Write code in c++ Program to a topological sort on a graph Program pl.docx
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Write code in c++ Program to a topological sort on a graph Program plan: Initializes main() function. Declare int type variables. Accept the number of vertices in the graph and the adjacency matrix of in-degree. Perform topological sort using for loops. Display the result. Solution Please find the required code and output below: #include <iostream> using namespace std; struct node_info { int no; int lv_time, st_time; }*q = NULL, *r = NULL; struct node { node_info *pt; node *next; }*top = NULL, *p = NULL, *np = NULL; int c = 0; void push(node_info *ptr) { np = new node; np->pt = ptr; np->next = NULL; if (top == NULL) { top = np; } else { np->next = top; top = np; } } node_info *pop() { if (top == NULL) { cout<<\"underflow\ \"; } else { p = top; top = top->next; return(p->pt); delete(p); } } void topologicalSort(int *v,int am[][7],int i) { q = new node_info; q->no = i; q->st_time = c; cout<<\"start time for node no \"<<q->no<<\":\"<<c<<endl; push(q); v[i] = 1; for (int j = 0; j < 7; j++) { if (am[i][j] == 0 || (am[i][j] == 1 && v[j] == 1)) continue; else if(am[i][j] == 1 && v[j] == 0) { c++; topologicalSort(v,am,j); } } c++; r = pop(); cout<<\"leave time for \"<<r->no<<\":\"<<c<<endl; return; } int main() { int v[7],am[7][7]; for (int i = 0; i < 7; i++) v[i] = 0; for (int i = 0; i < 7; i++) { cout<<\"enter the values for adjacency matrix row:\"<<i + 1<<endl; for(int j = 0; j < 7; j++) { cin>>am[i][j]; } } topologicalSort(v,am,0); } ----------------------------------------------------------------------------------------------- OUTPUT: .
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![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;
}](data:image/gif;base64,R0lGODlhAQABAIAAAAAAAP///yH5BAEAAAAALAAAAAABAAEAAAIBRAA7)
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Write code in C++ Write a program to perform a topological sort on a g.docx
Write code in C++
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Name : Shamshad Jamali
Roll No :22BS( IT ) 40
Subject : Programming .. Fundamental
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normal program
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while(quotient!=0){
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quotient = quotient / 2;
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printf(\"Equivalent binary value of decimal number %d: \",decimalNumber);
for(j = i -1 ;j> 0;j--)
printf(\"%d\",binaryNumber[j]);
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int data;
struct node* next;
};
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{
struct node* new_node =
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(*head) = new_node;
}
void printList(struct node *head)
{
struct node *temp = head;
while(temp != NULL)
{
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temp = temp->next;
}
}
void reverseList(struct node** head)
{
struct node* prev = NULL;
struct node* current = *head;
struct node* next;
while (current != NULL)
{
next = current->next;
current->next = prev;
prev = current;
current = next;
}
*head = prev;
}
int main()
{
struct node* head = NULL;
push(&head, 12);
push(&head, 14);
push(&head, 5);
push(&head, 8);
printf(\"\ Linked list Before Reversing :\ \");
printList(head);
reverseList(&head);
printf(\"\ Linked list After Reversing :\ \");
printList(head);
}
--------------------------------------------------------------------------------
OUTPUT:
Linked list Before Reversing :
8Â Â 5Â Â 14Â Â 12
Linked list After Reversing :
12Â Â 14Â Â 5Â Â 8
.
AnswerThe new program with the required constructor and with a te.pdf
Answer:
The new program with the required constructor and with a test is given below
Code:
#include
#include
using namespace std;
class TNode
{
public:
int data;
TNode *left;
TNode *right;
TNode(int n){ data = n; left = right = 0; }
TNode(int n, TNode *left, TNode *right)
{
data = n;
//Assign the left node address and right node address with the parameters passed
this->left = left;
this->right = right;
}
};
void printTree(TNode *t)
{
if (!t) return;
printTree(t->left);
cout << t->data << \" \";
printTree(t->right);
}
void search_for(int n, TNode *t)
{
cout << \"Searhing for \" << n << endl;
while (t)
{
cout << \"looking at \" << t->data << endl;
if (t->data == n)
{
cout << \"Found it!\";
return;
}
if (n < t->data)
{
cout << \"Going left\ \";
t = t->left;
}
else
{
cout << \"Going right\ \";
t = t->right;
}
}
cout << \"Can\\\'t find it\";
}
// Creates a two node tree
int main()
{
TNode *root = 0;
root = new TNode(5);
cout << root->data << endl;
TNode *one_node = new TNode(1);
TNode *three_node = new TNode(3);
//use the new constructor
TNode *two_node = new TNode(2, one_node, three_node);
root->left = new TNode(3);
printTree(root);
cout << endl;
search_for(3, root);
getchar();
return 0;
}
Solution
Answer:
The new program with the required constructor and with a test is given below
Code:
#include
#include
using namespace std;
class TNode
{
public:
int data;
TNode *left;
TNode *right;
TNode(int n){ data = n; left = right = 0; }
TNode(int n, TNode *left, TNode *right)
{
data = n;
//Assign the left node address and right node address with the parameters passed
this->left = left;
this->right = right;
}
};
void printTree(TNode *t)
{
if (!t) return;
printTree(t->left);
cout << t->data << \" \";
printTree(t->right);
}
void search_for(int n, TNode *t)
{
cout << \"Searhing for \" << n << endl;
while (t)
{
cout << \"looking at \" << t->data << endl;
if (t->data == n)
{
cout << \"Found it!\";
return;
}
if (n < t->data)
{
cout << \"Going left\ \";
t = t->left;
}
else
{
cout << \"Going right\ \";
t = t->right;
}
}
cout << \"Can\\\'t find it\";
}
// Creates a two node tree
int main()
{
TNode *root = 0;
root = new TNode(5);
cout << root->data << endl;
TNode *one_node = new TNode(1);
TNode *three_node = new TNode(3);
//use the new constructor
TNode *two_node = new TNode(2, one_node, three_node);
root->left = new TNode(3);
printTree(root);
cout << endl;
search_for(3, root);
getchar();
return 0;
}.
Doublylinklist
This document presents information on doubly linked lists including their representation, initialization, node creation, and various operations like insertion, deletion, and traversal. It discusses inserting and deleting nodes at the beginning or end of the list, as well as inserting before or after a specified node. Code examples are provided for initializing a doubly linked list and performing each operation.
L25-L26-Parameter passing techniques.pptx
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.
-- Task 2- Debugging a program with stacks- queues- and doubly-linked.docx
/* Task 2: Debugging a program with stacks, queues, and doubly-linked lists There are a number of errors in the following program. All errors are located in main() and structure definitions. Function declarations and definitions are correct! Locate all errors, fix them (as shown below), run the program and save its output as a comment at the end of the source file. Example: int num = 10; int *ptr; num = &ptr; // <== Error: Comment the line and write the correct line below // Write a short justification where appropriate // num = &ptr; // Error #1 ptr = # Name: */ #include #include #include #include #define DUMMY_TRAILER '\177' // octal ASCII code of the // last character in the ASCII table #define NUM_CITIES 10 typedef struct { char name[12]; int temperature[5]; } CITY; // Stack and Queue Node typedef struct node NODE; struct node { CITY city; node *next; }; // Doubly Linked List Node typedef struct d_node D_NODE; struct d_node { CITY city; NODE *forw; NODE *back; }; // Stack Functions NODE *push(NODE *stack, const CITY *pStu); NODE *pop(NODE **stack); // Queue Functions void enqueue(NODE **queue, NODE **rear, const CITY *pStu); NODE *dequeue(NODE **queue, NODE **rear); // Doubly Linked List Functions D_NODE *init_list(void); int insert(D_NODE *list, const CITY *pStu); void traverse_forw(D_NODE *list); void traverse_back(D_NODE *list); // Other Functions void printCity(const CITY *pCity); int main (void) { CITY cList[NUM_CITIES] = { {"Cupertino", {88, 89, 87, 85, 89}}, {"Flagstaff", {81, 80, 88, 89, 89}}, {"Los Angeles", {87, 88, 89, 89, 90}}, {"Philadelphia", {96, 99, 99, 90, 95}}, {"Phoenix", {106, 109, 109, 100, 105}}, {"Portland", {89, 90, 85, 89, 90}}, {"Reno", {108, 105, 109, 100, 108}}, {"Salem", {85, 90, 85, 89, 90}}, {"Tucson", {107, 100, 109, 100, 108}}, {"Yreka", {101, 109, 100, 108, 109}} }; NODE *stack = NULL; NODE *top = NULL; NODE *queue = NULL, *rear = NULL; NODE *front; D_NODE *list; list = init_list(); // build stack and queue with data from an array of CITY structures srand((unsigned int)time(NULL)); int count = rand() % 10; for ( int n = 0; n < count; n++) { int i = rand() % NUM_CITIES; int duplicate = insert(list, &cList[i]); if(duplicate) { // already in the list! push(stack, &cList[i]); enqueue(&queue, &rear, cList[i]); } } // display list printf("\nLIST contents (forwards):\n"); traverse_forw(list); printf("\nLIST contents (backwards):\n"); traverse_back(list); // display stack if (top) { printf("\nSTACK contents from top to bottom:\n"); while ((top = pop(stack))) { printCity(top->city); } } else printf ("Empty Stack!\n"); // display queue if (front) { printf("\nQUEUE contents from front to rear:\n"); while ((front = dequeue( queue, rear))) { printCity(front->city); } } else printf ("Empty Queue!\n"); return 0; } /*************************************************** Displays the fileds of a CIS_CLASS structure Pre pCls - a pointer to a CIS_CLASS structure Post */ void printCity(const CITY *pCity) { printf("%-20s %3d\n",.
