Im suppose to make a preOrder transversal method so that it performs.pdf
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Im suppose to make a preOrder transversal method so that it performs a preOrder transversal of the tree. For each node it traverses, it should call sb.append() to add a \"[\" the results of traversing the subtree rooted at that node, and then a \"]\". You should add a space before the results of the left and right child traversals, but only if the node exists. This is what i have so far public class BinarySearchTree> extends AbstractSet { protected Entry root; protected int size; /** * Initializes this BinarySearchTree object to be empty, to contain only * elements of type E, to be ordered by the Comparable interface, and to * contain no duplicate elements. */ public BinarySearchTree() { root = null; size = 0; } /** * Initializes this BinarySearchTree object to contain a shallow copy of a * specified BinarySearchTree object. The worstTime(n) is O(n), where n is the * number of elements in the specified BinarySearchTree object. * * @param otherTree - the specified BinarySearchTree object that this * BinarySearchTree object will be assigned a shallow copy of. */ public BinarySearchTree(BinarySearchTree otherTree) { root = copy(otherTree.root, null); size = otherTree.size; } /* Method to complete is here! */ public void preOrder(Entry ent, StringBuilder sb) { // TODO: Complete me! if(ent == null){ return; } if(ent.left != null){ sb.append(\"[\"); preOrder(ent.left, sb); } if(ent.right != null){ sb.append(\"[\"); preOrder(ent.right, sb); } sb.append(\"]\"); } But my preOrder method gives me these errors The following hint(s) may help you locate some ways in which your solution may be improved: --Pre-order traversal should process the root, then its left child (and exclude the non-existent right child)! Expected [t [o]], but was []] --Should just include the root\'s element when the tree only has 1 element! Expected [z], but was ] --Pre-order traversal should process the root, ignore the non-existent left child, then process the right child! Expected [w [z]]. but was []] --Pre-order traversal should process the root, the left child, and then the right child! Expected [9 [0] [d]], but was [][]] --Pre-order traversal should process the root, and then the left child. In processing the left child, it should process that node first, and then process its children! Expected [d [9 [d]]], but was [[]]] --Pre-order traversal should process the root, and then the right child. In processing the right child, it should process that node first, and then process its children! Expected [v [w [z]]], but was [[]]] --Pre-order traversal should process a node and then recursively process each of its children! Expected [v [d [9 [0] [d]] [j [g] [t [o]]]] [w [z]]], but was [[[][]][[][[]]]][[]]] --Pre-order traversal should process a node and then recursively process each of its children! Expected [v [d [9 [0] [d]] [j [g] [t]]]], but was [[[][]][[][]]]] Solution import java.util.Stack; /* Class containing left and right child of current node and key value*/ class Node { i.
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![Im suppose to make a preOrder transversal method so that it performs a preOrder transversal of
the tree.
For each node it traverses, it should call sb.append() to add a "[" the results of traversing the
subtree rooted at that node, and then a "]". You should add a space before the results of the left
and right child traversals, but only if the node exists.
This is what i have so far
public class BinarySearchTree> extends AbstractSet {
protected Entry root;
protected int size;
/**
* Initializes this BinarySearchTree object to be empty, to contain only
* elements of type E, to be ordered by the Comparable interface, and to
* contain no duplicate elements.
*/
public BinarySearchTree() {
root = null;
size = 0;
}
/**
* Initializes this BinarySearchTree object to contain a shallow copy of a
* specified BinarySearchTree object. The worstTime(n) is O(n), where n is the
* number of elements in the specified BinarySearchTree object.
*
* @param otherTree - the specified BinarySearchTree object that this
* BinarySearchTree object will be assigned a shallow copy of.
*/
public BinarySearchTree(BinarySearchTree otherTree) {
root = copy(otherTree.root, null);
size = otherTree.size;
}
/* Method to complete is here! */
public void preOrder(Entry ent, StringBuilder sb) {
// TODO: Complete me!
if(ent == null){
return;](https://image.slidesharecdn.com/imsupposetomakeapreordertransversalmethodsothatitperforms-230702112443-a9f88a5d/85/Im-suppose-to-make-a-preOrder-transversal-method-so-that-it-performs-pdf-1-320.jpg)
![}
if(ent.left != null){
sb.append("[");
preOrder(ent.left, sb);
}
if(ent.right != null){
sb.append("[");
preOrder(ent.right, sb);
}
sb.append("]");
}
But my preOrder method gives me these errors
The following hint(s) may help you locate some ways in which your solution may be improved:
--Pre-order traversal should process the root, then its left child (and exclude the non-existent
right child)! Expected [t [o]], but was []]
--Should just include the root's element when the tree only has 1 element! Expected [z], but was
]
--Pre-order traversal should process the root, ignore the non-existent left child, then process the
right child! Expected [w [z]]. but was []]
--Pre-order traversal should process the root, the left child, and then the right child! Expected [9
[0] [d]], but was [][]]
--Pre-order traversal should process the root, and then the left child. In processing the left child,
it should process that node first, and then process its children! Expected [d [9 [d]]], but was [[]]]
--Pre-order traversal should process the root, and then the right child. In processing the right
child, it should process that node first, and then process its children! Expected [v [w [z]]], but
was [[]]]
--Pre-order traversal should process a node and then recursively process each of its children!
Expected [v [d [9 [0] [d]] [j [g] [t [o]]]] [w [z]]], but was [[[][]][[][[]]]][[]]]
--Pre-order traversal should process a node and then recursively process each of its children!
Expected [v [d [9 [0] [d]] [j [g] [t]]]], but was [[[][]][[][]]]]
Solution](data:image/gif;base64,R0lGODlhAQABAIAAAAAAAP///yH5BAEAAAAALAAAAAABAAEAAAIBRAA7)
Recommended
Please write in C++ and should be able to compile and debug.Thank yo.pdf
Please write in C++ and should be able to compile and debug.Thank you
Create a class called BinaryTreeDeque. This binary tree implements all of the functionality of
the BinaryTree class, but instead of storing its values in a tree structure, the element values are
stored in a deque, using pointers to maintain the relationship between the binary tree and its node
values. Additionally: deque * getValues() - returns a copy of the deque the current array of
values (which should be in insertion order).
Solution
Following are steps to insert a new node in Complete Binary Tree.
1. If the tree is empty, initialize the root with new node.
2. Else, get the front node of the queue.
…….If the left child of this front node doesn’t exist, set the left child as the new node.
…….else if the right child of this front node doesn’t exist, set the right child as the new node.
3. If the front node has both the left child and right child, Dequeue() it.
4. Enqueue() the new node.
Code :-
-------------------------
binarytreedequeue.h
#ifndef BINARYTREEDEQUE_H
#define BINARYTREEDEQUE_H
// Program for linked implementation of complete binary tree
#include
#include
// For Queue Size
#define SIZE 50
// A tree node
struct node
{
int data;
struct node *right,*left;
};
// A queue node
struct Queue
{
int front, rear;
int size;
struct node* *array;
};
class BinaryTreeDeque
{
public:
BinaryTreeDeque();
struct node* newNode(int data);
struct Queue* createQueue(int size);
int isEmpty(struct Queue* queue);
int isFull(struct Queue* queue);
int hasOnlyOneItem(struct Queue* queue);
void Enqueue(struct node *root, struct Queue* queue);
struct node* Dequeue(struct Queue* queue);
struct node* getFront(struct Queue* queue);
int hasBothChild(struct node* temp);
void insert(struct node **root, int data, struct Queue* queue);
void display(struct Queue* queue);
node * getValues(struct Queue* queue);
};
#endif // BINARYTREEDEQUE_H
binarydequeue.cpp
---------------------------------------
#include \"binarytreedeque.h\"
BinaryTreeDeque::BinaryTreeDeque()
{
}
// A utility function to create a new tree node
struct node* BinaryTreeDeque::newNode(int data)
{
struct node* temp = (struct node*) malloc(sizeof( struct node ));
temp->data = data;
temp->left = temp->right = NULL;
return temp;
}
// A utility function to create a new Queue
struct Queue* BinaryTreeDeque::createQueue(int size)
{
struct Queue* queue = (struct Queue*) malloc(sizeof( struct Queue ));
queue->front = queue->rear = -1;
queue->size = size;
queue->array = (struct node**) malloc(queue->size * sizeof( struct node* ));
int i;
for (i = 0; i < size; ++i)
queue->array[i] = NULL;
return queue;
}
// Standard Queue Functions
int BinaryTreeDeque::isEmpty(struct Queue* queue)
{
return queue->front == -1;
}
int BinaryTreeDeque::isFull(struct Queue* queue)
{ return queue->rear == queue->size - 1; }
int BinaryTreeDeque::hasOnlyOneItem(struct Queue* queue)
{ return queue->front == queue->rear; }
void BinaryTreeDeque::Enqueue(struct node *root, s.
Once you have all the structures working as intended- it is time to co.docx
Once you have all the structures working as intended, it is time to compare the
performances of the 2. Use runBenchmark() to start.
a. First, generate an increasing number (N) of random integers (1000, 5000, 10000,
50000, 75000, 100000, 500000 or as far as your computing power will let you)
i. Time and print how long it takes to insert the random integers into an initially
empty BST. Do not print the tree.
You can get a random list using the java class Random, located in
java.util.Random. To test the speed of the insertion algorithm, you should use
the System.currentTimeMillis() method, which returns a long that contains the
current time (in milliseconds). Call System.currentTimeMillis() before and after
the algorithm runs and subtract the two times. (Instant.now() is an alternative
way of recording the time.)
ii. Time and print how long it takes to insert the same random integers into an
initially empty AVL tree. Do not print the tree.
iii. If T(N) is the time function, how does the growth of TBST(N) compare with the
growth of TAVL(N)?
Plot a simple graph to of time against N for the two types of BSTs to visualize
your results.
b. Second, generate a list of k random integers. k is also some large value.
i. Time how long it takes to search your various N-node BSTs for all k random
integers. It does not matter whether the search succeeds.
ii. Time how long it takes to search your N-node AVL trees for the same k random
integers.
iii. Compare the growth rates of these two search time-functions with a graph the
same way you did in part a.
public class BinarySearchTree<AnyType extends Comparable<? super AnyType>> {
protected BinaryNode<AnyType> root;
public BinarySearchTree() {
root = null;
}
/**
* Insert into the tree; duplicates are ignored.
*
* @param x the item to insert.
* @param root
* @return
*/
protected BinaryNode<AnyType> insert(AnyType x, BinaryNode<AnyType> root) {
// If the root is null, we've reached an empty leaf node, so we create a new node
// with the value x and return it
if (root == null) {
return new BinaryNode<>(x, null, null);
}
// Compare the value of x to the value stored in the root node
int compareResult = x.compareTo(root.element);
// If x is less than the value stored in the root node, we insert it into the left subtree
if (compareResult < 0) {
root.left = insert(x, root.left);
}
// If x is greater than the value stored in the root node, we insert it into the right subtree
else if (compareResult > 0) {
root.right = insert(x, root.right);
}
// If x is equal to the value stored in the root node, we ignore it since the tree does not allow duplicates
return root;
}
/**
* Counts the number of leaf nodes in this tree.
*
* @param t The root of the tree.
* @return
*/
private int countLeafNodes(BinaryNode<AnyType> root) {
// If the root is null, it means the tree is empty, so we return 0
if (root == null) {
return 0;
}
// If the root has no children, it means it is a leaf node, so we return 1
if (root.left == .
For the code below complete the preOrder() method so that it perform.pdf
For the code below complete the preOrder() method so that it performs a preOrder traversal of
the tree. For each node it traverses, it should call sb.append() to add a \"[\" the results of
traversing the subtree rooted at that node, and then a \"]\". You should add a space before the
results of the left and right child traversals, but only if the node exists.
code:
Solution
In preorder Nodes visited are in the order of
code:
package edu.buffalo.cse116;
import java.util.AbstractSet;
import java.util.Iterator;
public class BinarySearchTree> extends AbstractSet {
protected Entry root;
protected int size;
/**
* Initializes this BinarySearchTree object to be empty, to contain only
* elements of type E, to be ordered by the Comparable interface, and to
* contain no duplicate elements.
*/
public BinarySearchTree() {
root = null;
size = 0;
}
/**
* Initializes this BinarySearchTree object to contain a shallow copy of a
* specified BinarySearchTree object. The worstTime(n) is O(n), where n is the
* number of elements in the specified BinarySearchTree object.
*
* @param otherTree - the specified BinarySearchTree object that this
* BinarySearchTree object will be assigned a shallow copy of.
*/
public BinarySearchTree(BinarySearchTree otherTree) {
root = copy(otherTree.root, null);
size = otherTree.size;
}
/* Method to complete is here! */
public void preOrder(Entry ent, StringBuilder sb) {
preorder(root)
protected Entry copy(Entry p, Entry parent) {
if (p != null) {
Entry q = new Entry(p.element, parent);
q.left = copy(p.left, q);
q.right = copy(p.right, q);
return q;
}
return null;
} // method copy
@SuppressWarnings(\"unchecked\")
@Override
public boolean equals(Object obj) {
if (!(obj instanceof BinarySearchTree)) {
return false;
}
return equals(root, ((BinarySearchTree) obj).root);
}
public boolean equals(Entry p, Entry q) {
if ((p == null) || (q == null)) {
return p == q;
}
if (!p.element.equals(q.element)) {
return false;
}
if (equals(p.left, q.left) && equals(p.right, q.right)) {
return true;
}
return false;
}
/**
* Returns the size of this BinarySearchTree object.
*
* @return the size of this BinarySearchTree object.
*/
@Override
public int size() {
return size;
}
/**
* Iterator method will be implemented for a future
*
* @return an iterator positioned at the smallest element in this
* BinarySearchTree object.
*/
@Override
public Iterator iterator() {
throw new UnsupportedOperationException(\"Not implemented yet!\");
}
/**
* Determines if there is at least one element in this BinarySearchTree object
* that equals a specified element. The worstTime(n) is O(n) and
* averageTime(n) is O(log n).
*
* @param obj - the element sought in this BinarySearchTree object.
* @return true - if there is an element in this BinarySearchTree object that
* equals obj; otherwise, return false.
* @throws ClassCastException - if obj cannot be compared to the elements in
* this BinarySearchTree object.
* @throws NullPointerException - if obj is null.
*/
@Override
public .
Help please!!(Include your modified DList.java source code file in.pdf
Help please!!
(Include your modified DList.java source code file in your homework solution zip archive)
Using
whatever Java IDE you prefer, create a project and add DList.java and DListTest.java to it (these
files are provided
in the Week 6 Source zip archive). Modify DList to implement a method that removes all
occurrences of a specific
integer from the list. Here is the pseudocode:
Method removeAll(In: Integer pData) Returns Nothing
Define index variable i and initialize i to 0
While i < the size of this Dlist Do
If get(i) equals pData Then
remove(i)
Else
Increment i
End If
End While
End Method removeAll
Next, modify DListTest() to add test case 21 which tests that removeAll() works correctly.
WHat more do i need to provide to get help on this???
Solution
/* DListNode.java */
/**
* A DListNode is a node in a DList (doubly-linked list).
*/
class DListNode {
/**
* item references the item stored in the current node.
* prev references the previous node in the DList.
* next references the next node in the DList.
*
* DO NOT CHANGE THE FOLLOWING FIELD DECLARATIONS.
*/
public Object item;
protected DListNode prev;
protected DListNode next;
/**
* DListNode() constructor.
* @param i the item to store in the node.
* @param p the node previous to this node.
* @param n the node following this node.
*/
DListNode(Object i, DListNode p, DListNode n) {
item = i;
prev = p;
next = n;
}
}
/* DList.java */
/**
* A DList is a mutable doubly-linked list ADT. Its implementation is
* circularly-linked and employs a sentinel (dummy) node at the head
* of the list.
*
* DO NOT CHANGE ANY METHOD PROTOTYPES IN THIS FILE.
*/
class DList {
/**
* head references the sentinel node.
* size is the number of items in the list. (The sentinel node does not
* store an item.)
