Linear Algebra and the C Language/a042

/* ------------------------------------ */
/*  Save as:   c00a.c                   */
/* ------------------------------------ */
#include "v_a.h"
/* ------------------------------------ */
/* ------------------------------------ */
#define   RA R5
#define   CA C5
/* ------------------------------------ */
#define FACTOR_E        +1.E-0         
/* ------------------------------------ */
int main(void)
{
double  txy[8] ={
   1,      0,
   2,      3,
   3,      4,
   4,      0   };
   
double tA[RA*CA]={
/* x**2     y**2     x        y        e     */
  +1.00,   +0.00,   +0.00,   +0.00,   +0.00,
  +1.00,   +0.00,   +1.00,   +0.00,   +1.00,
  +4.00,   +9.00,   +2.00,   +3.00,   +1.00,   
  +9.00,  +16.00,   +3.00,   +4.00,   +1.00,
 +16.00,   +0.00,   +4.00,   +0.00,   +1.00
};

double tb[RA*C1]={
/* = 0   */
  +1.00, 
  +0.00, 
  +0.00, 
  +0.00, 
  +0.00, 
};

double **xy      =    ca_A_mR(txy,    i_mR(R4,C2));
double **A       =    ca_A_mR(tA,     i_mR(RA,CA));
double **b       =    ca_A_mR(tb,     i_mR(RA,C1));
double **Pinv    = Pinv_Rn_mR(A,      i_mR(CA,RA),FACTOR_E);            
double **Pinvb   =     mul_mR(Pinv,b, i_mR(CA,C1));          

  clrscrn();
  printf("\n");
  printf(" Find the coefficients a, b, c, d, e, of the curve \n\n");
  printf("     ax**2 + by**2 + cx + dy + e  = 0 \n\n");
  printf(" that passes through these four points.\n\n");
  printf("         x         y");
  p_mR(xy, S10,P0,C6);
  stop();
  
  clrscrn();
  printf(" Using the given points, we obtain this matrix.\n");
  printf("  (a = 1. This is my choice)\n\n");  
  printf(" A:");
  p_mR(A, S10,P2,C7);
  printf(" b:");
  p_mR(b, S10,P2,C7);
  printf(" Pinv = V invS_T U_T ");
  pE_mR(Pinv, S12,P4,C10); 
  stop();
  
  clrscrn();
  printf(" Solving this system yields a unique\n"
         " least squares solution, namely   \n\n");  
  printf(" Pinv b ");   
  p_mR(Pinvb, S10,P4,C10);
  printf(" The coefficients a, b, c, d, e, of the curve are: \n\n"
         "  %+.9f*x^2 %+.9f*y^2 %+.9f*x %+.9f*y %+.9f = 0\n\n"
            ,Pinvb[R1][C1],Pinvb[R2][C1],Pinvb[R3][C1],
             Pinvb[R4][C1],Pinvb[R5][C1]);      
  stop();  

  f_mR(xy);
  f_mR(b);
  f_mR(A);
  f_mR(Pinv);
  f_mR(Pinvb); 

  return 0;
}
/* ------------------------------------ */
/* ------------------------------------ */

 Find the coefficients a, b, c, d, e, of the curve 

     ax**2 + by**2 + cx + dy + e  = 0 

 that passes through these four points.

         x         y
        +1         +0 
        +2         +3 
        +3         +4 
        +4         +0 

 Press return to continue. 


 Using the given points, we obtain this matrix.
  (a = 1. This is my choice)

 A :
     +1.00      +0.00      +0.00      +0.00      +0.00 
     +1.00      +0.00      +1.00      +0.00      +1.00 
     +4.00      +9.00      +2.00      +3.00      +1.00 
     +9.00     +16.00      +3.00      +4.00      +1.00 
    +16.00      +0.00      +4.00      +0.00      +1.00 

 b :
     +1.00 
     +0.00 
     +0.00 
     +0.00 
     +0.00 

 Pinv = V * invS_T * U_T 
 +1.0000e+00  +4.2688e-14  -5.9459e-14  +3.4212e-14  -6.0507e-15 
 -1.6667e-01  +1.3889e-01  -3.3333e-01  +2.5000e-01  -5.5556e-02 
 -5.0000e+00  -3.3333e-01  +2.7142e-13  -1.5601e-13  +3.3333e-01 
 +1.1667e+00  -6.3889e-01  +1.3333e+00  -7.5000e-01  +5.5556e-02 
 +4.0000e+00  +1.3333e+00  -1.3500e-13  +7.7716e-14  -3.3333e-01 

 Press return to continue. 


 Solving this system yields a unique
 least squares solution, namely   

 Pinv * b 
   +1.0000 
   -0.1667 
   -5.0000 
   +1.1667 
   +4.0000 

 The coefficients a, b, c, d, e, of the curve are : 

  +1.000000000*x^2 -0.166666667*y^2 -5.000000000*x +1.166666667*y +4.000000000 = 0

 Press return to continue.