AnswerThe new program with the required constructor and with a te.pdf
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Answer: The new program with the required constructor and with a test is given below Code: #include #include using namespace std; class TNode { public: int data; TNode *left; TNode *right; TNode(int n){ data = n; left = right = 0; } TNode(int n, TNode *left, TNode *right) { data = n; //Assign the left node address and right node address with the parameters passed this->left = left; this->right = right; } }; void printTree(TNode *t) { if (!t) return; printTree(t->left); cout << t->data << \" \"; printTree(t->right); } void search_for(int n, TNode *t) { cout << \"Searhing for \" << n << endl; while (t) { cout << \"looking at \" << t->data << endl; if (t->data == n) { cout << \"Found it!\"; return; } if (n < t->data) { cout << \"Going left\ \"; t = t->left; } else { cout << \"Going right\ \"; t = t->right; } } cout << \"Can\\\'t find it\"; } // Creates a two node tree int main() { TNode *root = 0; root = new TNode(5); cout << root->data << endl; TNode *one_node = new TNode(1); TNode *three_node = new TNode(3); //use the new constructor TNode *two_node = new TNode(2, one_node, three_node); root->left = new TNode(3); printTree(root); cout << endl; search_for(3, root); getchar(); return 0; } Solution Answer: The new program with the required constructor and with a test is given below Code: #include #include using namespace std; class TNode { public: int data; TNode *left; TNode *right; TNode(int n){ data = n; left = right = 0; } TNode(int n, TNode *left, TNode *right) { data = n; //Assign the left node address and right node address with the parameters passed this->left = left; this->right = right; } }; void printTree(TNode *t) { if (!t) return; printTree(t->left); cout << t->data << \" \"; printTree(t->right); } void search_for(int n, TNode *t) { cout << \"Searhing for \" << n << endl; while (t) { cout << \"looking at \" << t->data << endl; if (t->data == n) { cout << \"Found it!\"; return; } if (n < t->data) { cout << \"Going left\ \"; t = t->left; } else { cout << \"Going right\ \"; t = t->right; } } cout << \"Can\\\'t find it\"; } // Creates a two node tree int main() { TNode *root = 0; root = new TNode(5); cout << root->data << endl; TNode *one_node = new TNode(1); TNode *three_node = new TNode(3); //use the new constructor TNode *two_node = new TNode(2, one_node, three_node); root->left = new TNode(3); printTree(root); cout << endl; search_for(3, root); getchar(); return 0; }.
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Lab Week 2 Game Programming.docx
This program implements a doubly linked list to print the list forward and backward. It defines a Node struct with data, next and prev pointers. It inserts nodes at the head of the list using InsertAtHead() and prints the list forward using Print() and backward using ReversePrint(). It initially inserts two nodes and prints the list both ways. It then inserts three more nodes at the head and prints the list again forward and backward.
Implement a function TNode copy_tree(TNode t) that creates a copy .pdf
Implement a function TNode* copy_tree(TNode* t) that creates a copy of the tree t and returns a
pointer to the root of the copy. Test this function by writing a main( that creates a tree and uses
the function to create a copy. Print both trees to verify they are the same.
Solution
// BSTree.h #ifndef BSTREE_H #define BSTREE_H #include using namespace std;
template struct TNode { TNode(const T &); T data; TNode* left, * right;
}; template TNode::TNode(const T & newItem) { data = newItem; left = right = NULL;
} template class BSTree { public: enum {T_OK = 0, T_DUPLICATE = -1,
T_NOT_FOUND = -2}; BSTree(); BSTree(const BSTree&); ~BSTree();
void clear(); int insert(const T&); int remove(const T&); int height() const; int
size() const; void printPreorder() const; void printInorder() const; void
printPostorder() const; const BSTree& operator=(const BSTree&); private:
TNode * root; void destroyTree(TNode*); int size(TNode*) const; int
height(TNode*) const; TNode* copyTree(TNode*); void preTrav(TNode*) const;
void inTrav(TNode*) const; void postTrav(TNode*) const; }; template
BSTree::BSTree() { root = NULL; } template BSTree::BSTree(const BSTree&
treeToCopy) { root = copyTree(treeToCopy.root); } template BSTree::~BSTree()
{ destroyTree(root); } template void BSTree::destroyTree(TNode* current) { if
(current != NULL) { destroyTree(current->left); destroyTree(current->right);
delete current; } } template void BSTree::clear() { destroyTree(root); root =
NULL; } template int BSTree::insert(const T& newItem) { TNode* current = root,
* parent = NULL; while (current != NULL && newItem != current->data) { parent
= current; if (newItem < current->data) current = current->left; else // if (newItem >
current->data) current = current->right; } if (current != NULL) return
T_DUPLICATE; else { TNode* newNode = new TNode(newItem); if (parent ==
NULL) root = newNode; else if (newNode->data < parent->data) parent-
>left = newNode; else parent->right = newNode; return T_OK; }
} template int BSTree::remove(const T& deleteItem) { TNode* delNode = root, *
delParent = NULL, * replNode, * replParent; while (delNode != NULL && deleteItem
!= delNode->data) { delParent = delNode; if (deleteItem < delNode->data)
delNode = delNode->left; else // if (newItem > current->data) delNode = delNode-
>right; } if (delNode != NULL) { if (delNode->left == NULL) replNode =
delNode->right; else if (!delNode->right) replNode = delNode->left; else {
replParent = delNode; replNode = delNode-> right; while (replNode->left !=
NULL) { replParent = replNode; replNode = replNode->left; }
if (replParent != delNode) { replParent->left = replNode->right;
replNode->right = delNode->right; } replNode->left = delNode->left; }
if (delParent == NULL) root = replNode; else { if (delParent->left ==
delNode) delParent->left = replNode; else delParent->right = replNode;
} delete delNode; return T_OK; } else return T_NOT_FOUND; }
template int BSTree::size() const { return size(root); } template int BSTree::size(.
This is a c++ binary search program I worked so far but still cant g.pdf
This is a c++ binary search program I worked so far but still cant get it right.
Can anyone help me? Big thanks!!
the client should not be modified
/*
*File: client.cpp
*Author: Yingwu Zhu
*Warning: do not change this file and use it as is.
*Last Modififcation: 10/21/2016
*/
#include
#include
#include
#include
#include
#include \"bst.h\"
using namespace std;
int main(int argc, char* argv[]) {
if (argc != 2) {
cout << \"Format: \" << argv[0] << \" [data file]\" << endl;
return 0;
}
ifstream fin(argv[1]);
int cases;
fin >> cases;
int passed = 0;
for (int i = 1; i <= cases; i++) {
cout << \"Checking test case #\" << i << \" ... \";
BST T;
set S;
int n, x;
fin >> n;
bool ok = true;
vector tmp;
int rem = 0;
for (int j = 0; j < n; j++) {
fin >> x;
T.Insert(x);
S.insert(x);
ok &= (T.Search(x) && T.RecurSearch(x));
if (tmp.empty())
tmp.push_back(x);
if (rand()%10 < 3) {
T.Erase(tmp[0]);
S.erase(tmp[0]);
ok &= (T.Search(tmp[0]) == false);
tmp.pop_back();
rem++;
}
}
if (rem
ok &= (T.MaxElement() == *S.rbegin());
while (!S.empty() && !T.Empty()) {
int x = T.MinElement();
ok &= (x == *S.begin());
T.Erase(x);
S.erase(S.begin());
}
cout << (ok ? \"Passed!\" : \"Failed\") << endl;
passed += ok;
}
cout << passed << \" of \" << cases << \" test cases have passed!\ \";
if (passed == cases)
cout << endl << \"Congratulations! Good to Submit Your Code\ \";
else if ((double)passed/cases >= 0.95)
cout << endl << \"You are almost there, but need to fix some bugs.\ \";
else
cout << endl << \"Your code needs a lot of fixes for submission\ \";
return 0;
}
=====================================================================
=======
//bst.h
#ifndef _BST_H_
#define _BST_H_
#include
using namespace std;
class BST {
private:
class Node {
public:
int data;
Node* left;
Node* right;
};
Node* root; //root node pointer
//you may add your auxiliary functions here!
public:
BST(); //constructor
~BST(); //destructor
bool Empty() const;
void Insert(int val);
int MinElement() const;
int MaxElement() const;
bool Search(int val) const;
bool RecurSearch(int val) const;
void Erase(int val);
};
#endif
=====================================================================
=======
/*
bst.cpp
Subject: Binary Seach Tree
Modification Time:10/26/2016
*/
#include
#include\"bst.h\"
using namespace std;
BST::BST(){
root= NULL;
}
BST::~BST(){
delete root;
delete root->left;
delete root->right;
}
bool BST::Empty() const {
return data;
}
void BST::Insert(int val){
Node* p= new Node;
p->data= val;
Node* cur = root;
if(val < cur->data){
cur = cur -> left;
}
else{
right->next=val;
}
}
int BST::MinElement() const {
Node* cur;
while(cur->left != NULL){
cur = cur->left;
}
return(cur->data);
}
int BST::MaxElement() const{
if(root==NULL){
return 0;
}
if(root->left>root->data){
root->data=root->right;
}
else if(root->right>root->data){
root->data=root->right;
}
return root->data;
}
bool BST:: Search(int val) const{
Node* cur;
cur = root;
while(cur!=NULL){
if(cur->data == val){
return true;
}
else if(cur->data .
How do you stop infinite loop Because I believe that it is making a.pdf
How do you stop infinite loop? Because I believe that it is making an infinite circular list.
c++ code:
Here is the list class:
#ifndef LIN_J_LIST
#define LIN_J_LIST
typedef unsigned int uint;
#include
#include
using namespace std;
/**
* a simplified generic singly linked list class to illustrate C++ concepts
* @author Jerry Lin
* @version 2/17/17
*/
template< typename Object >
class List
{
private:
/**
* A class to store data and provide a link to the next node in the list
*/
class Node
{
public:
/**
* The constructor
* @param data the data to be stored in this node
*/
explicit Node( const Object & data )
: data{ data }, next{ nullptr } {}
Object data;
Node * next;
};
public:
/**
* The constructor for an empty list
*/
List()
: size{ 0 }, first{ nullptr }, last{ nullptr } {}
/**
* the copy constructor-creates and copy the list
*/
List( List && rhs ) = delete;
List( const List & rhs )
{
count = 0;
size = 0;
if(rhs.size != 0)
{
push_front(rhs.front());
Node * current = rhs.first;
Node * tempNode = first;
size++;
while(current->next != nullptr)
{
current = current->next;
Node *newNode = new Node(current->data); //transfer the data. basic op.
count++;
tempNode->next = newNode;
last = newNode;
tempNode = tempNode->next;
size++;
}
}
// you document and implement this method
}
/**
* the operator= method-copies the list
*/
List & operator=( List && rhs) = delete;
List & operator=( const List & rhs )
{
count = 0;
size = 0;
if( size != 0 )
{
Node * current = first;
Node * temp;
while( current != nullptr )
{
temp = current;
current = current->next;
delete temp;
}
}
if(rhs.size!= 0)
{
push_front(rhs.front());
Node * current = rhs.first;
Node * tempNode = first; //create a temporary to store
size++;
while(current -> next != nullptr)
{
current = current->next;
Node *newNode = new Node(current->data); //transfer the data. basic op
count++;
tempNode->next = newNode;
last = newNode;
tempNode = tempNode -> next;
size++;
}
}
return *this;
// you document and implement this method
}
/**
* accessor
* @return count
*/
int get_count() const
{
return count;
}
/**
* The destructor that gets rid of everything that\'s in the list and
* resets it to empty. If the list is already empty, do nothing.
*/
~List()
{
if( size != 0 )
{
Node * current = first;
Node * temp;
while( current != nullptr )
{
temp = current;
current = current->next;
delete temp;
}
}
}
/**
* Put a new element onto the beginning of the list
* @param item the data the new element will contain
*/
void push_front( const Object& item )
{
Node * new_node = new Node( item );//basic op.
if(is_empty())
{
last = new_node;
}
else
{
new_node->next = first;
}
first = new_node;
size++;
/* you complete the rest */
}
/**
* Remove the element that\'s at the front of the list. Causes an
* assertion error if the list is empty.
*/
void pop_front()
{
assert( !is_empty() );
Node * temp = first;
if( first == last )
{
first = last = nullptr;
}
else
{
first = first->next;
}
delete temp;
size--;
}
/**
* Accessor to return the da.
1. Add a breadth-first (level-order) traversal function to the binar.pdf
1. Add a breadth-first (level-order) traversal function to the binary tree code.
2. Add a function to find the height of a tree.
3. Re-implement one of the depth-first traversal methods using a stack instead of recursion.
4. Add a link to each nodes parent node.
#include
#include
#include
using namespace std;
template < typename T >
class TreeNode
{
public:
T element; //
TreeNode < T > * left; //
TreeNode < T > * right; //
TreeNode * next;
TreeNode() //
{
left = NULL;
next = NULL;
}
TreeNode(T element) // Constructor
{
this->element = element;
left = NULL;
right = NULL;
}
};
template < typename T >
class BinaryTree
{
public:
BinaryTree();
BinaryTree(T elements[], int arraySize);
bool insert(T element);
void inorder();
void preorder();
void postorder();
int getSize();
bool search(T element);
void breadthFirstTraversal();
int depth();
private:
TreeNode < T > * root;
int size;
void inorder(TreeNode < T > * root);
void postorder(TreeNode < T > * root);
void preorder(TreeNode < T > * root);
bool search(T element, TreeNode < T > * root);
int depth(TreeNode * root);
};
template < typename T >
BinaryTree < T >::BinaryTree()
{
root = NULL;
size = 0;
}
template < typename T >
BinaryTree < T >::BinaryTree(T elements[], int arraySize)
{
root = NULL;
size = 0;
for (int i = 0; i < arraySize; i++)
{
insert(elements[i]);
}
}
template < typename T >
bool BinaryTree < T >::insert(T element)
{
if (root == NULL)
root = new TreeNode < T > (element); // Create a new root
else
{
// Locate the parent node
TreeNode < T > * parent = NULL;
TreeNode < T > * current = root;
while (current != NULL)
if (element < current->element)
{
parent = current;
current = current->left;
}
else if (element > current->element)
{
parent = current;
current = current->right;
}
else
return false; // Duplicate node not inserted
// Create the new node and attach it to the parent node
if (element < parent->element)
parent->left = new TreeNode < T > (element);
else
parent->right = new TreeNode < T > (element);
}
size++;
return true; // Element inserted
}
/* Inorder traversal */
template < typename T >
void BinaryTree < T >::inorder()
{
inorder(root);
}
/* Inorder traversal from a subtree */
template < typename T >
void BinaryTree < T >::inorder(TreeNode < T > * root)
{
if (root == NULL) return;
inorder(root->left);
cout << root->element << \" \";
inorder(root->right);
}
/* Postorder traversal */
template < typename T >
void BinaryTree < T >::postorder()
{
postorder(root);
}
/** Inorder traversal from a subtree */
template < typename T >
void BinaryTree < T >::postorder(TreeNode < T > * root)
{
if (root == NULL) return;
postorder(root->left);
postorder(root->right);
cout << root->element << \" \";
}
/* */
template < typename T >
void BinaryTree < T >::preorder()
{
preorder(root);
}
/* */
template < typename T >
void BinaryTree < T >::preorder(TreeNode < T > * root)
{
if (root == NULL) return;
cout << root->element << \" \";
preorder(root->left);
preorder(root->right);
}
/**/
template < typename .
#include stdafx.h #include iostream using namespace std;vo.docx
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
In the class we extensively discussed a node class called IntNode in.pdf
In the class we extensively discussed a node class called IntNode in which each linked list node
contains an integer and each node is linked forward to the next element in the list. Implement a
node class called DoublyIntNode in which each node contains an integer number and each node
is linked not only forward to the next element in the list but also backward to the previous
element in the list. Your implementation of DoublyIntNode should adapt and implement all the
methods of IntNode including the following methods:
Constructor for DoublyIntNode
addNodeAfter, removeNodeAfter
getData, setData
getForeLink, setForeLink
getBackLink, setBackLink
listCopy
listCopyWithTail
listLength
listPart
listPosition
listSearch
IntNode and Node classes:
public class IntNode{
private int data;
private IntNode link;
// methods here
}
public class Node {
private E data;
private Node link;
// methods here
}
Solution
program
#include
using namespace std;
struct list
{
struct list *prev;
int data;
struct list *next;
}
*node = NULL, *first = NULL, *last = NULL, *node1 = NULL, *node2 = NULL;
class Dlist
{
public:
void insert_beg() {
list *addBeg = new list;
cout << \"Enter value for the node:\" << endl;
cin >> addBeg->data;
if(first == NULL) {
addBeg->prev = NULL;
addBeg->next = NULL;
first = addBeg;
last = addBeg;
cout << \"Linked list Created!\" << endl;
}
else {
addBeg->prev = NULL;
first->prev = addBeg;
addBeg->next = first;
first = addBeg;
cout << \"Data Inserted at the beginning of the Linked list!\" << endl;
}
}
void insert_end() {
list *addEnd = new list;
cout << \"Enter value for the node:\" << endl;
cin >> addEnd->data;
if(first == NULL) {
addEnd->prev = NULL;
addEnd->next = NULL;
first = addEnd;
last = addEnd;
cout << \"Linked list Created!\" << endl;
}
else {
addEnd->next = NULL;
last->next = addEnd;
addEnd->prev = last;
last = addEnd;
cout << \"Data Inserted at the end of the Linked list!\" << endl;
}
}
void display() {
node = first;
cout << \"List of data in Linked list in Ascending order!\" << endl;
while(node != NULL) {
cout << node->data << endl;
node = node->next;
}
node = last;
cout << \"List of data in Linked list in Descending order!\" << endl;
while(node != NULL) {
cout << node->data << endl;
node = node->prev;
}
}
void del()
{
int count = 0, number, i;
node = node1 = node2 = first;
for(node = first; node != NULL; node = node->next)
cout << \"Enter value for the node:\" << endl;
count++;
display();
cout << count << \" nodes available here!\" << endl;
cout << \"Enter the node number which you want to delete:\" << endl;
cin >> number;
if(number != 1)
{
if(number < count && number > 0)
{
for(i = 2; i <= number; i++)
node = node->next;
for(i = 2; i <= number-1; i++)
node1 = node1->next;
for(i = 2; i <= number+1; i++)
node2 = node2->next;
node2->prev = node1;
node1->next = node2;
node->prev = NULL;
node->next = NULL;
node = NULL;
}
else if(number == count)
{
node = last;
last = node->prev;
last->next = NULL;
node = NULL;
}
else
cout << \"Invalid node.
