Write code in C++ Write a program to perform a topological sort on a g.docx
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Write code in C++ Write a program to perform a topological sort on a graph. *Use adjacency matrix Use these Variables: int L1, L2, L3, noVertices, adjMatrix[10][10], indegree[10], swap; *Add Comments Solution #include #include using namespace std; struct node_info { int no; int lv_time, st_time; }*q = NULL, *r = NULL; struct node { node_info *pt; node *next; }*top = NULL, *p = NULL, *np = NULL; int c = 0; void push(node_info *ptr) { np = new node; np->pt = ptr; np->next = NULL; if (top == NULL) { top = np; } else { np->next = top; top = np; } } node_info *pop() { if (top == NULL) { cout<<\"underflow\ \"; } else { p = top; top = top->next; return(p->pt); delete(p); } } void topo(int *v,int am[][7],int i) { q = new node_info; q->no = i; q->st_time = c; cout<<\"start time for node no \"< no<<\":\"< no<<\":\"< >am[i][j]; } } topo(v,am,0); getch(); } .
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![Write code in C++
Write a program to perform a topological sort on a graph.
*Use adjacency matrix
Use these Variables: int L1, L2, L3, noVertices, adjMatrix[10][10], indegree[10], swap;
*Add Comments
Solution
#include #include using namespace std; struct node_info { int no; int lv_time, st_time; }*q =
NULL, *r = NULL; struct node { node_info *pt; node *next; }*top = NULL, *p = NULL, *np =
NULL; int c = 0; void push(node_info *ptr) { np = new node; np->pt = ptr; np->next = NULL; if
(top == NULL) { top = np; } else { np->next = top; top = np; } } node_info *pop() { if (top ==
NULL) { cout<<"underflow "; } else { p = top; top = top->next; return(p->pt); delete(p); } }
void topo(int *v,int am[][7],int i) { q = new node_info; q->no = i; q->st_time = c; cout<<"start
time for node no "< no<<":"< no<<":"< >am[i][j]; } } topo(v,am,0); getch(); }](https://image.slidesharecdn.com/writecodeincwriteaprogramtoperformatopologicalsortonag-230205164728-6d6995b2/85/Write-code-in-C-Write-a-program-to-perform-a-topological-sort-on-a-g-docx-1-320.jpg)
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Program to a topological sort on a graph Program plan: Initializes main() function. Declare int type variables. Accept the number of vertices in the graph and the adjacency matrix of in-degree. Perform topological sort using for loops. Display the result.
Solution
Please find the required code and output below:
#include <iostream>
using namespace std;
struct node_info
{
int no;
int lv_time, st_time;
}*q = NULL, *r = NULL;
struct node
{
node_info *pt;
node *next;
}*top = NULL, *p = NULL, *np = NULL;
int c = 0;
void push(node_info *ptr)
{
np = new node;
np->pt = ptr;
np->next = NULL;
if (top == NULL)
{
top = np;
}
else
{
np->next = top;
top = np;
}
}
node_info *pop()
{
if (top == NULL)
{
cout<<\"underflow\ \";
}
else
{
p = top;
top = top->next;
return(p->pt);
delete(p);
}
}
void topologicalSort(int *v,int am[][7],int i)
{
q = new node_info;
q->no = i;
q->st_time = c;
cout<<\"start time for node no \"<<q->no<<\":\"<<c<<endl;
push(q);
v[i] = 1;
for (int j = 0; j < 7; j++)
{
if (am[i][j] == 0 || (am[i][j] == 1 && v[j] == 1))
continue;
else if(am[i][j] == 1 && v[j] == 0)
{
c++;
topologicalSort(v,am,j);
}
}
c++;
r = pop();
cout<<\"leave time for \"<<r->no<<\":\"<<c<<endl;
return;
}
int main()
{
int v[7],am[7][7];
for (int i = 0; i < 7; i++)
v[i] = 0;
for (int i = 0; i < 7; i++)
{
cout<<\"enter the values for adjacency matrix row:\"<<i + 1<<endl;
for(int j = 0; j < 7; j++)
{
cin>>am[i][j];
}
}
topologicalSort(v,am,0);
}
-----------------------------------------------------------------------------------------------
OUTPUT:
.
