Interpolation by the method of divided difference

Source code:

#include<stdio.h>

#include<conio.h>

#define max 20

void main()

{

 float table[max][max], xval, yval, product=1;

 int i,j, count;

 printf(" METHOD OF DIVIDED DIFFERENCCE  \n\n");

 printf("\n Enter no. of points:-  ");

 scanf("%d",&count);

 printf("\n Enter x and y values in tabular format - \n ");

 for(i=0; i<=2*count-2; i+=2)

  for(j=0; j<=1; j++)

   scanf("%f",&table[i][j]);

 printf("\n Enter x-value?  ");

 scanf("%f",&xval);

 /* Constructing divided difference table */

 for(j=2; j<=count; j++)

  for(i=j-1; i<=2*count-j-1; i+=2)

  {

   table[i][j]=table[i+1][j-1] - table[i-1][j-1];

   table[i][j]/=(table[i+j-1][0] - table[i-j+1][0]);

  }

 /* Calculation of y-value */

 yval = table[0][1] ;

 for(i=0,j=2; j<=count; j++)

 {

  product*=(xval-table[i][0]);

  yval+=product*table[j-1][j];

  i+=2;

 }

 product*=(xval-table[2*count-2][0]);

 /* printing divided difference table */

  printf("\n DIVIDED DIFFERENCE TABLE  \n\n");

 for(i=0; i<=2*count-2; i++)

 {

  if(i%2==0)

            printf("%2.3f %2.3f",table[i][0], table[i][1]);

  else

            printf("           ");

  for(j=2; j<=count; j++)

            if((j%2==0 && i%2==1) || (j%2==1 && i%2==0))

            {

             if(i>=j-1 && i<2*count-j)

                         printf(" %2.3f",table[i][j]);

            }

            else

             printf("     ");

  printf("\n");

 }

 printf("\n\n At x = %f, y = %f", xval, yval);

}

SAMPLE OUTPUT:

Evaluating the value of y at x = 5.604, using the method of divided difference,

given the following table of values:

 

x	5.600	5.602	5.605	5.607	5.608
y	0.7756	0.7768	0.7787	0.7799	0.7806