Please write the C++ code that would display the exact same output a.pdf
Please write the C++ code that would display the exact same output as provided. Thank you.
main.cpp
#include "TreeNode.h"
//GIVEN
void inorderTraversal(TreeNode * root);
//TODO
TreeNode * search(TreeNode * node,int data);
//GIVEN
void insert(TreeNode * node,int data);
//GIVEN
bool isNull(TreeNode * node){
return node==nullptr;
}
int main() {
// create the root, gets set to null
TreeNode * root;
// create the roor and set the data of root to 5
root= new TreeNode(5);
//create two additional nodes with data values 10 and 3
// these will go to the right and left of root respectively
TreeNode * ten= new TreeNode(10);
TreeNode * three= new TreeNode(3);
//connect ten to the right of root (which has data 5)
root->setRight(ten);
//connect three to the left of root (which has data 5)
root->setLeft(three);
// note this can also be done as
//root.setRight(new TreeNode(10));
//root.setLeft(new TreeNode(3);)
// add two more nodes
//the first has data value 7. So to keep the tree in insorted order, it should get attached as the
left node of ten
// the second has data value 4. So to keep the tree in insorted order, it should get attached as the
right node of three
ten->addLeft(7);
three->addRight(4);
std::cout<<"**************************************\n";
std::cout<<"Printing the Inorder Traversal\n";
inorderTraversal(root);
std::cout<<"**************************************\n";
std::cout<<"Searching for Node \n";
TreeNode* result = search(root,4);
if (result!=nullptr){
std::cout<<"Found "<getData()<<"\n";
}
else
std::cout<<"Not Found "<<4<<"\n";
result = search(root,1);
if (result!=nullptr){
std::cout<<"Found "<getData()<<"\n";
}
else
std::cout<<"Not Found "<<1<<"\n";
std::cout<<"**************************************\n";
std::cout<<"Inserting 6\n";
insert(root,6);
std::cout<<"**************************************\n";
std::cout<<"Printing the Inorder Traversal\n";
inorderTraversal(root);
}
// uses recursion
void inorderTraversal(TreeNode * node){
// exit case
if (isNull(node)) return;
if (node->isLeaf()){
std::cout<<"Printing Leaf Node "<getData()<<"\n";
return;}
// reached a node with no left, so print the node and travel right
if (!isNull(node->getLeft()))
// if there is a left path, then travel further on that
inorderTraversal(node->getLeft());
std::cout<<"Printing Node "<getData()<<"\n";
// save and travel the right path of the current node being processed
inorderTraversal(node->getRight());
}
// uses recursion
//TODO
TreeNode * search(TreeNode * node, int data){
// if the node is null return the null ptr
//if this nodes data equals the search date return found
// if not, if the left tree exists and search data less than
//this nodes data return the result of recursive all to search with left pointer
// if no left tree, but right tree exists and search data greater than
//this nodes data return the result of recursive all to search with right pointer
// if both above conditions not true return null ptr
}
//uses recursion
void insert(TreeNode * node,int .
Recommended
Lab Week 2 Game Programming.docx
This program implements a doubly linked list to print the list forward and backward. It defines a Node struct with data, next and prev pointers. It inserts nodes at the head of the list using InsertAtHead() and prints the list forward using Print() and backward using ReversePrint(). It initially inserts two nodes and prints the list both ways. It then inserts three more nodes at the head and prints the list again forward and backward.
Implement a function TNode copy_tree(TNode t) that creates a copy .pdf
Implement a function TNode* copy_tree(TNode* t) that creates a copy of the tree t and returns a
pointer to the root of the copy. Test this function by writing a main( that creates a tree and uses
the function to create a copy. Print both trees to verify they are the same.
Solution
// BSTree.h #ifndef BSTREE_H #define BSTREE_H #include using namespace std;
template struct TNode { TNode(const T &); T data; TNode* left, * right;
}; template TNode::TNode(const T & newItem) { data = newItem; left = right = NULL;
} template class BSTree { public: enum {T_OK = 0, T_DUPLICATE = -1,
T_NOT_FOUND = -2}; BSTree(); BSTree(const BSTree&); ~BSTree();
void clear(); int insert(const T&); int remove(const T&); int height() const; int
size() const; void printPreorder() const; void printInorder() const; void
printPostorder() const; const BSTree& operator=(const BSTree&); private:
TNode * root; void destroyTree(TNode*); int size(TNode*) const; int
height(TNode*) const; TNode* copyTree(TNode*); void preTrav(TNode*) const;
void inTrav(TNode*) const; void postTrav(TNode*) const; }; template
BSTree::BSTree() { root = NULL; } template BSTree::BSTree(const BSTree&
treeToCopy) { root = copyTree(treeToCopy.root); } template BSTree::~BSTree()
{ destroyTree(root); } template void BSTree::destroyTree(TNode* current) { if
(current != NULL) { destroyTree(current->left); destroyTree(current->right);
delete current; } } template void BSTree::clear() { destroyTree(root); root =
NULL; } template int BSTree::insert(const T& newItem) { TNode* current = root,
* parent = NULL; while (current != NULL && newItem != current->data) { parent
= current; if (newItem < current->data) current = current->left; else // if (newItem >
current->data) current = current->right; } if (current != NULL) return
T_DUPLICATE; else { TNode* newNode = new TNode(newItem); if (parent ==
NULL) root = newNode; else if (newNode->data < parent->data) parent-
>left = newNode; else parent->right = newNode; return T_OK; }
} template int BSTree::remove(const T& deleteItem) { TNode* delNode = root, *
delParent = NULL, * replNode, * replParent; while (delNode != NULL && deleteItem
!= delNode->data) { delParent = delNode; if (deleteItem < delNode->data)
delNode = delNode->left; else // if (newItem > current->data) delNode = delNode-
>right; } if (delNode != NULL) { if (delNode->left == NULL) replNode =
delNode->right; else if (!delNode->right) replNode = delNode->left; else {
replParent = delNode; replNode = delNode-> right; while (replNode->left !=
NULL) { replParent = replNode; replNode = replNode->left; }
if (replParent != delNode) { replParent->left = replNode->right;
replNode->right = delNode->right; } replNode->left = delNode->left; }
if (delParent == NULL) root = replNode; else { if (delParent->left ==
delNode) delParent->left = replNode; else delParent->right = replNode;
} delete delNode; return T_OK; } else return T_NOT_FOUND; }
template int BSTree::size() const { return size(root); } template int BSTree::size(.
This is a c++ binary search program I worked so far but still cant g.pdf
This is a c++ binary search program I worked so far but still cant get it right.
Can anyone help me? Big thanks!!
the client should not be modified
/*
*File: client.cpp
*Author: Yingwu Zhu
*Warning: do not change this file and use it as is.
*Last Modififcation: 10/21/2016
*/
#include
#include
#include
#include
#include
#include \"bst.h\"
using namespace std;
int main(int argc, char* argv[]) {
if (argc != 2) {
cout << \"Format: \" << argv[0] << \" [data file]\" << endl;
return 0;
}
ifstream fin(argv[1]);
int cases;
fin >> cases;
int passed = 0;
for (int i = 1; i <= cases; i++) {
cout << \"Checking test case #\" << i << \" ... \";
BST T;
set S;
int n, x;
fin >> n;
bool ok = true;
vector tmp;
int rem = 0;
for (int j = 0; j < n; j++) {
fin >> x;
T.Insert(x);
S.insert(x);
ok &= (T.Search(x) && T.RecurSearch(x));
if (tmp.empty())
tmp.push_back(x);
if (rand()%10 < 3) {
T.Erase(tmp[0]);
S.erase(tmp[0]);
ok &= (T.Search(tmp[0]) == false);
tmp.pop_back();
rem++;
}
}
if (rem
ok &= (T.MaxElement() == *S.rbegin());
while (!S.empty() && !T.Empty()) {
int x = T.MinElement();
ok &= (x == *S.begin());
T.Erase(x);
S.erase(S.begin());
}
cout << (ok ? \"Passed!\" : \"Failed\") << endl;
passed += ok;
}
cout << passed << \" of \" << cases << \" test cases have passed!\ \";
if (passed == cases)
cout << endl << \"Congratulations! Good to Submit Your Code\ \";
else if ((double)passed/cases >= 0.95)
cout << endl << \"You are almost there, but need to fix some bugs.\ \";
else
cout << endl << \"Your code needs a lot of fixes for submission\ \";
return 0;
}
=====================================================================
=======
//bst.h
#ifndef _BST_H_
#define _BST_H_
#include
using namespace std;
class BST {
private:
class Node {
public:
int data;
Node* left;
Node* right;
};
Node* root; //root node pointer
//you may add your auxiliary functions here!
public:
BST(); //constructor
~BST(); //destructor
bool Empty() const;
void Insert(int val);
int MinElement() const;
int MaxElement() const;
bool Search(int val) const;
bool RecurSearch(int val) const;
void Erase(int val);
};
#endif
=====================================================================
=======
/*
bst.cpp
Subject: Binary Seach Tree
Modification Time:10/26/2016
*/
#include
#include\"bst.h\"
using namespace std;
BST::BST(){
root= NULL;
}
BST::~BST(){
delete root;
delete root->left;
delete root->right;
}
bool BST::Empty() const {
return data;
}
void BST::Insert(int val){
Node* p= new Node;
p->data= val;
Node* cur = root;
if(val < cur->data){
cur = cur -> left;
}
else{
right->next=val;
}
}
int BST::MinElement() const {
Node* cur;
while(cur->left != NULL){
cur = cur->left;
}
return(cur->data);
}
int BST::MaxElement() const{
if(root==NULL){
return 0;
}
if(root->left>root->data){
root->data=root->right;
}
else if(root->right>root->data){
root->data=root->right;
}
return root->data;
}
bool BST:: Search(int val) const{
Node* cur;
cur = root;
while(cur!=NULL){
if(cur->data == val){
return true;
}
else if(cur->data .
How do you stop infinite loop Because I believe that it is making a.pdf
How do you stop infinite loop? Because I believe that it is making an infinite circular list.
c++ code:
Here is the list class:
#ifndef LIN_J_LIST
#define LIN_J_LIST
typedef unsigned int uint;
#include
#include
using namespace std;
/**
* a simplified generic singly linked list class to illustrate C++ concepts
* @author Jerry Lin
* @version 2/17/17
*/
template< typename Object >
class List
{
private:
/**
* A class to store data and provide a link to the next node in the list
*/
class Node
{
public:
/**
* The constructor
* @param data the data to be stored in this node
*/
explicit Node( const Object & data )
: data{ data }, next{ nullptr } {}
Object data;
Node * next;
};
public:
/**
* The constructor for an empty list
*/
List()
: size{ 0 }, first{ nullptr }, last{ nullptr } {}
/**
* the copy constructor-creates and copy the list
*/
List( List && rhs ) = delete;
List( const List & rhs )
{
count = 0;
size = 0;
if(rhs.size != 0)
{
push_front(rhs.front());
Node * current = rhs.first;
Node * tempNode = first;
size++;
while(current->next != nullptr)
{
current = current->next;
Node *newNode = new Node(current->data); //transfer the data. basic op.
count++;
tempNode->next = newNode;
last = newNode;
tempNode = tempNode->next;
size++;
}
}
// you document and implement this method
}
/**
* the operator= method-copies the list
*/
List & operator=( List && rhs) = delete;
List & operator=( const List & rhs )
{
count = 0;
size = 0;
if( size != 0 )
{
Node * current = first;
Node * temp;
while( current != nullptr )
{
temp = current;
current = current->next;
delete temp;
}
}
if(rhs.size!= 0)
{
push_front(rhs.front());
Node * current = rhs.first;
Node * tempNode = first; //create a temporary to store
size++;
while(current -> next != nullptr)
{
current = current->next;
Node *newNode = new Node(current->data); //transfer the data. basic op
count++;
tempNode->next = newNode;
last = newNode;
tempNode = tempNode -> next;
size++;
}
}
return *this;
// you document and implement this method
}
/**
* accessor
* @return count
*/
int get_count() const
{
return count;
}
/**
* The destructor that gets rid of everything that\'s in the list and
* resets it to empty. If the list is already empty, do nothing.
*/
~List()
{
if( size != 0 )
{
Node * current = first;
Node * temp;
while( current != nullptr )
{
temp = current;
current = current->next;
delete temp;
}
}
}
/**
* Put a new element onto the beginning of the list
* @param item the data the new element will contain
*/
void push_front( const Object& item )
{
Node * new_node = new Node( item );//basic op.
if(is_empty())
{
last = new_node;
}
else
{
new_node->next = first;
}
first = new_node;
size++;
/* you complete the rest */
}
/**
* Remove the element that\'s at the front of the list. Causes an
* assertion error if the list is empty.
*/
void pop_front()
{
assert( !is_empty() );
Node * temp = first;
if( first == last )
{
first = last = nullptr;
}
else
{
first = first->next;
}
delete temp;
size--;
}
/**
* Accessor to return the da.
1. Add a breadth-first (level-order) traversal function to the binar.pdf
1. Add a breadth-first (level-order) traversal function to the binary tree code.
2. Add a function to find the height of a tree.
3. Re-implement one of the depth-first traversal methods using a stack instead of recursion.
4. Add a link to each nodes parent node.
#include
#include
#include
using namespace std;
template < typename T >
class TreeNode
{
public:
T element; //
TreeNode < T > * left; //
TreeNode < T > * right; //
TreeNode * next;
TreeNode() //
{
left = NULL;
next = NULL;
}
TreeNode(T element) // Constructor
{
this->element = element;
left = NULL;
right = NULL;
}
};
template < typename T >
class BinaryTree
{
public:
BinaryTree();
BinaryTree(T elements[], int arraySize);
bool insert(T element);
void inorder();
void preorder();
void postorder();
int getSize();
bool search(T element);
void breadthFirstTraversal();
int depth();
private:
TreeNode < T > * root;
int size;
void inorder(TreeNode < T > * root);
void postorder(TreeNode < T > * root);
void preorder(TreeNode < T > * root);
bool search(T element, TreeNode < T > * root);
int depth(TreeNode * root);
};
template < typename T >
BinaryTree < T >::BinaryTree()
{
root = NULL;
size = 0;
}
template < typename T >
BinaryTree < T >::BinaryTree(T elements[], int arraySize)
{
root = NULL;
size = 0;
for (int i = 0; i < arraySize; i++)
{
insert(elements[i]);
}
}
template < typename T >
bool BinaryTree < T >::insert(T element)
{
if (root == NULL)
root = new TreeNode < T > (element); // Create a new root
else
{
// Locate the parent node
TreeNode < T > * parent = NULL;
TreeNode < T > * current = root;
while (current != NULL)
if (element < current->element)
{
parent = current;
current = current->left;
}
else if (element > current->element)
{
parent = current;
current = current->right;
}
else
return false; // Duplicate node not inserted
// Create the new node and attach it to the parent node
if (element < parent->element)
parent->left = new TreeNode < T > (element);
else
parent->right = new TreeNode < T > (element);
}
size++;
return true; // Element inserted
}
/* Inorder traversal */
template < typename T >
void BinaryTree < T >::inorder()
{
inorder(root);
}
/* Inorder traversal from a subtree */
template < typename T >
void BinaryTree < T >::inorder(TreeNode < T > * root)
{
if (root == NULL) return;
inorder(root->left);
cout << root->element << \" \";
inorder(root->right);
}
/* Postorder traversal */
template < typename T >
void BinaryTree < T >::postorder()
{
postorder(root);
}
/** Inorder traversal from a subtree */
template < typename T >
void BinaryTree < T >::postorder(TreeNode < T > * root)
{
if (root == NULL) return;
postorder(root->left);
postorder(root->right);
cout << root->element << \" \";
}
/* */
template < typename T >
void BinaryTree < T >::preorder()
{
preorder(root);
}
/* */
template < typename T >
void BinaryTree < T >::preorder(TreeNode < T > * root)
{
if (root == NULL) return;
cout << root->element << \" \";
preorder(root->left);
preorder(root->right);
}
/**/
template < typename .
#include stdafx.h #include iostream using namespace std;vo.docx
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
In the class we extensively discussed a node class called IntNode in.pdf
In the class we extensively discussed a node class called IntNode in which each linked list node
contains an integer and each node is linked forward to the next element in the list. Implement a
node class called DoublyIntNode in which each node contains an integer number and each node
is linked not only forward to the next element in the list but also backward to the previous
element in the list. Your implementation of DoublyIntNode should adapt and implement all the
methods of IntNode including the following methods:
Constructor for DoublyIntNode
addNodeAfter, removeNodeAfter
getData, setData
getForeLink, setForeLink
getBackLink, setBackLink
listCopy
listCopyWithTail
listLength
listPart
listPosition
listSearch
IntNode and Node classes:
public class IntNode{
private int data;
private IntNode link;
// methods here
}
public class Node {
private E data;
private Node link;
// methods here
}
Solution
program
#include
using namespace std;
struct list
{
struct list *prev;
int data;
struct list *next;
}
*node = NULL, *first = NULL, *last = NULL, *node1 = NULL, *node2 = NULL;
class Dlist
{
public:
void insert_beg() {
list *addBeg = new list;
cout << \"Enter value for the node:\" << endl;
cin >> addBeg->data;
if(first == NULL) {
addBeg->prev = NULL;
addBeg->next = NULL;
first = addBeg;
last = addBeg;
cout << \"Linked list Created!\" << endl;
}
else {
addBeg->prev = NULL;
first->prev = addBeg;
addBeg->next = first;
first = addBeg;
cout << \"Data Inserted at the beginning of the Linked list!\" << endl;
}
}
void insert_end() {
list *addEnd = new list;
cout << \"Enter value for the node:\" << endl;
cin >> addEnd->data;
if(first == NULL) {
addEnd->prev = NULL;
addEnd->next = NULL;
first = addEnd;
last = addEnd;
cout << \"Linked list Created!\" << endl;
}
else {
addEnd->next = NULL;
last->next = addEnd;
addEnd->prev = last;
last = addEnd;
cout << \"Data Inserted at the end of the Linked list!\" << endl;
}
}
void display() {
node = first;
cout << \"List of data in Linked list in Ascending order!\" << endl;
while(node != NULL) {
cout << node->data << endl;
node = node->next;
}
node = last;
cout << \"List of data in Linked list in Descending order!\" << endl;
while(node != NULL) {
cout << node->data << endl;
node = node->prev;
}
}
void del()
{
int count = 0, number, i;
node = node1 = node2 = first;
for(node = first; node != NULL; node = node->next)
cout << \"Enter value for the node:\" << endl;
count++;
display();
cout << count << \" nodes available here!\" << endl;
cout << \"Enter the node number which you want to delete:\" << endl;
cin >> number;
if(number != 1)
{
if(number < count && number > 0)
{
for(i = 2; i <= number; i++)
node = node->next;
for(i = 2; i <= number-1; i++)
node1 = node1->next;
for(i = 2; i <= number+1; i++)
node2 = node2->next;
node2->prev = node1;
node1->next = node2;
node->prev = NULL;
node->next = NULL;
node = NULL;
}
else if(number == count)
{
node = last;
last = node->prev;
last->next = NULL;
node = NULL;
}
else
cout << \"Invalid node.
Please write the C++ code that would display the exact same output a.pdf
Please write the C++ code that would display the exact same output as provided. Thank you.
main.cpp
#include "TreeNode.h"
//GIVEN
void inorderTraversal(TreeNode * root);
//TODO
TreeNode * search(TreeNode * node,int data);
//GIVEN
void insert(TreeNode * node,int data);
//GIVEN
bool isNull(TreeNode * node){
return node==nullptr;
}
int main() {
// create the root, gets set to null
TreeNode * root;
// create the roor and set the data of root to 5
root= new TreeNode(5);
//create two additional nodes with data values 10 and 3
// these will go to the right and left of root respectively
TreeNode * ten= new TreeNode(10);
TreeNode * three= new TreeNode(3);
//connect ten to the right of root (which has data 5)
root->setRight(ten);
//connect three to the left of root (which has data 5)
root->setLeft(three);
// note this can also be done as
//root.setRight(new TreeNode(10));
//root.setLeft(new TreeNode(3);)
// add two more nodes
//the first has data value 7. So to keep the tree in insorted order, it should get attached as the
left node of ten
// the second has data value 4. So to keep the tree in insorted order, it should get attached as the
right node of three
ten->addLeft(7);
three->addRight(4);
std::cout<<"**************************************\n";
std::cout<<"Printing the Inorder Traversal\n";
inorderTraversal(root);
std::cout<<"**************************************\n";
std::cout<<"Searching for Node \n";
TreeNode* result = search(root,4);
if (result!=nullptr){
std::cout<<"Found "<getData()<<"\n";
}
else
std::cout<<"Not Found "<<4<<"\n";
result = search(root,1);
if (result!=nullptr){
std::cout<<"Found "<getData()<<"\n";
}
else
std::cout<<"Not Found "<<1<<"\n";
std::cout<<"**************************************\n";
std::cout<<"Inserting 6\n";
insert(root,6);
std::cout<<"**************************************\n";
std::cout<<"Printing the Inorder Traversal\n";
inorderTraversal(root);
}
// uses recursion
void inorderTraversal(TreeNode * node){
// exit case
if (isNull(node)) return;
if (node->isLeaf()){
std::cout<<"Printing Leaf Node "<getData()<<"\n";
return;}
// reached a node with no left, so print the node and travel right
if (!isNull(node->getLeft()))
// if there is a left path, then travel further on that
inorderTraversal(node->getLeft());
std::cout<<"Printing Node "<getData()<<"\n";
// save and travel the right path of the current node being processed
inorderTraversal(node->getRight());
}
// uses recursion
//TODO
TreeNode * search(TreeNode * node, int data){
// if the node is null return the null ptr
//if this nodes data equals the search date return found
// if not, if the left tree exists and search data less than
//this nodes data return the result of recursive all to search with left pointer
// if no left tree, but right tree exists and search data greater than
//this nodes data return the result of recursive all to search with right pointer
// if both above conditions not true return null ptr
}
//uses recursion
void insert(TreeNode * node,int .
Binary Tree
The document provides C code examples for various binary tree operations and traversals including: finding the height/depth and size of a tree, deleting a tree by freeing its nodes, determining if two trees are identical, finding the minimum value in a BST, creating a mirror copy of a tree, computing the maximum depth, returning a pointer to the nth node of an inorder traversal, level order traversal, deleting a node from a BST, searching a BST, counting leaves, and iterative preorder, inorder and postorder traversals. For each operation, the code is presented and discussed.
Write code in c++ Program to a topological sort on a graph Program pl.docx
Write code in c++
Program to a topological sort on a graph Program plan: Initializes main() function. Declare int type variables. Accept the number of vertices in the graph and the adjacency matrix of in-degree. Perform topological sort using for loops. Display the result.
Solution
Please find the required code and output below:
#include <iostream>
using namespace std;
struct node_info
{
int no;
int lv_time, st_time;
}*q = NULL, *r = NULL;
struct node
{
node_info *pt;
node *next;
}*top = NULL, *p = NULL, *np = NULL;
int c = 0;
void push(node_info *ptr)
{
np = new node;
np->pt = ptr;
np->next = NULL;
if (top == NULL)
{
top = np;
}
else
{
np->next = top;
top = np;
}
}
node_info *pop()
{
if (top == NULL)
{
cout<<\"underflow\ \";
}
else
{
p = top;
top = top->next;
return(p->pt);
delete(p);
}
}
void topologicalSort(int *v,int am[][7],int i)
{
q = new node_info;
q->no = i;
q->st_time = c;
cout<<\"start time for node no \"<<q->no<<\":\"<<c<<endl;
push(q);
v[i] = 1;
for (int j = 0; j < 7; j++)
{
if (am[i][j] == 0 || (am[i][j] == 1 && v[j] == 1))
continue;
else if(am[i][j] == 1 && v[j] == 0)
{
c++;
topologicalSort(v,am,j);
}
}
c++;
r = pop();
cout<<\"leave time for \"<<r->no<<\":\"<<c<<endl;
return;
}
int main()
{
int v[7],am[7][7];
for (int i = 0; i < 7; i++)
v[i] = 0;
for (int i = 0; i < 7; i++)
{
cout<<\"enter the values for adjacency matrix row:\"<<i + 1<<endl;
for(int j = 0; j < 7; j++)
{
cin>>am[i][j];
}
}
topologicalSort(v,am,0);
}
-----------------------------------------------------------------------------------------------
OUTPUT:
.
This assignment and the next (#5) involve design and development of a.pdf
This assignment and the next (#5) involve design and development of a sequential
non contiguous and dynamic datastructure called LinkedList. A linked list object is
a container consisting of connected ListNode objects. As before, we are not going
to use pre-fabricated classes from the c++ library, but construct the LinkedList
ADT from scratch.
The first step is construction and testing of the ListNode class. A ListNode object
contains a data field and a pointer to the next ListNode object (note the recursive
definition).
#This assignment requires you to
1. Read the Assignment 4 Notes
2. Watch the Assignment 4 Support video
3. Implement the following methods of the ListNode class
-custom constructor
-setters for next pointer and data
4. Implement the insert and remove method in the main program
5. Scan the given template to find the above //TODO and implement the code
needed
//TODO in ListNodecpp.h file
public: ListNode(T idata, ListNode<T> * newNext);
public: void setNext(ListNode<T> * newNext);
public: void setData(T newData);
// TODO in main program
void remove(ListNode<int> * &front,int value)
void insert(ListNode<int> * &front,int value)
# The driver is given ListNodeMain.cpp is given to you that does the following
tests
1. Declares a pointer called front to point to a ListNode of datatype integer
2. Constructs four ListNodes with data 1,2,4 and adds them to form a linked
list.
3. Inserts ListNode with data 3 to the list
4. Removes node 1 and adds it back to test removing and adding the first
element
5. Removes node 3 to test removing a middle node
6. Removes node 4 to test removing the last node
7. Attempt to remove a non existent node
8. Remove all existing nodes to empty the list
9. Insert node 4 and then node 1 to test if insertions preserve order
10.Print the list
Main.cpp
#include <iostream>
#include "ListNodecpp.h"
// REMEMBER each ListNode has two parts : a data field
// and an address field. The address is either null or points to the next node
//in the chain
//Requires: integer value for searching, address of front
//Effects: traverses the list node chain starting from front until the end comparing search value
with listnode getData. Returns the original search value if found, if not adds +1 to indicate not
found
//Modifies: Nothing
int search(ListNode<int> * front, int value);
//Requires: integer value for inserting, address of front
//Effects: creates a new ListNode with value and inserts in proper position (increasing order)in
the chain. If chain is empty, adds to the beginning
//Modifies: front, if node is added at the beginning.
//Also changes the next pointer of the previous node to point to the
//newly inserted list node. the next pointer of the newly inserted pointer
//points to what was the next of the previous node.
//This way both previous and current links are adjusted
//******** NOTE the use of & in passing pointer to front as parameter -
// Why do you think this is needed ?**********
void insert(ListNode<int> * &fr.
C++ Please write the whole code that is needed for this assignment- wr.docx
Brianca plans to save $5,000, $1,000, and $42,000 a year over the next three years, respectively. How much would you need to deposit in one lump sum today to have the same amount as Brianca three years from now if you both earn 10.9 percent, compounded annually? $36,115 $35,192 $43,282 $41,635 $35,372
.
DATASTRUCTURES PPTS PREPARED BY M V BRAHMANANDA REDDY
This document discusses binary search trees and provides source code to implement binary search tree traversal operations. It explains that a binary search tree is a binary tree where the left child of a node is less than the parent and the right child is greater. It provides code for preorder, inorder, and postorder traversals of the binary search tree and outputs the traversal order for a sample tree.
Write a program that accepts an arithmetic expression of unsigned in.pdf
Write a program that accepts an arithmetic expression of unsigned integers in postfix notation
and builds the arithmetic expression tree that represents that expression. From that tree, the
corresponding fully parenthesized infix expression should be displayed and a file should be
generated that contains the three address format instructions. This topic is discussed in the week
4 reading in module 2, section II-B. The main class should create the GUI shown below:
Pressing the Construct Tree button should cause the tree to be constructed and using that tree, the
corresponding infix expression should be displayed and the three address instruction file should
be generated. The postfix expression input should not be required to have spaces between every
token. Note in the above example that 9+- are not separated by spaces. The above example
should produce the following output file containing the three address instructions:
Add R0 5 9
Sub R1 3
R0 Mul R2 2 3
Div R3 R1 R2
Inheritance should be used to define the arithmetic expression tree. At a minimum, it should
involve three classes: an abstract class for the tree nodes and two derived classes, one for
operand nodes and another for operator nodes. Other classes should be included as needed to
accomplish good object-oriented design. All instance data must be declared as private.
You may assume that the expression is syntactically correct with regard to the order of operators
and operands, but you should check for invalid tokens, such as characters that are not valid
operators or operands such as 2a, which are not valid integers. If an invalid token is detected a
RuntimeException should be thrown and caught by the main class and an appropriate error
message should be displayed. Three Adddress Generator Enter Postfix Expression 359 2 3
Construct Tree Infix Expression ((3-(5 9))/ (2* 3
Solution
#include
#include
#include
#include
#include
struct TREE //Structure to represent a tree
{
char data;
struct TREE *right,*left,*root;
};
typedef TREE tree; /* Stack Using Linked List */
struct Stack //Structure to represent Stack
{
struct TREE *data;
struct Stack *next,*head,*top; //Pointers to next node,head and top
};
typedef struct Stack node;
node *Nw; //Global Variable to represent node
void initialize(node *&n)
{
n = new node;
n -> next = n -> head = n -> top = NULL;
}
void create(node *n)
{
Nw = new node; //Create a new node
Nw -> next = NULL; //Initialize next pointer field
if(n -> head == NULL) //if first node
{
n -> head = Nw; //Initialize head
n -> top = Nw; //Update top
}
}
void push(node *n,tree *ITEM)
{
node *temp = n -> head;
if(n -> head == NULL) //if First Item is Pushed to Stack
{
create(n); //Create a Node
n -> head -> data = ITEM; //Insert Item to head of List
n -> head = Nw;
return; //Exit from Function
}
create(n); //Create a new Node
Nw -> data = ITEM; //Insert Item
while(temp -> next != NULL)
temp = temp -> next; //Go to Last Node
temp -> next = Nw; //Point New node
n -> top = Nw; //Upda.
DS Code (CWH).docx
The document discusses various operations on linked lists and arrays in C language like creation, traversal, insertion, deletion etc. It includes code snippets to implement array as an abstract data type, linear and binary search on arrays, creation and traversal of singly linked lists, different ways to insert and delete nodes from linked lists, operations on circular linked lists and basics of doubly linked lists.
#include iostream #include deque #include stdio.h scan.pdf
#include
#include
#include //scanf,printf
#include
#include //istringstream
using namespace std;
//maps
#include \"MapGeneric.h\"
#include \"MapTriple.h\"
#include \"MapAbsoluteValue.h\"
#include \"MapSquare.h\"
//filters
#include \"FilterGeneric.h\"
#include \"filterEven.h\"
#include \"filterPositive.h\"
#include \"filterPositiveTwoDigit.h\"
//reduce
#include \"ReduceGeneric.h\"
#include \"ReduceMinimum.h\"
#include \"ReduceGCD.h\"
int main()
{
deque L,L1,L2;
int i;
string line;
getline(cin,line);
if(line.size() == 0)
return 0;
line.erase(remove(line.begin(), line.end(), \',\'), line.end());
istringstream s(line);
while(s>>i)
{
L.push_back(i);
}
if(L.size() != 20)
return 0;
//converting L -> 3*|L| aka L1
MapTriple triple; MapAbsoluteValue abs;
L1 = triple.map(abs.map(L));
//converting L1 -> positive two digit integers -> even intergers aka L2
filterEven even; filterPositiveTwoDigit p2;
L2 = p2.filter(L1);
if(L2.size() <2)
return 0;
L2 = even.filter(L2);
if(L2.size() <2)
return 0;
//finding the minimum and GCD from L2 respectively
ReduceMinimum min; ReduceGCD gcd;
printf(\"%d %d\ \",min.reduce(L2),gcd.reduce(L2));
}
FilterGeneric.cpp
#include
#include
#include \"FilterGeneric.h\"
using namespace std;
deque FilterGeneric::filter(deque deq)
{
//calling recursive function with deq and its length
select(deq,deq.size()-1);
return filterDeq;
}
//int i is the index that will start from the length of deque till 0
void FilterGeneric::select(deque deq, int i)
{
if(i == 0)
{
bool val = f(deq[i]);
if(val)
filterDeq.push_front(deq[i]);
return;
}
bool val = f(deq[i]);
if(val)
filterDeq.push_front(deq[i]);
return select(deq,i-1);
}
FilterGeneric.h
#ifndef FILTERGENERIC_H
#define FILTERGENERIC_H
#include
class FilterGeneric
{
public:
std::deque filter(std::deque);
void select(std::deque,int);
protected:
//this deque will store the integers after filtration.
std::deque filterDeq;
private:
virtual bool f(int)=0;
};
#endif
MapAbsoluteValue.cpp
#include
#include
#include
#include \"MapGeneric.h\" //base
#include \"MapAbsoluteValue.h\"
using namespace std;
int MapAbsoluteValue::f(int num)
{
return abs(num);
}
MapAbsoluteValue.h
#ifndef MAPABSOLUTEVALUE_H
#define MAPABSOLUTEVALUE_H
#include
//this class will map a given input N, using the function map(input), to |N|
class MapAbsoluteValue: public MapGeneric
{
public:
//overwriting pure virtual function making N -> |N|
int f(int);
private:
};
#endif
MapGeneric.cpp
#include
#include
#include \"MapGeneric.h\"
using namespace std;
deque MapGeneric::map(deque deq)
{
//calling recursive function with deq and its length
calculate(deq,deq.size()-1);
return mapDeq;
}
//int i is the index that will start from the length of deque till 0
void MapGeneric::calculate(deque deq, int i)
{
if(i == 0)
{
int val = f(deq[i]);
mapDeq.push_front(val);
return;
}
int val = f(deq[i]);
mapDeq.push_front(val);
return calculate(deq,i-1);
}
MapGeneric.h
#ifndef MAPGENERIC_H
#define MAPGENERIC_H
#include
class MapGeneric
{
public:
std::deque ma.
#include iostream#include d_node.h #include d_nodel.h.docx
The document defines C++ functions for working with linked lists: getMax returns the maximum node, eraseValue removes the first occurrence of a target value, sum returns the sum of all node values, and outputReverse prints the list in reverse order. The main function creates a linked list of random numbers, prints it, calls sum and outputReverse, then removes nodes in descending order while printing. Templates are used to allow different data types.
DoublyList-cpp- #include -DoublyList-h- using namespace std- void Doub.pdf
DoublyList.cpp:
#include "DoublyList.h"
using namespace std;
void DoublyList::insertFront(int newData)
{
if (first == nullptr)
{
first = new DLLNode(newData, nullptr, nullptr);
last = first;
// Common error: Forgetting to reset pointer last.
}
else
{
first = new DLLNode(newData, nullptr, first);
first->getNext()->setPrev(first);
// Common error: Forgetting to connect pointer
// prev of what is now the second node to the
// new first node.
}
++count;
}
void DoublyList::printForward() const
{
DLLNode* current = first;
while (current != nullptr)
{
cout << current->getData() << " ";
current = current->getNext();
}
}
void DoublyList::printReverse() const
{
DLLNode* current = last;
while (current != nullptr)
{
cout << current->getData() << " ";
current = current->getPrev();
}
}
void DoublyList::clearList()
{
DLLNode* temp = first;
while (first != nullptr)
{
first = first->getNext();
delete temp;
temp = first;
}
last = nullptr;
// Don't forget to reset pointer last to nullptr.
count = 0;
}
DoublyList::~DoublyList()
{
if (first != nullptr)
clearList();
}
DoublyList.h
#ifndef DOUBLYLIST_H
#define DOUBLYLIST_H
#include <string>
#include <iostream>
class DLLNode
{
public:
DLLNode() : data(0), prev(nullptr), next(nullptr) {}
DLLNode(int theData, DLLNode* prevLink, DLLNode* nextLink)
: data(theData), prev(prevLink), next(nextLink) {}
int getData() const { return data; }
DLLNode* getPrev() const { return prev; }
DLLNode* getNext() const { return next; }
void setData(int theData) { data = theData; }
void setPrev(DLLNode* prevLink) { prev = prevLink; }
void setNext(DLLNode* nextLink) { next = nextLink; }
~DLLNode(){}
private:
int data; // To simplify, we are using only one piece of data.
