c program of gauss ELIMINITION method

/*Gauss Elimination */
# include <stdio.h>
# include <conio.h>
# include <math.h>
# define MAX 10
void main()
{
int i,j,n,k;
float mat[MAX][MAX],x[MAX],temp,pivot,sum=0;
clrscr();
printf("\t\t\t GAUSS ELIMINITION METHOD\n");
printf("-------------------------------------------------------------------\n");
printf("Enter No of Equtions : ");
scanf("%d",&n);
printf("Enter Coefficients of Eqution \n");
for(i=1;i<=n;i++)
for(j=1;j<=n;j++)
scanf("%f",&mat[i][j]);
printf("Enter Constant value\n");
for(i=1;i<=n;i++)
{
scanf("%f",&mat[i][n+1]);
x[i]=mat[i][n+1];
}
for(i=2;i<=n;i++)
{
for(j=i;j<=n;j++)
{
pivot=mat[j][i-1]/mat[i-1][i-1];
for(k=i-1;k<=n+1;k++)
mat[j][k]=mat[j][k]-pivot*mat[i-1][k];
}
}
printf("Eliminated matrix as :- \n");
for(i=1;i<=n;i++)
{
for(j=1;j<=n+1;j++)
printf("\t%.2f",mat[i][j]);
printf("\n");
}
for(i=1;i<=n;i++)
{
if(mat[i][i]==0)
{
printf("Since diagonal element become zero\n Hence solution is not possible\n");
exit(1);
}
}
printf("Solution : \n");
for(i=0;i {
sum=0;
for(j=n;j>n-i;j--)
sum=sum+mat[n-i][j];

x[n-i]=(mat[n-i][n+1]-sum*x[n])/mat[n-i][n-i];
printf("X%d = %4.2f\n",n-i,x[n-i]);
}

getch();
}

OUTPUT :-


GAUSS ELIMINITION METHOD
------------------------------------------------------------------------------
Enter No of Equtions : 3
Enter Coefficients of Eqution
4 3 -2
1 1 1
3 -2 1
Enter Constant value
5 3 2
Eliminated matrix as :-
4.00 3.00 -2.00 5.00
0.00 0.25 1.50 1.75
0.00 0.00 28.00 28.00
Solution :
X3 = 1.00
X2 = 1.00
X1 = 1.00

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C Program Of NA Using Trapezoidal Interpolation

/*Trapezoidal rule */
# include <stdio.h>
# include <conio.h>
# define MAX 20
void main()
{
int i,j,n;
float x[MAX],y[MAX],sum=0,h,res;
clrscr();
printf("Solution using Trapezoidal rule \n");
printf("Enter Number of data points : ");
scanf("%d",&n);
printf("Enter value of table value set by set\n");
for(i=1;i<=n;i++)
{
printf("Enter value of x[%d] and y[%d] ",i,i);
scanf("%f%f",&x[i],&y[i]);
}
h=x[2]-x[1];
printf("\nInterval :=%f\n",h);
sum=y[1]+y[n];
for(i=2;i<n;i++)
sum=sum+y[i]*2;

res=sum*(h/2);
printf ("Integral from %3.2f to %3.2f is : = %f",x[1],x[n],res);
getch();
}
OUTPUT :-
Solution using Trapezoidal rule
Enter Number of data points : 6
Enter value of table value set by set
Enter value of x[1] and y[1] 7.47 1.93
Enter value of x[2] and y[2] 7.48 1.95
Enter value of x[3] and y[3] 7.49 1.98
Enter value of x[4] and y[4] 7.50 2.01
Enter value of x[5] and y[5] 7.51 2.03
Enter value of x[6] and y[6] 7.52 2.06

Interval :=0.010000
Integral from 7.47 to 7.52 is : = 0.099652

C Program Of NA Using Simpson 1/3 Rule

/*Simpson 1/3 Rule */
# include <stdio.h>
# include <conio.h>
# define MAX 20
void main()
{
int i,j,n;
float x[MAX],y[MAX],sum=0,h,res;
clrscr();
printf("Solution of 1/(1+x)dx using Simpson's 1/3 rule \n");
printf("Enter Number of data points : ");
scanf("%d",&n);
printf("Enter value of table value set by set\n");
for(i=1;i<=n;i++)
{
scanf("%f",&x[i]);
y[i]=1/(x[i]+1);
}
printf("X \t\t Y\n");
for(i=1;i<=n;i++)
printf("%f\t%f\n",x[i],y[i]);
h=x[2]-x[1];
printf("\nSegment Interval `h':=0.25\n");
sum=sum+y[1]+y[n];
for(i=2;i<n;i++)
{
if(i%2==0)
sum=sum+y[i]*4;
else
sum=sum+y[i]*2;
}
res=sum*(0.25/3);
printf ("Integral from %3.2f to %3.2f is : = %f",x[1],x[n],res);
getch();
}
OUTPUT
Solution of 1/(1+x)dx using Simpson's 1/3 rule
Enter Number of data points : 5
Enter value of table value for x column
0 .25 .50 .75 1
X Y
0.000000 1.000000
0.250000 0.800000
0.500000 0.666667
0.750000 0.571429
1.000000 0.500000

Segment Interval `h':=0.25
Integral from 0.00 to 1.00 is : = 0.693254

C Program Of NA Using Simpson 3/8 Rule

/*Simpson 3/8 rule */
# include <stdio.h>
# include <conio.h>
# define MAX 20
void main()
{
int j,n=0;
float x[MAX],y[MAX],sum=0,h,res,lb,ub,i;
clrscr();
printf("Solution of 1/(1+x)dx using Simpson's 3/8 rule \n");
printf ("Enter initial value : ");
scanf("%f",&lb);
printf("Enter final value : ");
scanf("%f",&ub);
printf("Enter no. of Subinterval : ");
scanf("%f",&h);
h=(ub-lb)/h;
for(i=lb;i<=ub;i=i+h)
{
x[n]=i;
y[n]=1/(1+x[n]);
n++;
}
printf("No.\t X\t\tY=1/(1+x)\n");
printf("---------------------------------------\n");
sum=y[0]+y[n-1];
for(j=0;j<n;j++)
printf("%d\t%f\t%f\n",j,x[j],y[j]);
printf("---------------------------------------\n");
for(j=1;j<n-1;j++)
{
if(j%3==0)
sum=sum+y[j]*2;
else
sum=sum+y[j]*3;
}
res=((3*h)/8)*sum;
printf("\nIntegral from %f to %f when h = %f is = %f",x[0],x[n-1],h,res);
getch();
}

OUTPUT

Solution of 1/(1+x)dx using Simpson's 3/8 rule
Enter initial value : 6
Enter final value : 15
Enter no. of Subinterval : 8
No. X Y=1/(1+x)
---------------------------------------
0 6.000000 0.142857
1 7.125000 0.123077
2 8.250000 0.108108
3 9.375000 0.096386
4 10.500000 0.086957
5 11.625000 0.079208
6 12.750000 0.072727
7 13.875000 0.067227
8 15.000000 0.062500
---------------------------------------

Integral from 6.000000 to 15.000000 when h = 1.125000 is = 0.817303

C Program Of NA Using Runga-Kutta Mthod Of 4th Order

/* Runge-kutta method of 4th order to solve 10*dy/dx=(x*x)+(y*y)*/

# include <stdio.h>

# include <conio.h>

void main()

