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c00p.c
/* ------------------------------------ */
/*  Save as :   c00p.c                  */
/* ------------------------------------ */
#include "v_a.h"
/* ------------------------------------ */
/* ------------------------------------ */
#define   RA R5
#define   CA C5
#define   Cb C1 
/* ------------------------------------ */
/* ------------------------------------ */
void fun(void)
{
double   xy[8] ={
   1,      1,
   2,      4,
   3,      9,
   4,     16  };

   
double ab[RA*(CA+Cb)]={
/* x**2     y**2     x        y        e    =   0   */
  +1.00,   +0.00,   +0.00,   +0.00,   +0.00,   +1.00, 
  +1.00,   +1.00,   +1.00,   +1.00,   +1.00,   +0.00, 
  +4.00,  +16.00,   +2.00,   +4.00,   +1.00,   +0.00, 
  +9.00,  +81.00,   +3.00,   +9.00,   +1.00,   +0.00, 
 +16.00, +256.00,   +4.00,  +16.00,   +1.00,   +0.00 
};

double **XY = ca_A_mR(xy,i_mR(R4,C2));

double **Ab = ca_A_mR(ab,i_Abr_Ac_bc_mR(RA,CA,Cb));
double **A  = c_Ab_A_mR(Ab,i_mR(RA,CA));
double **b  = c_Ab_b_mR(Ab,i_mR(RA,Cb));

double **A_T        = i_mR(CA,RA);
double **A_TA       = i_mR(CA,CA); //         A_T*A
double **invA_TA    = i_mR(CA,CA); //     inv(A_T*A)
double **invA_TAA_T = i_mR(CA,RA); //     inv(A_T*A)*A_T

double **x          = i_mR(CA,Cb); // x = inv(A_T*A)*A_T*b

  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,S5,P0,C6);
  printf(" Using the given points, we obtain this matrix.\n");
  printf("  (a = 1. This is my choice)\n\n");  
  printf(" Ab :\n");
  printf("   x**2    y**2    x       y       e    =  0     ");
  p_mR(Ab,S7,P2,C6);
  stop();
  
  clrscrn();
  printf(" A_T :");
  p_mR(transpose_mR(A,A_T),S10,P2,C7);
  printf(" A_TA :");
  p_mR(mul_mR(A_T,A,A_TA),S10,P2,C7);
  stop();
  
  clrscrn();
  printf(" inv(A_TA) :");
  p_mR(inv_mR(A_TA,invA_TA),S10,P4,C7);  
  printf(" inv(A_TA)*A_T :");
  p_mR(mul_mR(invA_TA,A_T,invA_TAA_T),S10,P4,C7);
  printf("\n x = inv(A_TA)*A_T*b :");
  p_mR(mul_mR(invA_TAA_T,b,x),S10,P4,C7);
  stop();
  
  clrscrn();
  printf("\n x = inv(A_TA)*A_T*b :");
  p_mR(x,S10,P2,C7); 
  printf(" The coefficients a, b, c, d, e, of the curve are : \n\n"
         "  %+.2fx**2  %+.2fy  = 0\n\n"
            ,x[R1][C1],x[R4][C1]);        
  stop();  

  f_mR(XY);  
  f_mR(A);
  f_mR(b);
  f_mR(Ab);

  f_mR(A_T);
  f_mR(A_TA);       //         A_T*A
  f_mR(invA_TA);    //     inv(A_T*A)
  f_mR(invA_TAA_T); //     inv(A_T*A)*A_T
    
  f_mR(x); 
}
/* ------------------------------------ */
int main(void)
{
	
  fun();

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


 Trouver les coefficients a, b, c, d, e du conique,
      
      ax**2 + by**2 + cx + dy + e  = 0 
       qui passe par ces quatre points.    
         
      (x[1],y[1])  (x[2],y[2])  (x[3],y[3])  (x[4],y[4]) 
 En utilisant les quatre points nous obtenons la matrice.
 (a)x**2   (b)y**2   (c)x      (d)y        (e) = 0               
    x[1]**2   y[1]**2   x[1]      y[1]      1    0
    x[2]**2   y[2]**2   x[2]      y[2]      1    0
    x[3]**2   y[3]**2   x[3]      y[3]      1    0
    x[4]**2   y[4]**2   x[4]      y[4]      1    0
 Ce système à quatre lignes et cinq inconnus (a, b, c, d, e).
 C'est un système homogène, il a donc une infinité de solution.

 Pour trouver une solution j'ai choisi de poser que a = 1.
 Nous avons donc cinq lignes et cinq inconnus.

    1         0         0         0         0    1 
    x[1]**2   y[1]**2   x[1]      y[1]      1    0 
    x[2]**2   y[2]**2   x[2]      y[2]      1    0 
    x[3]**2   y[3]**2   x[3]      y[3]      1    0 
    x[4]**2   y[4]**2   x[4]      y[4]      1    0 
 Il suffit maintenant de résoudre le système. 


Exemple de sortie écran :
  -----------------------------------
 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    +1 
   +2    +4 
   +3    +9 
   +4   +16 

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

 Ab :
   x**2    y**2    x       y       e    =  0     
  +1.00   +0.00   +0.00   +0.00   +0.00   +1.00 
  +1.00   +1.00   +1.00   +1.00   +1.00   +0.00 
  +4.00  +16.00   +2.00   +4.00   +1.00   +0.00 
  +9.00  +81.00   +3.00   +9.00   +1.00   +0.00 
 +16.00 +256.00   +4.00  +16.00   +1.00   +0.00 

 Press return to continue. 


  -----------------------------------
 A_T :
     +1.00      +1.00      +4.00      +9.00     +16.00 
     +0.00      +1.00     +16.00     +81.00    +256.00 
     +0.00      +1.00      +2.00      +3.00      +4.00 
     +0.00      +1.00      +4.00      +9.00     +16.00 
     +0.00      +1.00      +1.00      +1.00      +1.00 

 A_TA :
   +355.00   +4890.00    +100.00    +354.00     +30.00 
  +4890.00  +72354.00   +1300.00   +4890.00    +354.00 
   +100.00   +1300.00     +30.00    +100.00     +10.00 
   +354.00   +4890.00    +100.00    +354.00     +30.00 
    +30.00    +354.00     +10.00     +30.00      +4.00 

 Press return to continue. 


  -----------------------------------
 inv(A_TA) :
   +1.0000    +0.0000    +0.0000    -1.0000    -0.0000 
   +0.0000    +0.0056    +0.6500    -0.2222    -0.4500 
   +0.0000    +0.6500   +82.5000   -27.2500   -59.4000 
   -1.0000    -0.2222   -27.2500   +10.1389   +19.2500 
   -0.0000    -0.4500   -59.4000   +19.2500   +44.2000 

 inv(A_TA)*A_T :
   +1.0000    +0.0000    +0.0000    +0.0000    +0.0000 
   +0.0000    -0.0167    +0.0500    -0.0500    +0.0167 
   +0.0000    -3.5000    +7.0000    -4.5000    +1.0000 
   -1.0000    +0.9167    -2.2500    +1.7500    -0.4167 
   +0.0000    +3.6000    -4.8000    +2.8000    -0.6000 


 x = inv(A_TA)*A_T*b :
   +1.0000 
   +0.0000 
   +0.0000 
   -1.0000 
   +0.0000 

 Press return to continue. 


  -----------------------------------
 x = inv(A_TA)*A_T*b :
     +1.00 
     +0.00 
     +0.00 
     -1.00 
     +0.00 

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

  +1.00x**2  -1.00y  = 0

 Press return to continue.