DATA STRUCTURES C This PPT is Dedicated to

DATA STRUCTURES ( C++ ) This PPT is Dedicated to my inner controller AMMA BHAGAVAN – ONENESS Founders. Developed by, S. V. G. REDDY, Associate professor, Dept. of CSE, GITAM UNIVERSITY. EDITED BY, M. Siva Naga Prasad student of M. tech(SE).

STACK USING ARRAYS • • Let us take an array a[5] and take a variable top points to -1. PUSH: Ø To INSERT the element in to stack using top. Ø Here we check for the OVERFLOW condition. POP: Ø To RETRIEVE elements from the stack using top. Ø Here we check for the UNDERFLOW condition. This concept is nothing but LIFO. SOURCE CODE: /* Program To Implement Stack using Array */ #include < iostream. h > #include < conio. h > #include < stdlib. h > #define MAX 10

class stack { private : int sp, a [ MAX ]; public : void init ( ); void push ( int ); void pop ( ); void display ( ); void count ( ); }; void stack : : init ( ) { sp = - 1; } void stack : : push ( int data) { if (sp = = ( MAX – 1 ) ) { cout<<"n STACK OVERFLOW. . . . n"; return; }

sp + + ; a [ sp ] = data; } void stack : : pop ( ) { if ( sp < 0 ) { cout<<"n STACK UNDERFLOW. . . n"; return; } cout<<"n POPED DATA IS : : : "<<a[sp]; sp - - ; } void stack : : display ( ) { cout << "n DATA PRESENT IN A STACK IS : : : n"; for ( int i = sp ; i > = 0 ; i - -) cout << a [ i ] <<"t"; } void stack : : count ( ) { cout<<"n NUMBER OF ELEMENTS IN A STACK ARE : : : "<<(sp+1); };

void main ( ) { stack ob; int data, ch; clrscr ( ); ob. init ( ); cout<<"n*****STACK OPERATIONS*****n"; cout<<"n 1. Push Operation"; cout<<"n 2. Pop Operation"; cout<<"n 3. Display Operation"; cout<<"n 4. Count Operation"; cout<<"n 5. Exit Operation"; cout<<"n*******************n"; do { cout<<"n ENTER YOUR CHOICE : : "; cin>>ch; switch ( ch ) { case 1: cout<<"n ENTER ELEMENT TO BE INSERTED : : : "; cin>>data; ob. push ( data ); break; case 2: ob. pop ( ); break; case 3: ob. display ( ); break; case 4: ob. count ( ); break; case 5: exit ( 0 ); defualt: cout<<"n. INVALID CHOICE "; } } while ( ch ! = 5 ); getch ( );

OUTPUT:

STACK USING LINKED LIST • • We will create a linked list and insert an element ‘ 10’ and address as ‘ 0’. using top for the first node. For second node insert data element ‘ 20’ and insert first node address at second node address field. For third node insert data element ‘ 30’ and insert second node address at third node address field. after thirty we will stop the process. If we want to print the elements 30, 20, 10 will be displayed, Thiss follows LIFO conceot.

Source code: #include<conio. h> #include<iostream. h> class st { public: struct node { int data; struct node *next; }*start, *temp, *top; st() { start=temp=top=NULL; } void create() { int d; cout<<"Enter data"; cin>>d; if(start==NULL) { start=new node; start->data=d;

start->next=NULL; top=start; } else { temp=new node; temp->data=d; temp->next=top; top=temp; } } void disp() { while(top!=NULL) { cout<<top->data<<"t"; top=top->next; } } }; void main() { st ob; int ch; clrscr();

while(ch) { cout<<"Enter ur choice"; cout<<"0 STOPn 1 CREATEn 2 READ"; cin>>ch; if(ch==1) ob. create(); else if(ch==2) ob. disp(); } } OUTPUT:

QUEUE USING ARRAYS • • Here we will take an array a[5], and two variables front, rear points to -1. WRITE: Ø Here will insert the element into the queue using rear variable. Ø Here check for the Overflow condition. READ: Ø Here we will retrieve the elements from the queue using front variable. Ø Here check for the Underflow condition. This follows the FIFO concept. rear -1 front 0 1 2 3 4

SOURCE CODE: /* Program To Implement Queue using Array */ #include< iostream. h > #include< conio. h > #include< process. h > #define MAX 10 class queue { private : int front, rear, a [ MAX ]; public : void init ( ); void write ( int ); void read ( ); void count ( ); void display ( ); }; void queue : : init ( ) { front = rear = - 1; } void queue : : write ( int data) { if ( rear = = ( MAX - 1 ) ) cout<<"n QUEUE IS OVERFLOW. . . "; else a [ + + rear ] = data; } void queue : : read ( ) { if( front = = rear ) cout<<"n QUEUE IS UNDERFLOW. . . "; else cout<<"n DELETED ELEMENT IN QUEUE IS : : "<<a[++front]; }

void queue : : count ( ) { cout<<"n NUMBER OF ELEMENTS IN A QUEUE ARE : : "<<(rear-front); } void queue : : display ( ) { cout<<"n ELEMENTS IN A QUEUE ARE: : "; for( int i = (front + 1); i < = rear; i + + ) cout<< a [ i ]<<"t"; } void main ( ) { queue ob; int ch, data; clrscr ( ); ob. init ( ); cout<<"n*****QUEUE OPERATIONS****n"; cout<<"n 1. Write "; cout<<"n 2. Read "; cout<<"n 3. Count"; cout<<"n 4. Display"; cout<<"n 5. Exit"; cout<<"*************n";

do { cout<<"n ENTER YOUR CHOICE : : "; cin>>ch; switch ( ch ) { case 1: cout<<"n ENTER ELEMENT TO BE INSERTED IN QUEUE : : "; cin>>data; ob. write ( data ); break; case 2: ob. read ( ); break; case 3: ob. count ( ); break; case 4: ob. display ( ); break; case 5: exit ( 0 ); break; default : cout<<"n INVALID CHOICE. . . "; } } while( ch ! = 5 ); getch ( ); }

OUTPUT:

Queue using linked list • • Here we will create linked list with ‘n’ nodes one after another 10, 20, 30 etc. If we try to print the elements it will display as 10, 20, 30. which follows FIFO concept.