DLLNode* prev;
DLLNode* next;
};
class DoublyList
{
public:
DoublyList() : first(nullptr), last(nullptr), count(0) {}
void insertFront(int newData);
void printForward() const;
void printReverse() const;
void rotateNodesRight(int);
void clearList();
~DoublyList();
private:
// Pointer to the first node in the list.
DLLNode*first;
// Pointer to the last node in the list.
DLLNode*last;
// Number of nodes in the list.
int count;
};
#endif
Main.cpp
#include "DoublyList.h"
#include <iostream>
#include <vector>
using namespace std;
int main()
{
vector<vector<int>> data = {
{25, 76, 35, 67, 15, 98},
{1, 2, 3, 4, 5, 6, 7, 8, 9},
{10, 20},
{34, 56, 78, 12, 89, 34, 76, 28, 54, 22, 41},
{123, 873, 619},
};
vector<int> nodesToRotate = { 2, 3, 1, 9, 2 };
{
DoublyList doublyList;
int vectorSize = static_cast<int>(data.size());
for (int i = 0; i < vectorSize; ++i)
{
int innerSize = static_cast<int>(data[i].size());
for (int j = innerSize - 1; j >= 0; --j)
doublyList.insertFront(data[i].at(j));
cout << "Rotate right: " << nodesToRotate[i] << "\n";
cout << " List is: ";
doublyList.printForward();
cout << "\n";
doublyList.rotateNodesRight(nodesToRotate[i]);
cout << "After rotating:";
cout << "\n Print forward: ";
doublyList.printForward();
cout << "\nPrint backwards: ";
doublyList.printReverse(.
Lab-2.4 101.pdf
The document contains C program code to implement three data structures using linked lists:
1) A stack using a singly linked list with push, pop, and display functions.
2) A queue using a singly linked list with enqueue, dequeue, and display functions.
3) A program to merge two sorted linked lists by comparing nodes and appending to the merged list.
Pseudocode and sample test cases are provided for each problem.
Assignment isPage 349-350 #4 and #5 Use the Linked List lab.pdf
Assignment is:
\"Page 349-350 #4 and #5 Use the \"Linked List lab\" you have been working on in class and add
the two functions the questions are asking you to develop: divideMid and divideAt. Be sure to
include comments Use meaningful identifier names (constants where appropriate) Turn in .cpp
file AND Turn in a \"print-screen\' of your output (press \"print-screen\' on keyboard, then
\'paste\' in MS-Word)\"
How do you solve QUESTION #4 in the book data structures using c++ by D.S. Malik in Visiual
Studios using the linked list below with what is being asked? Please need help
Linked list :
#include
#include
using namespace std;
struct nodeType
{
int info;
nodeType *link;
};
void createList(nodeType*& first, nodeType*& last);
void printList(nodeType*& first);
void insertFront(nodeType*& first);
void insertBack(nodeType*& last);
void deleteFirst(nodeType*& first);
void deleteLast(nodeType*& last, nodeType* first);
int main()
{
nodeType *first, *last;
int num;
createList(first, last);
int choice;
while(true)
{
cout<<\"1. Insert Front.\ 2. Insert Last.\ 3. Delete Front.\ 4. Delete Last.\ 5. Print List.\ 6. Exit.\
\";
cout<<\"Enter your choice: \";
cin>>choice;
switch(choice)
{
case 1: insertFront(first); break;
case 2: insertBack(last); break;
case 3: deleteFirst(first); break;
case 4: deleteLast(last, first); break;
case 5: printList(first); break;
case 6: return 0;
default: cout<<\"Invalid menu option. Try again.\"<>number;
while (number != -999)
{
newNode = new nodeType; // create new node
newNode->info = number;
newNode->link = NULL;
if (first == NULL)
{
first = newNode;
last = newNode;
}
else
{
last->link = newNode;
last = newNode;
}
cout<<\"Enter an integer (-999 to stop): \";
cin>>number;
} // end of while-loop
} // end of build list function
void deleteFirst(nodeType*& first)
{
nodeType *temp;
temp= first;
first= temp->link;
delete temp;
return;
}
void deleteLast(nodeType*& last, nodeType* current)
{
nodeType *temp;
while(current->link != NULL)
{
temp=current;
current=current->link;
}
temp=last;
current->link=NULL;
delete temp;
last = current;
return;
}
void insertFront(nodeType*& front)
{
int num;
cout<<\"\ Enter the number to insert: \";
cin>>num;
nodeType *newNode = new nodeType;
newNode->info=num;
newNode->link= front;
front= newNode;
return;
}
void insertBack(nodeType*& last)
{
int num;
cout<<\"\ Enter the number to insert: \";
cin>>num;
nodeType *newNode = new nodeType;
newNode->info=num;
newNode->link= NULL;
last->link= newNode;
last = newNode;
return;
}
void printList(nodeType*& first)
{
cout<<\"Inside printList...printing linked list...\ \"<info << \" \";
current = current->link;
}
cout<
#include
using namespace std;
struct nodeType
{
int info;
nodeType *link;
};
void createList(nodeType*& first, nodeType*& last);
void printList(nodeType*& first);
void insertFront(nodeType*& first);
void insertBack(nodeType*& last);
void deleteFirst(nodeType*& first);
void deleteLast(nodeType*& last, nodeType* first);
int main()
{
nodeType *firs.
I want help in the following C++ programming task. Please do coding .pdf
I want help in the following C++ programming task. Please do coding in C++.
Persistent Data Structures
A persistent data structure is one in which we can make changes (such as inserting and removing
elements) while still preserving the original data structure Of course, an easy way to do this is to
create a new copy of the entire data structure every time we perform an operation that changes
the data structure, but it is much more efficient to only make a copy of the smallest possible
portion of the data structure.
First, let\'s look at one of the easiest possible data structures — a stack, implemented using a
linked list.
Persistent Stack
Let\'s say that we have a stack S1, and we want to create a new stack S2, which is the result of
pushing a new element X onto S1, without changing S1 at all. This is easy — we just add a new
element that points to S1:
Starting with the stack S1:
We push X on S1, to get S2, leaving S1 as it was. So after the operation we have two versions of
the stack S1 the older version with elements A, B and C and S2 the newer version with elements
X, A, B and C.
What about popping? That works in the same way. Starting with our original S1 (elements A, B
and C), we can pop off the first element (A), to get S2:
What if we want to push something on to this new stack? It works in the same was as before – if
we push an X onto S2 above to get S3, we have
So for every operation you make a new version of the stack while reusing any existing stack
elements.
You will have to make a linked list of pointers that will point to the head of the different stack
versions.
Also each stack is also implemented as a linked list of elements SAB
Solution
#include #include #include #include #include #include #include #include #include
#include #include namespace dts { template > class PersistentSet { public:
PersistentSet(); PersistentSet(Func); bool add(const T&); bool add(T&&);
bool remove(const T& key); bool empty() const; size_t history_size() const;
class TreeIterator : public std::iterator, std::ptrdiff_t, const T*,
const T&> { using node = typename dts::PersistentSet< std::remove_cv_t,
Func>::Nodeptr; node itr; node nil; std::stack path; node
find_successor(node n) { n = n->rigth; if (n != nil) {
while (n->left != nil) { path.push(n); n = n->left;
} } else { n = path.top();
path.pop(); } return n; } public: explicit
TreeIterator(node n, node pnil) : nil(pnil) //begin { if (n == nil) itr
= nil; else { path.push(nil); while (n->left != nil)
{ path.push(n); n = n->left; } itr =
n; } } explicit TreeIterator(node pnil) // end : itr(pnil),
nil(pnil) { } TreeIterator& operator++ () { itr =
find_successor(itr); return *this; } TreeIterator operator++ (int)
{ TreeIterator tmp(*this); itr = find_successor(itr); return tmp;
} bool operator == (const TreeIterator& rhs) const { return itr ==
rhs.itr; } bool operator != (const TreeIterator& rhs) const {
return itr != rhs.itr; } const T& operator* () const { return itr-
>key; } const T& operator-> () c.
In C++ I need help with this method that Im trying to write fillLi.pdf
In C++ I need help with this method that I'm trying to write: fillListFromFile( ) : void -- this
method will add elements to the list from a file called linkedListData1.txt HINT: this is where
you will keep track of the number of elements in the list
I have all of the other methods, but I'm stuck on this one and I've added it as the last method in
the application. The code I have so far is below.
.cpp file:
.h file:
main:
#include
#include
#include
#include "Node.h"
using namespace std;
void displayList(Node*);
void insertFront(Node** head, int value);
void insertBack(Node** ptrToHead, int value);
void insertAfter(Node* previous, int value);
void deleteAtN(Node** ptrToHead, int n);
void deleteNode(Node** ptrToHead, int searchVal);
void clearList(Node**);
bool searchVal(Node** head, int value);
int countFrequencyOf(Node** ptrTohead, int value);
int size(Node** ptrTohead);
void displayFromFrontToN(Node** ptrToHead, int n);
void displayFromNToEnd(Node** ptrToHead, int n);
void insertInOrder(Node** head, Node* ptrToHead);
void displayInRows(Node** ptrToHead);
int main()
{
Node* node1 = new Node();
node1->setValue(10);
Node* node2 = new Node();
node2->setValue(20);
Node* node3 = new Node();
node3->setValue(30);
node1->next = node2;
node2->next = node3;
Node* head = node1;
displayList(head);
insertFront(&head, 5);
displayList(head);
insertBack(&head, 50);
displayList(head);
insertBack(&head, 60);
displayList(head);
insertAfter(node2, 25);
displayList(head);
deleteAtN(&head, 2);
displayList(head);
deleteNode(&head, 50);
displayList(head);
clearList(&head);
displayList(head);
insertFront(&head, 5);
insertFront(&head, 10);
insertFront(&head, 20);
insertBack(&head, 50);
searchVal(&head, 50);
displayList(head);
countFrequencyOf(&head, 25);
displayList(head);
insertBack(&head, 25);
insertFront(&head, 25);
countFrequencyOf(&head, 25);
displayList(head);
size(&head);
cout << "add one item to list" << endl;
insertBack(&head, 25);
size(&head);
displayList(head);
cout << "Display From Front To N" << endl;
displayFromFrontToN(&head, 4);
cout << "Display From N To End" << endl;
displayFromNToEnd(&head, 3);
cout << "Display In Rows" << endl;
displayInRows(&head);
insertFront(&head, 12);
insertFront(&head, 17);
insertFront(&head, 22);
insertBack(&head, 58);
cout << "Display In Rows" << endl;
displayInRows(&head);//call displayInRows()
return 0;
}//end main()
void displayList(Node* ptr) {
if (ptr == nullptr)
cout << "empty list" << endl;
else {
while (ptr != nullptr) {
cout << ptr->getValue() << endl;
ptr = ptr->next;
}
}
cout << "---------------------------" << endl;
}//end displayList()
void insertFront(Node** head, int value) {
Node* newNode = new Node();
newNode->setValue(value);
//cout<< newNode->getValue() << endl;
newNode->next = *head;
*head = newNode;
}
void insertBack(Node** ptrToHead, int value) {
Node* newNode = new Node();
newNode->setValue(value);
newNode->next = nullptr;
if (*ptrToHead == nullptr)
{
*ptrToHead = newNode;
}//end if
else
{
Node* last .
Modify this code to use multiple threads with the same data1.Modif.pdf
Modify this code to use multiple threads with the same data
1.Modify the main function to implement a loop that reads 10 integers from the console (user
input) and stores these numbers in a one-dimensional (1D) array (this code will go right after the
comment that says “Add code to perform any needed initialization or to process user input”).
You should use a global array for this.
2.Implement a separatepthread function function for each one of the following operations:
a.Count and print out how many of the entered numbers are negative. This function must be
named countNegatives
b.Calculate and print the average value of all the numbers entered. This function must be named
average
c.Print the numbers in reverse order from the order in which they were entered. This function
must be named reverse
3.Modify the main function to create one pthread for each one of the functions that you
implemented in (3) above (this code will go between the comment that says “TODO: Modify
according to assignment requirements” and the “if (rc)” check).
Compile your program and run it several times. If the output of your program is garbled, you
may need to add a small delay in between creating the next thread.
#include
#include
using namespace std;
void *routineName(void *arg)
{
// TODO: Add code that implements
// the thread\'s functionality
cout << \"Thread is running...\" << endl;
return 0;
}
int main()
{
pthread_t id;
int rc;
int ints;
for(int x; x<10; x++)
{
cout << \"Enter Integer: \" <>
}
rc = pthread_create(&id, NULL, routineName, NULL);
if (rc){
cout << \"ERROR; return code from pthread_create() is \" << rc << endl;
return -1;
}
pthread_exit(0);
}
Solution
#include
#include
#include
using namespace std;
int ints[10];
//Method to count negatives in the input array
void *countNegatives(void*){
int count = 0;
for(int i=0;i<10;i++){
if(ints[i]<0){
count++;
}
}
cout<<\"Number of negatives: \"<=0;i--){
cout<> ints[x];
}
//Creating pthread to countNegatives
rc = pthread_create(&t1, NULL, countNegatives, NULL);
if(rc){
cout << \"Error:unable to create thread,\" << rc << endl;
}
//Adding delay
sleep(1);
//Creating pthread to compute average
rc = pthread_create(&t2, NULL, average, NULL);
if(rc){
cout << \"Error:unable to create thread,\" << rc << endl;
}
//Adding delay
sleep(1);
//Creating pthread to print in reverse
rc = pthread_create(&t3, NULL, reverse, NULL);
if(rc){
cout << \"Error:unable to create thread,\" << rc << endl;
}
//Adding delay
sleep(1);
void* status;
//waiting for t1 to join
rc = pthread_join(t1,&status);
if(rc){
cout << \"Error:unable to join,\" << rc << endl;
}
//waiting for t2 to join
rc = pthread_join(t2,&status);
if(rc){
cout << \"Error:unable to join,\" << rc << endl;
}
//waiting for t3 to join
rc = pthread_join(t3,&status);
if(rc){
cout << \"Error:unable to join,\" << rc << endl;
}
pthread_exit(0);
}.
Doublylinklist
This document presents information on doubly linked lists including their representation, initialization, node creation, and various operations like insertion, deletion, and traversal. It discusses inserting and deleting nodes at the beginning or end of the list, as well as inserting before or after a specified node. Code examples are provided for initializing a doubly linked list and performing each operation.
Help with the following code1. Rewrite to be contained in a vecto.pdf
Help with the following code:
1. Rewrite to be contained in a vector object.
2. Using linked lists, redo to handle as many entries as required.
3. Add the ability to add or delete a new entry to the address book.
4. When the program terminates, write the data in the address book to a disk.
# include
# include
# include
using namespace std;
class addressbook
{
private:
char name[500];
long ph_num; //class addressType :public addressbook
char address[500];
int year;
int mounth;
int day;
char gender;
int d, Th, y;
char M, F, m, f;
// char freind[];
public:
addressbook();
void setName();
void getName();
void getphone();
long setphone();
void setAddress();
void getAddress();
float getAverage();
int setDay();
void getDay();
int setMounth();
void getMounth();
int setYear();
void getYear();
char setGender();
void getGender();
};
addressbook::addressbook() // constructor
{
name;
ph_num = 0;
year = 0;
mounth = 0;
day = 0;
}
int addressbook::setDay()
{
cout << \"\ \\tEnter the day:\ \\t\" << endl;
cin >> d;
day = (d >= 1 && d<32) ? d : 0;
return day;
}
int addressbook::setMounth()
{
cout << \"\ \\tEnter the mounth:\ \\t\" << endl;
cin >> m;
mounth = (Th >= 1 && Th<13) ? Th : 0;
return mounth;
}
int addressbook::setYear()
{
cout << \"\ \\tEnter the year :\ \\t\" << endl;
cin >> y;
year = (y >= 1 && y<2008) ? y : 0;
return year;
}
void addressbook::getMounth()
{
cout << mounth << \"/\";
}
void addressbook::getDay()
{
cout << \"\\tBIRTH DAY : \" << day << \"/\";
}
void addressbook::getYear()
{
cout << year;
}
char addressbook::setGender()
{
int flag = 0;
cout << \" \ \\tEnter the gender,(F)for female (M)for male:\ \\t\";
cin >> gender;
while (flag != 1)
{
if (gender == (\'F\') || gender == (\'M\') || gender == (\'m\') || gender == (\'f\'))
flag = 1;
else
{
cout << \"\\t error.....Try Again\ \\t\";
cin >> gender;
return gender;
}
}
}
void addressbook::getGender()
{
cout << \"\\tGender : \" << gender << endl;
}
void addressbook::getAddress()
{
cout << \"\\tAddress : \" << address << endl;
}
float addressbook::getAverage()
{
return 0.0f;
}
void addressbook::setAddress()
{
cout << \" \ \\tEnter the address:\ \\t\" << endl;
gets_s(address);
}
void addressbook::setName()
{
cout << \" \\tEnter the name:\ \\t\" << endl;
gets_s(name);
}
void addressbook::getName()
{
cout << \"\ \\tName : \" << name << \" \" << endl;
}
long addressbook::setphone()
{
cout << \" \ \\tEnter the person number:\ \\t\" << endl;
cin >> ph_num;
return ph_num;
}
void addressbook::getphone()
{
cout << \"\\tPerson # : \" << ph_num << endl;
}
/*class addressType :public addressbook
{
do that
};*/
int main()
{
int s;
cout << \" Enter number person max(500): \";
cin >> s;
addressbook book[10];
for (int i = 1; i<2; i++)
{
book[i].setName();
book[i].setphone();
book[i].setAddress();
book[i].setDay();
book[i].setMounth();
book[i].setYear();
book[i].setGender();
cout << \"\ \ ***********************************\ \";
}
for (int u = 0; u<2; u++)
{
book[u].getName();
book[u].getphone();
book[u].getAddre.
Ds 2 cycle
The document discusses the implementation of a binary search tree (BST) using both recursive and non-recursive traversal methods. It defines a BST node structure with left and right child pointers. Functions are included to insert nodes, perform inorder, preorder and postorder traversals recursively and non-recursively using stacks. Main inserts sample nodes into the tree and calls the various traversal functions to output the node values in the specified orders.
C++ adt c++ implementations
This document provides code examples for implementing common data structures in C++. It includes code for implementing a binary search tree with functions for insertion, removal, and traversal. It also includes code for implementing stacks, queues, and trees as abstract data types with functions for common operations like push, pop, enqueue, dequeue, and traversal. The code demonstrates how to create node structures to represent data elements and link them together to form the desired data structures.
Write a C program that reads the words the user types at the command.pdf
Write a C program that reads the words the user types at the command prompt (using the \'int
argc, char * argv[] and store each unique letter in a Binary Search Tree. When a duplicate is
encountered do not store the letter again and instead keep track of the count in the tree. Once the
Binary Search tree has been created print out the tree both inorder and reverse order. Also print
the highest and lowest alphabetically letter in the tree if any.
Solution
# include
# include
# include
typedef struct BST {
int data;
struct BST *lchild, *rchild;
} node;
void insert(node *, node *);
void preorder(node *);
findMinimum(struct node* root)
findMaximum(struct node* root)
void reverseLevelOrder(struct node* root)
void main(int argc,char argv) {
char argv = \'N\';
int key;
node *new_node, *root, *tmp, *parent;
node *get_node();
root = NULL;
clrscr();
printf(\"\ Program For Binary Search Tree \");
do {
printf(\"\ 1.Create\");
printf(\"\ 2.Search\");
printf(\"\ 3.Recursive Traversals\");
printf(\"\ 4.Exit\");
printf(\"\ Enter your choice :\");
scanf(\"%d\", &argc);
switch (argc) {
case 1:
do {
new_node = get_node();
printf(\"\ Enter The Element \");
scanf(\"%d\", &new_node->data);
if (root == NULL) /* Tree is not Created */
root = new_node;
else
insert(root, new_node);
printf(\"\ Want To enter More Elements?(y/n)\");
argv= getch();
} while (argv == \'y\');
break;
case 2:
if (root == NULL)
printf(\"Tree Is Not Created\");
else {
printf(\"\ The Preorder display : \");
preorder(root);
}
break;
}
} while (argv != 4);
}
/*
Get new Node
*/
node *get_node() {
node *temp;
temp = (node *) malloc(sizeof(node));
temp->lchild = NULL;
temp->rchild = NULL;
return temp;
}
/*
This function is for creating a binary search tree
*/
void insert(node *root, node *new_node) {
if (new_node->data < root->data) {
if (root->lchild == NULL)
root->lchild = new_node;
return newNode(key);
else
insert(root->lchild, new_node);
}
if (new_node->data > root->data) {
if (root->rchild == NULL)
root->rchild = new_node;
return newNode(key);
else
insert(root->rchild, new_node);
}
}
if (key == node->key)
{
(node->count)++;
return node;
}
/*
This function displays the tree in preorder fashion
*/
void preorder(node *temp) {
if (temp != NULL) {
printf(\"%d\", temp->data);
preorder(temp->lchild);
preorder(temp->rchild);
}
}
// Returns maximum value in a given Binary Tree
int findMaximum(struct node* root)
{
// Base case
if (root == NULL)
return INT_MAXIMUM;
// Return maximum of 3 values:
// 1) Root\'s data 2) Max in Left Subtree
// 3) Max in right subtree
int res = root->data;
int lres = findMaximum (root->lchild);
int rres = findMaximum (root->rchild);
if (lres > res)
res = lres;
if (rres > res)
res = rres;
return res;
}
// Returns minimum value in a given Binary Tree
int findMinimum(struct node* root)
{
// Base case
if (root == NULL)
return INT_MINIMUM;
// Return minimum of 3 values:
// 1) Root\'s data 2) Max in Left Subtree
// 3) Max in right subtree
int res = root->data;
int lres = findMinimum(r.
#include stdio.h #include stdlib.h #include unistd.h .pdf
#include
#include
#include
#include
#include
#include \"msecond.h\"
#include \"random_int.h\"
float total_time_spent_waiting = 0.0;
int total_number_of_meals = 0;
#define semaphore_create(s,v) sem_init( &s, 0, v )
#define semaphore_wait(s) sem_wait( &s )
#define semaphore_signal(s) sem_post( &s )
#define semaphore_release(s) sem_destroy( &s )
typedef sem_t semaphore;
semaphore chopstick[NUM_PHILOSOPHERS];
semaphore screen;
semaphore mutex;
int screen_row[NUM_PHILOSOPHERS] = { 6, 2, 2, 6, 10 };
int screen_col[NUM_PHILOSOPHERS] = { 31, 36, 44, 49, 40 };
int chopstick_row[5] = { 9, 4, 3, 4, 9 };
int chopstick_col[5] = { 35, 33, 40, 47, 45 };
char chopstick_sym[5] = { \'/\', \'\\\\\', \'|\', \'/\', \'\\\\\' };
for ( i = 0; i < NUM_PHILOSOPHERS; i++ )
{
position( screen_row[i], screen_col[i] );
printf( \"%d\", i );
position( screen_row[i] + 1, screen_col[i] - 1 );
printf( \"(%d)\", 0 );
position( chopstick_row[i], chopstick_col[i] );
printf( \"%c\", chopstick_sym[i] );
}
position_flush( 13, 1 );
}
void draw_thinking( int n )
{
semaphore_wait( screen );
position( screen_row[n], screen_col[n] );
position_flush( 13, 1 );
semaphore_signal( screen );
}
void draw_hungry( int n )
{
semaphore_wait( screen );
position( screen_row[n], screen_col[n] );
position_flush( 13, 1 );
semaphore_signal( screen );
}
void draw_eating( int n, int eat_count )
{
semaphore_wait( screen );
position( screen_row[n], screen_col[n] );
position( screen_row[n] + 1, screen_col[n] - 1 );
printf( \"(%d)\", eat_count );
position_flush( 13, 1 );
semaphore_signal( screen );
}
void draw_chopstick_up( int n )
{
semaphore_wait( screen );
position( chopstick_row[n], chopstick_col[n] );
printf( \"%c\", \' \' );
position_flush( 13, 1 );
semaphore_signal( screen );
}
void draw_chopstick_down( int n )
{
semaphore_wait( screen );
position( chopstick_row[n], chopstick_col[n] );
position_flush( 13, 1 );
semaphore_signal( screen );
}
void draw_done( int n )
{
semaphore_wait( screen );
position( screen_row[n], screen_col[n] );
printf( \"D\" );
position_flush( 13, 1 );
semaphore_signal( screen );
}
void chopsticks( int n )
{
if ( n % 2 == 0 ) {
semaphore_wait( chopstick[(n+1) % NUM_PHILOSOPHERS] );
draw_chopstick_up( (n+1) % NUM_PHILOSOPHERS );
semaphore_wait( chopstick[n] );
draw_chopstick_up( n );
} else {
semaphore_wait( chopstick[n] );
draw_chopstick_up( n );
semaphore_wait( chopstick[(n+1) % NUM_PHILOSOPHERS] );
draw_chopstick_up( (n+1) % NUM_PHILOSOPHERS );
}
}
void release_chopsticks( int n )
{
draw_chopstick_down( n );
semaphore_signal( chopstick[n] );
draw_chopstick_down( (n+1) % NUM_PHILOSOPHERS );
semaphore_signal( chopstick[(n+1) % NUM_PHILOSOPHERS] );
}
void philosopher( int *philosopher_data )
{
int start_time;
int eat_count = 0;
int total_hungry_time = 0;
int became_hungry_time;
int n = philosopher_data[0];
int duration = philosopher_data[1];
start_time = msecond();
while( msecond() - start_time < duration * 1000 )
{
became_hungry_time = msecond();
draw_hungry( n );
chopsticks( n );
t.
More Related Content
Similar to AnswerThe new program with the required constructor and with a te.pdf
Binary Tree
The document provides C code examples for various binary tree operations and traversals including: finding the height/depth and size of a tree, deleting a tree by freeing its nodes, determining if two trees are identical, finding the minimum value in a BST, creating a mirror copy of a tree, computing the maximum depth, returning a pointer to the nth node of an inorder traversal, level order traversal, deleting a node from a BST, searching a BST, counting leaves, and iterative preorder, inorder and postorder traversals. For each operation, the code is presented and discussed.
Write code in c++ Program to a topological sort on a graph Program pl.docx
Write code in c++
Program to a topological sort on a graph Program plan: Initializes main() function. Declare int type variables. Accept the number of vertices in the graph and the adjacency matrix of in-degree. Perform topological sort using for loops. Display the result.
Solution
Please find the required code and output below:
#include <iostream>
using namespace std;
struct node_info
{
int no;
int lv_time, st_time;
}*q = NULL, *r = NULL;
struct node
{
node_info *pt;
node *next;
}*top = NULL, *p = NULL, *np = NULL;
int c = 0;
void push(node_info *ptr)
{
np = new node;
np->pt = ptr;
np->next = NULL;
if (top == NULL)
{
top = np;
}
else
{
np->next = top;
top = np;
}
}
node_info *pop()
{
if (top == NULL)
{
cout<<\"underflow\ \";
}
else
{
p = top;
top = top->next;
return(p->pt);
delete(p);
}
}
void topologicalSort(int *v,int am[][7],int i)
{
q = new node_info;
q->no = i;
q->st_time = c;
cout<<\"start time for node no \"<<q->no<<\":\"<<c<<endl;
push(q);
v[i] = 1;
for (int j = 0; j < 7; j++)
{
if (am[i][j] == 0 || (am[i][j] == 1 && v[j] == 1))
continue;
else if(am[i][j] == 1 && v[j] == 0)
{
c++;
topologicalSort(v,am,j);
}
}
c++;
r = pop();
cout<<\"leave time for \"<<r->no<<\":\"<<c<<endl;
return;
}
int main()
{
int v[7],am[7][7];
for (int i = 0; i < 7; i++)
v[i] = 0;
for (int i = 0; i < 7; i++)
{
cout<<\"enter the values for adjacency matrix row:\"<<i + 1<<endl;
for(int j = 0; j < 7; j++)
{
cin>>am[i][j];
}
}
topologicalSort(v,am,0);
}
-----------------------------------------------------------------------------------------------
OUTPUT:
.
This assignment and the next (#5) involve design and development of a.pdf
This assignment and the next (#5) involve design and development of a sequential
non contiguous and dynamic datastructure called LinkedList. A linked list object is
a container consisting of connected ListNode objects. As before, we are not going
to use pre-fabricated classes from the c++ library, but construct the LinkedList
ADT from scratch.
The first step is construction and testing of the ListNode class. A ListNode object
contains a data field and a pointer to the next ListNode object (note the recursive
definition).
#This assignment requires you to
1. Read the Assignment 4 Notes
2. Watch the Assignment 4 Support video
3. Implement the following methods of the ListNode class
-custom constructor
-setters for next pointer and data
4. Implement the insert and remove method in the main program
5. Scan the given template to find the above //TODO and implement the code
needed
//TODO in ListNodecpp.h file
public: ListNode(T idata, ListNode<T> * newNext);
public: void setNext(ListNode<T> * newNext);
public: void setData(T newData);
// TODO in main program
void remove(ListNode<int> * &front,int value)
void insert(ListNode<int> * &front,int value)
# The driver is given ListNodeMain.cpp is given to you that does the following
tests
1. Declares a pointer called front to point to a ListNode of datatype integer
2. Constructs four ListNodes with data 1,2,4 and adds them to form a linked
list.
3. Inserts ListNode with data 3 to the list
4. Removes node 1 and adds it back to test removing and adding the first
element
5. Removes node 3 to test removing a middle node
6. Removes node 4 to test removing the last node
7. Attempt to remove a non existent node
8. Remove all existing nodes to empty the list
9. Insert node 4 and then node 1 to test if insertions preserve order
10.Print the list
Main.cpp
#include <iostream>
#include "ListNodecpp.h"
// REMEMBER each ListNode has two parts : a data field
// and an address field. The address is either null or points to the next node
//in the chain
//Requires: integer value for searching, address of front
//Effects: traverses the list node chain starting from front until the end comparing search value
with listnode getData. Returns the original search value if found, if not adds +1 to indicate not
found
//Modifies: Nothing
int search(ListNode<int> * front, int value);
//Requires: integer value for inserting, address of front
//Effects: creates a new ListNode with value and inserts in proper position (increasing order)in
the chain. If chain is empty, adds to the beginning
//Modifies: front, if node is added at the beginning.
//Also changes the next pointer of the previous node to point to the
//newly inserted list node. the next pointer of the newly inserted pointer
//points to what was the next of the previous node.
//This way both previous and current links are adjusted
//******** NOTE the use of & in passing pointer to front as parameter -
// Why do you think this is needed ?**********
void insert(ListNode<int> * &fr.
C++ Please write the whole code that is needed for this assignment- wr.docx
Brianca plans to save $5,000, $1,000, and $42,000 a year over the next three years, respectively. How much would you need to deposit in one lump sum today to have the same amount as Brianca three years from now if you both earn 10.9 percent, compounded annually? $36,115 $35,192 $43,282 $41,635 $35,372
.
DATASTRUCTURES PPTS PREPARED BY M V BRAHMANANDA REDDY
This document discusses binary search trees and provides source code to implement binary search tree traversal operations. It explains that a binary search tree is a binary tree where the left child of a node is less than the parent and the right child is greater. It provides code for preorder, inorder, and postorder traversals of the binary search tree and outputs the traversal order for a sample tree.
Write a program that accepts an arithmetic expression of unsigned in.pdf
Write a program that accepts an arithmetic expression of unsigned integers in postfix notation
and builds the arithmetic expression tree that represents that expression. From that tree, the
corresponding fully parenthesized infix expression should be displayed and a file should be
generated that contains the three address format instructions. This topic is discussed in the week
4 reading in module 2, section II-B. The main class should create the GUI shown below:
Pressing the Construct Tree button should cause the tree to be constructed and using that tree, the
corresponding infix expression should be displayed and the three address instruction file should
be generated. The postfix expression input should not be required to have spaces between every
token. Note in the above example that 9+- are not separated by spaces. The above example
should produce the following output file containing the three address instructions:
Add R0 5 9
Sub R1 3
R0 Mul R2 2 3
Div R3 R1 R2
Inheritance should be used to define the arithmetic expression tree. At a minimum, it should
involve three classes: an abstract class for the tree nodes and two derived classes, one for
operand nodes and another for operator nodes. Other classes should be included as needed to
accomplish good object-oriented design. All instance data must be declared as private.
You may assume that the expression is syntactically correct with regard to the order of operators
and operands, but you should check for invalid tokens, such as characters that are not valid
operators or operands such as 2a, which are not valid integers. If an invalid token is detected a
RuntimeException should be thrown and caught by the main class and an appropriate error
message should be displayed. Three Adddress Generator Enter Postfix Expression 359 2 3
Construct Tree Infix Expression ((3-(5 9))/ (2* 3
Solution
#include
#include
#include
#include
#include
struct TREE //Structure to represent a tree
{
char data;
struct TREE *right,*left,*root;
};
typedef TREE tree; /* Stack Using Linked List */
struct Stack //Structure to represent Stack
{
struct TREE *data;
struct Stack *next,*head,*top; //Pointers to next node,head and top
};
typedef struct Stack node;
node *Nw; //Global Variable to represent node
void initialize(node *&n)
{
n = new node;
n -> next = n -> head = n -> top = NULL;
}
void create(node *n)
{
Nw = new node; //Create a new node
Nw -> next = NULL; //Initialize next pointer field
if(n -> head == NULL) //if first node
{
n -> head = Nw; //Initialize head
n -> top = Nw; //Update top
}
}
void push(node *n,tree *ITEM)
{
node *temp = n -> head;
if(n -> head == NULL) //if First Item is Pushed to Stack
{
create(n); //Create a Node
n -> head -> data = ITEM; //Insert Item to head of List
n -> head = Nw;
return; //Exit from Function
}
create(n); //Create a new Node
Nw -> data = ITEM; //Insert Item
while(temp -> next != NULL)
temp = temp -> next; //Go to Last Node
temp -> next = Nw; //Point New node
n -> top = Nw; //Upda.
DS Code (CWH).docx
The document discusses various operations on linked lists and arrays in C language like creation, traversal, insertion, deletion etc. It includes code snippets to implement array as an abstract data type, linear and binary search on arrays, creation and traversal of singly linked lists, different ways to insert and delete nodes from linked lists, operations on circular linked lists and basics of doubly linked lists.
#include iostream #include deque #include stdio.h scan.pdf
#include
#include
#include //scanf,printf
#include
#include //istringstream
using namespace std;
//maps
#include \"MapGeneric.h\"
#include \"MapTriple.h\"
#include \"MapAbsoluteValue.h\"
#include \"MapSquare.h\"
//filters
#include \"FilterGeneric.h\"
#include \"filterEven.h\"
#include \"filterPositive.h\"
#include \"filterPositiveTwoDigit.h\"
//reduce
#include \"ReduceGeneric.h\"
#include \"ReduceMinimum.h\"
#include \"ReduceGCD.h\"
int main()
{
deque L,L1,L2;
int i;
string line;
getline(cin,line);
if(line.size() == 0)
return 0;
line.erase(remove(line.begin(), line.end(), \',\'), line.end());
istringstream s(line);
while(s>>i)
{
L.push_back(i);
}
if(L.size() != 20)
return 0;
//converting L -> 3*|L| aka L1
MapTriple triple; MapAbsoluteValue abs;
L1 = triple.map(abs.map(L));
//converting L1 -> positive two digit integers -> even intergers aka L2
filterEven even; filterPositiveTwoDigit p2;
L2 = p2.filter(L1);
if(L2.size() <2)
return 0;
L2 = even.filter(L2);
if(L2.size() <2)
return 0;
//finding the minimum and GCD from L2 respectively
ReduceMinimum min; ReduceGCD gcd;
printf(\"%d %d\ \",min.reduce(L2),gcd.reduce(L2));
}
FilterGeneric.cpp
#include
#include
#include \"FilterGeneric.h\"
using namespace std;
deque FilterGeneric::filter(deque deq)
{
//calling recursive function with deq and its length
select(deq,deq.size()-1);
return filterDeq;
}
//int i is the index that will start from the length of deque till 0
void FilterGeneric::select(deque deq, int i)
{
if(i == 0)
{
bool val = f(deq[i]);
if(val)
filterDeq.push_front(deq[i]);
return;
}
bool val = f(deq[i]);
if(val)
filterDeq.push_front(deq[i]);
return select(deq,i-1);
}
FilterGeneric.h
#ifndef FILTERGENERIC_H
#define FILTERGENERIC_H
#include
class FilterGeneric
{
public:
std::deque filter(std::deque);
void select(std::deque,int);
protected:
//this deque will store the integers after filtration.
std::deque filterDeq;
private:
virtual bool f(int)=0;
};
#endif
MapAbsoluteValue.cpp
#include
#include
#include
#include \"MapGeneric.h\" //base
#include \"MapAbsoluteValue.h\"
using namespace std;
int MapAbsoluteValue::f(int num)
{
return abs(num);
}
MapAbsoluteValue.h
#ifndef MAPABSOLUTEVALUE_H
#define MAPABSOLUTEVALUE_H
#include
//this class will map a given input N, using the function map(input), to |N|
class MapAbsoluteValue: public MapGeneric
{
public:
//overwriting pure virtual function making N -> |N|
int f(int);
private:
};
#endif
MapGeneric.cpp
#include
#include
#include \"MapGeneric.h\"
using namespace std;
deque MapGeneric::map(deque deq)
{
//calling recursive function with deq and its length
calculate(deq,deq.size()-1);
return mapDeq;
}
//int i is the index that will start from the length of deque till 0
void MapGeneric::calculate(deque deq, int i)
{
if(i == 0)
{
int val = f(deq[i]);
mapDeq.push_front(val);
return;
}
int val = f(deq[i]);
mapDeq.push_front(val);
return calculate(deq,i-1);
}
MapGeneric.h
#ifndef MAPGENERIC_H
#define MAPGENERIC_H
#include
class MapGeneric
{
public:
std::deque ma.
#include iostream#include d_node.h #include d_nodel.h.docx
The document defines C++ functions for working with linked lists: getMax returns the maximum node, eraseValue removes the first occurrence of a target value, sum returns the sum of all node values, and outputReverse prints the list in reverse order. The main function creates a linked list of random numbers, prints it, calls sum and outputReverse, then removes nodes in descending order while printing. Templates are used to allow different data types.
DoublyList-cpp- #include -DoublyList-h- using namespace std- void Doub.pdf
DoublyList.cpp:
#include "DoublyList.h"
using namespace std;
void DoublyList::insertFront(int newData)
{
if (first == nullptr)
{
first = new DLLNode(newData, nullptr, nullptr);
last = first;
// Common error: Forgetting to reset pointer last.
}
else
{
first = new DLLNode(newData, nullptr, first);
first->getNext()->setPrev(first);
// Common error: Forgetting to connect pointer
// prev of what is now the second node to the
// new first node.
}
++count;
}
void DoublyList::printForward() const
{
DLLNode* current = first;
while (current != nullptr)
{
cout << current->getData() << " ";
current = current->getNext();
}
}
void DoublyList::printReverse() const
{
DLLNode* current = last;
while (current != nullptr)
{
cout << current->getData() << " ";
current = current->getPrev();
}
}
void DoublyList::clearList()
{
DLLNode* temp = first;
while (first != nullptr)
{
first = first->getNext();
delete temp;
temp = first;
}
last = nullptr;
// Don't forget to reset pointer last to nullptr.
count = 0;
}
DoublyList::~DoublyList()
{
if (first != nullptr)
clearList();
}
DoublyList.h
#ifndef DOUBLYLIST_H
#define DOUBLYLIST_H
#include <string>
#include <iostream>
class DLLNode
{
public:
DLLNode() : data(0), prev(nullptr), next(nullptr) {}
DLLNode(int theData, DLLNode* prevLink, DLLNode* nextLink)
: data(theData), prev(prevLink), next(nextLink) {}
int getData() const { return data; }
DLLNode* getPrev() const { return prev; }
DLLNode* getNext() const { return next; }
void setData(int theData) { data = theData; }
void setPrev(DLLNode* prevLink) { prev = prevLink; }
void setNext(DLLNode* nextLink) { next = nextLink; }
~DLLNode(){}
private:
int data; // To simplify, we are using only one piece of data.
DLLNode* prev;
DLLNode* next;
};
class DoublyList
{
public:
DoublyList() : first(nullptr), last(nullptr), count(0) {}
void insertFront(int newData);
void printForward() const;
void printReverse() const;
void rotateNodesRight(int);
void clearList();
~DoublyList();
private:
// Pointer to the first node in the list.
DLLNode*first;
// Pointer to the last node in the list.
DLLNode*last;
// Number of nodes in the list.
int count;
};
#endif
Main.cpp
#include "DoublyList.h"
#include <iostream>
#include <vector>
using namespace std;
int main()
{
vector<vector<int>> data = {
{25, 76, 35, 67, 15, 98},
{1, 2, 3, 4, 5, 6, 7, 8, 9},
{10, 20},
{34, 56, 78, 12, 89, 34, 76, 28, 54, 22, 41},
{123, 873, 619},
};
vector<int> nodesToRotate = { 2, 3, 1, 9, 2 };
{
DoublyList doublyList;
int vectorSize = static_cast<int>(data.size());
for (int i = 0; i < vectorSize; ++i)
{
int innerSize = static_cast<int>(data[i].size());
for (int j = innerSize - 1; j >= 0; --j)
doublyList.insertFront(data[i].at(j));
cout << "Rotate right: " << nodesToRotate[i] << "\n";
cout << " List is: ";
doublyList.printForward();
cout << "\n";
doublyList.rotateNodesRight(nodesToRotate[i]);
cout << "After rotating:";
cout << "\n Print forward: ";
doublyList.printForward();
cout << "\nPrint backwards: ";
doublyList.printReverse(.
Lab-2.4 101.pdf
The document contains C program code to implement three data structures using linked lists:
1) A stack using a singly linked list with push, pop, and display functions.
2) A queue using a singly linked list with enqueue, dequeue, and display functions.
3) A program to merge two sorted linked lists by comparing nodes and appending to the merged list.
Pseudocode and sample test cases are provided for each problem.
Assignment isPage 349-350 #4 and #5 Use the Linked List lab.pdf
Assignment is:
\"Page 349-350 #4 and #5 Use the \"Linked List lab\" you have been working on in class and add
the two functions the questions are asking you to develop: divideMid and divideAt. Be sure to
include comments Use meaningful identifier names (constants where appropriate) Turn in .cpp
file AND Turn in a \"print-screen\' of your output (press \"print-screen\' on keyboard, then
\'paste\' in MS-Word)\"
How do you solve QUESTION #4 in the book data structures using c++ by D.S. Malik in Visiual
Studios using the linked list below with what is being asked? Please need help
Linked list :
#include
#include
using namespace std;
struct nodeType
{
int info;
nodeType *link;
};
void createList(nodeType*& first, nodeType*& last);
void printList(nodeType*& first);
void insertFront(nodeType*& first);
void insertBack(nodeType*& last);
void deleteFirst(nodeType*& first);
void deleteLast(nodeType*& last, nodeType* first);
int main()
{
nodeType *first, *last;
int num;
createList(first, last);
int choice;
while(true)
{
cout<<\"1. Insert Front.\ 2. Insert Last.\ 3. Delete Front.\ 4. Delete Last.\ 5. Print List.\ 6. Exit.\
\";
cout<<\"Enter your choice: \";
cin>>choice;
switch(choice)
{
case 1: insertFront(first); break;
case 2: insertBack(last); break;
case 3: deleteFirst(first); break;
case 4: deleteLast(last, first); break;
case 5: printList(first); break;
case 6: return 0;
default: cout<<\"Invalid menu option. Try again.\"<>number;
while (number != -999)
{
newNode = new nodeType; // create new node
newNode->info = number;
newNode->link = NULL;
if (first == NULL)
{
first = newNode;
last = newNode;
}
else
{
last->link = newNode;
last = newNode;
}
cout<<\"Enter an integer (-999 to stop): \";
cin>>number;
} // end of while-loop
} // end of build list function
void deleteFirst(nodeType*& first)
{
nodeType *temp;
temp= first;
first= temp->link;
delete temp;
return;
}
void deleteLast(nodeType*& last, nodeType* current)
{
nodeType *temp;
while(current->link != NULL)
{
temp=current;
current=current->link;
}
temp=last;
current->link=NULL;
delete temp;
last = current;
return;
}
void insertFront(nodeType*& front)
{
int num;
cout<<\"\ Enter the number to insert: \";
cin>>num;
nodeType *newNode = new nodeType;
newNode->info=num;
newNode->link= front;
front= newNode;
return;
}
void insertBack(nodeType*& last)
{
int num;
cout<<\"\ Enter the number to insert: \";
cin>>num;
nodeType *newNode = new nodeType;
newNode->info=num;
newNode->link= NULL;
last->link= newNode;
last = newNode;
return;
}
void printList(nodeType*& first)
{
cout<<\"Inside printList...printing linked list...\ \"<info << \" \";
current = current->link;
}
cout<
#include
using namespace std;
struct nodeType
{
int info;
nodeType *link;
};
void createList(nodeType*& first, nodeType*& last);
void printList(nodeType*& first);
void insertFront(nodeType*& first);
void insertBack(nodeType*& last);
void deleteFirst(nodeType*& first);
void deleteLast(nodeType*& last, nodeType* first);
int main()
{
nodeType *firs.
I want help in the following C++ programming task. Please do coding .pdf
I want help in the following C++ programming task. Please do coding in C++.
Persistent Data Structures
A persistent data structure is one in which we can make changes (such as inserting and removing
elements) while still preserving the original data structure Of course, an easy way to do this is to
create a new copy of the entire data structure every time we perform an operation that changes
the data structure, but it is much more efficient to only make a copy of the smallest possible
portion of the data structure.
First, let\'s look at one of the easiest possible data structures — a stack, implemented using a
linked list.
Persistent Stack
Let\'s say that we have a stack S1, and we want to create a new stack S2, which is the result of
pushing a new element X onto S1, without changing S1 at all. This is easy — we just add a new
element that points to S1:
Starting with the stack S1:
We push X on S1, to get S2, leaving S1 as it was. So after the operation we have two versions of
the stack S1 the older version with elements A, B and C and S2 the newer version with elements
X, A, B and C.
What about popping? That works in the same way. Starting with our original S1 (elements A, B
and C), we can pop off the first element (A), to get S2:
What if we want to push something on to this new stack? It works in the same was as before – if
we push an X onto S2 above to get S3, we have
So for every operation you make a new version of the stack while reusing any existing stack
elements.
You will have to make a linked list of pointers that will point to the head of the different stack
versions.
Also each stack is also implemented as a linked list of elements SAB
Solution
#include #include #include #include #include #include #include #include #include
#include #include namespace dts { template > class PersistentSet { public:
PersistentSet(); PersistentSet(Func); bool add(const T&); bool add(T&&);
bool remove(const T& key); bool empty() const; size_t history_size() const;
class TreeIterator : public std::iterator, std::ptrdiff_t, const T*,
const T&> { using node = typename dts::PersistentSet< std::remove_cv_t,
Func>::Nodeptr; node itr; node nil; std::stack path; node
find_successor(node n) { n = n->rigth; if (n != nil) {
while (n->left != nil) { path.push(n); n = n->left;
} } else { n = path.top();
path.pop(); } return n; } public: explicit
TreeIterator(node n, node pnil) : nil(pnil) //begin { if (n == nil) itr
= nil; else { path.push(nil); while (n->left != nil)
{ path.push(n); n = n->left; } itr =
n; } } explicit TreeIterator(node pnil) // end : itr(pnil),
nil(pnil) { } TreeIterator& operator++ () { itr =
find_successor(itr); return *this; } TreeIterator operator++ (int)
{ TreeIterator tmp(*this); itr = find_successor(itr); return tmp;
} bool operator == (const TreeIterator& rhs) const { return itr ==
rhs.itr; } bool operator != (const TreeIterator& rhs) const {
return itr != rhs.itr; } const T& operator* () const { return itr-
>key; } const T& operator-> () c.
In C++ I need help with this method that Im trying to write fillLi.pdf
In C++ I need help with this method that I'm trying to write: fillListFromFile( ) : void -- this
method will add elements to the list from a file called linkedListData1.txt HINT: this is where
you will keep track of the number of elements in the list
I have all of the other methods, but I'm stuck on this one and I've added it as the last method in
the application. The code I have so far is below.
.cpp file:
.h file:
main:
#include
#include
#include
#include "Node.h"
using namespace std;
void displayList(Node*);
void insertFront(Node** head, int value);
void insertBack(Node** ptrToHead, int value);
void insertAfter(Node* previous, int value);
void deleteAtN(Node** ptrToHead, int n);
void deleteNode(Node** ptrToHead, int searchVal);
void clearList(Node**);
bool searchVal(Node** head, int value);
int countFrequencyOf(Node** ptrTohead, int value);
int size(Node** ptrTohead);
void displayFromFrontToN(Node** ptrToHead, int n);
void displayFromNToEnd(Node** ptrToHead, int n);
void insertInOrder(Node** head, Node* ptrToHead);
void displayInRows(Node** ptrToHead);
int main()
{
Node* node1 = new Node();
node1->setValue(10);
Node* node2 = new Node();
node2->setValue(20);
Node* node3 = new Node();
node3->setValue(30);
node1->next = node2;
node2->next = node3;
Node* head = node1;
displayList(head);
insertFront(&head, 5);
displayList(head);
insertBack(&head, 50);
displayList(head);
insertBack(&head, 60);
displayList(head);
insertAfter(node2, 25);
displayList(head);
deleteAtN(&head, 2);
displayList(head);
deleteNode(&head, 50);
displayList(head);
clearList(&head);
displayList(head);
insertFront(&head, 5);
insertFront(&head, 10);
insertFront(&head, 20);
insertBack(&head, 50);
searchVal(&head, 50);
displayList(head);
countFrequencyOf(&head, 25);
displayList(head);
insertBack(&head, 25);
insertFront(&head, 25);
countFrequencyOf(&head, 25);
displayList(head);
size(&head);
cout << "add one item to list" << endl;
insertBack(&head, 25);
size(&head);
displayList(head);
cout << "Display From Front To N" << endl;
displayFromFrontToN(&head, 4);
cout << "Display From N To End" << endl;
displayFromNToEnd(&head, 3);
cout << "Display In Rows" << endl;
displayInRows(&head);
insertFront(&head, 12);
insertFront(&head, 17);
insertFront(&head, 22);
insertBack(&head, 58);
cout << "Display In Rows" << endl;
displayInRows(&head);//call displayInRows()
return 0;
}//end main()
void displayList(Node* ptr) {
if (ptr == nullptr)
cout << "empty list" << endl;
else {
while (ptr != nullptr) {
cout << ptr->getValue() << endl;
ptr = ptr->next;
}
}
cout << "---------------------------" << endl;
}//end displayList()
void insertFront(Node** head, int value) {
Node* newNode = new Node();
newNode->setValue(value);
//cout<< newNode->getValue() << endl;
newNode->next = *head;
*head = newNode;
}
void insertBack(Node** ptrToHead, int value) {
Node* newNode = new Node();
newNode->setValue(value);
newNode->next = nullptr;
if (*ptrToHead == nullptr)
{
*ptrToHead = newNode;
}//end if
else
{
Node* last .
Modify this code to use multiple threads with the same data1.Modif.pdf
Modify this code to use multiple threads with the same data
1.Modify the main function to implement a loop that reads 10 integers from the console (user
input) and stores these numbers in a one-dimensional (1D) array (this code will go right after the
comment that says “Add code to perform any needed initialization or to process user input”).
You should use a global array for this.
2.Implement a separatepthread function function for each one of the following operations:
a.Count and print out how many of the entered numbers are negative. This function must be
named countNegatives
b.Calculate and print the average value of all the numbers entered. This function must be named
average
c.Print the numbers in reverse order from the order in which they were entered. This function
must be named reverse
3.Modify the main function to create one pthread for each one of the functions that you
implemented in (3) above (this code will go between the comment that says “TODO: Modify
according to assignment requirements” and the “if (rc)” check).
Compile your program and run it several times. If the output of your program is garbled, you
may need to add a small delay in between creating the next thread.
#include
#include
using namespace std;
void *routineName(void *arg)
{
// TODO: Add code that implements
// the thread\'s functionality
cout << \"Thread is running...\" << endl;
return 0;
}
int main()
{
pthread_t id;
int rc;
int ints;
for(int x; x<10; x++)
{
cout << \"Enter Integer: \" <>
}
rc = pthread_create(&id, NULL, routineName, NULL);
if (rc){
cout << \"ERROR; return code from pthread_create() is \" << rc << endl;
return -1;
}
pthread_exit(0);
}
Solution
#include
#include
#include
using namespace std;
int ints[10];
//Method to count negatives in the input array
void *countNegatives(void*){
int count = 0;
for(int i=0;i<10;i++){
if(ints[i]<0){
count++;
}
}
cout<<\"Number of negatives: \"<=0;i--){
cout<> ints[x];
}
//Creating pthread to countNegatives
rc = pthread_create(&t1, NULL, countNegatives, NULL);
if(rc){
cout << \"Error:unable to create thread,\" << rc << endl;
}
//Adding delay
sleep(1);
//Creating pthread to compute average
rc = pthread_create(&t2, NULL, average, NULL);
if(rc){
cout << \"Error:unable to create thread,\" << rc << endl;
}
//Adding delay
sleep(1);
//Creating pthread to print in reverse
rc = pthread_create(&t3, NULL, reverse, NULL);
if(rc){
cout << \"Error:unable to create thread,\" << rc << endl;
}
//Adding delay
sleep(1);
void* status;
//waiting for t1 to join
rc = pthread_join(t1,&status);
if(rc){
cout << \"Error:unable to join,\" << rc << endl;
}
//waiting for t2 to join
rc = pthread_join(t2,&status);
if(rc){
cout << \"Error:unable to join,\" << rc << endl;
}
//waiting for t3 to join
rc = pthread_join(t3,&status);
if(rc){
cout << \"Error:unable to join,\" << rc << endl;
}
pthread_exit(0);
}.