{

int c=0;

float x,y0,xp,h,i,y1,m1,m2,m3,m4;

float f(float,float);

clrscr();

printf("Solution by Runge kutta 2nd order Method\n");

printf("Enter initial Boundry condition x,y : ");

scanf("%f%f",&x,&y0);

printf("Enter Value of X at which Y is required : ");

scanf("%f",&xp);

printf("Enter Interval ,h : ");

scanf("%f",&h);

printf(" No. X\t m1 \t m2\t m3 \t m4 \t Y\n");

printf("-------------------------------------------------------------------\n");

for(i=x;i<=xp;i=i+h)

{

c++;

m1=h*f(i,y0);

m2=h*f(i+(h/2),y0+(m1/2));

m3=h*f(i+(h/2),y0+(m2/2));

m4=h*f(i+h,y0+m3);

y1=y0+((m1+2*m2+2*m3+m4)/6);

printf("%2d %5.6f %5.6f %5.6f %5.6f %5.6f %5.6f\n",c,i,m1,m2,m3,m4,y1);

y0=y1;

}

getch();

}

float f(float x,float y)

{

return ((x*x)+(y*y))/10;

}

OUTPUT

Solution by Runge kutta 2nd order Method

Enter initial Boundry condition x,y : 0 1

Enter Value of X at which Y is required : .09

Enter Interval ,h : .03

No. X m1 m2 m3 m4 Y

-------------------------------------------------------------------------------

1 0.000000 0.003000 0.003010 0.003010 0.003021 1.003010

2 0.030000 0.003021 0.003033 0.003033 0.003047 1.006043

3 0.060000 0.003047 0.003062 0.003062 0.003079 1.009106

4 0.090000 0.003079 0.003097 0.003097 0.003117 1.012204

Program in C to insert a node at the last position in a given doubly linked list.

#include<stdio.h>

#include<conio.h>

#include<alloc.h>

#define null 0

void main()

{

struct dnode

{

struct dnode *prev;

int data;

struct dnode *next;

} *temp,*start,*end,*n;



int i;

char c;

clrscr();

start=end=null;

for(i=0;i<=1;i++)

{

if(start==null)

{

printf("enter data for the node:");

scanf("%d",&temp->data);

temp->next=null;

temp->prev=null;

start=temp;

end=temp;

}

else

{

while(1)

{

printf("are you want to add another node(y/Y):");

fflush(stdin);

scanf("%c",&c);

if(c=='y'||c=='Y')

{

n=(struct dnode*)malloc(sizeof(struct dnode));

printf("enter data for the node:");

scanf("%d",&n->data);

n->next=null;

n->prev=temp;

temp->next=n;

temp=n;

end=n;

}

else

break;

}

}

}



printf("the list you created is:\n");

temp=start;

while(temp->next!=null)

{

printf("%d\t",temp->data);

temp=temp->next;

}

printf("%d\n",temp->data);

printf("addition at last place\n");

{

n=(struct dnode *)malloc(sizeof(struct dnode));

printf("enter the data for the node:");

scanf("%d",&n->data);

n->prev=end;

n->next=null;

end->next=n;

end=n;

}

printf("after operation list is:\n");

temp=start;

while(temp->next!=null)

{

printf("%d\t",temp->data);

temp=temp->next;

}

printf("%d\n",temp->data);

getch();

}



OUTPUT:

Enter data for the node:45

Are you want to create another node(y/Y):Y



Enter data for the node:65

Are you want to create another node(y/Y):n

The list you created

45 65

Addition at the first place

Enter data for the node:22

After operation list is: 45 65 22

Program in C to insert a node at first position in a given doubly linked list.

#include<stdio.h>

#include<conio.h>

#include<alloc.h>

#define null 0

void main()

{

struct dnode

{

struct dnode *prev;

int data;

struct dnode *next;

} *temp,*start,*end,*n;



int i;

char c;

clrscr();

start=end=null;

for(i=0;i<=1;i++)

{

if(start==null)

{

printf("enter data for the node:");

scanf("%d",&temp->data);

temp->next=null;

temp->prev=null;

start=temp;

end=temp;

}

else

{

while(1)

{

printf("are you want to add another node(y/Y):");

fflush(stdin);

scanf("%c",&c);

if(c=='y'||c=='Y')

{

n=(struct dnode*)malloc(sizeof(struct dnode));

printf("enter data for the node:");

scanf("%d",&n->data);

n->next=null;

n->prev=temp;

temp->next=n;

temp=n;

end=n;

}

else

break;

}

}

}



printf("the list you created is:\n");

temp=start;

while(temp->next!=null)

{

printf("%d\t",temp->data);

temp=temp->next;

}

printf("%d\n",temp->data);

printf("addition at first place\n");

{

n=(struct dnode *)malloc(sizeof(struct dnode));

printf("enter the data for the node:");

scanf("%d",&n->data);

n->prev=null;

n->next=start;

start->prev=n;

start=n;

}

printf("after operation list is:\n");

temp=start;

while(temp->next!=null)

{

printf("%d\t",temp->data);

temp=temp->next;

}

printf("%d\n",temp->data);

getch();

}



OUTPUT:

Enter data for the node:45

Are you want to create another node(y/Y):Y



Enter data for the node:65

Are you want to create another node(y/Y):n

The list you created

45 65

Addition at the first place

Enter data for the node:22

After operation list is: 22 45 65

Program in C to subtract two polynomials.

#include<stdio.h>

#include<conio.h>

#include<stdlib.h>

#define NULL 0

void main()

{

struct poly

{

int coff;

int expo;

struct poly *link;

}*temp1,*start1,*temp2,*start2,*start3,*temp3;

int n,i,z=1,num;

char c;

clrscr();

start1=NULL;

printf("\tPOLYNOMIAL CREATION IN LINK LIST AND SUBTRACTION\n\n");

while(1)

{

if(start1==NULL)

{

temp1=(struct poly*)malloc(sizeof(struct poly));

printf("Enter Coefficient and Exponent for Node %d\n",z);

scanf("%d %d",&temp1->coff,&temp1->expo);

start1=temp1;

temp1->link=NULL;

}

else

{

temp1->link=(struct poly*)malloc(sizeof(struct poly));

temp1=temp1->link;

printf("Enter Coefficient and exponent for Node %d\n",z);

scanf("%d %d",&temp1->coff,&temp1->expo);

}

z++;

printf("Do you want to create another node\n");

fflush(stdin);

scanf("%c",&c);

if(c!='y')

{

temp1->link=NULL;

break;

}

}

start2=NULL;

temp2=NULL;

z=1;

while(1)

{

if(start2==NULL)

{

temp2=(struct poly*)malloc(sizeof(struct poly));

printf("Enter Coefficient and exponent for Node %d\n",z);

scanf("%d %d",&temp2->coff,&temp2->expo);

start2=temp2;

temp2->link=NULL;

}

else

{

temp2->link=(struct poly*)malloc(sizeof(struct poly));

temp2=temp2->link;

printf("Enter Coefficient and exponent for Node %d\n",z);

scanf("%d %d",&temp2->coff,&temp2->expo);

}

z++;

printf("Do you want to create another node\n");

fflush(stdin);

scanf("%c",&c);

if(c!='y')

{

temp2->link=NULL;

break;

}

}

// TARVERSING

temp1=NULL;

temp2=NULL;

temp1=start1;

temp2=start2;

printf("Traversal of Polynomial Linked List 1\n");

while(temp1!=NULL)

{

printf("%2dx^%d-",temp1->coff,temp1->expo);

temp1=temp1->link;

}

printf("\b \b");

printf("\nTraversal of Polynomial Linked List 2\n");

while(temp2!=NULL)

{

printf("%2dx^%d -",temp2->coff,temp2->expo);

temp2=temp2->link;

}

printf("\b \b");

// subtraction

temp1=NULL;

temp2=NULL;

temp1=start1;

temp2=start2;

temp3=NULL;

start3=NULL;

while(1)

{

if((temp1!=NULL)||(temp2!=NULL))

{

if(start3==NULL)

{

if((temp1->expo)==(temp2->expo))

{

temp3=(struct poly*)malloc(sizeof(struct poly));

temp3->coff= (temp1->coff) - (temp2->coff);

temp3->expo=temp1->expo;

start3=temp3;

temp1=temp1->link;

temp2=temp2->link;

temp3->link=NULL;

}

else if(temp1->expo>temp2->expo)

{

temp3=(struct poly*)malloc(sizeof(struct poly));

temp3->coff=temp1->coff;

temp3->expo=temp1->expo;

start3=temp3;

temp1=temp1->link;

temp3->link=NULL;

}

else

{

temp3=(struct poly*)malloc(sizeof(struct poly));

temp3->coff=temp2->coff;

temp3->expo=temp2->expo;

start3=temp3;

temp2=temp2->link;

temp3->link=NULL;

}

}

else

{

if(temp1->expo==temp2->expo)

{

temp3->link=(struct poly*)malloc(sizeof(struct poly));

temp3=temp3->link;

temp3->coff= (temp1->coff) - (temp2->coff);

temp3->expo=temp1->expo;

temp1=temp1->link;

temp2=temp2->link;

}

else if(temp1->expo>temp2->expo)

{

temp3->link=(struct poly*)malloc(sizeof(struct poly));

temp3=temp3->link;

temp3->coff=temp1->coff;

temp3->expo=temp1->expo;

temp1=temp1->link;

}

else

{

temp3->link=(struct poly*)malloc(sizeof(struct poly));

temp3=temp3->link;

temp3->coff=temp2->coff;

temp3->expo=temp2->expo;

temp2=temp2->link;

}

}

}

else

break;

}

temp3->link=NULL;

//traversing temp3

temp3=NULL;

temp3=start3;

printf("\nTraversal of Polynomial Linked List after subtracting Temp1 & Temp2\n");

while(temp3!=NULL)

{

printf("%2dx^%d - ",temp3->coff,temp3->expo);

temp3=temp3->link;

}

printf("\b\b");

printf("\nTHE END\n");

getch();

}



OUTPUT:

POLYNOMIAL CREATION IN LINK LIST AND SUBTRACTION



Enter Coefficient and exponent for Node 1

6

3

Do you want to create another node

Y

Enter Coefficient and exponent for Node 2

8

2

Do u want to create another node

N

Enter Coefficient and exponent for Node 1

7

3

Do you want to create another node

Y

Enter Coefficient and exponent for Node 2

5

2

Do u want to create another node

N

Traversal of Polynomial Linked List 1

6x^3 + 8x^2



Traversal of Polynomial Linked List 2

7x^3 + 5x^2



Traversal of Polynomial Linked List after adding Temp1 & Temp2

-1X^3 + 3X^2 THE END

Program in C to add two polynomials.

#include<stdio.h>

#include<conio.h>

#include<stdlib.h>

#define NULL 0

void main()

{

struct poly

{

int coff;

int expo;

struct poly *link;

}*temp1,*start1,*temp2,*start2,*start3,*temp3;

int n,i,z=1,num;

char c;

clrscr();

start1=NULL;

printf("\tPOLYNOMIAL CREATION IN LINK LIST AND ADDITION\n\n");

while(1)

{

if(start1==NULL)

{

temp1=(struct poly*)malloc(sizeof(struct poly));

printf("Enter Coefficient and Exponent for Node %d\n",z);

scanf("%d %d",&temp1->coff,&temp1->expo);

start1=temp1;

temp1->link=NULL;

}

else

{

temp1->link=(struct poly*)malloc(sizeof(struct poly));

temp1=temp1->link;

printf("Enter Coefficient and exponent for Node %d\n",z);

scanf("%d %d",&temp1->coff,&temp1->expo);

}

z++;

printf("Do you want to create another node\n");

fflush(stdin);

scanf("%c",&c);

if(c!='y')

{

temp1->link=NULL;

break;

}

}

start2=NULL;

temp2=NULL;

z=1;

while(1)

{

if(start2==NULL)

{

temp2=(struct poly*)malloc(sizeof(struct poly));

printf("Enter Coefficient and exponent for Node %d\n",z);

scanf("%d %d",&temp2->coff,&temp2->expo);

start2=temp2;

temp2->link=NULL;

}

else

{

temp2->link=(struct poly*)malloc(sizeof(struct poly));

temp2=temp2->link;

printf("Enter Coefficient and exponent for Node %d\n",z);

scanf("%d %d",&temp2->coff,&temp2->expo);

}

z++;

printf("Do you want to create another node\n");

fflush(stdin);

scanf("%c",&c);

if(c!='y')

{

temp2->link=NULL;

break;

}

}

// TARVERSING

temp1=NULL;

temp2=NULL;

temp1=start1;

temp2=start2;

printf("Traversal of Polynomial Linked List 1\n");

while(temp1!=NULL)

{

printf("%2dx^%d+",temp1->coff,temp1->expo);

temp1=temp1->link;

}

printf("\b\b");

printf("\nTraversal of Polynomial Linked List 2\n");

while(temp2!=NULL)

{

printf("%2dx^%d+",temp2->coff,temp2->expo);

temp2=temp2->link;

}

printf("\b\b");

// addition

temp1=NULL;

temp2=NULL;

temp1=start1;

temp2=start2;

temp3=NULL;

start3=NULL;

while(1)

{

if((temp1!=NULL)||(temp2!=NULL))

{

if(start3==NULL)

{

if((temp1->expo)==(temp2->expo))

{

temp3=(struct poly*)malloc(sizeof(struct poly));

temp3->coff= (temp1->coff) + (temp2->coff);

temp3->expo=temp1->expo;

start3=temp3;

temp1=temp1->link;

temp2=temp2->link;

temp3->link=NULL;

}

else if(temp1->expo>temp2->expo)

{

temp3=(struct poly*)malloc(sizeof(struct poly));

temp3->coff=temp1->coff;

temp3->expo=temp1->expo;

start3=temp3;

temp1=temp1->link;

temp3->link=NULL;

}

else

{

temp3=(struct poly*)malloc(sizeof(struct poly));

temp3->coff=temp2->coff;

temp3->expo=temp2->expo;

start3=temp3;

temp2=temp2->link;

temp3->link=NULL;

}

}

else

{

if(temp1->expo==temp2->expo)

