave. We even use the wave diffraction of light by reading interference patterns in X-ray crystallography. >> Particle Behaviour photoelectric effect, which describes the way electrons are excited and emitted from matter when they absorb the energy from light. In 1887, Heinrich Hertz observed that a charged object would create a bigger, faster spark if it was treated with ultraviolet light because the light was actually exciting the electrons. Further studies by other scientists showed that electrons really could be knocked out of a metal in response to a beam of light. For a while, scientists thought that the electrons were just absorbing the energy in the light wave, and then using that energy to jump out of the metal. The more energy the electrons could absorb, the more energy they could use to jump out. A photon is a nearly massless particle carrying a small amount of energy. We use photons to quantify, or measure the amount of, the energy in light and other electromagnetic waves. Each photon can excite only one electron at a time. When the intensity of light is increased, then the number of photons is increased, so a higher number of electrons are knocked loose from the metal. The energy of electrons does not change because the energy of each photon is still the same. Solution >> Wave Character Many of the things that light does are only explained sufficiently by thinking of light as a wave. Refraction and diffraction are two examples. Light refracts when it travels from one medium to another, because waves travel at different speeds through different media. In a similar way, light diffracts when it travels between or around objects, because obstacles make the light waves bend. So, obviously we\'re not wrong about light behaving like a wave. We even use the wave diffraction of light by reading interference patterns in X-ray crystallography. >> Particle Behaviour photoelectric effect, which describes the way electrons are excited and emitted from matter when they absorb the energy from light. In 1887, Heinrich Hertz observed that a charged object would create a bigger, faster spark if it was treated with ultraviolet light because the light was actually exciting the electrons. Further studies by other scientists showed that electrons really could be knocked out of a metal in response to a beam of light. For a while, scientists thought that the electrons were just absorbing the energy in the light wave, and then using that energy to jump out of the metal. The more energy the electr.

The solvent doesnt interfere with the purification processOil Bath.pdfaravlitraders2012

The solvent doesnt interfere with the purification process Oil Bath Assembly Mineral oil or silicone oil is typically used for oil baths in research labs for reactions that require heating/reflux temperatures up to 200 °C. Oil baths provide more uniform heat in comparison to other heating devices. A laboratory oil bath is made of an aluminum or stainless steel pan, a heavy porcelain dish or thick walled Pyrex® glass to withstand breakage and accidental spill. The electric heating coil (or oil bath on a hot plate) is controlled using a variable voltage controller (i.e., Variac). The voltage controller is adjusted to increase or decrease the temperature setting of the oil bath. Heating mantles should NEVER be plugged directly into an outlet. Bath Material Useful Range Flash Point Advantage/Disadvantage Risk Silicone Oil 25 °C to 230 °C 150 °C to 350 °C Acceptable MODERATE Water (rotary evaporator) 0 °C to 70 °C Acceptable. LOW Sand 25°C to 500+ °C NA Acceptable. LOW Hazards Hazards associated with the use of oil baths include hot temperatures and fire. Medical Attention and Emergency Treatment Use cold water and rinse the affected skin immediately for at least for 15 minutes; remove any hot oil residue without rubbing the skin. If the affected area is on the hand, wrap with a clean bandage to prevent further irritation and injury. DO NOT apply any ointments and avoid breaking blisters. Seek medical attention as necessary. Medical Attention and Emergency Treatment information is provided on the EHS website. Personal Protective Equipment (PPE) Slip-resistant insulated thermal gloves. Safety glasses with side shields or a face shield. Lab coat (and wear pants that cover the legs). Prudent Safety Practices Storage Spill Cleanup Granular clay absorbents, universal polypropylene sorbent pads, or other oil-specific absorbent pads can be used for cleanup of small incidental spills of both bath oil and vacuum pump oil. Solid waste from the cleanup can be scooped, bagged, and placed in a metal container or other compatible container. Oil-specific absorbents absorb only oil and not aqueous materials. For questions, contact EHS for proper cleanup and disposal procedures. Disposal of Used Oil advantages techniques are understandably used only for samples containing a significant amount of combustible or organic material as the matrix. With this in mind, let\'s look at the major advantages · The ability to decompose large sample sizes. · The need for little or no reagents. · The technique is relatively safe. · The ability to prepare samples containing volatile combustion elements such as sulfur, fluorine and chlorine (the Schöniger oxygen flask combustion technique is very popular in this case). · The technique lends itself to mass production. The technique of graphite furnace atomic absorption spectrometry (GFAA) incorporates sample ashing as part of an automatic measurement cycle. Trace analysts have learned a great deal about the loss of volatile componen.