SOURCE CODE: /* Program To Implement Queue using Linked List */ #include < iostream. h > #include< conio. h > #include < alloc. h > #define NULL 0 class node { int data; node *next; public: void create ( node *); void print ( node *); }; void node : : create (node *list) { cout<<"n ENTER THE INPUT NO : : "; cout<<"n TYPE 999 AT THE END : : "; cin>>list->data; if(list -> data = = 999) list->next = NULL; else { list -> next = new node; create( list -> next); } return;

void node : : print (node *list) { if( list -> next ! = 0) { cout<< list->data; cout<<"->"; } else return; print( list -> next); } void main ( ) { node *head, ob; clrscr ( ); head = new node; ob. create ( head ); cout<<"n QUEUE ELEMENTS ARE: : "; ob. print( head ); cout<<"999"; getch ( ); }

OUTPUT:

BINARY TREE USING RECURSION • • • A binary tree is a tree data structure in which each node has at most two children. Typically the child nodes are called left and right. Binary trees are commonly used to implement binary search trees and binary heaps. Starting at the root of a binary tree, there are three main steps that can be performed and the order in which they are performed define the traversal type. There are 3 types of traversals: Ø 1. Pre-Order Ø 2. In-Order Ø 3. Post-Order To traverse a non-empty binary tree in preorder, perform the following operations recursively at each node, starting with the root node: 1. Visit the root. 2. Traverse the left sub tree. 3. Traverse the right sub tree. To traverse a non-empty binary tree in in order, perform the following operations recursively at each node, starting with the root node: 1. Traverse the left sub tree. 2. Visit the root. 3. Traverse the right sub tree. To traverse a non-empty binary tree in post order, perform the following operations recursively at each node, starting with the root node: 1. Traverse the left sub tree. 2. Traverse the right sub tree. 3. Visit the root.

BINARY TREE: 15 7 22 Preorder: - 15, 7, 22 will be displayed. Post order: - 7, 22, 15 will be displayed. In order: - 7, 15, 22 will be displayed.

SOURCE CODE: /* Program To Implement Binary Tree Traversing */ #include < iostream. h > #include < conio. h > class bstree { public: struct node { int data; node *left; node *right; }*head; void create (node *); void inorder (node *); void preorder (node *); void postorder (node *); }; void* bstree: : create(node *list) { node *temp 1, *temp 2; int val; if(list = = NULL) { list = new node; cout<<"n. Enter Data Element: : "; cin>>list->data; list -> left = list -> right = NULL; } else

{ cout<<"n enter the data element"; cin>>val; temp 1 = list; while( temp 1 ! = NULL ) { temp 2 = temp 1; if(temp 1 -> data > val) temp 1 = temp 1 -> left; else temp 1 = temp 1 -> right; } if(temp 2 -> data > val) { temp 2 -> left = new node; temp 2 = temp 2 -> left; temp 2 -> data = val; temp 2 -> left = temp 2 -> right = NULL; } else { temp 2 -> right = new node; temp 2 = temp 2 -> right; temp 2 -> data = val; temp 2 -> left = temp 2 -> right = NULL; } } return (list); }

void bstree: : inorder(node *root) { if( ! root ) return; inorder( root -> left ); cout<<root->data<<"t"; inorder( root -> right ); } void bstree: : preorder(node*root) { if( ! root ) return; cout<<root->data<<”t”; preorder( root -> left ); preorder( root -> right); } void bstree: : postorder(node*root) { if( ! root) return; postorder( root -> left ); postorder( root -> right ); cout<<root->data<<”t”; }

void main ( ) { node n, *head; head = NULL; clrscr ( ); cout<<"n. Create A Binary Treen"; head=n. create ( head ); cout<<"n the inorder traversal gives the following nodes"; n. inorder ( head ); getch ( ); } OUTPUT:

BINARY SEARCH TREE 15 7 22 ØA tree having left child less than parent and right child grater than the parent. ØTraversals are same as binary tree.

SOURCE CODE: /* Program to implement Binary search tree */ #include < iostream. h > #include < conio. h > class btree { private : struct btreenode { btreenode *leftchild ; int data ; btreenode *rightchild ; } *root; public: btree ( ) ; void buildtree ( int num ) ; static void insert ( btreenode **sr, int num ) ; void traverse ( ) ; static void inorder ( btreenode *sr ) ; static void preorder ( btreenode *sr ) ; static void postorder ( btreenode *sr ) ; static void del ( btreenode *sr ) ; ~btree ( ) ; } ;

btree : : btree ( ) { root = NULL ; } void btree : : buildtree ( int num ) { insert ( &root, num ) ; } void btree : : insert ( btreenode **sr, int num ) { if ( *sr == NULL ) { *sr = new btreenode ; ( *sr ) -> leftchild = NULL ; ( *sr ) -> data = num ; ( *sr ) -> rightchild = NULL ; return ; } else // search the node to which new node will be attached { // if new data is less, traverse to left if ( num < ( *sr ) -> data ) insert ( & ( ( *sr ) -> leftchild ), num ) ; else // else traverse to right insert ( & ( ( *sr ) -> rightchild ), num ) ; } return ; }

void btree : : traverse( ) { cout << "n. IN - ORDER TRAVERSAL : : " ; inorder ( root ) ; cout << "n. PRE - ORDER TRAVERSAL : : " ; preorder ( root ) ; cout << "n. POST - ORDER TRAVERSAL : : " ; postorder ( root ) ; } void btree : : inorder ( btreenode *sr ) { if ( sr != NULL ) { inorder ( sr -> leftchild ) ; cout << "t" << sr -> data ; inorder ( sr -> rightchild ) ; } else return ; } void btree : : preorder ( btreenode *sr ) { if ( sr != NULL ) { // print the data of a node cout << "t" << sr -> data ; // traverse till leftchild is not NULL preorder ( sr -> leftchild ) ; // traverse till rightchild is not NULL preorder ( sr -> rightchild ) ; }

else return ; } void btree : : postorder ( btreenode *sr ) { if ( sr != NULL ) { postorder ( sr -> leftchild ) ; postorder ( sr -> rightchild ) ; cout << "t" << sr -> data ; } else return ; } btree : : ~btree( ) { del ( root ) ; } void btree : : del ( btreenode *sr ) { if ( sr != NULL ) { del ( sr -> leftchild ) ; del ( sr -> rightchild ) ; } delete sr ; }

void main( ) { btree bt ; int req, i = 1, num ; clrscr(); cout << "n SPECIFY THE NUMBER OF ITEMS TO BE INSERTED : : " ; cin >> req ; while ( i + + <= req ) { cout << "n ENTER THE DATA : : " ; cin >> num ; bt. buildtree ( num ) ; } bt. traverse( ) ; getch(); } OUTPUT:

SPARSE MATRIX AIM: Write a program in C++ to implement ADDITION and MULTIPLICTION of two SPARSE matrixes. THEORY: If a lot of elements from a matrix have a value 0 then the matrix is known as SPARSE MATRIX. If the matrix is sparse we must consider an alternate way of representing it rather the normal row major or column major arrangement. This is because if majority of elements of the matrix are 0 then an alternative through which we can store only the non-zero elements and keep intact the functionality of the matrix can save a lot of memory space. Example: Sparse matrix of dimension 7 x 7. COLUMNS 0 1 2 3 4 5 6 0 0 -5 0 0 0 1 0 4 0 0 7 2 0 0 9 0 0 ROWS 3 0 2 0 0 0 4 1 0 2 0 0 5 0 0 0 0 6 0 0 8 0 0

