Challenge

Project 1 - Creation of user defined data type to implement the user interfaces for working with ‘Set’ (Mathematical Set theory) using Linked List

RRanjith G12 Oct 2023

Interfaces for mathematical Set Theory are implemented using C programming language. User can give the input like two list of data and this program will give the output as the union and intersection of these two lists

Project1:

Aim:

To create interfaces for mathematical Set Theory implemented using C programming language.

Objective:

To get the union and intersection operation performed for the two sets of datas given by the user.
Implementation of interface for mathematical Set Theory using C programming.

Principle:

Intersection of two sets:
It is the set that contains all the elements that are common to both the sets.
Example X={1,2,3} and Y={2,3,4} so the intersection gives {2,3}
Union of two sets:
It is the set containing all the elements of the given sets.
Example X={1,2,3} and Y={2,3,4} so the union gives {1,2,3,4}

CODE :

#ifndef LIST_H_// macro is defined by list h//

#define LIST_H_// define list h//

#include //input output header file//

#include //standerd library header file//

#include //boolan data type header file

typedef struct node

{

int data;

struct node *next;

}Node;

static Node *head1,*head2,*head,*head_union,*head_intersection,*newnode,*tail,*res_union,*res_intersection;

Node *insert_node(Node *head_ref,int n);

void display_list(Node *head_ref);

Node *get_union(Node *head_pointer1,Node *head_pointer2);

Node *get_intersection(Node *head_pointer1,Node *head_pointer2);

int compare(Node *head,int data);

#endif

int main()

{

int first_list_nodes = 0,second_list_nodes = 0,first_list_items = 0,

second_list_items = 0;

printf("Enter the number of nodes in first list:");

fflush(stdout);

scanf("%d",&first_list_nodes);

printf("Enter %d the elements:",first_list_nodes);

for(int i = 0;i < first>

{

fflush(stdout);

scanf("%d",&first_list_items);

head1 = insert_node(head1,first_list_items);

}

display_list(head1);

printf("\n Enter the number of nodes in second list:");

fflush(stdout);

scanf("%d",&second_list_nodes);

printf("Enter %d the elements:",second_list_nodes);

for(int i = 0; i < second>

{

fflush(stdout);

scanf("%d",&second_list_items);

head2 = insert_node(head2,second_list_items);

}

display_list(head2);

res_union = get_union(head1,head2);

#ifdef DEBUG_MAIN

printf("\n Value of head1 and head2 :%d%d",head1,head2);)

#endif

#ifdef DEBUG_MAIN

printf("\n Value of head1 and head2 after performing union :%d%d\n",head1,head2);

printf("Value of res_union :%d\n",res_union);

#endif

printf("Here is the result of union:");

display_list(res_union);

res_intersection = get_intersection(head1,head2);

#ifdef DEBUG_MAIN

printf("\n Value of head1 and head2 after performing union:%d%d\n",head1,head2);

printf("Value of res_intersection:%d\n",res_intersection);

#endif

printf("Here is the result of intersection:");

display_list(res_intersection);

}

void display_list(Node *head_ref)

{

if(head_ref==0)

{

printf("List is empty.");

}

else

{

tail = head_ref;

while(tail != 0)

{

printf("%d",tail->data);

tail = tail->next;

}

tail = newnode;

}

Node *insert_node(Node *head_ref,int n)

{

printf("\nInserting elements at the beginning...\n");

newnode =(Node*) malloc(sizeof(Node));

printf("Address of created newnode :%d\n",newnode);

newnode->data = n;

printf("Inserted value is:%d\n",newnode->data);

newnode-> next = head_ref;

head_ref = newnode;

return head_ref;

}

Node *get_union(Node *head_pointer1, Node *head_pointer2)

{

while(head_pointer1 !=0)

{

head_union=insert_node(head_union,head_pointer1->data);

head_pointer1=head_pointer1->next;

}

while(head_pointer2 !=0)

{

if(!compare(head_union,head_pointer2->data))

head_union=insert_node(head_union,head_pointer2->data);

head_pointer2=head_pointer2->next;

}

return head_union;

}

int compare(Node* head,int data)

{

while(head!= NULL)

{

if(head->data==data)

return 1;

head=head->next;

}

return 0;

}

Node *get_intersection(Node *head_pointer1,Node *head_pointer2)

{

while(head_pointer1 !=0)

{

if (compare(head_pointer2,head_pointer1->data))

head_intersection=insert_node(head_intersection,head_pointer1->data);

head_pointer1=head_pointer1->next;

}

return head_intersection;

}

Output :

Enter the number of nodes in first list:4

Enter 4 the elements:2 4 6 8