You can use the below function for linear interpolation i.e.#inclu.pdf
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You can use the below function for linear interpolation i.e. #include int NearestIndex( double value, double *x, int len ) { double dist; int index; int i; index = -1; dist = Double_Max; for ( i = 0; i < len; i++ ) { double newDist = value - x[i]; if ( newDist > 0 && newDist < dist ) { dist = newDist; index = i; } } return index; } void linearinterpolation(double *x, int x_tem, double *y, double *xx, int xx_tem, double *yy) { double da, db, *slp, *inter; int i, Envectorindice; slp=(double *)calloc(x_tem,sizeof(double)); inter=(double *)calloc(x_tem,sizeof(double)); for(i = 0; i < x_tem; i++){ if(i Solution You can use the below function for linear interpolation i.e. #include int NearestIndex( double value, double *x, int len ) { double dist; int index; int i; index = -1; dist = Double_Max; for ( i = 0; i < len; i++ ) { double newDist = value - x[i]; if ( newDist > 0 && newDist < dist ) { dist = newDist; index = i; } } return index; } void linearinterpolation(double *x, int x_tem, double *y, double *xx, int xx_tem, double *yy) { double da, db, *slp, *inter; int i, Envectorindice; slp=(double *)calloc(x_tem,sizeof(double)); inter=(double *)calloc(x_tem,sizeof(double)); for(i = 0; i < x_tem; i++){ if(i.
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![You can use the below function for linear interpolation i.e.
#include
int NearestIndex( double value, double *x, int len )
{
double dist;
int index;
int i;
index = -1;
dist = Double_Max;
for ( i = 0; i < len; i++ ) {
double newDist = value - x[i];
if ( newDist > 0 && newDist < dist ) {
dist = newDist;
index = i;
}
}
return index;
}
void linearinterpolation(double *x, int x_tem, double *y, double *xx, int xx_tem, double *yy)
{
double da, db, *slp, *inter;
int i, Envectorindice;
slp=(double *)calloc(x_tem,sizeof(double));
inter=(double *)calloc(x_tem,sizeof(double));
for(i = 0; i < x_tem; i++){
if(i
Solution
You can use the below function for linear interpolation i.e.
#include
int NearestIndex( double value, double *x, int len )
{
double dist;
int index;
int i;](https://image.slidesharecdn.com/youcanusethebelowfunctionforlinearinterpolationi-230409040907-f637047b/85/You-can-use-the-below-function-for-linear-interpolation-i-e-inclu-pdf-1-320.jpg)
![index = -1;
dist = Double_Max;
for ( i = 0; i < len; i++ ) {
double newDist = value - x[i];
if ( newDist > 0 && newDist < dist ) {
dist = newDist;
index = i;
}
}
return index;
}
void linearinterpolation(double *x, int x_tem, double *y, double *xx, int xx_tem, double *yy)
{
double da, db, *slp, *inter;
int i, Envectorindice;
slp=(double *)calloc(x_tem,sizeof(double));
inter=(double *)calloc(x_tem,sizeof(double));
for(i = 0; i < x_tem; i++){
if(i](data:image/gif;base64,R0lGODlhAQABAIAAAAAAAP///yH5BAEAAAAALAAAAAABAAEAAAIBRAA7)
Recommended
(Parent reference for BST) Redefine TreeNode by adding a reference to.pdf
(Parent reference for BST) Redefine TreeNode by adding a reference to a node\'s parent, as
shown below\': Reimplement the insert and delete methods in the BST class to update the parent
for each node in the tree. Add the following new method in BST:/** Returns the node for the
specified element. * Returns null if the element is not in the tree. */private TreeNode getNode(E
element)/** Returns true if the node for the element is a leaf */private boolean isLeaf(E
element)/** Returns the path of elements from the specified element * to the root in an array
list. */public ArrayList getPath(E e) Write a test program that prompts the user to enter 10
integers, adds them to the tree, deletes the first integer from the tree, and displays the paths for
all leaf nodes. Here is a sample run:
Solution
#include
#include
void swap(char *x, char *y)
{
char temp;
temp = *x;
*x = *y;
*y = temp;
void permute(char *a, int l, int r)
{
int i;
if (l == r)
printf(\"%s\ \", a);
else
{
for (i = l; i <= r; i++)
{
swap((a+l), (a+i));
permute(a, l+1, r);
swap((a+l), (a+i)); //backtrack
}
}
}
int main()
{
char str[] = \"ABC\";
int n = strlen(str);
permute(str, 0, n-1);
return 0;
}
part ii radix sort:
c++
#include
using namespace std;
int getMax(int arr[], int n)
{
int mx = arr[0];
for (int i = 1; i < n; i++)
if (arr[i] > mx)
mx = arr[i];
return main()
void countSort(int arr[], int n, int exp
for (i = 0; i < n; i++)
count[ (arr[i]/exp)%10 ]++;
for (i = 1; i < 10; i++)
count[i] += count[i - 1];
for (i = n - 1; i >= 0; i--)
{
output[count[ (arr[i]/exp)%10 ] - 1] = arr[i];
count[ (arr[i]/exp)%10 ]--;
}
for (i = 0; i < n; i++)
arr[i] = output[i];
}
adixsort(int arr[], int n)
{
int m = getMax(arr, n);
for (int exp = 1; m/exp > 0; exp *= 10)
countSort(arr, n, exp);
}
void print(int arr[], int n)
{
for (int i = 0; i < n; i++)
cout << arr[i] << \" \";
}
int main()
{
int arr[] = {170, 45, 75, 90, 802, 24, 2, 66};
int n = sizeof(arr)/sizeof(arr[0]);
radixsort(arr, n);
print(arr, n);
return 0;
}
java:
import java.io.*;
import java.util.*;
class Radix {
static int getMax(int arr[], int n)
{
int mx = arr[0];
for (int i = 1; i < n; i++)
if (arr[i] > mx)
mx = arr[i];
return mx;
}
static void countSort(int arr[], int n, int exp)
{
int output[] = new int[n]; // output array
int i;
int count[] = new int[10];
Arrays.fill(count,0);
for (i = 0; i < n; i++)
count[ (arr[i]/exp)%10 ]++;
for (i = 1; i < 10; i++)
count[i] += count[i - 1];
for (i = n - 1; i >= 0; i--)
{
output[count[ (arr[i]/exp)%10 ] - 1] = arr[i];
count[ (arr[i]/exp)%10 ]--;
}
for (i = 0; i < n; i++)
arr[i] = output[i];
}
static void radixsort(int arr[], int n)
{
int m = getMax(arr, n);
for (int exp = 1; m/exp > 0; exp *= 10)
countSort(arr, n, exp);
}
static void print(int arr[], int n)
{
for (int i=0; i
#include
void swap(char *x, char *y)
{
char temp;
temp = *x;
*x = *y;
*y = temp;
void permute(char *a, int l, int r)
{
int i;
if (l == r)
printf(\"%s\ \", a);
else
{
for (i = l; i <= r; i++)
{
swap((a+l), (a+i));
permute(a, l+1, .
program#include iostreamusing namespace std;void calculatio.pdf
program:
#include
using namespace std;
void calculations(double[], int);
void output(double[], int);
int main()
{
// variables declared
const int arraysize = 50;
double numbers[arraysize];
// calling the functions
calculations(numbers, arraysize);
output(numbers, arraysize);
return 0;
}
void calculations(double nums[], int size) {
for (int index = 0; index < 25; index++)
nums[index] = index * index;
for (int index = 25; index < size; index++)
nums[index] = 3 * index;
}
void output(double nums[], int size) {
int count = 0;
for (int i = 0; i < size; i++) {
cout << nums[i] << \" \";
++count;
if (count % 10 == 0)
cout << endl;
}
}
Solution
program:
#include
using namespace std;
void calculations(double[], int);
void output(double[], int);
int main()
{
// variables declared
const int arraysize = 50;
double numbers[arraysize];
// calling the functions
calculations(numbers, arraysize);
output(numbers, arraysize);
return 0;
}
void calculations(double nums[], int size) {
for (int index = 0; index < 25; index++)
nums[index] = index * index;
for (int index = 25; index < size; index++)
nums[index] = 3 * index;
}
void output(double nums[], int size) {
int count = 0;
for (int i = 0; i < size; i++) {
cout << nums[i] << \" \";
++count;
if (count % 10 == 0)
cout << endl;
}
}.
DoublyList-cpp- #include -DoublyList-h- using namespace std- void Doub.pdf
DoublyList.cpp:
#include "DoublyList.h"
using namespace std;
void DoublyList::insertFront(int newData)
{
if (first == nullptr)
{
first = new DLLNode(newData, nullptr, nullptr);
last = first;
// Common error: Forgetting to reset pointer last.
}
else
{
first = new DLLNode(newData, nullptr, first);
first->getNext()->setPrev(first);
// Common error: Forgetting to connect pointer
// prev of what is now the second node to the
// new first node.
}
++count;
}
void DoublyList::printForward() const
{
DLLNode* current = first;
while (current != nullptr)
{
cout << current->getData() << " ";
current = current->getNext();
}
}
void DoublyList::printReverse() const
{
DLLNode* current = last;
while (current != nullptr)
{
cout << current->getData() << " ";
current = current->getPrev();
}
}
void DoublyList::clearList()
{
DLLNode* temp = first;
while (first != nullptr)
{
first = first->getNext();
delete temp;
temp = first;
}
last = nullptr;
// Don't forget to reset pointer last to nullptr.
count = 0;
}
DoublyList::~DoublyList()
{
if (first != nullptr)
clearList();
}
DoublyList.h
#ifndef DOUBLYLIST_H
#define DOUBLYLIST_H
#include <string>
#include <iostream>
class DLLNode
{
public:
DLLNode() : data(0), prev(nullptr), next(nullptr) {}
DLLNode(int theData, DLLNode* prevLink, DLLNode* nextLink)
: data(theData), prev(prevLink), next(nextLink) {}
int getData() const { return data; }
DLLNode* getPrev() const { return prev; }
DLLNode* getNext() const { return next; }
void setData(int theData) { data = theData; }
void setPrev(DLLNode* prevLink) { prev = prevLink; }
void setNext(DLLNode* nextLink) { next = nextLink; }
~DLLNode(){}
private:
int data; // To simplify, we are using only one piece of data.
DLLNode* prev;
DLLNode* next;
};
class DoublyList
{
public:
DoublyList() : first(nullptr), last(nullptr), count(0) {}
void insertFront(int newData);
void printForward() const;
void printReverse() const;
void rotateNodesRight(int);
void clearList();
~DoublyList();
private:
// Pointer to the first node in the list.
DLLNode*first;
// Pointer to the last node in the list.
DLLNode*last;
// Number of nodes in the list.
int count;
};
#endif
Main.cpp
#include "DoublyList.h"
#include <iostream>
#include <vector>
using namespace std;
int main()
{
vector<vector<int>> data = {
{25, 76, 35, 67, 15, 98},
{1, 2, 3, 4, 5, 6, 7, 8, 9},
{10, 20},
{34, 56, 78, 12, 89, 34, 76, 28, 54, 22, 41},
{123, 873, 619},
};
vector<int> nodesToRotate = { 2, 3, 1, 9, 2 };
{
DoublyList doublyList;
int vectorSize = static_cast<int>(data.size());
for (int i = 0; i < vectorSize; ++i)
{
int innerSize = static_cast<int>(data[i].size());
for (int j = innerSize - 1; j >= 0; --j)
doublyList.insertFront(data[i].at(j));
cout << "Rotate right: " << nodesToRotate[i] << "\n";
cout << " List is: ";
doublyList.printForward();
cout << "\n";
doublyList.rotateNodesRight(nodesToRotate[i]);
cout << "After rotating:";
cout << "\n Print forward: ";
doublyList.printForward();
cout << "\nPrint backwards: ";
doublyList.printReverse(.
Find attached the code to implement the following two functions- 1- In.pdf
Fill in the answer/essay formatted box with a TYPED and full solution including full calculator
syntax as shown to students in the applicable the class lecture video. Note: Answer alone will
result in zero credit if appropriate full calculator syntax is not provided that supports the correct
solution and please following the rounding rule given. No file uploads are allowed, you must
type out your full and complete solution. [Students, as mentioned already, cannot just use the
standard normal table in this course, but must show mastery of the appropriate calculator
function.] Question: Weights of 2 y r old females are normally distributed with mean of 26.0
pounds and standard deviation of 2.8 pounds. Give the weight of a 2 yr old female that is the 62
nd percentile. Give solution that is rounded to the nearest tenth..