A common way of representing non-zero elements of a sparse matrix is the 3 -tuple forms. In this form each non-zero element is stored in a row, with the 1 st and 2 nd element of this row containing the row and column in which the element is present in the original matrix. The 3 rd element in this row stores the actual value of the nonstore element. For example 3 -tuple representation of the sparse matrix as shown in below. int spmat[10][3]={ 7, 7, 9, 0, 3, -5, 1, 1, 4, 1, 6, 7, 2, 4, 9, 3, 1, 3, 3, 3, 2, 4, 0, 11, 4, 2, 2, 6, 2, 8 }

SOURCE CODE: /*Program to demonstrate addition and multiplication of Two Sparse Matrix */ #include < iostream. h > #include < conio. h > #define x 25 class sparce { private: int a [ x ], b [ x ], c [ x ], m, n, p, q; public: void init ( ); void input ( ); void add ( ); void mul ( ); void display ( int [25], int ); void convert( int [25], int ); }; void sparce : : init ( ) { int i, j; for(i = 0; i < x; i + + ) for( j = 0; j < x; j + +) c [ i ] [ j ] = 0; }

void sparce : : input() { int i, j; cout<<"n. Enter order Of First matrix: : "; cin>>m>>n; cout<<"n. Enter order Of Second matrix: : "; cin>>p>>q; cout<<"n. Enter"<<m*n<<"Elements Into First Matrixn"; for(i=0; i<m; i++) for( j = 0; j < n; j + + ) cin>> a[ i ] [ j ]; cout<<"n. Enter"<<p*q<<"Elements Into Second Matrixn"; for(i = 0; i < p ; i + + ) for ( j = 0; j < q ; j + + ) cin>>b [ i ] [ j ]; } void sparce : : add ( ) { int i, j; if( m = = p && n = = q ) { for( i = 0 ; i < m ; i + + ) for( j = 0; j < n; j + + ) c[ i ] [ j ] = a [ i ][ j ] + b [ i ] [ j ]; convert( c, m, n); } else cout<<"n. Addition Is Not Possible"; }

void sparce : : mul ( ) { int i, j, k; if(n = = p) { for( i = 0; i < m; i + +) for( j = 0; j < q; j + + ) for( k = 0; k < n; k + + ) c[ I ] [ j ] + = a [ I ] [ k ] * b [ k ] [ j ]; convert(c, m, n); } else cout<<"n Multiplecation Is Not Possible"; } void sparce : : display(int c[25], int m, int n) { int i, j; for( i = 0 ; i < m; i + + ) { for( j = 0 ; j < n ; j + + ) cout<<c [ i ] [ j ]<<"t"; cout<<"n"; } }

void sparce : : convert(int c[25], int m, int n) { int i, j, k = 1, t = 0; int sp[25]; for( i = 0 ; i < m ; i + +) for( j = 0 ; j < n ; j + + ) if(c [ i ] [ j ] ! = 0 ) { sp [ k ] [ 0 ] = i; sp [ k ] [ 1 ] = j; sp [ k ] [ 2 ] = c [ i ] [ j ]; k + + ; t + + ; } sp[ 0 ] = m; sp[ 0 ] [ 1 ] = n; sp[ 0 ] [ 2 ] = t; display( sp, k, 3); } void main ( ) { sparce ob; clrscr ( ); ob. init ( ); ob. input ( ); cout<<"n. Addition of Two Sparce Matrixn"; ob. add ( ); ob. init ( ); cout<<"n. Multiplecation Of Two Sparce Matrixn"; ob. mul ( ); getch ( ); }

OUTPUT:

INFIX TO POSTFIX CONVERTIONic • • • Suppose Q is an arithmetic expression written in infix notation. This algorithm finds the equivalent postfix expression P. Step 1. Push “(“ onto stack and add “)” to the end of Q. 2. Scan Q from left to right and repeat step 3 to 6 for each element of Q until the stack is empty. 3. If an operand is encountered , add it to p. 4. If a left parenthesis is encountered , push it onto stack. • 5 If an operator * is encountered , then: a. repeatedly pop from stack and top each operator (on the top of stack ) which has the same precedence or higher precedence than *. b. Add * to stack. • 6. If a right parenthesis is encountered , then: a. repeatedly from stack and add to P each operator (on the top of stack) until a left parenthesis is encountered. b. remove the left parenthesis [ Do not add the left parenthesis top] [End of if structure] [End of step 2 loop] • 7. Exit

(A+(B*C-(D/E^F)*G)*H) Symbol scanned 1 A 2 + 3 ( 4 B 5 * 6 C 7 8 ( 9 D 10 / 11 E 12 ^ 13 F 14 ) 15 * 16 G 17 ) 18 * 19 H 20 ) stack ( (+ (+( (+(* (+((+(-(/ (+(-(/^ (+((+(-* (+ (+* Expression P A AB AB ABC*D ABC*DEF ABC*DEF^/G*ABC*DEF^/G*-H*+

SOURCE CODE: /* Program To implement infix to postfix Expression */ #include < iostream. h > #include< process. h > #include < conio. h > char stack[30], postfix[30], infix[30]; int top = - 1; int pri( char x ) { int value; switch ( x ) { case ')': value=0; break; case '+': case '-': value=1; break; case '*': case '/': case '%': value=2; break; case '^': value=3; break; case '(': value=4; break; default: cout<<"INVALID EXPRESSION !!!!!!"; exit(1); } return value; }

void push ( char x ) { top = top + 1; stack [top] = x; } char stacktop ( ) { return stack [ top ]; } int isalnum (char x) { return ( (x>='0' && x<='9') ||( x>='a' && x<='z') || ( x>='A' && x<='Z')); } char pop( ) { return stack[top - - ]; }

void intopost(char infix[ ], char postfix[ ]) { int i, j=0; char c, pc; for ( i = 0; ( c = infix[ i ] ) != '' ; i + +) { if ( isalnum (c) ) postfix [ j + + ] = c; else { while ( top ! = - 1 && (pri (stacktop () ) >= pri (c) ) ) { If ( stacktop( ) = = '(' && c! = ')' ) break; if ( stacktop( ) = = '(' && c = =')' ) { pop () ; break; } pc = pop( ); if ( pc! = '(' ) postfix [ j + + ] = pc; else break; } if( c! = ')' ) push ( c ); } } while( top ! = -1 ) postfix[ j + + ] = pop( ); postfix [ j ] = '';

void main ( ) { clrscr ( ); cout<<"ENTER INFIX EXPRESSION : : nnttt"; cin>>infix; intopost( infix, postfix ); cout<<"POSTFIX EXPRESSION : : nnttt "; cout<<postfix; getch ( ); } OUTPUT:

POSTFIX EVALUATION THEORY: Reverse Polish notation is a mathematical notation wherein every operator follows all of its operands. It is also known as Postfix notation and is parenthesis free. In Reverse Polish notation the operators follow their operands; for instance, to add three and four, one would write “ 3 4 +” rather than “ 3 + 4”. If there are multiple operations, the operator is given immediately after its second operand; so the expression written “ 3 − 4 + 5” in conventional infix notation would be written “ 3 4 − 5 +” in RPN: first subtract 4 from 3, then add 5 to that. Infix Expression: Any expression in the standard form like "2*3 -4/5" is an Infix(In order) expression. Postfix Expression: The Postfix(Post order) form of the above expression is "23*45/-". Postfix Evaluation: In normal algebra we use the infix notation like a+b*c. The corresponding postfix notation is abc*+. The algorithm for the conversion is as follows: • Scan the Postfix string from left to right. • Initialize an empty stack. • If the scanned character is an operand, add it to the stack. If the scanned character is an operator, there will be at least two operands in the stack ØIf the scanned character is an Operator, then we store the top most element of the stack(top. Stack) in a variable temp. Pop the stack. Now evaluate top. Stack(Operator)temp. Let the result of this operation be ret. Val. Pop the stack and Push ret. Val into the stack. ØRepeat this step till all the characters are scanned. • After all characters are scanned, we will have only one element in the stack. Return top. Stack.

Example: Postfix String: 1 2 3 * + 4 -. Initially the Stack is empty. Now, the first three characters scanned are 1, 2 and 3, which are operands. Thus they will be pushed into the stack in that order. Stack Expression Next character scanned is "*", which is an operator. Thus, we pop the top two elements from the stack and perform the "*" operation with the two operands. The second operand will be the first element that is popped. Stack Expression The value of the expression(2*3) that has been evaluated(6) is pushed into the stack. Stack Expression

Next character scanned is "+", which is an operator. Thus, we pop the top two elements from the stack and perform the "+" operation with the two operands. The second operand will be the first element that is popped. Stack Expression The value of the expression(1+6) that has been evaluated(7) is pushed into the stack. Stack Expression Next character scanned is "4", which is added to the stack. Stack Expression Next character scanned is "-", which is an operator. Thus, we pop the top two elements from the stack and perform the "-" operation with the two operands. The second operand will be the first element that is popped.

The value of the expression(7 -4) that has been evaluated(3) is pushed into the stack. Stack Expression Now, since all the characters are scanned, the remaining element in the stack (there will be only one End result: Postfix String : 1 2 3 * + 4 - Result : 3

SOURCE CODE: /*Program To Evaluate Postfix Expression */ #include < iostream. h > #include < conio. h > #include < math. h > #include < string. h > class postfix { private: int stack[50], len, top; char post[50]; public: postfix ( ); void push ( int ); int pop ( ); int pfix ( ); }; void postfix : : postfix ( ) { top = - 1; } int postfix : : pfix ( ) { int a, b, i, temp; cout<<"n. Enter Postfix Expression: : "; cin>>post; len = strlen ( post ); post [ len] = '#';

for( i = 0 ; post [ i ] ! = '#' ; i + +) { if( post [ i ] <= '9' && post [ i ] >= '0') push( post [ i ] - 48); else { a = pop ( ); b = pop ( ); switch ( post [ i ]) { case '+': temp = b + a; break; case '-': temp = b - a; break; case '*': temp = b * a; break; case '/': temp = b/a; break; case '%': temp = b%a; break; case '^': temp = pow( b, a ); } push ( temp ); } } return( pop ( ) ); }

void postfix : : push( int x ) { stack[ + + top ] = x; } int postfix : : pop ( ) { int x = stack [ top ]; top- -; return x; } void main ( ) { int x; postfix ob; clrscr ( ); x=ob. pfix ( ); cout<<"n. Result Of Postfix Expression Ist"<<x; getch ( ); } OUTPUT:

Quick Sort 11 7 21 3 46 89 2 34 right pivot left 1)When pivot is at left end, 1)Compare a[pivot] with a[right] element if (a[pivot] < a[right]) then right-- else swap a[pivot] and a[right] 2)When pivot is at right end, Compare a[pivot] with a[left] element if (a[left] < a[pivot]) then left++ else swap a[left] and a[pivot]

STEP 1: 11 7 21 3 46 89 2 34 left, pivot right STEP 2: 2 7 21 3 46 89 11 left right, pivot 34 STEP 3: 2 7 21 3 46 89 11 left right, pivot 34 STEP 4: 2 7 21 3 46 89 11 left right, pivot 34 STEP 5: 2 7 11 3 46 89 21 left, pivot right 34 STEP 6: 2 7 11 3 46 89 left, pivot right 34 21

STEP 7: 2 7 11 3 46 89 left, pivot right 21 34 STEP 8: 2 7 11 3 46 89 left, pivot right 21 34 STEP 9: 2 7 3 11 46 89 left right, pivot 21 34 STEP 10: 2 7 3 11 46 89 left, right, pivot 21 34 • • • Here we will stop the main process as the left and right pointers are equal. Now see the elements left to ‘ 11’ are less than ‘ 11’ and elements right to ‘ 11’ are grater than ‘ 11’. Now divide the main list into 2 sub lists such as(2, 7, 3) and (46, 89, 21, 34) and do the same above process.

Source code • #include<stdio. h> #include<conio. h> #define MAXSIZE 500 void quick. Sort(int elements[], int maxsize); void sort(int elements[], int left, int right); int elements[MAXSIZE]; int main() { int i, maxsize; printf(“n. How many elements you want to sort: “); scanf(“%d”, &maxsize); printf(“n. Enter the values one by one: “); for (i = 0; i < maxsize; i++) { printf (“n. Enter element %i : ”, i); scanf(“%d”, &elements[i]); }

printf(“n. Array before sorting: n”); for (i = 0; i < maxsize; i++) printf(“[%i], “, elements[i]); printf (“n”); quick. Sort(elements, maxsize); printf(“n. Array after sorting: n”); for (i = 0; i < maxsize; i++) printf(“[%i], “, elements[i]); } void quick. Sort(int elements[], int maxsize) { sort(elements, 0, maxsize - 1); } void sort(int elements[], int left, int right) { int pivot, l, r; l = left; r = right; pivot = elements[left]; while (left < right) { while ((elements[right] >= pivot) && (left < right)) right—;

if (left != right) { elements[left] = elements[right]; left++; } while ((elements[left] <= pivot) && (left < right)) left++; if (left != right) { elements[right] = elements[left]; right—; } } elements[left] = pivot; pivot = left; left = l; right = r; if (left < pivot) sort(elements, left, pivot - 1); if (right > pivot) sort(elements, pivot + 1, right); }

Selection sort • Consider the elements as shown, 77 33 44 11 88 77 22 66 55 0 min i • Here min is compared with a[1] Ø as min is > a[1] Ø min=a[1] 33 0 min i • This min is compared with a[2] , as this is < a[2] Ø min is same that is 33 • This min is compared with a[3] , as this is > a[3] Ø min =a[3]. 11 • • 0 min I Now this is compared with a[4], a[5], a[6], a[7] as min is less than all of these min remains 33 At last swap min and a[i] like this continue the process with i=1, 2, 3……

SOURCE CODE: #include < iostream. h > #include < conio. h > class selsort {public : void sort(int *, int); }; void selsort: : sort(int *a, int n) { int i, j, x, min, temp; for( i = 0 ; i < ( n – 1 ) ; i + + ) { x = i; min = a [ i ]; for( j = i + 1; j < n; j + + ) { if( min > a [ j ] ) { min = a [ j ]; x = j; } } temp = a [ i ] ; a [ i ] = a [ x ]; } } void main( ) { int a[50], n, i; clrscr( ); a [ x ] = temp;

cout<<"n ENTER THE SIZE OF THE ARRAY: nt "; cin>>n; cout<<"n ENTER THE ELEMENTS: nt"; for( i = 0 ; i < n ; i + + ) cin>>a [ i ]; cout<<"n ELEMENTS BEFORE SORTING: nt"; for( i = 0 ; i < n ; i + + ) cout<<a[i]<<"t"; selsort obj; obj. sort(a, n); cout<<"n ELEMENTS AFTER SORTING ARE: nt"; for( i = 0 ; i < n ; i + + ) cout<<a[i]<<"t"; getch(); } OUTPUT:

LINEAR SEARCH 10 20 30 40 50 60 70 0 1 2 3 4 5 6 • Here we want to search for ‘ 50’. • So compare ’ 50’ with a[i] where i=0, 1, 2, 3, …. ØIf (a[i]==50) Then element is found at location i that is 4 Else i++ • Here the time complexity is O(n).

SOURCE CODE: #include < iostream. h > #include < conio. h > class lsearch { private: int a[50], n, count, key; public: void init ( ); void linear ( ); }; void lsearch: : init ( ) { count = 0; } void lsearch: : linear ( ) { int i; clrscr ( ); cout<<"n. ENTER SIZE OF AN ARRAY : : "; cin>>n; cout<<"nn. ENTER "<<n<<" ELEMENTS INTO AN ARRAY : : "; for( i = 0; i < n; i + +) cin>> a [ i ]; cout<<"nn. ENTER SEARCH ELEMENT : : "; cin>>key; cout<<"nn. ELEMENTS IN ARRAY ARE : n";

for( i = 0; i < n; i + +) cout<< a [ i ]<<"t"; for( i = 0; i < n; i + + ) if(a [ i ] = = key) { count + +; break; } if( count = = 1 ) cout<<"nn ELEMENT IS FOUND IN "<< ( i + 1)<<" LOCATION"; else cout<<"n. ELEMENT IS NOT FOUND. . "; } void main ( ) { lsearch ob; clrscr ( ); ob. init (); ob. linear ( ); getch ( ); }

OUTPUT:

BINARY SEARCH • Here elements must be in Ascending/Descending order. • Consider the elements in ascending order 7 11 15 23 46 64 71 83 low high here low=0 and high=7 Then calculate mid=(low+high)/2 • Let us search for k=71 • If (a[mid]==k) then element is found at ‘mid’ location • If(k<a[mid]) then high=mid-1 else low=mid+1 • Repeat the previous steps tell low and high are equal.

SOURCE CODE: /*Program To Implement Binary Search */ #include < iostream. h > #include < conio. h > class bsearch { private : int a[50], n , x; public : void binary ( ); }; void bsearch: : binary ( ) { int i, j, temp, mid, beg, end; beg = 0; cout<<"nn. ENTER THE SIZE OF THE ARRAY : : "; cin>>n; end = n - 1; cout<<"nn. ENTER THE ELEMENTS OF THE ARRAY : : "; for( i = 0; i < n; i + +) cin>>a[i]; cout<<"nn. ELEMENTS BEFORE SORTING ARE : : "; for( i = 0; i < n; i + + ) cout<< a [ i ]<<" ";

for( i = 0; i < n; i + + ) { for( j = i + 1; j < n; j + +) { if( a[ i ] > a[ j ] ) { temp = a [ i ]; a[ i ] = a[ j ]; a[ j ] = temp; } } } cout<<"nn. ELEMENTS AFTER SORTING ARE : : "; for( i = 0; i < n; i + + ) cout<<a[ i ]<<" "; cout<<"nn. ENTER THE ELEMENT TO BE SEARCHED : : "; cin>>x; while ( beg < = end ) { mid = ( beg + end ) / 2; if ( a [ mid ] = = x ) {

cout<<"n. SEARCHING IS SUCCESSFUL AND THE ELEMENTS IS PRESENT AT "<< ( mid + 1 )<<" LOCATION"; return; } else if(x<a[mid]) end = mid - 1; else beg = mid + 1; } cout<<"n SEARCH IS UNSUCCESSFUL"; } void main ( ) { bsearch obj; clrscr ( ); obj. binary ( ); getch ( ); } OUTPUT:

POLINOMIAL ADDITION AND MULTIPLICATION • 1 expression: 3 x 2+2 x+1 Store all the coefficients 1, 2, 3 into an array 1. • 1 expression: 2 x 2+1 x+2 Store all the coefficients 2, 1, 2 into an array 2. ADDITION: 3 x 2+2 x+1 2 x 2+1 x+2 5 x 2+3 x+3 Store the result expression coefficients in array 3

SOURCE CODE: /*Program To Demonstrate Addition And Multiplication Of Two Polynomial Expression */ #include < iostream. h > #include < conio. h > #define n 100 class poly { private: int a[n], b[n], add[n], mul[n], p, q, at; public: void init ( ); void input ( ); void process ( ); void display ( ); }; void poly : : init ( ) { int i; for( i = 0; i < n; i + + ) a[ i ] = b [ i ] = add[ i ] = mul[ i ] = 0; }

void poly : : input ( ) { int i; cout<<"n. Enter Degree Of First Polynomial: : "; cin>>p; cout<<"n. Enter Degree Of Second Polynomial: : "; cin>>q; cout<<"n. Enter Values First Polynomialn"; for( i = 0; i <= p; i + + ) { cout<<"n. Enter X^"<<i<<" Th Coefficient"; cin>>a[ i ]; } cout<<"n. Enter Values First Polynomialn"; for( i = 0; i <= q; i + + ) { cout<<"n. Enter X^"<<i<<" Th Coefficient"; cin>>b[ i ]; } }

void poly : : process ( ) { int i, j; if( p > q ) at = p; else at = q; for ( i = 0; i <= at; i + +) add[ i ] = a[ i ] + b[ i ]; for( i = 0; i <= p; i + + ) for( j = 0; j <= q; j + + ) mul [ i + j ] + = a [ i ] * b [ j ]; } void poly : : display ( ) { int i; cout<<"Addition Of Two Polynomial Expressions Arenn"; for( i = at; i >=0 ; i - -) cout<<add[i]<<"X^"<<i<<"+"; cout<<"nn. Multiplecation Of Two Polynomial Expressions Arenn"; for( i = p + q; i > = 0; i - -) cout<<mul[i]<<"X^"<< i <<"+"; }

void main() { poly ob; clrscr ( ); ob. init ( ); ob. input ( ); ob. process ( ); ob. display ( ); getch ( ); } OUTPUT:

SINGLE LINKED LIST THEORY: Figure shows a Linked List. Each item in the list is called a node and contain two fields, a data field and a next address field. The data field holds the actual element on the list. The next address field contains the address of the next node in the list. Such an address which is used to access a particular node, is known as a pointer. The entire linked list is accesses from an external pointer list, that points to the first node in the list. The next field of last node in the list contains a special value, known as NULL. The null pointer is used to signal the end of the list. The singly-linked list is the most basic of all the linked data structures. A singly-linked list is simply a sequence of dynamically allocated objects, each of which refers to its successor in the list. Despite this obvious simplicity, there are myriad implementation variations. The following code inserts a node after an existing node in a singly linked list. The diagram shows how it works. Inserting a node before an existing one cannot be done; instead, you have to locate it while keeping track of the previous node.

Similarly, we have functions for removing the node after a given node, and for removing a node from the beginning of the list. The diagram demonstrates the former. To find and remove a particular node, one must again keep track of the previous element.

SOURCE CODE: /*Program To Implement Single Linked list */ #include< stdio. h > #include < iostream. h > #include < conio. h > #include < process. h > #include< alloc. h > class slist { private: struct list { int data; struct list *next; }*start, *temp, *curr, *add, *tem, *addr; public: void init ( ); void create ( ); void disp ( ); list *search ( int ); void insert ( ); void del ( ); };

void slist : : init ( ) { start = temp = curr = NULL; } void slist: : create ( ) { char ch; temp = new list; cout<<"n ENTER THE DATA TO BE STORED n"; cin>> temp->data; temp->next = NULL; start = curr = temp; cout<<"n DO YOU WANT TO INSERT ANOTHER NODE (Y/N)"; cin>>ch; while( ch = = 'y' ) { temp = new list; cout<<"n ENTER DATA TO BE STORED: n"; cin>>temp->data; temp->next = NULL; curr->next = temp; curr = temp; cout<<"n DO YOU WANT TO INSERT ANOTHER NODE (Y/N): "; cin>>ch; } }

void slist : : disp ( ) { if( start = = NULL) cout<<"n LIST IS EMPTY"; else { cout<<"n DATA PRESENT IN A LIST IS n"; temp = start; while( temp -> next ! = NULL) { cout<<"|"<<temp->data<<"|"<<temp->next<<"|-->"; temp = temp -> next; } cout<<"|"<<temp->data<<"|"<<temp->next<<"|"; } } slist: : list *slist : : search( int key) { temp = start; while( temp -> next ! = NULL) { if( temp->data = = key ) return temp; else temp = temp->next; } if( temp->next = = NULL )

if( temp->data = = key ) return temp; else return NULL; } void slist: : insert ( ) { int key; cout<<"n ENTER DATA AFTER WHICH WE CAN INSERT NEW NODE: "; cin>>key; add=search(key); if( add = = NULL ) cout<<"n NODE IS NOT FOUND"; else { temp = new list; cout<<"n ENTER INSERTED ELEMENT"; cin>>temp->data; if( add->next = = NULL) { temp->next = NULL; add->next = temp; curr = temp; }

else { addr = add->next; add->next = temp; temp->next = addr; } } } void slist : : del ( ) { int key; cout<<"n ENTER NODE DATA SHOULD BE DELETE: n"; cin>>key; add = search ( key ); if( add = = NULL ) cout<<"n NODE IS NOT FOUNDn"; else if( curr = = add ) { curr = start; while( curr->next ! = NULL) { temp = curr; curr = curr->next; }

free ( curr ); curr = temp; curr->next = NULL; } else if( start = = add ) { temp = start; start = start->next; free( temp ); } else { tem = add->next; temp = start; while( temp-> next ! = add) temp = temp->next; temp->next = tem; free( add ); } } void main ( ) { slist ob; int key, ch; list *temp; clrscr ( ); cout<<"n * * * SINGLE LINKED LIST OPERATION * * * n";

cout<<"n 1. CREATE n 2. DISPLAY n 3. INSERT n 4. DELETE n 5. SEARCH n 6. EXIT n"; cout<<"n *************n"; do { cout<<"n ENTER YOUR CHOICE n"; cin>>ch; switch ( ch ) { case 1: ob. create ( ); break; case 2: ob. disp ( ); break; case 3: ob. insert ( ); break; case 4: ob. del ( ); break; case 5: cout<<"n ENTER THE ELEMENT TO SEARCH"; cin>>key; temp=ob. search(key); if( temp = = NULL) cout<<"n ELEMENT IS NOT FOUND n"; else cout<<"n ELEMENT IS FOUND n"; break; case 6: exit(0); default: cout<<"n INVALID CHOICE n"; } }while( ch ! = 0 ); getch ( ); }

OUTPUT:

SINGLE CIRCULAR LINKED LIST THEORY: The linked list that we have seen so far is often know as linear lists. The elements of such a linked list can be accessed, first by setting up a pointer pointing to the first node in the list and then traversing the entire list using this pointer. Although a linear linked list is a useful data structure, it has several shortcomings.

SOURCE CODE: /* Program to implement single circular linked list */ #include<iostream. h> #include<conio. h> #include<process. h> #include<alloc. h> class clist { private: struct list { int data; struct list *next; }*start, *temp, *curr, *add, *tem, *addr; public: void init(); void creat(); void display(); list *search(int); void insert(); void del(); }; void clist: : init() { start=temp=curr=NULL; }

void clist: : creat() { char ch; temp=new list; cout<<"n ENTER DATA TO BE STORED : : "; cin>>temp->data; cout<<"n. ADDRESS OF STARTING NODE : : "<<temp; temp->next=start; start=curr=temp; cout<<"n. DO YOU WANT TO INSERT ANOTHER NODE (y/n) : : "; cin>>ch; while(ch=='y') { temp=new list; cout<<"n ENTER DATA TO BE STORED : : "; cin>>temp->data; temp->next=start; curr->next=temp; curr=temp; cout<<"n. DO YOU WANT TO INSERT ANOTHER NODE (y/n) : : "; cin>>ch; } }

void clist: : display() { if(start==NULL) cout<<"n. LIST IS EMPTY. . . "; else cout<<"n. DATA PRESENT IN A LIST IS : : n"; temp=start; while(temp->next!=start) { cout<<"|"<<temp->data<<"|"<<temp->next<<"|-->"; temp=temp->next; } cout<<"|"<<temp->data<<"|"<<temp->next<<"|"; } clist: : list *clist: : search(int key) { temp=start; while(temp->next!=start) { if(temp->data==key) return temp; else temp=temp->next; }

if(temp->next==NULL) { if(temp->data==key) return temp; else return NULL; } void clist: : insert() { int key; cout<<"n ENTER DATA AFTER WHICH WE CAN INSERTED NEW NODE : : "; cin>>key; add=search(key); if(add==NULL) cout<<"n NODE IS NOT FOUND. . "; else { temp=new list; cout<<"n ENTER INSERTED ELEMENT : : "; cin>>temp->data;

if(add->next==start) { temp->next=start; add->next=temp; curr=temp; } else { addr=add->next; add->next=temp; temp->next=addr; } } } void clist: : del() { int key; cout<<"n. Enter node to deleted: "; cin>>key; add=search(key); if(add==NULL) cout<<"n. Node is not found"; else if(curr==add)

{ curr=start; while(curr->next!=start) { temp=curr; curr=curr->next; } free(curr); curr=temp; curr->next=start; } else if(start==add) { temp=start; start=start->next; free(temp); } else { tem=add->next; temp=start; while(temp->next!=add) temp=temp->next; temp->next=tem; free(add); } }

void main() { clist ob; int key, ch; clist: : list *temp; clrscr(); cout<<"n. CIRCULAR LINKED LIST n"; cout<<"n 1. Createn 2. Displayn 3. Insertn 4. Deleten 5. Searchn 6. Exitn"; do { cout<<"n. Enter your choice"; cin>>ch; switch(ch) { case 1: ob. creat(); break; case 2: ob. display(); break; case 3: ob. insert(); break; case 4: ob. del(); break;

case 5: cout<<"n. Enter search element"; cin>>key; temp=ob. search(key); if(temp==NULL) cout<<"n. Element is not found"; else cout<<"n. Element is found"; break; case 6: exit(0); default: cout<<"Invalid choice"; } }while(ch!=6); getch(); }

OUTPUT:

DOUBLE LINKED LIST AIM: Write a program in C++ to implement DOUBLE LINKED LIST THEORY: A two-way list is a linear collection of data elements, called nodes, where each node N is divided into three parts: • An item data field. • A pointer field next which contains the location of the next node in the list. • A pointer field prev which contains the location of the previous node in the list. The list requires two list pointer variables: FIRST, which points to the first node in the list, and LAST, which points to the last node in the list. The figure contains a schematic diagram of such a list. Observe that the null pointer appears in the next field of the last node in the list and also in the prev field of the first node in the list. Observe that, using the variable FIRST and the pointer field next, we can traverse a two-way list in the forward direction as before. On the other hand, using the variable LAST and the pointer field prev, we can also traverse the list in the backward direction.

OPERATION ON TWO-WAY LISTS: 1. Traversing. 2. Searching. 3. Deleting 4 Inserting

SOURCE CODE: /* Program to implement Double linked list */ #include<iostream. h> #include<conio. h> #include<process. h> #include<alloc. h> class dlist { private: struct list { int data; struct list *next, *prev; }*start, *temp, *curr, *addr, *tem; public: void init(); void creat(); void display(); list *search(int); void insert(); void del(); }; void dlist: : init() { start=temp=curr=NULL; }

void dlist: : creat() { char ch; temp=new list; cout<<"n. ENTER DATA TO BE STORED : : "; cin>>temp->data; cout<<"n STARTING NODE ADDRESS : : "<<temp<<"n"; temp->next=NULL; temp->prev=NULL; start=curr=temp; cout<<"n DO YOU WANT TO INSERT ANOTHER NODE (y/n) : : "; cin>>ch; while(ch=='y') { temp=new list; cout<<"n. ENTER DATA TO BE STORED : : "; cin>>temp->data; temp->next=NULL; temp->prev=curr; curr->next=temp; curr=temp; cout<<"n. DO YOU WANT TO INSERT ANOTHER NODE (y/n) : : "; cin>>ch; } }

void dlist: : display() { if(start==NULL) cout<<"n LIST IS EMPTY. . "; else { cout<<"n DATA PRESENT IN A LISTn: : : "; temp=start; while(temp->next!=NULL) { cout<<"|"<<temp->prev<<"|"<<temp->data<<"|"<<temp->next<<"|-->"; temp=temp->next; } cout<<"|"<<temp->prev<<"|"<<temp->data<<"|"<<temp->next<<"|"; } } dlist: : list *dlist: : search(int key) { temp=start; while(temp->next!=NULL) { if(temp->data==key)

return temp; else temp=temp->next; } if(temp->next==NULL) { if(temp->data==key) return temp; else return NULL; } void dlist: : insert() { int key; cout<<"n. ENTER DATA AFTER WHICH WE CAN INSERT A NEW NODE : : "; cin>>key; add=search(key); if(add==NULL) cout<<"n NODE IS NOT FOUND. . . "; else tem=new list;

cout<<"n ENTER ELEMENT TO BE SEARCHED : : "; cin>>tem->data; if(add->next==NULL) { tem->next=NULL; tem->prev=add; add->next=tem; curr=tem; } else { addr=add->next; add->next=tem; tem->next=addr; tem->prev=add; } } void dlist: : del() { int key; cout<<"n ENTER NODE DATA TO BE DELETED : : "; cin>>key; add=search(key); if(add==NULL) cout<<"n NODE IS NOT FOUND : : "; else

if(curr==add) { curr=start; while(curr->next!=NULL) { temp=curr; curr=curr->next; } free(curr); curr=temp; curr->next=NULL; } else if(start==NULL) { temp=start; start=start->next; free(temp); } else { tem=add->next; temp=start; while(temp->next!=add) temp=temp->next; temp->next=tem; free(add); } }

void main() { dlist ob; int key, ch; dlist: : list *temp; clrscr(); cout<<"*****DOUBLE LINKED LIST*****"; cout<<"n 1. Createn 2. Displayn 3. Insertn 4. Deleten 5. Searchn 6. Exitn"; do { cout<<"n. ENTER YOUR CHOICE : : "; cin>>ch; switch(ch) { case 1: ob. creat(); break; case 2: ob. display(); break; case 3: ob. insert(); break; case 4: ob. del(); break;

case 5: cout<<"n ENTER SEARCH ELEMENT : : "; cin>>key; temp=ob. search(key); if(temp==NULL) cout<<"n ELEMENT IS NOT FOUND. . "; else cout<<"n ELEMENT IS FOUND. . . "; break; case 6: exit(0); default: cout<<"n INVALID CHOICE. . "; } }while(ch!=6); getch(); }

OUTPUT:

GRAPH TRAVERSING: DEPTH FIRST SEARCH THEORY: • DFS is an uninformed search that progresses by expanding the first child node of the search tree that appears and thus going deeper and deeper until a goal node is found, or until it hits a node that has no children. Then the search backtracks, returning to the most recent node it hadn't finished exploring. In a non-recursive implementation, all freshly expanded nodes are added to a LIFO stack for exploration. • Space complexity of DFS is much lower than BFS (breadth-first search). It also lends itself much better to heuristic methods of choosing a likely-looking branch. Time complexity of both algorithms are proportional to the number of vertices plus the number of edges in the graphs they traverse (O(|V| + |E|)).

SOURCE CODE: /*Program To Implement Depth First Search */ #include < iostream. h > #include < conio. h > #define MAX 20 class depth { private: int a[MAX], visited[MAX]; int n, top; public: void init ( ); void input ( ); void dfs ( int ); }; void depth: : init ( ) { int i, j; for( i = 0; i < MAX; i + + ) { visited[ i ] = 0; for( j =0; j < MAX ; j + + ) a[ i ] [ j ] = 0; } top = - 1; } void depth: : input ( ) { int i, j; cout<<"n. ENTER NUMBER OF NODES IN A GRAPH : : ";

cin>>n; cout<<"n. ENTER ADJACENCY MATRIX FOR A GRAPH : : n"; for( i = 1; i <= n; i + +) for( j = 1; j <= n; j + + ) cin>>a[ i ][ j ]; } void depth: : dfs ( int v) { int i; visited[v] = 1; cout<<v<<"->"; for( i = 1; i <= n; i. I + + ) if( a [ v ] [ i ] = = 1 && visited [ i ] = = 0) dfs ( i ); } void main ( ) { depth ob; int start; clrscr ( ); ob. init ( ); ob. input ( ); cout<<"n. STARTING NODE FOR DFS TRAVERSING : : "; cin>>start; cout<<"n. DEPTH FIRST SEARCH TRAVERSING IS : : nn"; ob. dfs ( start ); getch ( ); }

OUTPUT:

GRAPH TRAVERSING: BREADTH FIRST THEORY: • BFS is an uninformed search method that aims to expand examine all nodes of a graph or combinations of sequence by systematically searching through every solution. In other words, it exhaustively searches the entire graph or sequence without considering the goal until it finds it. • From the standpoint of the algorithm, all child nodes obtained by expanding a node are added to a FIFO queue. In typical implementations, nodes that have not yet been examined for their neighbors are placed in some container (such as a queue or linked list) called "open" and then once examined are placed in the container "closed". Ø 1. 2. Algorithm for Breadth First Search Enqueue the root node. Dequeue a node and examine it. • If the element sought is found in this node, quit the search and return a result. • Otherwise enqueue any successors (the direct child nodes) that have not yet been discovered. 3. If the queue is empty, every node on the graph has been examined – quit the search and return "not found". 4. Repeat from Step 2.

SOURCE CODE: /*Program To Implement Breadth First Search */ #include < iostream. h > #include < conio. h > #define MAX 20 class breadth { private: int a[MAX], visited[MAX], queue[50]; int n, front, rear; public: void init ( ); void input ( ); void bfs ( ); }; void breadth: : init ( ) { int i, j; for( i = 0; i < MAX; i + + ) { visited [ i ] = 0; for( j = 0; j < MAX; j + + ) a[ i ] [ j ] = 0; } front = rear = - 1; }

void breadth: : input ( ) { int i, j; cout<<"n. ENTER NUMBER OF NODES IN A GRAPH : : "; cin>>n; cout<<"n. ENTER ADJACENCY MATRIX FOR A GRAPH : : n"; for( i = 1; i <= n; i + + ) for( j = 1; j <= n; j + + ) cin>>a[ i ][ j ]; } void breadth: : bfs ( ) { int i, start; cout<<"n. STARTING NODE FOR BFS TRAVERSING : : "; cin>>start; cout<<"n BREADTH FIRST SEARCH TRAVERSING IS: : n t"; cout<<start; visited[ start ] = 1; rear + +; front + +; queue[ rear ] = start; while(front <= rear) { start = queue[front]; front + +;

for( i =1; i <= n; i + + ) { if(a[ start ][ i ] = =1 && visited[ i ] = = 0) { cout<<"->"<<i; visited[ i ] =1; rear + +; queue [ rear ] = i; } } void main ( ) { breadth ob; int start; clrscr ( ); ob. init ( ); ob. input ( ); ob. bfs ( ); getch ( ); }

OUTPUT:

SHORTEST PATH FOR GRAPH THEORY: Shortest path is nothing but the path which lies between two nodes with the lowest cost. In a graph contain so many paths are existed between two nodes (source node to destination node) to choose the lowest cost path to reach from source node to destination node is nothing but shortest path algorithm. Shortest path algorithm was first proposed by E. W. DIJKSTRA. SOURCE CODE: /*Program To Implement Shortest Path for Graph */ #include < iostream. h > #include < conio. h > #define INF 9999 class stpath { private: int i, j, k; public: void spath(int [ ][20], int ); void display(int [ ][20], int ); };

void stpath: : spath(int a[ ][20], int n) { for( i = 0 ; i < n; I + + ) for( j = 0; j < n; j + + ) if(a[ i ] [ j ] = = 0) a[ i ][ j ] = INF; cout<<"n. ADJACENCY MATRIX OF COST OF EDGES ARE : : "; display( a, n ); for( k = 0; k < n; k + + ) for( i = 0; i < n; i + + ) for( j = 0; j < n; j + + ) if( a[ i ][ j ] > a[ i ] [ k] + a[ k ][ i ]) a[ i ][ j ] = a[ i ][ k ] + a[ k ][ j ]; cout<<"n. ADJACENCY MATRIX OF LOWEST COST OF EDGES ARE : : n"; display(a, n); } void stpath: : display(int a[ ] [20], int n) { for( i = 0; i < n; i + + ) { for( j = 0; j < n; j + + ) cout<<a[ i ][ j ]<<"t"; cout<<"n"; } }

void main() { int i, j , n , a[20]; stpath ob; clrscr(); cout<<"n. ENTER NUMBER OF NODES IN A GRAPH : : "; cin>>n; cout<<"n. ENTER ADJACENCY MATRIX : : n"; for( i = 0; i < n; i + + ) for( j = 0; j < n; j + + ) { cout<<"Enter "<<i+1<<" To "<<j+1<<" Node Distance"; cin>>a[ i ] [ j ]; } ob. spath(a, n); getch ( ); }

OUTPUT: