OPERATIONS ON COMPLEX NUMBERS USING FILES AS STORAGE

OPERATIONS ON COMPLEX NUMBERS USING FILES AS STORAGE

AIM:

To write a C++ program to perform addition of complex numbers using files.

ALGORITHM:

Step 1: Start the program.

Step 2: Create a class called COMPLEX and create its members: real and imaginary.

Step 3: Generate the random numbers by using rand() function.

Step 4: Write the randomly generated numbers in a file called “complex1.txt”.

Step 5: Add the two complex numbers.

Step 6: Store the Resultant complex number in another file called “result.txt”

Step 7: Display the resultant value.

Step 8: Stop the program.

PROGRAM:

#include<iostream.h>

#include<conio.h>

#include<fstream.h>

#include<stdlib.h>

#include<time.h>

class complex

{

public:

int real;

int imag;

complex(int r, int i)

{

real = r;

imag = i;

}

complex()

{

real = imag = 0;

}

void display(void);

};

void complex::display(void)

{

cout<<real<<((imag<0)?"-i":"+i")<<imag<<"\n";

}

void main()

{

clrscr();

ofstream ocom("complex1.txt");

float real,imag;

time_t ti;

srand((unsigned) time(&ti));

real = rand()%100;

imag = rand()%100;

ocom<<"("<<real<<((imag<0)?"-i":"+i")<<imag<<")"<<"+";

real = rand()%100;

imag = rand()%100;

ocom<<"("<<real<<((imag<0)?"-i":"+i")<<imag<<")"<<"\n";

ocom.close();

ifstream icom("complex1.txt");

char no,t,ch,op;

icom>>no;

icom>>real;

icom>>ch;

icom>>no;

icom>>imag;

imag=(ch=='+')?imag:-imag;

icom>>no;

icom>>op;

complex a(real,imag);

icom>>no;

icom>>real;

icom>>ch;

icom>>no;

icom>>imag;

imag=(ch=='+')?imag:-imag;

icom>>no;

icom>>op;

complex b(real,imag);

complex c;

switch(op)

{

case '+':

c.real = a.real+b.real;

c.imag = a.imag+b.imag;

break;

case '-':

c.real = a.real-b.real;

c.imag = a.imag-b.imag;

break;

case '*':

c.real = (a.real*b.real)-(a.imag*b.imag);

c.imag = (a.real*b.imag)+(a.imag*b.real);

break;

case '/':

float qt;

qt = b.real*b.real+b.imag*b.imag;

c.real = (a.real*b.real+a.imag*b.imag)/qt;

c.imag = (a.imag*b.real-a.real*b.imag)/qt;

break;

default:

cout<<"\n Invalid";

}

cout<<"\n complex 1:";

a.display();

cout<<"\n complex 2:";

b.display();

cout<<"\n Resultant complex:";

c.display();

ofstream out("result.txt");

out<<"("<<c.real<<((c.imag<0)?"-i":"+i")<<c.imag<<")";

out.close();

}

OUTPUT:

complex1.txt:

(0+i72)+(39+i71)

result.txt

(39+i143)

RESULT:

Thus the program for addition of complex numbers using files was executed.

DYNAMIC POLYMORPHISM AND RTTI

DYNAMIC POLYMORPHISM AND RTTI

AIM:

To write a C++ program to demonstrate a simple test application for dynamic polymorphism.

ALGORITHM:

Step 1: Start the program.

Step 2: Create a class called Point in order to locate the position of the shapes.

Step 3: Create a class called Shape and use virtual function to draw the corresponding shape.

Step 4: Create classes for shapes like Square, Rectangle, Triangle, Circle, Ellipses.

Step 5: Call all the shapes using objects, in main function.

Step 6: Shapes will be drawn with the given values.

Step 7: Shapes can also be drawn by using their address and display their location.

Step 8: Stop the program.

