Linear Algebra and the C Language/a0bg
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[8] ={
1, 4,
2, 5,
3, -7,
4, 5 };
double ab[RA*(CA+Cb)]={
/* x**2 y**2 x y e = 0 */
+1, +0, +0, +0, +0, +1,
+1, +16, +1, +4, +1, +0,
+4, +25, +2, +5, +1, +0,
+9, +49, +3, -7, +1, +0,
+16, +25, +4, +5, +1, +0
};
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 **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, 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();
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);
inv_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, e, 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, e, of the curve
ax**2 + by**2 + cx + dy + e = 0
that passes through these four points.
x y
+1 +4
+2 +5
+3 -7
+4 +5
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 +16.00 +1.00 +4.00 +1.00 +0.00
+4.00 +25.00 +2.00 +5.00 +1.00 +0.00
+9.00 +49.00 +3.00 -7.00 +1.00 +0.00
+16.00 +25.00 +4.00 +5.00 +1.00 +0.00
Press return to continue.
Q :
+0.0531 -0.0740 -0.3021 +0.8975 -0.3081
+0.0531 +0.3652 +0.4419 +0.4171 +0.7033
+0.2123 +0.3903 +0.6371 +0.0179 -0.6296
+0.4777 +0.6791 -0.5380 -0.1419 +0.0335
+0.8492 -0.4977 +0.1346 -0.0069 +0.1139
R :
+18.8414 +50.7923 +5.3074 +2.1761 +1.5922
+0.0000 +36.4300 +1.1919 -3.8300 +0.9368
-0.0000 -0.0000 +0.6405 +9.3923 +0.6756
+0.0000 +0.0000 +0.0000 +2.7168 +0.2863
-0.0000 -0.0000 -0.0000 +0.0000 +0.2210
Press return to continue.
Q_T :
+5.3074e-02 +5.3074e-02 +2.1230e-01 +4.7767e-01 +8.4919e-01
-7.3999e-02 +3.6520e-01 +3.9025e-01 +6.7906e-01 -4.9773e-01
-3.0208e-01 +4.4185e-01 +6.3712e-01 -5.3802e-01 +1.3462e-01
+8.9751e-01 +4.1711e-01 +1.7931e-02 -1.4185e-01 -6.8557e-03
-3.0810e-01 +7.0327e-01 -6.2960e-01 +3.3489e-02 +1.1386e-01
invR :
+5.3074e-02 -7.3999e-02 -3.0208e-01 +8.9751e-01 -3.0810e-01
+0.0000e+00 +2.7450e-02 -5.1082e-02 +2.1530e-01 -2.3912e-01
+0.0000e+00 -0.0000e+00 +1.5613e+00 -5.3975e+00 +2.2203e+00
+0.0000e+00 -0.0000e+00 +0.0000e+00 +3.6808e-01 -4.7684e-01
-0.0000e+00 +0.0000e+00 -0.0000e+00 +0.0000e+00 +4.5243e+00
Press return to continue.
Solving this system yields a unique
least squares solution, namely
x = invR * Q_T * b :
+1.00
+0.28
-6.00
+0.48
-1.39
The coefficients a, b, c, d, e, of the curve are :
+1.00x**2 +0.28y**2 -6.00x +0.48y -1.39 = 0
Press return to continue.
Copy and paste in Octave:
function xy = f (x,y)
xy = +1.000000000*x^2 +0.280303030*y^2 -6.000000000*x +0.477272727*y -1.393939394;
endfunction
f (+1,+4)
f (+2,+5)
f (+3,-7)
f (+4,+5)