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#includeiostream struct node { char value; struct no.pdfankitmobileshop235 #include
struct node
{
char value;
struct node *next;
};
class StringOfNode
{
struct node * head;
public:
StringOfNode()
{
head=NULL;
}
int size()
{
struct node *p=head;
int count=1;
while(p)
{
count++;
p=p->next;
}
return count;
}
struct node * begin()
{
return head;
}
char &operator[](int index)
{
int i=0;
struct node *p=head;
while(index!=i)
{
p=p->next;
i++;
}
return p->value;
}
void insertFront(char valuee)
{
struct node * temp=(struct node *)malloc(sizeof(struct node));
temp->value=valuee;
temp->next=NULL;
if(head==NULL)
head=temp;
else
{
temp->next=head;
head=temp;
}
}
void insert(char valuee, int pos)
{
struct node *p=head;
struct node * temp=(struct node *)malloc(sizeof(struct node));
temp->value=valuee;
temp->next=NULL;
int i=0;
if(p==NULL)
{
head=temp;
}
else
{
while(i!=pos-1)
{
p=p->next;
i++;
}
temp->next=p->next;
p->next=temp;
}
}
void insertBack(char valuee)
{
struct node *p=head;
struct node * temp=(struct node *)malloc(sizeof(struct node));
temp->value=valuee;
temp->next=NULL;
if(p==NULL)
{
head=temp;
}
else
{
while(p->next!=NULL)
p=p->next;
p->next=temp;
}
}
void print()
{
struct node *p=head;
while(p)
{
std::cout<value<<\"\ \";
p=p->next;
}
}
StringOfNode operator+(const StringOfNode& b)
{
StringOfNode *S;
struct node *p=head;
while(p->next!=NULL)
p=p->next;
p->next=S->begin();
}
};
int main()
{
StringOfNode *p1=new StringOfNode;
p1->insertBack(\'b\');
p1->insertBack(\'c\');
p1->insertBack(\'e\');
p1->insert(\'d\',2);
p1->insertFront(\'a\');
p1->print();
system(\"pause\");
return 0;
}
Solution
#include
struct node
{
char value;
struct node *next;
};
class StringOfNode
{
struct node * head;
public:
StringOfNode()
{
head=NULL;
}
int size()
{
struct node *p=head;
int count=1;
while(p)
{
count++;
p=p->next;
}
return count;
}
struct node * begin()
{
return head;
}
char &operator[](int index)
{
int i=0;
struct node *p=head;
while(index!=i)
{
p=p->next;
i++;
}
return p->value;
}
void insertFront(char valuee)
{
struct node * temp=(struct node *)malloc(sizeof(struct node));
temp->value=valuee;
temp->next=NULL;
if(head==NULL)
head=temp;
else
{
temp->next=head;
head=temp;
}
}
void insert(char valuee, int pos)
{
struct node *p=head;
struct node * temp=(struct node *)malloc(sizeof(struct node));
temp->value=valuee;
temp->next=NULL;
int i=0;
if(p==NULL)
{
head=temp;
}
else
{
while(i!=pos-1)
{
p=p->next;
i++;
}
temp->next=p->next;
p->next=temp;
}
}
void insertBack(char valuee)
{
struct node *p=head;
struct node * temp=(struct node *)malloc(sizeof(struct node));
temp->value=valuee;
temp->next=NULL;
if(p==NULL)
{
head=temp;
}
else
{
while(p->next!=NULL)
p=p->next;
p->next=temp;
}
}
void print()
{
struct node *p=head;
while(p)
{
std::cout<value<<\"\ \";
p=p->next;
}
}
StringOfNode operator+(const StringOfNode& b)
{
StringOfNode *S;
struct node *p=head;
while(p->next!=NULL)
p=p->next;
p->next=S->begin();
}
};
int main()
{
StringOfNode *p1=new StringOfNode;
p1->insertBack(\'b\');
p1->insertBack(\'c\');
p1->insertBack(\'e\');
p1->insert(\'d\',2.
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because my question from this book.page270 ,question 7.4
The implementation for Mergesort given in Section 7.4 takes an array as input and sorts that
array. At the beginning of Section 7.4 there is a simple pseudocode implementation for sorting a
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Solution
Linked list:
#include
#include
struct node
{
int data;
struct node* next;
};
struct node* SortedMerge(struct node* a, struct node* b);
void FrontBackSplit(struct node* source,
struct node** frontRef, struct node** backRef);
void MergeSort(struct node** headRef)
{
struct node* head = *headRef;
struct node* a;
struct node* b;
if ((head == NULL) || (head->next == NULL))
{
return;
}
FrontBackSplit(head, &a, &b);
MergeSort(&a);
MergeSort(&b);
*headRef = SortedMerge(a, b);
}
struct node* SortedMerge(struct node* a, struct node* b)
{
struct node* result = NULL;
if (a == NULL)
return(b);
else if (b==NULL)
return(a);
if (a->data <= b->data)
{
result = a;
result->next = SortedMerge(a->next, b);
}
else
{
result = b;
result->next = SortedMerge(a, b->next);
}
return(result);
}
void FrontBackSplit(struct node* source,
struct node** frontRef, struct node** backRef)
{
struct node* fast;
struct node* slow;
if (source==NULL || source->next==NULL)
{
*frontRef = source;
*backRef = NULL;
}
else
{
slow = source;
fast = source->next;
while (fast != NULL)
{
fast = fast->next;
if (fast != NULL)
{
slow = slow->next;
fast = fast->next;
}
}
*frontRef = source;
*backRef = slow->next;
slow->next = NULL;
}
}
void printList(struct node *node)
{
while(node!=NULL)
{
printf(\"%d \", node->data);
node = node->next;
}
}
void push(struct node** head_ref, int new_data)
{
struct node* new_node =
(struct node*) malloc(sizeof(struct node));
new_node->data = new_data;
new_node->next = (*head_ref);
(*head_ref) = new_node;
}
int main()
{
struct node* res = NULL;
struct node* a = NULL;
push(&a, 15);
push(&a, 10);
push(&a, 5);
push(&a, 20);
push(&a, 3);
push(&a, 2);
MergeSort(&a);
printf(\"\ Sorted Linked List is: \ \");
printList(a);
getchar();
return 0;
}
Time complexity:o(nlogn)
Array :
#include
#include
void merge(int arr[], int l, int m, int r)
{
int i, j, k;
int n1 = m - l + 1;
int n2 = r - m;
int L[n1], R[n2];
for (i = 0; i < n1; i++)
L[i] = arr[l + i];
for (j = 0; j < n2; j++)
R[j] = arr[m + 1+ j];
i = 0; // Initial index of first subarray
j = 0;
k = l;
while (i < n1 && j < n2)
{
if (L[i] <= R[j])
{
arr[k] = L[i];
i++;
}
else
{
arr[k] = R[j];
j++;
}
k++;
}
while (i < n1)
{
arr[k] = L[i];
i++;
k++;
}
while (j < n2)
{
arr[k] = R[j];
j++;
k++;
}
}
void mergeSort(int arr[], int l, int r)
{
if (l < r)
{
int m = l+(r-l)/2;
// Sort first and second halves
mergeSort(arr, l, m);
mergeSort(arr, m+1, r);
merge(arr, l, m, r);
}
}
void printArray(int A[], i.
C++ Program to Implement Singly Linked List #includei.pdf
C++ Program to Implement Singly Linked List #includei.pdfanupambedcovers /*
* C++ Program to Implement Singly Linked List
*/
#include
#include
#include
using namespace std;
/*
* Node Declaration
*/
struct node
{
int info;
struct node *next;
}*start;
/*
* Class Declaration
*/
class single_llist
{
public:
node* create_node(int);
void insert_begin();
void insert_pos();
void insert_last();
void delete_pos();
void sort();
void search();
void update();
void reverse();
void display();
single_llist()
{
start = NULL;
}
};
/*
* Main :contains menu
*/
main()
{
int choice, nodes, element, position, i;
single_llist sl;
start = NULL;
while (1)
{
cout<<\"1.Insert Node at beginning\"<>choice;
switch(choice)
{
case 1:
cout<<\"Inserting Node at Beginning: \"<info = value;
temp->next = NULL;
return temp;
}
}
/*
* Inserting element in beginning
*/
void single_llist::insert_begin()
{
int value;
cout<<\"Enter the value to be inserted: \";
cin>>value;
struct node *temp, *p;
temp = create_node(value);
if (start == NULL)
{
start = temp;
start->next = NULL;
}
else
{
p = start;
start = temp;
start->next = p;
}
cout<<\"Element Inserted at beginning\"<>value;
struct node *temp, *s;
temp = create_node(value);
s = start;
while (s->next != NULL)
{
s = s->next;
}
temp->next = NULL;
s->next = temp;
cout<<\"Element Inserted at last\"<>value;
struct node *temp, *s, *ptr;
temp = create_node(value);
cout<<\"Enter the postion at which node to be inserted: \";
cin>>pos;
int i;
s = start;
while (s != NULL)
{
s = s->next;
counter++;
}
if (pos == 1)
{
if (start == NULL)
{
start = temp;
start->next = NULL;
}
else
{
ptr = start;
start = temp;
start->next = ptr;
}
}
else if (pos > 1 && pos <= counter)
{
s = start;
for (i = 1; i < pos; i++)
{
ptr = s;
s = s->next;
}
ptr->next = temp;
temp->next = s;
}
else
{
cout<<\"Positon out of range\"<next;s !=NULL;s = s->next)
{
if (ptr->info > s->info)
{
value = ptr->info;
ptr->info = s->info;
s->info = value;
}
}
ptr = ptr->next;
}
}
/*
* Delete element at a given position
*/
void single_llist::delete_pos()
{
int pos, i, counter = 0;
if (start == NULL)
{
cout<<\"List is empty\"<>pos;
struct node *s, *ptr;
s = start;
if (pos == 1)
{
start = s->next;
}
else
{
while (s != NULL)
{
s = s->next;
counter++;
}
if (pos > 0 && pos <= counter)
{
s = start;
for (i = 1;i < pos;i++)
{
ptr = s;
s = s->next;
}
ptr->next = s->next;
}
else
{
cout<<\"Position out of range\"<>pos;
cout<<\"Enter the new value: \";
cin>>value;
struct node *s, *ptr;
s = start;
if (pos == 1)
{
start->info = value;
}
else
{
for (i = 0;i < pos - 1;i++)
{
if (s == NULL)
{
cout<<\"There are less than \"<next;
}
s->info = value;
}
cout<<\"Node Updated\"<>value;
struct node *s;
s = start;
while (s != NULL)
{
pos++;
if (s->info == value)
{
flag = true;
cout<<\"Element \"<next;
}
if (!flag)
cout<<\"Element \"<next == NULL)
{
return;
}
ptr1 = start;
ptr2 = ptr1->next;
ptr3 = ptr2->next;
ptr1->next = NULL;
ptr2->next = ptr1;
while (ptr3 != NULL)
{
ptr1 = ptr2;
ptr2 = ptr3;
ptr3 = ptr3->next;
ptr2->next = ptr1;
}
start = ptr2;
}
/*
* Display Elements of a link list
*/
vo.
Singly Linked List
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CPP Homework HelpC++ Homework Help I am Daniel N. I am a C++ Homework Expert at cpphomeworkhelp.com. I hold a Masters in Programming from the University of Leeds, UK. I have been helping students with their homework for the past 10 years. I solve homework related to C++.
Visit cpphomeworkhelp.com or email [email protected]. You can also call on +1 678 648 4277 for any assistance with C++ Homework.
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DoublyList-cpp- #include -DoublyList-h- using namespace std- void Doub.pdf
DoublyList-cpp- #include -DoublyList-h- using namespace std- void Doub.pdfaathiauto 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(.
Ch8a
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Zone Diameter Interpretation Examine the zone diameter data in the tab.docx
Zone Diameter Interpretation Examine the zone diameter data in the tab.docxkarlynwih Zone Diameter Interpretation
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b. The relative susceptibility of these two organisms cannot be determined from these data.
c. Pseudomonas and E. coli are equally susceptible to carbenicillin.
d. Pseudomonas is more susceptible to carbenicillin than E. coli .
I tried selecting answer choice A, but it was wrong...
.
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please make a dichotomous key of these bacteria
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#include
struct node
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int data;
struct node* left, *right;
};
void printGivenLevel(struct node* root, int level);
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int h = height(root);
int i;
for (i=1; i<=h; i++)
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printGivenLevel(root->left, level-1);
printGivenLevel(root->right, level-1);
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int lheight = height(node->left);
int rheight = height(node->right);
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node->right = NULL;
return(node);
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int main()
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struct node *root = newNode(1);
root->left = newNode(2);
root->right = newNode(3);
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root->left->right = newNode(5);
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import random
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n=100
#MC simulation block
pi_all=[]
#
for j in range(n):
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struct node
{
char value;
struct node *next;
};
class StringOfNode
{
struct node * head;
public:
StringOfNode()
{
head=NULL;
}
int size()
{
struct node *p=head;
int count=1;
while(p)
{
count++;
p=p->next;
}
return count;
}
struct node * begin()
{
return head;
}
char &operator[](int index)
{
int i=0;
struct node *p=head;
while(index!=i)
{
p=p->next;
i++;
}
return p->value;
}
void insertFront(char valuee)
{
struct node * temp=(struct node *)malloc(sizeof(struct node));
temp->value=valuee;
temp->next=NULL;
if(head==NULL)
head=temp;
else
{
temp->next=head;
head=temp;
}
}
void insert(char valuee, int pos)
{
struct node *p=head;
struct node * temp=(struct node *)malloc(sizeof(struct node));
temp->value=valuee;
temp->next=NULL;
int i=0;
if(p==NULL)
{
head=temp;
}
else
{
while(i!=pos-1)
{
p=p->next;
i++;
}
temp->next=p->next;
p->next=temp;
}
}
void insertBack(char valuee)
{
struct node *p=head;
struct node * temp=(struct node *)malloc(sizeof(struct node));
temp->value=valuee;
temp->next=NULL;
if(p==NULL)
{
head=temp;
}
else
{
while(p->next!=NULL)
p=p->next;
p->next=temp;
}
}
void print()
{
struct node *p=head;
while(p)
{
std::cout<value<<\"\ \";
p=p->next;
}
}
StringOfNode operator+(const StringOfNode& b)
{
StringOfNode *S;
struct node *p=head;
while(p->next!=NULL)
p=p->next;
p->next=S->begin();
}
};
int main()
{
StringOfNode *p1=new StringOfNode;
p1->insertBack(\'b\');
p1->insertBack(\'c\');
p1->insertBack(\'e\');
p1->insert(\'d\',2);
p1->insertFront(\'a\');
p1->print();
system(\"pause\");
return 0;
}
Solution
#include
struct node
{
char value;
struct node *next;
};
class StringOfNode
{
struct node * head;
public:
StringOfNode()
{
head=NULL;
}
int size()
{
struct node *p=head;
int count=1;
while(p)
{
count++;
p=p->next;
}
return count;
}
struct node * begin()
{
return head;
}
char &operator[](int index)
{
int i=0;
struct node *p=head;
while(index!=i)
{
p=p->next;
i++;
}
return p->value;
}
void insertFront(char valuee)
{
struct node * temp=(struct node *)malloc(sizeof(struct node));
temp->value=valuee;
temp->next=NULL;
if(head==NULL)
head=temp;
else
{
temp->next=head;
head=temp;
}
}
void insert(char valuee, int pos)
{
struct node *p=head;
struct node * temp=(struct node *)malloc(sizeof(struct node));
temp->value=valuee;
temp->next=NULL;
int i=0;
if(p==NULL)
{
head=temp;
}
else
{
while(i!=pos-1)
{
p=p->next;
i++;
}
temp->next=p->next;
p->next=temp;
}
}
void insertBack(char valuee)
{
struct node *p=head;
struct node * temp=(struct node *)malloc(sizeof(struct node));
temp->value=valuee;
temp->next=NULL;
if(p==NULL)
{
head=temp;
}
else
{
while(p->next!=NULL)
p=p->next;
p->next=temp;
}
}
void print()
{
struct node *p=head;
while(p)
{
std::cout<value<<\"\ \";
p=p->next;
}
}
StringOfNode operator+(const StringOfNode& b)
{
StringOfNode *S;
struct node *p=head;
while(p->next!=NULL)
p=p->next;
p->next=S->begin();
}
};
int main()
{
StringOfNode *p1=new StringOfNode;
p1->insertBack(\'b\');
p1->insertBack(\'c\');
p1->insertBack(\'e\');
p1->insert(\'d\',2.
L25-L26-Parameter passing techniques.pptx
L25-L26-Parameter passing techniques.pptxhappycocoman The document discusses different parameter passing techniques in C like pass by value, pass by reference, and how to pass arrays to functions. It provides examples of passing single and multi-dimensional arrays to functions using pass by value and pass by reference. It also discusses nesting of functions and using pointers as function arguments.
pleaase I want manual solution forData Structures and Algorithm An.pdf
pleaase I want manual solution forData Structures and Algorithm An.pdfwasemanivytreenrco51 pleaase I want manual solution for
Data Structures and Algorithm Analysis in C++, Third Edition
By Clifford A. Shaffer
because my question from this book.page270 ,question 7.4
The implementation for Mergesort given in Section 7.4 takes an array as input and sorts that
array. At the beginning of Section 7.4 there is a simple pseudocode implementation for sorting a
linked list using Mergesort. Implement both a linked list-based version of Mergesort and the
array-based version of Mergesort, and compare and analyze their running times.
Solution
Linked list:
#include
#include
struct node
{
int data;
struct node* next;
};
struct node* SortedMerge(struct node* a, struct node* b);
void FrontBackSplit(struct node* source,
struct node** frontRef, struct node** backRef);
void MergeSort(struct node** headRef)
{
struct node* head = *headRef;
struct node* a;
struct node* b;
if ((head == NULL) || (head->next == NULL))
{
return;
}
FrontBackSplit(head, &a, &b);
MergeSort(&a);
MergeSort(&b);
*headRef = SortedMerge(a, b);
}
struct node* SortedMerge(struct node* a, struct node* b)
{
struct node* result = NULL;
if (a == NULL)
return(b);
else if (b==NULL)
return(a);
if (a->data <= b->data)
{
result = a;
result->next = SortedMerge(a->next, b);
}
else
{
result = b;
result->next = SortedMerge(a, b->next);
}
return(result);
}
void FrontBackSplit(struct node* source,
struct node** frontRef, struct node** backRef)
{
struct node* fast;
struct node* slow;
if (source==NULL || source->next==NULL)
{
*frontRef = source;
*backRef = NULL;
}
else
{
slow = source;
fast = source->next;
while (fast != NULL)
{
fast = fast->next;
if (fast != NULL)
{
slow = slow->next;
fast = fast->next;
}
}
*frontRef = source;
*backRef = slow->next;
slow->next = NULL;
}
}
void printList(struct node *node)
{
while(node!=NULL)
{
printf(\"%d \", node->data);
node = node->next;
}
}
void push(struct node** head_ref, int new_data)
{
struct node* new_node =
(struct node*) malloc(sizeof(struct node));
new_node->data = new_data;
new_node->next = (*head_ref);
(*head_ref) = new_node;
}
int main()
{
struct node* res = NULL;
struct node* a = NULL;
push(&a, 15);
push(&a, 10);
push(&a, 5);
push(&a, 20);
push(&a, 3);
push(&a, 2);
MergeSort(&a);
printf(\"\ Sorted Linked List is: \ \");
printList(a);
getchar();
return 0;
}
Time complexity:o(nlogn)
Array :
#include
#include
void merge(int arr[], int l, int m, int r)
{
int i, j, k;
int n1 = m - l + 1;
int n2 = r - m;
int L[n1], R[n2];
for (i = 0; i < n1; i++)
L[i] = arr[l + i];
for (j = 0; j < n2; j++)
R[j] = arr[m + 1+ j];
i = 0; // Initial index of first subarray
j = 0;
k = l;
while (i < n1 && j < n2)
{
if (L[i] <= R[j])
{
arr[k] = L[i];
i++;
}
else
{
arr[k] = R[j];
j++;
}
k++;
}
while (i < n1)
{
arr[k] = L[i];
i++;
k++;
}
while (j < n2)
{
arr[k] = R[j];
j++;
k++;
}
}
void mergeSort(int arr[], int l, int r)
{
if (l < r)
{
int m = l+(r-l)/2;
// Sort first and second halves
mergeSort(arr, l, m);
mergeSort(arr, m+1, r);
merge(arr, l, m, r);
}
}
void printArray(int A[], i.
C++ Program to Implement Singly Linked List #includei.pdf
C++ Program to Implement Singly Linked List #includei.pdfanupambedcovers /*
* C++ Program to Implement Singly Linked List
*/
#include
#include
#include
using namespace std;
/*
* Node Declaration
*/
struct node
{
int info;
struct node *next;
}*start;
/*
* Class Declaration
*/
class single_llist
{
public:
node* create_node(int);
void insert_begin();
void insert_pos();
void insert_last();
void delete_pos();
void sort();
void search();
void update();
void reverse();
void display();
single_llist()
{
start = NULL;
}
};
/*
* Main :contains menu
*/
main()
{
int choice, nodes, element, position, i;
single_llist sl;
start = NULL;
while (1)
{
cout<<\"1.Insert Node at beginning\"<>choice;
switch(choice)
{
case 1:
cout<<\"Inserting Node at Beginning: \"<info = value;
temp->next = NULL;
return temp;
}
}
/*
* Inserting element in beginning
*/
void single_llist::insert_begin()
{
int value;
cout<<\"Enter the value to be inserted: \";
cin>>value;
struct node *temp, *p;
temp = create_node(value);
if (start == NULL)
{
start = temp;
start->next = NULL;
}
else
{
p = start;
start = temp;
start->next = p;
}
cout<<\"Element Inserted at beginning\"<>value;
struct node *temp, *s;
temp = create_node(value);
s = start;
while (s->next != NULL)
{
s = s->next;
}
temp->next = NULL;
s->next = temp;
cout<<\"Element Inserted at last\"<>value;
struct node *temp, *s, *ptr;
temp = create_node(value);
cout<<\"Enter the postion at which node to be inserted: \";
cin>>pos;
int i;
s = start;
while (s != NULL)
{
s = s->next;
counter++;
}
if (pos == 1)
{
if (start == NULL)
{
start = temp;
start->next = NULL;
}
else
{
ptr = start;
start = temp;
start->next = ptr;
}
}
else if (pos > 1 && pos <= counter)
{
s = start;
for (i = 1; i < pos; i++)
{
ptr = s;
s = s->next;
}
ptr->next = temp;
temp->next = s;
}
else
{
cout<<\"Positon out of range\"<next;s !=NULL;s = s->next)
{
if (ptr->info > s->info)
{
value = ptr->info;
ptr->info = s->info;
s->info = value;
}
}
ptr = ptr->next;
}
}
/*
* Delete element at a given position
*/
void single_llist::delete_pos()
{
int pos, i, counter = 0;
if (start == NULL)
{
cout<<\"List is empty\"<>pos;
struct node *s, *ptr;
s = start;
if (pos == 1)
{
start = s->next;
}
else
{
while (s != NULL)
{
s = s->next;
counter++;
}
if (pos > 0 && pos <= counter)
{
s = start;
for (i = 1;i < pos;i++)
{
ptr = s;
s = s->next;
}
ptr->next = s->next;
}
else
{
cout<<\"Position out of range\"<>pos;
cout<<\"Enter the new value: \";
cin>>value;
struct node *s, *ptr;
s = start;
if (pos == 1)
{
start->info = value;
}
else
{
for (i = 0;i < pos - 1;i++)
{
if (s == NULL)
{
cout<<\"There are less than \"<next;
}
s->info = value;
}
cout<<\"Node Updated\"<>value;
struct node *s;
s = start;
while (s != NULL)
{
pos++;
if (s->info == value)
{
flag = true;
cout<<\"Element \"<next;
}
if (!flag)
cout<<\"Element \"<next == NULL)
{
return;
}
ptr1 = start;
ptr2 = ptr1->next;
ptr3 = ptr2->next;
ptr1->next = NULL;
ptr2->next = ptr1;
while (ptr3 != NULL)
{
ptr1 = ptr2;
ptr2 = ptr3;
ptr3 = ptr3->next;
ptr2->next = ptr1;
}
start = ptr2;
}
/*
* Display Elements of a link list
*/
vo.
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Singly Linked ListEr. Ganesh Ram Suwal This C program implements functions for a singly linked list including inserting and deleting nodes from different positions as well as displaying the list. The main function uses a switch case to allow the user to choose an operation on the list such as inserting a new node at the beginning, end, or a specified position. Other operations include deleting nodes from the beginning, end, or a specified position. The length of the list is tracked and positions are validated. Pointers are used to traverse the list and add/remove nodes as needed.
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CPP Homework HelpC++ Homework Help I am Daniel N. I am a C++ Homework Expert at cpphomeworkhelp.com. I hold a Masters in Programming from the University of Leeds, UK. I have been helping students with their homework for the past 10 years. I solve homework related to C++.
Visit cpphomeworkhelp.com or email [email protected]. You can also call on +1 678 648 4277 for any assistance with C++ Homework.
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DoublyList-cpp- #include -DoublyList-h- using namespace std- void Doub.pdf
DoublyList-cpp- #include -DoublyList-h- using namespace std- void Doub.pdfaathiauto 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(.
Ch8a
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if (L = nil)
then
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else if ( ( subL = nil ) orelse
( hd(subL) = hd(L) ) )
then
groupdupes_helper( tl(L), hd(L)::subL, result )
else
groupdupes_helper( tl(L), hd(L)::nil, result @ [subL] )
(* groupdupes Wrapper Function *)
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import java.util.*;
public class ArrayListNames
{
publicstaticvoid main(String[] args)
{
int i;
ArrayList<String> name = new ArrayList<String>();
System.out.println(\"Enter 10 Names \");
for( i=0; i<10; i++ )
{
Scanner s =newScanner(System.in);
name add(s.next());
}
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Collections.shuffle(name);
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Object minimum = Collections.min(name);
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}
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#!/user/bin/python
def add(x, y):
return x + y
def subtract(x, y):
return x - y
def multiply(x, y):
return x * y
def divide(x, y):
return x / y
def exp(x, y):
return x ** y
def mod(x, y):
return x % y
print(\"Select any one operation what you want to perform.\")
print(\"1.Add\")
print(\"2.Subtract\")
print(\"3.Multiply\")
print(\"4.Divide\")
print(\"5.Exponent\")
print(\"6.Remainder\")
choice = input(\"Enter choice(1/2/3/4/5/6):\")
num1 = int(input(\"Enter first number: \"))
num2 = int(input(\"Enter second number: \"))
if choice == \'1\':
print(num1,\"+\",num2,\"=\", add(num1,num2))
elif choice == \'2\':
print(num1,\"-\",num2,\"=\", subtract(num1,num2))
elif choice == \'3\':
print(num1,\"*\",num2,\"=\", multiply(num1,num2))
elif choice == \'4\':
print(num1,\"/\",num2,\"=\", divide(num1,num2))
elif choice == \'5\':
print(num1,\"**\",num2,\"=\", exp(num1,num2))
elif choice == \'6\':
print(num1,\"%\",num2,\"=\", mod(num1,num2))
else:
print(\"Invalid input\")
.
write a topic about RFID in SCM and what you have learned about adopti.docx
write a topic about RFID in SCM and what you have learned about adopti.docxkarlynwih write a topic about RFID in SCM and what you have learned about adoption of RFID in SCM
Solution
RFID(Radio frequency identification) in SRM is a technology that incorporates the use of electromagnetic or electrostatic coupling in the radio frequency portion of the electromagnetic spectrum to uniquely identify an object, animal, or person. RFDI is coming into increasing use in industry as an alternative to the barcode.