PROGRAM:

#include<iostream.h>

#include<conio.h>

class Point

{

public:

            int x; int y;

            Point(){}

            Point(int tempX, int tempY)

            {

            x = tempX;

            y = tempY;

            }

            int GetX()

            {

            return x;

            }

            int GetY()

            {

            return y;

            }

            friend ostream & operator <<(ostream &tempout, Point &tempPoint)

            {

            tempout<<"("<<tempPoint.GetX()<<tempPoint.GetY()<<")";

            return tempout;

            }

};

class Shape

{

 Point position;

 public:

            Shape(){}

            virtual void draw()

            {

             cout<<"shape is drawm";

             }

};

class Square:public Shape

{

 Point leftbottom;

 int length;

 public:

  Square(){}

  Square(Point tleftbottom, int tlength)

  {

   leftbottom = tleftbottom;

   length = tlength;

  }

  void draw()

  {

   cout<<"Square is drawn at"<<leftbottom<<"and with length as :"<<length<<"\n";

  }

 };

class Rectangles : public Shape

{

 Point leftbottom, lefttop, rightbottom, righttop;

 public:

  Rectangles(){}

  Rectangles(Point tleftbottom, Point tlefttop, Point trightbottom, Point trighttop)

  {

   leftbottom = tleftbottom;

   lefttop = tlefttop;

   rightbottom = trightbottom;

   righttop = trighttop;

  }

  void draw()

  {

   cout<<"Rectangle is drawn at ("<<leftbottom<<", "<<rightbottom<<")"<<"and"<<"("<<lefttop<<","<<righttop<<")\n";

  }

};

class Triangle : public Shape

{

 Point avertex, bvertex, cvertex;

 public:

 Triangle(){}

 Triangle(Point tavertex, Point tbvertex, Point tcvertex)

 {

  avertex = tavertex;

  bvertex = tbvertex;

  cvertex = tcvertex;

  }

  void draw()

  {

   cout<<"Triangle is drawn at"<<avertex<<" "<<bvertex<<" "<<cvertex<<"\n";

   }

};

class Circle: public Shape

{

 Point center; int radius;

 public:

 Circle(){}

 Circle(Point tcenter, int tradius)

 {

  center = tcenter;

  radius = tradius;

 }

 void draw()

 {

  cout<<"Circle is drawn at"<<" " <<center<<" " <<"and the radius is: "<<radius<<"\n";

 }

};

class Ellipses: public Shape

{

 Point center;

 int radius; int angle;

 public:

 Ellipses(){}

 Ellipses(Point tcenter, int tradius, int tangle)

 {

  center = tcenter;

  radius = tradius;

  angle = tangle;

 }

 void draw()

 {

  cout<<"Ellipse is drawn at"<<center<<"and the radius is"<<radius<<"with an angle"<<angle<<"\n";

 }

};

void main()

{

clrscr();

cout<<"\n";

Point p1(10,20);

Point p2(3,2);

Square sq(p1,5);

sq.draw();

Rectangles rect(p1,p2,p1,p2);

rect.draw();

Circle c(p1,50);

c.draw();

Ellipses e(p2,34,23);

e.draw();

Triangle t(p1,p2,p1);

t.draw();

Shape *s;

s=&sq;

s->draw();

s=&rect;

s->draw();

s=&t;

s->draw();

getch();

}

OUTPUT:

Square is drawn at(1020)and with length as :5

Rectangle is drawn at ((1020), (1020))and((32),(32))

Circle is drawn at (1020) and the radius is: 50

Ellipse is drawn at(32)and the radius is34with an angle23

Triangle is drawn at(1020) (32) (1020)

Square is drawn at(1020)and with length as :5

Rectangle is drawn at ((1020), (1020))and((32),(32))

Triangle is drawn at(1020) (32) (1020)

RESULT:

Thus the C++ program to demonstrate a simple test application for dynamic polymorphism was executed.

MINIMUM SPANNING TREE

MINIMUM SPANNING TREE

AIM:

To find the minimum cost spanning tree in a graph using Prims algorithm.

ALGORITHM:

Step 1 : Start the process.

Step 2: Create two classes as POINT and ARC.

Step 3: Define a Graph class which represents the collection of Point and Arc objects.

Step 4: Write a method to find a minimum cost spanning tree in a graph.

Step 5: Enter the Number of nodes, source node, destination node, weight for the corresponding edge.

Step6: Repeat step5 until all vertices, edges and weight are defined.

Step 7: The edge does not form a cycle in the graph, for Minimum Spanning Tree.

Step 8: Thus set of edges is connected and form a Minimum Spanning Tree.

Step 9: Display the adjacency matrix for the graph and the minimum total path of all the edges. 

Step 10: Stop the process.

PROGRAM:

#include<iostream.h>

#include<conio.h>

#define MAX 50

#define TRUE 1

#define FALSE 0

#define MAXINT 250

class node

{

public:

int no;

node()

{}

node (int a)

{

no = a;

}

};

class arc

{

public:

int adj;

int weight;

arc()

{}

arc(int a)

{

adj = a;

}

};

class graph

{

public:

node nodes[MAX];

arc arcs[MAX][MAX];

graph(int n)

{

for(int i=1;i<=n;i++)

{

nodes[i].no = 0;

for(int j=1;j<=n;j++)

arcs[i][j].adj = FALSE;

}

}

void join(node n1, node n2, int w)

{

arcs[n1.no][n2.no].adj = w;

arcs[n2.no][n1.no].adj = w;

}

void displayadj(int n)

{

cout<<"\n The adjacency matrix....";

cout<<endl;

for(int i = 1; i<=n; i++)

{

for(int j = 1; j<=n;j++)

cout<<"\t"<<arcs[i][j].adj;

cout<<endl;

}

cout<<endl;

}

void shortpath(int n)

{

int lcost[20];

int clost[20],i,j,k,min;

for(i = 2; i<=n;i++)

{

lcost[i]=arcs[1][i].adj;

clost[i]=1;

}

cout<<"\n Minimum cost spanning tree edges are:\n";

for(i=2;i<=n;++i)

{

min=lcost[2];

k=2;

for(j=3;j<=n;++j)

if(lcost[j]<min)

{

min = lcost[j];

k=j;

}

cout<<"\n"<<k<<"<->"<<clost[k];

lcost[k]=MAXINT;

for(j=2;j<=n;++j)

if((arcs[k][j].adj<lcost[j])&&(lcost[j]<MAXINT))

{

lcost[j]=arcs[k][j].adj;

clost[j]=k;

}

}

}

};

int main()

{

int n;

clrscr();

cout<<"\n Enter total number of nodes...";

cin>>n;

graph g(n);

cout<<"\n Assigning number for each node...";

for(int i = 1;i<=n;i++)

g.nodes[i].no=i;

char ch ='y';

int w;

do

{

node a,b;

cout<<"\n Create path b/w the nodes";

cout<<"\n Enter the source node...";

cin>>a.no;

cout<<"\n Enter the destination node...";

cin >>b.no;

cout<<"\n Enter the weight:";

cin>>w;

g.join(a,b,w);

cout<<"\n Want to continue...[y]es or [n]:";

cin>>ch;

}while(ch=='y');

g.displayadj(n);

g.shortpath(n);

cin>>n;

getch();

return 0;

}

OUTPUT:

 Enter total number of nodes...4

 Assigning number for each node...

 Create path b/w the nodes

 Enter the source node...1

 Enter the destination node...2

 Enter the weight:1

 Want to continue...[y]es or [n]:y

 Create path b/w the nodes

 Enter the source node...1

 Enter the destination node...4

 Enter the weight:2

 Want to continue...[y]es or [n]:y

 Create path b/w the nodes

 Enter the source node...1

 Enter the destination node...3

 Enter the weight:2

 Want to continue...[y]es or [n]:y

 Create path b/w the nodes

 Enter the source node...2

 Enter the destination node...4

 Enter the weight:4

 Want to continue...[y]es or [n]:y

 Create path b/w the nodes

 Enter the source node...2

 Enter the destination node...3

 Enter the weight:5

 Want to continue...[y]es or [n]:y

 Create path b/w the nodes

Enter the source node...3

 Enter the destination node...4

 Enter the weight:3

 Want to continue...[y]es or [n]:n

 The adjacency matrix....

        0       1       2       2

        1       0       5       4

        2       5       0       3

        2       4       3       0

 Minimum cost spanning tree edges are:

2<->1

3<->1

4<->1

RESULT:

Thus the program to find the minimum cost spanning tree in a graph using Prims algorithm was executed.