*
* DO NOT CHANGE THE FOLLOWING FIELD DECLARATIONS.
*/
protected DListNode head;
protected int size;
/* DList invariants:
* 1) head != null.
* 2) For any DListNode x in a DList, x.next != null.
* 3) For any DListNode x in a DList, x.prev != null.
* 4) For any DListNode x in a DList, if x.next == y, then y.prev == x.
* 5) For any DListNode x in a DList, if x.prev == y, then y.next == x.
* 6) size is the number of DListNodes, NOT COUNTING the sentinel,
* that can be accessed from the sentinel (head) by a sequence of
* \"next\" references.
*/
/**
* newNode() calls the DListNode constructor. Use this class to allocate
* new DListNodes rather than calling the DListNode constructor directly.
* That way, only this method needs to be overridden if a subclass of DList
* wants to use a different kind of node.
* @param item the item to store in the node.
* @param prev the node previous to this node.
* @param next the node following this node.
*/
protected DListNode newNode(Object item, DListNode prev, DListNode next) {
return new DListNode(item, prev, next);
}
/**
* DList() constructor for an empty DList.
*/
public DList() {
// Your solution here.
head = newNode(null, head, head);
head.prev = head;
head.next = head;
si.
#include iostream using namespace std; const int nil = 0; cl.docx
#include iostream using namespace std; const int nil = 0; cl
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 .
Recommended
Please write in C++ and should be able to compile and debug.Thank yo.pdf
Please write in C++ and should be able to compile and debug.Thank you
Create a class called BinaryTreeDeque. This binary tree implements all of the functionality of
the BinaryTree class, but instead of storing its values in a tree structure, the element values are
stored in a deque, using pointers to maintain the relationship between the binary tree and its node
values. Additionally: deque * getValues() - returns a copy of the deque the current array of
values (which should be in insertion order).
Solution
Following are steps to insert a new node in Complete Binary Tree.
1. If the tree is empty, initialize the root with new node.
2. Else, get the front node of the queue.
…….If the left child of this front node doesn’t exist, set the left child as the new node.
…….else if the right child of this front node doesn’t exist, set the right child as the new node.
3. If the front node has both the left child and right child, Dequeue() it.
4. Enqueue() the new node.
Code :-
-------------------------
binarytreedequeue.h
#ifndef BINARYTREEDEQUE_H
#define BINARYTREEDEQUE_H
// Program for linked implementation of complete binary tree
#include
#include
// For Queue Size
#define SIZE 50
// A tree node
struct node
{
int data;
struct node *right,*left;
};
// A queue node
struct Queue
{
int front, rear;
int size;
struct node* *array;
};
class BinaryTreeDeque
{
public:
BinaryTreeDeque();
struct node* newNode(int data);
struct Queue* createQueue(int size);
int isEmpty(struct Queue* queue);
int isFull(struct Queue* queue);
int hasOnlyOneItem(struct Queue* queue);
void Enqueue(struct node *root, struct Queue* queue);
struct node* Dequeue(struct Queue* queue);
struct node* getFront(struct Queue* queue);
int hasBothChild(struct node* temp);
void insert(struct node **root, int data, struct Queue* queue);
void display(struct Queue* queue);
node * getValues(struct Queue* queue);
};
#endif // BINARYTREEDEQUE_H
binarydequeue.cpp
---------------------------------------
#include \"binarytreedeque.h\"
BinaryTreeDeque::BinaryTreeDeque()
{
}
// A utility function to create a new tree node
struct node* BinaryTreeDeque::newNode(int data)
{
struct node* temp = (struct node*) malloc(sizeof( struct node ));
temp->data = data;
temp->left = temp->right = NULL;
return temp;
}
// A utility function to create a new Queue
struct Queue* BinaryTreeDeque::createQueue(int size)
{
struct Queue* queue = (struct Queue*) malloc(sizeof( struct Queue ));
queue->front = queue->rear = -1;
queue->size = size;
queue->array = (struct node**) malloc(queue->size * sizeof( struct node* ));
int i;
for (i = 0; i < size; ++i)
queue->array[i] = NULL;
return queue;
}
// Standard Queue Functions
int BinaryTreeDeque::isEmpty(struct Queue* queue)
{
return queue->front == -1;
}
int BinaryTreeDeque::isFull(struct Queue* queue)
{ return queue->rear == queue->size - 1; }
int BinaryTreeDeque::hasOnlyOneItem(struct Queue* queue)
{ return queue->front == queue->rear; }
void BinaryTreeDeque::Enqueue(struct node *root, s.
Once you have all the structures working as intended- it is time to co.docx
Once you have all the structures working as intended, it is time to compare the
performances of the 2. Use runBenchmark() to start.
a. First, generate an increasing number (N) of random integers (1000, 5000, 10000,
50000, 75000, 100000, 500000 or as far as your computing power will let you)
i. Time and print how long it takes to insert the random integers into an initially
empty BST. Do not print the tree.
You can get a random list using the java class Random, located in
java.util.Random. To test the speed of the insertion algorithm, you should use
the System.currentTimeMillis() method, which returns a long that contains the
current time (in milliseconds). Call System.currentTimeMillis() before and after
the algorithm runs and subtract the two times. (Instant.now() is an alternative
way of recording the time.)
ii. Time and print how long it takes to insert the same random integers into an
initially empty AVL tree. Do not print the tree.
iii. If T(N) is the time function, how does the growth of TBST(N) compare with the
growth of TAVL(N)?
Plot a simple graph to of time against N for the two types of BSTs to visualize
your results.
b. Second, generate a list of k random integers. k is also some large value.
i. Time how long it takes to search your various N-node BSTs for all k random
integers. It does not matter whether the search succeeds.
ii. Time how long it takes to search your N-node AVL trees for the same k random
integers.
iii. Compare the growth rates of these two search time-functions with a graph the
same way you did in part a.
public class BinarySearchTree<AnyType extends Comparable<? super AnyType>> {
protected BinaryNode<AnyType> root;
public BinarySearchTree() {
root = null;
}
/**
* Insert into the tree; duplicates are ignored.
*
* @param x the item to insert.
* @param root
* @return
*/
protected BinaryNode<AnyType> insert(AnyType x, BinaryNode<AnyType> root) {
// If the root is null, we've reached an empty leaf node, so we create a new node
// with the value x and return it
if (root == null) {
return new BinaryNode<>(x, null, null);
}
// Compare the value of x to the value stored in the root node
int compareResult = x.compareTo(root.element);
// If x is less than the value stored in the root node, we insert it into the left subtree
if (compareResult < 0) {
root.left = insert(x, root.left);
}
// If x is greater than the value stored in the root node, we insert it into the right subtree
else if (compareResult > 0) {
root.right = insert(x, root.right);
}
// If x is equal to the value stored in the root node, we ignore it since the tree does not allow duplicates
return root;
}
/**
* Counts the number of leaf nodes in this tree.
*
* @param t The root of the tree.
* @return
*/
private int countLeafNodes(BinaryNode<AnyType> root) {
// If the root is null, it means the tree is empty, so we return 0
if (root == null) {
return 0;
}
// If the root has no children, it means it is a leaf node, so we return 1
if (root.left == .
For the code below complete the preOrder() method so that it perform.pdf
For the code below complete the preOrder() method so that it performs a preOrder traversal of
the tree. For each node it traverses, it should call sb.append() to add a \"[\" the results of
traversing the subtree rooted at that node, and then a \"]\". You should add a space before the
results of the left and right child traversals, but only if the node exists.
code:
Solution
In preorder Nodes visited are in the order of
code:
package edu.buffalo.cse116;
import java.util.AbstractSet;
import java.util.Iterator;
public class BinarySearchTree> extends AbstractSet {
protected Entry root;
protected int size;
/**
* Initializes this BinarySearchTree object to be empty, to contain only
* elements of type E, to be ordered by the Comparable interface, and to
* contain no duplicate elements.
*/
public BinarySearchTree() {
root = null;
size = 0;
}
/**
* Initializes this BinarySearchTree object to contain a shallow copy of a
* specified BinarySearchTree object. The worstTime(n) is O(n), where n is the
* number of elements in the specified BinarySearchTree object.
*
* @param otherTree - the specified BinarySearchTree object that this
* BinarySearchTree object will be assigned a shallow copy of.
*/
public BinarySearchTree(BinarySearchTree otherTree) {
root = copy(otherTree.root, null);
size = otherTree.size;
}
/* Method to complete is here! */
public void preOrder(Entry ent, StringBuilder sb) {
preorder(root)
protected Entry copy(Entry p, Entry parent) {
if (p != null) {
Entry q = new Entry(p.element, parent);
q.left = copy(p.left, q);
q.right = copy(p.right, q);
return q;
}
return null;
} // method copy
@SuppressWarnings(\"unchecked\")
@Override
public boolean equals(Object obj) {
if (!(obj instanceof BinarySearchTree)) {
return false;
}
return equals(root, ((BinarySearchTree) obj).root);
}
public boolean equals(Entry p, Entry q) {
if ((p == null) || (q == null)) {
return p == q;
}
if (!p.element.equals(q.element)) {
return false;
}
if (equals(p.left, q.left) && equals(p.right, q.right)) {
return true;
}
return false;
}
/**
* Returns the size of this BinarySearchTree object.
*
* @return the size of this BinarySearchTree object.
*/
@Override
public int size() {
return size;
}
/**
* Iterator method will be implemented for a future
*
* @return an iterator positioned at the smallest element in this
* BinarySearchTree object.
*/
@Override
public Iterator iterator() {
throw new UnsupportedOperationException(\"Not implemented yet!\");
}
/**
* Determines if there is at least one element in this BinarySearchTree object
* that equals a specified element. The worstTime(n) is O(n) and
* averageTime(n) is O(log n).
*
* @param obj - the element sought in this BinarySearchTree object.
* @return true - if there is an element in this BinarySearchTree object that
* equals obj; otherwise, return false.
* @throws ClassCastException - if obj cannot be compared to the elements in
* this BinarySearchTree object.
* @throws NullPointerException - if obj is null.
*/
@Override
public .
Help please!!(Include your modified DList.java source code file in.pdf
Help please!!
(Include your modified DList.java source code file in your homework solution zip archive)
Using
whatever Java IDE you prefer, create a project and add DList.java and DListTest.java to it (these
files are provided
in the Week 6 Source zip archive). Modify DList to implement a method that removes all
occurrences of a specific
integer from the list. Here is the pseudocode:
Method removeAll(In: Integer pData) Returns Nothing
Define index variable i and initialize i to 0
While i < the size of this Dlist Do
If get(i) equals pData Then
remove(i)
Else
Increment i
End If
End While
End Method removeAll
Next, modify DListTest() to add test case 21 which tests that removeAll() works correctly.
WHat more do i need to provide to get help on this???
Solution
/* DListNode.java */
/**
* A DListNode is a node in a DList (doubly-linked list).
*/
class DListNode {
/**
* item references the item stored in the current node.
* prev references the previous node in the DList.
* next references the next node in the DList.
*
* DO NOT CHANGE THE FOLLOWING FIELD DECLARATIONS.
*/
public Object item;
protected DListNode prev;
protected DListNode next;
/**
* DListNode() constructor.
* @param i the item to store in the node.
* @param p the node previous to this node.
* @param n the node following this node.
*/
DListNode(Object i, DListNode p, DListNode n) {
item = i;
prev = p;
next = n;
}
}
/* DList.java */
/**
* A DList is a mutable doubly-linked list ADT. Its implementation is
* circularly-linked and employs a sentinel (dummy) node at the head
* of the list.
*
* DO NOT CHANGE ANY METHOD PROTOTYPES IN THIS FILE.
*/
class DList {
/**
* head references the sentinel node.
* size is the number of items in the list. (The sentinel node does not
* store an item.)
*
* DO NOT CHANGE THE FOLLOWING FIELD DECLARATIONS.
*/
protected DListNode head;
protected int size;
/* DList invariants:
* 1) head != null.
* 2) For any DListNode x in a DList, x.next != null.
* 3) For any DListNode x in a DList, x.prev != null.
* 4) For any DListNode x in a DList, if x.next == y, then y.prev == x.
* 5) For any DListNode x in a DList, if x.prev == y, then y.next == x.
* 6) size is the number of DListNodes, NOT COUNTING the sentinel,
* that can be accessed from the sentinel (head) by a sequence of
* \"next\" references.
*/
/**
* newNode() calls the DListNode constructor. Use this class to allocate
* new DListNodes rather than calling the DListNode constructor directly.
* That way, only this method needs to be overridden if a subclass of DList
* wants to use a different kind of node.
* @param item the item to store in the node.
* @param prev the node previous to this node.
* @param next the node following this node.
*/
protected DListNode newNode(Object item, DListNode prev, DListNode next) {
return new DListNode(item, prev, next);
}
/**
* DList() constructor for an empty DList.
*/
public DList() {
// Your solution here.
head = newNode(null, head, head);
head.prev = head;
head.next = head;
si.
#include iostream using namespace std; const int nil = 0; cl.docx
#include iostream using namespace std; const int nil = 0; cl
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 .
ReversePoem.java ---------------------------------- public cl.pdf
ReversePoem.java :-
---------------------------------
public class ReversePoem {
/*This programs has you display a pessimistic poem from a list of phrases*/
// and then reverse the phrases to find another more optimistic poem.
public static void main(String[] args) throws Exception
{
//Queue object
MyQueue queue = new MyQueue<>();
//Stack object
MyStack stack = new MyStack<>();
//String buffer to apppend all Strings
StringBuffer sb = new StringBuffer();
// Create a single String object from the 16 Strings below
String set1 = \"I am part of a lost generation#and I refuse to believe that#\";
sb.append(set1);
String set2 = \"I can change the world#I realize this may be a shock but#\";
sb.append(set2);
String set3 = \"\'Happiness comes from within\'#is a lie, and#\";
sb.append(set3);
String set4 = \"\'Money will make me happy\'#So in 30 years I will tell my children#\";
sb.append(set4);
String set5 = \"they are not the most important thing in my life#\";
sb.append(set5);
String set6 = \"My employer will know that#I have my priorities straight because#\";
sb.append(set6);
String set7 = \"work#is more important than#family#I tell you this#\";
sb.append(set7);
String set8 = \"Once upon a time#Families stayed together#\";
sb.append(set8);
String set9 = \"but this will not be true in my era#\";
sb.append(set9);
String set10 = \"This is a quick fix society#Experts tell me#\";
sb.append(set10);
String set11 = \"30 years from now, I will be celebrating the 10th anniversary of my
divorce#\";
sb.append(set11);
String set12 = \"I do not concede that#I will live in a country of my own making#\";
sb.append(set12);
String set13 = \"In the future#Environmental destruction will be the norm#\";
sb.append(set13);
String set14 = \"No longer can it be said that#My peers and I care about this earth#\";
sb.append(set14);
String set15 = \"It will be evident that#My generation is apathetic and lethargic#\";
sb.append(set15);
String set16 = \"It is foolish to presume that#There is hope#\";
sb.append(set16);
String finalString = sb.toString();
String itmes[] = finalString.split(\"#\");
System.out.println(\"========== Original Phrase ==============\");
for(int i = 0 ; i < itmes.length;i++){
queue.enqueue(itmes[i]);
System.out.println(itmes[i]);
}
for(int i = 0 ; i < itmes.length;i++){
stack.push(queue.dequeue());
}
System.out.println(\"========== Reverse Phrase ==============\");
for(int i = 0 ; i < itmes.length;i++){
System.out.println(stack.pop());
}
/* You are given a list of phrases in Strings; the phrases
are separated by pound signs: \'#\':
1. Create a single String object from this list.
2. Then, split the String of phrases into an array of
phrases using the String split method.
3. Display a poem by walking through the array and
displaying each phrase one per line.
4. And, at the same time, place each phrase on a
MyQueue object using only the enqueue method.
5. After all the phrases have been placed on the queue,
transfer the phrases from the MyQueue object to a
MyS.
