Friday, 28 February 2014

Major Project Review Details

MAJOR PROJECT DETAILS
 ROLLNO                                             GUIDE
   1-25                                                     G.M sir
   26-50                                           K.SWAROOPARANI
   51-76                                           M.BAKI sir


         REVIEW DETAILS
1ST REVIEW  ---MARCH 17TH
2ND REVIEW---APRIL 7TH
3RD REVIEW---APRIL 30TH   (date may be varies)


Note: If the student id doing IEEE project ,You must submit base paper and abstract on first review.  

Sunday, 9 June 2013

MCA career After Completing .............:)

Jobs After MCA, MCA Career Prospects, After completing MCA Future, career opportunities after mca


MCA is a three year master degree program comprising of six semesters. This professional course is designed to meet the growing requirement of qualifiedprofessionals not only in IT industry, but also in other industries as well. The admission to the MCA program can be obtained after completion of the graduation course. On completion of the MCA course, the students will be able to have a deep knowledge on subjects like database management, statistics, algorithm design, mobile technologies, finance, programming languages, accounting, computational theory, Quality assurance/Control, network management, probability, organizational behavior, Project Management, etc.
After successful completion of MCA, the candidates will be able to get jobs in different industries like software engineering and development, hardware and networking, research and development, services and application support, etc…As far as software engineering and development is concerned, the candidate can get job as Trainee Programmer, software tester, software analyst, Software Engineer, software administrator or as a software programmer. The hardware and networking industry contain the physical elements of computer system (i.e.) designing, manufacturing and maintenance of computers. It also involves integration and installation of computer systems through networking. Research and development involves the crafting of circuits and chips and computer architecture. The service and application support industry includes application management and system integration, maintenance of software application and its installation and much more.
Initially, the candidate can begin his/her career as a junior programmer or trainee and he/she can grow at a faster pace and can become system analyst or can also become project leaders and the growth prospects is high up to top level management in many reputed organizations.

Top companies: Many top rated and widely recognized organizations in the IT industry namely Tata Consultancy Services (TCS), Infosys, Wipro, Cognizant Technology Services, IBM, Accenture offers good job opportunity and salary packages to fresh MCA candidates, which may be starting from 2.5 lakhs to 3.5 lakhs approximately and salary packages for some of other companies are even higher and negotiable depends on applicant’s profile.

Monday, 20 May 2013

Sunday, 10 March 2013

IV SEM QUESTION PAPERS ADDED....

Refer to SVU MCA Question papers tab to download 2012 IV sem and 2010 IV sem question papers....................

Thursday, 7 February 2013

.Net Lab Prints for IV Sem

For MCA IV SEM VB.Net  Lab programsPrints Click Below.
 
            Click here for download..............




For MCA IV SEM ASP.Net Lab Programs Prints Click Below.
 
            Click here for download..............

Friday, 28 December 2012

Invitation

We welcome u all jb college guys to make swagath-2k13 a grand success in our college premises on 24-01-2013 @ 4:30 pm

Saturday, 24 November 2012

Saturday, 3 November 2012

Feedback please ....

Hi friends... please give feedback about this blog.
..

Tuesday, 9 October 2012

Thursday, 4 October 2012

Wednesday, 12 September 2012

Tech-Mahindra Placement Papers 2011



The test comprised of 6 sections as.....
1. Non-Verbal Reasoning (10 ques):
Q1)  2,6,12,20,32,_
a) 42   b) 46    c)52     d)48
ans: 46

Q2) 3,323,3663,36663. which doesn't belongs to the series?
a) 3    b) 323    c) 3663    d) series is correct.
ans: 323(all other are divisible by 3)

Q3) & Q4) two questions on mirror images?

Q5) pig:sty::aeroplane:
a) airport
b)dockyard
C)hanger
d)harbour
ans:-hanger

Q6)find odd man out:
parrot, sparrow, kiwi, dove
ans: kiwi(all others can fly high)

Q7)find odd man out:
hawaii,  lakshadweep, paris,  java
ans: paris(others are islands)

Q8,9,10) they were based on the alphabetical analogy like
cat:dbu::dog:____
SECTION 2: Verbal Reasoning (12 QUES)
This section consisted of the questions on critical reasoning, blood relation, few analogies, 2-3 synonyms and antonyms.
SECTION 3:-Numerical Aptitude (12 QUES)

Q1)A deposits Rs. 5000 for six months and B deposited Rs. 6000 for 5 months. what will be the ratio of their profits?
a) 3:2  b)1:1   C)2:1    d)1:2

Q2) A and B deposited Rs. 4000 and 2000 for 6 months and 12 months resp. wat will be profit of B?
Q3) A,B,C divided their profit in the ratio 5:3:4. if the total profit for year was Rs. 1200 wat will b the share of B?
a) 400   b)500  c)300   d)200.
Q4)A complets a race in 4min20sec. B completes same distance of 2km in 4min34sec. wat more distance can A travel in the meanwhile?
failed to solve.....
similar kinds of question based on the topics of:
1. ages
2. time and distance.
3. time and work.
4. allegation and mixture (MOST IMP)
5. pipes ans cistrens.
6. stocks and shares.
7. simple and compound interest
8. problems on trains.
9. problems on boats.
even if u are solving the solved question s of
R.S.Aggrawal... it will be more than sufficient....
they wont ask too tricky questions as time alloted is pretty less.
SECTION 4: ENGLISH 1(22QUESTION)
consisted of various short passage and 4-5 questions were asked on each passage.
simplest of all hence should start first.. even the officials of tch mah. will tell u to start from this section only in their ppt.
u should complete this section in 10-12 min so that u will get lot of  time to devote to other sections.
passages were simple as in 10th -12th.
SECTION  5: ENGLISH 2 (22QUES)
this section mainly consist of fill in the blanks type ques were u were asked to fill correct preposition, verb, phrase, etc
for eg
Q1) the meeting was presided _____ the prime minister.
a) by    b)over   C) upon    d)at

Q2)he said exactly _____i was expecting.
A) as  b)which    c) like  d)that

Q3)gave few sentences and asked them to arrange in proper order.

Q4,5) same as above. (these question consumes time as we hav to consider all the given 5-6 sentence and then arrange them
in proper order)
similar kind of question were their in this section.

SECTION  6:-(22QUES)
it went same as above with few extra questions of analogies, synonyms and antonyms.
can be solved in 5-7 minutes if u have good English
Preparation from R.S. Aggrawal (QUANTI and Reasoning) is sufficient to crack the APTI.
u just need to be quick in your calculations.

Sunday, 6 November 2011

II sem DS lab programs


ARRAY OPERATIONS

SOURCE CODE:

import java.io.*;
class Arrayf
{
  int loc,item;
  int a[]=new int[20];
  int i,j,n;
  BufferedReader br=new BufferedReader(new InputStreamReader(System.in));
void create()throws IOException
{           
  System.out.print("enter array size:");
  n=Integer.parseInt(br.readLine());
  System.out.print("enter elements in array:");
  for(i=0;i<n;i++)
 {
  System.out.print("enter"+i+"element");
  a[i]=Integer.parseInt(br.readLine());
 }
}
void insert()throws IOException
{
  System.out.println("Enter location:");
  loc=Integer.parseInt(br.readLine());
  if(loc>=0&&loc<=n)
 {
   n=n+1;
   System.out.println("enter data:");
   item=Integer.parseInt(br.readLine());
   for(i=n;i>loc-1;i--)
  {
   a[i]=a[i-1];
  }
   a[loc-1]=item;
   System.out.print("......insertion is success");
 }
 else
  System.out.println("given location is out of arraybounadary");
}
void delete()throws IOException
{
  System.out.println("enter location to remove:");
  loc=Integer.parseInt(br.readLine());
   if((loc>=0)&&(loc<=n))
  
{
    for(i=loc-1;i<n;i++)
   {
    a[i]=a[i+1];
   }
   n--;
   System.out.println("element is deleted");
}
  else
  System.out.println("given location is outof arrayboundary");
}
void display()
{
  System.out.println("elements in array are:");
  for(i=0;i<n;i++)
  {
   System.out.println(a[i]+" ");
  }
  System.out.print(" ");
 }
}
class ArrOperation
{
 public static void main(String args[])throws IOException
 {
  int ch;
  Arrayf ar=new Arrayf();
  BufferedReader br=new BufferedReader(new InputStreamReader(System.in));
  System
    case 1:ar. .out.println("prog to perform an array operations");
  do
  {
   System.out.println("1.create array");
   System.out.println("2.insert array");
   System.out.println("3.delete array");
   System.out.println("4.display array");
   System.out.println("5.exit");
   System.out.println("enter ur choice");
   ch=Integer.parseInt(br.readLine());
  






switch(ch)
   {
create();
           break;
    case 2:ar.insert();
            break;
    case 3:ar.delete();
            break;
    case 4:ar.display();
                break;
    case 5:break;
    default:
       System.out.print("invalid choice");
    }
    }while(ch!=5);
   }
 }

























OUT PUT:
prog to perform an array operations
1.create array
2.insert array
3.delete array
4.display array
5.exit
enter ur choice:1
enter array size:3
enter elements in array
enter0element:2
enter1element:4
enter2element:6
1.create array
2.insert array
3.delete array
4.display array
5.exit
enter ur choice:2
Enter location:3
enter data:7
......insertion is successfull
1.create array
2.insert array
3.delete array
4.display array
5.exit
enter ur choice:4
elements in array are
2
4
7
6
1.create array
2.insert array
3.delete array
4.display array
5.exit
enter ur choice:5







//ADDITION,MULTIPLICATION OF POLYNOMIALS//

import java.io.*;
import java.lang.*;
import java.util.Scanner;
class polyad
{
     public static void main(String args[])throws Exception
     {
      Scanner ob=new Scanner(System.in);
      int m,n;
      System.out.print("\nEnter heighest degree of 1st polynomial :");
      m=ob.nextInt();
      System.out.print("\nEnter heighest degree of 2nd polynomial :");
      n=ob.nextInt();
      int hd=(m>n?m:n);
      int eq1[]=new int[hd+1];
      int eq2[]=new int[hd+1];
      int i,j;
      for(i=hd;i>=0;i--)
      eq1[i] = eq2[i] =0;
      if(m<n)
      {       for(i=m;i>0;i--)
              eq1[i]=0;      }
      else
      {       for(i=n;i>0;i--)
              eq2[i]=0;      }
      System.out.print("\n Enter coefficient value of 1st polynomial \n");
      for(i=m;i>0;i--)
      {
      System.out.print("\n Enter coeffvalue of "+i+" degree :");
      eq1[i]=ob.nextInt();
      }
      System.out.print("\n Enter coefficient value of 2nd polynomial \n");
      for(i=n;i>0;i--)
      {
      System.out.print(" \n Enter coeffvalue of "+i+" degree :");
      eq2[i]=ob.nextInt();
      }
      int sum[]=new int[hd+1];
      for(i=hd;i>=0;i--)
      {
      sum[i]=eq1[i]+eq2[i];
      }
      System.out.print("\n given two equations are----\n");
      System.out.print("\n \n 1st equation :");
      for(i=m;i>0;i--)
      {
      if(i==1)
          System.out.print(eq1[i] + "x+");
      else if(i==0)
          System.out.print(eq1[i]);
      else
          System.out.print(eq1[i]+"x^"+i+"+");
      }
      System.out.print("\n \n 2nd equation :");
      for(i=n;i>0;i--)
      {
      if(i==1)
          System.out.print(eq2[i]+"x+");
      else if(i==0)
          System.out.print(eq2[i]);
       else
          System.out.print(eq2[i]+"x^"+i+"+");
       }
       System.out.print("\n\nsum of two equations is---> ");
       for(i=hd;i>=0;i--)
       {
       if(i==1)
             System.out.print(sum[i]+"x+");
       else if(i==0)
             System.out.print(sum[i]);
       else
             System.out.print(sum[i]+"x^"+i+"+");
        }
        int mul[]=new int[hd+1];
        for(i=hd;i>=0;i--)
        mul[i]=mul[i]+eq1[i]*eq2[i];
        System.out.print("\n\nmultiplication of two polynomials is----->");
        for(i=hd;i>=0;i--)
        {
        if(i==1)
             System.out.print(mul[i]+"x+");
        else if(i==0)
             System.out.print(mul[i]);
        else
             System.out.print(mul[i]+"x^"+i+"+");
        }      
}
}


OUTPUT:

C:\Program Files\Java\jdk1.6.0_13\bin>javac polyad.java

C:\Program Files\Java\jdk1.6.0_13\bin>java polyad

Enter heighest degree of 1st polynomial :3

Enter heighest degree of 2nd polynomial :3

 Enter coefficient value of 1st polynomial

 Enter coeffvalue of 3 degree :4

 Enter coeffvalue of 2 degree :2

 Enter coeffvalue of 1 degree :4

 Enter coefficient value of 2nd polynomial

 Enter coeffvalue of 3 degree :5

 Enter coeffvalue of 2 degree :3

 Enter coeffvalue of 1 degree :2

 given two equations are----


 1st equation :4x^3+2x^2+4x+

 2nd equation :5x^3+3x^2+2x+

sum of two equations is---> 9x^3+5x^2+6x+0

multiplication of two polynomials is----->20x^3+6x^2+8x+0










// PROGARM IN JAVA USING QUEUES TO IMPLEMENT LIBRARY BOOK REQUESTS

import java.io.*;
class Node
{
            int sno;
            int bookid;
            String book name;
            Node next;
}
class QueueLink
{
            Node front=null;
            Node rear=null;
            void enQueue()throws IOException
            {
                        BufferedReader br=new BufferedReader(new InputStreamReader(System.in));
                        Node first = new Node();
                        System.out.print("\nEnter SNO: ");
                        first.sno=Integer.parseInt(br.readLine());
                        System.out.print("Enter book ID: ");
                        first.bookid=Integer.parseInt(br.readLine());
                        System.out.print("Enter Book Name: ");
                        first.bookname=br.readLine();
                        first.next=null;
                        if(front==null)
                        {front=rear=first;}
                        else
                        {                     
                                    rear.next=first;
                                    rear=rear.next;
                        }
            }
            void delQueue()
            {
                        if(front==null)
                        {System.out.println("Queue is empty..");}
                        else
                        {
                                    System.out.println("Removed book details are.. ");
                                    System.out.print(front.sno+"\t");
                                    System.out.print(front.bookid+"\t");
                                    System.out.print(front.bookname+"\t");
                       

            System.out.println(" ");
                                    front=front.next;
                        }
            }
            void display()
            {
                        if(front==null)
                        {System.out.println("Queue is empty");}
                        else
                        {
                                    Node temp=front;
                                    System.out.println("Issue Library book lists");
                                    System.out.println("S.no\t Book_ID \t Book_Name ");
                                    while(temp!=null)
                                    {
                                                System.out.print(temp.sno + "\t");
                                                System.out.print(temp.bookid+ "\t\t");
                                                System.out.print(temp.bookname+"\t");
                                                System.out.println("    ");
                                                temp=temp.next;
                                    }
}
}
}
class Library
{
            public static void main(String args[])throws IOException
            {
                        int sw;
                        QueueLink QL=new QueueLink();
                        BufferedReader br=new BufferedReader(new InputStreamReader(System.in));
                        do
                        {         

System.out.println("\n\t Queues to implement Library book request \n");
                                    System.out.println("1.Enter Book details");
                                    System.out.println("2.Remove Book Details");
                                    System.out.println("3.Display");
                                    System.out.println("4.Exit");
                                   



System.out.print("\nEnter your choice(1-4) :  ");
                                    sw=Integer.parseInt(br.readLine());
                                    switch(sw)
                                    {
                                                case 1: QL.enQueue();
                                                        break;
                                                case 2: QL.delQueue();
                                                        break;
                                                case 3: QL.display();
                                                        break;
                                                case 4: break;

                                                default:System.out.println("Invalid option enter only (1-4)");
                                    }
                        }while(sw!=4);
               }
}




























OUTPUT:
C:\Program Files\Java\jdk1.6.0_13\bin>javac Library.java
C:\Program Files\Java\jdk1.6.0_13\bin>java Library

         Queues to implement Library book request

1.Enter Book details
2.Remove Book Details
3.Display
4.Exit
Enter your choice(1-4) :  1

Enter SNO: 101
Enter book ID: 2217
Enter Book Name: DS structure

         Queues to implement Library book request

1.Enter Book details
2.Remove Book Details
3.Display
4.Exit

Enter your choice(1-4) :  3
Issue Library book lists
S.no     Book_ID         Book_Name
101     2217                 DS structure

         Queues to implement Library book request

1.Enter Book details
2.Remove Book Details
3.Display
4.Exit

Enter your choice(1-4) :  2
Removed book details are..
101     2217    DS structure

         Queues to implement Library book request

1.Enter Book details
2.Remove Book Details
3.Display
4.Exit
Enter your choice(1-4) :  4

/* Convert infix expression to postfix,prefix format*/
                                                 
        import java.util.*;
        class obj
           {
                char pst[] = new char[15];
                char inf[] = new char[15];
                char oprt[] = new char[15];
                char ods[] = new char[20];
       
                int i,n, top = 0,ptop=0, dsp = 0;
                  String s;
        void pushd(char x)
          {
           ods[dsp++] = x;
          }
          char popd()
          {
           return(ods[--dsp]);
          }
        void prefix()
          {
            char a;
            System.out.println( "Enter Expression in infix format  : ");
            Scanner in = new Scanner(System.in);
            s = in.nextLine();
            n = s.length();
            for(i=0;i<n;i++)
            inf[i] = s.charAt(i);
            System.out.print("\n THE PREFIX FORM IS : ");
            for(i=n;i>=0;i--)
              {
                a = inf[i];
                if (a != '+' && a != '-' && a != '*' && a != '/'
                             && a != '^' && a != '(' && a != ')')
                  pushd(a);
                 if (a == '+' || a == '-' || a == '*' || a == '/')
                 System.out.print(a);
                 if(a == '(')
                  continue;
                 if(a == ')')
               continue;
             }
            while(dsp != 0)
             System.out.print(popd());
            System.out.println("\n");
           }

        void push(char x)
            {
               oprt[top++] = x;
             }
       
               char pop()
                {
                     return(oprt[--top]);
                   }
       
        void insr(char x)
            {
                ptop = ptop + 1;
                pst[ptop] = x;
            }

        void display(int n)
            {
              for(i=0;i<n+1;i++)
              System.out.print(pst[i]);
            }
       
        void postfix()
            {
                Scanner in = new Scanner(System.in);
                System.out.print( "Enter Expression in infix format : ");
                s = in.nextLine();
                n = s.length();
                for(i=0;i<n;i++)
                inf[i] = s.charAt(i);
                i = 0;
                while (i < n)
                   {
                    if (inf[i] != '+' && inf[i] != '-' && inf[i] != '*'
                         && inf[i] != '/' && inf[i] != '^' && inf[i] != '('
                         && inf[i] != ')')
                    insr(inf[i]);
                    else
                      {
                        if (inf[i] == '+' || inf[i] == '-' || inf[i] == '*'
                             || inf[i] == '/')
                        push(inf[i]);  
                        else
                       {
                          if(inf[i]=='(')
                         push(inf[i]);
                    else
                        {
                     if(inf[i]==')')
                          {
                        while(top!= 0 && oprt[top-1]!='(')
                          insr(pop());
                          pop();
                         }
                      }
                  }
                 }
                 i++;               
              }

           while(top != 0)
           insr(pop());
           System.out.print("\n POSTFIX FORM IS : ");
           display(n);
           System.out.println("\n\n");
         }

     }

class infprepost
 {
   public static void main(String args[])
       {
            
              Scanner in = new Scanner(System.in);
              obj P = new obj();
              int ch;
             do
              {
               System.out.println("\n1. INFIX TO PREFIX");
               System.out.println("2. INFIX TO POSTFIX");
               System.out.println("3. EXIT");
               System.out.print("Enter Your Choice : ");
               ch = in.nextInt();
               switch(ch)
                  {
                    case 1 : P.prefix();
                             break;
                    case 2 : P.postfix();
                             break;
                    case 3 :
                                 System.out.println("This is the End of the Program");
                             break;
                    default :
                            System.out.println("Invalid Choice ? ");
                  }
               }
              while(ch != 3);
      }

}


































Output:

C:\Program Files\Java\jdk1.6.0_13\bin>javac infprepost.java

C:\Program Files\Java\jdk1.6.0_13\bin>java infprepost

1. INFIX TO PREFIX
2. INFIX TO POSTFIX
3. EXIT
Enter Your Choice : 1
Enter Expression in infix format  :
a*b+c/d

 THE PREFIX FORM IS : /+*abcd


1. INFIX TO PREFIX
2. INFIX TO POSTFIX
3. EXIT
Enter Your Choice : 2
Enter Expression in infix format : a*b+c/d

 POSTFIX FORM IS :  abcd/+*



1. INFIX TO PREFIX
2. INFIX TO POSTFIX
3. EXIT
Enter Your Choice : 3
This is the End of the Program















/* Heap tree traversal*/

#include <iostream.h>
#include <conio.h>

void cheap(int t[20],int n)
{
 int i,j,key, q;
 for(q=2;q<=n;++q)
 { i = q;
   key = t[q];
   j = (int) (i/2);
   while((i>1)  && (key > t[j]))
   {
     t[i] = t[j];
     i = j;
     j = (int) (i/2);
     if (j < 1)
      j = i;
  } t[i] = key;
  } }
 void heap(int t[20],int n)
 {
  int q, temp, i,j, key;
  cheap(t,n);
  for(q=n;q >= 2; --q)
  {
    temp = t[1];
    t[1] = t[q];
    t[q] = temp;
    i = 1;
    key = t[1];
    j = 2;
    if((j+1) < q)
      if(t[j+1] > t[j])
      j++;
      while(j <= (q-1) && (t[j] > key))
      {
       t[i] = t[j];
       i = j;
       j = 2 * i;
       if (j+1 < q)
            if (t[j+1] > t[j])
             j++;
            else
             if (j > n)
               j = n;
            t[i] = key;
          }  } }
void main(void)
{ int tree[20],i,n;
  clrscr();
  cout << "Enter the Number of Elements \n";
  cin >>n;
  cout << "Enter the Numbers \n";
  for(i=1;i<=n;i++)
   cin >> tree[i];
  heap(tree,n);
  cout << "\n\n   Heap Tree   ";
  cout<<"\n-------------------";
  cout<<"\nAscending Order: ";
  for(i=1;i<=n;i++)
   {cout << "---> " << tree[i];}
    cout<<"\n\nDecending Order: ";
  for(i=n;i>=1;i--)
   {cout << "---> " << tree[i];}

   getch();
}
























Output:

Enter the Number of Elements
5

Enter the Numbers

23

1

58

96

4


   Heap Tree
-------------------

Ascending Order: ---> 1---> 4---> 23---> 58---> 96


Decending Order: ---> 96---> 58---> 23---> 4---> 1



                                     














DOUBLE STACK:
#include<iostream.h>
#include<conio.h>
#include<stdlib.h>

class ds
 {
  private:
   int d[11],l,r,e;
  public:
   ds()
   { l=r=5; }

void PUSHA()
 {
  ++r;
  if(r>10)
    {     
      cout<<"DS Full-right side";
      return;
    }
  cout<<"Enter element(int) to insert";
  cin>>e;
  d[r]=e;
 }

void PUSHB()
 {
   --l;
   if(l<1)
     {
      cout<<"DS Full left side";
      return;
     }
  cout<< "Enter element(int) to insert";
  cin>>e;
  d[l]=e;
 }

void POPA()
 {
   if(r==5)
     {
      cout<<"DS Empty right side";
      return;
     }
   e=d[r];--r;
   cout<<"The deleted element is"<<e;
 }

void POPB()
 {
   if(l==5)
    {
      cout<<"DS Empty left side";
      return;
    }
   e=d[l];++l;
   cout<<"The deleted element is"<<e;
 }

void display()
 {
   cout<<"Elements in DS are\n";
   for(int i=l;i<=r;++i)
   cout<<d[i]<<" ";
 }

};

void main()
 {
   int o;
   ds odq;
   clrscr();
   do
    {
      cout<<"\n1.PUSHA Right\n";
      cout<<"2.PUSHB Left\n";
      cout<<"3.POPA Right\n";
      cout<<"4.POPB Left\n";
      cout<<"5.Display\n";
      cout<<"6.Exit\n";
      cout<<"Enter your choice  ";
      cin>>o;
      switch(o)
        {
         case 1: odq.PUSHA();
                 break;
         case 2: odq.PUSHB();
                         break;
         case 3: odq.POPA();
                 break;
         case 4: odq.POPB();
                 break;
         case 5: odq.display();
                 break;
         case 6: exit(1);
                 break;
         default: cout<<"Illegal choice";
        }
   }while(1);
}


































OUTPUT:

1.PUSHA Right
2.PUSHB Left
3.POPA Right
4.POPB Left
5.Display
6.Exit
Enter your choice  1
Enter element(int) to insert 10

1.PUSHA Right
2.PUSHB Left
3.POPA Right
4.POPB Left
5.Display
6.Exit
Enter your choice  2
Enter element(int) to insert 20

1.PUSHA Right
2.PUSHB Left
3.POPA Right
4.POPB Left
5.Display
6.Exit
Enter your choice  5
Elements in DS are
20 0 10
1.PUSHA Right
2.PUSHB Left
3.POPA Right
4.POPB Left
5.Display
6.Exit
Enter your choice  3
The deleted element is 10
1.PUSHA Right
2.PUSHB Left
3.POPA Right
4.POPB Left
5.Display
6.Exit
Enter your choice  5
Elements in DS are
20 0
// implementation of double queue using linked lists

#include<iostream.h>

#include<conio.h>

#include<stdlib.h>

 class dq

{ private:

    int d[11],l,r,e;

  public:

   dq() { l=r=5; }

  void insertright()

  { ++r;

  if(r>10) {cout<<"DQ Full-right side"; return;}

  cout<<"Enter element(int) to insert";

  cin>>e;

  d[r]=e;

  }

   void insertleft()

  { --l;

  if(l<1) {cout<<"DQ Full left side"; return;}

  cout<< "Enter element(int) to insert";

  cin>>e;

  d[l]=e;

  }



  void delright()

  {

  if(r==5) {cout<<"DQ Empty right side"; return;}

   e=d[r];--r;

   cout<<"The deleted element is"<<e;

  }

   void delleft()

  {  if(l==5) {cout<<"DQ Empty left side"; return;}

   e=d[l];++l;

   cout<<"The deleted element is"<<e;

  }

  void display()

  { cout<<"Elements in DQ are\n";

  for(int i=l;i<=r;++i)

    cout<<d[i]<<" ";

  }
 };

void main()

{ int o;

  dq odq;

  clrscr();

  do{

  cout<<"\n1.Insert Right\n";

    cout<<"2.Insert Left\n";

    cout<<"3.Delete Right\n";

    cout<<"4.Delete Left\n";

    cout<<"5.Display\n";

    cout<<"6.Exit\n";

    cout<<"Enter your choice  ";

    cin>>o;

    switch(o)

     { case 1: odq.insertright();break;

       case 2: odq.insertleft();break;

       case 3: odq.delright();break;

       case 4: odq.delleft();break;

       case 5: odq.display();break;

       case 6: exit(1);break;

       default: cout<<"Illegal choice";

      }

    }while(1);

 }











Output :

1.Insert Right
2.Insert Left
3.Delete Right
4.Delete Left
5.Display
6.Exit
Enter your choice  1
Enter element(int) to insert 25

1.Insert Right
2.Insert Left
3.Delete Right
4.Delete Left
5.Display
6.Exit
Enter your choice  2
Enter element(int) to insert 45

1.Insert Right
2.Insert Left
3.Delete Right
4.Delete Left
5.Display
6.Exit
Enter your choice   5
Elements in DQ are
45 0 25
1.Insert Right
2.Insert Left
3.Delete Right
4.Delete Left
5.Display
6.Exit
Enter your choice  3
The deleted element is25
1.Insert Right
2.Insert Left
3.Delete Right
4.Delete Left
5.Display
6.Exit

Enter your choice 6

// Department binary tree

#include<iostream.h>  // Exno= 13
#include<stdlib.h>
#include <string.h>
#include<conio.h>
struct college
{
char dept[20];
int status;
int i;
}bt[10];
int n,i,c,o,l,r;
char ch,op;
char dept[20];

void createt()
{
 i=1;ch='y' || 'Y';
 cout<<"\nEnter Root Node  : ";
 cin >> bt[i].dept;
 bt[i].status = 0;
 do
 {
 cout << "\nIs there Any Left Node (Y/N)? : ";
 cin >> op;
 if (op == 'y' || 'Y')
 {
 l = 2 * i;
 cout << "\nEnter the " <<l<< " Left Node : ";
 cin >> bt[l].dept;
 bt[l].status = 0;
 }

 cout << "\nIs there Any Right Node (Y/N)? : ";
 cin >> op;
 if (op == 'y' || 'Y')
 {
  r = 2 * i + 1;
  cout << "\nEnter the " <<r<< " Right Node : ";
  cin >> bt[r].dept;
  bt[r].status = 0;
 }
 c = i;
 cout << "\nDo you Want to Continue (Y/N) ? : ";
 cin >> ch;
 i++;
 }
 while(ch == 'y' || ch =='Y');
}

void insertt()
{                                                                                    
 n = c * 2 + 1;
 int x;
 cout << "\nEnter the New Department to Add in the TREE / COLLEGE\n";
 cin >>dept;
 i = 1;
 do
 {
  x =strcmp((char*)bt[i].dept,(char *)dept);
  if (x != 0)
  {
    strcpy(bt[n+1].dept,dept);
    bt[n+1].status = 0;
  }
  else
  {
  cout << "\nThis Department is Already Exists\n";
  break;
  }
  i++;
 }while(i<=n);
 n = i+1;
}
void deletet()
{
 n = c * 2 +1;
 i = 1;
 int x;
 cout << "\nEnter the Department Name to Delete in the TREE / COLLEGE\n";
 cin >>dept;
 do
 {
 x =strcmp((char*)bt[i].dept,(char *)dept);
 if (x == 0)
  {
    bt[i].status = 1;
    break;
  }
  i++;
 }while(i<=n);
}
void display()
{
 n = c * 2 + 2;
 i =0;
 cout << "\nThe List of Department names in the College\n";
 while(i<=n)
 {
  if (bt[i].status == 0)
  cout <<  "\t"  << bt[i].dept <<endl;
  i++;
 }}
void main()
{ clrscr();
 createt();
 do{
    cout<<"\n1.Insert the Department into the College tree";
    cout<<"\n2.Delete the Department from College tree";
    cout<<"\n3.Display all the Department in the College tree";
    cout<<"\n4.Exit";
    cout<<"\nEnter your option (1-4) ?";
    cin>>o;
    switch(o)
    {   case 1:insertt();break;
            case 2:deletet();break;
            case 3:display();break;
            case 4:exit(0);
    }
    }while(1);
}














Output:

Enter Root Node  : jbict
Is there Any Left Node (Y/N)? : y
Enter the 2 Left Node : mba
Is there Any Right Node (Y/N)? : y
Enter the 3 Right Node : mca
Do you Want to Continue (Y/N) ? : n

1.Insert the Department into the College tree
2.Delete the Department from College tree
3.Display all the Department in the College tree
4.Exit
Enter your option (1-4) ? 1
Enter the New Department to Add in the TREE / COLLEGE

b.tech

1.Insert the Department into the College tree
2.Delete the Department from College tree
3.Display all the Department in the College tree
4.Exit
Enter your option (1-4) ? 3

The List of Department names in the College

        jbict
        mba
        mca
        b.tech

1.Insert the Department into the College tree
2.Delete the Department from College tree
3.Display all the Department in the College tree
4.Exit
Enter your option (1-4) ? 2

Enter the Department Name to Delete in the TREE / COLLEGE
mba

1.Insert the Department into the College tree
2.Delete the Department from College tree
3.Display all the Department in the College tree
4.Exit

// Priority queue implementation on airline reservation

#include <iostream.h>    
#include <conio.h>

struct pass

  char name[20];
  long double p;

};
void main(void)
{
  pass q[10], temp;
  int i,j,n;
  long mile;
  int years,seq;
  long double pri;
  clrscr();
  cout << "\nEnter How many Passengers? : ";
  cin >> n;
  for(i=1;i<=n;i++)
  {
   cout << "\nEnter the  "<< i << " Passenger Name : ";
   cin >> q[i].name;
   cout << "\nEnter the  "<< i << " Mileage : ";
   cin >> mile;
   cout << "\nEnter the  "<< i << " Years : ";
   cin >> years;
   cout << "\nEnter the  "<< i << " Sequence : ";
   cin >> seq;
   pri = mile / 1000 + years - seq;
   q[i].p = pri;
  }
  for(i=1;i<=n;i++)
  for(j=1;j<n;j++)
  {
   if (q[j].p > q[j+1].p)
   {
             temp = q[j];
            q[j] = q[j+1];
            q[j+1] = temp;
   }
}



  cout << "\n\nPassenger Name    ||" ;
  cout <<"Waiting Customer - Priority \n";
  cout << "--------------------------";
  cout<<"----------------------------------";
  for(i=1;i<=n;i++)
   cout << endl<<"\t"<<q[i].name <<"\t\t"<< q[i].p<< endl;
  getch();
 }




































Output :

Enter How many Passengers? : 3

Enter the  1 Passenger Name : a

Enter the  1 Mileage : 2000

Enter the  1 Years : 5

Enter the  1 Sequence : 1

Enter the  2 Passenger Name : b

Enter the  2 Mileage : 300

Enter the  2 Years : 10

Enter the  2 Sequence : 2

Enter the  3 Passenger Name : c

Enter the  3 Mileage : 900

Enter the  3 Years : 3

Enter the  3 Sequence : 3


Passenger Name          ||Waiting Customer - Priority
---------------------          ---------------------------------------
        c                            0

        a                            6

        b                            8










// Library Database Using Linked Lists
        
#include <iostream.h>
#include <dos.h>
#include <conio.h>
#include <process.h>

int i=0;

struct link
{
 int acno;
 char title[20];
 int cop;
 int status;
 struct link *next;
};

struct link *item, *temp, *first;

struct link2
{
 int bno;
 int acno;
 char bname[20];
 char doi[20];
 char dor[20];
 int status;
 struct link2 *next;
};
struct link2 *item2, *temp2, *first2;

int ch;

library();
issue();
libsearch();
borrow();
retun();

void main(void)
{
  first = NULL;
  first2 = NULL;
  pos: clrscr();
  gotoxy(15,4);cout<< "   1. CREATING LIBRARY BOOKS  ";
  gotoxy(15,5);cout<< "   2. ISSUING LIBRARY BOOKS ";
  gotoxy(15,6);cout<< "   3. RETURN LIBRARY BOOKS ";
  gotoxy(15,7);cout<< "   4. BORROWER DETAILS ";
  gotoxy(15,8);cout<< "   5. LIBRARY SEARCH ";
  gotoxy(15,9);cout<< "   6. Exit  ";
  gotoxy(15,10);cout<< "Enter Your Choice:  " ;
  cin>> ch;
  switch(ch)
  {
  case 1 : library();
               break;
  case 2 : issue();
               break;
  case 3 : retun();
               break;
  case 4 : borrow();
               break;
  case 5 : libsearch();
               break;
  case 6 : exit(0);
               break;
  default : { gotoxy(10,19);
              cout<< "Your choice is out of Range ";
                  goto pos;
                }
   }
  goto pos;
}

library()
{
  if (first == NULL)
  {
     item = new link;
     first = item;
  }
  else
  {
  temp = new link;
  item->next = temp;
  item = temp;
  }
  cout<<endl<<"Enter the Book Accession Number:";
  cin >> item->acno;
  cout<<endl<<"Enter the Book Title :   ";
  cin >> item->title;
  cout<<endl<<"Enter the No. of Copies of the Book:";
  cin >> item->cop;
  item->status = 0;
  item->next = NULL;
  return(0);
}

libsearch()
{
 int an;
 item = first;
 cout<<"Enter the Book Accession Number to Search:";
 cin >> an;

 do
 {
  if(item->acno == an && item->status == 0)
  {
  cout<<"\n"<<"The Accession number of the Book ="<<item->acno<<"\n";
  cout<<"\n"<<"The Title of the Book = " <<item->title<<"\n";
  cout<<"\n"<<"The Number of copies of the book = "<<item->cop<<"\n";
  break;
  }
  item = item->next;
 }while(item != NULL);
 if (item == NULL)
 cout << "Book Not Found ";
 getch();
return(0);
}

issue()
{
 int assn;
 item = first;
 cout<<"Enter the Book Accession Number to Issue :";
 cin >> assn;
 do
 {
  if(item->acno == assn)
  {
  i++;
  if (first2 == NULL)
  {
     item2 = new link2;
     first2 = item2;
  }
  else
  {
  temp2 = new link2;
  item2->next = temp2;
  item2 = temp2;
  }
  item2->acno = assn;
  item->cop = item->cop - 1;
  cout<<endl<<"Enter the Borrower A/C Number  =  ";
  cin >> item2->bno;
  cout<<endl<<"Enter the Borrower Name  =  ";
  cin >> item2->bname;
  cout<<endl<<"Enter the Date of Issue =  ";
  cin >> item2->doi;
  item2->status = 0;
  if ( i > 5)
  item->status = 1;
  item2->next = NULL;
  break;
  }
  item = item->next;
 }while(item != NULL);
 if (item == NULL)
 cout << "Book Not Found ";
 getch();
return(0);
}

retun()
{
 int assn;
 item = first;
 item2 = first2;
 cout<<endl<<"Enter the Book Accession Number to Return :";
 cin >> assn;
 do
 {
 if(item->acno == assn)
 {
 do
 {
  if(item2->acno == assn)
  {
  i--;
  cout<<endl<<"The Borrower Name  =  " <<item2->bname;
  cout<<endl<<"Enter Borrower A/C Number="<<item2->bno;
  cout<<endl<<"Enter the Date of Return =  ";
  cin >> item2->dor;
  item2->status = 1;
  item->cop = item->cop + 1;
  if ( i < 5)
  item->status = 0;
  break;
  }
  item2 = item2->next;
 }while(item2 != NULL);
 if (item2 == NULL)
 cout << "Book Not Found ";
 }
 item = item->next;
 }while(item != NULL);
 getch();
return(0);
}
borrow()
{
 item2 = first2;
 while(item2 != NULL)
 {
 if(item2->status == 0)
 {
  cout<<"\n"<<"The Borrower Number for the Book = "<<item2->bno<<"\n";
  cout<<"\n"<<"The Accession number of the Book = "<<item2->acno<<"\n";
  cout<<"\n"<<"The Borrwer Name for the Book = "<<item2->bname<<"\n";
  cout<<"\n"<<"The Date of Issue  = "<<item2->doi<<"\n";
 }
  item2 = item2->next;
 }
 cout<<endl<<"The Number of Books in his/her A/c = "<<i<<"\n";
 getch();
return(0);
}









Output:


                 1. CREATING LIBRARY BOOKS
                 2. ISSUING LIBRARY BOOKS
                 3. RETURN LIBRARY BOOKS
                 4. BORROWER DETAILS
                 5. LIBRARY SEARCH
                 6. Exit

              Enter Your Choice:  1

Enter the Book Accession Number: 105

Enter the Book Title:   FS

Enter the No. of Copies of the Book: 50

                 1. CREATING LIBRARY BOOKS
                 2. ISSUING LIBRARY BOOKS
                 3. RETURN LIBRARY BOOKS
                 4. BORROWER DETAILS
                 5. LIBRARY SEARCH
                 6. Exit

              Enter Your Choice:  2

Enter the Book Accession Number to Issue: 105

Enter the Borrower A/C Number = 101

Enter the Borrower Name = Harun

Enter the Date of Issue = 30-06-2011

                 1. CREATING LIBRARY BOOKS
                 2. ISSUING LIBRARY BOOKS
                 3. RETURN LIBRARY BOOKS
                 4. BORROWER DETAILS
                 5. LIBRARY SEARCH
                 6. Exit

              Enter Your Choice:  3

Enter the Book Accession Number to Return: 105

The Borrower Name = Harun


Enter Borrower A/C Number =101

Enter the Date of Return = 1-07-2011

                 1. CREATING LIBRARY BOOKS
                 2. ISSUING LIBRARY BOOKS
                 3. RETURN LIBRARY BOOKS
                 4. BORROWER DETAILS
                 5. LIBRARY SEARCH
                 6. Exit

              Enter Your Choice:  4

The Borrower Number for the Book = 101

The Accession number of the Book = 105

The Borrower Name for the Book = Harun

The Date of Issue = 30-06-2011

The Number of Books in his/her A/c = 1

                 1. CREATING LIBRARY BOOKS
                 2. ISSUING LIBRARY BOOKS
                 3. RETURN LIBRARY BOOKS
                 4. BORROWER DETAILS
                 5. LIBRARY SEARCH
                 6. Exit

              Enter Your Choice:  5

Enter the Book Accession Number to Search: 105

The Accession number of the Book =105

The Title of the Book = FS

The Number of copies of the book = 50





// print groceries shop information using linked list

#include<iostream.h>
#include<conio.h>
class node
{
public:
int itemcode,stock;
char itemname[20];
float price,vat;
node *next;
};
class grocery
{
node *x,*y,*start;
float amount;
public:
grocery()
{
x=NULL;
y=NULL;
start=NULL;
amount=0;
}
void create();
void display();
void search();
};
void grocery::create()
{
x=new node;
cout<<"\n";
cout<<"enter itemcode"<<"\n";
cin>>x->itemcode;
cout<<"enter item name"<<"\n";
cin>>x->itemname;
cout<<"enter stock"<<"\n";
cin>>x->stock;
cout<<"enter price"<<"\n";
cin>>x->price;
cout<<"enter vat"<<"\n";
cin>>x->vat;
amount=amount+((x->stock*x->price)+(x->stock*(x->price*x->vat/100)));
x->next=NULL;
start=x;
y=x;
cout<<"enter 1 to enter another node";
int ch;
cin>>ch;
while(ch==1)
{
x=new node;
cout<<"\n";
cout<<"enter itemcode"<<"\n";
cin>>x->itemcode;
cout<<"enter item name"<<"\n";
cin>>x->itemname;
cout<<"enter stock"<<"\n";
cin>>x->stock;
cout<<"enter price"<<"\n";
cin>>x->price;
cout<<"enter vat"<<"\n";
cin>>x->vat;
amount=amount+((x->stock*x->price)+(x->stock*(x->price*x->vat/100)));
x->next=NULL;
y->next=x;
y=x;
cout<<"enter 1 to enter another node";
cin>>ch;
}
}
void grocery::display()
{
node *temp;
temp=new node;
for(temp=start;temp!=NULL;temp=temp->next)
{
cout<<"\n"<<"\n";
cout<<"ITEM CODE ::"<<temp->itemcode<<"\n";
cout<<"ITEM NAME::"<<temp->itemname<<"\n";
cout<<"STOCK::"<<temp->stock<<"\n";
cout<<"PRICE::"<<temp->price<<"\n";
cout<<"VAT::"<<temp->vat<<"\n";
}
cout<<"********************"<<"\n";
cout<<"total amount is::"<<amount<<”\n”;
cout<<"********************"<<"\n";
}
void grocery::search()
{
int code;
node *temp;
cout<<"enter item code to search";
cin>>code;
for(temp=start;temp!=NULL;temp=temp->next)
{
if(temp->itemcode==code)
{
cout<<"\n"<<"\n";
cout<<"ITEM CODE ::"<<temp->itemcode<<"\n";
cout<<"ITEM NAME::"<<temp->itemname<<"\n";
cout<<"STOCK::"<<temp->stock<<"\n";
cout<<"PRICE::"<<temp->price<<"\n";
cout<<"VAT::"<<temp->vat<<"\n";
}
}
}
void main()
{
clrscr();
grocery g1;
int ch;
do
{
cout<<"\n"<<" MAHESH GROCERIS SHOP"<<"\n"<<"\n";
cout<<"1.Create item information"<<"\n";
cout<<"2.Display item information"<<"\n";
cout<<"3.search item"<<"\n";
cout<<"4.Exit"<<"\n";
cout<<"enter your choice";
cin>>ch;
switch(ch)
{
case 1:g1.create();break;
case 2:g1.display();break;
case 3:g1.search();break;
case 4:break;
}
}while(ch!=4);
}








Output:

 MAHESH GROCERIS SHOP

1.Create item information
2.Display item information
3.search item
4.Exit
enter your choice 1

enter itemcode
1001
enter item name
hamam
enter stock
100
enter price
20
enter vat
3.50
enter 1 to enter another node 1
 enter itemcode
1002
enter item name
colgate
enter stock
1000
enter price
15
enter vat
2.05
enter 1 to enter another node 2

 MAHESH GROCERIS SHOP

1.Create item information
2.Display item information
3.search item
4.Exit

enter your choice 2

ITEM CODE ::1001
ITEM NAME::hamam
STOCK::100
PRICE::20
VAT::3.5

ITEM CODE ::1002
ITEM NAME::colgate
STOCK::1000
PRICE::15
VAT::2.05

**********************
   total amount is::17377.5
**********************

 MAHESH GROCERIS SHOP

1.Create item information
2.Display item information
3.search item
4.Exit

enter your choice 3

enter item code to search1002

ITEM CODE ::1002
ITEM NAME::colgate
STOCK::1000
PRICE::15
VAT::2.05

 MAHESH GROCERIS SHOP

1.Create item information
2.Display item information
3.search item
4.Exit
enter your choice 4









//program for AVL tree operations using  c++
#include<iostream.h>
#include<stdlib.h>
#define TRUE 1
#define FALSE 0
#define NULL 0
class AVL;
class AVLNODE
{
            friend class AVL;
            private:
                        int data;
                        AVLNODE *left,*right;
                        int bf;
};
class AVL
 {
            private:
                        AVLNODE *root;
            public:
                        AVLNODE *loc,*par;
                        AVL()
                        {
                                    root=NULL;
                        }
                                    int insert(int);
                        void displayitem();
                        void display(AVLNODE *);
                        void removeitem(int);
                        void remove1(AVLNODE *,AVLNODE *,int);
                        void remove2(AVLNODE *,AVLNODE *,int);
                        void search(int x);
                        void search1(AVLNODE *,int);
};
int AVL::insert(int x)
{
            AVLNODE *a,*b,*c,*f,*p,*q,*y,*clchild,*crchild;
            int found,unbalanced;
            int d;
            if(!root)   //special case empty tree
            {          y=new AVLNODE;
                        y->data=x;
                        root=y;
                        root->bf=0;
                        root->left=root->right=NULL;
                        return TRUE;  }
            //phase 1:locate insertion point for x.a keeps track of the most
            // recent node with balance factor +/-1,and f is the parent of a
            // q follows p through the tree.
            f=NULL;
            a=p=root;
            q=NULL;
            found=FALSE;
            while(p&&!found)
            {                 //search for insertion point for x
                        if(p->bf)
                        {
                                    a=p;
                                    f=q;
                        }
                        if(x<p->data)    //take left branch
                        {
                                    q=p;
                                    p=p->left;
                        }
                        else if(x>p->data)
                        {
                                    q=p;
                                    p=p->right;
                        }
                        else
                        {
                                    y=p;
                                    found=TRUE;
                        }
            }               //end while
            //phase 2:insert and rebalance.x is not in the tree and
            // may be inserted as the appropriate child of q.
            if(!found)
            {
                        y = new AVLNODE;
                        y->data=x;
                        y->left=y->right=NULL;
                        y->bf=0;
                        if(x<q->data)    //insert as left child
                        q->left=y;
                        else
                        q->right=y;    //insert as right child
                        //adjust balance factors of nodes on path from a to q
                        //note that by the definition of a,all nodes on this
                        //path must have balance factors of 0 and so will change
                        //to +/- d=+1 implies that x is inserted in the left
                        // subtree of a d=-1 implies
                        //to that x inserted in the right subtree of a.


if(x>a->data)
                        {
                                    p=a->right;
                                    b=p;
                                    d=-1;
                        }
                        else
                        {
                                    p=a->left;
                                    b=p;
                                    d=1;
                        }
                        while(p!=y)
                        if(x>p->data)          //height of  right increases by 1
                        {
                                    p->bf=-1;
                                    p=p->right;
                        }
                        else                 //height of left increases by 1
                        {
                                    p->bf=1;
                                    p=p->left;
                        }
                                                                                                                        //is tree unbalanced
                        unbalanced=TRUE;
                        if(!(a->bf)||!(a->bf+d))
                        {                     //tree still balanced
                                    a->bf+=d;
                                    unbalanced=FALSE;
                        }
                        if(unbalanced)            //tree unbalanced,determine rotation type
                        {
                                    if(d==1)
                                    {                              //left imbalance
                                                if(b->bf==1)      //rotation type LL
                                                {
                                                            a->left=b->right;
                                                            b->right=a;
                                                            a->bf=0;
                                                            b->bf=0;
                                                }
                                                else                   //rotation type LR
                                                {
                                                            c=b->right;
                                                            b->right=c->left;
                                                            a->left=c->right;
                                                            c->left=b;
                                                            c->right=a;


switch(c->bf)
                                                            {
                                                                        case 1: a->bf=-1;  //LR(b)
                                                                                    b->bf=0;
                                                                                    break;
                                                                        case -1:b->bf=1;  //LR(c)
                                                                                    a->bf=0;
                                                                                    break;
                                                                        case 0: b->bf=0;  //LR(a)
                                                                                    a->bf=0;
                                                                                    break;
                                                            }
                                                            c->bf=0;
                                                            b=c;                //b is the new root
                                                }                                //end of LR
                                    }                                          //end of left imbalance
                                                else                                           //right imbalance
                                                {
                                    if(b->bf==-1)                     //rotation type RR
                                                {
                                                            a->right=b->left;
                                                            b->left=a;
                                                            a->bf=0;
                                                            b->bf=0;
                                                }
                                                else    //rotation type LR
                                                {
                                                            c=b->right;
                                                            b->right=c->left;
                                                            a->right=c->left;
                                                            c->right=b;
                                                            c->left=a;
                                                            switch(c->bf)
                                                            {
                                                                        case 1: a->bf=-1;  //LR(b)
                                                                                    b->bf=0;
                                                                                    break;
                                                                        case -1:b->bf=1;  //LR(c)
                                                                                    a->bf=0;
                                                                                    break;
                                                                        case 0: b->bf=0;  //LR(a)
                                                                                    a->bf=0;
                                                                                    break;
                                                            }
                                                            c->bf=0;
                                                            b=c; //b is the new root
                                                } //end of LR
                                                }
//subtree with root b has been rebalanced and is the new subtree

if(!f)
                                                 root=b;
                                    else
                                                                                                                                                            if(a==f->left)
                                                            f->left=b;
                                     else
                                                                                                                                                              if(a==f->right)
                                                                        f->right=b;
                        }                                                  //end of if unbalanced
                        return TRUE;
            }                                                               //end of if(!found)
            return FALSE;
}                                                                            //end of AVL INSERTION

void AVL::displayitem()
{
            display(root);
}
void AVL::display(AVLNODE *temp)
{
            if(temp==NULL)
            return;
            cout<<temp->data<<" ";
            display(temp->left);
            display(temp->right);
}
void AVL::removeitem(int x)
{
            search(x);
            if(loc==NULL)
            {
                        cout<<"\nitem is not in tree";
                        return;
            }
            if(loc->right!=NULL&&loc->left!=NULL)
            remove1(loc,par,x);
            else
            remove2(loc,par,x);
}
void AVL::remove1(AVLNODE *l,AVLNODE *p,int x)
{
            AVLNODE *ptr,*save,*suc,*psuc;
            ptr=l->right;
            save=l;
            while(ptr->left!=NULL)
            {
                        save=ptr;
                        ptr=ptr->left;
            }
            suc=ptr;
            psuc=save;
            remove2(suc,psuc,x);
            if(p!=NULL)
                        if(l==p->left)
                                    p->left=suc;
                        else
                                    p->right=suc;
            else
                        root=l;
             suc->left=l->left;
             suc->right=l->right;
              return;
}
void AVL::remove2(AVLNODE *s,AVLNODE *p,int x)
{
            AVLNODE *child;
            if(s->left==NULL && s->right==NULL)
                        child=NULL;
            else if(s->left!=NULL)
                        child=s->left;
            else
                        child=s->right;
            if(p!=NULL)
                        if(s==p->left)
                                    p->left=child;
                        else
                                    p->right=child;
            else
                        root=child;

}
void AVL::search(int x)
{
            search1(root,x);
}
void AVL::search1(AVLNODE *temp,int x)
{
       AVLNODE *ptr,*save;
       int flag;
       if(temp==NULL)
       {
                        cout<<"\nthe tree is empty";
                        return;
       }
       if(temp->data==x)
       {
                        cout<<"\nthe item is root and is found";
                        par=NULL;
                        loc=temp;
                        par->left=NULL;
                        par->right=NULL;
                        return;       }
       if( x < temp->data)
       {
                        ptr=temp->left;
                        save=temp;
       }
       else
       {
                        ptr=temp->right;
                        save=temp;
       }
       while(ptr!=NULL)
       {
                        if(x==ptr->data)
                        {       flag=1;
                                    cout<<"\nitemfound";
                                    loc=ptr;
                                    par=save;

                        }
                        if(x<ptr->data)
                        ptr=ptr->left;
                        else
                        ptr=ptr->right;
       }
       if(flag!=1)
       {
                        cout<<"item is not there in tree";
                        loc=NULL;
                        par=NULL;
                        cout<<loc;
                        cout<<par;
       }
}

main()
{
            AVL a;
            int x,y,c;
        char ch;   
            do
            {
                        cout<<"\n1.insert";
                        cout<<"\n2.display";
                        cout<<"\n3.delete";
                        cout<<"\n4.search";
                        cout<<"\n5.exit";
                        cout<<"\nEnter u r choice to perform on AVL tree";
                        cin>>c;
                       

switch(c)
                        {
                                    case 1:cout<<"\nEnter an element to insert into tree";
                                                cin>>x;
                                                a.insert(x);
                                                break;
                                    case 2:a.displayitem(); break;
                                    case 3:cout<<"\nEnter an item to deletion";
                                           cin>>y;
                                           a.removeitem(y);
                                           break;
                                    case 4:cout<<"\nEnter an element to search";
                                                cin>>c;
                                                a.search(c);
                                                break;
                                    case 5:exit(0);  break;
                              default :cout<<"\nInvalid option try again";
                        }
                        cout<<"\ndo u want to continue";
                        cin>>ch;
            }
            while(ch=='y'||ch=='Y');
}









































Output:

1.insert
2.display
3.delete
4.search
5.exit
Enter u r choice to perform on AVL tree  1

Enter an element to insert into tree  20

do u want to continue  y

1.insert
2.display
3.delete
4.search
5.exit
Enter u r choice to perform on AVL tree  1

Enter an element to insert into tree  1

do u want to continue  y

1.insert
2.display
3.delete
4.search
5.exit
Enter u r choice to perform on AVL tree  2
20 1
do u want to continue  y

1.insert
2.display
3.delete
4.search
5.exit
Enter u r choice to perform on AVL tree  4
Enter an element to search  1

itemfound
do u want to continue  y

1.insert
2.display
3.delete
4.search
5.exit
Enter u r choice to perform on AVL tree  3
Enter an item to deletion  1

itemfound
do u want to continue  y

1.insert
2.display
3.delete
4.search
5.exit
Enter u r choice to perform on AVL tree  5






























// double linked list in c++

#include<iostream.h>

#include<alloc.h>

#include<stdlib.h>

#include<conio.h>

#define new(p) p=(struct node*) malloc(sizeof(struct node))

#define read(p) cin>>p

class dll

{ private:

   struct node

     { int info;

       struct node* right,*left;

     }*h,*p,*t,*k,*l;

  public:

   dll()

    { h=l=0;}

   void create()

    { cout<<"Enter elements at end give 0  ";

      new(p);h=p;

      read(p->info);

      while(p->info!=0)

       { new(k);

                 read(k->info);

                 p->right=k;

                 k->left=p;

                 p=k;
       }

      l=p;

      l->right=0;

      deletelast();

    }

  void insertfirst()

 { cout<<"Enter value to insert";

   new(t);

   cin>>t->info;

   t->right=h;

   h->left=t;

   h=t;

 }

 void insertlast()

 { t=h;

   while(t->right!=0) t=t->right;

   new(p);

   cout<<"Enter value to insert";

   cin>>p->info;

   t->right=p;

   p->left=t;
   p->right=0;
 }
 void insertmiddle()

 { int e;

   new(t);

   cout<<"Enter value to insert";

   cin>>t->info;

   cout<<"Enter after which node you want to insert";

   cin>>e;

   p=h;

   while(p->info!=e)

    { if(p==0) {cout<<"Element not found"; return;}

      p=p->right;

    }

   t->right=p->right;

   p->right->left=t;

   p->right=t;

   t->left=p;

 }
 void insert()

 { int o;

   do

    { cout<<"1.insertfirst\n";

      cout<<"2.insertlast\n";

      cout<<"3.insertmiddle\n";
      cout<<"4.Exit";

      cout<<"Enter your option";

      cin>>o;

      switch(o)

      { case 1: insertfirst();break;

                case 2: insertlast();break;

                case 3: insertmiddle();break;

                case 4: return;
      }

    }while(1);

 }

 void deletefirst()

 {  p=h->right;

    free(h);

    h=p;
 }

 void deletelast()

 { p=h;  while(p->right!=0)  p=p->right;

   t=h;  while(t->right!=p)  t=t->right;

   free(p);

   t->right=0;
 }
 void deletemiddle()

 { int e;

   cout<<"enter node to be deleted";

   cin>>e;

   p=h;

   while(p->info!=e)

    { if(p==0) cout<<"Element not found";

      p=p->right;

    }
   t=h;

   while(t->right!=p) t=t->right;

   t->right=p->right;

   p->right->left=t;

   free(p);
 }
void del()

 { int o;

   do

    { cout<<"1.Delete First\n";

      cout<<"2.Delete Last\n";

      cout<<"3.Delete Middle\n";

      cout<<"4.Exit";

      cout<<"Enter your option";

      cin>>o;

      switch(o)

      { case 1: deletefirst();break;

                case 2: deletelast();break;

                case 3: deletemiddle();break;
                case 4: return;
     }
    }while(1);
 }
 void disp()
   {
     cout<<"Elements in the list are   ";
     p=h;
    while(p!=0)
      {
          cout<<p->info<<" ";
           p=p->right;

      }      }        }

  Void main()
{
dll l;
 int o;
 clrscr();
  l.create();
 do
  {
    cout<<"\n";
    cout<<"  1.Insertion\n";
    cout<<"  2.Deletion\n";
    cout<<"  3.Display\n";
    cout<<"  4.Exit\n";
    cout<<"  Enter your option:";
    cin>>o;
    switch(o)
    {
       case 1: l.insert();break;
      case 2: l.del();break;
      case 3: l.disp();break;
      case 4: exit(1);
    }
  }while(1);
 }







 Output:

Enter elements at end give 0 10
23
56
89
0

  1.Insertion
  2.Deletion
  3.Display
  4.Exit
  Enter your option: 1
1.insertfirst
2.insertlast
3.insertmiddle
4.Exit

Enter your option 1
Enter value to insert 20
1.insertfirst
2.insertlast
3.insertmiddle
4.Exit

Enter your option 2
Enter value to insert 50
1.insertfirst
2.insertlast
3.insertmiddle
4.Exit

Enter your option 4

  1.Insertion
  2.Deletion
  3.Display
  4.Exit
  Enter your option: 3
Elements in the list are   20 10 23 56 89 50
  1.Insertion
  2.Deletion
  3.Display
  4.Exit
  Enter your option: 2
1.Delete First
2.Delete Last
3.Delete Middle
4.Exit
Enter your option 1
1.Delete First
2.Delete Last
3.Delete Middle
4.ExitEnter your option 4

  1.Insertion
  2.Deletion
  3.Display
  4.Exit
  Enter your option: 3
Elements in the list are   10 23 56 89 50
  1.Insertion
  2.Deletion
  3.Display
  4.Exit
  Enter your option: 4


























//Binary search trees using arrays 

#include<iostream.h>
#include<stdlib.h>
#include<conio.h>
int bst[50],e,r,p,pa;
void createt()
{
 cout<<"Enter elements at end give 0\n";
 up1:r=1;
     cin>>e;
     if(e==0) return;
 up2:if(bst[r]==0) {bst[r]=e; goto up1;}
     if(bst[r]>e) {r=2*r; goto up2;}
     if(bst[r]<e) {r=2*r+1; goto up2;}
}
int search(int e)
{
r=1;
up3: if(bst[r]==0) {return(0);}
        if(bst[r]==e) {return(1);}
        if(bst[r]>e) {pa=r;r=2*r; goto up3;}
        if(bst[r]<e) {pa=r;r=2*r+1; goto up3;}

}
void searcht()
{cout<<"enter element to search";
 cin>>e;
 if(search(e)) cout<<"Element found";
 else cout<<"Element not found";
}
void insertt()
{cout<<"Enter element to insert";
 cin>>e;
 if(search(e))
  {cout<<"Element already exists"; return;}
 else
 {if(e>bst[pa]) bst[2*pa+1]=e;
  else bst[2*pa]=e;
 }
 cout<<"The element is inserted successfully";
}
void swap(int a,int b)
{int t=a;a=b;b=t;}
void delet(int e)
{ if(search(e))
   {if((bst[2*r]==0)&&(bst[2*r+1]==0)) {bst[r]=0;return;}
    if(bst[2*r]==0) {bst[2*pa+1]=bst[2*r+1];bst[2*r+1]=0;return;}
    if(bst[2*r+1]==0){bst[2*pa]=bst[2*r];bst[2*r]=0;return;}
    p=2*r+1;
    while(bst[2*p]!=0) p=2*p;
    swap(bst[r],bst[p]);
    delet(bst[p-1]);
   }
 else {cout<<"Element not existing";return;}
}
void deletet()
{cout<<"Enter the element you want to delete";
 cin>>e;
 delet(e);
}
void preorder(int r)
{if(bst[r]==0) return;
 cout<<" "<<bst[r];
 preorder(2*r);
 preorder(2*r+1);
}
void inorder(int r)
{ if(bst[r]==0) return;
  inorder(2*r);
  cout<<" "<<bst[r];
  inorder(2*r+1);
}
void postorder(int r)
{ if(bst[r]==0) return;
  postorder(2*r);
  postorder(2*r+1);
  cout<<" "<<bst[r];
}
void displayt()
{
 int o;
 do{ r=1;
  cout<<"\n1.preorder";
  cout<<"\n2.inorder";
  cout<<"\n3.postorder";
  cout<<"\n4.return";
  cout<<"\n5.Enter your option";
  cin>>o;
  switch(o)
    { case 1: preorder(r);break;
      case 2: inorder(r);break;
      case 3: postorder(r);break;
      case 4: return;
     }
    }while(1);
}
void main()
{
 int o;
 clrscr();
 createt();
 do{
    cout<<"\n1.Insert an element into tree";
    cout<<"\n2.Delete an element from tree";  
    cout<<"\n3.Display elements in tree";
    cout<<"\n4.Search for an element in tree";
    cout<<"\n5.Exit";
    cout<<"\nEnter your option";
    cin>>o;
    switch(o)
    { case 1:insertt();break;
      case 2:deletet();break;
      case 3:displayt();break;
      case 4:searcht();break;
      case 5:exit(1);
    }
    }while(1);
}




















Output:

Enter elements at end give 0
1
2
3
4
0

1.Insert an element into tree
2.Delete an element from tree
3.Display elements in tree
4.Search for an element in tree
5.Exit
Enter your option 3

1.preorder
2.inorder
3.postorder
4.return
5.Enter your option 3
 4 3 2 1
1.preorder
2.inorder
3.postorder
4.return
5.Enter your option 4

1.Insert an element into tree
2.Delete an element from tree
3.Display elements in tree
4.Search for an element in tree
5.Exit
Enter your option 4
enter element to search 4
Element found
1.Insert an element into tree
2.Delete an element from tree
3.Display elements in tree
4.Search for an element in tree
5.Exit





Enter your option
1
Enter element to insert 10
The element is inserted successfully
1.Insert an element into tree
2.Delete an element from tree
3.Display elements in tree
4.Search for an element in tree
5.Exit
Enter your option 2
Enter the element you want to delete  4
1.Insert an element into tree
2.Delete an element from tree
3.Display elements in tree
4.Search for an element in tree
5.Exit
Enter your option 5




























// quick sort in java

import java.util.*; 

class ss
{
    int temp,left,right,p,i,n;
    int[] a = new int[100];
            Scanner in = new Scanner(System.in);
            public void  reading(int n)
            {
            for(i=0;i<n;i++)
    {
     System.out.print("Enter Element of at a[" +  i   +  "]  :  ");
     a[i] = in.nextInt();
     }
   }
 public void sort(int beg, int end)
  {
    if( beg < end)
              {
               left = beg;
               right = end;
               p = a[beg];
              do
              {
               while((a[left] <= p) && (left < end))
               left = left + 1;
               while((a[right] >= p) && (right > beg))
               right = right - 1;
               if(left < right)
               {
                 temp = a[left];
                 a[left] = a[right];
                 a[right] = temp;
               }
               else
               break;
               }while(left < right);
               a[beg] = a[right];
               a[right] = p;
               sort(beg, right-1);
               sort(right+1, end);
               }
  }



public void display(int n)
{
  System.out.println("The Resultant Sorted Array is  : ");
  for(i=0;i<n;i++)
  System.out.println(a[i]);
}
}
class Quick
{
public static void main(String args[])
{
  ss  obj = new ss();
  Scanner in = new Scanner(System.in);
  int n;
  System.out.print("Enter the Array Size  : ");
  n = in.nextInt();
  obj.reading(n);
  obj.sort(0,n-1);
  obj.display(n);
}
 }

























Output:

C:\Program Files\Java\jdk1.6.0_13\bin>javac Quick.java

C:\Program Files\Java\jdk1.6.0_13\bin>java Quick
Enter the Array Size  : 5
Enter Element of at a[0]  :  12
Enter Element of at a[1]  :  6
Enter Element of at a[2]  :  4
Enter Element of at a[3]  :  89
Enter Element of at a[4]  :  3
The Resultant Sorted Array is  :
3
4
6
12
89


























//BFS using c++

#define SIZE 20
#define TRUE 1
#define FALSE 0
#include<iostream.h>
#include<conio.h>
#include<stdlib.h>
int g[SIZE][SIZE];
int visit[SIZE];
int q[SIZE],front,rear,n;
void main()
{
int v1,v2;
char ans='y';
void create();
void bfs(int);
clrscr();
create();
clrscr();
cout<<"\n the adjacency matrix for the graph";
for(v1=0;v1<n;v1++)
{
cout<<"\n";
for(v2=0;v2<n;v2++)
cout<<"\t"<<g[v1][v2];
}
getch();
do
{
for(v1=0;v1<n;v1++)
visit[v1]=FALSE;
clrscr();
cout<<"\n enter the vertex from which you want to traverse";
cin>>v1;
if(v1>=n)
cout<<"\n invalid vertex";
else
{
cout<<"\n Breadth first Search of a graph is...........";
bfs(v1);
getch();
}
cout<<"\n do u want traverse from any other node";
cin>>ans;
}while(ans=='y');
}
void create()
{
int v1,v2;
char ans='y';
cout<<"\n\t\t this program is to create a graph";
cout<<"\n enter no of nodes";
cin>>n;
for(v1=0;v1<n;v1++)
for(v2=0;v2<n;v2++)
g[v1][v2]=FALSE;
cout<<"\n enter the vertex number starting from 0";
do
{
cout<<"\n enter the vertices v1 and v2";
cin>>v1>>v2;
if(v1>=n||v2>=n)
cout<<"\n invalid vertex value";
else
{
g[v1][v2]=TRUE;
g[v2][v1]=TRUE;
}
cout<<"\n add more edges(Y/N)";
cin>>ans;
}while(ans=='y');
}
void bfs(int v1)
{
int v2;
visit[v1]=TRUE;
front=rear=-1;
q[++rear]=v1;
while(front!=rear)
{
v1=q[++front];
cout<<"\n"<<v1;
for(v2=0;v2<n;v2++)
if(g[v1][v2]==TRUE && visit[v2]==FALSE)
{
q[++rear]=v2;
visit[v2]=TRUE;
}
}
}

Output:

                 this program is to create a graph
 enter no of nodes 4

 enter the vertex number starting from 0
 enter the vertices v1 and v2 0 1

 add more edges(Y/N) y

 enter the vertices v1 and v2 0 3

 add more edges(Y/N) y

 enter the vertices v1 and v2 3 2

 add more edges(Y/N) n

 the adjacency matrix for the graph
        0       1       0       1
        1       0       0       0
        0       0       0       1
        1       0       1       0


 enter the vertex from which you want to traverse 0

 Breadth first Search of a graph is...........
0
1
3
2
do u want traverse from any other node y

 enter the vertex from which you want to traverse 1

 Breadth first Search of a graph is...........
1
0
3
2
 do u want traverse from any other node  n



//DFS using c++

#define SIZE 20
#define TRUE 1
#define FALSE 0
#include<iostream.h>
#include<conio.h>
#include<stdio.h>
#include<stdlib.h>
int g[SIZE][SIZE];
int visit[SIZE];
int n;
void main()
{
int v1,v2;
char ans='y';
void create();
void dfs(int);
clrscr();
create();
clrscr();
cout<<"\n the adjacency matrix for the graph";
for(v1=0;v1<n;v1++)
{
cout<<"\n";
for(v2=0;v2<n;v2++)
cout<<"\t"<<g[v1][v2];
}
getch();
do
{
for(v1=0;v1<n;v1++)
visit[v1]=FALSE;
clrscr();
cout<<"\n enter the vertex from which you want to traverse";
cin>>v1;
if(v1>=SIZE)
cout<<"\n invalid vertex";
else
{
cout<<"\n Deapth first Search of a graph is...........";
dfs(v1);
getch();
}
cout<<"\n do u want traverse from any other node";
cin>>ans;


}while(ans=='y');
}
void create()
{
int v1,v2,ch,flag;
char ans='y';
cout<<"\n\t\t this program is to create a graph";
flushall();
for(v1=0;v1<n;v1++)
for(v2=0;v2<n;v2++)
g[v1][v2]=FALSE;
cout<<"\n enter no of nodes";
cin>>n;
cout<<"\n  the vertex number starting from 0";
do
{
cout<<"\n enter the vertices v1 and v2";
cin>>v1>>v2;
if(v1>=n||v2>=n)
cout<<"\n invalid vertex value";
else
{
g[v1][v2]=TRUE;
g[v2][v1]=TRUE;
}
cout<<"\n add more edges(Y/N)";
cin>>ans;
}while(ans=='y');
}
void dfs(int v1)
{
int v2;
cout<<"\n"<<v1;
visit[v1]=TRUE;
for(v2=0;v2<SIZE;v2++)
if(g[v1][v2]==TRUE && visit[v2]==FALSE)
dfs(v2);
}








Output:

                 this program is to create a graph
 enter no of nodes 4

  the vertex number starting from 0
 enter the vertices v1 and v2 0 1

 add more edges(Y/N) y

 enter the vertices v1 and v2 0 3

 add more edges(Y/N) n

 the adjacency matrix for the graph
        0       1       0       1
        1       0       0       0
        0       0       0       0
        1       0       0       0

 enter the vertex from which you want to traverse 0

 Deapth first Search of a graph is...........
0
1
3
 do u want traverse from any other node y

 enter the vertex from which you want to traverse 1

 Deapth first Search of a graph is...........
1
0
3
 do u want traverse from any other node n









//PRIMS ALGORITHM
#include<iostream.h>
#include<conio.h>
#define SIZE 20
#define INF 32767
void prims(int g[][SIZE],int nodes)
{
int select[SIZE],i,j,k,min_dist,v1,v2,tot=0;
for(i=0;i<nodes;i++)
select[i]=0;
cout<<"\n\n minimum spanning tree is...........";
select[0]=1;
cout<<"\n\n source vertex \t destination vertex \t weight";
for(k=1;k<nodes;k++)
{
min_dist=INF;
for(i=0;i<nodes;i++)
{
for(j=0;j<nodes;j++)
{
if(g[i][j]&&((select[j])||(!select[j])))
{
if(g[i][j]<min_dist)
{
min_dist=g[i][j];
v1=i;v2=j;
}
}
}
}
cout<<"\t"<<v1<<"\t\t"<<v2<<"\t\t"<<min_dist<<"\n";
select[v1]=select[v2]=1;
tot=tot+min_dist;
cout<<"\n\n\t Total path length is :: "<<tot;
}
}
void main()
{
int g[SIZE][SIZE],nodes,v1,v2,length,i,j,n;
clrscr();
cout<<"\n\tprims algorithm \n enter no of nodes in the graph";
cin>>nodes;
cout<,"\n enter no of edges in the graph;
cin>>n;
for(i=0;i<nodes;i++)
for(j=0;j<nodes;j++)
g[i][j]=0;
cout<<"\n enter edges and weights";
for(i=0;i<n;i++)
{
cout<<"\n enter edge by v1 and v2";
cin>>v1>>v2;
cout<<"\n enter weights";
cin>>length;
g[v1][v2]=g[v2][v1]=length;
}
getch();
cout<<"\n\t";
clrscr();
prims(g,nodes);
getch();
}






























OUTPUT:

        prims algorithm
 enter no of nodes in the graph 5

 enter no of edges in the graph 6

 enter edges and weights
 enter edge by v1 and v2 0 1

 enter weights 2

 enter edge by v1 and v2 0 4

 enter weights 4

 enter edge by v1 and v2 1 4

 enter weights 1

 enter edge by v1 and v2 1 2

 enter weights 4
enter edge by v1 and v2 0 2

 enter weights 3
enter edge by v1 and v2 0 4

 enter weights 4

 minimum spanning tree is...........

 source vertex   destination vertex      weight 1               4
1


         Total path length is :: 1      1               4               1


         Total path length is :: 2      1               4               1


         Total path length is :: 3      1               4               1


         Total path length is :: 4

/*  program for towers of Honai */

#include<iostream.h>
 #include<conio.h>

 int n;
 char a,b,c;
 void move(int n, char ta,char tb,char tc)
 {
if(n>=1)
{
move(n-1,ta,tc,tb);
               cout<<ta<<"---->"<<tb;
               cout<<endl;
               move(n-1,tc,tb,ta);}
 }


 main()
 {
   clrscr();
   cout<<"Enter n";
   cin>>n;
   move(n,'a','b','c');
   getch();
 }





















Output:

Enter n 4

a---->c
a---->b
c---->b
a---->c
b---->a
b---->c
a---->c
a---->b
c---->b
c---->a
b---->a
c---->b
a---->c
a---->b
c---->b



























/* program for Newton’s equation */

#include<stdio.h>
#include<conio.h>
#include<iostream.h>
#include<stdlib.h>
#include<math.h>
#define f(x) (x*x-2)
#define d(x) (2*x)

void main()
{
int istep;
double a,b,x0,dx;
double dl=0.0000001;
clrscr();
dx=b-a;
x0=a-(f(a)/d(x0));
istep=0;
while(fabs(dx)>dl)
{
dx=f(x0)/d(x0);
cout<<"xo="<<x0<<"\n";
x0-=dx;
istep++;
}
cout<<the square root algorithm"<<"\n";
cout<<"Newton's method converged in <<istep<<"steps"<<"\n";
cout<<"the square root is "<<"\n";
printf("%16.241f\n",x0);
cout<<"\n";
getch();
}














Output:

xo=1.033333

xo=1.484409

xo=1.415873

xo=1.414215

xo=1.414214

the square root algorithm
Newton's method converged in 5steps
the square root is
1.414214




























/* program for merge sort */

#include<iostream.h>
#include<stdio.h>
#include<conio.h>
#include<fstream.h>
int a[20],b[20],i,n,j,m,s=0;
class customer
{
public:
int cnum,accno,bal;
char cname[20],branch[20];
void getdata();
void input();
void sort();
void display();
};
void customer::getdata()
{
cout<<"\n enter customer number:";
cin>>cnum;
cout<<"\n enter customer name:";
cin>>cname;
cout<<"\n enter account number:";
cin>>accno;
cout<<"\n enter initial balance:";
cin>>bal;
cout<<"\n enter branch name:";
cin>>branch;
}
void customer::display()
{
cout<<"\n"<<cnum<<"\t"<<cname<<"\t"<<accno<<"\t"<<bal<<"\t"<<branch<<endl;
}
void customer::sort()
{
customer bk,s1;
fstream bank1,bsort;
bank1.open("bank.dat",ios::in);
for(i=0;i<s;i++)
{
bank1.read((char*)&bk,sizeof(bk));
a[i]=bk.cnum;
b[i]=i;
}


for(i=0;i<s;i++)
{
for(j=i+1;j<s;j++)
{
if(a[i]>=a[j])
{
int t=a[i];
a[i]=a[j];
a[j]=t;
int t1=b[i];
b[i]=b[j];
b[j]=t1;
}
}
}
bank1.close();
bank1.open("bank.dat",ios::in);
bsort.open("bsort.dat",ios::out|ios::trunc);
cout<<"\n\n all customer details of the bank";
cout<<"\n number\tname\tnumber\tbalance\tbranch";
cout<<"\n ----------------------------------------------";
for(i=0;i<s;i++)
{
bank1.seekg(b[i]*sizeof(s1),ios::beg);
bank1.read((char*)&bk,sizeof(bk));
bsort<<"\n"<<bk.cnum<<"\t"<<bk.cname<<"\t"<<bk.accno<<"\t"<<bk.bal<<"\t"<<bk.branch<<endl;
bk.display();
}
bank1.close();
bsort.close();
}
void customer::input()
{
customer b1,b2;
fstream b1out,b2out,bank;
cout<<"\n enter branch1 customer details";
cout<<"\n enter how many customer you want to add:";
cin>>n;
b1out.open("branch1.dat",ios::out);
bank.open("bank.dat",ios::out);
for(i=0;i<n;i++)
{
b1.getdata();


bank.write((char*)&b1,sizeof(b1));
b1out<<b1.cnum<<"\t"<<b1.cname<<"\t"<<b1.accno<<"\t"<<b1.bal<<"\t"<<b1.branch<<endl;
}
bank.close();
bank.open("bank.dat",ios::app);
cout<<"\n enter branch2 customer details";
cout<<"\n enter how many customer you want to add:";
cin>>m;
s=m+n;
b2out.open("branch2.dat",ios::out|ios::trunc);
for(i=0;i<m;i++)
{
b2.getdata();
bank.write((char*)&b2,sizeof(b2));
b2out<<b2.cnum<<"\t"<<b2.cname<<"\t"<<b2.accno<<"\t"<<b2.bal<<"\t"<<b2.branch<<endl;
}
b1out.close();
b2out.close();
bank.close();
}
void main()
{
clrscr();
customer c;
c.input();
c.sort();
getch();
}
















Output:

 enter branch1 customer details
 enter how many customer you want to add:1

 enter customer number: 101

 enter customer name:honey

 enter account number: 1001

 enter initial balance: 5000

 enter branch name: tirupati

 enter branch2 customer details
 enter how many customer you want to add: 1

 enter customer number: 201

 enter customer name: venki

 enter account number: 501

 enter initial balance: 3000

 enter branch name: chittoor


 all customer details of the bank
 number name    number  balance branch
 ----------------------------------------------
201     venki   501     3000    chittoor

101     honey   1001    5000    irupati





     


/* java package for sorting techniques */
import java.io.*;
public class sortings
{
public static void main(String args[])throws IOException
{
int n,i,j,temp,count;
int arr[]=new int[10];
do
{
BufferedReader br=new BufferedReader(new InputStreamReader(System.in));
System.out.println("\n\n\t\t\t sorting techniques");
System.out.println("\t\t\t_____________________________");
System.out.println("1.Bubble Sort");
System.out.println("2.Insertion Sort");
System.out.println("3.Selection Sort");
System.out.println("4.Shell Sort");
System.out.println("\n Enter your choice");
n=Integer.parseInt(br.readLine());
switch(n)
{
case 1: System.out.println("Enter the data of 5 Elements:");
        for(i=0;i<5;i++)
        {
           arr[i]=Integer.parseInt(br.readLine());

        }
        count=0;j=0;
        for(i=0;i<5;i++)
           {
              for(j=0;j<5;j++)
                {
                      count++;
                      if(arr[j]>arr[j+1])
                      {
                      temp=arr[j];
                      arr[j]=arr[j+1];
                      arr[j+1]=temp;
                      }
               }
         }
       System.out.println("\n Complexity of Bubble sort:"+count);
       System.out.println("\n  Bubble sort order.........");
       for(i=1;i<=5;i++)
         {
                System.out.println(arr[i]);
         }
         break;
case 2:System.out.println("\n Enter the data of 5 Elements:");
       for(i=0;i<5;i++)
        {
               arr[i]=Integer.parseInt(br.readLine());
        }
        count=0;
        for(i=1;i<5;i++)
         {
              for(j=0;j<i;j++)
                 {
                     count++;
                     if(arr[j]>arr[i])
                      {
                           temp=arr[i];
                           arr[i]=arr[j];
                           arr[j]=temp;
                      }
                }
        }
       System.out.println("\n Complexity  of Insertion sort:"+count);
       System.out.println("\n insert sort order......");
       for(i=0;i<5;i++)
        {
              System.out.println(arr[i]);
        }
         break;
case 3:System.out.println("Enter the data of 5 Elements:");
        for(i=0;i<5;i++)
        {
              arr[i]=Integer.parseInt(br.readLine());
        }
        count=0;j=0;
        for(i=0;i<5;i++)
        {
              for(j=i+1;j<5;j++)
                {
                   count++;
                   if(arr[i]>arr[j])
                    {
                          temp=arr[i];
                          arr[i]=arr[j];
                              arr[j]=temp;
                    }
               }
        }
        System.out.println("\n  Complexity of Selection sort:"+count);
        System.out.print("\n select sort order....");  
        for(i=0;i<5;i++)
         {
               System.out.println(arr[i]);
         }
         break;
case 4: System.out.println("\n Enter the data of 5 Elements");
        for(i=0;i<5;i++)
        {
              arr[i]=Integer.parseInt(br.readLine());
        }
        int k,m,mid;
        count=0;
        for(m=5/2;m>0;m/=2)
        {
            for(j=m;j<5;j++)
              {
                   for(i=j-m;i>=0;i-=m)
                    {
                        count++;
                        if(arr[i+m]>=arr[i])
                        break;
                        else
                         {
                            mid=arr[i];
                            arr[i]=arr[i+m];
                            arr[i+m]=mid;
                         }
                     }
              }
        }
        System.out.println("\n Complexity of Shell sort:"+count);
        System.out.println("\n Shell sort order.....");
        for(i=0;i<5;i++)
        {
             System.out.println(arr[i]);
        }
        break;
         }
         }while(n<=4);
       }
      } 
                                        


Output:


C:\Program Files\Java\jdk1.6.0_13\bin>javac sortings.java

C:\Program Files\Java\jdk1.6.0_13\bin>java sortings


                         sorting techniques
                        _____________________________
1.Bubble Sort
2.Insertion Sort
3.Selection Sort
4.Shell Sort

 Enter your choice
1
Enter the data of 5 Elements:
1
9
4
6
2

 Complexity of Bubble sort:25

  Bubble sort order.........
1
2
4
6
9


                         sorting techniques
                        _____________________________
1.Bubble Sort
2.Insertion Sort
3.Selection Sort
4.Shell Sort

 Enter your choice
2



 Enter the data of 5 Elements:
5
8
1
3
4

 Complexity  of Insertion sort:10

 insert sort order......
1
3
4
5
8


                         sorting techniques
                        _____________________________
1.Bubble Sort
2.Insertion Sort
3.Selection Sort
4.Shell Sort

 Enter your choice
3
Enter the data of 5 Elements:
7
2
9
4
1

  Complexity of Selection sort:10

 select sort order....1
2
4
7
9


                      



                       sorting techniques 
                     ____________________________
1.Bubble Sort
2.Insertion Sort
3.Selection Sort
4.Shell Sort

 Enter your choice
4

 Enter the data of 5 Elements
2
9
1
8
6

 Complexity of Shell sort:10

 Shell sort order.....
1
2
6
8
9


                         sorting techniques
                        _____________________________
1.Bubble Sort
2.Insertion Sort
3.Selection Sort
4.Shell Sort

 Enter your choice
5