In inline heapsort we use only single array throughout the program t.pdf
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In inline heapsort we use only single array throughout the program to decrease the running time, where as, in simple heap sort we use external data structure memory usage of more than O(1) at a time. Please consider the below inline heap program which prints out the running time, which is lesser to the the simple heap sort. //program class heapsort { private final int array[]; // The array being sorted heapsort(int array[], int N) { this.array = array; int Last = N-1; // its children are at elements 2*I+1 and 2*I+2. // phase 1: form heap // Construct heap bottom-up, starting with small trees just above leaves // and coalescing into larger trees near the root. for( int Top = Last/2; Top >= 0; Top-- ) { adjust(Top, Last); } // phase 2: use heap to sort // Move top element (largest) out of heap, swapping with last element // and changing the heap boundary, until only one element remains. while( Last > 0 ) { swap(0, Last); adjust(0, --Last); } } /** * adjust(Top, Last) adjusts the tree between Top and Last **/ void adjust(int Top, int Last) { int TopVal = array[Top]; // Set aside top of heap int Parent, Child; for( Parent = Top; ; Parent = Child ) // Iterate down through tree { Child = 2*Parent+1; // Child means left child if( Child > Last ) break; // Left child non-existent if( Child+1 <= Last // Right child exists && array[Child] < array[Child+1] ) // and right child is larger Child++; // Child is the larger child if( TopVal >= array[Child] ) break; // Location for TopVal found array[Parent] = array[Child]; // Move larger child up in tree } array[Parent] = TopVal; // Install TopVal in place } //swapping void swap(int i, int j) { int temp = array[i]; array[i] = array[j]; array[j] = temp; } public static void main(String[] args) { int N = 0; // number of elements in array int array[] = {43,65,12,89,11,5,1,88,123,999,233,18,553,332,144,98,452,7663,111,23,654,8735,998}; //initialising array N=array.length; System.err.println(\"Sorting started\"); long startTime = System.nanoTime(); new heapsort(array, N); // sorting long endTime = System.nanoTime(); long totalTime = endTime - startTime; for( int i = 0; i < N; i++ ) { System.out.print(array[i] + \" \"); } System.out.println(); System.out.println(\"Time in nanos\"+totalTime); } } //runn the above program for inline heap sort. Solution In inline heapsort we use only single array throughout the program to decrease the running time, where as, in simple heap sort we use external data structure memory usage of more than O(1) at a time. Please consider the below inline heap program which prints out the running time, which is lesser to the the simple heap sort. //program class heapsort { private final int array[]; // The array being sorted heapsort(int array[], int N) { this.array = array; int Last = N-1; // its children are at elements 2*I+1 and 2*I+2. // phase 1: form heap // Construct heap bottom-up, starting with small trees just above leaves // and coalescing into larger trees near the root. for( int T.
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![In inline heapsort we use only single array throughout the program to decrease the running time,
where as, in simple heap sort we use external data structure memory usage of more than O(1) at
a time.
Please consider the below inline heap program which prints out the running time, which is lesser
to the the simple heap sort.
//program
class heapsort
{
private final int array[]; // The array being sorted
heapsort(int array[], int N)
{
this.array = array;
int Last = N-1;
// its children are at elements 2*I+1 and 2*I+2.
// phase 1: form heap
// Construct heap bottom-up, starting with small trees just above leaves
// and coalescing into larger trees near the root.
for( int Top = Last/2; Top >= 0; Top-- )
{
adjust(Top, Last);
}
// phase 2: use heap to sort
// Move top element (largest) out of heap, swapping with last element
// and changing the heap boundary, until only one element remains.
while( Last > 0 )
{
swap(0, Last);
adjust(0, --Last);
}
}
/**
* adjust(Top, Last) adjusts the tree between Top and Last](https://image.slidesharecdn.com/ininlineheapsortweuseonlysinglearraythroughouttheprogramt-230412115754-0a24d41d/85/In-inline-heapsort-we-use-only-single-array-throughout-the-program-t-pdf-1-320.jpg)
![**/
void adjust(int Top, int Last)
{
int TopVal = array[Top]; // Set aside top of heap
int Parent, Child;
for( Parent = Top; ; Parent = Child ) // Iterate down through tree
{
Child = 2*Parent+1; // Child means left child
if( Child > Last )
break; // Left child non-existent
if( Child+1 <= Last // Right child exists
&& array[Child] < array[Child+1] ) // and right child is larger
Child++; // Child is the larger child
if( TopVal >= array[Child] )
break; // Location for TopVal found
array[Parent] = array[Child]; // Move larger child up in tree
}
array[Parent] = TopVal; // Install TopVal in place
}
//swapping
void swap(int i, int j)
{
int temp = array[i];
array[i] = array[j];
array[j] = temp;
}
public static void main(String[] args)
{
int N = 0; // number of elements in array
int array[] =
{43,65,12,89,11,5,1,88,123,999,233,18,553,332,144,98,452,7663,111,23,654,8735,998};
//initialising array
N=array.length;
System.err.println("Sorting started");](data:image/gif;base64,R0lGODlhAQABAIAAAAAAAP///yH5BAEAAAAALAAAAAABAAEAAAIBRAA7)
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identically. Specify, design and implement a class for a double-ended queue. Include operations
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operations for adding and deleting items. Use doubly linked list for your implementation. Call
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//package com.java2novice.ds.queue;
public class DequeDblLinkedListImpl {
private LinkedNode front;
private LinkedNode rear;
public void insertFront(T item){
//add element at the beginning of the queue
System.out.println(\"adding at front: \"+item);
LinkedNode nd = new LinkedNode();
nd.setValue(item);
nd.setNext(front);
if(front != null) front.setPrev(nd);
if(front == null) rear = nd;
front = nd;
}
public void insertRear(T item){
//add element at the end of the queue
System.out.println(\"adding at rear: \"+item);
LinkedNode nd = new LinkedNode();
nd.setValue(item);
nd.setPrev(rear);
if(rear != null) rear.setNext(nd);
if(rear == null) front = nd;
rear = nd;
}
public void removeFront(){
if(front == null){
System.out.println(\"Deque underflow!! unable to remove.\");
return;
}
//remove an item from the beginning of the queue
LinkedNode tmpFront = front.getNext();
if(tmpFront != null) tmpFront.setPrev(null);
if(tmpFront == null) rear = null;
System.out.println(\"removed from front: \"+front.getValue());
front = tmpFront;
}
public int size(){
if(front == null && rear ==null){
return 0;
}
//remove an item from the beginning of the queue
LinkedNode tmpFront = front;
LinkedNode tmpRear = rear;
int i=0;
while(tmpFront != rear)
{
i++;
tmpFront=tmpFront.getNext();
}
i++;
return i;
}
public void removeRear(){
if(rear == null){
System.out.println(\"Deque underflow!! unable to remove.\");
return;
}
//remove an item from the beginning of the queue
LinkedNode tmpRear = rear.getPrev();
if(tmpRear != null) tmpRear.setNext(null);
if(tmpRear == null) front = null;
System.out.println(\"removed from rear: \"+rear.getValue());
rear = tmpRear;
}
public static void main(String a[]){
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deque.insertFront(67);
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deque.insertFront(29);
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deque.removeFront();
deque.removeFront();
deque.insertRear(43);
deque.insertRear(83);
deque.insertRear(84);
deque.insertRear(546);
deque.insertRear(356);
deque.removeRear();
deque.removeRear();
deque.removeRear();
deque.removeRear();
deque.removeFront();
deque.removeFront();
deque.removeFront();
}
}
class LinkedNode{
private LinkedNode prev;
private LinkedNode next;
private T value;
public LinkedNode getPrev() {
return prev;
}
public void setPrev(LinkedNode prev) {
this.prev = prev;
}
public LinkedNode getNext() {
return next;
}
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Solution
//package com.java2novice.ds.queue;
public class DequeDblLinkedListImpl {
private LinkedNode front;
private LinkedNode rear;
public void insertFront(T item){
//add element at the beginning of the queue
System.out.println(\"adding at front: \"+item);
LinkedNode nd = new LinkedNode();
nd.setValue(item);
nd.setNext(front);
if(front != null) front.setPrev(nd);
if(front == null) rear = nd;
front = nd;
}
public void insertRear(T item){
//add element at the end of the queue
System.out.println(\"adding at rear: \"+item);
LinkedNode nd = new LinkedNode();
nd.setValue(item);
nd.setPrev(rear);
if(rear != null) rear.setNext(nd);
if(rear == null) front = nd;
rear = nd;
}
public void removeFront(){
if(front == null){
System.out.println(\"Deque underflow!! unable to remove.\");
return;
}
//remove an item from the beginning of the queue
LinkedNode tmpFront = front.getNext();
if(tmpFront != null) tmpFront.setPrev(null);
if(tmpFront == null) rear = null;
System.out.println(\"removed from front: \"+front.getValue());
front = tmpFront;
}
public int size(){
if(front == null && rear ==null){
return 0;
}
//remove an item from the beginning of the queue
LinkedNode tmpFront = front;
LinkedNode tmpRear = rear;
int i=0;
while(tmpFront != rear)
{
i++;
tmpFront=tmpFront.getNext();
}
i++;
return i;
}
public void removeRear(){
if(rear == null){
System.out.println(\"Deque underflow!! unable to remove.\");
return;
}
//remove an item from the beginning of the queue
LinkedNode tmpRear = rear.getPrev();
if(tmpRear != null) tmpRear.setNext(null);
if(tmpRear == null) front = null;
System.out.println(\"removed from rear: \"+rear.getValue());
rear = tmpRear;
}
public static void main(String a[]){
DequeDblLinkedListImpl deque = new DequeDblLinkedListImpl();
deque.insertFront(34);
deque.insertFront(67);
System.out.println(\"Size of the queue : \"+deque.size());
deque.insertFront(29);
deque.insertFront(765);
deque.removeFront();
deque.removeFront();
deque.removeFront();
deque.insertRear(43);
deque.insertRear(83);
deque.insertRear(84);
deque.insertRear(546);
deque.insertRear(356);
deque.removeRear();
deque.removeRear();
deque.removeRear();
deque.removeRear();
deque.removeFront();
deque.removeFront();
deque.removeFront();
}
}
class LinkedNode{
private LinkedNode prev;
private LinkedNode next;
private T value;
public LinkedNode getPrev() {
return prev;
}
public void setPrev(LinkedNode prev) {
this.prev = prev;
}
public LinkedNode getNext() {
return next;
}
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the ages of 22-35 you know that:
• the probability a man is a ballet dancer is 0.245.
• the probability a woman is a musical performer is 0.365
• the probability a man is a musical performer, given that he\'s a ballet dancer is 0.250.
• the probability a woman is an actress is 0.550.
• the probability a woman is an actress, given that she\'s a musical performer is .785.
Compute the following probabilities:
A. P (man is a ballet dancer and a musical performer)
B. P (woman is an actress and a musical performer)
C. P (woman is an actress or a musical performer)
Solution
=0.245*(1-0.365)=0.155
b. P (woman is an actress and a musical performer)
=0.365*0.55=0.2
c. P (woman is an actress or a musical performer)
=P(woman is an actress)+P(musical performer)-P(actress or performer)
=0.55+0.365-0.785*0.365=0.628.
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Please make it to clear 12-95. The basketball passed through the hoop even though it barely
cleared the hands of the player B who attempted to block it. Neglecting the size of the ball,
determine the magnitude vn of its initial velocity and the height h of the ball when it passes over
player B 10 ft 5 ft Prob. 12-95
Solution
Consider the motion along horizontal direction
Vox = initial velocity in horizontal direction = Va Cos30
X = horizontal distance = 25 + 3 = 28 ft
t = time taken
t = X / Vox = 28/(Va Cos30) eq-1
consider the motion in vertical direction
Y = vertical displacement = 10 - 7 = 3 ft
a = acceleration = - 32.2
Voy = initial velocity = Va Sin30
t = time taken
using the equation
Y = Voy t + (0.5) a t2
3 = (Va Sin30) (28/(Va Cos30)) + (0.5) (- 32.2) (28/(Va Cos30))2
Va = 35.8 m/s
t\' = time taken to reach B = 25 / (Va Cos30) = 25 / (35.8 Cos30) = 0.81 sec
h = 7 + Voy t\' + (0.5) a t\'2
h = 7 + (35.8 Sin30) (0.81) + (0.5) (- 32.2) (0.81)2
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4. The size of instructions can be fixed or variable. What are advantage and disadvantage of
fixed compared to variable? point) eft 2 20 0 C-4131-28 Add suction 131-26 nvaction 125-2 PC
egister daiess mancion 120-16 date 1 regnter 2 nstructien 31-00 eg stcrs Read U ALLu dela tlata
dvts 02 rstend In the above figure (single-cycle implementation), current instruction is SW S1,
4(S2) (SW: Store Word; current addrss for this instruction is 100 in decimal; S1 and $2 are
initially 8). For redundancy, you should write \"X\" instead of 0 or . You should write the reason
for cach question as well. 5. What is the value of RegDst? (1 point) 6. What is the value of
RegWrite? (I point) 7. What is the value after sign-extension? 1 point) 8. What is the value of
MemRead? (1 point) 9. What is the value of MemtoReg? (1 point) In the above figure, current
instruction is ADDI $1. S2, #-1(which is minus l) (ADDI is ADD Immediate; current address for
this instruction s 100 in decimal and S2 are initially1). For redundancy, you should write “X\',
instead of 0 or 1 . You should write the reason for each question as well. 10. What is the value
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value after sign extension? Write the value in 32-bit binary. (I point) 12. What is the value of
Reg Write? (1 point) 13. What is the value of MemtoReg? (1 point) 2/4
Solution
import java.util.AbstractSet; import java.util.Arrays; import java.util.Iterator; import
java.util.NoSuchElementException; /** * Implementation of an abstract set using an array-
based binary tree. This is * used to help teach binary tree\'s and will have more details explained
in * future lectures. * * @author William J. Collins * @author Matthew Hertz * @param
Data type (which must be Comparable) of the elements in this tree. */ public class
ArrayBinaryTree> extends AbstractSet { /** Entry in the data store where the root node can be
found. */ private static final int ROOT = 0; /** Array used to store the nodes which consist of
this binary tree. */ protected Node[] store; /** Number of elements within the tree. */
protected int size; /** * Initializes this ArrayBinaryTree object to be empty. This creates the
array * in which items will be stored. */ @SuppressWarnings(\"unchecked\") public
ArrayBinaryTree() { store = new Node[63]; size = 0; } /** * Initializes this
ArrayBinaryTree object to contain a shallow copy of a * specified ArrayBinaryTree object.
The worstTime(n) is O(n), where n is the * number of elements in the specified
ArrayBinaryTree object. * * @param otherTree The tree which will be copied to create our
new tree, */ @SuppressWarnings(\"unchecked\") public ArrayBinaryTree(ArrayBinaryTree
otherTree) { store = (Node[]) Arrays.copyOf(otherTree.store, otherTree.store.length); size =
otherTree.size; } public int countLeaves() { } /** * Returns the size of this
ArrayBinaryTree object. * * @return the size of this ArrayBinaryTree object. */
@Overrid.
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public boolean isFull(); //Method to determine whether the list is full.
public int listSize(); //Method to return the number of elements in the list.
public int maxListSize(); //Method to return the maximum size of the list.
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public void clearList(); //Method to remove all the elements from the list.
public int search(int searchItem); //Method to determine whether searchItem is in the list.
public void remove(int removeItem); //Method to remove an item from the list.
}
//Class: ArrayListClass implements
//Interface: ArrayListADT
public abstract class ArrayListClass implements ArrayListADT {
protected int length; //to store the length of the list
protected int maxSize; //to store the maximum size of the list
protected int[] list; //array to hold the list elements
//Default constructor
public ArrayListClass() {
maxSize = 100;
length = 0;
list = new int[maxSize];
}
//Alternate Constructor
public ArrayListClass(int size) {
if(size <= 0) {
System.err.println(\"The array size must be positive. Creating an array of size 100.\");
maxSize = 100;
}
else
maxSize = size;
length = 0;
list = new int[maxSize];
}
public boolean isEmpty() {
return (length == 0);
}
public boolean isFull() {
return (length == maxSize);
}
public int listSize() {
return length;
}
public int maxListSize() {
return maxSize;
}
public void print() {
for (int i = 0; i < length; i++)
System.out.print(list[i] + \" \");
System.out.println();
}
public boolean isItemAtEqual(int location, int item) {
if (location < 0 || location >= length) {
System.err.println(\"The location of the item to be compared is out of range.\");
return false;
}
return list[location]== item;
}
public void clearList() {
for (int i = 0; i < length; i++)
list[i] = 0;
length = 0;
System.gc(); //invoke the Java garbage collector
}
public void removeAt(int location) {
if (location < 0 || location >= length)
System.err.println(\"The location of the item to be removed is out of range.\");
else {
for(int i = location; i < length - 1; i++)
list[i] .
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ReversePoem.java ---------------------------------- public cl.pdf
ReversePoem.java :-
---------------------------------
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/*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
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sb.append(set1);
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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.
Describe a data structure that supports both removeMin() and rem.pdf
Describe a data structure that supports both removeMin() and removeMax() with O(log(n))
complexity. You need to show the implementation of insert(), removeMin(), and removeMax(),
and prove their complexity. Hint: one can build on top of the heap.
/*IF YOU WRITE CODE, PLEASE PROVIDE COMMENTS*/
I need to understand
Describe a data structure that supports both removeMin() and removeMax() with O(log(n))
complexity. You need to show the implementation of insert(), removeMin(), and removeMax(),
and prove their complexity. Hint: one can build on top of the heap.
/*IF YOU WRITE CODE, PLEASE PROVIDE COMMENTS*/
I need to understand
Describe a data structure that supports both removeMin() and removeMax() with O(log(n))
complexity. You need to show the implementation of insert(), removeMin(), and removeMax(),
and prove their complexity. Hint: one can build on top of the heap.
/*IF YOU WRITE CODE, PLEASE PROVIDE COMMENTS*/
I need to understand
Solution
The idea here is to use the concept of two binary heaps along with the concept of list(doubly
linked list).
The doubly linked list contains all input items and indexes of corresponding min and max heap
nodes. The nodes of min and max heaps store addresses of nodes of doubly linked list. The root
node of min heap stores the address of minimum item in doubly linked list. Similarly, root of
max heap stores address of maximum item in doubly linked list. Following are the details of
operations.
1) findMax(): We get the address of maximum value node from root of Max Heap. So this is a
O(1) operation.
1) findMin(): We get the address of minimum value node from root of Min Heap. So this is a
O(1) operation.
3) removeMin(): We get the address of minimum value node from root of Min Heap. We use this
address to find the node in doubly linked list. From the doubly linked list, we get node of Max
Heap. We delete node from all three. We can delete a node from doubly linked list in O(1) time.
delete() operations for max and min heaps take O(Logn) time.
4) removeMax(): is similar to deleteMin()
5) Insert() We always insert at the beginning of linked list in O(1) time. Inserting the address in
Max and Min Heaps take O(Logn) time. So overall complexity is O(Logn)
#include
#include
#include
// A node of doubly linked list
struct LNode
{
int data;
int minHeapIndex;
int maxHeapIndex;
struct LNode *next, *prev;
};
// Structure for a doubly linked list
struct List
{
struct LNode *head;
};
// Structure for min heap
struct MinHeap
{
int size;
int capacity;
struct LNode* *array;
};
// Structure for max heap
struct MaxHeap
{
int size;
int capacity;
struct LNode* *array;
};
// The required data structure
struct CDS
{
struct MinHeap* minHeap;
struct MaxHeap* maxHeap;
struct List* list;
};
// A utility function to create a new List node
struct LNode* newLNode(int data)
{
struct LNode* node =
(struct LNode*) malloc(sizeof(struct LNode));
node->minHeapIndex = node->maxHeapIndex = -1;
node->data = data;
node->prev = node->next = NULL;
retu.
Write a program that accepts an arithmetic expression of unsigned in.pdf
Write a program that accepts an arithmetic expression of unsigned integers in postfix notation
and builds the arithmetic expression tree that represents that expression. From that tree, the
corresponding fully parenthesized infix expression should be displayed and a file should be
generated that contains the three address format instructions. This topic is discussed in the week
4 reading in module 2, section II-B. The main class should create the GUI shown below:
Pressing the Construct Tree button should cause the tree to be constructed and using that tree, the
corresponding infix expression should be displayed and the three address instruction file should
be generated. The postfix expression input should not be required to have spaces between every
token. Note in the above example that 9+- are not separated by spaces. The above example
should produce the following output file containing the three address instructions:
Add R0 5 9
Sub R1 3
R0 Mul R2 2 3
Div R3 R1 R2
Inheritance should be used to define the arithmetic expression tree. At a minimum, it should
involve three classes: an abstract class for the tree nodes and two derived classes, one for
operand nodes and another for operator nodes. Other classes should be included as needed to
accomplish good object-oriented design. All instance data must be declared as private.
You may assume that the expression is syntactically correct with regard to the order of operators
and operands, but you should check for invalid tokens, such as characters that are not valid
operators or operands such as 2a, which are not valid integers. If an invalid token is detected a
RuntimeException should be thrown and caught by the main class and an appropriate error
message should be displayed. Three Adddress Generator Enter Postfix Expression 359 2 3
Construct Tree Infix Expression ((3-(5 9))/ (2* 3
Solution
#include
#include
#include
#include
#include
struct TREE //Structure to represent a tree
{
char data;
struct TREE *right,*left,*root;
};
typedef TREE tree; /* Stack Using Linked List */
struct Stack //Structure to represent Stack
{
struct TREE *data;
struct Stack *next,*head,*top; //Pointers to next node,head and top
};
typedef struct Stack node;
node *Nw; //Global Variable to represent node
void initialize(node *&n)
{
n = new node;
n -> next = n -> head = n -> top = NULL;
}
void create(node *n)
{
Nw = new node; //Create a new node
Nw -> next = NULL; //Initialize next pointer field
if(n -> head == NULL) //if first node
{
n -> head = Nw; //Initialize head
n -> top = Nw; //Update top
}
}
void push(node *n,tree *ITEM)
{
node *temp = n -> head;
if(n -> head == NULL) //if First Item is Pushed to Stack
{
create(n); //Create a Node
n -> head -> data = ITEM; //Insert Item to head of List
n -> head = Nw;
return; //Exit from Function
}
create(n); //Create a new Node
Nw -> data = ITEM; //Insert Item
while(temp -> next != NULL)
temp = temp -> next; //Go to Last Node
temp -> next = Nw; //Point New node
n -> top = Nw; //Upda.