There is BinarySearchTree class. When removing a node from a BST, we.pdf
There is BinarySearchTree class. When removing a node from a BST, we can replace it with
either its predecessor or its successor. Complete the predecessor() method
so that it returns the node that comes immediately before e. Hint: the work needed to find the
predecessor is a mirror image of what is required to find a node\'s
successor, so use successor() as a resource.
BinarySearchTree class:
Solution
import java.util.AbstractSet;
import java.util.Iterator;
public class BinarySearchTree> extends AbstractSet {
protected BTNode root;
protected int size;
/**
* Initializes this BinarySearchTree object to be empty, to contain only
* elements of type E, to be ordered by the Comparable interface, and to
* contain no duplicate elements.
*/
public BinarySearchTree() {
root = null;
size = 0;
}
/**
* Initializes this BinarySearchTree object to contain a shallow copy of a
* specified BinarySearchTree object. The worstTime(n) is O(n), where n is the
* number of elements in the specified BinarySearchTree object.
*
* @param otherTree - the specified BinarySearchTree object that this
* BinarySearchTree object will be assigned a shallow copy of.
*/
public BinarySearchTree(BinarySearchTree otherTree) {
root = copy(otherTree.root, null);
size = otherTree.size;
}
protected BTNode copy(BTNode p, BTNode parent) {
if (p != null) {
BTNode q = new BTNode(p.element, parent);
q.left = copy(p.left, q);
q.right = copy(p.right, q);
return q;
}
return null;
} // method copy
/**
* Finds the predecessor of a specified BTNode object in this BinarySearchTree. The
worstTime(n) is O(n) and
* averageTime(n) is constant.
*
* @param e - the BTNode object whose successor is to be found.
* @return the successor of e, if e has a successor; otherwise, return null.
*/
protected BTNode predecessor(BTNode e) {
if (e == null) {
return null;
} else if (e.left != null) {
// successor is leftmost BTNode in right subtree of e
BTNode p = e.left;
while (p.right != null) {
p = p.right;
}
return p;
} // e has a right child
else {
// go up the tree to the right as far as possible, then go up
// to the left.
BTNode p = e.parent;
BTNode ch = e;
while ((p != null) && (ch == p.left)) {
ch = p;
p = p.parent;
} // while
return p;
} // e has no left child
// method successor
}
@SuppressWarnings(\"unchecked\")
@Override
public boolean equals(Object obj) {
if (!(obj instanceof BinarySearchTree)) {
return false;
}
return equals(root, ((BinarySearchTree) obj).root);
} // method 1-parameter equals
public boolean equals(BTNode p, BTNode q) {
if ((p == null) || (q == null)) {
return p == q;
}
if (!p.element.equals(q.element)) {
return false;
}
if (equals(p.left, q.left) && equals(p.right, q.right)) {
return true;
}
return false;
}
/**
* Returns the size of this BinarySearchTree object.
*
* @return the size of this BinarySearchTree object.
*/
@Override
public int size() {
return size;
}
/**
* Iterator method will be implemented for a future
*
* @return an iterator positioned at the smallest element in this
* Bina.
Write a program that displays an AVL tree along with its balance fac.docx
Write a program that displays an AVL tree along with its balance fac
In java , I want you to implement a Data Structure known as a Doubly.pdf
In java , I want you to implement a Data Structure known as a Doubly-Ended-Queue. it is a
“fair” data structure in that it implements a FIFO (First In First Out ) behavior. As such, it is
often used to implement various wait lists in computer systems. For example, jobs waiting to use
the CPU, jobs waiting for a printer, jobs waiting to be placed into RAM for execution. In short,
whenever we want a fair strategy for waiting we use queues.
A DEQUE (Doubly-ended-queue) is a related data structure. Although similar to a Queue, it
differs in that it allows for insertions AND deletions from either end of the list (both the front
and the rear).
Your implementation MUST use a doubly-linked-list implementation. You may not use a static
array implementation.
Thus, a Deque is a List but it is one which only concerns itself with the first and last positions for
any insertion or deletion. The 6 operations supported are :
public void insertFront( int item ) - insert the given item (as a node) into the first position of the
Deque.
public void insertRear( int item ) - insert the given item (as a node) into the last position of the
Deque.
public int deleteFront( ) - delete and return the element stored in the first node of the Deque.
public int deletRear( ) – delete and return the element stored in the last node of the Deque.
public boolean isempty( ) - returns true if the Deque is currently empty or false if it is not.
public void printDeque( ) - print the integers from the list, one per line, from the first element
through to the last in order.
Classes
Your program must implement the following 3 classes. public class dequeDriver
This class will contain your program’s main method. It will need to declare a deque object and
process input as indicated below.
Your program should prompt the user for the path of an input file. It should open the file for
input and process it line by line. Each line of the input file will have one of the following forms.
PR
IF
IR
DF
DR
The meanings of each input is as follows:
PR - print the current contents of the deque from front to rear using the printDeque( ) method of
the deque object.
IF - insert the given int value into the front of the deque.
IR - insert the given int value into the rear of the deque.
DF - delete the front value from the deque.
DR – delete the rear element of the deque.
Below is an example input file that your program should be able to process.
PR
IF 4
IF 5
IF 6
IR 7
PR
DR
PR
DF
PR
The output for the input file shown above is :
EMPTY DEQUE
----- Front -----
6
5
4
7
----- Rear -----
----- Front -----
6
5
4
----- Rear -----
----- Front -----
5
4
----- Rear -----
public class dequeNode
This class will implement the linked nodes that will be used to implement the deque itself.
It should have the following protected data members.
protected dequeNode next; // next pointer to next node
protected dequeNode prev; // previous pointer to previous node
protected int val; // the integer value stored within the dequeNod.
Need Help with this Java Assignment. Program should be done in JAVA .pdf
Need Help with this Java Assignment. Program should be done in JAVA please
Complete required scripts based on eclipse and troubleshoot the scripts until the tasks are done.
1. Implement a binary search tree (write the code for it), for a height of 2 or higher. Please show
the output if you can Thank you in advance!
Solution
I have compiled this code and checked for errors, worked fine.
Also after the code I have pasted the Output transcript to show what User Inputs I gave and what
all outputs I got.
// Java Program to Implement Binary Tree
import java.util.Scanner;
/* Define Class BinaryTreeNode */
class BinaryTreeNode
{
BinaryTreeNode left, right;
int data;
/* Constructor */
public BinaryTreeNode()
{
left = null;
right = null;
data = 0;
}
/* Constructor */
public BinaryTreeNode(int n)
{
left = null;
right = null;
data = n;
}
/* Function to set left node */
public void setLeft(BinaryTreeNode n)
{
left = n;
}
/* Function to set right node */
public void setRight(BinaryTreeNode n)
{
right = n;
}
/* Function to get left node */
public BinaryTreeNode getLeft()
{
return left;
}
/* Function to get right node */
public BinaryTreeNode getRight()
{
return right;
}
/* Function to set data to node */
public void setData(int d)
{
data = d;
}
/* Function to get data from node */
public int getData()
{
return data;
}
}
/* Define anothe Class Operations to perform required operationns on Binary Tree*/
class Operations
{
private BinaryTreeNode root;
/* Constructor */
public Operations()
{
root = null;
}
/* Function to check if tree is empty */
public boolean isEmpty()
{
return root == null;
}
/* Functions to insert data */
public void insert(int data)
{
root = insert(root, data);
}
/* Function to insert data recursively */
private BinaryTreeNode insert(BinaryTreeNode node, int data)
{
if (node == null)
node = new BinaryTreeNode(data);
else
{
if (node.getRight() == null)
node.right = insert(node.right, data);
else
node.left = insert(node.left, data);
}
return node;
}
/* Function to count number of nodes */
public int countNodes()
{
return countNodes(root);
}
/* Function to count number of nodes recursively */
private int countNodes(BinaryTreeNode r)
{
if (r == null)
return 0;
else
{
int l = 1;
l += countNodes(r.getLeft());
l += countNodes(r.getRight());
return l;
}
}
/* Function to search for an element */
public boolean search(int val)
{
return search(root, val);
}
/* Function to search for an element recursively */
private boolean search(BinaryTreeNode r, int val)
{
if (r.getData() == val)
return true;
if (r.getLeft() != null)
if (search(r.getLeft(), val))
return true;
if (r.getRight() != null)
if (search(r.getRight(), val))
return true;
return false;
}
/* Function for inorder traversal */
public void inorder()
{
inorder(root);
}
private void inorder(BinaryTreeNode r)
{
if (r != null)
{
inorder(r.getLeft());
System.out.print(r.getData() +\" \");
inorder(r.getRight());
}
}
/* Function for preorder traversal */
public void preorder()
{
preorder(root);
}
priva.
Write a C++ program that implements a binary search tree (BST) to man.pdf
Write a C++ program that implements a binary search tree (BST) to manage a number of integer
items with different priorities. In order to do so. you will need to maintain a queue as an item in
the tree. Each queue item is an integer. All items in a given queue will have the same priority. As
a new item with a user-specified priority comes in, it should be enqueued in the queue with items
of that priority. A new queue may need to created in case a queue with that priority does not
exist. Use dynamic memory to manage nodes in the queue and BST. Implement the following:
Constructor. Destructor. Overloaded copy constructor. Overloaded assignment operator.
Overloaded \"==\" operator to check if two trees are equal. A tree t1 is equal to t2 if for every
queue in t1, a corresponding queue with the same priority exists in t2 with the same set of
elements (the order of the elements in the queue docs not matter). Overloaded \"
Solution
program using namespace std;
#define width_unit 5
class BStree
{
private:
class Node
{
public:
int data;
Node *left, *right;
Node(int d=0)
:data(d), left(NULL), right(NULL) {}
};
Node *root;
Node * trav(int, Node * &);
void chop(Node * N);
void copy(Node * N);
void print(ostream &, Node *, int) const;
void print(Node *, int) const;
public:
BStree(void);
~BStree(void);
bool find(int);
void insert(int);
void remove(int);
bool empty(void) const;
Tree(const BStree &);
const BStree & operator=(const BStree &);
friend ostream & operator<<(ostream &, const BStree &);
};
BStree::BStree(void)
{
root=NULL;
}
bool Tree::empty(void) const
{
return !root;
}
BStree::Node * BStree::trav(int foo, Node * & par)
{
Node * curr=root;
par=NULL;
while(curr && curr->data != foo)
{
par=curr;
if(foo < curr->data)
curr=curr->left;
else
curr=curr->right;
}
return curr;
}
bool BStree::find(int foo)
{
Node * par=NULL;
Node * curr=trav(foo, par);
return curr;
}
void BStree::insert(int foo)
{
Node * par=NULL;
Node * curr=trav(foo,par);
if(!curr)
{
curr= new Node(foo);
if(!par)
root=curr;
else if(foo < par->data)
par->left=curr;
else
par->right=curr;
}
}
void BStree::remove(const int foo)
{
Node * par=NULL;
Node * curr=trav(foo,par);
if(curr)
{
if(curr->left && curr->right)
{
Node * tmp=curr;
par=curr;
curr=curr->left;
while(curr->right)
{
par=curr;
curr=curr->right;
}
tmp->data=curr->data;
}
Node *tmp=(curr->left ? curr->left : curr->right);
if(!par)
root=tmp;
else if(par->data < curr->data)
par->right=tmp;
else
par->left=tmp;
delete curr;
}
}
void BStree::chop(Node *N)
{
if(N)
{
chop(N->left);
chop(N->right);
delete N;
}
}
BStree::~BStree(void)
{
chop(root);
}
BStree::BStree(const BStree & T)
{
root=NULL;
copy(T.root);
}
void BStree::copy(Node * N)
{
if(N)
{
insert(N->data);
copy(N->left);
copy(N->right);
}
}
const BStree & BStree::operator=(const BStree & T)
{
if(this != &T)
{
chop(root);
root=NULL;
copy(T.root);
}
return *this;
}
void BStree::print(ostream & ost, Node * curr, int level) const
{
if(curr)
{
print(ost,curr->right,level+1);
ost<data<le.
L 17 ct1120
Class lecture of Data Structure and Algorithms and Python.
Stack, Queue, Tree, Python, Python Code, Computer Science, Data, Data Analysis, Machine Learning, Artificial Intellegence, Deep Learning, Programming, Information Technology, Psuedocide, Tree, pseudocode, Binary Tree, Binary Search Tree, implementation, Binary search, linear search, Binary search operation, real-life example of binary search, linear search operation, real-life example of linear search, example bubble sort, sorting, insertion sort example, stack implementation, queue implementation, binary tree implementation, priority queue, binary heap, binary heap implementation
A deque (pronounced deck) is an ordered set of items from which item.pdf
A deque (pronounced deck) is an ordered set of items from which items may be deleted at either
end and into which items may be inserted at either end. Call the two ends left and right. This is
an access-restricted structure since no insertions or deletions can happen other than at the ends.
Implement a deque as a doubly-linked circular list with a header. Write InsertRight and
DeleteLeft.
Solution
code:
#include
using namespace std;
class Node
{
public:
int data;
Node* next;
Node* prev;
};
class Deque
{
private:
Node* front;
Node* rear;
int count;
public:
Deque()
{
front = NULL;
rear = NULL;
count = 0;
}
bool isEmpty()
{
if(count == 0)
return true;
else
return false;
}
int DeleteLeft()
{
if ( isEmpty() ) {
cout<<\"deque is empty... can\'t delete\'\";
return -1;
}
int ret = front->data;
Node* tmp = front;
if ( front->next != NULL )
{
front = front->next;
front->prev = NULL;
}
else
{
front = NULL;
}
count--;
delete tmp;
return ret;
}
void InsertRight(int element)
{
Node* tmp = new Node();
tmp->data = element;
tmp->next = NULL;
tmp->prev = NULL;
if ( isEmpty() ) {
front = rear = tmp;
}
else {
rear->next = tmp;
tmp->prev = rear;
rear = tmp;
}
count++;
}
};
int main()
{
Deque q;
if ( q.isEmpty() )
{
cout << \"Deque is empty\" << endl;
}
q.InsertRight(100);
q.InsertRight(200);
q.InsertRight(300);
Deque q1;
if ( q1.isEmpty() )
{
cout << \"Deque is empty\" << endl;
}
q1.InsertRight(100);
q1.InsertRight(200);
q1.InsertRight(300);
cout << q1.DeleteLeft() << endl;
cout << q1.DeleteLeft() << endl;
cout << q1.DeleteLeft() << endl;
}.
AvlTree.h#ifndef AVL_TREE_H#define AVL_TREE_H#include d.docx
AvlTree.h
#ifndef AVL_TREE_H
#define AVL_TREE_H
#include "dsexceptions.h"
#include <iostream> // For NULL
#include <queue> // For level order printout
#include <vector>
#include <algorithm> // For max() function
using namespace std;
// AvlTree class
//
// CONSTRUCTION: with ITEM_NOT_FOUND object used to signal failed finds
//
// ******************PUBLIC OPERATIONS*********************
// Programming Assignment Part I
// bool empty( ) --> Test for empty tree @ root
// int size( ) --> Quantity of elements in tree
// int height( ) --> Height of the tree (null == -1)
// void insert( x ) --> Insert x
// void insert( vector<T> ) --> Insert whole vector of values
// void remove( x ) --> Remove x (unimplemented)
// bool contains( x ) --> Return true if x is present
// Comparable findMin( ) --> Return smallest item
// Comparable findMax( ) --> Return largest item
// boolean isEmpty( ) --> Return true if empty; else false
// void printTree( ) --> Print tree in sorted (in) order
// void printPreOrder( ) --> Print tree in pre order
// void printPostOrder( ) --> Print tree in post order
// void printInOrder( ) --> Print tree in *in* order
// Programming Assignment Part II (microassignment)
// void makeEmpty( ) --> Remove and delete all items
// void ~AvlTree( ) --> Big Five Destructor
// AvlTree(const AvlTree &other) --> BigFive Copy Constructor
// AvlTree(const AvlTree &&other) --> BigFive Move Constructor
// AvlTree &operator= ( AvlTree & other ) --> Big Five Copy *assignment* operator
// AvlTree &operator= ( AvlTree && other ) --> Big Five Move *assignment* operator
// void printLevelOrder( ) --> Print tree in LEVEL order :-)
// ******************ERRORS********************************
// Throws UnderflowException as warranted
template <typename Comparable>
class AvlTree
{
public:
/**
* Basic constructor for an empty tree
*/
AvlTree( ) : root( NULL )
{
//cout << " [d] AvlTree constructor called. " << endl;
}
/**
* Vector of data initializer (needed for move= operator rvalue)
*/
AvlTree( vector<Comparable> vals ) : root( NULL )
{
insert(vals);
}
//*******************************************************************************************
// START AVL TREES PART II - TODO: Implement
// Other functions to look for include: clone, makeEmpty, printLevelOrder
/**
* Destroy contents of the AvlTree object - Big Five Destructor
*/
~AvlTree( )
{
//cout << " [d] AvlTree Destructor called. " << endl;
// Empty out the AVL Tree and destroy all nodes (makeEmpty?)