Q1 Consider the below omp_trap1.c implantation, modify the code so t.pdf
Q1 Consider the below omp_trap1.c implantation, modify the code so that: i. The time used by
the parallel block is timed using the OpenMP function omp_get_wtime(). The syntax is double
omp_get_wtime ( void ) It returns the number of seconds that have elapsed since some time in
the past. ii. Now on a system with at least two cores, time the program with a. one thread and a
large value of n, and b. two threads and the same value of n. Explain what happens? Then Do the
same measurements for the omp_trap2.c code.
1. How does its performance compare with the performance of the omp_trap1.c ?
2. Explain your answers.
3. provide with your answers the code and detailed screenshots showing how did you compile
and run the code and times for each run:
/* File: omp_trap1.c * Purpose: Estimate definite integral (or area under curve) using trapezoidal
rule. * Input: a, b, n * Output: estimate of integral from a to b of f(x) * using n trapezoids. * *
Compile: gcc -g -Wall -fopenmp -o omp_trap1 omp_trap1.c * Usage: ./omp_trap1 * * Notes: *
1. The function f(x) is hardwired. * 2. In this version, each thread explicitly computes the
integral over its assigned subinterval, a critical directive is used for the global sum. * 3. This
version assumes that n is evenly divisible by the number of threads #include #include #include
#include void Usage(char* prog_name); double f(double x); /* Function we're integrating */
double Local_trap ( double a , double b , i n t n ) ; int main(int argc, char* argv[]) { double
global_result = 0.0; /* Store result in global_result */ double a, b; /* Left and right endpoints */
int n; /* Total number of trapezoids */ int thread_count; if (argc != 2) Usage(argv[0]);
thread_count = strtol(argv[1], NULL, 10); printf("Enter a, b, and n\n"); scanf("%lf %lf %d", &a,
&b, &n); if (n % thread_count != 0) Usage(argv[0]); double global_result = 0.0; # pragma omp p
a r a l l e l num_threads ( t h r e a d _ c o u n t ) { # pragma omp c r i t i c a l global_result +=
Local_trap ( double a , double b , i n t n ) ; } printf("With n = %d trapezoids, our estimate\n", n);
printf("of the integral from %f to %f = %.14e\n", a, b, global_result); return 0; } /* main */ /*----
---------------------------------------------------------------- * Function: Usage * Purpose: Print
command line for function and terminate * In arg: prog_name */ void Usage(char* prog_name) {
fprintf(stderr, "usage: %s \n", prog_name); fprintf(stderr, " number of trapezoids must be evenly
divisible by\n"); fprintf(stderr, " number of threads\n"); exit(0); } /* Usage */ /*---------------------
--------------------------------------------- * Function: f * Purpose: Compute value of function to be
integrated * Input arg: x * Return val: f(x) */ double f(double x) { double return_val; return_val
= x*x; return return_val; } /* f */ /*------------------------------------------------------------------ *
Function: Trap * Purpose: Use trapezoidal rule to estimate definite integral * Input args.
Suppose the population of interest in a particular study is all Queen'.docx
Suppose the population of interest in a particular study is all Queen's students. The researchers email a questionnaire to 20 randomly selected students that are black. 20 that are white, 20 that are asian, 20 that are hispanic, and 20 that don't fall into any of these categories (other). What type of sampling is this? Stratified Random Sampling Simple Random Sampling Voluntary Response Sampling Multistage Random Sampling
.
Suppose the Federal Reserve increases deposits at financial institutio.docx
Suppose the Federal Reserve increases deposits at financial institutions by $47 billion through its open market operations. If the reserve requirement for all deposits is 16%, what is the maximum impact the Fed's actions can have on total deposits?
.
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In C Programming- Create a program that converts a number from decimal.docx
In C Programming:
Create a program that converts a number from decimal to binary. A stack MUST be used to complete the program. Create an implementation of a stack. An array or linked list can be used as the underlying structure. The following stack functions MUST be implemented into the program:
• Void pop()
• Void push(int data)
• Int top()
• Int isEmpty()
The program should take decimal input from the user, and convert it to binary, and simply print it out.
Example: Enter a number to convert to binary: 29 11101
Solution
#include<stdio.h>
#include<conio.h>
#include<process.h>
#define MAX 10
typedef struct stack
{
int data[MAX];
int top;
}stack;
//---------------------
int empty(stack *s)
{
if(s->top==-1)
return(1);
return(0);
}
//----------------------
int full(stack *s)
{
if(s->top==MAX-1)
return(1);
return(0);
}
//-----------------------
void push(stack *s,int x)
{
s->top=s->top+1;
s->data[s->top]=x;
}
//-----------------------
int pop(stack *s)
{
int x;
x=s->data[s->top];
s->top=s->top-1;
return(x);
}
//------------------------
void main()
{
stack s;
int num;
s.top=-1;
printf(\"nEnter decimal number:\");
scanf(\"%d\",&num);
while((num!=0))
{
if(!full(&s))
{
push(&s,num%2);
num=num/2;
}
else
{
printf(\"nStack overflow\");
exit(0);
}
}
printf(\"n\");
while(!empty(&s))
{
num=pop(&s);
printf(\"%d\",num);
}
}
.