{

temp3->link=(struct poly*)malloc(sizeof(struct poly));

temp3=temp3->link;

temp3->coff= (temp1->coff) + (temp2->coff);

temp3->expo=temp1->expo;

temp1=temp1->link;

temp2=temp2->link;

}

else if(temp1->expo>temp2->expo)

{

temp3->link=(struct poly*)malloc(sizeof(struct poly));

temp3=temp3->link;

temp3->coff=temp1->coff;

temp3->expo=temp1->expo;

temp1=temp1->link;

}

else

{

temp3->link=(struct poly*)malloc(sizeof(struct poly));

temp3=temp3->link;

temp3->coff=temp2->coff;

temp3->expo=temp2->expo;

temp2=temp2->link;

}

}

}

else

break;

}

temp3->link=NULL;

//traversing temp3

temp3=NULL;

temp3=start3;

printf("\nTraversal of Polynomial Linked List after adding Temp1 & Temp2\n");

while(temp3!=NULL)

{

printf("%2dx^%d + ",temp3->coff,temp3->expo);

temp3=temp3->link;

} printf("\b \b");

getch();

}

OUTPUT:

POLYNOMIAL CREATION IN LINK LIST AND ADDITION



Enter Coefficient and exponent for Node 1

6

3

Do you want to create another node

Y

Enter Coefficient and exponent for Node 2

8

2

Do u want to create another node

N

Enter Coefficient and exponent for Node 1

7

3

Do you want to create another node

Y

Enter Coefficient and exponent for Node 2

5

2

Do u want to create another node

N

Traversal of Polynomial Linked List 1

6x^3 + 8x^2



Traversal of Polynomial Linked List 2

7x^3 + 5x^2



Traversal of Polynomial Linked List after adding Temp1 & Temp2

13X^3 +13X^2

program in C to store any polynomial using linked list.

#include<stdio.h>

#include<conio.h>

#include<stdlib.h>

#define NULL 0

void main()

 {

  struct poly

   {

    int coff;

    int expo;

    struct poly *link;

   }*temp,*start,*n1,*prev;

  int n,i,z=1,num;

  char c;

  clrscr();

  start=NULL;

  n1=NULL;

  printf("\tPOLYNOMIAL CREATION IN LINK LIST AND TRAVERSAL\n\n");

  while(1)

   {

    if(start==NULL)

     {

                temp=(struct poly*)malloc(sizeof(struct poly));

                printf("Enter Coefficient and exponent for Node %d\n",z);

                scanf("%d %d",&temp->coff,&temp->expo);

                start=temp;

                temp->link=NULL;

       }

     else

     {

      temp->link=(struct poly*)malloc(sizeof(struct poly));

      temp=temp->link;

      printf("Enter Coefficient and exponent for Node %d\n",z);

      scanf("%d %d",&temp->coff,&temp->expo);

     }

       z++;

      printf("Do you want to create another node\n");

      fflush(stdin);

      scanf("%c",&c);

      if(c!='y')

       {

                temp->link=NULL;

                break;

       }

     }

   // TARVERSING

  n1=start;

  printf("Traversal\n");

  while(n1!=NULL)

   {

    printf(" %2dx^%d  + ",n1->coff ,n1->expo);

    n1=n1->link;

   }

   printf("\b\b\nTHE END\n");

 getch();

}

Output:

                 POLYNOMIAL CREATION IN LINK LIST AND TRAVERSAL

Enter Coefficient and exponent for Node 1

6

3

Do you want to create another node

Y

Enter Coefficient and exponent for Node 2

8

2

Do u want to create another node

N

Traversal

6x^3  + 8x^2

THE END

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C Program for Circular Queue implementation through Array

//Program for Circular Queue implementation through Array
#include <stdio.h>
#include<conio.h>
#define MAXSIZE 5
int cq[MAXSIZE]={0};
int front,rear;
void main()
{
void add(int);
void del();
void display();
int ch,i,num;
front = -1;
rear = -1;
clrscr();
printf("\nProgram for Circular Queue demonstration through array");
while(1)
{
printf("\n\nMAIN MENU\n1.INSERTION\n2.DELETION\n3.DISPLAY\n4.EXIT");
printf("\n\nENTER YOUR CHOICE : ");
scanf("%d",&ch);
switch(ch)
{
case 1:
printf("\n\nENTER THE QUEUE ELEMENT : ");
scanf("%d",&num);
add(num);
break;
case 2:
del();
break;
case 3:
   display();
    break;
case 4:
    exit(0);
default: printf("\n\nInvalid Choice . ");
}

}
}


void add(int item)
{
if(front ==(rear+1)%MAXSIZE)
{
printf("\n\nCIRCULAR QUEUE IS OVERFLOW");
}
else
{
 if(front==-1)
   front=rear=0;
   else
   rear=(rear+1)%MAXSIZE;
   cq[rear]=item;
   printf("\n\nRear = %d    Front = %d ",rear,front);
}
}


void del()
{
int a;
if(front == -1)
{
printf("\n\nCIRCULAR QUEUE IS UNDERFLOW");
}
else
{
a=cq[front];
 cq[front]=0;
if(front==rear)
front=rear=-1;
else
 front = (front+1)%MAXSIZE;
printf("\n\nDELETED ELEMENT FROM QUEUE IS : %d ",a);
printf("\n\nRear = %d    Front = %d ",rear,front);

}
}

void display()
{
 int i;
 for(i=0;i<MAXSIZE;i++)
   printf("%d\t",cq[i]);
}

C Program to implement NEWTON'S BACKWARD METHOD OF INTEROLATION

/*
  Program to implement NEWTON'S BACKWARD METHOD OF INTEROLATION.
      ----------------------------------------

*/

#include<stdio.h>
#include<conio.h>
#include<math.h>
#include<process.h>
#include<conio.h>

void main()
{
int n;                          // no. of terms.
int i,j,k;                      // Loop variables
float mx[10];                   // 'X' array limit 9
float my[10];                   // 'Y' array limit 9
float x;                        // User Query for what value of X
float x0=0;                     //
float y0;                       // Value coressponding to x0.
float sum=0;                      // Calculated value for coressponding X.
float h;                        // Calc. Section
float fun=0;                      //
float p;                        // Calc. Section
float diff[20][20];             // to store Y
float y1,y2,y3,y4;              // Formulae variables.

clrscr();
// Input section.

printf("\t\t !! NEWTON'S GREGORY BACKWARD INTERPOLATION FORMULA !! ");
printf("\n\n Enter the no. of terms -> ");
scanf("%d",&n);

// Input Sequel for array X
printf("\n\n Enter the value in the form of x -> ");
// Input loop for X.
for(i=0;i<n;i++)
   {
   printf("\n Enter the value of x%d -> ",i+1);
   scanf("%f",&mx[i]);
   }

// Input sequel for array Y.
printf("\n\n Enter the value in the form of y -> ");
// Input loop for Y.
for(i=0;i<n;i++)
   {
   printf("\n Enter the value of y%d -> ",i+1);
   scanf("%f",&my[i]);
   }

// Inputting the required value quarry
printf("\n\n Enter the value of x for ");
printf("\n which u want the value of y -> ");
scanf("%f",&x);