The advantage of RFDI is that it does not require direct contact or line of sight scanning.
It consists of three components - 1) Antenna, 2. Transreciever, 3. Transponder
The term supply chain covers all possible processes involved in the flow of goods from manufacturing to customers including manufacturing, distribution and transportaion. RFDI technology is risen to become a revolutionary elament in supply chain management. It is not replacement for barcodes. RFDI is an innovative technology that will make the supply chain faster and efficient with less human error.
There is a common trend when integrating the supply chain for many companies and provides a good model for better understanding of what the adoption of RFDI technology can do for supply chain network.
there are many factors that maybe considered for adoption in RFDI in SCM like
information technology - complexity, organization compatibility.
Industrial organization - firm performance is enabled or constrained by industry structure.
Resource based view- presence of resource that meet certain conditions, such as value, rarity, imperfect, imitability and lack of substituability.
.
Write up a simple c-program to find the median of any array-Solution#i.docx
Write up a simple c-program to find the median of any array-Solution#i.docxkarlynwih Write up a simple c-program to find the median of any array?
Solution
#include <stdio.h>
#include <conio.h>
void sort (int[],int);
float median (int A[], int nElem);
int main()
{int A[14]={1,2,3,4,4,2,2,1,2,4,1,4,4,4},nElem=14,i,m;
float fm;
printf(\"The array:\ \");
for(i=0;i<nElem;i++)
printf(\"%d \",A[i]);
printf(\"\ \ \");
sort(A,nElem);
printf(\"The array sorted is:\ \");
for(i=0;i<nElem;i++)
printf(\"%d \",A[i]);
printf(\"\ \ \");
fm=median(A,nElem);
printf(\"The median is %.2f\ \",fm);
getch();
return 0;
}
void sort (int A[], int nElem)
{int i,j,t;
for(i=0;i<nElem-1;i++)
for(j=i+1;j<nElem;j++)
if(A[i]>A[j])
{t=A[i];
A[i]=A[j];
A[j]=t;
}
}
float median (int A[], int nElem)
{float m;
if(nElem%2==0)
return((float)(A[nElem/2-1]+A[nElem/2])/2.);
else
return A[nElem/2];
}
.
Write an assembly language program in the Pep-8 simulator that corresp.docx
Write an assembly language program in the Pep-8 simulator that corresp.docxkarlynwih Write an assembly language program in the Pep/8 simulator that corresponds to the following C++ program:
#include <iostream>
using namespace std;
int num1;
int num2;
int main() {
cin >> num1 >> num2;
cout << num2 << endl << num1 << endl;
return 0;
}
Solution
c++ Program:
#include<iostream>
using namespace std;
int num1;
int num2;
int main()
{
cin>>num1>>num2;
cout<<num2<<endl<<num1<<endl;
return 0;
}
Assembly Language:
BR main
Sum1: .WORD 0x0000
Sum2: .WORD 0x0000
Num1: .BLOCK 2
Num2: .BLOCK 2
Main: LDA sum1,sum2,d ; ‘d’ = direct
DECI num1,d
ADDA num1,d
STA sum1,d
DEC2 num2,d
STA sum2,d
DECO sum1,d
DECO sum2,d
STOP
.END
.
Write the for structure in JAVA coding to read and display all element.docx
Write the for structure in JAVA coding to read and display all element.docxkarlynwih Write the for structure in JAVA coding to read and display all elements in the third row of array x.
Solution
import java.util.Scanner;
public class Program1 {
public static void main(String[] args) {
Scanner sc= new Scanner(System.in);
System.out.print(\"Enter number of elements: \");
int n= sc.nextInt();
//creating an array of element n
int []x = new int[n];
//now reading array elements
System.out.println(\"Enter \"+n+\" elements: \");
for(int i=0; i<n; i++)
x[i] = sc.nextInt();
// now displaying using enhanced for loop
System.out.println();
System.out.println(\"Array elemets are: \");
for(int e: x)
System.out.print(e+\" \");
System.out.println();
}
}
/*
Sample Output:
Enter number of elements: 6
Enter 6 elements:
4
8
9
23
12
15
Array elemets are:
4 8 9 23 12 15
*/
.
Write the formal description of the following state machine (M) What.docx
Write the formal description of the following state machine (M) What.docxkarlynwih Write the formal description of the following state machine (M) What is the language recognized by M;L(M)?
Solution
the language that is accepted is: 0 * 10 * 10 *
explanation:
start node contains the self loop with value \'0\' i.e it may iterate as many times as it needed zero or more number of times
* represents zero or more number of times
\' s 0 \' \': it must pass through \'1\' no other choice
it reaches s 1
when it is on s 1 it has self loop with \'0\'
so it will iterate zero or more number of times i.e 0 *
\' s 1 \' \': it must pass through \'1\' no other choice
it reaches s 2
when it is on s2 it has self loop with \'0\'
so it will iterate zero or more number of times i.e 0 *
s 2 is the end state
so we cannot go from here an further
.
Write the C++ code for a function getInput which will read in an unkno.docx
Write the C++ code for a function getInput which will read in an unkno.docxkarlynwih Write the C++ code for a function getInput which will read in an unknown of values from a file and store them in a one-dimensional array of integers. The function stores, in a reference parameter, the actual number of values it reads. The input file is declared and opened in main(), so the input filestream variable is needed as a function parameter. Write a C ++ function to find the minimum in a designated row of a 2-dimensional array which has 10 columns. The function parameters are the array, the row index of the row in which the minimum is to be found, and a reference parameter for the minimum value.
Solution
4.
#include<iostream.h>
#include<conio.h>
void min_row(int a[4][10],int row)
{
int m=a[row][0];
for(int i=1;i<10;i++)
{
if(a[row][i]<m)
m=a[row][i];
}
cout<<\"minimum number in row \"<<row<<\" is \"<<m;
}
void main()
{
int arr[4][10],i,j,r;
clrscr();
cout<<\"Enter Element to the matrix\";
for(i=0;i<4;i++)
{
cout<<\" Row \"<<i<<\" :\";
for(j=0;j<10;j++)
{
cout<<\"Column \"<<j<<\":\";
cin>>arr[i][j];
}
}
cout<<\"\ Enter the row number\";
cin>>r;
min_row(arr,r);
getch();
}
.
Write the balanced reaction where thiosulfate and protons are the only.docx
Write the balanced reaction where thiosulfate and protons are the only.docxkarlynwih Write the balanced reaction where thiosulfate and protons are the only reactants; and water, elemental sulfur, sulfur trioxide, and water are the only products. (Hint: this easier if you start with the balanced half-reactions). How many electrons are transferred in the balanced oxidation half reaction? How many electrons are transferred in the balanced reduction half reaction? How many electrons are transferred in the overall reaction?
Solution
S2O3^2- (aq) + H^+ -------------> S + SO3 + H2O
Oxidation half reaction                                                                               reduction half reaction
S2O3^2- (aq) ------------> 2SO3(g)Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â S2O3^2- ------------> 2S
S2O3^2- (aq)+3H2O ------------> 2SO3(g)Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â S2O3^2- ------------> 2S+ 3H2O
S2O3^2- (aq)+3H2O ------------> 2SO3(g)+ 6H^+(aq) Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â S2O3^2- + 6H^+ (aq) ------------> 2S+ 3H2O
S2O3^2- (aq)+3H2O ------------> 2SO3(g)+ 6H^+(aq) + 8e^- Â Â Â Â Â Â Â Â S2O3^2- + 6H^+ (aq) +4e^- ------------> 2S+ 3H2O
S2O3^2- (aq)+3H2O ------------> 2SO3(g)+ 6H^+(aq) + 8e^- Â Â Â Â Â Â Â ( 8 electrons are transferred)
2S2O3^2- + 12H^+ (aq) +8e^- ------------> 4S+ 6H2OÂ Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â ( 8 electrons are transferred)
------------------------------------------------------------------------------------------------------
3S2O3^2- (aq) +6H^+ (aq) --------------> 2So3(g) + 4S +3H2O overall reaction
.
Write functions odd and even- which takes a list of symbols L- and pro.docx
Write functions odd and even- which takes a list of symbols L- and pro.docxkarlynwih Write functions odd and even, which takes a list of symbols L, and produces a sublist of L containing symbols located at odd and even positions repectively. E.g., (odd \'(a b c d)) = (a c) (even \'(a b c d)) = (b d) (odd \'(a)) = (a) (even \'(a)) = ()
Solution
//just assumed that the size of array is 10 you can chnage it accordingly
void oddAndEven(char arr[10])
{
char sublistodd[5],sublisteven[5];
int j,k;
for(int i=0; i <10 ; i++)
{
if(i/2 == 0 && i<=4)
{
sublisteven[i] = arr[i];
j = i;
j=j+1;
}
else if(i/2 != 0 && i<=4)
{
sublistodd[i] = arr[i];
k = i;
k++;
}
else if(i/2 == 0 && i>4 && j<5)
{
sublisteven[j] = arr[i];
j++;
}
else if(i/2 !=0 && i>4 && k<5)
{
sublistodd[k] = arr[i];
k++;
}
}
}
.
Write about the two sources of energy Write about the two sources of.docx
Write about the two sources of energy Write about the two sources of.docxkarlynwih Write about the two sources of energy
Write about the two sources of energy
Solution
Two sources of energy can be solar energy and wind energy
Solar Energy
Sun is the primary source of all energy. Light given by Sun, sunlight, is a clean and renewable source of energy. It is a sustainable resource which means it doesn\'t run out, but it can be maintained because the sun is always giving off energy. Coal or gas are not categorized as sustainable or renewable because once they are used, there is none left. More and more people have been using clean, renewable energy such as solar, wind, geothermal steam and others. It is called \'Green Power\' .
Some Advantages of Solar Energy
Some Disadvantages of Solar Energy
Wind Energy
Another source of renewable energy is wind. Wind can be described as the natural movement of air across the land or sea. To produce electricity, wind is used to turn the blades of a wind mill turns the shaft to which they are attached. This movement of the windmill shaft through a pump or generator produces electricity.
Advantages
Disadvantages
.
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