Linear Algebra and the C Language/a0bf
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[8] ={
1, -8,
2, 2,
3, 1,
4, 2 };
double ab[RA*(CA+Cb)]={
/* x**2 y**2 x y e = 0 */
+1, +0, +0, +0, +0, +1,
+1, +64, +1, -8, +1, +0,
+4, +4, +2, +2, +1, +0,
+9, +1, +3, +1, +1, +0,
+16, +4, +4, +2, +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 -8
+2 +2
+3 +1
+4 +2
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 +64.00 +1.00 -8.00 +1.00 +0.00
+4.00 +4.00 +2.00 +2.00 +1.00 +0.00
+9.00 +1.00 +3.00 +1.00 +1.00 +0.00
+16.00 +4.00 +4.00 +2.00 +1.00 +0.00
Press return to continue.
Q :
+0.0531 -0.0068 -0.2458 +0.1164 +0.9608
+0.0531 +0.9973 -0.0277 -0.0429 +0.0022
+0.2123 +0.0357 +0.7522 +0.6136 +0.1066
+0.4777 -0.0452 +0.4432 -0.7365 +0.1759
+0.8492 -0.0454 -0.4203 +0.2563 -0.1858
R :
+18.8414 +8.1204 +5.3074 +2.1761 +1.5922
+0.0000 +63.7421 +0.7515 -8.0429 +0.9424
-0.0000 -0.0000 +1.1253 +1.3284 +0.7475
+0.0000 +0.0000 +0.0000 +1.3462 +0.0905
+0.0000 +0.0000 +0.0000 -0.0000 +0.0989
Press return to continue.
Q_T :
+5.3074e-02 +5.3074e-02 +2.1230e-01 +4.7767e-01 +8.4919e-01
-6.7614e-03 +9.9728e-01 +3.5707e-02 -4.5164e-02 -4.5430e-02
-2.4581e-01 -2.7670e-02 +7.5217e-01 +4.4320e-01 -4.2025e-01
+1.1638e-01 -4.2860e-02 +6.1361e-01 -7.3649e-01 +2.5628e-01
+9.6082e-01 +2.1987e-03 +1.0664e-01 +1.7589e-01 -1.8579e-01
invR :
+5.3074e-02 -6.7614e-03 -2.4581e-01 +1.1638e-01 +9.6082e-01
+0.0000e+00 +1.5688e-02 -1.0477e-02 +1.0407e-01 -1.6551e-01
+0.0000e+00 +0.0000e+00 +8.8867e-01 -8.7694e-01 -5.9111e+00
-0.0000e+00 -0.0000e+00 -0.0000e+00 +7.4285e-01 -6.7975e-01
-0.0000e+00 -0.0000e+00 -0.0000e+00 +0.0000e+00 +1.0107e+01
Press return to continue.
Solving this system yields a unique
least squares solution, namely
x = invR * Q_T * b :
+1.00
-0.14
-6.00
-0.57
+9.71
The coefficients a, b, c, d, e, of the curve are :
+1.00x**2 -0.14y**2 -6.00x -0.57y +9.71 = 0
Press return to continue.
Copy and paste in Octave:
function xy = f (x,y)
xy = +1.000000000*x^2 -0.144444444*y^2 -6.000000000*x -0.566666667*y +9.711111111;
endfunction
f (1, -8)
f (2, 2)
f (3, 1)
f (4, 2)