// Calculation and processing section.
h=mx[1]-mx[0];
for(i=0;i<n-1;i++)
   diff[i][1]=my[i+1]-my[i];

for(j=2;j<=4;j++)
   for(i=0;i<n-j;i++)
      diff[i][j]=diff[i+1][j-1]-diff[i][j-1];
i=0;

while(!mx[i]>x)
     i++;

x0=mx[i];
sum=0;

y0=my[i];
fun=1;

p=(x-x0)/h;
sum=y0;

for(k=1;k<=4;k++)
   {
   fun=(fun*(p-(k-1)))/k;
   sum=sum+fun*diff[i][k];
   }

// Outut Section
printf("\n When x = %6.4f , y = %6.8f",x,sum);

// Invoke user watch halt function
printf("\n\n\n\t\t\t !! PRESS ENTER TO EXIT !! ");
getch();
}

C Program to implement NEWTON'S FORWARD METHOD OF INTEROLATION

/*
  Program to implement NEWTON'S FORWARD METHOD OF INTEROLATION.
      --------------------------------------
*/

#include<stdio.h>
#include<conio.h>
#include<math.h>
#include<process.h>
#include<conio.h>

void main()
{
int n;                          // no. of terms.
int i,j;                        // Loop variables
float ax[10];                   // 'X' array limit 9
float ay[10];                   // 'Y' array limit 9
float x;                        // User Query for what value of X
float y=0;                      // Calculated value for coressponding X.
float h;                        // Calc. Section
float p;                        // Calc. Section
float diff[20][20];             // to store Y
float y1,y2,y3,y4;              // Formulae variables.

clrscr();
printf("\t\t !! NEWTON'S GREGORY FORWARD INTERPOLATION FORMULA !! ");
// Input section.
printf("\n\n Enter the no. of terms -> ");
scanf("%d",&n);

// Input Sequel for array X
printf("\n\n Enter the value in the form of x -> ");
// Input loop for X.
for(i=0;i<n;i++)
   {
   printf("\n Enter the value of x%d -> ",i+1);
   scanf("%f",&ax[i]);
   }

// Input sequel for array Y.
printf("\n\n Enter the value in the form of y -> ");
// Input loop for Y.
for(i=0;i<n;i++)
   {
   printf("\n Enter the value of y%d -> ",i+1);
   scanf("%f",&ay[i]);
   }

// Inputting the required value quarry
printf("\n\n Enter the value of x for ");
printf("\n which u want the value of y -> ");
scanf("%f",&x);

// Calculation and processing section.
h=ax[1]-ax[0];

// for 1.
for(i=0;i<n-1;i++)
   diff[i][1]=ay[i+1]-ay[i];
// for 2, 3, 4
for(j=2;j<=4;j++)
   for(i=0;i<n-j;i++)
      diff[i][j]=diff[i+1][j-1]-diff[i][j-1];

do
 {
 i++;
 }
 while(ax[i]<x);
i--;

p=(x-ax[i])/h;
y1=p*diff[i-1][1];                                            // 1
y2=p*(p+1)*diff[i-1][2]/2;                                    // 2
y3=p*(p+1)*(p-1)*diff[i-2][3]/6;                              // 3
y4=(p+2)*(p+1)*p*(p-1)*diff[i-3][4]/24;

// Taking sum
y=ay[i]+y1+y2+y3+y4;

// Outut Section
printf("\n When x = %6.4f , y = %6.8f",x,y);

// Invoke user watch halt function
printf("\n\n\n\t\t\t !! PRESS ENTER TO EXIT !! ");


getch();
}

C Program to Implement REGULAR FALSI (FALSE POSITION) METHOD

/*
  Program to Implement REGULAR FALSI (FALSE POSITION) METHOD.
  EQN -> 3*x + sin(x) - exp(x) = 0.
*/

#include<stdio.h>
#include<conio.h>
#include<math.h>
#include<string.h>
#include<process.h>

// Formulae Declaration
#define EPS 0.00005
#define f(x)  3*x + sin(x) - exp(x)
int n;

void FAL_POS();

void main()
{
clrscr();
printf("\n Solution by ITERATION METHOD ");
printf("\n\n Equation is -> x*x*x - 2*x + 1 = 0\n");
printf("\n Enter the no. of iterations ");
scanf("%d",&n);
FAL_POS();
getch();
}

void FAL_POS()
 {
 long float x1,x2,x0;
 float f1,f2,f0;
 int itr=n,i;
 for(x1=0.0;;)
    {
    f1=f(x1);
    if(f1>0)
      break;
    else
      x1=x1+0.1;
    }
 x0=x1-0.1;
 f0=f(x0);
 printf("\n\t\t----------------------------------------------------");
 printf("\n \t\tITERATION\t x2\t\t\t   F(X)\n");
 printf("\n\t\t----------------------------------------------------\n");
 for(i=0;i<itr;i++)
    {
    x2=x0-((x1-x0)/(f1-f0))*f0;
    f2=f(x2);
    if(f0*f2>0)
      {
      x1=x2;
      f1=f2;
      }
    else
      {
      x0=x2;
      f0=f2;
      }
    if(fabs(f(2))>EPS)
      printf("\n\t\t %d \t\t %f \t\t %f \n",i+1,x2,f2);
    }
 printf("\n\n\n\t\t----------------------------------------------------");
 printf("\n\t\t ROOT  = %f ",x2);
 printf("\n\t\t----------------------------------------------------");
}

C Program to Implement NEWTON RAPHSON METHOD

/*
  Program to Implement NEWTON RAPHSON METHOD to solve eqn.
  EQN -> 3*x - cos(x) - 1 = 0.
*/

#include<stdio.h>
#include<conio.h>
#include<math.h>
#include<string.h>
#include<process.h>

// Formulae Declaration
#define f(x)   3*x - cos(x) - 1
#define df(x)  3+sin(x)
int n;

void NEW_RAP();

void main()
{
clrscr();
printf("\n Solution by NEWTON RAPHSON METHOD ");
printf("\n\n Equation is -> 3*x - cos(x) - 1 = 0\n");
printf("\n Enter the no. of iterations ");
scanf("%d",&n);
NEW_RAP();
getch();
}

void NEW_RAP()
 {
 float x1,x0;
 float f1,f0;
 float df0;
 int i=1,itr;
 float EPS,error;

 /* finding an approximate Root of a given equation, having +ve value*/
 for(x1=0.0;;x1+=0.01)
    {
    f1=f(x1);
    if(f1>0)
      break;
    }
 itr=n;
 x0=x1-0.01;
 f0=f(x0);
 printf("\n Enter the Maximum possible error: ");
 scanf("%f",&EPS);
 if(fabs(f0)>f1)
   printf("\n\t\t The root is near to %.4f\n\n\n",x1);
 if(f1>fabs(x0))
   printf("\n\t\t The root is near to %.4f\n\n\n",x0);
 x0=(x0+x1)/2;
 for(;i<=itr;i++)
    {
    f0=f(x0);
    df0=df(x0);
    x1=x0-(f0/df0);
    printf("\n\t\t The %d approximation to the root is : %f",i,x1);
    error=fabs(x1-x0);
    if(error<EPS)
      break;
    x0=x1;
    }
 if(error>EPS)
   {
   printf("\n\n\t NOTE :-");
   printf(" The No. of Iterartions are not sufficient. ");
   }
 printf("\n\n\n\t\t----------------------------------------------------");
 printf("\n\t\t ROOT  = %.4f ",x1);
 printf("\n\t\t----------------------------------------------------");
}

Program to implement LANGRANGE'S INTERPOLATION FORMULA in C

/*

  Program to implement LANGRANGE'S INTERPOLATION FORMULA in C.

      ----------------------------------------------

*/

#include<stdio.h>

#include<conio.h>

#include<math.h>

#include<process.h>

#include<conio.h>

void main()

{

int n;                          // no. of terms.

int i,j;                        // Loop variables

float ax[100];                   // 'X' array limit 9

float ay[100];                   // 'Y' array limit 9

float x=0;                        // User Query for what value of X

float y=0;                      // Calculated value for coressponding X.

float nr,dr;

clrscr();

printf("\t\t !! LANGRANGE'S INTERPOLATION FORMULA !! ");

// Input section.

printf("\n\n Enter the no. of terms -> ");

scanf("%d",&n);

// Input Sequel for array X

printf("\n\n Enter the value in the form of x -> ");

// Input loop for X.

for(i=0;i<n;i++)

   {

   printf("\n Enter the value of x%d -> ",i+1);

   scanf("%f",&ax[i]);

   }

// Input sequel for array Y.

printf("\n\n Enter the value in the form of y -> ");

// Input loop for Y.

for(i=0;i<n;i++)

   {

   printf("\n Enter the value of y%d -> ",i+1);

   scanf("%f",&ay[i]);

   }

// Inputting the required value quarry

printf("\n\n Enter the value of x for ");

printf("\n which u want the value of y -> ");

scanf("%f",&x);

// Calculation and processing section.

for(i=0;i<n;i++)

   {

   nr=1;

   dr=1;

   for(j=0;j<n;j++)

      {

      if(j!=i)

{

nr=nr*(x-ax[j]);

dr=dr*(ax[i]-ax[j]);

}

      y+=(nr/dr)*ay[i];

      }

   }

// Outut Section

printf("\n When x = %5.2f , y = %5.2f",x,y);

// Invoke user watch halt function

printf("\n\n\n\t\t\t !! PRESS ENTER TO EXIT !! ");

getch();

}

C Program to Implement ITERATION METHOD

/*
  Program to Implement ITERATION METHOD.
  EQN -> x*x*x - 2*x + 1 = 0.
*/

#include<stdio.h>
#include<conio.h>
#include<math.h>

#define EPS 0.00005
#define F(x) (x*x*x + 1)/2
#define f(x)  x*x*x - 2*x + 1

int n;

void iter();

void main()
{
clrscr();
printf("\n Solution by ITERATION METHOD ");
printf("\n\n Equation is -> x*x*x - 2*x + 1 = 0\n");
printf("\n Enter the no. of iterations ");
scanf("%d",&n);
iter();
getch();
}

void iter()
 {
 int i=0;
 float x1,x2,x0;
 float f1,f2,f0,error;
 for(x1=1; ;x1++)
    {
    f1=f(x1);
    if(f1>0)
      break;
    }
 for(x0=x1-1; ;x0--)
    {
    f0=f(x0);
    if(f0<0)
      break;
    }
 x2=(x0+x1)/2;
 printf("\n\n\t\t The 1 approximatrion to the root is : %f",x2);
 for(;i<n-1;i++)
    {
    f2=F(x2);
    printf("\n\n\t\t The %d approximatrion to the root is : %f",i+2,f2);
    x2=F(x2);
    error=fabs(f2-f1);
    if(error<EPS)
       break;
    f1=f2;
    }
 if(error>EPS)
   printf("\n\n\t NOTE :- The no. of iterations are not sufficient.");
 printf("\n\n\n\t\t -----------------------------------------------");
 printf("\n\t\t ROOT  = %.4f (Correct to 4 Decimal places)",f2);
 printf("\n\t\t -----------------------------------------------");
}

C Program to implement GAUSS' FORWARD INTERPOLATION FORMULA

/*
  Program to implement GAUSS' FORWARD INTERPOLATION FORMULA.
      ---------------------------------
*/

#include<stdio.h>
#include<conio.h>
#include<math.h>
#include<process.h>
#include<conio.h>

void main()
{
int n;                          // no. of terms.
int i,j;                        // Loop variables
float ax[10];                   // 'X' array limit 9
float ay[10];                   // 'Y' array limit 9
float x;                        // User Query for what value of X
float nr,dr;
float y=0;                      // Calculated value for coressponding X.
float h;                        // Calc. Section
float p;                        // Calc. Section
float diff[20][20];             // to store Y
float y1,y2,y3,y4;              // Formulae variables.

clrscr();
printf("\t\t !! GAUSS' FORWARD INTERPOLATION FORMULA !! ");
// Input section.
printf("\n\n Enter the no. of terms -> ");
scanf("%d",&n);

// Input Sequel for array X
printf("\n\n Enter the value in the form of x -> ");
// Input loop for X.
for(i=0;i<n;i++)
   {
   printf("\n Enter the value of x%d -> ",i+1);
   scanf("%f",&ax[i]);
   }

// Input sequel for array Y.
printf("\n\n Enter the value in the form of y -> ");
// Input loop for Y.
for(i=0;i<n;i++)
   {
   printf("\n Enter the value of y%d -> ",i+1);
   scanf("%f",&ay[i]);
   }

// Inputting the required value quarry
printf("\n\n Enter the value of x for ");
printf("\n which u want the value of y -> ");
scanf("%f",&x);

// Calculation and processing section.
h=ax[1]-ax[0];
for(i=0;i<n-1;i++)
   diff[i][1]=ay[i+1]-ay[i];

for(j=2;j<=4;j++)
   for(i=0;i<n-j;i++)
      diff[i][j]=diff[i+1][j-1]-diff[i][j-1];

i=0;
do
 {
 i++;
 }
while(ax[i]<x);

i--;
p=(x-ax[i])/h;
y1=p*diff[i][1];
y2=p*(p-1)*diff[i-1][2]/2;
y3=(p+1)*p*(p-1)*diff[i-2][3]/6;
y4=(p+1)*p*(p-1)*(p-2)*diff[i-3][4]/24;

// Taking sum
y=ay[i]+y1+y2+y3+y4;

// Outut Section
printf("\n When x = %6.4f , y = %6.8f",x,y);

// Invoke user watch halt function
printf("\n\n\n\t\t\t !! PRESS ENTER TO EXIT !! ");
getch();
}

c Program to implement GAUSS' BACKWARD INTERPOLATION FORMULA

/*
  Program to implement GAUSS' BACKWARD INTERPOLATION FORMULA
*/

#include<stdio.h>
#include<conio.h>
#include<math.h>
#include<process.h>
#include<conio.h>

void main()
{
int n;                          // no. of terms.
int i,j;                        // Loop variables
float ax[10];                   // 'X' array limit 9
float ay[10];                   // 'Y' array limit 9
float x;                        // User Query for what value of X
float y=0;                      // Calculated value for coressponding X.
float h;                        // Calc. Section
float p;                        // Calc. Section
float diff[20][20];             // to store Y
float y1,y2,y3,y4;              // Formulae variables.

clrscr();
printf("\t\t !! GAUSS' FORWARD INTERPOLATION FORMULA !! ");
// Input section.
printf("\n\n Enter the no. of terms -> ");
scanf("%d",&n);

// Input Sequel for array X
printf("\n\n Enter the value in the form of x -> ");
// Input loop for X.
for(i=0;i<n;i++)
   {
   printf("\n Enter the value of x%d -> ",i+1);
   scanf("%f",&ax[i]);
   }

// Input sequel for array Y.
printf("\n\n Enter the value in the form of y -> ");
// Input loop for Y.
for(i=0;i<n;i++)
   {
   printf("\n Enter the value of y%d -> ",i+1);
   scanf("%f",&ay[i]);
   }

// Inputting the required value quarry
printf("\n\n Enter the value of x for ");
printf("\n which u want the value of y -> ");
scanf("%f",&x);

// Calculation and processing section.
h=ax[1]-ax[0];
for(i=0;i<n-1;i++)
   diff[i][1]=ay[i+1]-ay[i];

for(j=2;j<=4;j++)
   for(i=0;i<n-j;i++)
      diff[i][j]=diff[i+1][j-1]-diff[i][j-1];

i=0;
do
 {
 i++;
 }
while(ax[i]<x);

i--;
p=(x-ax[i])/h;
y1=p*diff[i-1][1];
y2=p*(p+1)*diff[i-1][2]/2;
y3=(p+1)*p*(p-1)*diff[i-2][3]/6;
y4=(p+2)*(p+1)*p*(p-1)*diff[i-3][4]/24;

// Taking sum
y=ay[i]+y1+y2+y3+y4;

// Outut Section
printf("\n When x = %6.4f , y = %6.4f",x,y);

// Invoke user watch halt function
printf("\n\n\n\t\t\t !! PRESS ENTER TO EXIT !! ");
getch();
}

c Program to Implement BISECTION METHOD

/*
  Program to Implement BISECTION METHOD.
  EQN -> x*log10(x) - 1.2
*/

#include<stdio.h>
#include<conio.h>
#include<math.h>

#define F(x)  (x)*log10(x)-1.2
int n;
float root=1;

void bisect();

void main()
{
clrscr();
printf("\n Solution by BISECTION METHOD ");
printf("\n Equation is -> x*log(x) - 1.2 = 0\n");
printf("\n Enter the no. of iterations ");
scanf("%d",&n);
bisect();
getch();
}

void bisect()
 {
 int count;
 float x1,x2,x0;
 float f1,f2,f0;
 // Finding x2
 for(x2=1; ;x2++)
    {
    f2=F(x2);
    if(f2>0)
      {
      break;
      }
    }
 printf("\n x2 = %d ",x2);
 // Finding x1
 for(x1=x2-1; ;x1--)
    {
    f1=F(x1);
    if(f1<0)
      {
      break;
      }
    }
 printf("\n x2 = %d ",x2);
 printf("\n ----------------------------------------------------");
 printf("\n \tIteration No.    \t -        ROOTS ");
 printf("\n ----------------------------------------------------\n");
 for(count=1;count<=n;count++)
    {
    x0=(x1+x2)/2;
    f0=F(x0);
    if(f0==0)
      {
      root=x0;
      }
    if(f0*f1<0)
      {
      x2=x0;
      f2=f0;
      }
    else
      {
      x1=x0;
      f1=f0;
      }
    printf("\n \t ITERATION %d \t : \t %f",count,x0);
    if( fabs((x1-x2)/2) < 0.00000005)
      {
      printf("\n -----------------------------------------------");
      printf("\n \t \t \tROOT       = %f ",x0);
      printf("\n \t \t \tITERATIONS = %d ",count);
      printf("\n -----------------------------------------------");
      exit(0);
      }
    }
printf("\n ----------------------------------------------------");
printf("\n \t \t    ROOT       = %7.4f ",x0);
printf("\n \t \t    ITERATIONS = %d ",count-1);
printf("\n ----------------------------------------------------");
}

tips nd tricks section added soon

Tips nd tricks section added soon. Pls give ur comment nd suggestion to improve this blog.

Program to implement LAPLACE - EVERETT'S INTERPOLATION FORMULA in C

#include<stdio.h>
#include<conio.h>
#include<math.h>
#include<process.h>
#include<conio.h>

void main()
{
int n;                          // no. of terms.
int i,j;                        // Loop variables
float ax[10];                   // 'X' array limit 9
float ay[10];                   // 'Y' array limit 9
float x;                        // User Query for what value of X
float nr,dr;
float y=0;                      // Calculated value for coressponding X.
float h;                        // Calc. Section
float p,q;                      // Calc. Section
float diff[20][20];             // to store Y
float y1,y2,y3,y4;              // Formulae variables.
float py1,py2,py3,py4;

clrscr();
printf("\t\t !! LAPALCE-EVERETT'S INTERPOLATION FORMULA !! ");
// Input section.
printf("\n\n Enter the no. of terms -> ");
scanf("%d",&n);

// Input Sequel for array X
printf("\n\n Enter the value in the form of x -> ");
// Input loop for X.
for(i=0;i<n;i++)
   {
   printf("\n Enter the value of x%d -> ",i+1);
   scanf("%f",&ax[i]);
   }

// Input sequel for array Y.
printf("\n\n Enter the value in the form of y -> ");
// Input loop for Y.
for(i=0;i<n;i++)
   {
   printf("\n Enter the value of y%d -> ",i+1);
   scanf("%f",&ay[i]);
   }

// Inputting the required value quarry
printf("\n\n Enter the value of x for ");
printf("\n which u want the value of y -> ");
scanf("%f",&x);

// Calculation and processing section.
h=ax[1]-ax[0];
for(i=0;i<n-1;i++)
   diff[i][1]=ay[i+1]-ay[i];

for(j=2;j<=4;j++)
   for(i=0;i<n-j;i++)
      diff[i][j]=diff[i+1][j-1]-diff[i][j-1];

i=0;
do
 {
 i++;
 }
while(ax[i]<x);

i--;
p=(x-ax[i])/h;
q=1-p;
y1=q*(ay[i]);
y2=q*(q*q-1)*diff[i-1][2]/4;
y3=q*(q*q-1)*(q*q-4)*diff[i-2][4]/120;
py1=p*(ay[i+1]);
py2=p*(p*p-1)*diff[i][2]/6;
py3=p*(p*p-1)*(p*p-4)*diff[i-1][4]/120;

// Taking sum
y=ay[i]+y1+y2+y3+py1+py2+py3;

// Outut Section
printf("\n When x = %6.2f , y = %6.8f",x,y);

// Invoke user watch halt function
printf("\n\n\n\t\t\t !! PRESS ENTER TO EXIT !! ");
getch();
}

program in to implement Gauss Seidal method

# include <stdio.h>
# include <conio.h>
# include <math.h>
# define MAXIT 50
# define EPS .000001
void main()
{
float a[10][10],b[10],x[10],sum,x0[10];
int i,j,n,count=0;
clrscr();
printf("Enter Number of equation : ");
scanf("%d",&n);
printf("Enter Coefficients row by row \n");
printf("One row in each line\n");
for(i=1;i<=n;i++)
for(j=1;j<=n;j++)
scanf("%f",&a[i][j]);
printf("Enter vector elements\n");
for(i=1;i<=n;i++)
{
x0[i]=x[i]=0;
scanf("%f",&b[i]);
}
printf("N\t");
for(i=1;i<=n;i++)
printf("X%d \t",i);
printf("\n");
printf("-------------------------------------------------------------------\n");
s: printf("%d\t",count+1);
for(i=1;i<=n;i++)
{
sum=0;
for(j=1;j<=n;j++)
{
if(i==j)
continue;
else
sum=sum+(a[i][j]*x[j]);
}
x[i]=(b[i]-sum)/a[i][i];
printf("%f \t",x[i]);
}
count++;
if(fabs(x[n]-x0[n])>EPS && count<MAXIT)
{
for(i=1;i<=n;i++)
x0[i]=x[i];
printf("\n");
goto s;
}
else
{
printf("\n----------------------------------------------\nSolution : \n");
for(i=1;i<=n;i++)
printf("X%d = %3.4f\t",i,x[i]);
}
getch();
}

Newton Divided difference formula

# include <stdio.h>
# include <conio.h>
# define MAX 20
void main()
{
int i,j,n;
float x[MAX],fx[MAX][MAX],xp,u,sum=0;
printf("Newton Divided Difference formula \n");
printf("Enter No. of data points : ");
scanf("%d",&n);
printf("Enter values of X and f(x) set by set each at once \n");
for(i=1;i<=n;i++)
scanf("%f%f",&x[i],&fx[i][1]);
printf("Enter value of X at which interpolation is required : ");
scanf("%f",&xp);
for(i=2;i<=n;i++)
for(j=1;j<=n-i+1;j++)
fx[j][i]=(fx[j+1][i-1]-fx[j][i-1])/(x[j+i-1]-x[j]);
printf("-------------------------------------------------------------------\n");
printf("X F(x) ");
for(i=1;i<n;i++)
printf("y%d ",i);
printf("\n-----------------------------------------------------------------\n");
for(i=1;i<=n;i++)
{
printf("%4.2f\t ",x[i]);
for(j=1;j<=n-i+1;j++)
printf("%4.2f ",fx[i][j]);
printf("\n");
}
printf("\n-----------------------------------------------------------------\n");
sum=fx[1][1];
for(i=2;i<=n;i++)
{
u=1;
for(j=1;j<i;j++)
u=u*(xp-x[j]);
u=u*fx[1][i];
sum=sum+u;
}
printf("For %3.2f Interpolated value = %4.2f",xp,sum);
getch();
}

first experience of blogging

This blog is for all computer learners. my goal to make this blog for c beginners and data structure students. i will try my best to update this on daily basis, so plz visit this . and say that you like this or not. and how can i make this better. thanx.

Program to implement BESSELS INTERPOLATION FORMULA in C

/*
  Program to implement BESSELS INTERPOLATION FORMULA in C.
      ------------------------------------
*/

#include<stdio.h>
#include<conio.h>
#include<math.h>
#include<process.h>
#include<conio.h>

void main()
{
int n;                          // no. of terms.
int i,j;                        // Loop variables
float ax[10];                   // 'X' array limit 9
float ay[10];                   // 'Y' array limit 9
float x;                        // User Query for what value of X
float y;                      // Calculated value for coressponding X.
float h;                        // Calc. Section
float p;                        // Calc. Section
float diff[20][20];             // to store Y
float y1,y2,y3,y4;              // Formulae variables.

clrscr();
printf("\t\t !! BESSELS INTERPOLATION FORMULA !! ");
// Input section.
printf("\n\n Enter the no. of terms -> ");
scanf("%d",&n);

// Input Sequel for array X
printf("\n\n Enter the value in the form of x -> ");
// Input loop for X.
for(i=0;i<n;i++)
   {
   printf("\n Enter the value of x%d -> ",i+1);
   scanf("%f",&ax[i]);
   }

// Input sequel for array Y.
printf("\n\n Enter the value in the form of y -> ");
// Input loop for Y.
for(i=0;i<n;i++)
   {
   printf("\n Enter the value of y%d -> ",i+1);
   scanf("%f",&ay[i]);
   }

// Inputting the required value quarry
printf("\n\n Enter the value of x for ");
printf("\n which u want the value of y -> ");
scanf("%f",&x);

// Calculation and processing section.
h=ax[1]-ax[0];
for(i=0;i<n-1;i++)
   diff[i][1]=ay[i+1]-ay[i];

for(j=2;j<=4;j++)
   for(i=0;i<n-j;i++)
      diff[i][j]=diff[i+1][j-1]-diff[i][j-1];

i=0;
do
 {
 i++;
 }
while(ax[i]<x);

i--;
p=(x-ax[i])/h;
y1=p*diff[i][1];
y2=p*(p-1)*(diff[i][2]+diff[i-1][2])/4;
y3=p*(p-1)*(p-0.5)*diff[i-1][3]/6;
y4=(p+1)*p*(p-1)*(p-2)*(diff[i-2][4]+diff[i-1][4])/48;

// Taking sum
y=ay[i]+y1+y2+y3+y4;

// Outut Section
printf("\n When x = %6.2f , y = %6.8f",x,y);

// Invoke user watch halt function
printf("\n\n\n\t\t\t !! PRESS ENTER TO EXIT !! ");
getch();
}

program to create two way linked list

program to create two way linked list.

#include<stdio.h>

#include<conio.h>

#include<alloc.h>

#define NULL 0

void main()

{

struct dnode

{

struct dnode *prev;

int data;

struct dnode *next;

}*start,*temp,*k,*p,*pre;

int i,c;

char ch;

clrscr();

start=NULL;

for(i=0;i<=1;i++)

{

if(start==NULL)

{

temp=(struct dnode*)malloc(sizeof(struct dnode));

printf("\nEnter the value for node: -");

scanf("%d",&temp->data);

start= temp;

temp->prev=NULL;

temp->next=NULL;

}

else

{

while(1)

{

printf("\nDo you want to create one more node(Y/N):- ");

fflush(stdin);

scanf("%c",&ch);

if(ch=='y' || ch=='Y')

{

temp->next=(struct dnode*)malloc(sizeof(struct dnode));

printf("\nEnter the value of  node: -");

k=temp->next;

scanf("%d",&k->data);

k->prev=temp;

k->next=NULL;

temp=k;

}

else

{

temp->next=NULL;

break;

}

}

}

}

//For forward Displaying the linked list.

printf("\n\n\nBy forward Displaying: -");

printf("\nNodes of linked list are-\n");

temp=start;

while(temp!=NULL)

{

printf("%4d",temp->data);

temp=temp->next;

}

//For backward Displaying the linked list.

printf("\n\n\nBy backward Displaying:");

printf("\nNodes of linked list are-\n");

temp=k;

while(temp!=NULL)

{

printf("%4d",temp->data);

temp=temp->prev;

}

getch();

}

output:

Enter the value for node: -14

Do you want to create one more node(Y/N):- y

Enter the value of  node: -63

Do you want to create one more node(Y/N):- y

Enter the value of  node: -87

Do you want to create one more node(Y/N):- y

Enter the value of  node: -69

Do you want to create one more node(Y/N):- y

Enter the value of  node: -100

Do you want to create one more node(Y/N):- n

By forward Displaying: -

Nodes of linked list are-

  14  63  87  69 100

By backward Displaying:

Nodes of linked list are-

 100  69  87  63 14