DATA STRUCTURE USING C & C++
The document describes data structures and their implementation using C and C++. It includes:
1) An introduction to data structures and their use in organizing data efficiently.
2) Requirements for using data structures in C and C++ including hardware requirements like RAM and software requirements like Turbo C++.
3) Examples of programs to implement basic data structures like linked lists, stacks, queues and sorting algorithms like insertion sort, bubble sort, quick sort and merge sort.
DSU C&C++ Practical File Diploma
The document describes data structures and their implementation using C and C++. It includes:
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2) Requirements for using data structures in C and C++ including hardware requirements like RAM and software requirements like Turbo C++.
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This document discusses binary search trees and provides source code to implement binary search tree traversal operations. It explains that a binary search tree is a binary tree where the left child of a node is less than the parent and the right child is greater. It provides code for preorder, inorder, and postorder traversals of the binary search tree and outputs the traversal order for a sample tree.
pleaase I want manual solution forData Structures and Algorithm An.pdf
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.
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import sys
import numpy as np # an alias for the namespace
import random
import matplotlib.pyplot as plt
#setting MC parameters
NumP=10000 # generate how many random point in side the square
n=100
#MC simulation block
pi_all=[]
#
for j in range(n):
cordx=[];cordy=[] # initialize the two empty coordinate lists
inside=0
for i in range(0,NumP): #generate random coordinates
cordx.append(random.random())
cordy.append(random.random())
r2 = cordx[i]*cordx[i]+cordy[i]*cordy[i]
if r2 <= 1:
inside=inside+1
# write control to count inside
pit= 4.0*inside/NumP
#End of sampling
pi_all.append(pit)
#print out result
print("pi: ","{:10.6f}".format(pit))
plt.hist(pi_all) # print hist of the sampling
# plot the random pair: cordx,cordy graphically for the last MC simulation
Symbol=10
LineW = 3
Lsp=100
f = plt.figure()
f.set_figwidth(8)
f.set_figheight(8)
plt.xlim(0,1)
plt.ylim(0,1)
plt.xlabel('X Cordinate')
plt.ylabel('Y Cordinate')
plt.scatter(cordx, cordy, s=Symbol,marker='o')
#plot circle
x = np.linspace(0,1,Lsp)
y = np.sqrt(1. - x*x)
plt.plot(x, y,'-r',color='red', linestyle='-', lw=LineW)
plt.show()
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The document describes exercises on structures and strings in C programming. It defines a structure called _PLAYER to store details of players like name, date of birth, height and weight. It then shows how to define the structure, read and print records of multiple players from an array of _PLAYER structures. Functions are defined to find the tallest player, print names in descending order of height, check character cases, convert case of characters in a string, and compute operations on structures like distance between points.
#include iostream #include deque #include stdio.h scan.pdf
#include
#include
#include //scanf,printf
#include
#include //istringstream
using namespace std;
//maps
#include \"MapGeneric.h\"
#include \"MapTriple.h\"
#include \"MapAbsoluteValue.h\"
#include \"MapSquare.h\"
//filters
#include \"FilterGeneric.h\"
#include \"filterEven.h\"
#include \"filterPositive.h\"
#include \"filterPositiveTwoDigit.h\"
//reduce
#include \"ReduceGeneric.h\"
#include \"ReduceMinimum.h\"
#include \"ReduceGCD.h\"
int main()
{
deque L,L1,L2;
int i;
string line;
getline(cin,line);
if(line.size() == 0)
return 0;
line.erase(remove(line.begin(), line.end(), \',\'), line.end());
istringstream s(line);
while(s>>i)
{
L.push_back(i);
}
if(L.size() != 20)
return 0;
//converting L -> 3*|L| aka L1
MapTriple triple; MapAbsoluteValue abs;
L1 = triple.map(abs.map(L));
//converting L1 -> positive two digit integers -> even intergers aka L2
filterEven even; filterPositiveTwoDigit p2;
L2 = p2.filter(L1);
if(L2.size() <2)
return 0;
L2 = even.filter(L2);
if(L2.size() <2)
return 0;
//finding the minimum and GCD from L2 respectively
ReduceMinimum min; ReduceGCD gcd;
printf(\"%d %d\ \",min.reduce(L2),gcd.reduce(L2));
}
FilterGeneric.cpp
#include
#include
#include \"FilterGeneric.h\"
using namespace std;
deque FilterGeneric::filter(deque deq)
{
//calling recursive function with deq and its length
select(deq,deq.size()-1);
return filterDeq;
}
//int i is the index that will start from the length of deque till 0
void FilterGeneric::select(deque deq, int i)
{
if(i == 0)
{
bool val = f(deq[i]);
if(val)
filterDeq.push_front(deq[i]);
return;
}
bool val = f(deq[i]);
if(val)
filterDeq.push_front(deq[i]);
return select(deq,i-1);
}
FilterGeneric.h
#ifndef FILTERGENERIC_H
#define FILTERGENERIC_H
#include
class FilterGeneric
{
public:
std::deque filter(std::deque);
void select(std::deque,int);
protected:
//this deque will store the integers after filtration.