}
/**
* Copy other to new object - Big Five Copy Constructor
*/
AvlTree( const AvlTree &other ) : root( NULL )
{
cout << " [d] Copy Constructor Called." << endl;
// Copy contents of other to ourselves (maybe clone?)
// Get a deep copy of other's tree
}
/* ...
I need help with this maze gui that I wrote in java, I am trying to .pdf
I need help with this maze gui that I wrote in java, I am trying to find the shortest path but I dont
know how to implement the FindshortestPath Method.
using a BREADTH FIRST SEARCH and the stradgey below to ONLY FIND THE SHORTEST
PATH, the method \"FindshortestPath(); is the only part you need to add code too.
I was given this as a stragedy to solve the findshortestpathMethod():
Create two LinkedLists of MazeMap objects. One will be the list of map objects for cells at
distance x from the center. The other will be the list of map objects for cells at distance x + 1.
Initialize the list at distance x to contain a MazeMap object which refers to the center cell of the
maze. Since this map object refers to the center cell, it does not need to refer to any other
MazeMap object, so its MazeMap reference can be set to null. Set the center cell’s visited field
to true to indicate that this cell has already been added to the map.
Set up a loop to run until the maze entrance is mapped. This loop does the following:
Set up a loop to process each map object from the list at distance x. This loop does the following:
i.Get the cell from the map object
ii.If this cell is the entrance cell, set the entrance and break out of the loop!
iii.Check each direction from this cell and for an accessible neighbor cell
Get a reference to that neighbor cell
If the neighbor cell has not already been visited
Set the cell’s visited field to true so its only processed once
Create a MazeMap object that refers to this neighbor cell
Make this MazeMap object refer to the current map object being processed – ie the cell one step
closer to center
Add this MazeMap object to the list at distance x + 1
List at distance x has been processed. The list at distance x + 1 is now complete, so make the list
at distance x refer to list at distance x + 1. Then create a new empty list for the list at distance x +
1.
At this point, the entrance points at the map object which refers to cell 1,1. That map object also
refers to a map object which is one step closer to the center. Write a loop that traverses this trail
of references until it reaches the end (a null reference). For each map object:
Get the cell referred to.
Call the drawCircle method with a color, the cell’s row and column, and SMALL.
Call the delay method with a SHORT delay. This slows down the drawing process so you can
watch the path being displayed.
Advance to the next map object in the list.
here is the code so far*********
package mazepackage;
import java.util.LinkedList;
import java.util.Random;
import javafx.scene.paint.Color;
public class Maze {
private int N; // dimension of maze
private MazeCell[][] maze;
private Random rand = new Random();
// Used to signal the maze has been solved
private boolean done;
// time in milliseconds (from currentTimeMillis()) when we can draw again
// used to control the frame rate
private long nextDraw = -1;
private MazeApp mf;
// Define constants for the circle size
private final double BI.
Given the following codepackage data1;import java.util.;p.pdf
Given the following code
package data1;
import java.util.*;
/*public class BST> implementing a set ADT and containing the following:
* private inner class BinaryNode representing a node with a (possibly) left child and a (possibly)
right child
** instance fields BinaryNode root, int size
* contains/insert/remove method - w/ O(height) complexity
** size O(1), isEmpty O(1), clear O(1),
*** findMin, findMax, findHeight methods - w/ O(height) complexity
**** toString() method printing set (in-order traversal) and the tree (BFS traversal)
**
*/
publicclass BST> {
private BinaryNode root;
privateint size;
//Search opertion of Set ADT
publicboolean contains(E value) {
return contains(root, value);
}
//Insert opertion of Set ADT
publicvoid insert(E value) {
root = insert(root, value);
size++;
}
//Remove operation of Set ADT
publicvoid remove(E value){
root = remove(root, value);
size--;
}
publicint size(){
return size;
}
publicboolean isEmpty(){
return root == null;
}
publicvoid clear(){
root = null;
size = 0;
}
public E findMin(){
return findMin(root);
}
public E findMax(){
return findMax(root);
}
publicint findHeight(){
return root.height();
}
public String toString(){
if(root == null)
return "set {} stored in an empty tree.\n";
String elements = root.inOrder().toString();
return "set {" + elements.substring(1,elements.length()-1) + "} stored in tree \n" + root;
}
private E findMin(BinaryNode node){
if(node == null)
returnnull;
if(node.left == null)
return node.element;
return findMin(node.left);
}
private E findMax(BinaryNode node){
if(node == null)
returnnull;
if(node.right == null)
return node.element;
return findMax(node.right);
}
privateboolean contains(BinaryNode node, E value) {
if (node == null)//node represents an empty substree
returnfalse;
int comparisonResult = value.compareTo(node.element);
if(comparisonResult < 0)//if value is less than root's value
return contains(node.left, value);
if(comparisonResult > 0)//if value is greater than root's value
return contains(node.right, value);
returntrue;//successful search
}
private BinaryNode insert(BinaryNode node, E value) {
if (node == null)//base case: empty tree
returnnew BinaryNode<>(value);
if (value.compareTo(node.element) < 0)
node.left = insert(node.left, value);//insert recursively to left-subtree
elseif (value.compareTo(node.element) > 0)
node.right = insert(node.right, value);//insert recursively to right-subtree
else {//duplicate value cannot be inserted in a BST implementing the Set ADT
size--;//insertion failed!
//System.out.print("BST.insert: Warning: " + value + " is already stored in the tree \n" + node);
}
return node;
}
private BinaryNode remove(BinaryNode node, E value) {
if (node == null) {//base case: empty tree
System.out.println("BST.remove: Warning: " + value + " doesn't exist in the tree.");
size++;//removal failed!
return node;
}
if (value.compareTo(node.element) < 0)
node.left = remove(node.left, value);
elseif (value.compareTo(node.element) > 0)
node.right = remov.
JAVA OOP project; desperately need help asap im begging.Been stuck.pdf
JAVA OOP project; desperately need help asap im begging.
Been stuck on this for a week since my OOP skills are not good at all. Due in a couple hours so
im desperate. Codes needed will be provided at the end, thank you.
Codes:
DotChaser.java:
import java.util.*;
public class DotChaser {
public static Random rand = new Random(System.currentTimeMillis());
/**
* A "Thing" moves in a grid world. A TypeA Thing randomly
* decides to turn left or right (or not turn) every "round",
* and, afterward, takes a step forward. A TypeB Thing
* only considers making a random turn every 10th round.
*
* A STATIC CLASS? OH NO! GET IT OUT OF HERE!
*/
public static class Thing {
// dir: 0=North, 1=East, 2=South, 3=West.
// timeSinceLast: this is only important for "TypeB" Things.
public int row, col, dir, timeSinceLast;
public char lab = 'r';
public boolean isTypeB;
}
/**
* YOU'LL NEED TO PUT THIS SOMEWHERE ELSE
* HINT: WOULDN'T IT BE NICE TO HAVE A LIST OR QUEUE SO THAT
* WE DON'T HAVE TO USE NODES HERE?
* This class is for linked lists of Thing's
*/
public static class Node {
public Thing data;
public Node next;
}
// EEEEEK! STATIC METHODS!!! PLEASE FIND THEM A BETTER HOME.
public static void rightTurn(Thing t) {
t.dir = (t.dir + 1) % 4;
}
public static void leftTurn(Thing t) {
t.dir = (t.dir + 3) % 4;
}
public static void maybeTurn(Thing t) {
int i = rand.nextInt(3);
if (t.isTypeB) {
t.timeSinceLast++;
if (t.timeSinceLast == 10) {
t.timeSinceLast = 0;
if (i == 1) {
rightTurn(t);
}
if (i == 2) {
leftTurn(t);
}
}
} else {
if (i == 1) {
rightTurn(t);
}
if (i == 2) {
leftTurn(t);
}
}
}
public static void step(Thing t) {
final int[] dc = {
0, 1, 0, -1
}, dr = {
1, 0, -1, 0
};
t.row += dr[t.dir];
t.col += dc[t.dir];
}
/**
* This static method is ok :)
*/
public static void main(String[] args) {
int N = 200;
if( args.length != 0 )
N = Integer.parseInt(args[0]);
// INSTEAD OF A NODE, CREATE SOMETHING MORE USER-FRIENDLY.
Node L = null;
int count = 0;
while( true ) {
// Every N rounds, add another typeA and typeB Thing.
if( count % N == 0 ) {
// Add a typeA thing to the list.
// (GEE, THAT'S A LOT OF CODE FOR JUST CREATING ONE THING)
Thing tA = new Thing();
tA.row = 45;
tA.col = 50;
Node nA = new Node();
nA.data = tA;
nA.next = L;
L = nA;
// Add a typeB thing to the list
Thing tB = new Thing();
tB.row = 55;
tB.col = 50;
tB.lab = 'b';
tB.isTypeB = true;
Node nB = new Node();
nB.data = tB;
nB.next = L;
L = nB;
}
// Print out each thing.
// (SEEMS LIKE A NICE PRINTALL() METHOD CALL WOULD WORK HERE)
// (SEEMS LIKE A toString() METHOD IN THE CLASS WOULD ALSO BE NICE)
for( Node T = L; T != null; T = T.next )
System.out.println(T.data.row + " " + T.data.col + " " + T.data.lab);
System.out.println("done");
System.out.flush();
// Move each thing.
// (SEEMS LIKE A NICE MOVEALL() METHOD CALL WOULD WORK HERE)
for( Node T = L; T != null; T = T.next ) {
maybeTurn(T.data);
step(T.data);
}
count++;
}
}
}
Lab2_Tester.java
import org.junit.Test;
import static org.junit.Assert.assertEqual.
Modify this code to do an Insert function for an AVL tree, instead o.pdf
Match the cost behavior that is described with the proper cost type. The unit cost of an item is
$30 per unit when activity is 10,000 units and $15 per unit when activity is 20,000 units. The
total cost of an item is $30,000 when activity is 10,000 units and $30,000 when activity is 20,000
units. The total cost of an item is $30,000 when activity is 10,000 units and $50,000 when
activity is 20,000 units. 1. Variable cost 2. Fixed Cost The unit cost of an item is $30 per unit
when activity is 10,000 units and $20 per unit when activity is 20,000 units 3. Mixed Cost The
total cost of an item is $30,000 when activity is 10,000 units and $60,000 when activity is 20,000
units. The unit cost of an item is $30 per unit when activity is 10,000 units and $30 per unit when
activity is 20,000 units.
Solution
Variable cost is constant or same in per unit but varies in total with change in output
Fixed cost is constant in total but varies in per unit [ that is doesnot change with change in
output]
Mixed cost is neither constant in total nor in per unit
A)Fixed cost ,since same in total [10000*30 = 300000 or 15*20000 = 300000]
B)Fixed cost ,since same in total
C)Mixed cost,neither constant in per unit nor in total
D)Mixed cost ,neither constant in per unit nor in total
E)Variable cost ,constant in per unit [30000/10000 or 60000/20000 = $ 3 per unit]
F)Variable cost .
c).
In this lab, you will be given a simple code for a min Heap, and you.pdf
In this lab, you will be given a simple code for a min Heap, and youre going to complete some of
the methods to handle some edge cases. Tasks for today: There are mainly two tasks, and the
logic of both of them is provided in their bodies: 1) Fully implement AddItem(). 2) Fully
implement HeapUp() Use the following start-up code. The main code and a demo class are
provided as .cs files. Use them to test your code. Once you complete the above two tasks, upload
the source code for your MinHeap class to Blackboard. class BinaryHeap : IEnumerable where T
: IComparable { private T[] array; private int count; // Number of elements (nodes) public
BinaryHeap(int size) { array = new T[size]; count = 0; } public T GetItem(int index) { return
array[index]; } private void SetItem(int index, T value) { while (index >= array.Length)
Grow(array.Length * 2); array[index] = value; } private void Grow(int newsize) {
Array.Resize(ref array, newsize); } // Indices of left and right children private int
LeftChildIndex(int pos) { return 2 * pos + 1; } private int RightChildIndex(int pos) { return 2 *
pos + 2; } private int GetParentIndex(int pos) => (pos - 1) / 2; private T GetRightChild(int pos)
=> array[RightChildIndex(pos)]; private T GetLeftChild(int pos) => array[LeftChildIndex(pos)];
private T GetParent(int pos) => array[GetParentIndex(pos)]; // "Has" methods to determine if the
indices exist private bool HasLeftChild(int pos) { if (LeftChildIndex(pos) < count) return true;
else return false; } private bool HasRightChild(int pos) { if (RightChildIndex(pos) < count)
return true; else return false; } private bool IsRoot(int pos) => pos == 0; // (true if element is
root) // Swap, uh, swaps two values given two indicies. This should be private but I originally
had it public for some reason. private void Swap(int position1, int position2) { T first =
array[position1]; array[position1] = array[position2]; array[position2] = first; } public void
AddItem(T value) { // This is part of la //Insert Logic // If the tree is empty, insert at the bottom
(it does that already) // if not, insert at the end, // From the end you either need to swap to the
root, and keep minheapify (Heapdown) // or, you should probably implement move up (for lab 6,
you need to implement // for that you run a while loop, check if the current position both isn't the
root, and is higher priority than a parent (in this case that probably means it's a lower value) // if
it is, swap with parent, and keep doing that until it's its in the array[count] = value; count++;
ReCalculateUp(); } private void HeapDown() { //CompareTo //this.CompareTo(value) returns <
0 if this < value //this.CompareTo(value) returns >0 if this > value int index = 0; while
(HasLeftChild(index)) { var smallerIndex = LeftChildIndex(index); if (HasRightChild(index)
&& (GetRightChild(index).CompareTo(GetLeftChild(index))<0) ) //there's a set of ( ) around the
right expression that are redundant but hopefully easier to read { smallerI.
I have a .java program that I need to modify so that it1) reads i.pdf
I have a .java program that I need to modify so that it:
1) reads in a file called books.txt (i will provide a few lines of the text file below)
2) outputs a statement every time there's a collision (see the example output below in the original
prompt)
THIS IS THE ORIGINAL PROMPT:
Part A: Modify the attached code to build and customize a AVL tree of Book nodes. Using an
input file, insert Book nodes and detect imbalance. If imbalance is true, then call the proper
rotation function as discussed in the lecture to fix the imbalance condition.
