Linear Algebra and the C Language/a0bb
Install and compile this file in your working directory.
/* ------------------------------------ */
/* Save as : c00b.c */
/* ------------------------------------ */
#include "v_a.h"
/* ------------------------------------ */
/* ------------------------------------ */
#define RA R5
#define CA C5
#define Cb C1
/* ------------------------------------ */
/* ------------------------------------ */
int main(void)
{
double xy[6] ={
1, -1,
2, -9,
3, -8 };
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, +1, +1, -1, +1, +0,
+4, +81, +2, -9, +1, +0,
+9, +64, +3, -8, +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 -1
+2 -9
+3 -8
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 +1.00 +1.00 -1.00 +1.00 +0.00
+4.00 +81.00 +2.00 -9.00 +1.00 +0.00
+9.00 +64.00 +3.00 -8.00 +1.00 +0.00
Press return to continue.
Q :
+0.1005 -0.1835 -0.3374 -0.6212 +0.6757
+0.0000 +0.0202 -0.0143 +0.7368 +0.6757
+0.1005 -0.1634 +0.9344 -0.1904 +0.2323
+0.4020 +0.8993 +0.0633 -0.1202 +0.1056
+0.9045 -0.3612 -0.0944 +0.1436 -0.1478
R :
+9.9499 +90.5539 +3.6181 -10.9549 +1.4071
+0.0000 +49.5882 +0.5518 -5.0413 +0.3748
-0.0000 -0.0000 +0.7776 -0.7482 +0.9032
-0.0000 -0.0000 -0.0000 +0.1234 -0.1670
-0.0000 -0.0000 -0.0000 +0.0000 +0.1900
Press return to continue.
Q_T :
+1.0050e-01 +0.0000e+00 +1.0050e-01 +4.0202e-01 +9.0453e-01
-1.8353e-01 +2.0166e-02 -1.6337e-01 +8.9933e-01 -3.6116e-01
-3.3742e-01 -1.4312e-02 +9.3436e-01 +6.3264e-02 -9.4444e-02
-6.2115e-01 +7.3679e-01 -1.9038e-01 -1.2020e-01 +1.4359e-01
+6.7567e-01 +6.7567e-01 +2.3226e-01 +1.0557e-01 -1.4780e-01
invR :
+1.0050e-01 -1.8353e-01 -3.3742e-01 -6.2115e-01 +6.7567e-01
-0.0000e+00 +2.0166e-02 -1.4312e-02 +7.3679e-01 +6.7567e-01
-0.0000e+00 +0.0000e+00 +1.2861e+00 +7.7945e+00 +7.3667e-01
-0.0000e+00 +0.0000e+00 -0.0000e+00 +8.1005e+00 +7.1180e+00
-0.0000e+00 +0.0000e+00 -0.0000e+00 -0.0000e+00 +5.2623e+00
Press return to continue.
Solving this system yields a unique
least squares solution, namely
x = invR * Q_T * b :
+1.00
+1.00
+1.44
+10.56
+7.11
The coefficients a, b, c, d of the curve are :
+1.00x**2 +1.00y**2 +1.44x +10.56y +7.11 = 0
Press return to continue.
Copy and paste in Octave:
function xy = f (x,y)
xy = +0.999999988*x^2 +1.000000219*y^2 +1.444445879*x +10.555557922*y +7.111111809;
endfunction
f (+1,-1)
f (+2,-9)
f (+3,-8)