std::deque filterDeq;
private:
virtual bool f(int)=0;
};
#endif
MapAbsoluteValue.cpp
#include
#include
#include
#include \"MapGeneric.h\" //base
#include \"MapAbsoluteValue.h\"
using namespace std;
int MapAbsoluteValue::f(int num)
{
return abs(num);
}
MapAbsoluteValue.h
#ifndef MAPABSOLUTEVALUE_H
#define MAPABSOLUTEVALUE_H
#include
//this class will map a given input N, using the function map(input), to |N|
class MapAbsoluteValue: public MapGeneric
{
public:
//overwriting pure virtual function making N -> |N|
int f(int);
private:
};
#endif
MapGeneric.cpp
#include
#include
#include \"MapGeneric.h\"
using namespace std;
deque MapGeneric::map(deque deq)
{
//calling recursive function with deq and its length
calculate(deq,deq.size()-1);
return mapDeq;
}
//int i is the index that will start from the length of deque till 0
void MapGeneric::calculate(deque deq, int i)
{
if(i == 0)
{
int val = f(deq[i]);
mapDeq.push_front(val);
return;
}
int val = f(deq[i]);
mapDeq.push_front(val);
return calculate(deq,i-1);
}
MapGeneric.h
#ifndef MAPGENERIC_H
#define MAPGENERIC_H
#include
class MapGeneric
{
public:
std::deque ma.
Chapter 1 Programming Fundamentals Assignment.docx
Name : Shamshad Jamali
Roll No :22BS( IT ) 40
Subject : Programming .. Fundamental
Submit To :Sir Umair Sheikh
In C Programming create a program that converts a number from decimal.docx
In C Programming create a program that converts a number from decimal to binary. You must use a stack to complete this assignment. • Void pop() • Void push(int data) • Int top() • Int isEmpty() You must create an implementation of a stack. You may use either an array or a linked list as the underlying structure. You will, at a minimum, need to implement the following stack functions: Your program should take decimal input from the user, and convert it to binary, and simply print it out.
Solution
normal program
#include<stdio.h>
int main(){
long int decimalNumber,remainder,quotient;
int binaryNumber[100],i=1,j;
printf(\"Enter any decimal number: \");
scanf(\"%ld\",&decimalNumber);
quotient = decimalNumber;
while(quotient!=0){
binaryNumber[i++]= quotient % 2;
quotient = quotient / 2;
}
printf(\"Equivalent binary value of decimal number %d: \",decimalNumber);
for(j = i -1 ;j> 0;j--)
printf(\"%d\",binaryNumber[j]);
return 0;
}
using stack
.
coding in C- Create a function called reverseList that takes the head.docx
coding in C;
Create a function called reverseList that takes the head of a linked list, reverses the order of all the nodes. For example, if the list contained 1, 2, 3, 4 in its nodes, the list will now contain 4, 3, 2, 1.
please add a main test.
Solution
Please find the required program below:
#include<stdio.h>
#include<stdlib.h>
struct node
{
int data;
struct node* next;
};
void push(struct node** head, int new_element)
{
struct node* new_node =
(struct node*) malloc(sizeof(struct node));
new_node->data = new_element;
new_node->next = (*head);
(*head) = new_node;
}
void printList(struct node *head)
{
struct node *temp = head;
while(temp != NULL)
{
printf(\"%d \", temp->data);
temp = temp->next;
}
}
void reverseList(struct node** head)
{
struct node* prev = NULL;
struct node* current = *head;
struct node* next;
while (current != NULL)
{
next = current->next;
current->next = prev;
prev = current;
current = next;
}
*head = prev;
}
int main()
{
struct node* head = NULL;
push(&head, 12);
push(&head, 14);
push(&head, 5);
push(&head, 8);
printf(\"\ Linked list Before Reversing :\ \");
printList(head);
reverseList(&head);
printf(\"\ Linked list After Reversing :\ \");
printList(head);
}
--------------------------------------------------------------------------------
OUTPUT:
Linked list Before Reversing :
8Â Â 5Â Â 14Â Â 12
Linked list After Reversing :
12Â Â 14Â Â 5Â Â 8
.