An object of class StatCalc can be used to compute several simp.pdfaravlitraders2012

/* An object of class StatCalc can be used to compute several simple statistics for a set of numbers. Numbers are entered into the dataset using the enter(double) method. Methods are provided to return the following statistics for the set of numbers that have been entered: The number of items, the sum of the items, the average, the standard deviation, the maximum, and the minimum. */ public class Stats { private int count; // Number of numbers that have been entered. private double sum; // The sum of all the items that have been entered. private double squareSum; // The sum of the squares of all the items. private double max = Double.NEGATIVE_INFINITY; // Largest item seen. private double min = Double.POSITIVE_INFINITY; // Smallest item seen. int[] myList; public int getCount() { System.out.println(myList.size()); count = myList.size(). // Return number of items that have been entered. return count; } public double getSum() { int sum = 0; for (int i = 0; i < myList.length; i++) { sum +=mylist[i]; } return sum; } public double getMean() { int sum = 0; for (int i = 0; i < myList.length; i++) { sum +=mylist[i]; } if(count > 0) return sum / count; else return -1; } public double getMin() { int smallest = myList[0]; int largetst = myList[0]; for(int i=1; i< mylist.length; i++) { if(mylist[i] > largetst) largetst = numbers[i]; else if (mylist[i] < smallest) smallest = numbers[i]; } return smallest; } public double getMax() { int smallest = myList[0]; int largetst = myList[0]; for(int i=1; i< mylist.length; i++) { if(mylist[i] > largetst) largetst = numbers[i]; else if (mylist[i] < smallest) smallest = numbers[i]; } return largetst; } stats(int [] data) { if(data.count <= count) { if(myList.size() > 0) System.arraycopy( myList, 0, data, 0, count ); } } stats() { Scanner s = new Scanner(System.in); int count = s.nextInt(); s.nextLine(); // throw away the newline. int [] numbers = new int[count]; Scanner numScanner = new Scanner(s.nextLine()); for (int i = 0; i < count; i++) { if (numScanner.hasNextInt()) { numbers[i] = numScanner.nextInt(); } else { System.out.println(\"You didn\'t provide enough numbers\"); break; } } System.arraycopy( myList, 0, numbers, 0, count ); } stats(int size) { setValue(size); } private void setValue(int size) { count = size; myList = new int[size]; } } // end class StatCalc public class SimpleStats { public static void main(String[] args) { Stat calc; // Computes stats for numbers entered by user. calc = new Stat(100); int arry[] = {20.20.40,30.80,,70,4,5,6,7,10} calc.setvalue(array); double item; // One number entered by the user. TextIO.putln(\"Enter your numbers. Enter 0 to end.\"); TextIO.putln(); do { TextIO.put(\"? \"); item = TextIO.getlnDouble(); if (item != 0) calc.enter(item); } while ( item != 0 ); TextIO.putln(\"\ Statistics about your calc:\ \"); TextIO.putln(\" Count: \" + calc.getCount()); TextIO.putln(\" Sum: \" + calc.getSum()); TextIO.putln(\" Minimum: \" + calc.getMin()); TextIO.putln(\" Maximum: \" + ca.

