Linear Algebra and the C Language/a042
Install and compile this file in your working directory.
/* ------------------------------------ */
/* Save as : c00a.c */
/* ------------------------------------ */
#include "v_a.h"
/* ------------------------------------ */
/* ------------------------------------ */
#define RA R5
#define CA C5
/* ------------------------------------ */
#define FACTOR_E +1.E-0
/* ------------------------------------ */
int main(void)
{
double xy[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(xy,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 = i_mR(CA,RA);
double **Pinvb = 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 ");
Pinv_Rn_mR(A,Pinv,FACTOR_E);
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 ");
mul_mR(Pinv,b,Pinvb);
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;
}
/* ------------------------------------ */
/* ------------------------------------ */
Screen output example:
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.
Copy and paste in Octave:
function xy = f (x,y)
xy = +1.000000000*x^2 -0.166666667*y^2 -5.000000000*x +1.166666667*y +4.000000000;
endfunction
f (+1,+0)
f (+2,+3)
f (+3,+4)
f (+4,+0)