AnswerThe new program with the required constructor and with a te.pdf
Answer:
The new program with the required constructor and with a test is given below
Code:
#include
#include
using namespace std;
class TNode
{
public:
int data;
TNode *left;
TNode *right;
TNode(int n){ data = n; left = right = 0; }
TNode(int n, TNode *left, TNode *right)
{
data = n;
//Assign the left node address and right node address with the parameters passed
this->left = left;
this->right = right;
}
};
void printTree(TNode *t)
{
if (!t) return;
printTree(t->left);
cout << t->data << \" \";
printTree(t->right);
}
void search_for(int n, TNode *t)
{
cout << \"Searhing for \" << n << endl;
while (t)
{
cout << \"looking at \" << t->data << endl;
if (t->data == n)
{
cout << \"Found it!\";
return;
}
if (n < t->data)
{
cout << \"Going left\ \";
t = t->left;
}
else
{
cout << \"Going right\ \";
t = t->right;
}
}
cout << \"Can\\\'t find it\";
}
// Creates a two node tree
int main()
{
TNode *root = 0;
root = new TNode(5);
cout << root->data << endl;
TNode *one_node = new TNode(1);
TNode *three_node = new TNode(3);
//use the new constructor
TNode *two_node = new TNode(2, one_node, three_node);
root->left = new TNode(3);
printTree(root);
cout << endl;
search_for(3, root);
getchar();
return 0;
}
Solution
Answer:
The new program with the required constructor and with a test is given below
Code:
#include
#include
using namespace std;
class TNode
{
public:
int data;
TNode *left;
TNode *right;
TNode(int n){ data = n; left = right = 0; }
TNode(int n, TNode *left, TNode *right)
{
data = n;
//Assign the left node address and right node address with the parameters passed
this->left = left;
this->right = right;
}
};
void printTree(TNode *t)
{
if (!t) return;
printTree(t->left);
cout << t->data << \" \";
printTree(t->right);
}
void search_for(int n, TNode *t)
{
cout << \"Searhing for \" << n << endl;
while (t)
{
cout << \"looking at \" << t->data << endl;
if (t->data == n)
{
cout << \"Found it!\";
return;
}
if (n < t->data)
{
cout << \"Going left\ \";
t = t->left;
}
else
{
cout << \"Going right\ \";
t = t->right;
}
}
cout << \"Can\\\'t find it\";
}
// Creates a two node tree
int main()
{
TNode *root = 0;
root = new TNode(5);
cout << root->data << endl;
TNode *one_node = new TNode(1);
TNode *three_node = new TNode(3);
//use the new constructor
TNode *two_node = new TNode(2, one_node, three_node);
root->left = new TNode(3);
printTree(root);
cout << endl;
search_for(3, root);
getchar();
return 0;
}.
Doublylinklist
This document presents information on doubly linked lists including their representation, initialization, node creation, and various operations like insertion, deletion, and traversal. It discusses inserting and deleting nodes at the beginning or end of the list, as well as inserting before or after a specified node. Code examples are provided for initializing a doubly linked list and performing each operation.
L25-L26-Parameter passing techniques.pptx
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.
-- Task 2- Debugging a program with stacks- queues- and doubly-linked.docx
/* Task 2: Debugging a program with stacks, queues, and doubly-linked lists There are a number of errors in the following program. All errors are located in main() and structure definitions. Function declarations and definitions are correct! Locate all errors, fix them (as shown below), run the program and save its output as a comment at the end of the source file. Example: int num = 10; int *ptr; num = &ptr; // <== Error: Comment the line and write the correct line below // Write a short justification where appropriate // num = &ptr; // Error #1 ptr = # Name: */ #include #include #include #include #define DUMMY_TRAILER '\177' // octal ASCII code of the // last character in the ASCII table #define NUM_CITIES 10 typedef struct { char name[12]; int temperature[5]; } CITY; // Stack and Queue Node typedef struct node NODE; struct node { CITY city; node *next; }; // Doubly Linked List Node typedef struct d_node D_NODE; struct d_node { CITY city; NODE *forw; NODE *back; }; // Stack Functions NODE *push(NODE *stack, const CITY *pStu); NODE *pop(NODE **stack); // Queue Functions void enqueue(NODE **queue, NODE **rear, const CITY *pStu); NODE *dequeue(NODE **queue, NODE **rear); // Doubly Linked List Functions D_NODE *init_list(void); int insert(D_NODE *list, const CITY *pStu); void traverse_forw(D_NODE *list); void traverse_back(D_NODE *list); // Other Functions void printCity(const CITY *pCity); int main (void) { CITY cList[NUM_CITIES] = { {"Cupertino", {88, 89, 87, 85, 