Doublylinklist
This document presents information on doubly linked lists including their representation, initialization, node creation, and various operations like insertion, deletion, and traversal. It discusses inserting and deleting nodes at the beginning or end of the list, as well as inserting before or after a specified node. Code examples are provided for initializing a doubly linked list and performing each operation.
Help with the following code1. Rewrite to be contained in a vecto.pdf
Help with the following code:
1. Rewrite to be contained in a vector object.
2. Using linked lists, redo to handle as many entries as required.
3. Add the ability to add or delete a new entry to the address book.
4. When the program terminates, write the data in the address book to a disk.
# include
# include
# include
using namespace std;
class addressbook
{
private:
char name[500];
long ph_num; //class addressType :public addressbook
char address[500];
int year;
int mounth;
int day;
char gender;
int d, Th, y;
char M, F, m, f;
// char freind[];
public:
addressbook();
void setName();
void getName();
void getphone();
long setphone();
void setAddress();
void getAddress();
float getAverage();
int setDay();
void getDay();
int setMounth();
void getMounth();
int setYear();
void getYear();
char setGender();
void getGender();
};
addressbook::addressbook() // constructor
{
name;
ph_num = 0;
year = 0;
mounth = 0;
day = 0;
}
int addressbook::setDay()
{
cout << \"\ \\tEnter the day:\ \\t\" << endl;
cin >> d;
day = (d >= 1 && d<32) ? d : 0;
return day;
}
int addressbook::setMounth()
{
cout << \"\ \\tEnter the mounth:\ \\t\" << endl;
cin >> m;
mounth = (Th >= 1 && Th<13) ? Th : 0;
return mounth;
}
int addressbook::setYear()
{
cout << \"\ \\tEnter the year :\ \\t\" << endl;
cin >> y;
year = (y >= 1 && y<2008) ? y : 0;
return year;
}
void addressbook::getMounth()
{
cout << mounth << \"/\";
}
void addressbook::getDay()
{
cout << \"\\tBIRTH DAY : \" << day << \"/\";
}
void addressbook::getYear()
{
cout << year;
}
char addressbook::setGender()
{
int flag = 0;
cout << \" \ \\tEnter the gender,(F)for female (M)for male:\ \\t\";
cin >> gender;
while (flag != 1)
{
if (gender == (\'F\') || gender == (\'M\') || gender == (\'m\') || gender == (\'f\'))
flag = 1;
else
{
cout << \"\\t error.....Try Again\ \\t\";
cin >> gender;
return gender;
}
}
}
void addressbook::getGender()
{
cout << \"\\tGender : \" << gender << endl;
}
void addressbook::getAddress()
{
cout << \"\\tAddress : \" << address << endl;
}
float addressbook::getAverage()
{
return 0.0f;
}
void addressbook::setAddress()
{
cout << \" \ \\tEnter the address:\ \\t\" << endl;
gets_s(address);
}
void addressbook::setName()
{
cout << \" \\tEnter the name:\ \\t\" << endl;
gets_s(name);
}
void addressbook::getName()
{
cout << \"\ \\tName : \" << name << \" \" << endl;
}
long addressbook::setphone()
{
cout << \" \ \\tEnter the person number:\ \\t\" << endl;
cin >> ph_num;
return ph_num;
}
void addressbook::getphone()
{
cout << \"\\tPerson # : \" << ph_num << endl;
}
/*class addressType :public addressbook
{
do that
};*/
int main()
{
int s;
cout << \" Enter number person max(500): \";
cin >> s;
addressbook book[10];
for (int i = 1; i<2; i++)
{
book[i].setName();
book[i].setphone();
book[i].setAddress();
book[i].setDay();
book[i].setMounth();
book[i].setYear();
book[i].setGender();
cout << \"\ \ ***********************************\ \";
}
for (int u = 0; u<2; u++)
{
book[u].getName();
book[u].getphone();
book[u].getAddre.
Ds 2 cycle
The document discusses the implementation of a binary search tree (BST) using both recursive and non-recursive traversal methods. It defines a BST node structure with left and right child pointers. Functions are included to insert nodes, perform inorder, preorder and postorder traversals recursively and non-recursively using stacks. Main inserts sample nodes into the tree and calls the various traversal functions to output the node values in the specified orders.
C++ adt c++ implementations
This document provides code examples for implementing common data structures in C++. It includes code for implementing a binary search tree with functions for insertion, removal, and traversal. It also includes code for implementing stacks, queues, and trees as abstract data types with functions for common operations like push, pop, enqueue, dequeue, and traversal. The code demonstrates how to create node structures to represent data elements and link them together to form the desired data structures.
Write a C program that reads the words the user types at the command.pdf
Write a C program that reads the words the user types at the command prompt (using the \'int
argc, char * argv[] and store each unique letter in a Binary Search Tree. When a duplicate is
encountered do not store the letter again and instead keep track of the count in the tree. Once the
Binary Search tree has been created print out the tree both inorder and reverse order. Also print
the highest and lowest alphabetically letter in the tree if any.
Solution
# include
# include
# include
typedef struct BST {
int data;
struct BST *lchild, *rchild;
} node;
void insert(node *, node *);
void preorder(node *);
findMinimum(struct node* root)
findMaximum(struct node* root)
void reverseLevelOrder(struct node* root)
void main(int argc,char argv) {
char argv = \'N\';
int key;
node *new_node, *root, *tmp, *parent;
node *get_node();
root = NULL;
clrscr();
printf(\"\ Program For Binary Search Tree \");
do {
printf(\"\ 1.Create\");
printf(\"\ 2.Search\");
printf(\"\ 3.Recursive Traversals\");
printf(\"\ 4.Exit\");
printf(\"\ Enter your choice :\");
scanf(\"%d\", &argc);
switch (argc) {
case 1:
do {
new_node = get_node();
printf(\"\ Enter The Element \");
scanf(\"%d\", &new_node->data);
if (root == NULL) /* Tree is not Created */
root = new_node;
else
insert(root, new_node);
printf(\"\ Want To enter More Elements?(y/n)\");
argv= getch();
} while (argv == \'y\');
break;
case 2:
if (root == NULL)
printf(\"Tree Is Not Created\");
else {
printf(\"\ The Preorder display : \");
preorder(root);
}
break;
}
} while (argv != 4);
}
/*
Get new Node
*/
node *get_node() {
node *temp;
temp = (node *) malloc(sizeof(node));
temp->lchild = NULL;
temp->rchild = NULL;
return temp;
}
/*
This function is for creating a binary search tree
*/
void insert(node *root, node *new_node) {
if (new_node->data < root->data) {
if (root->lchild == NULL)
root->lchild = new_node;
return newNode(key);
else
insert(root->lchild, new_node);
}
if (new_node->data > root->data) {
if (root->rchild == NULL)
root->rchild = new_node;
return newNode(key);
else
insert(root->rchild, new_node);
}
}
if (key == node->key)
{
(node->count)++;
return node;
}
/*
This function displays the tree in preorder fashion
*/
void preorder(node *temp) {
if (temp != NULL) {
printf(\"%d\", temp->data);
preorder(temp->lchild);
preorder(temp->rchild);
}
}
// Returns maximum value in a given Binary Tree
int findMaximum(struct node* root)
{
// Base case
if (root == NULL)
return INT_MAXIMUM;
// Return maximum of 3 values:
// 1) Root\'s data 2) Max in Left Subtree
// 3) Max in right subtree
int res = root->data;
int lres = findMaximum (root->lchild);
int rres = findMaximum (root->rchild);
if (lres > res)
res = lres;
if (rres > res)
res = rres;
return res;
}
// Returns minimum value in a given Binary Tree
int findMinimum(struct node* root)
{
// Base case
if (root == NULL)
return INT_MINIMUM;
// Return minimum of 3 values:
// 1) Root\'s data 2) Max in Left Subtree
// 3) Max in right subtree
int res = root->data;
int lres = findMinimum(r.
Similar to AnswerThe new program with the required constructor and with a te.pdf (20)
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This assignment and the next (#5) involve design and development of a.pdf
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C++ Please write the whole code that is needed for this assignment- wr.docx
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DATASTRUCTURES PPTS PREPARED BY M V BRAHMANANDA REDDY
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Write a program that accepts an arithmetic expression of unsigned in.pdf
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Assignment isPage 349-350 #4 and #5 Use the Linked List lab.pdf
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I want help in the following C++ programming task. Please do coding .pdf
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In C++ I need help with this method that Im trying to write fillLi.pdf
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Modify this code to use multiple threads with the same data1.Modif.pdf
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Write a C program that reads the words the user types at the command.pdf
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More from nipuns1983
#include stdio.h #include stdlib.h #include unistd.h .pdf
#include
#include
#include
#include
#include
#include \"msecond.h\"
#include \"random_int.h\"
float total_time_spent_waiting = 0.0;
int total_number_of_meals = 0;
#define semaphore_create(s,v) sem_init( &s, 0, v )
#define semaphore_wait(s) sem_wait( &s )
#define semaphore_signal(s) sem_post( &s )
#define semaphore_release(s) sem_destroy( &s )
typedef sem_t semaphore;
semaphore chopstick[NUM_PHILOSOPHERS];
semaphore screen;
semaphore mutex;
int screen_row[NUM_PHILOSOPHERS] = { 6, 2, 2, 6, 10 };
int screen_col[NUM_PHILOSOPHERS] = { 31, 36, 44, 49, 40 };
int chopstick_row[5] = { 9, 4, 3, 4, 9 };
int chopstick_col[5] = { 35, 33, 40, 47, 45 };
char chopstick_sym[5] = { \'/\', \'\\\\\', \'|\', \'/\', \'\\\\\' };
for ( i = 0; i < NUM_PHILOSOPHERS; i++ )
{
position( screen_row[i], screen_col[i] );
printf( \"%d\", i );
position( screen_row[i] + 1, screen_col[i] - 1 );
printf( \"(%d)\", 0 );
position( chopstick_row[i], chopstick_col[i] );
printf( \"%c\", chopstick_sym[i] );
}
position_flush( 13, 1 );
}
void draw_thinking( int n )
{
semaphore_wait( screen );
position( screen_row[n], screen_col[n] );
position_flush( 13, 1 );
semaphore_signal( screen );
}
void draw_hungry( int n )
{
semaphore_wait( screen );
position( screen_row[n], screen_col[n] );
position_flush( 13, 1 );
semaphore_signal( screen );
}
void draw_eating( int n, int eat_count )
{
semaphore_wait( screen );
position( screen_row[n], screen_col[n] );
position( screen_row[n] + 1, screen_col[n] - 1 );
printf( \"(%d)\", eat_count );
position_flush( 13, 1 );
semaphore_signal( screen );
}
void draw_chopstick_up( int n )
{
semaphore_wait( screen );
position( chopstick_row[n], chopstick_col[n] );
printf( \"%c\", \' \' );
position_flush( 13, 1 );
semaphore_signal( screen );
}
void draw_chopstick_down( int n )
{
semaphore_wait( screen );
position( chopstick_row[n], chopstick_col[n] );
position_flush( 13, 1 );
semaphore_signal( screen );
}
void draw_done( int n )
{
semaphore_wait( screen );
position( screen_row[n], screen_col[n] );
printf( \"D\" );
position_flush( 13, 1 );
semaphore_signal( screen );
}
void chopsticks( int n )
{
if ( n % 2 == 0 ) {
semaphore_wait( chopstick[(n+1) % NUM_PHILOSOPHERS] );
draw_chopstick_up( (n+1) % NUM_PHILOSOPHERS );
semaphore_wait( chopstick[n] );
draw_chopstick_up( n );
} else {
semaphore_wait( chopstick[n] );
draw_chopstick_up( n );
semaphore_wait( chopstick[(n+1) % NUM_PHILOSOPHERS] );
draw_chopstick_up( (n+1) % NUM_PHILOSOPHERS );
}
}
void release_chopsticks( int n )
{
draw_chopstick_down( n );
semaphore_signal( chopstick[n] );
draw_chopstick_down( (n+1) % NUM_PHILOSOPHERS );
semaphore_signal( chopstick[(n+1) % NUM_PHILOSOPHERS] );
}
void philosopher( int *philosopher_data )
{
int start_time;
int eat_count = 0;
int total_hungry_time = 0;
int became_hungry_time;
int n = philosopher_data[0];
int duration = philosopher_data[1];
start_time = msecond();
while( msecond() - start_time < duration * 1000 )
{
became_hungry_time = msecond();
draw_hungry( n );
chopsticks( n );
t.
Thalamus acts as the gateway to the cerebral cortex. It is situated .pdf
Thalamus acts as the gateway to the cerebral cortex. It is situated in the diencephalon The
pineal gland is located within the epithalamus, while the pituitary gland is attached to the
hypothalamus via the infundibulum Limbic system is associated with emotion, memory,
olfaction and also with drives and autonomic regulation Hippocampus of the limbic system
transfers short term memory to long term memory
Solution
Thalamus acts as the gateway to the cerebral cortex. It is situated in the diencephalon The
pineal gland is located within the epithalamus, while the pituitary gland is attached to the
hypothalamus via the infundibulum Limbic system is associated with emotion, memory,
olfaction and also with drives and autonomic regulation Hippocampus of the limbic system
transfers short term memory to long term memory.
ADH causes increased permeability of DCT and collecting tubule to wa.pdf
ADH causes increased permeability of DCT and collecting tubule to water that leads to increase
in water re absorption that leads to decrease in water excretion ADH secreted by hypothalamus
and storeed in posterior pituitary gland Syndrome of inappropriate ADH secretion caused by
decreasing the production of ADH from posterior pituitary or ectopic sources Clinical features
are increasing blood volume,decresed urinary excretion of water that leads to overhydration
Diabetes insipidus complete or partial failure to ADH secretion,polyurea,polydipsia are seen as
clinical features. Diuretics are drugs used to increase water excretion.
Solution
ADH causes increased permeability of DCT and collecting tubule to water that leads to increase
in water re absorption that leads to decrease in water excretion ADH secreted by hypothalamus
and storeed in posterior pituitary gland Syndrome of inappropriate ADH secretion caused by
decreasing the production of ADH from posterior pituitary or ectopic sources Clinical features
are increasing blood volume,decresed urinary excretion of water that leads to overhydration
Diabetes insipidus complete or partial failure to ADH secretion,polyurea,polydipsia are seen as
clinical features. Diuretics are drugs used to increase water excretion..
There are several solutions for treating chronically mentally ill pa.pdf
There are several solutions for treating chronically mentally ill patients. These are helpful for
treating chronically mentally ill patients of our society.
Partnership model: in which the one is treated individually and allow solving the problems by
oneself.
Continuity of care: mentally ill patients need extensive care so they will begin to behave
normally if the care is prolonged. This may happen because of emotional attachment.
Education for patient: this is also a very important part. Education with creativity helps the
patients to be busy in activities and gradual storming the brain by this method is very productive
for treatment of patients.
Safety measures: mentally ill patients are not well aware about threats and safety will bring long
life to them.
Evolve surviving skills: there are many necessary skills are needed to develop in such patients.
These skills bring positivity in such patients and rise livingness in them.
Communication capability: this needs to enhance in each mentally ill patients.
Social network: making social network of mentally ill patients included normal people too, can
bring lot of positivity among them.
Solution
There are several solutions for treating chronically mentally ill patients. These are helpful for
treating chronically mentally ill patients of our society.
Partnership model: in which the one is treated individually and allow solving the problems by
oneself.
Continuity of care: mentally ill patients need extensive care so they will begin to behave
normally if the care is prolonged. This may happen because of emotional attachment.
Education for patient: this is also a very important part. Education with creativity helps the
patients to be busy in activities and gradual storming the brain by this method is very productive
for treatment of patients.
Safety measures: mentally ill patients are not well aware about threats and safety will bring long
life to them.
Evolve surviving skills: there are many necessary skills are needed to develop in such patients.
These skills bring positivity in such patients and rise livingness in them.
Communication capability: this needs to enhance in each mentally ill patients.
Social network: making social network of mentally ill patients included normal people too, can
bring lot of positivity among them..
The water is split providing the electrons for PSII, which then exci.pdf
The water is split providing the electrons for PSII, which then excites the electrons to the higher
energy state by absorbing photons from the sunlight. At the higher energy state, these electrons
are accepted by a electron acceptor that through series of electron transport chain donates the
electrons to the PSI. In doing so it generates ATP from ADP and Pi.
Then the electrons are again excited to higher energy state from PSI with absorption of photon
from the sunlight. At higher energy state, the elctrons are accepted by electron acceptor. This
donates the electron transport chain through seroes of protein, which produces NADPH in this
process.
By this way the energy from the sunlight is converted into the chemical energy in the NADPH
reducing equivalents.
Solution
The water is split providing the electrons for PSII, which then excites the electrons to the higher
energy state by absorbing photons from the sunlight. At the higher energy state, these electrons
are accepted by a electron acceptor that through series of electron transport chain donates the
electrons to the PSI. In doing so it generates ATP from ADP and Pi.
Then the electrons are again excited to higher energy state from PSI with absorption of photon
from the sunlight. At higher energy state, the elctrons are accepted by electron acceptor. This
donates the electron transport chain through seroes of protein, which produces NADPH in this
process.
By this way the energy from the sunlight is converted into the chemical energy in the NADPH
reducing equivalents..
The Sarbanes-Oxlet act (SOX) was primarily enacted following the Enr.pdf
The Sarbanes-Oxlet act (SOX) was primarily enacted following the Enron and other scandals. It
basically sought to remove the compliance and audit gaps that may lead to the same kind or
scandals in future. This is considering the fact that these scandals lead to losses worth trillion
dollars to the investors and the general public and hurts the image of corporates and government
identically.
The SOX was enacted for US corporations and sought to tighten accounting, financial reporting
and audit practices to strengthen the basic foundations that may lead to humongous
consequences. However the basic difference was that this act only was implemented to the
corporations and management alike.
The following were the major points which impacted corporations financial reports in a way that
set a benchmark for all future compliances:
This was basically set up to oversee the funtioning of independent audit firms which do
compliance and financial audit for the specific companies. It was taken care that there the
auditors be registered and there be specific set and defined processes which should be followed.
All systems for process, compliance, quality control etc. were put in place to prevent any further
slippage and mistake.
