Linear Algebra and the C Language/a0bc


Install and compile this file in your working directory.

/* ------------------------------------ */
/*  Save as :   c00c.c                  */
/* ------------------------------------ */
#include "v_a.h"
/* ------------------------------------ */
/* ------------------------------------ */
#define   RA R5
#define   CA C5
#define   Cb C1 
/* ------------------------------------ */
/* ------------------------------------ */
int main(void)
{
double xy[6] ={
   1,     10,
   2,      1,
   3,    -10    };

   
double ab[RA*(CA+Cb)]={
/* x**2    y**2    x       y            =  0   */
  +1,     +0,     +0,     +0,     +0,     +1,   
  +0,     +1,     +0,     +0,     +0,     +1,   
  +1,   +100,     +1,    +10,     +1,     +0,   
  +4,     +1,     +2,     +1,     +1,     +0,   
  +9,   +100,     +3,    -10,     +1,     +0,          
}; 

double **XY = ca_A_mR(xy,i_mR(R3,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 **Q    = i_mR(RA,CA);
double **R    = i_mR(CA,CA);

double **invR = i_mR(CA,CA);
double **Q_T  = i_mR(CA,RA);


double **invR_Q_T = i_mR(CA,RA);
double **x        = i_mR(CA,Cb); // x = invR * Q_T * b

  clrscrn();
  printf("\n");
  printf(" Find the coefficients a, b, c, d   of a circle  \n\n");
  printf("     ax**2 + ay**2 + bx + cy + d  = 0            \n\n");
  printf(" that passes through these three XY.             \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            =  0     ");
  p_mR(Ab,S7,P2,C6);
  stop();

    
  clrscrn();
  QR_mR(A,Q,R);    
  printf(" Q :");
  p_mR(Q,S10,P4,C6); 
  printf(" R :");
  p_mR(R,S10,P4,C6);
  stop();

  clrscrn();
  transpose_mR(Q,Q_T);   
  printf(" Q_T :");
  pE_mR(Q_T,S12,P4,C6); 
  invgj_mR(R,invR); 
  printf(" invR :");
  pE_mR(invR,S12,P4,C6);
  stop();
  
  clrscrn();
  printf(" Solving this system yields a unique\n"
         " least squares solution, namely   \n\n");
  mul_mR(invR,Q_T,invR_Q_T);
  mul_mR(invR_Q_T,b,x);
  printf(" x = invR * Q_T * b :");
  p_mR(x,S10,P2,C6);
  printf(" The coefficients a, b, c, d  of the curve are : \n\n"
         "  %+.2fx**2 %+.2fy**2 %+.2fx %+.2fy %+.2f = 0\n\n"
            ,x[R1][C1],x[R2][C1],x[R3][C1],x[R4][C1],x[R5][C1]);  
  
  stop();

  f_mR(XY);  
  f_mR(A);
  f_mR(b);
  f_mR(Ab);
  f_mR(Q);
  f_mR(Q_T);
  f_mR(R);
  f_mR(invR);  
  f_mR(invR_Q_T); 
  f_mR(x); 

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


Screen output example:
 Find the coefficients a, b, c, d   of a circle  

     ax**2 + ay**2 + bx + cy + d  = 0            

 that passes through these three XY.             

    x     y
   +1   +10 
   +2    +1 
   +3   -10 

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

 Ab :
   x**2    y**2    x       y            =  0     
  +1.00   +0.00   +0.00   +0.00   +0.00   +1.00 
  +0.00   +1.00   +0.00   +0.00   +0.00   +1.00 
  +1.00 +100.00   +1.00  +10.00   +1.00   +0.00 
  +4.00   +1.00   +2.00   +1.00   +1.00   +0.00 
  +9.00 +100.00   +3.00  -10.00   +1.00   +0.00 

 Press return to continue. 


 Q :
   +0.1005    -0.1023    -0.3709    +0.9175    -0.0082 
   +0.0000    +0.0101    -0.0043    -0.0095    -0.9999 
   +0.1005    +0.9068    +0.3402    +0.2277    +0.0055 
   +0.4020    -0.3993    +0.7909    +0.2310    -0.0096 
   +0.9045    +0.0881    -0.3481    -0.2299    +0.0046 

 R :
   +9.9499  +100.9058    +3.6181    -7.6383    +1.4071 
   -0.0000   +99.0960    +0.3725    +7.7879    +0.5956 
   -0.0000    +0.0000    +0.8777    +7.6742    +0.7830 
   +0.0000    -0.0000    +0.0000    +4.8075    +0.2288 
   +0.0000    +0.0000    +0.0000    +0.0000    +0.0005 

 Press return to continue. 


 Q_T :
 +1.0050e-01  +0.0000e+00  +1.0050e-01  +4.0202e-01  +9.0453e-01 
 -1.0234e-01  +1.0091e-02  +9.0678e-01  -3.9927e-01  +8.8069e-02 
 -3.7088e-01  -4.2823e-03  +3.4023e-01  +7.9089e-01  -3.4810e-01 
 +9.1750e-01  -9.5115e-03  +2.2771e-01  +2.3104e-01  -2.2993e-01 
 -8.1722e-03  -9.9989e-01  +5.5283e-03  -9.6144e-03  +4.5668e-03 

 invR :
 +1.0050e-01  -1.0234e-01  -3.7088e-01  +9.1750e-01  -8.1722e-03 
 +0.0000e+00  +1.0091e-02  -4.2823e-03  -9.5115e-03  -9.9989e-01 
 +0.0000e+00  -0.0000e+00  +1.1393e+00  -1.8187e+00  -9.9009e+02 
 +0.0000e+00  -0.0000e+00  +0.0000e+00  +2.0801e-01  -9.9012e+01 
 -0.0000e+00  +0.0000e+00  -0.0000e+00  +0.0000e+00  +2.0802e+03 

 Press return to continue. 


 Solving this system yields a unique
 least squares solution, namely   

 x = invR * Q_T * b :
     +1.00 
     +1.00 
   +996.00 
   +100.00 
  -2097.00 

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

  +1.00x**2 +1.00y**2 +996.00x +100.00y -2097.00 = 0

 Press return to continue.


Copy and paste in Octave:
function xy = f (x,y)
  xy = +1.00*x^2 +1.00*y^2 +996.00*x +100.00*y -2097.00;
endfunction

f (+1,+10)  
f (+2,+1)  
f (+3,-10)