USING JAVAThere are at least two types of nearly sorted array.pdf
USING JAVA:
There are at least two types of \"nearly sorted\" arrays: k-exchanges and k-distance, where k is an
integer. An array is called a k-exchanges array if the array becomes sorted by performing at most
k exchanges. For example, sorting.java uses 100-exchanges for nearly sorted arrays. An array is
called a k-distance array if, for each element, the distance from its initial position to its position
in the sorted array is no more than k. A k-distance array can be created from a sorted array by
randomly shuffling each subarray of size k. www.sorting-algorithms.com uses 1-distance for
nearly sorted arrays.compare the performances of quicksort (choosing the best quicksort you
have), heapsort, mergesort, natural mergesort, and insertion sort on k-exchanges and k-distances
arrays for various k values. Create a method called task4, which takes k = 10, 20, 30, ..., until
quicksort becomes a clear winner, create 100 k-distance arrays of size 10,000,000, run in turn the
chosen sorting methods on each array, and collect running times. Then for k = 100, 200, 300, ...,
until quicksort becomes a clear winner, create 100 k-exchange arrays of size 10,000,000, run in
turn the chosen sorting methods on each array, and collect running times.
I have the code for all the algorithms done correctly, so just call them by quicksort(), heapsort(),
mergesort(), naturalmergesort(), insertionsort(). I can figure out how to deal with differing sizes
but what I don\'t understand is how to create k-distance and k-exchanges nearly sorted arrays.
Solution
void insertionSort(int A[], int size)
{
int i, key, j;
for (i = 1; i < size; i++)
{
key = A[i];
j = i-1;
/* Move elements of A[0..i-1], that are greater than key, to one
position ahead of their current position.
This loop will run at most k times */
while (j >= 0 && A[j] > key)
{
A[j+1] = A[j];
j = j-1;
}
A[j+1] = key;
}
}
he inner loop will run at most k times. To move every element to its correct place, at most k
elements need to be moved. So overall complexity will be O(nk)
We can sort such arrays more efficiently with the help of Heap data structure. Following is the
detailed process that uses Heap.
1) Create a Min Heap of size k+1 with first k+1 elements. This will take O(k) time (See this
GFact)
2) One by one remove min element from heap, put it in result array, and add a new element to
heap from remaining elements.
Removing an element and adding a new element to min heap will take Logk time. So overall
complexity will be O(k) + O((n-k)*logK)
#include
using namespace std;
// Prototype of a utility function to swap two integers
void swap(int *x, int *y);
// A class for Min Heap
class MinHeap
{
int *harr; // pointer to array of elements in heap
int heap_size; // size of min heap
public:
// Constructor
MinHeap(int a[], int size);
// to heapify a subtree with root at given index
void MinHeapify(int );
// to get index of left child of node at index i
int left(int i) { return (2*i + 1); }
// to get index of right chil.
Heap Tree.pdf
A heap is a special tree-based data structure where the tree is a complete binary tree. There are two types of heaps: a max-heap where the root node has the greatest key, and a min-heap where the root node has the minimum key. Operations on heaps include insertion and deletion of nodes in O(log N) time and peeking the max/min element. Heapify is the process of building the heap from an array by comparing nodes to ensure the heap property is maintained.
please tell me what lines do i alter to make this stack a queue. tel.pdf
please tell me what lines do i alter to make this stack a queue. tell me what to change it to
#include
using namespace std;
const int MAXSIZE = 100; // Maximum size of the stack
class Stack {
private:
int top; // Index of the top element in the stack
int data[MAXSIZE]; // Array to hold the stack elements
public:
Stack () {
top = -1; // Initialize the top index to -1 to indicate an empty stack
}
bool push (int val) {
if (top == MAXSIZE - 1) { // Stack is full
return false;
}
top++; // Increment the top index
data[top] = val; // Add the new element to the top of the stack
return true;
}
bool pop() {
if (top == -1) { // Stack is empty
return false;
}
top--; // Decrement the top index to remove the top element
return true;
}
bool peek(int& val) {
if (top == -1) { // Stack is empty
return false;
}
val = data[top]; // Return the top element without removing it.
return true;
}
void display(ostream& out) {
if (top == -1) { // Stack is empty
out << "Empty stack" << endl;
return;
}
for (int i = top; i >= 0; i--) { // Loop through the stack elements from top to bottom
out << data[i] << "";
}
out << endl;
}
};
int main() {
Stack S;
int tmp;
S.push (5);
S. peek (tmp);
cout << tmp << endl;
S.push (33);
S.push (1);
S.push (7);
S.peek (tmp);
cout << tmp << endl;
S.push (33) ;
S.push (12);
S.display ( cout );
S.pop ( );
S.display ( cout );
S.push (14);
S.display ( cout );
S.pop ( ) ;
S.pop ( );
S.display ( cout );
S.pop ( );
S. pop ( ) ;
S.display ( cout );
S.peek (tmp);
cout << tmp << endl;
}
#include
using namespace std;
const int MAXSIZE = 100; // Maximum size of the stack
class Stack {
private:
int top; // Index of the top element in the stack
int data[MAXSIZE]; // Array to hold the stack elements
public:
Stack () {
top = -1; // Initialize the top index to -1 to indicate an empty stack
}
bool push (int val) {
if (top == MAXSIZE - 1) { // Stack is full
return false;
}
top++; // Increment the top index
data[top] = val; // Add the new element to the top of the stack
return true;
}
bool pop() {
if (top == -1) { // Stack is empty
return false;
}
top--; // Decrement the top index to remove the top element
return true;
}
bool peek(int& val) {
if (top == -1) { // Stack is empty
return false;
}
val = data[top]; // Return the top element without removing it.
return true;
}
void display(ostream& out) {
if (top == -1) { // Stack is empty
out << "Empty stack" << endl;
return;
}
for (int i = top; i >= 0; i--) { // Loop through the stack elements from top to bottom
out << data[i] << "";
}
out << endl;
}
};
int main() {
Stack S;
int tmp;
S.push (5);
S. peek (tmp);
cout << tmp << endl;
S.push (33);
S.push (1);
S.push (7);
S.peek (tmp);
cout << tmp << endl;
S.push (33) ;
S.push (12);
S.display ( cout );
S.pop ( );
S.display ( cout );
S.push (14);
S.display ( cout );
S.pop ( ) ;
S.pop ( );
S.display ( cout );
S.pop ( );
S. pop ( ) ;
S.display ( cout );
S.peek (tmp);
cout << tmp << endl;
}
#include
using namespace std;
const int MAXSIZE = 100; // Maximum size of the stack
cl.
Merge Sort java with a few details, please include comments if possi.pdf
Merge Sort java with a few details, please include comments if possible Generate an array with
100 random integers between 0 and 1000 Print the original array to the screen Use your merge
sort algorithm to sort the array from lowest to highest Print the sorted array to the screen
Solution
/* Class name MergeSortRandomNumber.java */
import java.util.Random;
public class MergeSortRandomNumber{
private int[] array;
private int[] tempMergArr;
private int length;
public static void main(String[] args) {
int inputArr[] =new int[10];// Storing array elements default 10 we can modify depends on
requirement
Random r = new Random();//Create Object for Random Predefined class
System.out.println(\"Random Numbers\");//Printing Random Numbers
for(int i=0;i<10;i++)
{
int Low = 0;//Start from 0
int High = 100;//Up to 100
int Result = r.nextInt(High-Low) + Low;
System.out.print(Result);//Prining on the console random numbers
System.out.print(\" \");
inputArr[i]=Result;
}
System.out.println(\"\");
System.out.println(\"Input Array\");//Printing input array on the console start
for(int i=0;i<10;i++)
{
System.out.print(inputArr[i]);//
System.out.print(\" \");
} //Printing input array on the console end
System.out.println(\"\");
System.out.println(\"After Merge sort from low to high\");
//merge Sort Main logic started
RandomNumber mms = new RandomNumber();
mms.sort(inputArr);//Passing our input array calling sort() method
for(int i:inputArr){
System.out.print(i);//printing Final values from low to high
System.out.print(\" \");
}
}
public void sort(int inputArr[]) {
this.array = inputArr;
this.length = inputArr.length;
this.tempMergArr = new int[length];
doMergeSort(0, length - 1);//Calling Merge sort main logic
}
private void doMergeSort(int lowerIndex, int higherIndex) {
if (lowerIndex < higherIndex) {
int middle = lowerIndex + (higherIndex - lowerIndex) / 2;
// Below step sorts the left side of the array
doMergeSort(lowerIndex, middle);
// Below step sorts the right side of the array
doMergeSort(middle + 1, higherIndex);
// Now merge both sides
mergeParts(lowerIndex, middle, higherIndex);
}
}
private void mergeParts(int lowerIndex, int middle, int higherIndex) {
//Compare elements at First and Second,
//and move smaller element at Merged array
for (int i = lowerIndex; i <= higherIndex; i++) {
tempMergArr[i] = array[i];
}
int i = lowerIndex;
int j = middle + 1;
int k = lowerIndex;
while (i <= middle && j <= higherIndex) {
if (tempMergArr[i] <= tempMergArr[j]) {
array[k] = tempMergArr[i];
i++;
} else {
array[k] = tempMergArr[j];
j++;
}
k++;
}
while (i <= middle) {
array[k] = tempMergArr[i];
k++;
i++;
}
}
}
=============================
Expected Out put
Random Numbers
19 95 97 32 86 36 48 34 46 71
Input Array
19 95 97 32 86 36 48 34 46 71
After Merge sort from low to high
19 32 34 36 46 48 71 86 95 97.
Quick and Heap Sort with examples
The document describes two sorting algorithms: quicksort and heapsort. Quicksort is a divide and conquer algorithm that works by selecting a pivot element and partitioning the array around it, recursively sorting the subarrays. The performance depends on how well balanced the partitions are. Heapsort uses a binary heap data structure to sort an array in-place. It works by building a max heap from the array and then removing elements from the heap one by one.
Unit III Heaps.ppt
Heap sort is based on the priority queue data structure implemented using a heap. Specifically, heap sort uses a max-heap or min-heap to sort elements in ascending or descending order respectively.
The key steps are:
1) Build a max-heap from the input array in O(n) time
2) Repeatedly remove the max/min element from the heap and put it in its sorted position in the array. This takes O(nlogn) time.
So the overall time complexity of heap sort is O(nlogn). It is an in-place sorting algorithm that does not require additional storage.
Heapsort using Heap
Heapsort algorithm using heap data structure explain time complexity, how to use it , why we should use heap
Heaps
The document discusses different types of heaps and heap algorithms. It describes binary min-heaps and max-heaps, including their properties and implementations using arrays. Basic heap operations like insert, delete, and build heap are explained along with their time complexities. Applications of heaps like priority queues and selection algorithms are covered. More advanced heap types like leftist heaps, skew heaps and binomial queues are also mentioned.
Implement the additional 5 methods as indicated in the LinkedList fi.pdf
Implement a minimum spanning tree using Boruvka\'s algorithm.
Implement a minimum spanning tree using Boruvka\'s algorithm.
Solution
The psuedo code is as follows:
1 Graph input = /* read input */;
2 UnionFind uf (input.getNodes ());
3 Workset ws = input.getNodes();
4 foreach (Node n : ws) {
5 Edge e = minWeight (input.getOutEdges (n));
6 Node rep1 = uf.find (e.src);
7 Node rep2 = uf.find (e.dst);
8 if (rep1 != rep2) {
9 e.markInMST();
10 Node rep = uf.union (rep1, rep2);
11 ws.add (rep)
12 }
13 }.
Here is the code given in the instructionsclass AVL {.pdf
Here is the code given in the instructions:
class AVL {
Node root;
private class Node {
int data;
Node left, right;
int height;
private Node(int D, Node L, Node R, int H) {
data=D;
left=L;
right=R;
height=H; // user has to set height
} // of constructor Node
} //of Node
static private void UpdateHeight(Node T) {
if (T ==null) return;
else T.height = Math.max(HEIGHT(T.left),HEIGHT(T.right)) + 1;
} // UPdate Height
static private int HEIGHT(Node T) {
if (T== null ) return(-1);
else return (T.height);
}
// we need to more our right child up
static private Node LeftRotate(Node T) {
System.out.println(\"left rotate with data \" + T.data);
Node Tr;
Tr=T.right; // right child
T.right=Tr.left; // our right child IS NOW hhh
Tr.left=T; // move T down to the left
UpdateHeight(Tr.left); // update the height of T
UpdateHeight(Tr); // update hte height of the new root
return Tr;
} // of LeftRotate
// we move our immediate left node up.
// in doing so, we are now immediat right of our left child
static private Node RightRotate(Node T) {
Node Tl;
System.out.println(\"left rotate with data \" + T.data);
Tl=T.left; // left child
T.left=Tl.right; // our left child is now what our left was pointing to
Tl.right=T; // move T down to the right
UpdateHeight(Tl.right); // update the height of T
UpdateHeight(Tl); // update hte height of the new root
return Tl;
} // of RightRotate
public AVL() {
root=null;
} // of constructor AVL
// method to allow external entity to insert an element into the tree
public void insert(int D)
{
root=insert_internal(D, root);
} // of insert
private Node insert_internal(int D, Node T) {
if (T==null) return( new Node (D, null, null, 0));
if (T.data == D) return T; // the data is already in there
if ( D < T. data ) // go left
{ if (T.left == null)
{
T.left = insert_internal(D,null);
UpdateHeight(T);
}
else { //interior node
T.left = insert_internal(D, T.left);
}
} // of go left
else // D goes to the right
{ if (T.right == null)
{
T.right = insert_internal(D,null);
UpdateHeight(T);
}
else { //interior node
T.right = insert_internal(D, T.right);
}
} // of go right
// now we have to figure out if things are out of
// balance
int diff= HEIGHT(T.right) - HEIGHT(T.left);
System.out.println(\"difference is \" + diff + \" data is \" + T.data);
if ( Math.abs(diff) <= 1) // we are good to go at this level
{
UpdateHeight(T);
return(T);
}
// only if diff is bigger than 1
if ( diff > 1)// right leaning
{ // look at right child and figure out how it is leaning
} // of right leaning
else // left leaning
{ // look at left child to see how it is leaning
Node child = T.left;
int cdiff;
cdiff = HEIGHT(child.right) - HEIGHT(child.left);
System.out.println(\"cdiff is \" + cdiff);
if ( cdiff < 0 ) // left left lean
{ T=RightRotate(T); }
else
{
System.out.println(\"SHAUN\");
preorder_internal(T);
T.left = LeftRotate(T.left);
System.out.println(\"SHAUN\");
preorder_internal(T);
T=RightRotate(T);
}
} // of left leaning
// at this point we have rotated, so we need to.
y = 22Solutiony = 22.pdf
This short document defines the variable y and assigns it the value of 22. No further calculations or explanations are provided.
More Related Content
Similar to In inline heapsort we use only single array throughout the program t.pdf
In Java code, implement the HeapSort algorithm. Your function must t.pdf
In a random sample of male and female graduates of the New York School for the Arts between
the ages of 22-35 you know that:
• the probability a man is a ballet dancer is 0.245.
• the probability a woman is a musical performer is 0.365
• the probability a man is a musical performer, given that he\'s a ballet dancer is 0.250.
• the probability a woman is an actress is 0.550.
• the probability a woman is an actress, given that she\'s a musical performer is .785.
Compute the following probabilities:
A. P (man is a ballet dancer and a musical performer)
B. P (woman is an actress and a musical performer)
C. P (woman is an actress or a musical performer)
Solution
=0.245*(1-0.365)=0.155
b. P (woman is an actress and a musical performer)
=0.365*0.55=0.2
c. P (woman is an actress or a musical performer)
=P(woman is an actress)+P(musical performer)-P(actress or performer)
=0.55+0.365-0.785*0.365=0.628.
Please review my code (java)Someone helped me with it but i cannot.pdf
Please make it to clear 12-95. The basketball passed through the hoop even though it barely
cleared the hands of the player B who attempted to block it. Neglecting the size of the ball,
determine the magnitude vn of its initial velocity and the height h of the ball when it passes over
player B 10 ft 5 ft Prob. 12-95
Solution
Consider the motion along horizontal direction
Vox = initial velocity in horizontal direction = Va Cos30
X = horizontal distance = 25 + 3 = 28 ft
t = time taken
t = X / Vox = 28/(Va Cos30) eq-1
consider the motion in vertical direction
Y = vertical displacement = 10 - 7 = 3 ft
a = acceleration = - 32.2
Voy = initial velocity = Va Sin30
t = time taken
using the equation
Y = Voy t + (0.5) a t2
3 = (Va Sin30) (28/(Va Cos30)) + (0.5) (- 32.2) (28/(Va Cos30))2
Va = 35.8 m/s
t\' = time taken to reach B = 25 / (Va Cos30) = 25 / (35.8 Cos30) = 0.81 sec
h = 7 + Voy t\' + (0.5) a t\'2
h = 7 + (35.8 Sin30) (0.81) + (0.5) (- 32.2) (0.81)2
h = 10.94 ft.
4. The size of instructions can be fixed or variable. What are advant.pdf
4. The size of instructions can be fixed or variable. What are advantage and disadvantage of
fixed compared to variable? point) eft 2 20 0 C-4131-28 Add suction 131-26 nvaction 125-2 PC
egister daiess mancion 120-16 date 1 regnter 2 nstructien 31-00 eg stcrs Read U ALLu dela tlata
dvts 02 rstend In the above figure (single-cycle implementation), current instruction is SW S1,
4(S2) (SW: Store Word; current addrss for this instruction is 100 in decimal; S1 and $2 are
initially 8). For redundancy, you should write \"X\" instead of 0 or . You should write the reason
for cach question as well. 5. What is the value of RegDst? (1 point) 6. What is the value of
RegWrite? (I point) 7. What is the value after sign-extension? 1 point) 8. What is the value of
MemRead? (1 point) 9. What is the value of MemtoReg? (1 point) In the above figure, current
instruction is ADDI $1. S2, #-1(which is minus l) (ADDI is ADD Immediate; current address for
this instruction s 100 in decimal and S2 are initially1). For redundancy, you should write “X\',
instead of 0 or 1 . You should write the reason for each question as well. 10. What is the value
(immediate value) of instruction [15-0)? Write the value in 6-bit binary (1 point) 11. What is the
value after sign extension? Write the value in 32-bit binary. (I point) 12. What is the value of
Reg Write? (1 point) 13. What is the value of MemtoReg? (1 point) 2/4
Solution
import java.util.AbstractSet; import java.util.Arrays; import java.util.Iterator; import
java.util.NoSuchElementException; /** * Implementation of an abstract set using an array-
based binary tree. This is * used to help teach binary tree\'s and will have more details explained
in * future lectures. * * @author William J. Collins * @author Matthew Hertz * @param
Data type (which must be Comparable) of the elements in this tree. */ public class
ArrayBinaryTree> extends AbstractSet { /** Entry in the data store where the root node can be
found. */ private static final int ROOT = 0; /** Array used to store the nodes which consist of
this binary tree. */ protected Node[] store; /** Number of elements within the tree. */
protected int size; /** * Initializes this ArrayBinaryTree object to be empty. This creates the
array * in which items will be stored. */ @SuppressWarnings(\"unchecked\") public
ArrayBinaryTree() { store = new Node[63]; size = 0; } /** * Initializes this
ArrayBinaryTree object to contain a shallow copy of a * specified ArrayBinaryTree object.
The worstTime(n) is O(n), where n is the * number of elements in the specified
ArrayBinaryTree object. * * @param otherTree The tree which will be copied to create our
new tree, */ @SuppressWarnings(\"unchecked\") public ArrayBinaryTree(ArrayBinaryTree
otherTree) { store = (Node[]) Arrays.copyOf(otherTree.store, otherTree.store.length); size =
otherTree.size; } public int countLeaves() { } /** * Returns the size of this
ArrayBinaryTree object. * * @return the size of this ArrayBinaryTree object. */
@Overrid.
2.(Sorted list array implementation)This sorted list ADT discussed .pdf
2.(Sorted list: array implementation)This sorted list ADT discussed in class should be extended
by the addition of two new methods:
//Interface: ArrayListADT
//works for int
public interface ArrayListADT {
public boolean isEmpty(); //Method to determine whether the list is empty.
public boolean isFull(); //Method to determine whether the list is full.
public int listSize(); //Method to return the number of elements in the list.
public int maxListSize(); //Method to return the maximum size of the list.
public void print(); //Method to output the elements of the list.
public boolean isItemAtEqual(int location, int item); //Method to determine whether item is
the same as the item in the list at location.
public void insertAt(int location, int insertItem); //Method to insert insertItem in the list at
the position
public void insertEnd(int insertItem); //Method to insert insertItem at the end of the list.
public void removeAt(int location); //Method to remove the item from the list at location.
public int retrieveAt(int location); //Method to retrieve the element from the list at location.
public void replaceAt(int location, int repItem); //Method to replace the element in the list at
location with repItem.
public void clearList(); //Method to remove all the elements from the list.
public int search(int searchItem); //Method to determine whether searchItem is in the list.
public void remove(int removeItem); //Method to remove an item from the list.
}
//Class: ArrayListClass implements
//Interface: ArrayListADT
public abstract class ArrayListClass implements ArrayListADT {
protected int length; //to store the length of the list
protected int maxSize; //to store the maximum size of the list
protected int[] list; //array to hold the list elements
//Default constructor
public ArrayListClass() {
maxSize = 100;
length = 0;
list = new int[maxSize];
}
//Alternate Constructor
public ArrayListClass(int size) {
if(size <= 0) {
System.err.println(\"The array size must be positive. Creating an array of size 100.\");
maxSize = 100;
}
else
maxSize = size;
length = 0;
list = new int[maxSize];
}
public boolean isEmpty() {
return (length == 0);
}
public boolean isFull() {
return (length == maxSize);
}
public int listSize() {
return length;
}
public int maxListSize() {
return maxSize;
}
public void print() {
for (int i = 0; i < length; i++)
System.out.print(list[i] + \" \");
System.out.println();
}
public boolean isItemAtEqual(int location, int item) {
if (location < 0 || location >= length) {
System.err.println(\"The location of the item to be compared is out of range.\");
return false;
}
return list[location]== item;
}
public void clearList() {
for (int i = 0; i < length; i++)
list[i] = 0;
length = 0;
System.gc(); //invoke the Java garbage collector
}
public void removeAt(int location) {
if (location < 0 || location >= length)
System.err.println(\"The location of the item to be removed is out of range.\");
else {
for(int i = location; i < length - 1; i++)
list[i] .
Ds sorting
The document discusses various sorting algorithms including bubble sort, insertion sort, selection sort, merge sort, and quick sort. It provides descriptions of how each algorithm works and examples of the steps involved in sorting an array using these algorithms. Pseudocode for merge sort is included to demonstrate how it recursively divides the array into halves and then merges the sorted halves.
Data structures and algorithms lab10
This document summarizes a lab on data structures and algorithms that focuses on heaps and heap sort. It introduces heaps as almost complete binary trees stored in an array that obey the max-heap or min-heap property. Basic heap operations like insertion and deletion are described along with filtering an element up or down the heap to maintain the property. Heap sort is then introduced as using a heap to iteratively extract and place the minimum element. The objectives are to implement functions for getting a node's parent and children in a heap as well as implement and test heap sort.
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.
Describe a data structure that supports both removeMin() and rem.pdf
Describe a data structure that supports both removeMin() and removeMax() with O(log(n))
complexity. You need to show the implementation of insert(), removeMin(), and removeMax(),
and prove their complexity. Hint: one can build on top of the heap.
/*IF YOU WRITE CODE, PLEASE PROVIDE COMMENTS*/
I need to understand
Describe a data structure that supports both removeMin() and removeMax() with O(log(n))
complexity. You need to show the implementation of insert(), removeMin(), and removeMax(),
and prove their complexity. Hint: one can build on top of the heap.
/*IF YOU WRITE CODE, PLEASE PROVIDE COMMENTS*/
I need to understand
Describe a data structure that supports both removeMin() and removeMax() with O(log(n))
complexity. You need to show the implementation of insert(), removeMin(), and removeMax(),
and prove their complexity. Hint: one can build on top of the heap.
/*IF YOU WRITE CODE, PLEASE PROVIDE COMMENTS*/
I need to understand
Solution
The idea here is to use the concept of two binary heaps along with the concept of list(doubly
linked list).
The doubly linked list contains all input items and indexes of corresponding min and max heap
nodes. The nodes of min and max heaps store addresses of nodes of doubly linked list. The root
node of min heap stores the address of minimum item in doubly linked list. Similarly, root of
max heap stores address of maximum item in doubly linked list. Following are the details of
operations.
1) findMax(): We get the address of maximum value node from root of Max Heap. So this is a
O(1) operation.
1) findMin(): We get the address of minimum value node from root of Min Heap. So this is a
O(1) operation.
3) removeMin(): We get the address of minimum value node from root of Min Heap. We use this
address to find the node in doubly linked list. From the doubly linked list, we get node of Max
Heap. We delete node from all three. We can delete a node from doubly linked list in O(1) time.
delete() operations for max and min heaps take O(Logn) time.
4) removeMax(): is similar to deleteMin()
5) Insert() We always insert at the beginning of linked list in O(1) time. Inserting the address in
Max and Min Heaps take O(Logn) time. So overall complexity is O(Logn)
#include
#include
#include
// A node of doubly linked list
struct LNode
{
int data;
int minHeapIndex;
int maxHeapIndex;
struct LNode *next, *prev;
};
// Structure for a doubly linked list
struct List
{
struct LNode *head;
};
// Structure for min heap
struct MinHeap
{
int size;
int capacity;
struct LNode* *array;
};
// Structure for max heap
struct MaxHeap
{
int size;
int capacity;
struct LNode* *array;
};
// The required data structure
struct CDS
{
struct MinHeap* minHeap;
struct MaxHeap* maxHeap;
struct List* list;
};
// A utility function to create a new List node
struct LNode* newLNode(int data)
{
struct LNode* node =
(struct LNode*) malloc(sizeof(struct LNode));
node->minHeapIndex = node->maxHeapIndex = -1;
node->data = data;
node->prev = node->next = NULL;
retu.
Write a program that accepts an arithmetic expression of unsigned in.pdf
Write a program that accepts an arithmetic expression of unsigned integers in postfix notation
and builds the arithmetic expression tree that represents that expression. From that tree, the
corresponding fully parenthesized infix expression should be displayed and a file should be
generated that contains the three address format instructions. This topic is discussed in the week
4 reading in module 2, section II-B. The main class should create the GUI shown below:
Pressing the Construct Tree button should cause the tree to be constructed and using that tree, the
corresponding infix expression should be displayed and the three address instruction file should
be generated. The postfix expression input should not be required to have spaces between every
token. Note in the above example that 9+- are not separated by spaces. The above example
should produce the following output file containing the three address instructions:
Add R0 5 9
Sub R1 3
R0 Mul R2 2 3
Div R3 R1 R2
Inheritance should be used to define the arithmetic expression tree. At a minimum, it should
involve three classes: an abstract class for the tree nodes and two derived classes, one for
operand nodes and another for operator nodes. Other classes should be included as needed to
accomplish good object-oriented design. All instance data must be declared as private.
You may assume that the expression is syntactically correct with regard to the order of operators
and operands, but you should check for invalid tokens, such as characters that are not valid
operators or operands such as 2a, which are not valid integers. If an invalid token is detected a
RuntimeException should be thrown and caught by the main class and an appropriate error
message should be displayed. Three Adddress Generator Enter Postfix Expression 359 2 3
Construct Tree Infix Expression ((3-(5 9))/ (2* 3
Solution
#include
#include
#include
#include
#include
struct TREE //Structure to represent a tree
{
char data;
struct TREE *right,*left,*root;
};
typedef TREE tree; /* Stack Using Linked List */
struct Stack //Structure to represent Stack
{
struct TREE *data;
struct Stack *next,*head,*top; //Pointers to next node,head and top
};
typedef struct Stack node;
node *Nw; //Global Variable to represent node
void initialize(node *&n)
{
n = new node;
n -> next = n -> head = n -> top = NULL;
}
void create(node *n)
{
Nw = new node; //Create a new node
Nw -> next = NULL; //Initialize next pointer field
if(n -> head == NULL) //if first node
{
n -> head = Nw; //Initialize head
n -> top = Nw; //Update top
}
}
void push(node *n,tree *ITEM)
{
node *temp = n -> head;
if(n -> head == NULL) //if First Item is Pushed to Stack
{
create(n); //Create a Node
n -> head -> data = ITEM; //Insert Item to head of List
n -> head = Nw;
return; //Exit from Function
}
create(n); //Create a new Node
Nw -> data = ITEM; //Insert Item
while(temp -> next != NULL)
temp = temp -> next; //Go to Last Node
temp -> next = Nw; //Point New node
n -> top = Nw; //Upda.
USING JAVAThere are at least two types of nearly sorted array.pdf
USING JAVA:
There are at least two types of \"nearly sorted\" arrays: k-exchanges and k-distance, where k is an
integer. An array is called a k-exchanges array if the array becomes sorted by performing at most
k exchanges. For example, sorting.java uses 100-exchanges for nearly sorted arrays. An array is
called a k-distance array if, for each element, the distance from its initial position to its position
in the sorted array is no more than k. A k-distance array can be created from a sorted array by
randomly shuffling each subarray of size k. www.sorting-algorithms.com uses 1-distance for
nearly sorted arrays.compare the performances of quicksort (choosing the best quicksort you
have), heapsort, mergesort, natural mergesort, and insertion sort on k-exchanges and k-distances
arrays for various k values. Create a method called task4, which takes k = 10, 20, 30, ..., until
quicksort becomes a clear winner, create 100 k-distance arrays of size 10,000,000, run in turn the
chosen sorting methods on each array, and collect running times. Then for k = 100, 200, 300, ...,
until quicksort becomes a clear winner, create 100 k-exchange arrays of size 10,000,000, run in
turn the chosen sorting methods on each array, and collect running times.
I have the code for all the algorithms done correctly, so just call them by quicksort(), heapsort(),
mergesort(), naturalmergesort(), insertionsort(). I can figure out how to deal with differing sizes
but what I don\'t understand is how to create k-distance and k-exchanges nearly sorted arrays.
Solution
void insertionSort(int A[], int size)
{
int i, key, j;
for (i = 1; i < size; i++)
{
key = A[i];
j = i-1;
/* Move elements of A[0..i-1], that are greater than key, to one
position ahead of their current position.
This loop will run at most k times */
while (j >= 0 && A[j] > key)
{
A[j+1] = A[j];
j = j-1;
}
A[j+1] = key;
}
}
he inner loop will run at most k times. To move every element to its correct place, at most k
elements need to be moved. So overall complexity will be O(nk)
We can sort such arrays more efficiently with the help of Heap data structure. Following is the
detailed process that uses Heap.
1) Create a Min Heap of size k+1 with first k+1 elements. This will take O(k) time (See this
GFact)
2) One by one remove min element from heap, put it in result array, and add a new element to
heap from remaining elements.
Removing an element and adding a new element to min heap will take Logk time. So overall
complexity will be O(k) + O((n-k)*logK)
#include
using namespace std;
// Prototype of a utility function to swap two integers
void swap(int *x, int *y);
// A class for Min Heap
class MinHeap
{
int *harr; // pointer to array of elements in heap
int heap_size; // size of min heap
public:
// Constructor
MinHeap(int a[], int size);
// to heapify a subtree with root at given index
void MinHeapify(int );
// to get index of left child of node at index i
int left(int i) { return (2*i + 1); }
// to get index of right chil.
Heap Tree.pdf
A heap is a special tree-based data structure where the tree is a complete binary tree. There are two types of heaps: a max-heap where the root node has the greatest key, and a min-heap where the root node has the minimum key. Operations on heaps include insertion and deletion of nodes in O(log N) time and peeking the max/min element. Heapify is the process of building the heap from an array by comparing nodes to ensure the heap property is maintained.
please tell me what lines do i alter to make this stack a queue. tel.pdf
please tell me what lines do i alter to make this stack a queue. tell me what to change it to
#include
using namespace std;
const int MAXSIZE = 100; // Maximum size of the stack
class Stack {
private:
int top; // Index of the top element in the stack
int data[MAXSIZE]; // Array to hold the stack elements
public:
Stack () {
top = -1; // Initialize the top index to -1 to indicate an empty stack
}
bool push (int val) {
if (top == MAXSIZE - 1) { // Stack is full
return false;
}
top++; // Increment the top index
data[top] = val; // Add the new element to the top of the stack
return true;
}
bool pop() {
if (top == -1) { // Stack is empty
return false;
}
top--; // Decrement the top index to remove the top element
return true;
}
bool peek(int& val) {
if (top == -1) { // Stack is empty
return false;
}
val = data[top]; // Return the top element without removing it.
return true;
}
void display(ostream& out) {
if (top == -1) { // Stack is empty
out << "Empty stack" << endl;
return;
}
for (int i = top; i >= 0; i--) { // Loop through the stack elements from top to bottom
out << data[i] << "";
}
out << endl;
}
};
int main() {
Stack S;
int tmp;
S.push (5);
S. peek (tmp);
cout << tmp << endl;
S.push (33);
S.push (1);
S.push (7);
S.peek (tmp);
cout << tmp << endl;
S.push (33) ;
S.push (12);
S.display ( cout );
S.pop ( );
S.display ( cout );
S.push (14);
S.display ( cout );
S.pop ( ) ;
S.pop ( );
S.display ( cout );
S.pop ( );
S. pop ( ) ;
S.display ( cout );
S.peek (tmp);
cout << tmp << endl;
}
#include
using namespace std;
const int MAXSIZE = 100; // Maximum size of the stack
class Stack {
private:
int top; // Index of the top element in the stack
int data[MAXSIZE]; // Array to hold the stack elements
public:
Stack () {
top = -1; // Initialize the top index to -1 to indicate an empty stack
}
bool push (int val) {
if (top == MAXSIZE - 1) { // Stack is full
return false;
}
top++; // Increment the top index
data[top] = val; // Add the new element to the top of the stack
return true;
}
bool pop() {
if (top == -1) { // Stack is empty
return false;
}
top--; // Decrement the top index to remove the top element
return true;
}
bool peek(int& val) {
if (top == -1) { // Stack is empty
return false;
}
val = data[top]; // Return the top element without removing it.
return true;
}
void display(ostream& out) {
if (top == -1) { // Stack is empty
out << "Empty stack" << endl;
return;
}
for (int i = top; i >= 0; i--) { // Loop through the stack elements from top to bottom
out << data[i] << "";
}
out << endl;
}
};
int main() {
Stack S;
int tmp;
S.push (5);
S. peek (tmp);
cout << tmp << endl;
S.push (33);
S.push (1);
S.push (7);
S.peek (tmp);
cout << tmp << endl;
S.push (33) ;
S.push (12);
S.display ( cout );
S.pop ( );
S.display ( cout );
S.push (14);
S.display ( cout );
S.pop ( ) ;
S.pop ( );
S.display ( cout );
S.pop ( );
S. pop ( ) ;
S.display ( cout );
S.peek (tmp);
cout << tmp << endl;
}
#include
using namespace std;
const int MAXSIZE = 100; // Maximum size of the stack
cl.
Merge Sort java with a few details, please include comments if possi.pdf
Merge Sort java with a few details, please include comments if possible Generate an array with
100 random integers between 0 and 1000 Print the original array to the screen Use your merge
sort algorithm to sort the array from lowest to highest Print the sorted array to the screen
Solution
/* Class name MergeSortRandomNumber.java */
import java.util.Random;
public class MergeSortRandomNumber{
private int[] array;
private int[] tempMergArr;
private int length;
public static void main(String[] args) {
int inputArr[] =new int[10];// Storing array elements default 10 we can modify depends on
requirement
Random r = new Random();//Create Object for Random Predefined class
System.out.println(\"Random Numbers\");//Printing Random Numbers
for(int i=0;i<10;i++)
{
int Low = 0;//Start from 0
int High = 100;//Up to 100
int Result = r.nextInt(High-Low) + Low;
System.out.print(Result);//Prining on the console random numbers
System.out.print(\" \");
inputArr[i]=Result;
}
System.out.println(\"\");
System.out.println(\"Input Array\");//Printing input array on the console start
for(int i=0;i<10;i++)
{
System.out.print(inputArr[i]);//
System.out.print(\" \");
} //Printing input array on the console end
System.out.println(\"\");
System.out.println(\"After Merge sort from low to high\");
//merge Sort Main logic started
RandomNumber mms = new RandomNumber();
mms.sort(inputArr);//Passing our input array calling sort() method
for(int i:inputArr){
System.out.print(i);//printing Final values from low to high
System.out.print(\" \");
}
}
public void sort(int inputArr[]) {
this.array = inputArr;
this.length = inputArr.length;
this.tempMergArr = new int[length];
doMergeSort(0, length - 1);//Calling Merge sort main logic
}
private void doMergeSort(int lowerIndex, int higherIndex) {
if (lowerIndex < higherIndex) {
int middle = lowerIndex + (higherIndex - lowerIndex) / 2;
// Below step sorts the left side of the array
doMergeSort(lowerIndex, middle);
// Below step sorts the right side of the array
doMergeSort(middle + 1, higherIndex);
// Now merge both sides
mergeParts(lowerIndex, middle, higherIndex);
}
}
private void mergeParts(int lowerIndex, int middle, int higherIndex) {
//Compare elements at First and Second,
//and move smaller element at Merged array
for (int i = lowerIndex; i <= higherIndex; i++) {
tempMergArr[i] = array[i];
}
int i = lowerIndex;
int j = middle + 1;
int k = lowerIndex;
while (i <= middle && j <= higherIndex) {
if (tempMergArr[i] <= tempMergArr[j]) {
array[k] = tempMergArr[i];
i++;
} else {
array[k] = tempMergArr[j];
j++;
}
k++;
}
while (i <= middle) {
array[k] = tempMergArr[i];
k++;
i++;
}
}
}
=============================
Expected Out put
Random Numbers
19 95 97 32 86 36 48 34 46 71
Input Array
19 95 97 32 86 36 48 34 46 71
After Merge sort from low to high
19 32 34 36 46 48 71 86 95 97.
Quick and Heap Sort with examples
The document describes two sorting algorithms: quicksort and heapsort. Quicksort is a divide and conquer algorithm that works by selecting a pivot element and partitioning the array around it, recursively sorting the subarrays. The performance depends on how well balanced the partitions are. Heapsort uses a binary heap data structure to sort an array in-place. It works by building a max heap from the array and then removing elements from the heap one by one.
Unit III Heaps.ppt
Heap sort is based on the priority queue data structure implemented using a heap. Specifically, heap sort uses a max-heap or min-heap to sort elements in ascending or descending order respectively.
The key steps are:
1) Build a max-heap from the input array in O(n) time
2) Repeatedly remove the max/min element from the heap and put it in its sorted position in the array. This takes O(nlogn) time.
So the overall time complexity of heap sort is O(nlogn). It is an in-place sorting algorithm that does not require additional storage.
Heapsort using Heap
Heapsort algorithm using heap data structure explain time complexity, how to use it , why we should use heap
Heaps
The document discusses different types of heaps and heap algorithms. It describes binary min-heaps and max-heaps, including their properties and implementations using arrays. Basic heap operations like insert, delete, and build heap are explained along with their time complexities. Applications of heaps like priority queues and selection algorithms are covered. More advanced heap types like leftist heaps, skew heaps and binomial queues are also mentioned.
Implement the additional 5 methods as indicated in the LinkedList fi.pdf
Implement a minimum spanning tree using Boruvka\'s algorithm.
Implement a minimum spanning tree using Boruvka\'s algorithm.
Solution
The psuedo code is as follows:
1 Graph input = /* read input */;
2 UnionFind uf (input.getNodes ());
3 Workset ws = input.getNodes();
4 foreach (Node n : ws) {
5 Edge e = minWeight (input.getOutEdges (n));
6 Node rep1 = uf.find (e.src);
7 Node rep2 = uf.find (e.dst);
8 if (rep1 != rep2) {
9 e.markInMST();
10 Node rep = uf.union (rep1, rep2);
11 ws.add (rep)
12 }
13 }.
Here is the code given in the instructionsclass AVL {.pdf
Here is the code given in the instructions:
class AVL {
Node root;
private class Node {
int data;
Node left, right;
int height;
private Node(int D, Node L, Node R, int H) {
data=D;
left=L;
right=R;
height=H; // user has to set height
} // of constructor Node
} //of Node
static private void UpdateHeight(Node T) {
if (T ==null) return;
else T.height = Math.max(HEIGHT(T.left),HEIGHT(T.right)) + 1;
} // UPdate Height
static private int HEIGHT(Node T) {
if (T== null ) return(-1);
else return (T.height);
}
// we need to more our right child up
static private Node LeftRotate(Node T) {
System.out.println(\"left rotate with data \" + T.data);
Node Tr;
Tr=T.right; // right child
T.right=Tr.left; // our right child IS NOW hhh
Tr.left=T; // move T down to the left
UpdateHeight(Tr.left); // update the height of T
UpdateHeight(Tr); // update hte height of the new root
return Tr;
} // of LeftRotate
// we move our immediate left node up.
// in doing so, we are now immediat right of our left child
static private Node RightRotate(Node T) {
Node Tl;
System.out.println(\"left rotate with data \" + T.data);
Tl=T.left; // left child
T.left=Tl.right; // our left child is now what our left was pointing to
Tl.right=T; // move T down to the right
UpdateHeight(Tl.right); // update the height of T
UpdateHeight(Tl); // update hte height of the new root
return Tl;
} // of RightRotate
public AVL() {
root=null;
} // of constructor AVL
// method to allow external entity to insert an element into the tree
public void insert(int D)
{
root=insert_internal(D, root);
} // of insert
private Node insert_internal(int D, Node T) {
if (T==null) return( new Node (D, null, null, 0));
if (T.data == D) return T; // the data is already in there
if ( D < T. data ) // go left
{ if (T.left == null)
{
T.left = insert_internal(D,null);
UpdateHeight(T);
}
else { //interior node
T.left = insert_internal(D, T.left);
}
} // of go left
else // D goes to the right
{ if (T.right == null)
{
T.right = insert_internal(D,null);
UpdateHeight(T);
}
else { //interior node
T.right = insert_internal(D, T.right);
}
} // of go right
// now we have to figure out if things are out of
// balance
int diff= HEIGHT(T.right) - HEIGHT(T.left);
System.out.println(\"difference is \" + diff + \" data is \" + T.data);
if ( Math.abs(diff) <= 1) // we are good to go at this level
{
UpdateHeight(T);
return(T);
}
// only if diff is bigger than 1
if ( diff > 1)// right leaning
{ // look at right child and figure out how it is leaning
} // of right leaning
else // left leaning
{ // look at left child to see how it is leaning
Node child = T.left;
int cdiff;
cdiff = HEIGHT(child.right) - HEIGHT(child.left);
System.out.println(\"cdiff is \" + cdiff);
if ( cdiff < 0 ) // left left lean
{ T=RightRotate(T); }
else
{
System.out.println(\"SHAUN\");
preorder_internal(T);
T.left = LeftRotate(T.left);
System.out.println(\"SHAUN\");
preorder_internal(T);
T=RightRotate(T);
}
} // of left leaning
// at this point we have rotated, so we need to.
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In Java code, implement the HeapSort algorithm. Your function must t.pdf
In Java code, implement the HeapSort algorithm. Your function must t.pdf
Please review my code (java)Someone helped me with it but i cannot.pdf
Please review my code (java)Someone helped me with it but i cannot.pdf
4. The size of instructions can be fixed or variable. What are advant.pdf
4. The size of instructions can be fixed or variable. What are advant.pdf
2.(Sorted list array implementation)This sorted list ADT discussed .pdf
2.(Sorted list array implementation)This sorted list ADT discussed .pdf
ReversePoem.java ---------------------------------- public cl.pdf
ReversePoem.java ---------------------------------- public cl.pdf
Describe a data structure that supports both removeMin() and rem.pdf
Describe a data structure that supports both removeMin() and rem.pdf
Write a program that accepts an arithmetic expression of unsigned in.pdf
Write a program that accepts an arithmetic expression of unsigned in.pdf
USING JAVAThere are at least two types of nearly sorted array.pdf
USING JAVAThere are at least two types of nearly sorted array.pdf
please tell me what lines do i alter to make this stack a queue. tel.pdf
please tell me what lines do i alter to make this stack a queue. tel.pdf
Merge Sort java with a few details, please include comments if possi.pdf
Merge Sort java with a few details, please include comments if possi.pdf
Implement the additional 5 methods as indicated in the LinkedList fi.pdf
Implement the additional 5 methods as indicated in the LinkedList fi.pdf
Here is the code given in the instructionsclass AVL {.pdf
Here is the code given in the instructionsclass AVL {.pdf
More from anuragperipheral
y = 22Solutiony = 22.pdf
This short document defines the variable y and assigns it the value of 22. No further calculations or explanations are provided.
what ways are quorum sensing and the two component signaling system .pdf
what ways are quorum sensing and the two component signaling system in bacteria similar?
First we will see What is quorum sensing (QS). QS is a cell to cell communication in bacteria
that involve signaling molecules or autoinducers (AIs).