The answers can be found below as discussedPart 1)The human gen.pdfaravlitraders2012

The answers can be found below as discussed: Part 1) The human genome is highly vast in nature and hence there are some complicating factors in the study of human genetics: 1: Size and packing: The human genome is enoromously large in linear seqeunce which makes it highly difficult to be analysed as a whole. Moreover, the DNA is wounded in complex primary, secondary and tertiary structures which makes is difficult to isolate intact i.e. formation of nucleosomes over histones, solenoid formation, high writhe and twist etc. The human genome is highly compact and along with it, chemical modifications such as acetylation/methylation/sumoylation/phosphorylation of the DNA-associated proteins also occur. Similarly, post-transcriptional modifications of mRNA also occur only in eukaryotes. Compensation: These structural features of the human genome have been well exploited for their analysis. The size cannot be reduced directly but the complexity of packing can be used by action of relaxation enzymes such as topoisomerases. Part 2) The coding sequences: The human genome is not continuously coding in nature, but it has been differentiated into stretches of coding euchromatin exons and non-coding intervening heterochromatin intron sequences. This makes the linearity of the coding sequence highly compromised and a single mechanism of linear transcription or translation cannot take place. Compensation: The human genome contains the intron sequences only in the DNA. These sequences are effectively removed out of the DNA after transcription by mRNA splicing and thus only exons are remained in the mRNA. Thus, this ensures that only exons can reach the translation process. 3. Polymorphism: The human genome shows very high degree of single nucleotide polymorphism (SNP) which shows pattern of inheritance and can be transitted form one person to another genetically. These SNPs are crucial for providing discrete phenotypic features to individuals which can be difficult to assess and identify due to large size of human genome. Compensation: Although there occur SNPs in human genome, the modern techniques such as next-generation seqeuncing have made it possible to effectively, efficiently and quickly screen the whole genome for such targets. Answer part 2) Variations in Mendelian genetics in humans can be described as below: 1. Co-dominance: Unlike Mendelian genetics, the blood group system of humans show co- dominance between the allele A and allele B and both of these alleles are completely dominant over the allele O. This makes occurance of various phenotypes in blood groups i.e. A, B, AB and O. 2. Incomplete dominance: On contrary to Mendel\'s pattern of inheritance, the texture of hair in human is inherited genetically following a pattern of incomplete dominance. This deciphers that an offspring of parents having straight and curly hair will have softly curled hair. 3. Quantitative traits: Unlike Mendel\'s patterns of inheritance, some genes show quantitative.

Solution (3)The smallest respiratory bronchioles subdivide into t.pdfaravlitraders2012

An asset manager of a hedge fund or a mutual funds is the one who di.pdfaravlitraders2012

An asset manager of a hedge fund or a mutual funds is the one who directly handles the client money and is responsible for acting in an ethical manner while discharging their duties. Some of the major ways in which asset managers breach their fiduciary duties are as follows : (i) Asset Managers generally have access to private information of companies\' profits, next business move, major public announcements which have direct effect on the stock price. In order to make their judgement about stocks right, fund managers use manupulative techniques to inflate stock prices with the help of insider information. (ii) Fund managers in normal practice met out prefrential treatment to clients whose asset volume being managed is higher than the rest of the group. In this way only a group of clients gets all the attention of the fund manager. (iii) Fund managers usually undertake complex financial transactions and disclose less of these to the clients and regulators. (iv) Fund managers engage in activities which more oftem result in conflict of interest. Conflicts of interest can be in the form of raise larger funds when insufficient opportunities exist, taking over undisclosed and inappropriate risk in the investment etc. Solution An asset manager of a hedge fund or a mutual funds is the one who directly handles the client money and is responsible for acting in an ethical manner while discharging their duties. Some of the major ways in which asset managers breach their fiduciary duties are as follows : (i) Asset Managers generally have access to private information of companies\' profits, next business move, major public announcements which have direct effect on the stock price. In order to make their judgement about stocks right, fund managers use manupulative techniques to inflate stock prices with the help of insider information. (ii) Fund managers in normal practice met out prefrential treatment to clients whose asset volume being managed is higher than the rest of the group. In this way only a group of clients gets all the attention of the fund manager. (iii) Fund managers usually undertake complex financial transactions and disclose less of these to the clients and regulators. (iv) Fund managers engage in activities which more oftem result in conflict of interest. Conflicts of interest can be in the form of raise larger funds when insufficient opportunities exist, taking over undisclosed and inappropriate risk in the investment etc..