Must read AVLNode data from a text file
Create a Book Object; and an AVL node object to be inserted into the AVL tree
At each insert, detect imbalance and fix the AVL tree
Report each imbalance detection and the node where it occurred; and output the message:
Example Output:
Imbalance condition occurred at inserting ISBN 12345; fixed in LeftRight Rotation
Imbalance condition occurred at inserting ISBN 87654; fixed in Left Rotation
Imbalance condition occurred at inserting ISBN 974321; fixed in RightLeft Rotation
class AVLNode {
Book bookObject;
int height;
AVLNode leftPtr;
AVLNode rightPtr;
}
BELOW IS THE CODE I NEED TO MODIFY:
// AVL Binary search tree implementation in Java
// Author: AlgorithmTutor
// data structure that represents a node in the tree
class Node {
int data; // holds the key
Node parent; // pointer to the parent
Node left; // pointer to left child
Node right; // pointer to right child
int bf; // balance factor of the node
public Node(int data) {
this.data = data;
this.parent = null;
this.left = null;
this.right = null;
this.bf = 0;
}
}
public class AVLTree {
private Node root;
public AVLTree() {
root = null;
}
private void printHelper(Node currPtr, String indent, boolean last) {
// print the tree structure on the screen
if (currPtr != null) {
System.out.print(indent);
if (last) {
System.out.print("R----");
indent += " ";
} else {
System.out.print("L----");
indent += "| ";
}
System.out.println(currPtr.data + "(BF = " + currPtr.bf + ")");
printHelper(currPtr.left, indent, false);
printHelper(currPtr.right, indent, true);
}
}
private Node searchTreeHelper(Node node, int key) {
if (node == null || key == node.data) {
return node;
}
if (key < node.data) {
return searchTreeHelper(node.left, key);
}
return searchTreeHelper(node.right, key);
}
private Node deleteNodeHelper(Node node, int key) {
// search the key
if (node == null) return node;
else if (key < node.data) node.left = deleteNodeHelper(node.left, key);
else if (key > node.data) node.right = deleteNodeHelper(node.right, key);
else {
// the key has been found, now delete it
// case 1: node is a leaf node
if (node.left == null && node.right == null) {
node = null;
}
// case 2: node has only one child
else if (node.left == null) {
Node temp = node;
node = node.right;
}
else if (node.right == null) {
Node temp = node;
node = node.left;
}
// case 3: has both children
else {
Node temp = minimum(node.right);
node.data = temp.data;
node.right = .
Unit-2.Arrays and Strings.pptx.................
,,,,,,jhbhhhhhvggvvcccccccfghhhhhhhhhhhhhgggggggggyyyyy hhbbhhhhgggyhhsyddhdhhddhsiejdhajshdhdhdhdhdhdudufjdhyeyeg hhdhdhdhdhdhdhdhdh d bhsuhabuesjjsoeijshshdhsjajsijehfdhhdhdhdhdhhdxbbxxhxhhdja hi in najajajhwhr7fhhhhhhhhhhhhhhhhhuhhuwbshdhhddh8rhedi8fnf
Give an explanation of the pedigree below. SolutionThe inherit.pdf
Give an explanation of the pedigree below.
Solution
The inheritance is an recessive pattern of inheritance as only few progeny from one mating are
effected. This is due to recessive inheritance where both copies of the recessive allele are
required to cause the disease phenotype.
It is Autosomal recessive pattern of inheritance. It is not X linked recessive because, the infected
male child from 2nd F1 generation male (5th child ) donot recieve X chromosome from the
infected father. Hence it is due to autosomal trasmission..
elena ents you to impact into MAt ab SolutionThree fu.pdf
elena ents you to impact into MAt ab
Solution
Three functions---> fprintf-Write data to text file.
fread-Read data from binary file
fopen-Open file or obtain information about open files.
Interface Elements--->mech_interface_crate_transfer
mech_interface_dspring_damper
mech_interface_hyd_cylinder
Keywords--->Clear-Removes variables from memory
Lookfor-Searches help entries for a keyword.
delete-Deletes a file
Path-Display search path..
More Related Content
Similar to Im suppose to make a preOrder transversal method so that it performs.pdf
ReversePoem.java ---------------------------------- public cl.pdf
ReversePoem.java :-
---------------------------------
public class ReversePoem {
/*This programs has you display a pessimistic poem from a list of phrases*/
// and then reverse the phrases to find another more optimistic poem.
public static void main(String[] args) throws Exception
{
//Queue object
MyQueue queue = new MyQueue<>();
//Stack object
MyStack stack = new MyStack<>();
//String buffer to apppend all Strings
StringBuffer sb = new StringBuffer();
// Create a single String object from the 16 Strings below
String set1 = \"I am part of a lost generation#and I refuse to believe that#\";
sb.append(set1);
String set2 = \"I can change the world#I realize this may be a shock but#\";
sb.append(set2);
String set3 = \"\'Happiness comes from within\'#is a lie, and#\";
sb.append(set3);
String set4 = \"\'Money will make me happy\'#So in 30 years I will tell my children#\";
sb.append(set4);
String set5 = \"they are not the most important thing in my life#\";
sb.append(set5);
String set6 = \"My employer will know that#I have my priorities straight because#\";
sb.append(set6);
String set7 = \"work#is more important than#family#I tell you this#\";
sb.append(set7);
String set8 = \"Once upon a time#Families stayed together#\";
sb.append(set8);
String set9 = \"but this will not be true in my era#\";
sb.append(set9);
String set10 = \"This is a quick fix society#Experts tell me#\";
sb.append(set10);
String set11 = \"30 years from now, I will be celebrating the 10th anniversary of my
divorce#\";
sb.append(set11);
String set12 = \"I do not concede that#I will live in a country of my own making#\";
sb.append(set12);
String set13 = \"In the future#Environmental destruction will be the norm#\";
sb.append(set13);
String set14 = \"No longer can it be said that#My peers and I care about this earth#\";
sb.append(set14);
String set15 = \"It will be evident that#My generation is apathetic and lethargic#\";
sb.append(set15);
String set16 = \"It is foolish to presume that#There is hope#\";
sb.append(set16);
String finalString = sb.toString();
String itmes[] = finalString.split(\"#\");
System.out.println(\"========== Original Phrase ==============\");
for(int i = 0 ; i < itmes.length;i++){
queue.enqueue(itmes[i]);
System.out.println(itmes[i]);
}
for(int i = 0 ; i < itmes.length;i++){
stack.push(queue.dequeue());
}
System.out.println(\"========== Reverse Phrase ==============\");
for(int i = 0 ; i < itmes.length;i++){
System.out.println(stack.pop());
}
/* You are given a list of phrases in Strings; the phrases
are separated by pound signs: \'#\':
1. Create a single String object from this list.
2. Then, split the String of phrases into an array of
phrases using the String split method.
3. Display a poem by walking through the array and
displaying each phrase one per line.
4. And, at the same time, place each phrase on a
MyQueue object using only the enqueue method.
5. After all the phrases have been placed on the queue,
transfer the phrases from the MyQueue object to a
MyS.
There is BinarySearchTree class. When removing a node from a BST, we.pdf
There is BinarySearchTree class. When removing a node from a BST, we can replace it with
either its predecessor or its successor. Complete the predecessor() method
so that it returns the node that comes immediately before e. Hint: the work needed to find the
predecessor is a mirror image of what is required to find a node\'s
successor, so use successor() as a resource.
BinarySearchTree class:
Solution
import java.util.AbstractSet;
import java.util.Iterator;
public class BinarySearchTree> extends AbstractSet {
protected BTNode root;
protected int size;
/**
* Initializes this BinarySearchTree object to be empty, to contain only
* elements of type E, to be ordered by the Comparable interface, and to
* contain no duplicate elements.
*/
public BinarySearchTree() {
root = null;
size = 0;
}
/**
* Initializes this BinarySearchTree object to contain a shallow copy of a
* specified BinarySearchTree object. The worstTime(n) is O(n), where n is the
* number of elements in the specified BinarySearchTree object.
*
* @param otherTree - the specified BinarySearchTree object that this
* BinarySearchTree object will be assigned a shallow copy of.
*/
public BinarySearchTree(BinarySearchTree otherTree) {
root = copy(otherTree.root, null);
size = otherTree.size;
}
protected BTNode copy(BTNode p, BTNode parent) {
if (p != null) {
BTNode q = new BTNode(p.element, parent);
q.left = copy(p.left, q);
q.right = copy(p.right, q);
return q;
}
return null;
} // method copy
/**
* Finds the predecessor of a specified BTNode object in this BinarySearchTree. The
worstTime(n) is O(n) and
* averageTime(n) is constant.
*
* @param e - the BTNode object whose successor is to be found.
* @return the successor of e, if e has a successor; otherwise, return null.
*/
protected BTNode predecessor(BTNode e) {
if (e == null) {
return null;
} else if (e.left != null) {
// successor is leftmost BTNode in right subtree of e
BTNode p = e.left;
while (p.right != null) {
p = p.right;
}
return p;
} // e has a right child
else {
// go up the tree to the right as far as possible, then go up
// to the left.
BTNode p = e.parent;
BTNode ch = e;
while ((p != null) && (ch == p.left)) {
ch = p;
p = p.parent;
} // while
return p;
} // e has no left child
// method successor
}
@SuppressWarnings(\"unchecked\")
@Override
public boolean equals(Object obj) {
if (!(obj instanceof BinarySearchTree)) {
return false;
}
return equals(root, ((BinarySearchTree) obj).root);
} // method 1-parameter equals
public boolean equals(BTNode p, BTNode q) {
if ((p == null) || (q == null)) {
return p == q;
}
if (!p.element.equals(q.element)) {
return false;
}
if (equals(p.left, q.left) && equals(p.right, q.right)) {
return true;
}
return false;
}
/**
* Returns the size of this BinarySearchTree object.
*
* @return the size of this BinarySearchTree object.
*/
@Override
public int size() {
return size;
}
/**
* Iterator method will be implemented for a future
*
* @return an iterator positioned at the smallest element in this
* Bina.
Write a program that displays an AVL tree along with its balance fac.docx
Write a program that displays an AVL tree along with its balance fac
In java , I want you to implement a Data Structure known as a Doubly.pdf
In java , I want you to implement a Data Structure known as a Doubly-Ended-Queue. it is a
“fair” data structure in that it implements a FIFO (First In First Out ) behavior. As such, it is
often used to implement various wait lists in computer systems. For example, jobs waiting to use
the CPU, jobs waiting for a printer, jobs waiting to be placed into RAM for execution. In short,
whenever we want a fair strategy for waiting we use queues.
A DEQUE (Doubly-ended-queue) is a related data structure. Although similar to a Queue, it
differs in that it allows for insertions AND deletions from either end of the list (both the front
and the rear).
Your implementation MUST use a doubly-linked-list implementation. You may not use a static
array implementation.
Thus, a Deque is a List but it is one which only concerns itself with the first and last positions for
any insertion or deletion. The 6 operations supported are :
public void insertFront( int item ) - insert the given item (as a node) into the first position of the
Deque.
public void insertRear( int item ) - insert the given item (as a node) into the last position of the
Deque.
public int deleteFront( ) - delete and return the element stored in the first node of the Deque.
public int deletRear( ) – delete and return the element stored in the last node of the Deque.
public boolean isempty( ) - returns true if the Deque is currently empty or false if it is not.
public void printDeque( ) - print the integers from the list, one per line, from the first element
through to the last in order.
Classes
Your program must implement the following 3 classes. public class dequeDriver
This class will contain your program’s main method. It will need to declare a deque object and
process input as indicated below.
Your program should prompt the user for the path of an input file. It should open the file for
input and process it line by line. Each line of the input file will have one of the following forms.
PR
IF
IR
DF
DR
The meanings of each input is as follows:
PR - print the current contents of the deque from front to rear using the printDeque( ) method of
the deque object.
IF - insert the given int value into the front of the deque.
IR - insert the given int value into the rear of the deque.
DF - delete the front value from the deque.
DR – delete the rear element of the deque.
Below is an example input file that your program should be able to process.
PR
IF 4
IF 5
IF 6
IR 7
PR
DR
PR
DF
PR
The output for the input file shown above is :
EMPTY DEQUE
----- Front -----
6
5
4
7
----- Rear -----
----- Front -----
6
5
4
----- Rear -----
----- Front -----
5
4
----- Rear -----
public class dequeNode
This class will implement the linked nodes that will be used to implement the deque itself.
It should have the following protected data members.
protected dequeNode next; // next pointer to next node
protected dequeNode prev; // previous pointer to previous node
protected int val; // the integer value stored within the dequeNod.
Need Help with this Java Assignment. Program should be done in JAVA .pdf
Need Help with this Java Assignment. Program should be done in JAVA please
Complete required scripts based on eclipse and troubleshoot the scripts until the tasks are done.
1. Implement a binary search tree (write the code for it), for a height of 2 or higher. Please show
the output if you can Thank you in advance!
Solution
I have compiled this code and checked for errors, worked fine.
Also after the code I have pasted the Output transcript to show what User Inputs I gave and what
all outputs I got.
// Java Program to Implement Binary Tree
import java.util.Scanner;
/* Define Class BinaryTreeNode */
class BinaryTreeNode
{
BinaryTreeNode left, right;
int data;
/* Constructor */
public BinaryTreeNode()
{
left = null;
right = null;
data = 0;
}
/* Constructor */
public BinaryTreeNode(int n)
{
left = null;
right = null;
data = n;
}
/* Function to set left node */
public void setLeft(BinaryTreeNode n)
{
left = n;
}
/* Function to set right node */
public void setRight(BinaryTreeNode n)
{
right = n;
}
/* Function to get left node */
public BinaryTreeNode getLeft()
{
return left;
}
/* Function to get right node */
public BinaryTreeNode getRight()
{
return right;
}
/* Function to set data to node */
public void setData(int d)
{
data = d;
}
/* Function to get data from node */
public int getData()
{
return data;
}
}
/* Define anothe Class Operations to perform required operationns on Binary Tree*/
class Operations
{
private BinaryTreeNode root;
/* Constructor */
public Operations()
{
root = null;
}
/* Function to check if tree is empty */
public boolean isEmpty()
{
return root == null;
}
/* Functions to insert data */
public void insert(int data)
{
root = insert(root, data);
}
/* Function to insert data recursively */
private BinaryTreeNode insert(BinaryTreeNode node, int data)
{
if (node == null)
node = new BinaryTreeNode(data);
else
{
if (node.getRight() == null)
node.right = insert(node.right, data);
else
node.left = insert(node.left, data);
}
return node;
}
/* Function to count number of nodes */
public int countNodes()
{
return countNodes(root);
}
/* Function to count number of nodes recursively */
private int countNodes(BinaryTreeNode r)
{
if (r == null)
return 0;
else
{
int l = 1;
l += countNodes(r.getLeft());
l += countNodes(r.getRight());
return l;
}
}
/* Function to search for an element */
public boolean search(int val)
{
return search(root, val);
}
/* Function to search for an element recursively */
private boolean search(BinaryTreeNode r, int val)
{
if (r.getData() == val)
return true;
if (r.getLeft() != null)
if (search(r.getLeft(), val))
return true;
if (r.getRight() != null)
if (search(r.getRight(), val))
return true;
return false;
}
/* Function for inorder traversal */
public void inorder()
{
inorder(root);
}
private void inorder(BinaryTreeNode r)
{
if (r != null)
{
inorder(r.getLeft());
System.out.print(r.getData() +\" \");
inorder(r.getRight());
}
}
/* Function for preorder traversal */
public void preorder()
{
preorder(root);
}
priva.
Write a C++ program that implements a binary search tree (BST) to man.pdf
Write a C++ program that implements a binary search tree (BST) to manage a number of integer
items with different priorities. In order to do so. you will need to maintain a queue as an item in
the tree. Each queue item is an integer. All items in a given queue will have the same priority. As
a new item with a user-specified priority comes in, it should be enqueued in the queue with items
of that priority. A new queue may need to created in case a queue with that priority does not
exist. Use dynamic memory to manage nodes in the queue and BST. Implement the following:
Constructor. Destructor. Overloaded copy constructor. Overloaded assignment operator.
Overloaded \"==\" operator to check if two trees are equal. A tree t1 is equal to t2 if for every
queue in t1, a corresponding queue with the same priority exists in t2 with the same set of
elements (the order of the elements in the queue docs not matter). Overloaded \"
Solution
program using namespace std;
#define width_unit 5
class BStree
{
private:
class Node
{
public:
int data;
Node *left, *right;
Node(int d=0)
:data(d), left(NULL), right(NULL) {}
};
Node *root;
Node * trav(int, Node * &);
void chop(Node * N);
void copy(Node * N);
void print(ostream &, Node *, int) const;
void print(Node *, int) const;
public:
BStree(void);
~BStree(void);
bool find(int);
void insert(int);
void remove(int);
bool empty(void) const;
Tree(const BStree &);
const BStree & operator=(const BStree &);
friend ostream & operator<<(ostream &, const BStree &);
};
BStree::BStree(void)
{
root=NULL;
}
bool Tree::empty(void) const
{
return !root;
}
BStree::Node * BStree::trav(int foo, Node * & par)
{
Node * curr=root;
par=NULL;
while(curr && curr->data != foo)
{
par=curr;
if(foo < curr->data)
curr=curr->left;
else
curr=curr->right;
}
return curr;
}
bool BStree::find(int foo)
{
Node * par=NULL;
Node * curr=trav(foo, par);
return curr;
}
void BStree::insert(int foo)
{
Node * par=NULL;
Node * curr=trav(foo,par);
if(!curr)
{
curr= new Node(foo);
if(!par)
root=curr;
else if(foo < par->data)
par->left=curr;
else
par->right=curr;
}
}
void BStree::remove(const int foo)
{
Node * par=NULL;
Node * curr=trav(foo,par);
if(curr)
{
if(curr->left && curr->right)
{
Node * tmp=curr;
par=curr;
curr=curr->left;
while(curr->right)
{
par=curr;
curr=curr->right;
}
tmp->data=curr->data;
}
Node *tmp=(curr->left ? curr->left : curr->right);
if(!par)
root=tmp;
else if(par->data < curr->data)
par->right=tmp;
else
par->left=tmp;
delete curr;
}
}
void BStree::chop(Node *N)
{
if(N)
{
chop(N->left);
chop(N->right);
delete N;
}
}
BStree::~BStree(void)
{
chop(root);
}
BStree::BStree(const BStree & T)
{
root=NULL;
copy(T.root);
}
void BStree::copy(Node * N)
{
if(N)
{
insert(N->data);
copy(N->left);
copy(N->right);
}
}
const BStree & BStree::operator=(const BStree & T)
{
if(this != &T)
{
chop(root);
root=NULL;
copy(T.root);
}
return *this;
}
void BStree::print(ostream & ost, Node * curr, int level) const
{
if(curr)
{
print(ost,curr->right,level+1);
ost<data<le.
L 17 ct1120
Class lecture of Data Structure and Algorithms and Python.
Stack, Queue, Tree, Python, Python Code, Computer Science, Data, Data Analysis, Machine Learning, Artificial Intellegence, Deep Learning, Programming, Information Technology, Psuedocide, Tree, pseudocode, Binary Tree, Binary Search Tree, implementation, Binary search, linear search, Binary search operation, real-life example of binary search, linear search operation, real-life example of linear search, example bubble sort, sorting, insertion sort example, stack implementation, queue implementation, binary tree implementation, priority queue, binary heap, binary heap implementation
A deque (pronounced deck) is an ordered set of items from which item.pdf
A deque (pronounced deck) is an ordered set of items from which items may be deleted at either
end and into which items may be inserted at either end. Call the two ends left and right. This is
an access-restricted structure since no insertions or deletions can happen other than at the ends.
Implement a deque as a doubly-linked circular list with a header. Write InsertRight and
DeleteLeft.
Solution
code:
#include
using namespace std;
class Node
{
public:
int data;
Node* next;
Node* prev;
};
class Deque
{
private:
Node* front;
Node* rear;
int count;
public:
Deque()
{
front = NULL;
rear = NULL;
count = 0;
}
bool isEmpty()
{
if(count == 0)
return true;
else
return false;
}
int DeleteLeft()
{
if ( isEmpty() ) {
cout<<\"deque is empty... can\'t delete\'\";
return -1;
}
int ret = front->data;
Node* tmp = front;
if ( front->next != NULL )
{
front = front->next;
front->prev = NULL;
}
else
{
front = NULL;
}
count--;
delete tmp;
return ret;
}
void InsertRight(int element)
{
Node* tmp = new Node();
tmp->data = element;
tmp->next = NULL;
tmp->prev = NULL;
if ( isEmpty() ) {
front = rear = tmp;
}
else {
rear->next = tmp;
tmp->prev = rear;
rear = tmp;
}
count++;
}
};
int main()
{
Deque q;
if ( q.isEmpty() )
{
cout << \"Deque is empty\" << endl;
}
q.InsertRight(100);
q.InsertRight(200);
q.InsertRight(300);
Deque q1;
if ( q1.isEmpty() )
{
cout << \"Deque is empty\" << endl;
}
q1.InsertRight(100);
q1.InsertRight(200);
q1.InsertRight(300);
cout << q1.DeleteLeft() << endl;
cout << q1.DeleteLeft() << endl;
cout << q1.DeleteLeft() << endl;
}.
AvlTree.h#ifndef AVL_TREE_H#define AVL_TREE_H#include d.docx
AvlTree.h
#ifndef AVL_TREE_H
#define AVL_TREE_H
#include "dsexceptions.h"
#include <iostream> // For NULL
#include <queue> // For level order printout
#include <vector>
#include <algorithm> // For max() function
using namespace std;
// AvlTree class
//
// CONSTRUCTION: with ITEM_NOT_FOUND object used to signal failed finds
//
// ******************PUBLIC OPERATIONS*********************
// Programming Assignment Part I
// bool empty( ) --> Test for empty tree @ root
// int size( ) --> Quantity of elements in tree
// int height( ) --> Height of the tree (null == -1)
// void insert( x ) --> Insert x
// void insert( vector<T> ) --> Insert whole vector of values
// void remove( x ) --> Remove x (unimplemented)
// bool contains( x ) --> Return true if x is present
// Comparable findMin( ) --> Return smallest item
// Comparable findMax( ) --> Return largest item
// boolean isEmpty( ) --> Return true if empty; else false
// void printTree( ) --> Print tree in sorted (in) order
// void printPreOrder( ) --> Print tree in pre order
// void printPostOrder( ) --> Print tree in post order
// void printInOrder( ) --> Print tree in *in* order
// Programming Assignment Part II (microassignment)
// void makeEmpty( ) --> Remove and delete all items
// void ~AvlTree( ) --> Big Five Destructor
// AvlTree(const AvlTree &other) --> BigFive Copy Constructor
// AvlTree(const AvlTree &&other) --> BigFive Move Constructor
// AvlTree &operator= ( AvlTree & other ) --> Big Five Copy *assignment* operator
// AvlTree &operator= ( AvlTree && other ) --> Big Five Move *assignment* operator
// void printLevelOrder( ) --> Print tree in LEVEL order :-)
// ******************ERRORS********************************
// Throws UnderflowException as warranted
template <typename Comparable>
class AvlTree
{
public:
/**
* Basic constructor for an empty tree
*/
AvlTree( ) : root( NULL )
{
//cout << " [d] AvlTree constructor called. " << endl;
}
/**
* Vector of data initializer (needed for move= operator rvalue)
*/
AvlTree( vector<Comparable> vals ) : root( NULL )
{
insert(vals);
}
//*******************************************************************************************
// START AVL TREES PART II - TODO: Implement
// Other functions to look for include: clone, makeEmpty, printLevelOrder
/**
* Destroy contents of the AvlTree object - Big Five Destructor
*/
~AvlTree( )
{
//cout << " [d] AvlTree Destructor called. " << endl;
// Empty out the AVL Tree and destroy all nodes (makeEmpty?)
}
/**
* Copy other to new object - Big Five Copy Constructor
*/
AvlTree( const AvlTree &other ) : root( NULL )
{
cout << " [d] Copy Constructor Called." << endl;
// Copy contents of other to ourselves (maybe clone?)
// Get a deep copy of other's tree
}
/* ...
I need help with this maze gui that I wrote in java, I am trying to .pdf
I need help with this maze gui that I wrote in java, I am trying to find the shortest path but I dont
know how to implement the FindshortestPath Method.
using a BREADTH FIRST SEARCH and the stradgey below to ONLY FIND THE SHORTEST
PATH, the method \"FindshortestPath(); is the only part you need to add code too.
I was given this as a stragedy to solve the findshortestpathMethod():
Create two LinkedLists of MazeMap objects. One will be the list of map objects for cells at
distance x from the center. The other will be the list of map objects for cells at distance x + 1.
Initialize the list at distance x to contain a MazeMap object which refers to the center cell of the
maze. Since this map object refers to the center cell, it does not need to refer to any other
MazeMap object, so its MazeMap reference can be set to null. Set the center cell’s visited field
to true to indicate that this cell has already been added to the map.
Set up a loop to run until the maze entrance is mapped. This loop does the following:
Set up a loop to process each map object from the list at distance x. This loop does the following:
i.Get the cell from the map object
ii.If this cell is the entrance cell, set the entrance and break out of the loop!
iii.Check each direction from this cell and for an accessible neighbor cell
Get a reference to that neighbor cell
If the neighbor cell has not already been visited
Set the cell’s visited field to true so its only processed once
Create a MazeMap object that refers to this neighbor cell
Make this MazeMap object refer to the current map object being processed – ie the cell one step
closer to center
Add this MazeMap object to the list at distance x + 1
List at distance x has been processed. The list at distance x + 1 is now complete, so make the list
at distance x refer to list at distance x + 1. Then create a new empty list for the list at distance x +
1.
At this point, the entrance points at the map object which refers to cell 1,1. That map object also
refers to a map object which is one step closer to the center. Write a loop that traverses this trail
of references until it reaches the end (a null reference). For each map object:
Get the cell referred to.
Call the drawCircle method with a color, the cell’s row and column, and SMALL.
Call the delay method with a SHORT delay. This slows down the drawing process so you can
watch the path being displayed.
Advance to the next map object in the list.
here is the code so far*********
package mazepackage;
import java.util.LinkedList;
import java.util.Random;
import javafx.scene.paint.Color;
public class Maze {
private int N; // dimension of maze
private MazeCell[][] maze;
private Random rand = new Random();
// Used to signal the maze has been solved
private boolean done;
// time in milliseconds (from currentTimeMillis()) when we can draw again
// used to control the frame rate
private long nextDraw = -1;
private MazeApp mf;
// Define constants for the circle size
private final double BI.
Given the following codepackage data1;import java.util.;p.pdf
Given the following code
package data1;
import java.util.*;
/*public class BST> implementing a set ADT and containing the following:
* private inner class BinaryNode representing a node with a (possibly) left child and a (possibly)
right child
** instance fields BinaryNode root, int size
* contains/insert/remove method - w/ O(height) complexity
** size O(1), isEmpty O(1), clear O(1),
*** findMin, findMax, findHeight methods - w/ O(height) complexity
**** toString() method printing set (in-order traversal) and the tree (BFS traversal)
**
*/
publicclass BST> {
private BinaryNode root;
privateint size;
//Search opertion of Set ADT
publicboolean contains(E value) {
return contains(root, value);
}
//Insert opertion of Set ADT
publicvoid insert(E value) {
root = insert(root, value);
size++;
}
//Remove operation of Set ADT
publicvoid remove(E value){
root = remove(root, value);
size--;
}
publicint size(){
return size;
}
publicboolean isEmpty(){
return root == null;
}
publicvoid clear(){
root = null;
size = 0;
}
public E findMin(){
return findMin(root);
}
public E findMax(){
return findMax(root);
}
publicint findHeight(){
return root.height();
}
public String toString(){
if(root == null)
return "set {} stored in an empty tree.\n";
String elements = root.inOrder().toString();
return "set {" + elements.substring(1,elements.length()-1) + "} stored in tree \n" + root;
}
private E findMin(BinaryNode node){
if(node == null)
returnnull;
if(node.left == null)
return node.element;
return findMin(node.left);
}
private E findMax(BinaryNode node){
if(node == null)
returnnull;
if(node.right == null)
return node.element;
return findMax(node.right);
}
privateboolean contains(BinaryNode node, E value) {
if (node == null)//node represents an empty substree
returnfalse;
int comparisonResult = value.compareTo(node.element);
if(comparisonResult < 0)//if value is less than root's value
return contains(node.left, value);
if(comparisonResult > 0)//if value is greater than root's value
return contains(node.right, value);
returntrue;//successful search
}
private BinaryNode insert(BinaryNode node, E value) {
if (node == null)//base case: empty tree
returnnew BinaryNode<>(value);
if (value.compareTo(node.element) < 0)
node.left = insert(node.left, value);//insert recursively to left-subtree
elseif (value.compareTo(node.element) > 0)
node.right = insert(node.right, value);//insert recursively to right-subtree
else {//duplicate value cannot be inserted in a BST implementing the Set ADT
size--;//insertion failed!
//System.out.print("BST.insert: Warning: " + value + " is already stored in the tree \n" + node);
}
return node;
}
private BinaryNode remove(BinaryNode node, E value) {
if (node == null) {//base case: empty tree
System.out.println("BST.remove: Warning: " + value + " doesn't exist in the tree.");
size++;//removal failed!
return node;
}
if (value.compareTo(node.element) < 0)
node.left = remove(node.left, value);
elseif (value.compareTo(node.element) > 0)
node.right = remov.
JAVA OOP project; desperately need help asap im begging.Been stuck.pdf
JAVA OOP project; desperately need help asap im begging.
Been stuck on this for a week since my OOP skills are not good at all. Due in a couple hours so
im desperate. Codes needed will be provided at the end, thank you.
Codes:
DotChaser.java:
import java.util.*;
public class DotChaser {
public static Random rand = new Random(System.currentTimeMillis());
/**
* A "Thing" moves in a grid world. A TypeA Thing randomly
* decides to turn left or right (or not turn) every "round",
* and, afterward, takes a step forward. A TypeB Thing
* only considers making a random turn every 10th round.
*
* A STATIC CLASS? OH NO! GET IT OUT OF HERE!
*/
public static class Thing {
// dir: 0=North, 1=East, 2=South, 3=West.
// timeSinceLast: this is only important for "TypeB" Things.
public int row, col, dir, timeSinceLast;
public char lab = 'r';
public boolean isTypeB;
}
/**
* YOU'LL NEED TO PUT THIS SOMEWHERE ELSE
* HINT: WOULDN'T IT BE NICE TO HAVE A LIST OR QUEUE SO THAT
* WE DON'T HAVE TO USE NODES HERE?
* This class is for linked lists of Thing's
*/
public static class Node {
public Thing data;
public Node next;
}
// EEEEEK! STATIC METHODS!!! PLEASE FIND THEM A BETTER HOME.
public static void rightTurn(Thing t) {
t.dir = (t.dir + 1) % 4;
}
public static void leftTurn(Thing t) {
t.dir = (t.dir + 3) % 4;
}
public static void maybeTurn(Thing t) {
int i = rand.nextInt(3);
if (t.isTypeB) {
t.timeSinceLast++;
if (t.timeSinceLast == 10) {
t.timeSinceLast = 0;
if (i == 1) {
rightTurn(t);
}
if (i == 2) {
leftTurn(t);
}
}
} else {
if (i == 1) {
rightTurn(t);
}
if (i == 2) {
leftTurn(t);
}
}
}
public static void step(Thing t) {
final int[] dc = {
0, 1, 0, -1
}, dr = {
1, 0, -1, 0
};
t.row += dr[t.dir];
t.col += dc[t.dir];
}
/**
* This static method is ok :)
*/
public static void main(String[] args) {
int N = 200;
if( args.length != 0 )
N = Integer.parseInt(args[0]);
// INSTEAD OF A NODE, CREATE SOMETHING MORE USER-FRIENDLY.
Node L = null;
int count = 0;
while( true ) {
// Every N rounds, add another typeA and typeB Thing.
if( count % N == 0 ) {
// Add a typeA thing to the list.
// (GEE, THAT'S A LOT OF CODE FOR JUST CREATING ONE THING)
Thing tA = new Thing();
tA.row = 45;
tA.col = 50;
Node nA = new Node();
nA.data = tA;
nA.next = L;
L = nA;
// Add a typeB thing to the list
Thing tB = new Thing();
tB.row = 55;
tB.col = 50;
tB.lab = 'b';
tB.isTypeB = true;
Node nB = new Node();
nB.data = tB;
nB.next = L;
L = nB;
}
// Print out each thing.