AIs come in different forms depending on bacterial species.
As one would expect, the molecular mechanism of QS is dependent on species, but there are four
main characteristics that govern all QS systems:
At low bacterial cell density (LBCD), AIs diffuse away and their cellular concentration is
approximately the same as the environment.
At high bacterial cell density (HBCD), there is a cumulative production of AIs, which leads to a
local high (environment) concentration. This is detected by the cells which in turn trigger
different kinds of responses.
AIs are detected by receptors that are either transmembrane molecules or are present in the
cytoplasm.
Detection of AIs leads to further production of AIs (feed forward loop) in addition to activating
other genes.
Qouram sensing and two component signaling system are inter connected and we will check how
it is connected: Cell-density-dependent gene expression appears to be widely spread in bacteria.
This quorum-sensing phenomenon has been well established in Gram-negative bacteria, where
N-acyl homoserine lactones are the diffusible communication molecules that modulate cell-
density-dependent phenotypes. Similarly, a variety of processes are known to be regulated in a
cell-density- or growth-phase-dependent manner in Gram-positive bacteria. Examples of such
quorum-sensing modes in Gram-positive bacteria are the development of genetic competence in
Bacillus subtilis andStreptococcus pneumoniae, the virulence response in Staphylococcus aureus,
and the production of antimicrobial peptides by several species of Gram-positive bacteria
including lactic acid bacteria. Cell-density-dependent regulatory modes in these systems appear
to follow a common theme, in which the signal molecule is a post-translationally processed
peptide that is secreted by a dedicated ATP-binding-cassette exporter. This secreted peptide
pheromone functions as the input signal for a specific sensor component of a two-component
signal-transduction system. Moreover, genetic linkage of the common elements involved results
in autoregulation of peptide-pheromone production.
What advantages would an enzyme coupled receptor have over a G-coupled receptor?
To understand the advantage of Enzyme-linked receptors over G-Protein Linked Receptors, we
have to see how these both types of receptors function in cellular level.
G-Protein Linked Receptors
G-protein-linked receptors bind a ligand and activate a membrane protein called a G-protein. The
activated G-protein then interacts with either an ion channel or an enzyme in the membrane. All
G-protein-linked receptors have seven transmembrane domains, but each receptor has its own
specific extracellular domain and G-protein-binding site.
Cell signaling using G.
Average.javaimport java.util.Scanner; import java.util.Arra.pdf
// Average.java
import java.util.Scanner;
import java.util.ArrayList;
public class Average
{
private double mean;
private int [] data;
public Average()
{
Scanner scan = new Scanner(System.in);
//Create an array with 5 elements.
data = new int[5];
//input the data into the array
for (int i = 0; i < data.length; i++)
{
System.out.print(\"Enter your score for test #\" + (i +1) + \": \");
data[i] = scan.nextInt();
}
}
public void calculateMean()
{
double sum = 0; //Accumulator
//calculate the sum of the test scores in the data array.
for (int i = 0; i < data.length; i++)
sum += data[i];
mean = (Double)sum/data.length;
}
public String toString()
{
String result = \"\ The data stored in array: \";
for (int i = 0; i < data.length; i++)
result = result + data[i] + \" \";
result = result + \"\ \" + \"Average: \" + mean + \"\ \";
return result;
}
}
// AverageDriver.java
public class AverageDriver
{
public static void main(String[] args)
{
Average avg = new Average();
avg.calculateMean();
System.out.println(avg.toString());
}
}
/*
output:
Enter your score for test #1: 87
Enter your score for test #2: 90
Enter your score for test #3: 92
Enter your score for test #4: 77
Enter your score for test #5: 89
The data stored in array: 87 90 92 77 89
Average: 87.0
*/
Solution
// Average.java
import java.util.Scanner;
import java.util.ArrayList;
public class Average
{
private double mean;
private int [] data;
public Average()
{
Scanner scan = new Scanner(System.in);
//Create an array with 5 elements.
data = new int[5];
//input the data into the array
for (int i = 0; i < data.length; i++)
{
System.out.print(\"Enter your score for test #\" + (i +1) + \": \");
data[i] = scan.nextInt();
}
}
public void calculateMean()
{
double sum = 0; //Accumulator
//calculate the sum of the test scores in the data array.
for (int i = 0; i < data.length; i++)
sum += data[i];
mean = (Double)sum/data.length;
}
public String toString()
{
String result = \"\ The data stored in array: \";
for (int i = 0; i < data.length; i++)
result = result + data[i] + \" \";
result = result + \"\ \" + \"Average: \" + mean + \"\ \";
return result;
}
}
// AverageDriver.java
public class AverageDriver
{
public static void main(String[] args)
{
Average avg = new Average();
avg.calculateMean();
System.out.println(avg.toString());
}
}
/*
output:
Enter your score for test #1: 87
Enter your score for test #2: 90
Enter your score for test #3: 92
Enter your score for test #4: 77
Enter your score for test #5: 89
The data stored in array: 87 90 92 77 89
Average: 87.0
*/.
C. 0.20 M NaCl Solution C..pdf
This document describes a 0.20 M NaCl solution. A 0.20 M NaCl solution contains 0.20 moles of sodium chloride dissolved in 1 liter of water. This concentration of sodium chloride solution is commonly used in chemistry experiments and labs.
yep in water which is higly polar will dissolve i.pdf
yep in water which is higly polar will dissolve ionic NaCl but not nonpolar steric
acid yes
Solution
yep in water which is higly polar will dissolve ionic NaCl but not nonpolar steric
acid yes.
When thinking of E or Z, the first thing that com.pdf
When thinking of E or Z, the first thing that comes to mind is the double bond of
the alkene. From there, you look at the atom attach to the double bond of the alkene. There are 4
atoms: 1) Carbon (top left) : bonded to 1 carbon and 2 hydrogen 2) Carbon (bottom left) : bonded
to 2 Oxygen [treat double bond here as two bonds] and 1 Hydrogen 3) Carbon (top right) :
bonded to 2 carbon and 1 hydrogen 4) Hydrogen (bottom right) : bonded to nothing. On each
side of the double bond, rank the priority (with respect to molecular mass) Bottom left >> Top
left Top right >> bottom right since the two groups are trans each other, this is an E isomer
Solution
When thinking of E or Z, the first thing that comes to mind is the double bond of
the alkene. From there, you look at the atom attach to the double bond of the alkene. There are 4
atoms: 1) Carbon (top left) : bonded to 1 carbon and 2 hydrogen 2) Carbon (bottom left) : bonded
to 2 Oxygen [treat double bond here as two bonds] and 1 Hydrogen 3) Carbon (top right) :
bonded to 2 carbon and 1 hydrogen 4) Hydrogen (bottom right) : bonded to nothing. On each
side of the double bond, rank the priority (with respect to molecular mass) Bottom left >> Top
left Top right >> bottom right since the two groups are trans each other, this is an E isomer.
Since it contains oxygen double bonded to a carbo.pdf
Since it contains oxygen double bonded to a carbon atom , it may beCaboxylic acid
or carbonyl compound. So the characteristic IRabsorption band is aroud 1700 cm -1
Solution
Since it contains oxygen double bonded to a carbon atom , it may beCaboxylic acid
or carbonyl compound. So the characteristic IRabsorption band is aroud 1700 cm -1.
salicylic acid alcohol (R-OH) and carboxylic aci.pdf
salicylic acid: alcohol (R-OH) and carboxylic acid (R-COOH) acetylsalicylic acid
(aspirin): carboxylic acid and ester (R-COOR)
Solution
salicylic acid: alcohol (R-OH) and carboxylic acid (R-COOH) acetylsalicylic acid
(aspirin): carboxylic acid and ester (R-COOR).
no it doesnot .pdf
This very short document does not seem to provide any clear information that can be summarized in 3 sentences or less. It contains the phrase "no it doesnot" repeated but no other context or details.
It would be Ruthenium (II) Sulfide Ruthenium is .pdf
It would be Ruthenium (II) Sulfide Ruthenium is a transition metal, meaning that it
can have multiple oxidation numbers. Sulfur always takes the charge of -2, so from that we see
that there is a 1 to 1 relation between Ru and S in the empirical formula, so that means that
Ruthenium has the opposite charge, +2.
Solution
It would be Ruthenium (II) Sulfide Ruthenium is a transition metal, meaning that it
can have multiple oxidation numbers. Sulfur always takes the charge of -2, so from that we see
that there is a 1 to 1 relation between Ru and S in the empirical formula, so that means that
Ruthenium has the opposite charge, +2..
ketones do not give tollens test.. hence no react.pdf
ketones do not give tollens test.. hence no reaction
Solution
ketones do not give tollens test.. hence no reaction.
for IONIC the metal name followed by nonmetal io.pdf
for IONIC: the metal name followed by nonmetal ion (-ide) for ex: NaCl is Sodium
Chloride or MgS is Magnesium Sulfide
Solution
for IONIC: the metal name followed by nonmetal ion (-ide) for ex: NaCl is Sodium
Chloride or MgS is Magnesium Sulfide.
pH = 4-log2.4 = 3.62 pOH 14 - pH = 14-3.62 = 10.3.pdf
pH = 4-log2.4 = 3.62 pOH 14 - pH = 14-3.62 = 10.38 concentration of OH- = 10^(-
14) / (2.4*10^(-4)) = 4.167* 10^(-11)
Solution
pH = 4-log2.4 = 3.62 pOH 14 - pH = 14-3.62 = 10.38 concentration of OH- = 10^(-
14) / (2.4*10^(-4)) = 4.167* 10^(-11).
This suggest that dissociation of FeCl3 is 100 .pdf
This suggest that dissociation of FeCl3 is 100 % to form 4 ions (1Fe+3Cl). Other
dissociation also may be possible like FeCl3---->(FeCl2) + Cl- FeCl3---->(FeCl)2+ + 2Cl- so
number of ions is averaged considering dissociation constants which is 3.4
Solution
This suggest that dissociation of FeCl3 is 100 % to form 4 ions (1Fe+3Cl). Other
dissociation also may be possible like FeCl3---->(FeCl2) + Cl- FeCl3---->(FeCl)2+ + 2Cl- so
number of ions is averaged considering dissociation constants which is 3.4.
Total probability of dot received P = (37)(78).pdf
Total probability of dot received P = (3/7)*(7/8)+(4/7)*(1/8)=0.4464 probability
that dot sentP1 = 3/7=0.4285 so P(dot sent | dot received).= P1/P=0.4285/.4464=0.96
Solution
Total probability of dot received P = (3/7)*(7/8)+(4/7)*(1/8)=0.4464 probability
that dot sentP1 = 3/7=0.4285 so P(dot sent | dot received).= P1/P=0.4285/.4464=0.96.
Environmental contaminants of recent concern are .pdf
Environmental contaminants of recent concern are pharmaceuticals, estrogens and
other endocrine disrupting chemicals (EDC) such as degradation products of surfactants, algal
and cyanobacterial toxins, disinfection by-products (DBPs) and metalloids. In addition,
pesticides (especially their transformation products), microorganisms, and humic substances
(HS), in their function as vehicles for contaminants and as precursors for by-products in water
treatment, traditionally play an important role. The present status of the application of LC-MS
techniques for these water constituents are discussed and examples of application are given.
Solid-phase extraction with various non-selective materials in combination with liquid
chromatography (LC) on reversed-phase columns have been the most widely used methods for
sample preconcentration and separation for different compound classes like pesticides,
pharmaceuticals or estrogens. Electrospray ionization (ESI) and atmospheric pressure ionization
(APCI) are the most frequently used ionization techniques for polar and ionic compounds, as
well as for less polar non-ionic ones. The facilities of LC-MS have been successfully
demonstrated for different compound classes. Polar compounds from pharmaceuticals used as
betablockers, iodinated X-ray contrast media, or estrogens have been determined without
derivatization down to ultratrace concentrations. LC-MS can be viewed as a prerequisite for the
determination of algal and cyanobacterial toxins and the homologues and oligomers of
alkylphenol ethoxylates and their metabolites. Tandem mass spectrometric techniques and the
use of diagnostic ions reveal their usefulness for compound-class specific screening and
unknown identification, and are also valid for the analysis of pesticides and especially for their
transformation products.
Solution
Environmental contaminants of recent concern are pharmaceuticals, estrogens and
other endocrine disrupting chemicals (EDC) such as degradation products of surfactants, algal
and cyanobacterial toxins, disinfection by-products (DBPs) and metalloids. In addition,
pesticides (especially their transformation products), microorganisms, and humic substances
(HS), in their function as vehicles for contaminants and as precursors for by-products in water
treatment, traditionally play an important role. The present status of the application of LC-MS
techniques for these water constituents are discussed and examples of application are given.
Solid-phase extraction with various non-selective materials in combination with liquid
chromatography (LC) on reversed-phase columns have been the most widely used methods for
sample preconcentration and separation for different compound classes like pesticides,
pharmaceuticals or estrogens. Electrospray ionization (ESI) and atmospheric pressure ionization
(APCI) are the most frequently used ionization techniques for polar and ionic compounds, as
well as for less polar non-ionic ones. The facilities of .
C-H polar covalent bond Solution .pdf
The document discusses C-H polar covalent bonds. C-H bonds have some ionic character due to the difference in electronegativity between carbon and hydrogen. Hydrogen is more electronegative so it pulls the electrons in the bond slightly toward itself, giving the bond some polar character though it remains primarily covalent.
To insert an element at given position in arrayInsert(stack s1, s.pdf
To insert an element at given position in array:
Insert(stack s1, stack s2, int value, int pos)
//Inserts a value into the array at specified position.
//Assume s1 is a stack which holds the values, and s2 is an empty stack.
while(!s1.empty()) //Till the stack is empty.
s2.push(s1.pop()) //Pop element from s1, and push it in s2.
//By now, all elements are placed in reverse order from s1 to s2.
for i = 1 to pos - 1 //Till the pos-1
s1.push(s2.pop()) //Remove elements from s2, and push them to s1.
s1.push(value) //Now push your element in specified position pos.
while(!s2.empty()) //Push all the remaining elements in s2 to s1.
s1.push(s2.pop())
To delete an element at given position in array:
Delete(stack s1, stack s2, int pos)
//Deletes a value into the array at specified position.
//Assume s1 is a stack which holds the values, and s2 is an empty stack.
while(!s1.empty()) //Till the stack is empty.
s2.push(s1.pop()) //Pop element from s1, and push it in s2.
//By now, all elements are placed in reverse order from s1 to s2.
for i = 1 to pos - 1 //Till the pos-1
s1.push(s2.pop()) //Remove elements from s2, and push them to s1.
s1.pop(value) //Now pop your element in specified position pos.
while(!s2.empty()) //Push all the remaining elements in s2 to s1.
s1.push(s2.pop())
To delete a given element in array:
DeleteVal(stack s1, stack s2, int value)
//Deletes a value from the array.
//Assume s1 is a stack which holds the values, and s2 is an empty stack.
while(!s1.empty()) //Till the stack is empty.
s2.push(s1.pop()) //Pop element from s1, and push it in s2.
//By now, all elements are placed in reverse order from s1 to s2.
ele = s2.peek();
while(ele != value || !s2.empty()) //Till you found the element to be deleted.
s1.push(s2.pop()) //Remove elements from s2, and push them to s1.
ele = s2.peek()
s2.pop(value) //If element is found, delete it.
while(!s2.empty()) //Push all the remaining elements in s2 to s1.
s1.push(s2.pop())
If you have any queries with this, just get back to me.
Solution
To insert an element at given position in array:
Insert(stack s1, stack s2, int value, int pos)
//Inserts a value into the array at specified position.
//Assume s1 is a stack which holds the values, and s2 is an empty stack.
while(!s1.empty()) //Till the stack is empty.
s2.push(s1.pop()) //Pop element from s1, and push it in s2.
//By now, all elements are placed in reverse order from s1 to s2.
for i = 1 to pos - 1 //Till the pos-1
s1.push(s2.pop()) //Remove elements from s2, and push them to s1.
s1.push(value) //Now push your element in specified position pos.
while(!s2.empty()) //Push all the remaining elements in s2 to s1.
s1.push(s2.pop())
To delete an element at given position in array:
Delete(stack s1, stack s2, int pos)
//Deletes a value into the array at specified position.
//Assume s1 is a stack which holds the values, and s2 is an empty stack.
while(!s1.empty()) //Till the stack is empty.
s2.push(s1.pop()) //Pop element from s1, and push it in s2.
//By .
Their physical properties are so dofferent just because theintermole.pdf
Their physical properties are so dofferent just because theintermolecular or interatomic forces are
so different.
Solution
Their physical properties are so dofferent just because theintermolecular or interatomic forces are
so different..
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what ways are quorum sensing and the two component signaling system .pdf
what ways are quorum sensing and the two component signaling system .pdf
Average.javaimport java.util.Scanner; import java.util.Arra.pdf
Average.javaimport java.util.Scanner; import java.util.Arra.pdf
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yep in water which is higly polar will dissolve i.pdf
When thinking of E or Z, the first thing that com.pdf
When thinking of E or Z, the first thing that com.pdf
Since it contains oxygen double bonded to a carbo.pdf
Since it contains oxygen double bonded to a carbo.pdf
salicylic acid alcohol (R-OH) and carboxylic aci.pdf
salicylic acid alcohol (R-OH) and carboxylic aci.pdf
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It would be Ruthenium (II) Sulfide Ruthenium is .pdf
ketones do not give tollens test.. hence no react.pdf
ketones do not give tollens test.. hence no react.pdf
for IONIC the metal name followed by nonmetal io.pdf
for IONIC the metal name followed by nonmetal io.pdf
pH = 4-log2.4 = 3.62 pOH 14 - pH = 14-3.62 = 10.3.pdf
pH = 4-log2.4 = 3.62 pOH 14 - pH = 14-3.62 = 10.3.pdf
This suggest that dissociation of FeCl3 is 100 .pdf
This suggest that dissociation of FeCl3 is 100 .pdf
Total probability of dot received P = (37)(78).pdf
Total probability of dot received P = (37)(78).pdf
Environmental contaminants of recent concern are .pdf
Environmental contaminants of recent concern are .pdf
To insert an element at given position in arrayInsert(stack s1, s.pdf
To insert an element at given position in arrayInsert(stack s1, s.pdf
Their physical properties are so dofferent just because theintermole.pdf
Their physical properties are so dofferent just because theintermole.pdf
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www.StokeNewingtonHistory.com
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In inline heapsort we use only single array throughout the program t.pdf
- 1. In inline heapsort we use only single array throughout the program to decrease the running time, where as, in simple heap sort we use external data structure memory usage of more than O(1) at a time. Please consider the below inline heap program which prints out the running time, which is lesser to the the simple heap sort. //program class heapsort { private final int array[]; // The array being sorted heapsort(int array[], int N) { this.array = array; int Last = N-1; // its children are at elements 2*I+1 and 2*I+2. // phase 1: form heap // Construct heap bottom-up, starting with small trees just above leaves // and coalescing into larger trees near the root. for( int Top = Last/2; Top >= 0; Top-- ) { adjust(Top, Last); } // phase 2: use heap to sort // Move top element (largest) out of heap, swapping with last element // and changing the heap boundary, until only one element remains. while( Last > 0 ) { swap(0, Last); adjust(0, --Last); } } /** * adjust(Top, Last) adjusts the tree between Top and Last
- 2. **/ void adjust(int Top, int Last) { int TopVal = array[Top]; // Set aside top of heap int Parent, Child; for( Parent = Top; ; Parent = Child ) // Iterate down through tree { Child = 2*Parent+1; // Child means left child if( Child > Last ) break; // Left child non-existent if( Child+1 <= Last // Right child exists && array[Child] < array[Child+1] ) // and right child is larger Child++; // Child is the larger child if( TopVal >= array[Child] ) break; // Location for TopVal found array[Parent] = array[Child]; // Move larger child up in tree } array[Parent] = TopVal; // Install TopVal in place } //swapping void swap(int i, int j) { int temp = array[i]; array[i] = array[j]; array[j] = temp; } public static void main(String[] args) { int N = 0; // number of elements in array int array[] = {43,65,12,89,11,5,1,88,123,999,233,18,553,332,144,98,452,7663,111,23,654,8735,998}; //initialising array N=array.length; System.err.println("Sorting started");
- 3. long startTime = System.nanoTime(); new heapsort(array, N); // sorting long endTime = System.nanoTime(); long totalTime = endTime - startTime; for( int i = 0; i < N; i++ ) { System.out.print(array[i] + " "); } System.out.println(); System.out.println("Time in nanos"+totalTime); } } //runn the above program for inline heap sort. Solution In inline heapsort we use only single array throughout the program to decrease the running time, where as, in simple heap sort we use external data structure memory usage of more than O(1) at a time. Please consider the below inline heap program which prints out the running time, which is lesser to the the simple heap sort. //program class heapsort { private final int array[]; // The array being sorted heapsort(int array[], int N) { this.array = array; int Last = N-1; // its children are at elements 2*I+1 and 2*I+2. // phase 1: form heap // Construct heap bottom-up, starting with small trees just above leaves // and coalescing into larger trees near the root.
- 4. for( int Top = Last/2; Top >= 0; Top-- ) { adjust(Top, Last); } // phase 2: use heap to sort // Move top element (largest) out of heap, swapping with last element // and changing the heap boundary, until only one element remains. while( Last > 0 ) { swap(0, Last); adjust(0, --Last); } } /** * adjust(Top, Last) adjusts the tree between Top and Last **/ void adjust(int Top, int Last) { int TopVal = array[Top]; // Set aside top of heap int Parent, Child; for( Parent = Top; ; Parent = Child ) // Iterate down through tree { Child = 2*Parent+1; // Child means left child if( Child > Last ) break; // Left child non-existent if( Child+1 <= Last // Right child exists && array[Child] < array[Child+1] ) // and right child is larger Child++; // Child is the larger child if( TopVal >= array[Child] ) break; // Location for TopVal found array[Parent] = array[Child]; // Move larger child up in tree } array[Parent] = TopVal; // Install TopVal in place }
- 5. //swapping void swap(int i, int j) { int temp = array[i]; array[i] = array[j]; array[j] = temp; } public static void main(String[] args) { int N = 0; // number of elements in array int array[] = {43,65,12,89,11,5,1,88,123,999,233,18,553,332,144,98,452,7663,111,23,654,8735,998}; //initialising array N=array.length; System.err.println("Sorting started"); long startTime = System.nanoTime(); new heapsort(array, N); // sorting long endTime = System.nanoTime(); long totalTime = endTime - startTime; for( int i = 0; i < N; i++ ) { System.out.print(array[i] + " "); } System.out.println(); System.out.println("Time in nanos"+totalTime); } } //runn the above program for inline heap sort.