This takes care of the utmost factor of independence of the audit firms. It seeks to minimize the
conflict of interest that may arise in the event that a particular audit firm auditing the company is
already providing some said non-audit service to the company. This is taken care of by this
provision.
This basically takes care of micro-reporting of sorts. There are enhanced reporting of all
financial transactions i.e. off-balance-sheet transactions, any stock transactions of corporate
management etc. The point also mandates strict internal controls for disclosures, and accuracy of
the financial reports. It specifies audit controls and enhanced reviews by the SEC or its agents.
This includes measures related to help and restore investor confidence. This paves way for
investor confidence in reporting of securities analyst. This also mandates disclosures of conflict
of interests is any and code of conduct for these analysts.
This consists of sections which require the SEC with the Comptroller General to report specific
findings. These reports include the effect of credit rating agencies, whether there are any
securities violations, and whether any investment banks assist the companeis to manipulate
financials. ALso reporting related to effects of consolidation of public accounting firms are
provided in this.
This seeks to provide whistle-blower protection for any information that they reveal. It mandates
penalties i.e. criminal in sense, for alteration and manipulation of financial records.
This is expecially a step ahead of point 8 in terms of the penalties. It recommends stronger and
stricter sentences and implies that any failure to properly certify financial reports is a crimilThis
section increases the criminal penalties associated.
The integral has a pole at z=-1Integral region is around a circle .pdf
The integral has a pole at z=-1
Integral region is around a circle in the first quadrant with unit radius,
z=1 to z=i means
angle 0 to pi/2
Using residue theorem
the integral is 2pii (log (-1+1))
= -infty
Solution
The integral has a pole at z=-1
Integral region is around a circle in the first quadrant with unit radius,
z=1 to z=i means
angle 0 to pi/2
Using residue theorem
the integral is 2pii (log (-1+1))
= -infty.
SolutionThis encryption scheme is not considered to be a secure o.pdf
Solution
:
This encryption scheme is not considered to be a secure one as the key used for encryption is
very weak. It will be easy for Eve to crack the key and decrypt the message. The security
provided here is very limited and it would not be possible for Alice to implement other elements
of security to her encryption method.
It would not be possible to add digital signatures to the encrypted message due to which it lacks
repudiation. Authentication problem is also there with this weak encryption method.
There is no surity of message integrity. Before reaching the destination it is possible that
message gets modified to some extent or fully, of which the receiver or the sender will not be
able to know.
Sharing the key is also very dangerous as using any of electronic medium or magnetic medium is
not safe. Due to insecure network the sender or the receiver may never know that the messages
are being tapped. Sharing it in personal is only a safer method, but here distance and time
becomes an obstacle in many cases.
Alice used a method which have very limited security and thus insecure to send any message
over the network. By modifying her method of encryption she could easily convert it into a
healthy communication process.Strong keys will help her a lot and transferring it to the sender in
a secure manner will add to it..
Smart Beta ia a rather elusive term in modern finance sometimes know.pdf
Smart Beta ia a rather elusive term in modern finance sometimes known as advanced
beta,alternate beta or startegy indices.firstly theorized by Harry Markowitz , the first smart beta
ETF was launched in 2003.A smart Beta ETF is a type of exchange-traded fund that uses
alternative index construction rules instead of the typical cap-weighted index strategy, in a
transparent way. It takes into account factors such as size, value and volatility. It utilizes both
passive and active methods of investing ... passive because it follows an index, but active
because it considers alternative factors. Smart Beta ETFs are ideal for investors hoping to
maximize their income and returns and minimize risk.
\"Smart\" refers to the use of an alternative methodology rather than following an index\'s size-
based (market-cap) allocations.Its Evolution began with its namer\' Professional services firm
Towers Watson .\"Smart beta strategies have seen huge inflows in recent years. According to
ETF provider Invesco Powershares, from 2010 to 2017, Smart Beta ETFs took in more than 21%
of U.S. equity ETF inflows, representing 12% of total ETF industry assets.
Shifting On to the barriers for allocating smart beta in a investor`s portfolio
investors have consistently reported that \"determining the best strategy or combination of
strategies and managing unintended factor biases\" are the biggest concerns relating to
implementation of a smart beta strategy.Finding the best variation is the issue of major concern
for investors.
Stating that excess returns mean extra risks for a smart beta strategy is not really a criticism of
smart beta. Rather, it is a criticism of smart beta systems that claim to produce excess returns
without assuming extra risks
Smart beta is no different, regardless of how it is marketed.
smart beta managers don\'t investigate the underlying economic logic behind performance, which
can lead to problematic allocations.A smart beta construction is likely to be focused on past
performance, because that is how factors are identified. Factors that show the strongest
correlation with active management performance tend to be favored. That is as deep as the
diagnosis goes with many smart beta mutual funds and exchange-traded funds (ETFs).
One of the glorious things about market-cap-weighted indexes is that they\'re comparatively easy
to run. Once you own all the stocks you want in their market weights, you generally don\'t need
to touch the portfolio except to reinvest dividends, handle corporate actions and deal with index
changes.
Smart-beta strategies are holding securities in different weights, adjusting their portfolios to
match their \"smart\" strategies, and often rebalancing regularly. All of that means that ETFs
tracking smart-beta indexes often have a much more difficult time than their vanilla brethren.
Apart from these short comings, investors need to have a deep insight of these matters while
adding advanced beta to their portfolios.
1.. Not all s.
Ques-1 What is the differential diagnosis of this patients infectin.pdf
Ques-1: What is the differential diagnosis of this patients infecting organism?
Germ Tube Test is a screening diagnostic test. The differential diagnosis of this patient is obtain
some urine samples for laboratory cultures in carbohydrate medium as \"pathogenic organism
can reproduce by budding\" means it is \"pathogenic fungi\" infected at the surgical region of
cervix. Therefore, it will produce germ tube with fungal spores upon cell cultures & culture will
produce \"yellow precipitate\" when examined with I2-KI & NaOH (iodoform test) due to
alcohol formation as a metabolic byproduct. The morphology of diagnostic samples with
infected pathogen can be observed under the microscope.
Ques-2. The organism was subsequently shown to form germ tubes. What is the organism?
The organism was subsequently shown to form germ tubes depicting that the infected pathogen
is “Candida albicans\", a fungal pathogen
Ques-3. Is this organism part of the normal flora in humans?
Yes, this species in humans mainly located normal commensual in the regions of mouth, gut and
vagina. However, in the above surgical procedures of cervix, transition of the organism has
occurred from commensalism to virulence by germ tube formation with hyperthermia (increased
temperature).
Solution
Ques-1: What is the differential diagnosis of this patients infecting organism?
Germ Tube Test is a screening diagnostic test. The differential diagnosis of this patient is obtain
some urine samples for laboratory cultures in carbohydrate medium as \"pathogenic organism
can reproduce by budding\" means it is \"pathogenic fungi\" infected at the surgical region of
cervix. Therefore, it will produce germ tube with fungal spores upon cell cultures & culture will
produce \"yellow precipitate\" when examined with I2-KI & NaOH (iodoform test) due to
alcohol formation as a metabolic byproduct. The morphology of diagnostic samples with
infected pathogen can be observed under the microscope.
Ques-2. The organism was subsequently shown to form germ tubes. What is the organism?
The organism was subsequently shown to form germ tubes depicting that the infected pathogen
is “Candida albicans\", a fungal pathogen
Ques-3. Is this organism part of the normal flora in humans?
Yes, this species in humans mainly located normal commensual in the regions of mouth, gut and
vagina. However, in the above surgical procedures of cervix, transition of the organism has
occurred from commensalism to virulence by germ tube formation with hyperthermia (increased
temperature)..
package employeeType.employee;public abstract class Employee { .pdf
package employeeType.employee;
public abstract class Employee {
private String firstName;
private String lastName;
private char middleInitial;
private boolean fulltime;
private char gender;
private int employeeNum;
public Employee(String firstName, String lastName, char middleInitial,
char gender, int employeeNum, boolean fulltime) {
super();
this.firstName = firstName;
this.lastName = lastName;
this.middleInitial = middleInitial;
this.fulltime = fulltime;
this.gender = gender;
this.employeeNum = employeeNum;
}
public int getEmployeeNumber() {
return this.employeeNum;
}
public void setEmployeeNumber(int empNum) {
this.employeeNum = empNum;
}
public String getFirstName() {
return firstName;
}
String getLastName() {
return lastName;
}
char getMiddleInitial() {
return middleInitial;
}
public char getGender() {
return gender;
}
public void setFirstName(String fn) {
this.firstName = fn;
}
public void setLastName(String ln) {
this.lastName = ln;
}
public void setMiddleI(char m) {
this.middleInitial = m;
}
public void setGender(char g) {
this.gender = g;
}
public boolean equals(Object obj) {
// TODO Auto-generated method stub
if (obj instanceof Employee) {
Employee employee = (Employee) obj;
if (employee.getEmployeeNumber() == this.getEmployeeNumber())
return true;
else
return false;
} else
return false;
}
public String toString() {
// TODO Auto-generated method stub
return getEmployeeNumber() + \"\ \" + getFirstName() + \" \"
+ getLastName() + \"\ Gender:\" + getGender() + \"\ Status:\"
+ ((fulltime) ? \"Full Time\" : \"Part Time\");
}
public abstract double calculateWeeklyPay();
public abstract void annualRaise();
public abstract double holidayBonus();
public abstract void resetWeek();
}
package employeeType.subTypes;
import java.text.DecimalFormat;
import employeeType.employee.Employee;
public class HourlyEmployee extends Employee {
private double wage;
private double hoursWorked;
public HourlyEmployee(String firstName, String lastName,
char middleInitial, char gender, int employeeNum, boolean fulltime,
double wage) {
super(firstName, lastName, middleInitial, gender, employeeNum, fulltime);
// TODO Auto-generated constructor stub
this.wage = wage;
this.hoursWorked = 0.0d;
}
public void increaseHours(double hours) {
if (hours > 0)
this.hoursWorked += hours;
else
System.out.println(\"Error! Invalid hours\");
}
public String toString() {
// TODO Auto-generated method stub
return super.toString() + \"\ Wage: \" + wage + \"\ Hours Worked: \"
+ hoursWorked;
}
public double calculateWeeklyPay() {
double pay;
if (hoursWorked > 40) {
pay = (40 * wage) + (2 * (hoursWorked - 40) * wage);
} else {
pay = wage * hoursWorked;
}
return pay;
}
public void annualRaise() {
DecimalFormat decimalFormat = new DecimalFormat(\"#.##\");
wage = Double.valueOf(decimalFormat.format(wage * 0.05));
}
public double holidayBonus() {
return 40 * wage;
}
public void resetWeek() {
this.hoursWorked = 0;
}
}
package employeeType.subTypes;
import java.text.DecimalFormat;
import emplo.
Option 1Project management software Project management software .pdf
Option 1:
Project management software: Project management software assists employees, administrators,
and teams to manage team goals and long-term projects, and coordinate individual tasks. Project
management software does this with a range of tools to manage workloads, monitor productivity,
and allocate resources. Task interdependence is a large component of project management
software as it is largely used to coordinate assignments that are related to a sequence of tasks or a
larger project. Users use project management software to keep track of multiple projects, track a
team’s or individual’s progress, and analyze team productivity. These products can be used in
virtually any industry that requires individuals to organize projects, tasks within projects, and
goals for weekly, and monthly, or annual projects. Users usually have the ability to break
projects down into individual tasks, create timelines for project completion, analyze productivity,
and communicate with other team members. Project management software is closely related to
task management software, but differs in scale and scope. Project management software allows
users to manage a team of individuals, balance projects, and document effectiveness. Task
management software simply outlines individual efforts, breaks down components within a
project, determines due dates, and allows for self-management. To qualify for inclusion in the
Project Management software category, a product must:
Advantages of project management softwares:
With project management software, you can:
Example:
Redbooth, Inc., is a privately held computer software company that was founded in Barcelona,
Spain in 2008. It moved its headquarters in Redwood City, California. Dan Schoenbaum joined
the company as CEO in 2011. The company was able to secure seed funding, angel investments,
and Series A and B Funding of almost $20 million. The company rebranded in January 2014 and
continues to provide a solution where collaboration experience is deeply integrated in how
people track, manage and do work.
Redbooth provides a centralized team workspace to communicate, share, and collaborate,
resulting in maximum productivity and high performance. It has been included in Inc.com’s top
5000 fastest-growing privately held companies in the US at #89 and #10 on the software industry
list.
Redbooth is an online project management tool and collaboration platform that offers deep
integrated features of enterprise messaging, file sharing, search, in-stream task management,
video conferencing and voice, centralized into an intuitive virtual workspace. Its award-winning
platform enables teams and organizations to collaborate, communicate, and achieve increased
productivity. Companies that use it has experienced increased team productivity that has lead to
dramatically positive business results. It is offering a highly differentiated and innovative
product that is also broadly applicable to many types of businesses.
Features, Benef.
import java.util.Scanner;public class InputIntegers{public sta.pdf
import java.util.Scanner;
public class InputIntegers
{
public static void main(String args[])
{
Scanner s=new Scanner(System.in);
System.out.println(\"enter n :\");
int n=s.nextInt();
int a[]=new int[n];
System.out.println(\"enter n integers :\");
for(int i=0;i
Solution
import java.util.Scanner;
public class InputIntegers
{
public static void main(String args[])
{
Scanner s=new Scanner(System.in);
System.out.println(\"enter n :\");
int n=s.nextInt();
int a[]=new int[n];
System.out.println(\"enter n integers :\");
for(int i=0;i.
The main drawback of Troutons rule is that it i.pdf
The main drawback of Trouton\'s rule is that it is not applied for the liquids that
contains \"hydrogen bonding\" . In the given molecules HF and CH3OH both have the hydrogen
bonding in their solutions. Therefore - Toulene (C6H5CH3) only contains the value that is
closely related to Troutons value. Hence answer is C6H5CH2
Solution
The main drawback of Trouton\'s rule is that it is not applied for the liquids that
contains \"hydrogen bonding\" . In the given molecules HF and CH3OH both have the hydrogen
bonding in their solutions. Therefore - Toulene (C6H5CH3) only contains the value that is
closely related to Troutons value. Hence answer is C6H5CH2.
Solid to liquid, or liquid to gas change absorbs .pdf
Solid to liquid, or liquid to gas change absorbs heat. (a) melting (b) vaporization
Solution
Solid to liquid, or liquid to gas change absorbs heat. (a) melting (b) vaporization.
RNase P catalyzes the Mg2+-dependent 5-maturati.pdf
RNase P catalyzes the Mg2+-dependent 5\'-maturation of precursor tRNAs.
Biochemical studies on the bacterial holoenzyme, composed of one catalytic RNase P RNA
(RPR) and one RNase P protein (RPP), have helped understand the pleiotropic roles (including
substrate/Mg2+ binding) by which a protein could facilitate RNA catalysis. As a model for
uncovering the functional coordination among multiple proteins that aid an RNA catalyst, we use
archaeal RNase P, which comprises one catalytic RPR and at least four RPPs. Exploiting our
previous finding that these archaeal RPPs function as two binary RPP complexes (POP5•RPP30
and RPP21•RPP29), we prepared recombinant RPP pairs from three archaea and established
interchangeability of subunits through homologous/heterologous assemblies. Our finding that
archaeal POP5•RPP30 reconstituted with bacterial and organellar RPRs suggests functional
overlap of this binary complex with the bacterial RPP and highlights their shared recognition of a
phylogenetically-conserved RPR catalytic core, whose minimal attributes we further defined
through deletion mutagenesis. Moreover, single-turnover kinetic studies revealed that while
POP5•RPP30 is solely responsible for enhancing the RPR’s rate of precursor tRNA cleavage (by
60-fold), RPP21•RPP29 contributes to increased substrate affinity (by 16-fold). Collectively,
these studies provide new perspectives on the functioning and evolution of an ancient, catalytic
ribonucleoprotein. INTRODUCTION RNase P, a ribonucleoprotein (RNP), is the
endoribonuclease that catalyzes the removal of 5\'-leaders in precursor tRNAs (pre-tRNAs) in all
three domains of life (1–4). The bacterial variant is composed of 1 catalytic RNase P RNA
(RPR) and 1 RNase P protein (RPP) cofactor. Eukaryal (nuclear) RNase P from yeast and human
contain 1 RPR with 9 and 10 RPPs, respectively. Intermediate in complexity, the archaeal
holoenzyme is associated with 1 RPR and at least 4 RPPs (POP5, RPP30, RPP21 and RPP29),
which are homologous to eukaryal RPPs. Several years after the remarkable finding that the
bacterial RPR is a true RNA enzyme in the presence of Mg2+ and monovalent ions (5), archaeal
and eukaryal RPRs were also shown to be catalytically active in vitro (6,7). This common
attribute of evolutionarily divergent RPRs was anticipated from their shared ancestry, attested by
sequence and structural similarity of their putative catalytic core (8–15). However, dramatic
variations (106-fold) in catalytic potential indicate that not all RPRs are equal: activity of RPRs
from bacteria?>?archaea?>?eukarya (5–7). Although RPRs display activity in vitro without
RPPs, they are dependent on their cognate protein cofactors for cellular function. Interestingly,
there is an inverse relationship between RPR activity and RNP composition—the protein:RNA
mass ratio is 70% in eukaryal and 50% in archaeal RNase P compared to 10% in their bacterial
counterpart. Thus, elucidating the intimate cooperation between the RNA and.
pH of a solution is directly proportional to mola.pdf
pH of a solution is directly proportional to molarity,volume and pH itself...so
(V1M1pH1+V2M2pH2)/(v1 m1 +v2 m2 ) =ph....USE PH OF HCL AND HI03
Solution
pH of a solution is directly proportional to molarity,volume and pH itself...so
(V1M1pH1+V2M2pH2)/(v1 m1 +v2 m2 ) =ph....USE PH OF HCL AND HI03.
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n = number present before orbital =3 l= 0 for s, .pdf
n = number present before orbital =3 l= 0 for s, 1 for p, 2 for d... so l=1 m= -l to l =
-1, 0 , 1 s = +0.5 or -0.5
Solution
n = number present before orbital =3 l= 0 for s, 1 for p, 2 for d... so l=1 m= -l to l =
-1, 0 , 1 s = +0.5 or -0.5.
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