// (SEEMS LIKE A NICE PRINTALL() METHOD CALL WOULD WORK HERE)
// (SEEMS LIKE A toString() METHOD IN THE CLASS WOULD ALSO BE NICE)
for( Node T = L; T != null; T = T.next )
System.out.println(T.data.row + " " + T.data.col + " " + T.data.lab);
System.out.println("done");
System.out.flush();
// Move each thing.
// (SEEMS LIKE A NICE MOVEALL() METHOD CALL WOULD WORK HERE)
for( Node T = L; T != null; T = T.next ) {
maybeTurn(T.data);
step(T.data);
}
count++;
}
}
}
Lab2_Tester.java
import org.junit.Test;
import static org.junit.Assert.assertEqual.
Modify this code to do an Insert function for an AVL tree, instead o.pdf
Match the cost behavior that is described with the proper cost type. The unit cost of an item is
$30 per unit when activity is 10,000 units and $15 per unit when activity is 20,000 units. The
total cost of an item is $30,000 when activity is 10,000 units and $30,000 when activity is 20,000
units. The total cost of an item is $30,000 when activity is 10,000 units and $50,000 when
activity is 20,000 units. 1. Variable cost 2. Fixed Cost The unit cost of an item is $30 per unit
when activity is 10,000 units and $20 per unit when activity is 20,000 units 3. Mixed Cost The
total cost of an item is $30,000 when activity is 10,000 units and $60,000 when activity is 20,000
units. The unit cost of an item is $30 per unit when activity is 10,000 units and $30 per unit when
activity is 20,000 units.
Solution
Variable cost is constant or same in per unit but varies in total with change in output
Fixed cost is constant in total but varies in per unit [ that is doesnot change with change in
output]
Mixed cost is neither constant in total nor in per unit
A)Fixed cost ,since same in total [10000*30 = 300000 or 15*20000 = 300000]
B)Fixed cost ,since same in total
C)Mixed cost,neither constant in per unit nor in total
D)Mixed cost ,neither constant in per unit nor in total
E)Variable cost ,constant in per unit [30000/10000 or 60000/20000 = $ 3 per unit]
F)Variable cost .
c).
In this lab, you will be given a simple code for a min Heap, and you.pdf
In this lab, you will be given a simple code for a min Heap, and youre going to complete some of
the methods to handle some edge cases. Tasks for today: There are mainly two tasks, and the
logic of both of them is provided in their bodies: 1) Fully implement AddItem(). 2) Fully
implement HeapUp() Use the following start-up code. The main code and a demo class are
provided as .cs files. Use them to test your code. Once you complete the above two tasks, upload
the source code for your MinHeap class to Blackboard. class BinaryHeap : IEnumerable where T
: IComparable { private T[] array; private int count; // Number of elements (nodes) public
BinaryHeap(int size) { array = new T[size]; count = 0; } public T GetItem(int index) { return
array[index]; } private void SetItem(int index, T value) { while (index >= array.Length)
Grow(array.Length * 2); array[index] = value; } private void Grow(int newsize) {
Array.Resize(ref array, newsize); } // Indices of left and right children private int
LeftChildIndex(int pos) { return 2 * pos + 1; } private int RightChildIndex(int pos) { return 2 *
pos + 2; } private int GetParentIndex(int pos) => (pos - 1) / 2; private T GetRightChild(int pos)
=> array[RightChildIndex(pos)]; private T GetLeftChild(int pos) => array[LeftChildIndex(pos)];
private T GetParent(int pos) => array[GetParentIndex(pos)]; // "Has" methods to determine if the
indices exist private bool HasLeftChild(int pos) { if (LeftChildIndex(pos) < count) return true;
else return false; } private bool HasRightChild(int pos) { if (RightChildIndex(pos) < count)
return true; else return false; } private bool IsRoot(int pos) => pos == 0; // (true if element is
root) // Swap, uh, swaps two values given two indicies. This should be private but I originally
had it public for some reason. private void Swap(int position1, int position2) { T first =
array[position1]; array[position1] = array[position2]; array[position2] = first; } public void
AddItem(T value) { // This is part of la //Insert Logic // If the tree is empty, insert at the bottom
(it does that already) // if not, insert at the end, // From the end you either need to swap to the
root, and keep minheapify (Heapdown) // or, you should probably implement move up (for lab 6,
you need to implement // for that you run a while loop, check if the current position both isn't the
root, and is higher priority than a parent (in this case that probably means it's a lower value) // if
it is, swap with parent, and keep doing that until it's its in the array[count] = value; count++;
ReCalculateUp(); } private void HeapDown() { //CompareTo //this.CompareTo(value) returns <
0 if this < value //this.CompareTo(value) returns >0 if this > value int index = 0; while
(HasLeftChild(index)) { var smallerIndex = LeftChildIndex(index); if (HasRightChild(index)
&& (GetRightChild(index).CompareTo(GetLeftChild(index))<0) ) //there's a set of ( ) around the
right expression that are redundant but hopefully easier to read { smallerI.
I have a .java program that I need to modify so that it1) reads i.pdf
I have a .java program that I need to modify so that it:
1) reads in a file called books.txt (i will provide a few lines of the text file below)
2) outputs a statement every time there's a collision (see the example output below in the original
prompt)
THIS IS THE ORIGINAL PROMPT:
Part A: Modify the attached code to build and customize a AVL tree of Book nodes. Using an
input file, insert Book nodes and detect imbalance. If imbalance is true, then call the proper
rotation function as discussed in the lecture to fix the imbalance condition.
Must read AVLNode data from a text file
Create a Book Object; and an AVL node object to be inserted into the AVL tree
At each insert, detect imbalance and fix the AVL tree
Report each imbalance detection and the node where it occurred; and output the message:
Example Output:
Imbalance condition occurred at inserting ISBN 12345; fixed in LeftRight Rotation
Imbalance condition occurred at inserting ISBN 87654; fixed in Left Rotation
Imbalance condition occurred at inserting ISBN 974321; fixed in RightLeft Rotation
class AVLNode {
Book bookObject;
int height;
AVLNode leftPtr;
AVLNode rightPtr;
}
BELOW IS THE CODE I NEED TO MODIFY:
// AVL Binary search tree implementation in Java
// Author: AlgorithmTutor
// data structure that represents a node in the tree
class Node {
int data; // holds the key
Node parent; // pointer to the parent
Node left; // pointer to left child
Node right; // pointer to right child
int bf; // balance factor of the node
public Node(int data) {
this.data = data;
this.parent = null;
this.left = null;
this.right = null;
this.bf = 0;
}
}
public class AVLTree {
private Node root;
public AVLTree() {
root = null;
}
private void printHelper(Node currPtr, String indent, boolean last) {
// print the tree structure on the screen
if (currPtr != null) {
System.out.print(indent);
if (last) {
System.out.print("R----");
indent += " ";
} else {
System.out.print("L----");
indent += "| ";
}
System.out.println(currPtr.data + "(BF = " + currPtr.bf + ")");
printHelper(currPtr.left, indent, false);
printHelper(currPtr.right, indent, true);
}
}
private Node searchTreeHelper(Node node, int key) {
if (node == null || key == node.data) {
return node;
}
if (key < node.data) {
return searchTreeHelper(node.left, key);
}
return searchTreeHelper(node.right, key);
}
private Node deleteNodeHelper(Node node, int key) {
// search the key
if (node == null) return node;
else if (key < node.data) node.left = deleteNodeHelper(node.left, key);
else if (key > node.data) node.right = deleteNodeHelper(node.right, key);
else {
// the key has been found, now delete it
// case 1: node is a leaf node
if (node.left == null && node.right == null) {
node = null;
}
// case 2: node has only one child
else if (node.left == null) {
Node temp = node;
node = node.right;
}
else if (node.right == null) {
Node temp = node;
node = node.left;
}
// case 3: has both children
else {
Node temp = minimum(node.right);
node.data = temp.data;
node.right = .
Unit-2.Arrays and Strings.pptx.................
,,,,,,jhbhhhhhvggvvcccccccfghhhhhhhhhhhhhgggggggggyyyyy hhbbhhhhgggyhhsyddhdhhddhsiejdhajshdhdhdhdhdhdudufjdhyeyeg hhdhdhdhdhdhdhdhdh d bhsuhabuesjjsoeijshshdhsjajsijehfdhhdhdhdhdhhdxbbxxhxhhdja hi in najajajhwhr7fhhhhhhhhhhhhhhhhhuhhuwbshdhhddh8rhedi8fnf
Similar to Im suppose to make a preOrder transversal method so that it performs.pdf (20)
ReversePoem.java ---------------------------------- public cl.pdf
ReversePoem.java ---------------------------------- public cl.pdf
There is BinarySearchTree class. When removing a node from a BST, we.pdf
There is BinarySearchTree class. When removing a node from a BST, we.pdf
Write a program that displays an AVL tree along with its balance fac.docx
Write a program that displays an AVL tree along with its balance fac.docx
In java , I want you to implement a Data Structure known as a Doubly.pdf
In java , I want you to implement a Data Structure known as a Doubly.pdf
Need Help with this Java Assignment. Program should be done in JAVA .pdf
Need Help with this Java Assignment. Program should be done in JAVA .pdf
Write a C++ program that implements a binary search tree (BST) to man.pdf
Write a C++ program that implements a binary search tree (BST) to man.pdf
A deque (pronounced deck) is an ordered set of items from which item.pdf
A deque (pronounced deck) is an ordered set of items from which item.pdf
AvlTree.h#ifndef AVL_TREE_H#define AVL_TREE_H#include d.docx
AvlTree.h#ifndef AVL_TREE_H#define AVL_TREE_H#include d.docx
I need help with this maze gui that I wrote in java, I am trying to .pdf
I need help with this maze gui that I wrote in java, I am trying to .pdf
Given the following codepackage data1;import java.util.;p.pdf
Given the following codepackage data1;import java.util.;p.pdf
JAVA OOP project; desperately need help asap im begging.Been stuck.pdf
JAVA OOP project; desperately need help asap im begging.Been stuck.pdf
Modify this code to do an Insert function for an AVL tree, instead o.pdf
Modify this code to do an Insert function for an AVL tree, instead o.pdf
In this lab, you will be given a simple code for a min Heap, and you.pdf
In this lab, you will be given a simple code for a min Heap, and you.pdf
I have a .java program that I need to modify so that it1) reads i.pdf
I have a .java program that I need to modify so that it1) reads i.pdf
More from aniarihant
Give an explanation of the pedigree below. SolutionThe inherit.pdf
Give an explanation of the pedigree below.
Solution
The inheritance is an recessive pattern of inheritance as only few progeny from one mating are
effected. This is due to recessive inheritance where both copies of the recessive allele are
required to cause the disease phenotype.
It is Autosomal recessive pattern of inheritance. It is not X linked recessive because, the infected
male child from 2nd F1 generation male (5th child ) donot recieve X chromosome from the
infected father. Hence it is due to autosomal trasmission..
elena ents you to impact into MAt ab SolutionThree fu.pdf
elena ents you to impact into MAt ab
Solution
Three functions---> fprintf-Write data to text file.
fread-Read data from binary file
fopen-Open file or obtain information about open files.
Interface Elements--->mech_interface_crate_transfer
mech_interface_dspring_damper
mech_interface_hyd_cylinder
Keywords--->Clear-Removes variables from memory
Lookfor-Searches help entries for a keyword.
delete-Deletes a file
Path-Display search path..
Ecosystem ecologyCompare and contrast abiotic gradients in a tempe.pdf
Ecosystem ecology
Compare and contrast abiotic gradients in a temperate forest and a desert.
Solution
ANS:
Ecosystem: It explains the interactions between the population which present in the community
and the communities physical surroundings or abiotic factors.
In ecosystem all living organisms are lived in a similar area along with the nonliving materials.
Ex: In an African savanna the greater Gorongosa ecosystem that is home to this lion and
waterbuck and cattle egret. And the larger ecosystems and landscapes are embedded within
biomes.
Abiotic Factors:
Abiotic Factors are a non living factor, which mainly influences or affects an ecosystem.
Ex: sun light, temperature, Rocks, Air masses, climate and chemical environment etc...
Compare and contrast abiotic gradients in a temperate forest and a desert
Abiotic gradients in a temperate forest:
The soil is highly fertile and rich in humus. So it affects plant and animal communities that live
in them.
Winter season in temperate forest is between cold to moderate, but the summer season is warm.
Temperature affects temperate forests as well, suitable temperature is not maintained.
Sunlight is sufficient to grow plants and execute photosynthesis.
Pine trees are grown highly in Sandy-soiled forests.
Ex: Australasia and southern South America and Northern Hemisphere\'s deciduous forests.
Abiotic gradients in Desert:
In desert precipitation is very low.
Desert Soils are rich in minerals, but the organic matter present in the soil is so poor.
Top soil is so little, due to the high winds.
Temperature is too high in day time and low at night time.
High temperature in deserts results dry environment.
Ex: Sahara, Kalahari..
Compare and contrast ClearCase and ClearQuest.SolutionClearcas.pdf
Compare and contrast ClearCase and ClearQuest.
Solution
Clearcase
1.It is an ancsentary of software tools which assists software configuration management of a
source code.
2.it was an alignment deceit tool.
3.it is a version steering system.
4.it assists branching and labelling.
5.it is able to satisfy the high binary files and high reposity sizes.
Clearquest
1.it is a fault recording tool.
2.it offers the planned events.
3.it is a fault voucher application .if any trouble is identified , it was noted in an understandable
quest.so that the engineer handles that noted fault.
4.it is a error tracing system yet it was arranged to behave as a CRM tool..
Assume that you have a garden and some pea plants have solid leaves a.pdf
Assume that you have a garden and some pea plants have solid leaves and others have striped
leaves. You conduct a series of crosses ((a) through (e)l and obtain the results given in the table
Define gene symbols and give the possible genotypes of the parents of each cross.
Solution
If we consider solid S is dominant over striped s ,then SS ans Ss will be dominant and ss will be
striped.Accordingly,
a) solid * striped :
parental Solid will be Ss and parental striped will be ss.The progeny will be both solid and
striped.
b) solid * solid
Both parental genotype will be SS and SS.The progeny will be all solid.
c) striped * striped
Both parental genotype will be ss.The progeny will be all striped.
d) solid * solid
Both parental genotype will be solid Ss and in progeny both solid and striped will be formed.
e) solid * striped
Parental genotypes will be solid SS and striped ss :The progeny will be Ss and all solid..
According to NASA GISS measurements, the years 2005 and 2010 are tied.pdf
According to NASA GISS measurements, the years 2005 and 2010 are tied for the warmest
years in the instrumental record dating back over 100 years. Refer to figure 3.7 and explain what
role the sun played in setting these records. The sun played little to no role. According to the
figure, solar activity was near record lows so the atmosphere would likely have cooled if the sun
were responsible for climate change. The sun played a large role. According to the figure, solar
activity was at record highs so the atmosphere would likely have cooled if the sun were
responsible for climate change Both A and B are correct.
Solution
Answer:.
A drawer contains 6 blue and 4 white socks. Two of the socks are chos.pdf
A drawer contains 6 blue and 4 white socks. Two of the socks are chosen at random without
replacement. What is the probability that the 2 socks are the same color?
Solution
THERE ARE TOTAL 10 SOCKS
2 ARE CHOOSEN AT RANDON
PROBABILITY OF THE SOCKS BEING OF RED COLOR = 2/6= 1/3
PROBABILITY OF THE SOCKS BEING OF WHITE = 2/4 = 1/2
SO PROBABILITY OF BEING OF SAME COLOR = 1/3+1/2=0.83.
Bill is outside on a snowy day without a warm coat. His skin is al.pdf
* Bill is outside on a snowy day without a warm coat. His skin is ale and chille and he begins to
shiver. Explain how these responses are attempts of his body to maintain homeostasis of
temperature. What other responses might raise his body temperature?
* Arrange in order the events that occur in the production of a fever of 39.4oC (103oF) and
recovery from this state
A. 39.4oC temperature is reached and maintained.
B. Prostaglandins (PGs) cause the hypothalamic \"thermostat\" to be reet from normal 37.5oC
(98.6oF) to a higher temperature such as 39.4oC (103oF).
C. Vasoconstriction and shivering occur as heat is conserved in order to raise body temperature
to the 39.4oC (103oF) level ddirected by the hypothalamus.
D. Source of pyrogens is removed (for example, bacteria are killed by antibiotics), lowering
hypothalamic \"thermostat\" to 37.5oC (98.6oF).
E. \"Crisis\" occurs. Obeying hypothalamic orders, the body shifts to heat loss mechanisms such
as sweating and vasodilation, returning body temperature to normal.
F. Infectious organisms cause phagocytes to release interleukin-1, which then causes
hypothalamic release of PGs.
Solution
answer:To maintain homeostasis of temperature,in the body circulatory system is very important
to warm and maintain homeostasis temperature.it runs blood to all body parts with oxygen filled
in blood cells.the blood flows through the artery,veins and aorta to maintain supply nutrients to
all cells.it is used to get the carbondioxide out of the system.it also transports the harmones to the
targeted cells.all these functions are helpful to maintain cooling or heat of the homeostasis
temperature of the body..........
our cardiovascular system reduces blood flow to our skin surfaces and extremitiesby blood
vessel constriction.this tells the nervous system to warm up the body....bearing jackets, exposes
to hot weather and drinking alocohol can increase body temperature.........
2) C , F,B,A,D ,E,.
As a network administrator, what are some of the options you have fo.pdf
As a network administrator, what are some of the options you have for providing secure file
transfer?
Solution
as a network adminstrator he should consider all the scenarios while provding secure file transfer
1)choose the file transfer protocol which is suitable for your network,chooisng the best protocol
will give you fexible security features
2)encrypt the file before sending it through a network it will give you additional security features
to your data
3)install a firewall so that there will be no chance of accessing data from unauthorized port
number
4)make sure that the mft(managed file transfer should support all internal and extrenal security
mandates.
A concert has three pieces of music to be played before intermission.pdf
A concert has three pieces of music to be played before intermission. The time taken to play each
piece has a normal distribution. Assume that the three times are independent of one another. The
mean times are 15, 30, and 20 minutes, respectively and the standard deviations are 1, 2, and 1.5
minutes, respectively. What is the probability that this part of the concert takes at most one hour?
Are there reasons to question the independence assumption? Explain
Solution.
5 of 50 students are female. If I randomly selects 10 students witho.pdf
5 of 50 students are female. If I randomly selects 10 students without replacement, what is the
probability that one student is female?
Need to find P(x = 1), x is the number of female student, female student is the wanted outcome.
Use:
N = The population size
R = The number of wanted outcomes in the population
n = The sample size
x = The number of wanted outcomes in the sample
Solution.
Which if statement below tests if letter holds R (letter is a char .pdf
Which if statement below tests if letter holds R? (letter is a char variable)
if ( letter == \"R\")
if ( letter >= \'R\')
if ( letter == R)
if ( letter = \'R\')
if ( letter == \'R\')
What are if statements used for in programs?
Repeating commands
Storing data
Numeric calculations
Numeric casts
Making decisions
The following if statement tests the rainfall in New York’s Central Park during the months of
June, July and August.
if (low <= rain && rain <= high)
System.out.println(\"Rainfall amount is normal.\");
else
System.out.println(\"Rainfall amount is abnormal.\");
It could be replaced with:
I.
if (rain >= low)
{
if (rain <= high)
System.out.println(\"Rainfall amount is normal.\");
}
else
System.out.println(\"Rainfall amount is abnormal.\");
II.
if (rain >= low)
{
if (rain <= high)
System.out.println(\"Rainfall amount is normal.\");
else
System.out.println(\"Rainfall amount is abnormal.\");
}
else
System.out.println(\"Rainfall amount is abnormal.\");
III.
if (rain >= low)
System.out.println(\"Rainfall amount is normal.\");
else if (rain <= high)
System.out.println(\"Rainfall amount is normal.\");
else
System.out.println(\"Rainfall amount is abnormal.\");
I only
II only
III only
II or III
I, II or III
What is output by the following code?
int x = 36 % 8;
if (x >= 10)
System.out.println( 1);
else if (x >= 8)
System.out.println( 2);
else if (x >= 6)
System.out.println( 3);
else if ( x >= 4)
System.out.println( 4);
else
System.out.println( 5);
1
2
3
4
5
Consider the code:
if ( y == 0 || x * y > 10)
Which of the following is an example of short circuit evaluation?
if x * y > 10 is false it evaluates y ==0
if x * y > 10 is false it doesn\'t evaluate y ==0
if y == 0 is false it doesn\'t evaluate x * y > 10
if y == 0 is true it doesn\'t evaluate x * y > 10
if y == 0 is false it evaluates x * y > 10
The following truth table matches which boolean condition?
A && ( A || B)
A || ( !A && !B)
A && ( A && B)
!A && ( A || !B)
A || ( A || B)
Consider the code:
if (a < b && c != d)
Which of the following is an example of short circuit evaluation?
if a < b is true it doesn\'t evaluate c != d
if a < b is false it doesn\'t evaluate c != d
if c != d is false it evaluates a < b
if c != d is true it doesn\'t evaluate a < b
if a < b is true it evaluates c != d
! ( x < y && w == z) is the same as which boolean expression?
x <= y && w == z
x >= y || w != z
x <= y || w != z
x <= y && w != z
x < y && w != z
Assume that x and y are boolean variables and have been properly initialized.
!(x || y) || (x || y)
The result of evaluating the expression above is best described as:
always true
always false
true only when x is true and y is true
true only when x and y have the same value
true only when x and y have different values
What is output to the screen by the following code?
int c = 2;
while (c < 6)
{
System.out.print((int)Math.pow (-1, c)+\" \");
c++;
}
-1 1 -1 1 -1 1 -1
1 -1 1 -1
-1 1 -1 1 -1 1
1 1 1 1 1 1
-1 -1 -1 -1 -1 -1
How many times will the following loop repeat?.
Write a assembly program for IA-32 to count up the number of even num.pdf
Write a assembly program for IA-32 to count up the number of even numbers in the memory
locations from 00000 to 000FF.
Solution
DATA SEGMENT
A DW 1,2,3,4,5,6,7,8,9,10,11,12,13,14,15 (00000 to 000FF)
DATA ENDS
CODE SEGMENT
ASSUME DS:DATA,CS:CODE
START:
MOV AX,DATA
MOV DS,AX
LEA SI,A
MOV DX,0000 (starting from the address 0000)
MOV BL,02
MOV CL,10
Lone:MOV AX,WORD PTR[SI]
DIV BL
CMP AH,00
JNZ Ltwo
INC DH
JMP Lthree
Ltwo:INC DL
Lthree:
ADD SI,2
DEC CL
CMP CL,00
JNZ Lone
MOV AH,4CH
INT 21H
CODE ENDS
END START.
Which of the following is a weighted average of some elements having .pdf
Which of the following is a weighted average of some elements having two or more different
atomic masses, which are called isotopes atomic number, atomic weight The principal bond of
common table salt NaCl(Sodium Chloride) is ionic covalent metallic, van der Waals.
Solution
5) Atomic Number
6) Ionic
7)Covalent
8)Substitutional Atom
9)Increases
10)Atomic packing Factor.
Which of the following statements isare true about the graph of the .pdf
Which of the following statements is/are true about the graph of the function f(x) = log_2 x? I.
The range of f is (-infinity,infinity) II. The x-intercept of the function is (1,0) III. The graph of f
is increasing throughout its domain A. I and II only B. I and III only C. II and III only D. I, II,
and 111 are all true E. I, II, and III are all false
Solution
y = log2(x)
The function exists for x>0 and it exist for all values of y
So, range of f (- inf , inf)
X intercept : y=0 ; log2(x) =0
x= 2^0 = 1
(1, 0)
Graph is increasing for all values of x
So, Option D all options are true.
35. 2H H2 has E0 of -0.42 and NoihN2 has Eo of +0.74. Which is the .pdf
35. 2H /H2 has E0 of -0.42 and Noi/hN2 has Eo of +0.74. Which is the electron acceptor? a. 2H
b. H2 c. NO3 d. N2 36. Which of the following rotates? a. Type IV pili b. Flagella c. ATPase d.
ABC transporter
Solution
The answer for 36 is C. ATPase complex.
WE have to answer one question at a time, so post the other one as a seperate question, then I can
help you..
6. A more in depth look at how to recognize “true adaptation”. Skim .pdf
6. A more in depth look at how to recognize “true adaptation”. Skim Chapter 10,
a. How does he suggest adaptation should be studied? Are his methods in agreement with
defining an adaptation as outlined on the Evo 101 website.
b. Describe the examples Ridley uses to illustrate why adaptations are not perfect.
c. How we can determine whether features are due to some constraint (such as a trade off) or an
adaptation?
d. In this chapter Ridley introduces the idea of a morphospace and what it can tell us about
adaptation. Be able to summarize his ideas. e. Ridley elsewhere in his evolutionary text states,
“Pluralism is appropriate for the study of evolution, not of adaptation. “
Solution
Adaptation is the evolutionary process whereby an organism becomes better able to live in its
habitat or habitats. Adaptation can be studied that the laden with the compilation over the use of
best methadological approach. Instead of prioritizing the method, we should know which are best
suited for specific and certain adaptations. exploitative adaptation are the best adaptations that
utilizes all of the methods of adaptations. So that it become relevant to all the adaptations both
theoretical and practical by discussing avant-grade film. Ridley uses teh examples are the of the
HIV virus portrays an example of adaptation. he uses this example to contribute to evolution by
portraying how virus can adapt and mutate. he studies that the HIV ties in perfectly with natural
selection and evolution. Nearly all drugs currently used to treat this disease stop working after a
period of usage and as a result cause AIDS. Ridley also gave the example on
fecundity. fecundity is defined as, “the ability to have children, usually lots of them with ease”.
Fecundity is also when females produce more offspring than can survive. A population or
species might have the potential to evolve, but other factors like the movement of genes among
populations and/or trait interactions make evolutionary change difficult despite ongoing
selection. Both fundamental limits and other forms of constraints can prevent populations and
species from adapting to new environments. Constraints restrict species for living under such
kind of environmental conditions. The limits andother contrains also drive biodiversity there
might be a few common species adapted to a wide range of conditions, rather than a diversity of
species, the majority restricted to a narrow range of ecological conditions. Explanations for
evolutionary constraints can be divided into two categories: those that reflect the nature of
genetic variation required for evolution and adaptation, and those that have their origin in the
ecological processes to which populations and species are exposed.
What does antibody titer meanSolution It is a method for t.pdf
What does \"antibody titer\" mean?
Solution
It is a method for the measurement of the antigens and antibodies participating in primary,
secondary and tertiary reactions. Measurement may be in terms of mass or units or titre. The
antibody titre of a serum is the highest dilution of serum that show an observable reaction with
the antigen in the perticular test. The titre of serum influenced by the nature and quantity of the
antigen and the type and conditions of test..
What role do regular expressions typically play in language compiler.pdf
What role do regular expressions typically play in language compilers and interpreters?
Solution
Answer:
Regular expressions are mathematical symbolism which represent the set of strings of a
particular language in a compiler design. First of all comes the SYMBOLS : Symbols can be any
character a , b, ,c , d ...z , any numeric 1 , 2 , 3 , 4... or any special symbol * , + , . In regular
language \"phi\" represents empty set. + between the symbols represents the closure properties
like if we have (a+b) , it means strings of a\'s and b\'s . If (a+b)* , means in the star(*) , we can
assign its value ranging from 1 to any number. Thus we can say that regular expressions plays a
very important role as it gives the symbols and strings of language a pattern..
What is the role of ribosomes in an operon What is the role of.pdf
What is the role of ribosomes in an operon?
What is the role of ribosomes in an operon?
Solution
Prokaryotes contain tRNA, mRNA, 16S rRNA, 23S r RNA, and 5S rRNA are meticulously
organized as co-transcribed operon. Operon system of prokaryotes gene expression has alpha
operon- ribosome binding site where ribosome is going to bind to regulate \"operon genes\" to
synthesize proteins. Thereofre, for example, ribosomes are going to promote tranlation to
synthesize proteins after gene expression. There is an attenuator meticulously generate in mRNA
sequence during operon expression is going to be bypassed by the ribosome so that operons are
regulatedby ribosomes.
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- 1. Im suppose to make a preOrder transversal method so that it performs a preOrder transversal of the tree. For each node it traverses, it should call sb.append() to add a "[" the results of traversing the subtree rooted at that node, and then a "]". You should add a space before the results of the left and right child traversals, but only if the node exists. This is what i have so far public class BinarySearchTree> extends AbstractSet { protected Entry root; protected int size; /** * Initializes this BinarySearchTree object to be empty, to contain only * elements of type E, to be ordered by the Comparable interface, and to * contain no duplicate elements. */ public BinarySearchTree() { root = null; size = 0; } /** * Initializes this BinarySearchTree object to contain a shallow copy of a * specified BinarySearchTree object. The worstTime(n) is O(n), where n is the * number of elements in the specified BinarySearchTree object. * * @param otherTree - the specified BinarySearchTree object that this * BinarySearchTree object will be assigned a shallow copy of. */ public BinarySearchTree(BinarySearchTree otherTree) { root = copy(otherTree.root, null); size = otherTree.size; } /* Method to complete is here! */ public void preOrder(Entry ent, StringBuilder sb) { // TODO: Complete me! if(ent == null){ return;
- 2. } if(ent.left != null){ sb.append("["); preOrder(ent.left, sb); } if(ent.right != null){ sb.append("["); preOrder(ent.right, sb); } sb.append("]"); } But my preOrder method gives me these errors The following hint(s) may help you locate some ways in which your solution may be improved: --Pre-order traversal should process the root, then its left child (and exclude the non-existent right child)! Expected [t [o]], but was []] --Should just include the root's element when the tree only has 1 element! Expected [z], but was ] --Pre-order traversal should process the root, ignore the non-existent left child, then process the right child! Expected [w [z]]. but was []] --Pre-order traversal should process the root, the left child, and then the right child! Expected [9 [0] [d]], but was [][]] --Pre-order traversal should process the root, and then the left child. In processing the left child, it should process that node first, and then process its children! Expected [d [9 [d]]], but was [[]]] --Pre-order traversal should process the root, and then the right child. In processing the right child, it should process that node first, and then process its children! Expected [v [w [z]]], but was [[]]] --Pre-order traversal should process a node and then recursively process each of its children! Expected [v [d [9 [0] [d]] [j [g] [t [o]]]] [w [z]]], but was [[[][]][[][[]]]][[]]] --Pre-order traversal should process a node and then recursively process each of its children! Expected [v [d [9 [0] [d]] [j [g] [t]]]], but was [[[][]][[][]]]] Solution
- 3. import java.util.Stack; /* Class containing left and right child of current node and key value*/ class Node { int data; Node left, right; public Node(int item) { data = item; left = right = null; } } /* Class to print the inorder traversal */ class BinaryTree { Node root; void preorder() { if (root == null) { return; } //keep the nodes in the path that are waiting to be visited Stack stack = new Stack(); Node node = root; //first node to be visited will be the left one while (node != null) { stack.push(node); node = node.left; } // traverse the tree
- 4. while (stack.size() > 0) { // visit the top node node = stack.pop(); System.out.print(node.data + " "); if (node.right != null) { node = node.right; // the next node to be visited is the leftmost while (node != null) { stack.push(node); node = node.left; } } } } public static void main(String args[]) { /* creating a binary tree and entering the nodes */ BinaryTree tree = new BinaryTree(); tree.root = new Node(1); tree.root.left = new Node(2); tree.root.right = new Node(3); tree.root.left.left = new Node(4); tree.root.left.right = new Node(5); tree.preorder(); } }