Can someone help me to fix the code please package dlist i.pdf
Can someone help me to fix the code please?
package dlist;
import java.util.Iterator;
public class DList implements Iterable<String> {
private static class DListNode {
public String data;
public DListNode next;
public DListNode previous;
}
private DListNode nil;
private int size;
public DList() {
nil = new DListNode();
nil.previous = nil;
nil.next = nil;
nil.data = null;
size = 0;
}
public void addFirst(String elem) {
DListNode newNode = new DListNode();
newNode.data = elem;
newNode.next = nil.next;
newNode.previous = nil;
nil.next.previous = newNode;
nil.next = newNode;
size++;
}
public void addLast(String elem) {
DListNode newNode = new DListNode();
newNode.data = elem;
newNode.next = nil;
newNode.previous = nil.previous;
nil.previous.next = newNode;
nil.previous = newNode;
size++;
}
public String getFirst() {
if (size == 0) {
throw new RuntimeException("DList is empty");
}
return nil.next.data;
}
public String getLast() {
if (size == 0) {
throw new RuntimeException("DList is empty");
}
return nil.previous.data;
}
public String removeFirst() {
if (size == 0) {
throw new RuntimeException("DList is empty");
}
DListNode first = nil.next;
first.next.previous = nil;
nil.next = first.next;
size--;
return first.data;
}
public String removeLast() {
if (size == 0) {
throw new RuntimeException("DList is empty");
}
DListNode last = nil.previous;
last.previous.next = nil;
nil.previous = last.previous;
size--;
return last.data;
}
public String get(int index) {
if (index < 0 || index >= size) {
throw new IndexOutOfBoundsException();
}
DListNode node = nil.next;
for (int i = 0; i < index; i++) {
node = node.next;
}
return node.data;
}
public String set(int index, String value) {
if (index < 0 || index >= size) {
throw new IndexOutOfBoundsException();
}
DListNode node = nil.next;
for (int i = 0; i < index; i++) {
node = node.next;
}
String oldValue = node.data;
node.data = value;
return oldValue;
}
public boolean contains(Object obj) {
DListNode node = nil.next;
while (node != nil) {
if (node.data.equals(obj)) {
return true;
}
node = node.next;
}
return false;
}
public int size() {
return size;
}
public int indexOf(Object obj)
{
int index = 0;
DListNode node = nil.next;
while (node != nil)
{
if (node.data.equals(obj))
{
return index;
}
index++;
node = node.next;
}
return -1;
}
@Override
public Iterator<String> iterator()
{
return new DListIterator();
}
private class DListIterator implements Iterator<String>
{
private DListNode pointer;
public DListIterator()
{
if(nil.next==nil)
{
pointer = nil;
}
else
pointer=nil.next;
}
public boolean hasNext()
{
return pointer != nil;
}
}
}.
Recommended
(Parent reference for BST) Redefine TreeNode by adding a reference to.pdf
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shown below\': Reimplement the insert and delete methods in the BST class to update the parent
for each node in the tree. Add the following new method in BST:/** Returns the node for the
specified element. * Returns null if the element is not in the tree. */private TreeNode getNode(E
element)/** Returns true if the node for the element is a leaf */private boolean isLeaf(E
element)/** Returns the path of elements from the specified element * to the root in an array
list. */public ArrayList getPath(E e) Write a test program that prompts the user to enter 10
integers, adds them to the tree, deletes the first integer from the tree, and displays the paths for
all leaf nodes. Here is a sample run:
Solution
#include
#include
void swap(char *x, char *y)
{
char temp;
temp = *x;
*x = *y;
*y = temp;
void permute(char *a, int l, int r)
{
int i;
if (l == r)
printf(\"%s\ \", a);
else
{
for (i = l; i <= r; i++)
{
swap((a+l), (a+i));
permute(a, l+1, r);
swap((a+l), (a+i)); //backtrack
}
}
}
int main()
{
char str[] = \"ABC\";
int n = strlen(str);
permute(str, 0, n-1);
return 0;
}
part ii radix sort:
c++
#include
using namespace std;
int getMax(int arr[], int n)
{
int mx = arr[0];
for (int i = 1; i < n; i++)
if (arr[i] > mx)
mx = arr[i];
return main()
void countSort(int arr[], int n, int exp
for (i = 0; i < n; i++)
count[ (arr[i]/exp)%10 ]++;
for (i = 1; i < 10; i++)
count[i] += count[i - 1];
for (i = n - 1; i >= 0; i--)
{
output[count[ (arr[i]/exp)%10 ] - 1] = arr[i];
count[ (arr[i]/exp)%10 ]--;
}
for (i = 0; i < n; i++)
arr[i] = output[i];
}
adixsort(int arr[], int n)
{
int m = getMax(arr, n);
for (int exp = 1; m/exp > 0; exp *= 10)
countSort(arr, n, exp);
}
void print(int arr[], int n)
{
for (int i = 0; i < n; i++)
cout << arr[i] << \" \";
}
int main()
{
int arr[] = {170, 45, 75, 90, 802, 24, 2, 66};
int n = sizeof(arr)/sizeof(arr[0]);
radixsort(arr, n);
print(arr, n);
return 0;
}
java:
import java.io.*;
import java.util.*;
class Radix {
static int getMax(int arr[], int n)
{
int mx = arr[0];
for (int i = 1; i < n; i++)
if (arr[i] > mx)
mx = arr[i];
return mx;
}
static void countSort(int arr[], int n, int exp)
{
int output[] = new int[n]; // output array
int i;
int count[] = new int[10];
Arrays.fill(count,0);
for (i = 0; i < n; i++)
count[ (arr[i]/exp)%10 ]++;
for (i = 1; i < 10; i++)
count[i] += count[i - 1];
for (i = n - 1; i >= 0; i--)
{
output[count[ (arr[i]/exp)%10 ] - 1] = arr[i];
count[ (arr[i]/exp)%10 ]--;
}
for (i = 0; i < n; i++)
arr[i] = output[i];
}
static void radixsort(int arr[], int n)
{
int m = getMax(arr, n);
for (int exp = 1; m/exp > 0; exp *= 10)
countSort(arr, n, exp);
}
static void print(int arr[], int n)
{
for (int i=0; i
#include
void swap(char *x, char *y)
{
char temp;
temp = *x;
*x = *y;
*y = temp;
void permute(char *a, int l, int r)
{
int i;
if (l == r)
printf(\"%s\ \", a);
else
{
for (i = l; i <= r; i++)
{
swap((a+l), (a+i));
permute(a, l+1, .
program#include iostreamusing namespace std;void calculatio.pdf
program:
#include
using namespace std;
void calculations(double[], int);
void output(double[], int);
int main()
{
// variables declared
const int arraysize = 50;
double numbers[arraysize];
// calling the functions
calculations(numbers, arraysize);
output(numbers, arraysize);
return 0;
}
void calculations(double nums[], int size) {
for (int index = 0; index < 25; index++)
nums[index] = index * index;
for (int index = 25; index < size; index++)
nums[index] = 3 * index;
}
void output(double nums[], int size) {
int count = 0;
for (int i = 0; i < size; i++) {
cout << nums[i] << \" \";
++count;
if (count % 10 == 0)
cout << endl;
}
}
Solution
program:
#include
using namespace std;
void calculations(double[], int);
void output(double[], int);
int main()
{
// variables declared
const int arraysize = 50;
double numbers[arraysize];
// calling the functions
calculations(numbers, arraysize);
output(numbers, arraysize);
return 0;
}
void calculations(double nums[], int size) {
for (int index = 0; index < 25; index++)
nums[index] = index * index;
for (int index = 25; index < size; index++)
nums[index] = 3 * index;
}
void output(double nums[], int size) {
int count = 0;
for (int i = 0; i < size; i++) {
cout << nums[i] << \" \";
++count;
if (count % 10 == 0)
cout << endl;
}
}.
DoublyList-cpp- #include -DoublyList-h- using namespace std- void Doub.pdf
DoublyList.cpp:
#include "DoublyList.h"
using namespace std;
void DoublyList::insertFront(int newData)
{
if (first == nullptr)
{
first = new DLLNode(newData, nullptr, nullptr);
last = first;
// Common error: Forgetting to reset pointer last.
}
else
{
first = new DLLNode(newData, nullptr, first);
first->getNext()->setPrev(first);
// Common error: Forgetting to connect pointer
// prev of what is now the second node to the
// new first node.
}
++count;
}
void DoublyList::printForward() const
{
DLLNode* current = first;
while (current != nullptr)
{
cout << current->getData() << " ";
current = current->getNext();
}
}
void DoublyList::printReverse() const
{
DLLNode* current = last;
while (current != nullptr)
{
cout << current->getData() << " ";
current = current->getPrev();
}
}
void DoublyList::clearList()
{
DLLNode* temp = first;
while (first != nullptr)
{
first = first->getNext();
delete temp;
temp = first;
}
last = nullptr;
// Don't forget to reset pointer last to nullptr.
count = 0;
}
DoublyList::~DoublyList()
{
if (first != nullptr)
clearList();
}
DoublyList.h
#ifndef DOUBLYLIST_H
#define DOUBLYLIST_H
#include <string>
#include <iostream>
class DLLNode
{
public:
DLLNode() : data(0), prev(nullptr), next(nullptr) {}
DLLNode(int theData, DLLNode* prevLink, DLLNode* nextLink)
: data(theData), prev(prevLink), next(nextLink) {}
int getData() const { return data; }
DLLNode* getPrev() const { return prev; }
DLLNode* getNext() const { return next; }
void setData(int theData) { data = theData; }
void setPrev(DLLNode* prevLink) { prev = prevLink; }
void setNext(DLLNode* nextLink) { next = nextLink; }
~DLLNode(){}
private:
int data; // To simplify, we are using only one piece of data.
DLLNode* prev;
DLLNode* next;
};
class DoublyList
{
public:
DoublyList() : first(nullptr), last(nullptr), count(0) {}
void insertFront(int newData);
void printForward() const;
void printReverse() const;
void rotateNodesRight(int);
void clearList();
~DoublyList();
private:
// Pointer to the first node in the list.
DLLNode*first;
// Pointer to the last node in the list.
DLLNode*last;
// Number of nodes in the list.
int count;
};
#endif
Main.cpp
#include "DoublyList.h"
#include <iostream>
#include <vector>
using namespace std;
int main()
{
vector<vector<int>> data = {
{25, 76, 35, 67, 15, 98},
{1, 2, 3, 4, 5, 6, 7, 8, 9},
{10, 20},
{34, 56, 78, 12, 89, 34, 76, 28, 54, 22, 41},
{123, 873, 619},
};
vector<int> nodesToRotate = { 2, 3, 1, 9, 2 };
{
DoublyList doublyList;
int vectorSize = static_cast<int>(data.size());
for (int i = 0; i < vectorSize; ++i)
{
int innerSize = static_cast<int>(data[i].size());
for (int j = innerSize - 1; j >= 0; --j)
doublyList.insertFront(data[i].at(j));
cout << "Rotate right: " << nodesToRotate[i] << "\n";
cout << " List is: ";
doublyList.printForward();
cout << "\n";
doublyList.rotateNodesRight(nodesToRotate[i]);
cout << "After rotating:";
cout << "\n Print forward: ";
doublyList.printForward();
cout << "\nPrint backwards: ";
doublyList.printReverse(.
Find attached the code to implement the following two functions- 1- In.pdf
Fill in the answer/essay formatted box with a TYPED and full solution including full calculator
syntax as shown to students in the applicable the class lecture video. Note: Answer alone will
result in zero credit if appropriate full calculator syntax is not provided that supports the correct
solution and please following the rounding rule given. No file uploads are allowed, you must
type out your full and complete solution. [Students, as mentioned already, cannot just use the
standard normal table in this course, but must show mastery of the appropriate calculator
function.] Question: Weights of 2 y r old females are normally distributed with mean of 26.0
pounds and standard deviation of 2.8 pounds. Give the weight of a 2 yr old female that is the 62
nd percentile. Give solution that is rounded to the nearest tenth..
Can someone help me to fix the code please package dlist i.pdf
Can someone help me to fix the code please?
package dlist;
import java.util.Iterator;
public class DList implements Iterable<String> {
private static class DListNode {
public String data;
public DListNode next;
public DListNode previous;
}
private DListNode nil;
private int size;
public DList() {
nil = new DListNode();
nil.previous = nil;
nil.next = nil;
nil.data = null;
size = 0;
}
public void addFirst(String elem) {
DListNode newNode = new DListNode();
newNode.data = elem;
newNode.next = nil.next;
newNode.previous = nil;
nil.next.previous = newNode;
nil.next = newNode;
size++;
}
public void addLast(String elem) {
DListNode newNode = new DListNode();
newNode.data = elem;
newNode.next = nil;
newNode.previous = nil.previous;
nil.previous.next = newNode;
nil.previous = newNode;
size++;
}
public String getFirst() {
if (size == 0) {
throw new RuntimeException("DList is empty");
}
return nil.next.data;
}
public String getLast() {
if (size == 0) {
throw new RuntimeException("DList is empty");
}
return nil.previous.data;
}
public String removeFirst() {
if (size == 0) {
throw new RuntimeException("DList is empty");
}
DListNode first = nil.next;
first.next.previous = nil;
nil.next = first.next;
size--;
return first.data;
}
public String removeLast() {
if (size == 0) {
throw new RuntimeException("DList is empty");
}
DListNode last = nil.previous;
last.previous.next = nil;
nil.previous = last.previous;
size--;
return last.data;
}
public String get(int index) {
if (index < 0 || index >= size) {
throw new IndexOutOfBoundsException();
}
DListNode node = nil.next;
for (int i = 0; i < index; i++) {
node = node.next;
}
return node.data;
}
public String set(int index, String value) {
if (index < 0 || index >= size) {
throw new IndexOutOfBoundsException();
}
DListNode node = nil.next;
for (int i = 0; i < index; i++) {
node = node.next;
}
String oldValue = node.data;
node.data = value;
return oldValue;
}
public boolean contains(Object obj) {
DListNode node = nil.next;
while (node != nil) {
if (node.data.equals(obj)) {
return true;
}
node = node.next;
}
return false;
}
public int size() {
return size;
}
public int indexOf(Object obj)
{
int index = 0;
DListNode node = nil.next;
while (node != nil)
{
if (node.data.equals(obj))
{
return index;
}
index++;
node = node.next;
}
return -1;
}
@Override
public Iterator<String> iterator()
{
return new DListIterator();
}
private class DListIterator implements Iterator<String>
{
private DListNode pointer;
public DListIterator()
{
if(nil.next==nil)
{
pointer = nil;
}
else
pointer=nil.next;
}
public boolean hasNext()
{
return pointer != nil;
}
}
}.
1- please please please don't send me an incomplete program this code.docx
1- please please please don't send me an incomplete program this code has an error please solve it.(Find attached the code to implement the following two functions:
- Insert in Interval Heap
- Display minimum and maximum values
The interval heap is implemented as a one-dimensional array. The even indices store the minimum heap while odd indices store the max heap.
You can extend this code to add the function for deleting minimum or maximum values.)
#include <stdio.h>
void swap(int a[], int n1, int n2) {
int temp = a[n1];
a[n1] = a[n2];
a[n2] = temp;
}
int get_parent(int n) {
if (!(n % 4)) return (n - 4) / 2;
if (!(n % 2)) return (n - 2) / 2;
if (((n - 1) % 4)) return (n - 3) / 2;
else return (n - 1) / 2;
}
void heapify_max(int a[], int n) {
int parent = get_parent(n);
while (n > 1 && a[n] > a[parent]) {
swap(a, n, parent);
n = parent;
parent = get_parent(n);
}
}
void heapify_min(int a[], int n) {
int parent = get_parent(n);
while (n > 0 && a[n] < a[parent]) {
swap(a, n, parent);
n = parent;
parent = get_parent(n);
}
}
void insert_intervalheap(int a[], int *pos, int val) {
int n = *pos, parent, parent1;
(*pos)++;
a[n] = val;
if (n == 0) return;
if (n == 1) {
if (a[0] > a[1]) swap(a, 0, 1);
return;
}
if (!(n % 2)) {
parent = get_parent(n);
if (a[n] < a[parent]) {
swap(a, n, parent);
heapify_min(a, parent);
}
} else {
parent1 = get_parent(n) + 1;
if (a[n] < a[parent]) {
swap(a, n, parent);
heapify_min(a, parent);
}
if (a[n] > a[parent1]) {
swap(a, n, parent1);
heapify_max(a, parent1);
}
}
}
int get_min(int a[], int n) {
return (a[0]);
}
int get_max(int a[], int n) {
if (n > 1) {
return (a[1]);
} else {
return (a[0]);
}
}
int delete_min(int a[], int *pos) {
if (*pos == 0) {
printf("Interval heap is empty. Cannot delete minimum value.\n");
return -1;
}
int min = a[0];
*pos -= 1;
a[0] = a[*pos];
heapify_min(a, 0);
return min;
}
int delete_max(int a[], int *pos) {
if (*pos == 0) {
printf("Interval heap is empty. Cannot delete maximum value.\n");
return -1;
}
if (*pos == 1) {
int max = a[1];
*pos -= 1;
return max;
}
int max = a[1], n = *pos;
if (a[n-1] > max) {
a[1] = a[n-1];
heapify_max(a, 1);
}
else {
a[1] = a[n-2];
heapify_max(a, 1);
}
*pos -= 1;
return max;
}
int main() {
int heap[MAX_SIZE];
int size = 0;
insert_intervalheap(heap, &size, 10);
insert_intervalheap(heap, &size, 15);
insert_intervalheap(heap, &size, 20);
insert_intervalheap(heap, &size, 25);
insert_intervalheap(heap, &size, 30);
insert_intervalheap(heap, &size, 35);
insert_intervalheap(heap, &size, 40);
printf("Interval heap: ");
for (int i = 0; i < size; i++) {
printf("%d ", heap[i]);
}
printf("\n");
int min = delete_min(heap, &size);
printf("Minimum value: %d\n", min);
int max = delete_max(heap, &size);
printf("Maximum value: %d\n", max);
printf("Interval heap after deletion: ");
for (int i = 0; i < size; i++) {
printf("%d ", heap[i]);
}
printf("\n");
return 0;
}
2- Find a.
#include stdafx.h using namespace std; #include stdlib.h.docx
#include stdafx.h using namespace std; #include stdlib.h
SaveI need help with this maze gui that I wrote in java, I am tryi.pdf
Save
I need help with this maze gui that I wrote in java, I am trying to find the shortest path but I dont
know how to implement the FindshortestPath Method.
I was given this as a stragedy to solve the findshortestpathMethod():
Create two LinkedLists of MazeMap objects. One will be the list of map objects for cells at
distance x from the center. The other will be the list of map objects for cells at distance x + 1.
Initialize the list at distance x to contain a MazeMap object which refers to the center cell of the
maze. Since this map object refers to the center cell, it does not need to refer to any other
MazeMap object, so its MazeMap reference can be set to null. Set the center cell’s visited field
to true to indicate that this cell has already been added to the map.
Set up a loop to run until the maze entrance is mapped. This loop does the following:
Set up a loop to process each map object from the list at distance x. This loop does the following:
i.Get the cell from the map object
ii.If this cell is the entrance cell, set the entrance and break out of the loop!
iii.Check each direction from this cell and for an accessible neighbor cell
Get a reference to that neighbor cell
If the neighbor cell has not already been visited
Set the cell’s visited field to true so its only processed once
Create a MazeMap object that refers to this neighbor cell
Make this MazeMap object refer to the current map object being processed – ie the cell one step
closer to center
Add this MazeMap object to the list at distance x + 1
List at distance x has been processed. The list at distance x + 1 is now complete, so make the list
at distance x refer to list at distance x + 1. Then create a new empty list for the list at distance x +
1.
At this point, the entrance points at the map object which refers to cell 1,1. That map object also
refers to a map object which is one step closer to the center. Write a loop that traverses this trail
of references until it reaches the end (a null reference). For each map object:
Get the cell referred to.
Call the drawCircle method with a color, the cell’s row and column, and SMALL.
Call the delay method with a SHORT delay. This slows down the drawing process so you can
watch the path being displayed.
Advance to the next map object in the list.
here is the code so far*********
here is my code: ******mazeApp*******
package mazepackage;
import javafx.application.Application;
import javafx.application.Platform;
import javafx.scene.Group;
import javafx.scene.Scene;
import javafx.scene.canvas.Canvas;
import javafx.scene.canvas.GraphicsContext;
import javafx.scene.control.TextInputDialog;
import javafx.scene.paint.Color;
import javafx.stage.Stage;
public class MazeApp extends Application
{
// default canvas size is DEFAULT_SIZE-by-DEFAULT_SIZE
private static final int DEFAULT_SIZE = 768;
private int width = DEFAULT_SIZE;
private int height = DEFAULT_SIZE;
// The graphics context is needed to enable drawing on the canvas
private Graphics.
Need done for Date Structures please! 4-18 LAB- Sorted number list imp.pdf
Need done for Date Structures please!
4.18 LAB: Sorted number list implementation with linked lists
Step 1: Inspect the Node.h file
Inspect the class declaration for a doubly-linked list node in Node.h. Access Node.h by clicking
on the orange arrow next to main.cpp at the top of the coding window. The Node class has three
member variables:
a double data value,
a pointer to the next node, and
a pointer to the previous node.
Each member variable is protected. So code outside of the class must use the provided getter and
setter member functions to get or set a member variable.
Node.h is read only, since no changes are required.
Step 2: Implement the Insert() member function
A class for a sorted, doubly-linked list is declared in SortedNumberList.h. Implement the
SortedNumberList class's Insert() member function. The function must create a new node with
the parameter value, then insert the node into the proper sorted position in the linked list. Ex:
Suppose a SortedNumberList's current list is 23 47.25 86, then Insert(33.5) is called. A new node
with data value 33.5 is created and inserted between 23 and 47.25, thus preserving the list's
sorted order and yielding: 23 35.5 47.25 86
Step 3: Test in develop mode
Code in main() takes a space-separated list of numbers and inserts each into a SortedNumberList.
The list is displayed after each insertion. Ex: If input is
then output is:
Try various program inputs, ensuring that each outputs a sorted list.
Step 4: Implement the Remove() member function
Implement the SortedNumberList class's Remove() member function. The function takes a
parameter for the number to be removed from the list. If the number does not exist in the list, the
list is not changed and false is returned. Otherwise, the first instance of the number is removed
from the list and true is returned.
Uncomment the commented-out part in main() that reads a second input line and removes
numbers from the list. Test in develop mode to ensure that insertion and removal both work
properly, then submit code for grading. Ex: If input is
then output is:
main.cpp
#include <iostream>
#include <string>
#include <vector>
#include "Node.h"
#include "SortedNumberList.h"
using namespace std;
void PrintList(SortedNumberList& list);
vector<string> SpaceSplit(string source);
int main(int argc, char *argv[]) {
// Read the line of input numbers
string inputLine;
getline(cin, inputLine);
// Split on space character
vector<string> terms = SpaceSplit(inputLine);
// Insert each value and show the sorted list's contents after each insertion
SortedNumberList list;
for (auto term : terms) {
double number = stod(term);
cout << "List after inserting " << number << ": " << endl;
list.Insert(number);
PrintList(list);
}
/*
// Read the input line with numbers to remove
getline(cin, inputLine);
terms = SpaceSplit(inputLine);
// Remove each value
for (auto term : terms) {
double number = stod(term);
cout << "List after removing " << number << ": " << endl;
list.Remove(number.
C++ code, please help! RESPOND W COMPLETED CODE PLEASE, am using V.pdf
C++ code, please help! RESPOND W/ COMPLETED CODE PLEASE, am using Visual Studio
Code.
Error message:
ld: Undefined symbols:
parseName(std::__1::basic_stringstream, std::__1::allocator>&), referenced from:
parseAssignments(std::__1::basic_stringstream, std::__1::allocator>&) in module-a187a7.o
SymbolTable::init(), referenced from:
_main in module-a187a7.o
parseAssignments(std::__1::basic_stringstream, std::__1::allocator>&) in module-a187a7.o
SymbolTable::insert(std::__1::basic_string, std::__1::allocator>, int), referenced from:
parseAssignments(std::__1::basic_stringstream, std::__1::allocator>&) in module-a187a7.o
SubExpression::parse(std::__1::basic_stringstream, std::__1::allocator>&), referenced from:
_main in module-a187a7.o
clang: error: linker command failed with exit code 1 (use -v to see invocation)
[Done] exited with code=1 in 1.478 seconds
divide.h
class Divide : public SubExpression
{
public:
//define the default construtor
Divide(Expression* left, Expression* right) : SubExpression(left, right)
{
}
//define the function evaluate()
int evaluate()
{
//divide the value of left and value of the right
//and return the value.
return left->evaluate() / right->evaluate();
}
};
expression.h
// Expression
class Expression
{
public:
//declare a virtual function evaluate()
virtual int evaluate() = 0;
};
literal.h
//Operand
class Literal : public Operand
{
public:
//define the construtor
Literal(int value)
{
this->value = value;
}
//define the function evaluate()
//returns the value
int evaluate()
{
return value;
}
private:
int value;
};
minus.h
//define the class Minus subclass of the SubExpression
class Minus : public SubExpression
{
public:
//define the default construtor
Minus(Expression* left, Expression* right) : SubExpression(left, right)
{
}
//define the function evaluate()
int evaluate()
{
//subtract the value of right from the value of the left
//and return the value.
return left->evaluate() - right->evaluate();
}
};
module.cpp
#include
#include
#include
#include
#include
using namespace std;
#include "expression.h"
#include "subexpression.h"
#include "symboltable.h"
#include "parse.h"
//create an object of SymbolTable
SymbolTable symbolTable;
//prototype of the function
void parseAssignments(stringstream& in);
//define main function
int main()
{
// declare the variables
Expression* expression;
char paren, comma;
string line;
// create an input file stream
ifstream fin("input.txt");
// check, if the file is not opened
//then display a error message
if (!fin.is_open())
perror("error while opening file");
//use a loop, to read the content from the file
while (getline(fin, line))
{
symbolTable.init();
if (!fin)
break;
stringstream in(line, ios_base::in);
in >> paren;
cout << line << " ";
expression = SubExpression::parse(in);
in >> comma;
//call the function
parseAssignments(in);
//Display the result
int result = expression->evaluate();
cout << "Value = " << result << endl;
}
system("pause");
return 0;
}
//definition of the function p.
#include stdafx.h #include iostream using namespace std;vo.docx
#include stdafx.h #include iostream using namespace std;vo
Task based Programming with OmpSs and its Application
Charla de la Dra. Rosa M. Badia Sala del Barcelona Supercomputing Center impartida el 4 de Noviembre en la Facultad de Informática
To complete the task, you need to fill in the missing code. I’ve inc.pdf
To complete the task, you need to fill in the missing code. I’ve included code to create an
Iterator. An Iterator is an object that iterates over another object – in this case, a circular linked
list. You can use the .next() method to advance the Iterator to the next item (the first time you
call it, the iterator will travel to node at index 0). Using iterator’s .remove() removes the node the
iterator is currently at. Say that we had a CircularLinkedList that looked like this: A ==> B ==>
C ==> D ==> E ==> Calling .next() three times will advance the iterator to index 2. Calling
.remove() once will remove the node at index 2. A ==> B ==> D ==> E ==> Calling .remove()
once more will remove the node now at index 2. A ==> B ==> E ==> The Iterator methods
handle wrapping around the CircularLinkedList. Be sure to create the iterator using l.iterator()
and after you’ve added all the nodes to the list, where l is your CircularLinkedList
Solution
import java.util.Iterator;
class CircularLinkedList implements Iterable {
// Your variables
// You can include a reference to a tail if you want
Node head;
int size; // BE SURE TO KEEP TRACK OF THE SIZE
// implement this constructor
public CircularLinkedList() {
head=null;
}
// writing helper functions for add and remove, like the book did can help
// but remember, the last element\'s next node will be the head!
// attach a node to the end of the list
// Be sure to handle the adding to an empty list
// always returns true
public boolean add(E e) {
Node newNode=new Node(e);
if(size==0){
head=newNode;
}
else{
Node last=getNode(size-1);
last.next=newNode;
}
newNode.next=head; //last element node is set to head
size++;
return true;
}
// need to handle
// out of bounds
// empty list
// adding to front
// adding to middle
// adding to \"end\"
// REMEMBER TO INCREMENT THE SIZE
public boolean add(int index, E e){
if(index>size) return false;
Node tmp=new Node(e);
if(index==0){
tmp.next=head;
Node last=getNode(size-1);
head=tmp;
last.next=head;
}
else {
Node curr=getNode(index-1);
tmp.next=curr.next;
curr.next=tmp;
}
size++;
return true;
}
// I highly recommend using this helper method
// Return Node found at the specified index
// be sure to handle out of bounds cases
private Node getNode(int index ) {
Node prev=head;
for(int i=0;isize) {
e=null;
}
else if(index==0){
e = head.getElement();
Node last=getNode(size-1);
head=head.next;
last.next=head;
size--;
}
else{
Node prev=getNode(index-1);
Node curr=getNode(index);
e=curr.getElement();
prev.next=curr.next;
size--;
}
return e;
}
// Turns your list into a string
// Useful for debugging
public String toString(){
Node current = head;
StringBuilder result = new StringBuilder();
if(size == 0){
return \"\";
}
if(size == 1) {
return head.getElement().toString();
}
else{
do{
result.append(current.getElement());
result.append(\" ==> \");
current = current.next;
} while(current != head);
}
return result.toString();
}
public Iterator iterator() {
return new ListIterator();
}
/.
HinduArabic decimal numbers use the digits 0 1 2 3 4 5.pdf
Hindu-Arabic decimal numbers use the digits 0, 1, 2, 3, 4, 5, 6, 7, 8, 9. We form decimal numbers
creating strings with these digits, for example 102, 88, 12.
1 0
Recall that, for example, 12 is 1x10 + 2x10
Binary numbers use only 0 and 1. We can create numbers by creating strings with 0s and 1s.
Example:
0
1 is expressed as 1 (1x2 )1
0
2 is expressed as 10 (1x2 + 0x2 )
1 0
3 is expressed as 11 (1x2 + 1x2 )
The first 5 integers are 1, 10, 11, 100, 101
The first 8 integers are expressed as 1, 10, 11, 100, 101, 110, 111, 1000
The first 16 integers are expressed as 1, 10, 11, 100, 101, 110, 111, 1000, 1001, 1010, 1011,
1100, 1101, 1110, 1111, 10000
Write a non-member method binaryNumberSequence(int n) that produces the sequence of the
first N integers.
The method receives as parameter a number n, and returns a Queue<String> with n strings
representing the first n integers converted to binary.
The elements in the queue come in the increasing sequence order of the number. For example, a
call to binaryNumberSequence(5) returns Q = {1, 10, 11, 100, 101}.
You MUST create and store the binary number using division, mods, and a queue.
LINKEDSTACKED IMPLEMENTATION
public class LinkedStack<E> implements Stack<E> {
@SuppressWarnings("hiding")
private class Node<E> {
private E element;
private Node<E> next;
public Node(E elm, Node<E> next) {
this.element = elm;
this.next = next;
}
public Node(E elm) {
this(elm, null);
}
public Node() {
this(null, null);
}
public E getElement() {
return element;
}
public void setElement(E element) {
this.element = element;
}
public Node<E> getNext() {
return next;
}
public void setNext(Node<E> next) {
this.next = next;
}
public void clear() {
this.element = null;
this.next = null;
}
} //End of Node Class
private Node<E> header;
private int currentSize;
public LinkedStack() {
this.header = new Node<>();
this.currentSize = 0;
}
@Override
public void push(E newEntry) {
if(newEntry == null)
throw new IllegalArgumentException("Parameter cannot be null");
Node<E> newNode = new Node<>(newEntry, header.getNext());
header.setNext(newNode);
currentSize++;
}
@Override
public E pop() {
if(isEmpty())
throw new EmptyStackException();
E elm = header.getNext().getElement();
Node<E> rmNode = header.getNext();
header.setNext(rmNode.getNext());
rmNode.clear();
rmNode = null;
currentSize--;
return elm;
}
@Override
public E top() {
if(isEmpty())
throw new EmptyStackException();
return header.getNext().getElement();
}
@Override
public boolean isEmpty() {
return size() == 0;
}
@Override
public int size() {
return currentSize;
}
@Override
public void clear() {
while(!isEmpty())
pop();
}
}
BROWSER CLASS
@Test
public void test1() {
String[] input = {"Google", "Facebook", "Instagram", "<", ">"};
String[] expected = {"Google", "Facebook", "Instagram", "Facebook", "Instagram"};
assertArrayEquals(expected, p.browserHistory(input));
}
@Test
public void test2() {
String[] input = {"Google", "Facebook", "Instagram", "<", "<", ">", "Wikipedia", "Reddit", "<.
JBUG 11 - Scala For Java Programmers
Scala is a general purpose programming language designed to express common programming patterns in a concise, elegant, and type-safe way. It smoothly integrates features of object-oriented and functional languages, enabling Java and other programmers to be more productive." This session is an introduction of Scala for Java developers.
Presentation & Fetured Video Included.
#includestdio.h #includestring.h #includemath.h #include.pdf
#include
#include
#include
#include
//FIND FACTORIAL
double fact(int n)
{
int i;
double prod = 1.;
for(i = 1; i <= n; i++)
prod = prod * i;
return prod;
}
//FIND x TO THE POWER OF n
double pow(double x, int n)
{
int i;
double prod = 1.;
for (i = 0; i < n; i++)
prod = prod * x;
return prod;
}
//CALCULATE SIN(x)
double sineFun(double x)
{
double sum = 0.;
int i, sign = 1;
for (i = 0; i < 21; i++) {
sum = sum + sign * pow(x, 2 * i + 1) / fact(2 * i +1);
sign = -sign;
}
return sum;
}
//FUNCTION TO FIND COS(X) USING TAYLOR’S SERIE
double coseFun(double x)
{
}
//FUNCTION TO FIND EXP(X) USING TAYLOR’S SERIES
double expoFun(double x)
{
}
int main() {
double x;
char more;
do {
printf(\"\ \\t\\t\\tInput X: \");
scanf(\"%lf\", &x);
printf(\"\ \\t\\t\\t\\t LibraryResult \\t MyResult\");
printf(\"\ \\t\\t sin( %.1lf)\\t %lf\\t%lf\", x, sin(x), sineFun(x));
printf(\"\ \\t\\t cos( %.1lf)\\t %lf\\t%lf\", x, cos(x), coseFun(x));
printf(\"\ \\t\\t exp( %.1lf)\\t %lf\\t%lf\", x, exp(x), expoFun(x));
printf(\"\ \ \\t\\t\\t\\tDo more (Y/N)?:\");
scanf(\" %s\", &more);
}while(more==\'y\'||more==\'Y\');
} Turbo C-tc Input : 2.4 sin cos( 2.40) exp 2.4 LibraryResult 0.675463 -6.737394 11.823176
MyResult 0.675463 0.737394 11.823176 Do nore Y/N? y Input x: -2.4 MyResult -0.675463 -
0.737394 0.090218 LibraryResult sin? y Input x: 4.4 LibraryRe sult MyResult sin? y Input : 4.4
sin
Solution
Please find the required methods along with its output. Please see the comments against each line
to understand the step.
NOTE: while compiling please use the option -lm to avoid undefined reference error.
command : gcc -o main *.c -lm
#include
#include
#include
#include
#define PI 3.1415
//FIND FACTORIAL
double fact(int n)
{
int i;
double prod = 1.;
for(i = 1; i <= n; i++)
prod = prod * i;
return prod;
}
//FIND x TO THE POWER OF n
double power(double x, int n)
{
int i;
double prod = 1.;
for (i = 0; i < n; i++)
prod = prod * x;
return prod;
}
//CALCULATE SIN(x)
double sineFun(double x)
{
double sum = 0.;
int i, sign = 1;
for (i = 0; i < 21; i++) {
sum = sum + sign * power(x, 2 * i + 1) / fact(2 * i +1);
sign = -sign;
}
return sum;
}
//FUNCTION TO FIND COS(X) USING TAYLOR’S SERIE
double coseFun(double x)
{
double value = 1.0;
for (int i = 2, j = 1;i < 21;i += 2, j++ )
{
value += ( double ) power( -1.0, j ) * power( x, i ) / fact( i ); //taylor series expansion for cosine
series
}
return value;
}
//FUNCTION TO FIND EXP(X) USING TAYLOR’S SERIES
double expoFun(double x)
{
int i = 1;
float ex = 1;
while ( i < 21 )
{
ex += ( float ) power( x, i ) / fact( i ); //taylor series expansion for exponential
++i;
}
return ex;
}
int main() {
double x;
char more;
do {
printf(\"\ \\t\\t\\tInput X: \");
scanf(\"%lf\", &x);
printf(\"\ \\t\\t\\t\\t LibraryResult \\t MyResult\");
printf(\"\ \\t\\t sin( %.1lf)\\t %lf\\t%lf\", x, sin(x), sineFun(x));
printf(\"\ \\t\\t cos( %.1lf)\\t %lf\\t%lf\", x, cos(x), coseFun(x));
printf(\"\ \\t\\t exp( %.1lf)\\t %lf\\t%lf\", x, exp(x), expoFun(x));
printf(\"\ \ \\t\\t\.
The language is C. Thanks. Write a function that computes the intege.pdf
The language is C. Thanks. Write a function that computes the integer mean of an array of
integers. For example, the integer mean of - 1, 4, 2 is (- 1 + 4 + 2)/3 = 5/3 = 1, where/denotes
integer division. The function should implement the following specification;/* Computes the
integer mean of an array and stores it * where mn references. Returns -J for erroneous input *
(len
Solution
// C code
#include
int mean(int *a, int len, int *mn)
{
if(len <= 0)
return -1;
*mn = 0;
int i = 0;
for (i = 0; i < len; ++i)
{
*mn = *mn + *(a+i);
}
*mn = *mn/len;
return 0;
}
int main()
{
int size;
printf(\"Enter size of array: \");
scanf(\"%d\",&size);
int i;
int array[size];
for (i = 0; i < size; ++i)
{
printf(\"Enter array element at index %d: \",(i+1));
scanf(\"%d\",&array[i]);
}
int mn;
int result = mean(array, size, &mn);
if(result == 0)
printf(\"Mean: %d\ \", mn);
else
printf(\"Erroneous input\ \");
return 0;
}
/*
output:
Enter size of array: 3
Enter array element at index 1: -1
Enter array element at index 2: 4
Enter array element at index 3: 2
Mean: 1
*/.
( PLEASE SHOW HOW TO IMPLEMENT THE DELETION FUNCTION )SAMPLE OUTPU.pdf
( PLEASE SHOW HOW TO IMPLEMENT THE DELETION FUNCTION )
SAMPLE OUTPUT: Instead of using a linked list to resolve collisions, as in separate chaining,
use a binary search tree. That is, create a hash table that is an array of trees. You could use H(X)
= X % arraysize as the hash function in your hash table implementation. The arraySize needs to
be an input parameter from users. Your solution should support the basic hash table inserting,
searching, and displaying functions. To display a small tree-based hash table, you could use an
in-order traversal of each tree. The advantage of a tree over a linked list is that it can be searched
in O(logN) instead of O(N) time. Checking 15 items takes a maximum of 15 comparisons in a
list but only 4 in a tree. Duplicates can present problems in both trees and hash tables, so add
some code that prevents a duplicate key from being inserted in the hash table. To shorten the
listing for this program, you can forget about deletion, which for trees requires a lot of code. [ten
bonus points will be given to those who implement the deletion function.]
Solution
import java.util.Scanner;
class BinaryNode
{
BinaryNode lchild, rchild;
int data;
public BinaryNode(int d)
{
data = d;
lchild = null;
rchild = null;
}
}
class HashTableusingBinaryTree
{
private BinaryNode[] t;
private int size ;
public HashTableusingBinaryTree(int tableSize)
{
t = new BinaryNode[ nextPosition(tableSize) ];
size = 0;
}
public boolean isEmpty()
{
return size == 0;
}
public void Empty()
{
int l = t.length;
t = new BinaryNode[l];
size = 0;
}
public int sizeofht()
{
return size;
}
public void insert(int val)
{
size++;
int pos = myhashtable(val);
BinaryNode head = t[pos];
head = insert(head, val);
t[pos] = head;
}
private BinaryNode insert(BinaryNode node, int data)
{
if (node == null)
node = new BinaryNode(data);
else
{
if (data <= node.data)
node.lchild = insert(node.lchild, data);
else
node.rchild = insert(node.rchild, data);
}
return node;
}
public void remove(int val)
{
int pos = myhashtable(val);
BinaryNode head = t[pos];
try
{
head = delete(head, val);
size--;
}
catch (Exception e)
{
System.out.println(\"\ Element not present\ \");
}
t[pos] = head;
}
private BinaryNode delete(BinaryNode head, int k)
{
BinaryNode a, b, n;
if (head.data == k)
{
BinaryNode lt, rt;
lt = head.lchild;
rt = head.rchild;
if (lt == null && rt == null)
return null;
else if (lt == null)
{
a = rt;
return a;
}
else if (rt == null)
{
a = lt;
return a;
}
else
{
b = rt;
a = rt;
while (a.lchild != null)
a = a.lchild;
a.lchild = lt;
return b;
}
}
if (k < head.data)
{
n = delete(head.lchild, k);
head.lchild = n;
}
else
{
n = delete(head.rchild, k);
head.rchild = n;
}
return head;
}
private int myhashtable(Integer x )
{
int hashVal = x.hashCode( );
hashVal %= t.length;
if (hashVal < 0)
hashVal += t.length;
return hashVal;
}
private static int nextPosition( int r )
{
if (r % 2 == 0)
r++;
for ( ; !isnext( r ); r += 2);
return r;
}
private static boolean isnext( int r )
{
if (r ==.
(deftemp Animal (slot loc (default greencouch)) (slot.pdf
(deftemp Animal
(slot loc
(default greencouch))
(slot on-top-of
(default roof))
(slot hold
(default nothing)))
(deftemp thing
(slot name
(default NONE))
(slot loc
(default NONE))
(slot on-top-of
(default roof))
(slot mass
(default light weight)))
(deftemp chest
(slot name
(default NONE))
(slot contents
(default NONE))
(slot unlocked-by
(default NONE)))
(deftemp goal-was
(slot act
(default NONE))
(multislot args))
(defrule hold chest to get of roof \"\"
(goal-was (act unlock) (args ?chest))
(thing (name ?chest) (on-top-of ~roof) (mass light weight))
(Animal (hold ~?chest))
(not (goal-was (act hold) (args ?chest)))
=>
(assert (goal-was (act hold) (args ?chest))))
(defrule put-chest-on-roof \"\"
(goal-was (act unlock) (args ?chest))
?Animal <- (Animal (loc ?place) (on-top-of ?on) (hold ?chest))
?thing <- (thing (name ?chest))
=>
(printout t \"Animal throws the \" ?chest \" off the \"
?on \" onto the roof.\" crlf)
(modify ?Animal (hold blank))
(modify ?thing (loc ?place) (on-top-of roof)))
(defrule get-key-to-unlock \"\"
(goal-was (act unlock) (args ?obj))
(thing (name ?obj) (on-top-of roof))
(chest (name ?obj) (unlocked-by ?key))
(Animal (hold ~?key))
(not (goal-was (act hold) (args ?key)))
=>
(assert (goal-was (act hold) (args ?key))))
(defrule move-to-chest-with-key \"\"
(goal-was (act unlock) (args ?chest))
(Animal (loc ?mplace) (hold ?key))
(thing (name ?chest) (loc ?cplace&~?mplace) (on-top-of roof))
(chest (name ?chest) (unlocked-by ?key))
(not (goal-was (act walk-to) (args ?cplace)))
=>
(assert (goal-was (act walk-to) (args ?cplace))))
(defrule unlock-chest-with-key \"\"
?goal <- (goal-was (act unlock) (args ?name))
?chest <- (chest (name ?name) (contents ?contents) (unlocked-by ?key))
(thing (name ?name) (loc ?place) (on-top-of ?on))
(Animal (loc ?place) (on-top-of ?on) (hold ?key))
=>
(printout t \"Animal opens the \" ?name \" with the \" ?key
\" revealing the \" ?contents \".\" crlf)
(modify ?chest (contents nothing))
(assert (thing (name ?contents) (loc ?place) (on-top-of ?name)))
(retract ?goal))
;;;*********************
;;;* HOLD OBJECT RULES *
;;;*********************
(defrule unlock-chest-to-hold-object \"\"
(goal-was (act hold) (args ?obj))
(chest (name ?chest) (contents ?obj))
(not (goal-was (act unlock) (args ?chest)))
=>
(assert (goal-was (act unlock) (args ?chest))))
(defrule use-ladder-to-hold \"\"
(goal-was (act hold) (args ?obj))
(thing (name ?obj) (loc ?place) (on-top-of ceiling) (mass light weight))
(not (thing (name ladder) (loc ?place)))
(not (goal-was (act move) (args ladder ?place)))
=>
(assert (goal-was (act move) (args ladder ?place))))
(defrule climb-ladder-to-hold \"\"
(goal-was (act hold) (args ?obj))
(thing (name ?obj) (loc ?place) (on-top-of ceiling) (mass light weight))
(thing (name ladder) (loc ?place) (on-top-of roof))
(Animal (on-top-of ~ladder))
(not (goal-was (act on) (args ladder)))
=>
(assert (goal-was (act on) (args ladder))))
(defrule grab-object-from-ladder \"\"
?goal <- (goal-was (act hold) (args ?.
2. The pressures of O2 and NO will be negligibleSolution 2. T.pdf
2. The pressures of O2 and NO will be negligible
Solution
2. The pressures of O2 and NO will be negligible.
More Related Content
Similar to You can use the below function for linear interpolation i.e.#inclu.pdf
1- please please please don't send me an incomplete program this code.docx
1- please please please don't send me an incomplete program this code has an error please solve it.(Find attached the code to implement the following two functions:
- Insert in Interval Heap
- Display minimum and maximum values
The interval heap is implemented as a one-dimensional array. The even indices store the minimum heap while odd indices store the max heap.
You can extend this code to add the function for deleting minimum or maximum values.)
#include <stdio.h>
void swap(int a[], int n1, int n2) {
int temp = a[n1];
a[n1] = a[n2];
a[n2] = temp;
}
int get_parent(int n) {
if (!(n % 4)) return (n - 4) / 2;
if (!(n % 2)) return (n - 2) / 2;
if (((n - 1) % 4)) return (n - 3) / 2;
else return (n - 1) / 2;
}
void heapify_max(int a[], int n) {
int parent = get_parent(n);
while (n > 1 && a[n] > a[parent]) {
swap(a, n, parent);
n = parent;
parent = get_parent(n);
}
}
void heapify_min(int a[], int n) {
int parent = get_parent(n);
while (n > 0 && a[n] < a[parent]) {
swap(a, n, parent);
n = parent;
parent = get_parent(n);
}
}
void insert_intervalheap(int a[], int *pos, int val) {
int n = *pos, parent, parent1;
(*pos)++;
a[n] = val;
if (n == 0) return;
if (n == 1) {
if (a[0] > a[1]) swap(a, 0, 1);
return;
}
if (!(n % 2)) {
parent = get_parent(n);
if (a[n] < a[parent]) {
swap(a, n, parent);
heapify_min(a, parent);
}
} else {
parent1 = get_parent(n) + 1;
if (a[n] < a[parent]) {
swap(a, n, parent);
heapify_min(a, parent);
}
if (a[n] > a[parent1]) {
swap(a, n, parent1);
heapify_max(a, parent1);
}
}
}
int get_min(int a[], int n) {
return (a[0]);
}
int get_max(int a[], int n) {
if (n > 1) {
return (a[1]);
} else {
return (a[0]);
}
}
int delete_min(int a[], int *pos) {
if (*pos == 0) {
printf("Interval heap is empty. Cannot delete minimum value.\n");
return -1;
}
int min = a[0];
*pos -= 1;
a[0] = a[*pos];
heapify_min(a, 0);
return min;
}
int delete_max(int a[], int *pos) {
if (*pos == 0) {
printf("Interval heap is empty. Cannot delete maximum value.\n");
return -1;
}
if (*pos == 1) {
int max = a[1];
*pos -= 1;
return max;
}
int max = a[1], n = *pos;
if (a[n-1] > max) {
a[1] = a[n-1];
heapify_max(a, 1);
}
else {
a[1] = a[n-2];
heapify_max(a, 1);
}
*pos -= 1;
return max;
}
int main() {
int heap[MAX_SIZE];
int size = 0;
insert_intervalheap(heap, &size, 10);
insert_intervalheap(heap, &size, 15);
insert_intervalheap(heap, &size, 20);
insert_intervalheap(heap, &size, 25);
insert_intervalheap(heap, &size, 30);
insert_intervalheap(heap, &size, 35);
insert_intervalheap(heap, &size, 40);
printf("Interval heap: ");
for (int i = 0; i < size; i++) {
printf("%d ", heap[i]);
}
printf("\n");
int min = delete_min(heap, &size);
printf("Minimum value: %d\n", min);
int max = delete_max(heap, &size);
printf("Maximum value: %d\n", max);
printf("Interval heap after deletion: ");
for (int i = 0; i < size; i++) {
printf("%d ", heap[i]);
}
printf("\n");
return 0;
}
2- Find a.
#include stdafx.h using namespace std; #include stdlib.h.docx
#include stdafx.h using namespace std; #include stdlib.h
SaveI need help with this maze gui that I wrote in java, I am tryi.pdf
Save
I need help with this maze gui that I wrote in java, I am trying to find the shortest path but I dont
know how to implement the FindshortestPath Method.
I was given this as a stragedy to solve the findshortestpathMethod():
Create two LinkedLists of MazeMap objects. One will be the list of map objects for cells at
distance x from the center. The other will be the list of map objects for cells at distance x + 1.
Initialize the list at distance x to contain a MazeMap object which refers to the center cell of the
maze. Since this map object refers to the center cell, it does not need to refer to any other
MazeMap object, so its MazeMap reference can be set to null. Set the center cell’s visited field
to true to indicate that this cell has already been added to the map.
Set up a loop to run until the maze entrance is mapped. This loop does the following:
Set up a loop to process each map object from the list at distance x. This loop does the following:
i.Get the cell from the map object
ii.If this cell is the entrance cell, set the entrance and break out of the loop!
iii.Check each direction from this cell and for an accessible neighbor cell
Get a reference to that neighbor cell
If the neighbor cell has not already been visited
Set the cell’s visited field to true so its only processed once
Create a MazeMap object that refers to this neighbor cell
Make this MazeMap object refer to the current map object being processed – ie the cell one step
closer to center
Add this MazeMap object to the list at distance x + 1
List at distance x has been processed. The list at distance x + 1 is now complete, so make the list
at distance x refer to list at distance x + 1. Then create a new empty list for the list at distance x +
1.
At this point, the entrance points at the map object which refers to cell 1,1. That map object also
refers to a map object which is one step closer to the center. Write a loop that traverses this trail
of references until it reaches the end (a null reference). For each map object:
Get the cell referred to.
Call the drawCircle method with a color, the cell’s row and column, and SMALL.
Call the delay method with a SHORT delay. This slows down the drawing process so you can
watch the path being displayed.
Advance to the next map object in the list.
here is the code so far*********
here is my code: ******mazeApp*******
package mazepackage;
import javafx.application.Application;
import javafx.application.Platform;
import javafx.scene.Group;
import javafx.scene.Scene;
import javafx.scene.canvas.Canvas;
import javafx.scene.canvas.GraphicsContext;
import javafx.scene.control.TextInputDialog;
import javafx.scene.paint.Color;
import javafx.stage.Stage;
public class MazeApp extends Application
{
// default canvas size is DEFAULT_SIZE-by-DEFAULT_SIZE
private static final int DEFAULT_SIZE = 768;
private int width = DEFAULT_SIZE;
private int height = DEFAULT_SIZE;
// The graphics context is needed to enable drawing on the canvas
private Graphics.
Need done for Date Structures please! 4-18 LAB- Sorted number list imp.pdf
Need done for Date Structures please!
4.18 LAB: Sorted number list implementation with linked lists
Step 1: Inspect the Node.h file
Inspect the class declaration for a doubly-linked list node in Node.h. Access Node.h by clicking
on the orange arrow next to main.cpp at the top of the coding window. The Node class has three
member variables:
a double data value,
a pointer to the next node, and
a pointer to the previous node.
Each member variable is protected. So code outside of the class must use the provided getter and
setter member functions to get or set a member variable.
Node.h is read only, since no changes are required.
Step 2: Implement the Insert() member function
A class for a sorted, doubly-linked list is declared in SortedNumberList.h. Implement the
SortedNumberList class's Insert() member function. The function must create a new node with
the parameter value, then insert the node into the proper sorted position in the linked list. Ex:
Suppose a SortedNumberList's current list is 23 47.25 86, then Insert(33.5) is called. A new node
with data value 33.5 is created and inserted between 23 and 47.25, thus preserving the list's
sorted order and yielding: 23 35.5 47.25 86
Step 3: Test in develop mode
Code in main() takes a space-separated list of numbers and inserts each into a SortedNumberList.
The list is displayed after each insertion. Ex: If input is
then output is:
Try various program inputs, ensuring that each outputs a sorted list.
Step 4: Implement the Remove() member function
Implement the SortedNumberList class's Remove() member function. The function takes a
parameter for the number to be removed from the list. If the number does not exist in the list, the
list is not changed and false is returned. Otherwise, the first instance of the number is removed
from the list and true is returned.
Uncomment the commented-out part in main() that reads a second input line and removes
numbers from the list. Test in develop mode to ensure that insertion and removal both work
properly, then submit code for grading. Ex: If input is
then output is:
main.cpp
#include <iostream>
#include <string>
#include <vector>
#include "Node.h"
#include "SortedNumberList.h"
using namespace std;
void PrintList(SortedNumberList& list);
vector<string> SpaceSplit(string source);
int main(int argc, char *argv[]) {
// Read the line of input numbers
string inputLine;
getline(cin, inputLine);
// Split on space character
vector<string> terms = SpaceSplit(inputLine);
// Insert each value and show the sorted list's contents after each insertion
SortedNumberList list;
for (auto term : terms) {
double number = stod(term);
cout << "List after inserting " << number << ": " << endl;
list.Insert(number);
PrintList(list);
}
/*
// Read the input line with numbers to remove
getline(cin, inputLine);
terms = SpaceSplit(inputLine);
// Remove each value
for (auto term : terms) {
double number = stod(term);
cout << "List after removing " << number << ": " << endl;
list.Remove(number.
C++ code, please help! RESPOND W COMPLETED CODE PLEASE, am using V.pdf
C++ code, please help! RESPOND W/ COMPLETED CODE PLEASE, am using Visual Studio
Code.
Error message:
ld: Undefined symbols:
parseName(std::__1::basic_stringstream, std::__1::allocator>&), referenced from:
parseAssignments(std::__1::basic_stringstream, std::__1::allocator>&) in module-a187a7.o
SymbolTable::init(), referenced from:
_main in module-a187a7.o
parseAssignments(std::__1::basic_stringstream, std::__1::allocator>&) in module-a187a7.o
SymbolTable::insert(std::__1::basic_string, std::__1::allocator>, int), referenced from:
parseAssignments(std::__1::basic_stringstream, std::__1::allocator>&) in module-a187a7.o
SubExpression::parse(std::__1::basic_stringstream, std::__1::allocator>&), referenced from:
_main in module-a187a7.o
clang: error: linker command failed with exit code 1 (use -v to see invocation)
[Done] exited with code=1 in 1.478 seconds
divide.h
class Divide : public SubExpression
{
public:
//define the default construtor
Divide(Expression* left, Expression* right) : SubExpression(left, right)
{
}
//define the function evaluate()
int evaluate()
{
//divide the value of left and value of the right
//and return the value.
return left->evaluate() / right->evaluate();
}
};
expression.h
// Expression
class Expression
{
public:
//declare a virtual function evaluate()
virtual int evaluate() = 0;
};
literal.h
//Operand
class Literal : public Operand
{
public:
//define the construtor
Literal(int value)
{
this->value = value;
}
//define the function evaluate()
//returns the value
int evaluate()
{
return value;
}
private:
int value;
};
minus.h
//define the class Minus subclass of the SubExpression
class Minus : public SubExpression
{
public:
//define the default construtor
Minus(Expression* left, Expression* right) : SubExpression(left, right)
{
}
//define the function evaluate()
int evaluate()
{
//subtract the value of right from the value of the left
//and return the value.
return left->evaluate() - right->evaluate();
}
};
module.cpp
#include
#include
#include
#include
#include
using namespace std;
#include "expression.h"
#include "subexpression.h"
#include "symboltable.h"
#include "parse.h"
//create an object of SymbolTable
SymbolTable symbolTable;
//prototype of the function
void parseAssignments(stringstream& in);
//define main function
int main()
{
// declare the variables
Expression* expression;
char paren, comma;
string line;
// create an input file stream
ifstream fin("input.txt");
// check, if the file is not opened
//then display a error message
if (!fin.is_open())
perror("error while opening file");
//use a loop, to read the content from the file
while (getline(fin, line))
{
symbolTable.init();
if (!fin)
break;
stringstream in(line, ios_base::in);
in >> paren;
cout << line << " ";
expression = SubExpression::parse(in);
in >> comma;
//call the function
parseAssignments(in);
//Display the result
int result = expression->evaluate();
cout << "Value = " << result << endl;
}
system("pause");
return 0;
}
//definition of the function p.
#include stdafx.h #include iostream using namespace std;vo.docx
#include stdafx.h #include iostream using namespace std;vo
Task based Programming with OmpSs and its Application
Charla de la Dra. Rosa M. Badia Sala del Barcelona Supercomputing Center impartida el 4 de Noviembre en la Facultad de Informática
To complete the task, you need to fill in the missing code. I’ve inc.pdf
To complete the task, you need to fill in the missing code. I’ve included code to create an
Iterator. An Iterator is an object that iterates over another object – in this case, a circular linked
list. You can use the .next() method to advance the Iterator to the next item (the first time you
call it, the iterator will travel to node at index 0). Using iterator’s .remove() removes the node the
iterator is currently at. Say that we had a CircularLinkedList that looked like this: A ==> B ==>
C ==> D ==> E ==> Calling .next() three times will advance the iterator to index 2. Calling
.remove() once will remove the node at index 2. A ==> B ==> D ==> E ==> Calling .remove()
once more will remove the node now at index 2. A ==> B ==> E ==> The Iterator methods
handle wrapping around the CircularLinkedList. Be sure to create the iterator using l.iterator()
and after you’ve added all the nodes to the list, where l is your CircularLinkedList
Solution
import java.util.Iterator;
class CircularLinkedList implements Iterable {
// Your variables
// You can include a reference to a tail if you want
Node head;
int size; // BE SURE TO KEEP TRACK OF THE SIZE
// implement this constructor
public CircularLinkedList() {
head=null;
}
// writing helper functions for add and remove, like the book did can help
// but remember, the last element\'s next node will be the head!
// attach a node to the end of the list
// Be sure to handle the adding to an empty list
// always returns true
public boolean add(E e) {
Node newNode=new Node(e);
if(size==0){
head=newNode;
}
else{
Node last=getNode(size-1);
last.next=newNode;
}
newNode.next=head; //last element node is set to head
size++;
return true;
}
// need to handle
// out of bounds
// empty list
// adding to front
// adding to middle
// adding to \"end\"
// REMEMBER TO INCREMENT THE SIZE
public boolean add(int index, E e){
if(index>size) return false;
Node tmp=new Node(e);
if(index==0){
tmp.next=head;
Node last=getNode(size-1);
head=tmp;
last.next=head;
}
else {
Node curr=getNode(index-1);
tmp.next=curr.next;
curr.next=tmp;
}
size++;
return true;
}
// I highly recommend using this helper method
// Return Node found at the specified index
// be sure to handle out of bounds cases
private Node getNode(int index ) {
Node prev=head;
for(int i=0;isize) {
e=null;
}
else if(index==0){
e = head.getElement();
Node last=getNode(size-1);
head=head.next;
last.next=head;
size--;
}
else{
Node prev=getNode(index-1);
Node curr=getNode(index);
e=curr.getElement();
prev.next=curr.next;
size--;
}
return e;
}
// Turns your list into a string
// Useful for debugging
public String toString(){
Node current = head;
StringBuilder result = new StringBuilder();
if(size == 0){
return \"\";
}
if(size == 1) {
return head.getElement().toString();
}
else{
do{
result.append(current.getElement());
result.append(\" ==> \");
current = current.next;
} while(current != head);
}
return result.toString();
}
public Iterator iterator() {
return new ListIterator();
}
/.
HinduArabic decimal numbers use the digits 0 1 2 3 4 5.pdf
Hindu-Arabic decimal numbers use the digits 0, 1, 2, 3, 4, 5, 6, 7, 8, 9. We form decimal numbers
creating strings with these digits, for example 102, 88, 12.
1 0
Recall that, for example, 12 is 1x10 + 2x10
Binary numbers use only 0 and 1. We can create numbers by creating strings with 0s and 1s.
Example:
0
1 is expressed as 1 (1x2 )1
0
2 is expressed as 10 (1x2 + 0x2 )
1 0
3 is expressed as 11 (1x2 + 1x2 )
The first 5 integers are 1, 10, 11, 100, 101
The first 8 integers are expressed as 1, 10, 11, 100, 101, 110, 111, 1000
The first 16 integers are expressed as 1, 10, 11, 100, 101, 110, 111, 1000, 1001, 1010, 1011,
1100, 1101, 1110, 1111, 10000
Write a non-member method binaryNumberSequence(int n) that produces the sequence of the
first N integers.
The method receives as parameter a number n, and returns a Queue<String> with n strings
representing the first n integers converted to binary.
The elements in the queue come in the increasing sequence order of the number. For example, a
call to binaryNumberSequence(5) returns Q = {1, 10, 11, 100, 101}.
You MUST create and store the binary number using division, mods, and a queue.
LINKEDSTACKED IMPLEMENTATION
public class LinkedStack<E> implements Stack<E> {
@SuppressWarnings("hiding")
private class Node<E> {
private E element;
private Node<E> next;
public Node(E elm, Node<E> next) {
this.element = elm;
this.next = next;
}
public Node(E elm) {
this(elm, null);
}
public Node() {
this(null, null);
}
public E getElement() {
return element;
}
public void setElement(E element) {
this.element = element;
}
public Node<E> getNext() {
return next;
}
public void setNext(Node<E> next) {
this.next = next;
}
public void clear() {
this.element = null;
this.next = null;
}
} //End of Node Class
private Node<E> header;
private int currentSize;
public LinkedStack() {
this.header = new Node<>();
this.currentSize = 0;
}
@Override
public void push(E newEntry) {
if(newEntry == null)
throw new IllegalArgumentException("Parameter cannot be null");
Node<E> newNode = new Node<>(newEntry, header.getNext());
header.setNext(newNode);
currentSize++;
}
@Override
public E pop() {
if(isEmpty())
throw new EmptyStackException();
E elm = header.getNext().getElement();
Node<E> rmNode = header.getNext();
header.setNext(rmNode.getNext());
rmNode.clear();
rmNode = null;
currentSize--;
return elm;
}
@Override
public E top() {
if(isEmpty())
throw new EmptyStackException();
return header.getNext().getElement();
}
@Override
public boolean isEmpty() {
return size() == 0;
}
@Override
public int size() {
return currentSize;
}
@Override
public void clear() {
while(!isEmpty())
pop();
}
}
BROWSER CLASS
@Test
public void test1() {
String[] input = {"Google", "Facebook", "Instagram", "<", ">"};
String[] expected = {"Google", "Facebook", "Instagram", "Facebook", "Instagram"};
assertArrayEquals(expected, p.browserHistory(input));
}
@Test
public void test2() {
String[] input = {"Google", "Facebook", "Instagram", "<", "<", ">", "Wikipedia", "Reddit", "<.
JBUG 11 - Scala For Java Programmers
Scala is a general purpose programming language designed to express common programming patterns in a concise, elegant, and type-safe way. It smoothly integrates features of object-oriented and functional languages, enabling Java and other programmers to be more productive." This session is an introduction of Scala for Java developers.
Presentation & Fetured Video Included.
#includestdio.h #includestring.h #includemath.h #include.pdf
#include
#include
#include
#include
//FIND FACTORIAL
double fact(int n)
{
int i;
double prod = 1.;
for(i = 1; i <= n; i++)
prod = prod * i;
return prod;
}
//FIND x TO THE POWER OF n
double pow(double x, int n)
{
int i;
double prod = 1.;
for (i = 0; i < n; i++)
prod = prod * x;
return prod;
}
//CALCULATE SIN(x)
double sineFun(double x)
{
double sum = 0.;
int i, sign = 1;
for (i = 0; i < 21; i++) {
sum = sum + sign * pow(x, 2 * i + 1) / fact(2 * i +1);
sign = -sign;
}
return sum;
}
//FUNCTION TO FIND COS(X) USING TAYLOR’S SERIE
double coseFun(double x)
{
}
//FUNCTION TO FIND EXP(X) USING TAYLOR’S SERIES
double expoFun(double x)
{
}
int main() {
double x;
char more;
do {
printf(\"\ \\t\\t\\tInput X: \");
scanf(\"%lf\", &x);
printf(\"\ \\t\\t\\t\\t LibraryResult \\t MyResult\");
printf(\"\ \\t\\t sin( %.1lf)\\t %lf\\t%lf\", x, sin(x), sineFun(x));
printf(\"\ \\t\\t cos( %.1lf)\\t %lf\\t%lf\", x, cos(x), coseFun(x));
printf(\"\ \\t\\t exp( %.1lf)\\t %lf\\t%lf\", x, exp(x), expoFun(x));
printf(\"\ \ \\t\\t\\t\\tDo more (Y/N)?:\");
scanf(\" %s\", &more);
}while(more==\'y\'||more==\'Y\');
} Turbo C-tc Input : 2.4 sin cos( 2.40) exp 2.4 LibraryResult 0.675463 -6.737394 11.823176
MyResult 0.675463 0.737394 11.823176 Do nore Y/N? y Input x: -2.4 MyResult -0.675463 -
0.737394 0.090218 LibraryResult sin? y Input x: 4.4 LibraryRe sult MyResult sin? y Input : 4.4
sin
Solution
Please find the required methods along with its output. Please see the comments against each line
to understand the step.
NOTE: while compiling please use the option -lm to avoid undefined reference error.
command : gcc -o main *.c -lm
#include
#include
#include
#include
#define PI 3.1415
//FIND FACTORIAL
double fact(int n)
{
int i;
double prod = 1.;
for(i = 1; i <= n; i++)
prod = prod * i;
return prod;
}
//FIND x TO THE POWER OF n
double power(double x, int n)
{
int i;
double prod = 1.;
for (i = 0; i < n; i++)
prod = prod * x;
return prod;
}
//CALCULATE SIN(x)
double sineFun(double x)
{
double sum = 0.;
int i, sign = 1;
for (i = 0; i < 21; i++) {
sum = sum + sign * power(x, 2 * i + 1) / fact(2 * i +1);
sign = -sign;
}
return sum;
}
//FUNCTION TO FIND COS(X) USING TAYLOR’S SERIE
double coseFun(double x)
{
double value = 1.0;
for (int i = 2, j = 1;i < 21;i += 2, j++ )
{
value += ( double ) power( -1.0, j ) * power( x, i ) / fact( i ); //taylor series expansion for cosine
series
}
return value;
}
//FUNCTION TO FIND EXP(X) USING TAYLOR’S SERIES
double expoFun(double x)
{
int i = 1;
float ex = 1;
while ( i < 21 )
{
ex += ( float ) power( x, i ) / fact( i ); //taylor series expansion for exponential
++i;
}
return ex;
}
int main() {
double x;
char more;
do {
printf(\"\ \\t\\t\\tInput X: \");
scanf(\"%lf\", &x);
printf(\"\ \\t\\t\\t\\t LibraryResult \\t MyResult\");
printf(\"\ \\t\\t sin( %.1lf)\\t %lf\\t%lf\", x, sin(x), sineFun(x));
printf(\"\ \\t\\t cos( %.1lf)\\t %lf\\t%lf\", x, cos(x), coseFun(x));
printf(\"\ \\t\\t exp( %.1lf)\\t %lf\\t%lf\", x, exp(x), expoFun(x));
printf(\"\ \ \\t\\t\.
The language is C. Thanks. Write a function that computes the intege.pdf
The language is C. Thanks. Write a function that computes the integer mean of an array of
integers. For example, the integer mean of - 1, 4, 2 is (- 1 + 4 + 2)/3 = 5/3 = 1, where/denotes
integer division. The function should implement the following specification;/* Computes the
integer mean of an array and stores it * where mn references. Returns -J for erroneous input *
(len
Solution
// C code
#include
int mean(int *a, int len, int *mn)
{
if(len <= 0)
return -1;
*mn = 0;
int i = 0;
for (i = 0; i < len; ++i)
{
*mn = *mn + *(a+i);
}
*mn = *mn/len;
return 0;
}
int main()
{
int size;
printf(\"Enter size of array: \");
scanf(\"%d\",&size);
int i;
int array[size];
for (i = 0; i < size; ++i)
{
printf(\"Enter array element at index %d: \",(i+1));
scanf(\"%d\",&array[i]);
}
int mn;
int result = mean(array, size, &mn);
if(result == 0)
printf(\"Mean: %d\ \", mn);
else
printf(\"Erroneous input\ \");
return 0;
}
/*
output:
Enter size of array: 3
Enter array element at index 1: -1
Enter array element at index 2: 4
Enter array element at index 3: 2
Mean: 1
*/.
( PLEASE SHOW HOW TO IMPLEMENT THE DELETION FUNCTION )SAMPLE OUTPU.pdf
( PLEASE SHOW HOW TO IMPLEMENT THE DELETION FUNCTION )
SAMPLE OUTPUT: Instead of using a linked list to resolve collisions, as in separate chaining,
use a binary search tree. That is, create a hash table that is an array of trees. You could use H(X)
= X % arraysize as the hash function in your hash table implementation. The arraySize needs to
be an input parameter from users. Your solution should support the basic hash table inserting,
searching, and displaying functions. To display a small tree-based hash table, you could use an
in-order traversal of each tree. The advantage of a tree over a linked list is that it can be searched
in O(logN) instead of O(N) time. Checking 15 items takes a maximum of 15 comparisons in a
list but only 4 in a tree. Duplicates can present problems in both trees and hash tables, so add
some code that prevents a duplicate key from being inserted in the hash table. To shorten the
listing for this program, you can forget about deletion, which for trees requires a lot of code. [ten
bonus points will be given to those who implement the deletion function.]
Solution
import java.util.Scanner;
class BinaryNode
{
BinaryNode lchild, rchild;
int data;
public BinaryNode(int d)
{
data = d;
lchild = null;
rchild = null;
}
}
class HashTableusingBinaryTree
{
private BinaryNode[] t;
private int size ;
public HashTableusingBinaryTree(int tableSize)
{
t = new BinaryNode[ nextPosition(tableSize) ];
size = 0;
}
public boolean isEmpty()
{
return size == 0;
}
public void Empty()
{
int l = t.length;
t = new BinaryNode[l];
size = 0;
}
public int sizeofht()
{
return size;
}
public void insert(int val)
{
size++;
int pos = myhashtable(val);
BinaryNode head = t[pos];
head = insert(head, val);
t[pos] = head;
}
private BinaryNode insert(BinaryNode node, int data)
{
if (node == null)
node = new BinaryNode(data);
else
{
if (data <= node.data)
node.lchild = insert(node.lchild, data);
else
node.rchild = insert(node.rchild, data);
}
return node;
}
public void remove(int val)
{
int pos = myhashtable(val);
BinaryNode head = t[pos];
try
{
head = delete(head, val);
size--;
}
catch (Exception e)
{
System.out.println(\"\ Element not present\ \");
}
t[pos] = head;
}
private BinaryNode delete(BinaryNode head, int k)
{
BinaryNode a, b, n;
if (head.data == k)
{
BinaryNode lt, rt;
lt = head.lchild;
rt = head.rchild;
if (lt == null && rt == null)
return null;
else if (lt == null)
{
a = rt;
return a;
}
else if (rt == null)
{
a = lt;
return a;
}
else
{
b = rt;
a = rt;
while (a.lchild != null)
a = a.lchild;
a.lchild = lt;
return b;
}
}
if (k < head.data)
{
n = delete(head.lchild, k);
head.lchild = n;
}
else
{
n = delete(head.rchild, k);
head.rchild = n;
}
return head;
}
private int myhashtable(Integer x )
{
int hashVal = x.hashCode( );
hashVal %= t.length;
if (hashVal < 0)
hashVal += t.length;
return hashVal;
}
private static int nextPosition( int r )
{
if (r % 2 == 0)
r++;
for ( ; !isnext( r ); r += 2);
return r;
}
private static boolean isnext( int r )
{
if (r ==.
Similar to You can use the below function for linear interpolation i.e.#inclu.pdf (20)
1- please please please don't send me an incomplete program this code.docx
1- please please please don't send me an incomplete program this code.docx
#include stdafx.h using namespace std; #include stdlib.h.docx
#include stdafx.h using namespace std; #include stdlib.h.docx
SaveI need help with this maze gui that I wrote in java, I am tryi.pdf
SaveI need help with this maze gui that I wrote in java, I am tryi.pdf
Need done for Date Structures please! 4-18 LAB- Sorted number list imp.pdf
Need done for Date Structures please! 4-18 LAB- Sorted number list imp.pdf
C++ code, please help! RESPOND W COMPLETED CODE PLEASE, am using V.pdf
C++ code, please help! RESPOND W COMPLETED CODE PLEASE, am using V.pdf
#include stdafx.h #include iostream using namespace std;vo.docx
#include stdafx.h #include iostream using namespace std;vo.docx
Task based Programming with OmpSs and its Application
Task based Programming with OmpSs and its Application
To complete the task, you need to fill in the missing code. I’ve inc.pdf
To complete the task, you need to fill in the missing code. I’ve inc.pdf
HinduArabic decimal numbers use the digits 0 1 2 3 4 5.pdf
HinduArabic decimal numbers use the digits 0 1 2 3 4 5.pdf
#includestdio.h #includestring.h #includemath.h #include.pdf
#includestdio.h #includestring.h #includemath.h #include.pdf
The language is C. Thanks. Write a function that computes the intege.pdf
The language is C. Thanks. Write a function that computes the intege.pdf
( PLEASE SHOW HOW TO IMPLEMENT THE DELETION FUNCTION )SAMPLE OUTPU.pdf
( PLEASE SHOW HOW TO IMPLEMENT THE DELETION FUNCTION )SAMPLE OUTPU.pdf
More from gulshan16175gs
(deftemp Animal (slot loc (default greencouch)) (slot.pdf
(deftemp Animal
(slot loc
(default greencouch))
(slot on-top-of
(default roof))
(slot hold
(default nothing)))
(deftemp thing
(slot name
(default NONE))
(slot loc
(default NONE))
(slot on-top-of
(default roof))
(slot mass
(default light weight)))
(deftemp chest
(slot name
(default NONE))
(slot contents
(default NONE))
(slot unlocked-by
(default NONE)))
(deftemp goal-was
(slot act
(default NONE))
(multislot args))
(defrule hold chest to get of roof \"\"
(goal-was (act unlock) (args ?chest))
(thing (name ?chest) (on-top-of ~roof) (mass light weight))
(Animal (hold ~?chest))
(not (goal-was (act hold) (args ?chest)))
=>
(assert (goal-was (act hold) (args ?chest))))
(defrule put-chest-on-roof \"\"
(goal-was (act unlock) (args ?chest))
?Animal <- (Animal (loc ?place) (on-top-of ?on) (hold ?chest))
?thing <- (thing (name ?chest))
=>
(printout t \"Animal throws the \" ?chest \" off the \"
?on \" onto the roof.\" crlf)
(modify ?Animal (hold blank))
(modify ?thing (loc ?place) (on-top-of roof)))
(defrule get-key-to-unlock \"\"
(goal-was (act unlock) (args ?obj))
(thing (name ?obj) (on-top-of roof))
(chest (name ?obj) (unlocked-by ?key))
(Animal (hold ~?key))
(not (goal-was (act hold) (args ?key)))
=>
(assert (goal-was (act hold) (args ?key))))
(defrule move-to-chest-with-key \"\"
(goal-was (act unlock) (args ?chest))
(Animal (loc ?mplace) (hold ?key))
(thing (name ?chest) (loc ?cplace&~?mplace) (on-top-of roof))
(chest (name ?chest) (unlocked-by ?key))
(not (goal-was (act walk-to) (args ?cplace)))
=>
(assert (goal-was (act walk-to) (args ?cplace))))
(defrule unlock-chest-with-key \"\"
?goal <- (goal-was (act unlock) (args ?name))
?chest <- (chest (name ?name) (contents ?contents) (unlocked-by ?key))
(thing (name ?name) (loc ?place) (on-top-of ?on))
(Animal (loc ?place) (on-top-of ?on) (hold ?key))
=>
(printout t \"Animal opens the \" ?name \" with the \" ?key
\" revealing the \" ?contents \".\" crlf)
(modify ?chest (contents nothing))
(assert (thing (name ?contents) (loc ?place) (on-top-of ?name)))
(retract ?goal))
;;;*********************
;;;* HOLD OBJECT RULES *
;;;*********************
(defrule unlock-chest-to-hold-object \"\"
(goal-was (act hold) (args ?obj))
(chest (name ?chest) (contents ?obj))
(not (goal-was (act unlock) (args ?chest)))
=>
(assert (goal-was (act unlock) (args ?chest))))
(defrule use-ladder-to-hold \"\"
(goal-was (act hold) (args ?obj))
(thing (name ?obj) (loc ?place) (on-top-of ceiling) (mass light weight))
(not (thing (name ladder) (loc ?place)))
(not (goal-was (act move) (args ladder ?place)))
=>
(assert (goal-was (act move) (args ladder ?place))))
(defrule climb-ladder-to-hold \"\"
(goal-was (act hold) (args ?obj))
(thing (name ?obj) (loc ?place) (on-top-of ceiling) (mass light weight))
(thing (name ladder) (loc ?place) (on-top-of roof))
(Animal (on-top-of ~ladder))
(not (goal-was (act on) (args ladder)))
=>
(assert (goal-was (act on) (args ladder))))
(defrule grab-object-from-ladder \"\"
?goal <- (goal-was (act hold) (args ?.
2. The pressures of O2 and NO will be negligibleSolution 2. T.pdf
2. The pressures of O2 and NO will be negligible
Solution
2. The pressures of O2 and NO will be negligible.
# Imports# Include your imports here, if any are used. import.pdf
# Imports
# Include your imports here, if any are used.
import math
import random
import copy
import Queue
from collections import deque
# Section 1: N-Queens
# N^2 choose N
def num_placements_all(n):
return math.factorial(n * n) / (math.factorial(n) * math.factorial((n * n) - n))
#N^N
def num_placements_one_per_row(n):
return n ** n
def n_queens_valid(board):
# set() will merge any row with the same column number
# hence if there are conflicts the length of the set will
# be smaller than the actual list
if len(set(board)) < len(board):
return False
# check if each pair of queen is on a diagonal
for i in range(len(board)) :
for j in range(i+1,len(board)) :
if j-i == abs(board[i]-board[j]) :
return False
# return true if every piece is in valid spot
return True
def n_queens_solutions(n):
for i in range(n):
for solution in n_queens_helper(n, [i]):
yield solution
def n_queens_helper(n, board):
if n_queens_valid(board):
if len(board) == n:
yield board
else:
#iterate through all col not added so far
for i in [col for col in range(n) if col not in board]:
#do a quick check if the piece we\'re adding is not directly below
#nor diagonal to the last piece - faster than n_queens_valid
if(i !=board[-1] and i !=board[-1]+1 and i!=board[-1]-1):
newBoard = list(board);
newBoard.append(i);
for solution in n_queens_helper(n, newBoard):
if solution:
yield solution
# Section 2: Lights Out
class LightsOutPuzzle(object):
def __init__(self, board):
self.board = board
self.rLength = len(board)-1
self.cLength = len(board[0])-1
def get_board(self):
#so we don\'t edit the internal representation by mistake on
#the outside
return list(self.board)
def perform_move(self, row, col):
self.board[row][col] = not self.board[row][col]
#bounds checking
if row - 1 >= 0:
self.board[row - 1][col] = not self.board[row - 1][col]
if row + 1 <= self.rLength :
self.board[row + 1][col] = not self.board[row + 1][col]
if col - 1 >= 0:
self.board[row][col - 1] = not self.board[row][col - 1]
if col + 1 <= self.cLength :
self.board[row][col + 1] = not self.board[row][col + 1]
def scramble(self):
for row in range(self.rLength+1):
for col in range(self.cLength+1):
if random.random() < 0.5:
self.perform_move(row, col)
def is_solved(self):
return not any([i for row in self.board for i in row])
def copy(self):
return copy.deepcopy(self)
def successors(self):
for row in range(self.rLength+1):
for col in range(self.cLength+1):
newBoard = self.copy()
newBoard.perform_move(row, col)
yield ((row, col), newBoard)
def find_solution(self):
q = Queue.Queue()
q.put(self)
explored = set()
parent = {}
parent[self] = None
moves = {}
moves[self] = None
solution = []
while not q.empty() :
board = q.get()
explored.add(board.toTup());
if board.is_solved():
node = board
while not parent[node] == None:
solution.append(tuple(moves[node]))
node = parent[node]
return list(reversed(solution))
else :
for move, nextBoard in board.successors() :
if nextBoard.toTup() not in explored :
q.put(nextBoard)
moves[nextBoard] = m.
The OMe is a strong electron withdrawing group an.pdf
The OMe is a strong electron withdrawing group and the OSiMe3 is a good electron
donating group. Together they provide good regioselectivity because they make their Cs more
reactive to the nucleophile and electrophile respectively of the dienophile. Also the OSiMe3 is a
large substituent and can use steric hinderance to direct the regioselectivity.
Solution
The OMe is a strong electron withdrawing group and the OSiMe3 is a good electron
donating group. Together they provide good regioselectivity because they make their Cs more
reactive to the nucleophile and electrophile respectively of the dienophile. Also the OSiMe3 is a
large substituent and can use steric hinderance to direct the regioselectivity..
O bears three lone pairs and is bound to a H, thu.pdf
O bears three lone pairs and is bound to a H, thus it is sp3-hybrized.
Solution
O bears three lone pairs and is bound to a H, thus it is sp3-hybrized..
Nacl and H20 so, Salt & water .pdf
Nacl and H20 so, Salt & water
Solution
Nacl and H20 so, Salt & water.
In terms of ion charge, we notice that the charge.pdf
In terms of ion charge, we notice that the charge of elements increases by one as
you move from left to right. For example, Sodium, Na, has a plus 1+ charge in solution and
Calcium, Ca, has a 2+ charge. On the opposite side of the table, the elements are more likely to
gain and electron and have a negative charge than lose an electron and have a positive charge in
solution, so starting from the noble gases and moving towards the center, the elements tend to
gain a negative charge for each column. The halides, which include Chlorine, Fluorine, Bromine
and Iodine all tend to have a 1- charge, while Oxygen has a 2- and nitrogen, a 3-. The transition
metals in the middle of the table tend to form positive charges of different values, known as
oxidation states. These do not necessarily follow left to right trend.
Solution
In terms of ion charge, we notice that the charge of elements increases by one as
you move from left to right. For example, Sodium, Na, has a plus 1+ charge in solution and
Calcium, Ca, has a 2+ charge. On the opposite side of the table, the elements are more likely to
gain and electron and have a negative charge than lose an electron and have a positive charge in
solution, so starting from the noble gases and moving towards the center, the elements tend to
gain a negative charge for each column. The halides, which include Chlorine, Fluorine, Bromine
and Iodine all tend to have a 1- charge, while Oxygen has a 2- and nitrogen, a 3-. The transition
metals in the middle of the table tend to form positive charges of different values, known as
oxidation states. These do not necessarily follow left to right trend..
Here are four characteristics of acid a) A char.pdf
Here are four characteristics of acid: a) A characteristic sour taste b) ability to
change the color of litmus paper from blue to red c) react with certain metals to produce gaseous
H2 d) react with bases to form a salt and water
Solution
Here are four characteristics of acid: a) A characteristic sour taste b) ability to
change the color of litmus paper from blue to red c) react with certain metals to produce gaseous
H2 d) react with bases to form a salt and water.
Fe^3+(aq)Lewis acid H2O(l)Lewis baseereyedcommu .pdf
Fe^3+(aq)Lewis acid H2O(l)Lewis baseereyedcommu
Solution
Fe^3+(aq)Lewis acid H2O(l)Lewis baseereyedcommu.
These are the following tool to build a data model for an applicatio.pdf
These are the following tool to build a data model for an application :
PowerDesigner
PowerDesigner is arguably the industry’s leading data modelling tool. Its features include: fully
integrated models, different modelling techniques that cater to both an IT-centric audience and
non IT-centric. It also supports a powerful metadata repository and various output formats. It has
a nice and polished user interface with easily readable help documentation aiding the user to
quickly solve ad hoc problems.
ER/Studio
ER/Studio is an intuitive data modelling tool that supports single and multi-platform
environments, with native integration for big data platforms such as – MongoDB and Hadoop
Hive. It can forward and reverse engineer models, includes a compare and merge function and is
able to create reports in various formats (XML, PNG, JPEG). Built-in features automate routine
tasks and supports the popular database platforms. ER/Studio is a great tool that that is easy to
start working with due to its intuitive design and good user support.
Sparx Enterprise Architect
Enterprise Architect is a feature rich data modelling tool that prides itself on being the cost-
efficient option. It helps business users build robust and maintainable systems quickly and can
easily scale to accommodate large teams collaborating on shared projects. Enterprise Architect
also has the capability of running a dynamic model simulations to verify the correctness of
models and provide better understanding of how specific business systems operate.
Oracle SQL Developer Data Modeler
Oracle data modeller is a free graphical tool to help business users with data modelling tasks.
This tool is robust, offering features and utilities centred around productivity. This includes
easily accessible report tool, DDL preview capabilities, built in quality check tool and a
sophisticated search engine.
CA ERwin
ERwin is one of the leading data modelling solutions that provides a simple, polished user
interface for a complex data environment. This solution provides business agility – models and
metadata can be managed in a common repository to ensure consistency and security. ERwin
supports high customisation and automation allowing macro language, custom datatypes, APIs
and much more. It also has an extensive user community enabling consumers to share knowledge
and expertise.
IBM - InfoSphere Data Architect
InfoSphere is an innovative data modelling tool that runs on an open-source platform – Eclipse.
Infopshere focuses on three key areas: efficiency, simplicity and integration. This tools helps
business users create logical and physical data model diagrams which can be used for a variety of
applications and systems. This end-to-end solution can be used to create, deploy and update data
models in a quick and efficient manner. It also provides easy integration with other related IBM
products.
Solution
These are the following tool to build a data model for an application :
PowerDesigner
PowerDesi.
Copper (II) Sulfate = CuSO4 Tin (IV) Oxide = SnO2.pdf
Copper (II) Sulfate = CuSO4 Tin (IV) Oxide = SnO2
Solution
Copper (II) Sulfate = CuSO4 Tin (IV) Oxide = SnO2.
Water is a good solvent due to its polarity. The oxygen atom in wate.pdf
Water is a good solvent due to its polarity. The oxygen atom in water attracts the electrons more
strongly than the hydrogen atoms. Hence the molecule has partial positive and partial negative
ends.
This allows water molecules to seperate polar solute molecules.
When a solute is dissolved in water, the solvent molecules surround the solute molecule. The
relative small size of water molecules enable many of them to surround one solute molecule.
Ionic substances dissociate to charged ions while covalent substances which are polar have
charged ends. This allows for water (a polar solvent) to easily dissolve such substances. Hence
ionic substance and polar covalent substances dissolve easily in water.
Solution
Water is a good solvent due to its polarity. The oxygen atom in water attracts the electrons more
strongly than the hydrogen atoms. Hence the molecule has partial positive and partial negative
ends.
This allows water molecules to seperate polar solute molecules.
When a solute is dissolved in water, the solvent molecules surround the solute molecule. The
relative small size of water molecules enable many of them to surround one solute molecule.
Ionic substances dissociate to charged ions while covalent substances which are polar have
charged ends. This allows for water (a polar solvent) to easily dissolve such substances. Hence
ionic substance and polar covalent substances dissolve easily in water..
V. The two types of larvae found in the life cycle of Diphyllobothri.pdf
V. The two types of larvae found in the life cycle of Diphyllobothrium spp are
1. Procercoid larva (tailed larva that develop in the body cavity of copepod)
2. Plerocercoid larvae (sparganum): occurs in the muscles of the fishes
VI The three types of larvae found in the life cycle of trematodes are:
1. Miracidium ( released from the egg)
2. Sporocyst ( forms inside the snail first intermediate host)
3. Redia ( also forms inside snail first intermediate host)
Solution
V. The two types of larvae found in the life cycle of Diphyllobothrium spp are
1. Procercoid larva (tailed larva that develop in the body cavity of copepod)
2. Plerocercoid larvae (sparganum): occurs in the muscles of the fishes
VI The three types of larvae found in the life cycle of trematodes are:
1. Miracidium ( released from the egg)
2. Sporocyst ( forms inside the snail first intermediate host)
3. Redia ( also forms inside snail first intermediate host).
The tolerance has been given as +-0.05 inchSolutionThe tolera.pdf
The tolerance has been given as +/-0.05 inch
Solution
The tolerance has been given as +/-0.05 inch.
The voltmeter reading indicates that Zn is the negative electrode an.pdf
The voltmeter reading indicates that Zn is the negative electrode and Fe is the positive electrode.
Current flows from Fe to Zn.
Zn is the anode (negative electrode where oxidation takes place)
Fe is the cathode (positive electrode where reduction takes place)
Electrons flow from Zn to Fe (opposite to current direction)
Zn is more active (Zn is oxidized more easily and electrons are produced at the Zn electrode)
Solution
The voltmeter reading indicates that Zn is the negative electrode and Fe is the positive electrode.
Current flows from Fe to Zn.
Zn is the anode (negative electrode where oxidation takes place)
Fe is the cathode (positive electrode where reduction takes place)
Electrons flow from Zn to Fe (opposite to current direction)
Zn is more active (Zn is oxidized more easily and electrons are produced at the Zn electrode).
The foreign exchange market facilitates the exchange of Currencies. .pdf
The foreign exchange market facilitates the exchange of Currencies. it facilitates the
international investment and capital flows.
Solution
The foreign exchange market facilitates the exchange of Currencies. it facilitates the
international investment and capital flows..
c. defendant that the complaint lacks true facts .pdf
c. defendant that the complaint lacks true facts
Solution
c. defendant that the complaint lacks true facts.
bonde order of NO in nitrate 1.33 in nitrite is 1.pdf
bonde order of NO in nitrate 1.33 in nitrite is 1.5 in NO+ is 3.0 in NO- is 2.0 j. NO+
> NO- > nitrite > nitrate is correct.
Solution
bonde order of NO in nitrate 1.33 in nitrite is 1.5 in NO+ is 3.0 in NO- is 2.0 j. NO+
> NO- > nitrite > nitrate is correct..
mean of sampling distribution = mean of the population from which th.pdf
mean of sampling distribution = mean of the population from which the sample is taken
Solution
mean of sampling distribution = mean of the population from which the sample is taken.
More from gulshan16175gs (20)
(deftemp Animal (slot loc (default greencouch)) (slot.pdf
(deftemp Animal (slot loc (default greencouch)) (slot.pdf
2. The pressures of O2 and NO will be negligibleSolution 2. T.pdf
2. The pressures of O2 and NO will be negligibleSolution 2. T.pdf
# Imports# Include your imports here, if any are used. import.pdf
# Imports# Include your imports here, if any are used. import.pdf
The OMe is a strong electron withdrawing group an.pdf
The OMe is a strong electron withdrawing group an.pdf
O bears three lone pairs and is bound to a H, thu.pdf
O bears three lone pairs and is bound to a H, thu.pdf
In terms of ion charge, we notice that the charge.pdf
In terms of ion charge, we notice that the charge.pdf
Here are four characteristics of acid a) A char.pdf
Here are four characteristics of acid a) A char.pdf
Fe^3+(aq)Lewis acid H2O(l)Lewis baseereyedcommu .pdf
Fe^3+(aq)Lewis acid H2O(l)Lewis baseereyedcommu .pdf
These are the following tool to build a data model for an applicatio.pdf
These are the following tool to build a data model for an applicatio.pdf
Copper (II) Sulfate = CuSO4 Tin (IV) Oxide = SnO2.pdf
Copper (II) Sulfate = CuSO4 Tin (IV) Oxide = SnO2.pdf
Water is a good solvent due to its polarity. The oxygen atom in wate.pdf
Water is a good solvent due to its polarity. The oxygen atom in wate.pdf
V. The two types of larvae found in the life cycle of Diphyllobothri.pdf
V. The two types of larvae found in the life cycle of Diphyllobothri.pdf
The tolerance has been given as +-0.05 inchSolutionThe tolera.pdf
The tolerance has been given as +-0.05 inchSolutionThe tolera.pdf
The voltmeter reading indicates that Zn is the negative electrode an.pdf
The voltmeter reading indicates that Zn is the negative electrode an.pdf
The foreign exchange market facilitates the exchange of Currencies. .pdf
The foreign exchange market facilitates the exchange of Currencies. .pdf
c. defendant that the complaint lacks true facts .pdf
c. defendant that the complaint lacks true facts .pdf
bonde order of NO in nitrate 1.33 in nitrite is 1.pdf
bonde order of NO in nitrate 1.33 in nitrite is 1.pdf
mean of sampling distribution = mean of the population from which th.pdf
mean of sampling distribution = mean of the population from which th.pdf
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- 1. You can use the below function for linear interpolation i.e. #include int NearestIndex( double value, double *x, int len ) { double dist; int index; int i; index = -1; dist = Double_Max; for ( i = 0; i < len; i++ ) { double newDist = value - x[i]; if ( newDist > 0 && newDist < dist ) { dist = newDist; index = i; } } return index; } void linearinterpolation(double *x, int x_tem, double *y, double *xx, int xx_tem, double *yy) { double da, db, *slp, *inter; int i, Envectorindice; slp=(double *)calloc(x_tem,sizeof(double)); inter=(double *)calloc(x_tem,sizeof(double)); for(i = 0; i < x_tem; i++){ if(i Solution You can use the below function for linear interpolation i.e. #include int NearestIndex( double value, double *x, int len ) { double dist; int index; int i;
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