89}}, {"Flagstaff", {81, 80, 88, 89, 89}}, {"Los Angeles", {87, 88, 89, 89, 90}}, {"Philadelphia", {96, 99, 99, 90, 95}}, {"Phoenix", {106, 109, 109, 100, 105}}, {"Portland", {89, 90, 85, 89, 90}}, {"Reno", {108, 105, 109, 100, 108}}, {"Salem", {85, 90, 85, 89, 90}}, {"Tucson", {107, 100, 109, 100, 108}}, {"Yreka", {101, 109, 100, 108, 109}} }; NODE *stack = NULL; NODE *top = NULL; NODE *queue = NULL, *rear = NULL; NODE *front; D_NODE *list; list = init_list(); // build stack and queue with data from an array of CITY structures srand((unsigned int)time(NULL)); int count = rand() % 10; for ( int n = 0; n < count; n++) { int i = rand() % NUM_CITIES; int duplicate = insert(list, &cList[i]); if(duplicate) { // already in the list! push(stack, &cList[i]); enqueue(&queue, &rear, cList[i]); } } // display list printf("\nLIST contents (forwards):\n"); traverse_forw(list); printf("\nLIST contents (backwards):\n"); traverse_back(list); // display stack if (top) { printf("\nSTACK contents from top to bottom:\n"); while ((top = pop(stack))) { printCity(top->city); } } else printf ("Empty Stack!\n"); // display queue if (front) { printf("\nQUEUE contents from front to rear:\n"); while ((front = dequeue( queue, rear))) { printCity(front->city); } } else printf ("Empty Queue!\n"); return 0; } /*************************************************** Displays the fileds of a CIS_CLASS structure Pre pCls - a pointer to a CIS_CLASS structure Post */ void printCity(const CITY *pCity) { printf("%-20s %3d\n",.
Q1 Consider the below omp_trap1.c implantation, modify the code so t.pdf
Q1 Consider the below omp_trap1.c implantation, modify the code so that: i. The time used by
the parallel block is timed using the OpenMP function omp_get_wtime(). The syntax is double
omp_get_wtime ( void ) It returns the number of seconds that have elapsed since some time in
the past. ii. Now on a system with at least two cores, time the program with a. one thread and a
large value of n, and b. two threads and the same value of n. Explain what happens? Then Do the
same measurements for the omp_trap2.c code.
1. How does its performance compare with the performance of the omp_trap1.c ?
2. Explain your answers.
3. provide with your answers the code and detailed screenshots showing how did you compile
and run the code and times for each run:
/* File: omp_trap1.c * Purpose: Estimate definite integral (or area under curve) using trapezoidal
rule. * Input: a, b, n * Output: estimate of integral from a to b of f(x) * using n trapezoids. * *
Compile: gcc -g -Wall -fopenmp -o omp_trap1 omp_trap1.c * Usage: ./omp_trap1 * * Notes: *
1. The function f(x) is hardwired. * 2. In this version, each thread explicitly computes the
integral over its assigned subinterval, a critical directive is used for the global sum. * 3. This
version assumes that n is evenly divisible by the number of threads #include #include #include
#include void Usage(char* prog_name); double f(double x); /* Function we're integrating */
double Local_trap ( double a , double b , i n t n ) ; int main(int argc, char* argv[]) { double
global_result = 0.0; /* Store result in global_result */ double a, b; /* Left and right endpoints */
int n; /* Total number of trapezoids */ int thread_count; if (argc != 2) Usage(argv[0]);
thread_count = strtol(argv[1], NULL, 10); printf("Enter a, b, and n\n"); scanf("%lf %lf %d", &a,
&b, &n); if (n % thread_count != 0) Usage(argv[0]); double global_result = 0.0; # pragma omp p
a r a l l e l num_threads ( t h r e a d _ c o u n t ) { # pragma omp c r i t i c a l global_result +=
Local_trap ( double a , double b , i n t n ) ; } printf("With n = %d trapezoids, our estimate\n", n);
printf("of the integral from %f to %f = %.14e\n", a, b, global_result); return 0; } /* main */ /*----
---------------------------------------------------------------- * Function: Usage * Purpose: Print
command line for function and terminate * In arg: prog_name */ void Usage(char* prog_name) {
fprintf(stderr, "usage: %s \n", prog_name); fprintf(stderr, " number of trapezoids must be evenly
divisible by\n"); fprintf(stderr, " number of threads\n"); exit(0); } /* Usage */ /*---------------------
--------------------------------------------- * Function: f * Purpose: Compute value of function to be
integrated * Input arg: x * Return val: f(x) */ double f(double x) { double return_val; return_val
= x*x; return return_val; } /* f */ /*------------------------------------------------------------------ *
Function: Trap * Purpose: Use trapezoidal rule to estimate definite integral * Input args.
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Discover the Simplified Electron and Muon Model: A New Wave-Based Approach to Understanding Particles delves into a groundbreaking theory that presents electrons and muons as rotating soliton waves within oscillating spacetime. Geared towards students, researchers, and science buffs, this book breaks down complex ideas into simple explanations. It covers topics such as electron waves, temporal dynamics, and the implications of this model on particle physics. With clear illustrations and easy-to-follow explanations, readers will gain a new outlook on the universe's fundamental nature.
LAND USE LAND COVER AND NDVI OF MIRZAPUR DISTRICT, UP
This Dissertation explores the particular circumstances of Mirzapur, a region located in the
core of India. Mirzapur, with its varied terrains and abundant biodiversity, offers an optimal
environment for investigating the changes in vegetation cover dynamics. Our study utilizes
advanced technologies such as GIS (Geographic Information Systems) and Remote sensing to
analyze the transformations that have taken place over the course of a decade.
The complex relationship between human activities and the environment has been the focus
of extensive research and worry. As the global community grapples with swift urbanization,
population expansion, and economic progress, the effects on natural ecosystems are becoming
more evident. A crucial element of this impact is the alteration of vegetation cover, which plays a
significant role in maintaining the ecological equilibrium of our planet.Land serves as the foundation for all human activities and provides the necessary materials for
these activities. As the most crucial natural resource, its utilization by humans results in different
'Land uses,' which are determined by both human activities and the physical characteristics of the
land.
The utilization of land is impacted by human needs and environmental factors. In countries
like India, rapid population growth and the emphasis on extensive resource exploitation can lead
to significant land degradation, adversely affecting the region's land cover.
Therefore, human intervention has significantly influenced land use patterns over many
centuries, evolving its structure over time and space. In the present era, these changes have
accelerated due to factors such as agriculture and urbanization. Information regarding land use and
cover is essential for various planning and management tasks related to the Earth's surface,
providing crucial environmental data for scientific, resource management, policy purposes, and
diverse human activities.
Accurate understanding of land use and cover is imperative for the development planning
of any area. Consequently, a wide range of professionals, including earth system scientists, land
and water managers, and urban planners, are interested in obtaining data on land use and cover
changes, conversion trends, and other related patterns. The spatial dimensions of land use and
cover support policymakers and scientists in making well-informed decisions, as alterations in
these patterns indicate shifts in economic and social conditions. Monitoring such changes with the
help of Advanced technologies like Remote Sensing and Geographic Information Systems is
crucial for coordinated efforts across different administrative levels. Advanced technologies like
Remote Sensing and Geographic Information Systems
9
Changes in vegetation cover refer to variations in the distribution, composition, and overall
structure of plant communities across different temporal and spatial scales. These changes can
occur natural.
Your Skill Boost Masterclass: Strategies for Effective Upskilling
Your Skill Boost Masterclass: Strategies for Effective UpskillingExcellence Foundation for South Sudan
Strategies for Effective Upskilling is a presentation by Chinwendu Peace in a Your Skill Boost Masterclass organisation by the Excellence Foundation for South Sudan on 08th and 09th June 2024 from 1 PM to 3 PM on each day.Exploiting Artificial Intelligence for Empowering Researchers and Faculty, In...
Exploiting Artificial Intelligence for Empowering Researchers and Faculty, In...Dr. Vinod Kumar Kanvaria
Exploiting Artificial Intelligence for Empowering Researchers and Faculty,
International FDP on Fundamentals of Research in Social Sciences
at Integral University, Lucknow, 06.06.2024
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Executive Directors Chat Leveraging AI for Diversity, Equity, and Inclusion
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Azure Interview Questions and Answers PDF By ScholarHat
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The simplified electron and muon model, Oscillating Spacetime: The Foundation...
The simplified electron and muon model, Oscillating Spacetime: The Foundation...
LAND USE LAND COVER AND NDVI OF MIRZAPUR DISTRICT, UP
LAND USE LAND COVER AND NDVI OF MIRZAPUR DISTRICT, UP
Your Skill Boost Masterclass: Strategies for Effective Upskilling
Your Skill Boost Masterclass: Strategies for Effective Upskilling
Exploiting Artificial Intelligence for Empowering Researchers and Faculty, In...
Exploiting Artificial Intelligence for Empowering Researchers and Faculty, In...
Write code in c++ Program to a topological sort on a graph Program pl.docx
- 1. Write code in c++ Program to a topological sort on a graph Program plan: Initializes main() function. Declare int type variables. Accept the number of vertices in the graph and the adjacency matrix of in-degree. Perform topological sort using for loops. Display the result. Solution Please find the required code and output below: #include <iostream> using namespace std; struct node_info { int no; int lv_time, st_time; }*q = NULL, *r = NULL; struct node { node_info *pt; node *next; }*top = NULL, *p = NULL, *np = NULL; int c = 0; void push(node_info *ptr) { np = new node; np->pt = ptr; np->next = NULL; if (top == NULL) { top = np; }
- 2. 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; }
- 3. 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: