Linear Algebra and the C Language/a0bh
Install and compile this file in your working directory.
/* ------------------------------------ */
/* Save as : c00d.c */
/* ------------------------------------ */
#include "v_a.h"
/* ------------------------------------ */
/* ------------------------------------ */
#define RA R5
#define CA C5
#define Cb C1
/* ------------------------------------ */
/* ------------------------------------ */
int main(void)
{
double xy[8] ={
1, 2,
2, -8,
3, -8,
4, -3 };
double ab[RA*(CA+Cb)]={
/* x**2 y**2 x y e = 0 */
+1, +0, +0, +0, +0, +1,
+1, +4, +1, +2, +1, +0,
+4, +64, +2, -8, +1, +0,
+9, +64, +3, -8, +1, +0,
+16, +9, +4, -3, +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 +2
+2 -8
+3 -8
+4 -3
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 +4.00 +1.00 +2.00 +1.00 +0.00
+4.00 +64.00 +2.00 -8.00 +1.00 +0.00
+9.00 +64.00 +3.00 -8.00 +1.00 +0.00
+16.00 +9.00 +4.00 -3.00 +1.00 +0.00
Press return to continue.
Q :
+0.0531 -0.0369 -0.3184 +0.6481 +0.6888
+0.0531 +0.0166 +0.9276 +0.3589 +0.0879
+0.2123 +0.7087 +0.1022 -0.4485 +0.4909
+0.4777 +0.5240 -0.1642 +0.4552 -0.5129
+0.8492 -0.4707 +0.0287 -0.2069 +0.1172
R :
+18.8414 +52.0130 +5.3074 -7.9612 +1.5922
+0.0000 +74.7238 +1.1234 -8.4165 +0.7786
-0.0000 -0.0000 +0.7543 +2.2650 +0.8943
+0.0000 -0.0000 +0.0000 +1.2851 +0.1587
+0.0000 +0.0000 +0.0000 +0.0000 +0.1832
Press return to continue.
Q_T :
+5.3074e-02 +5.3074e-02 +2.1230e-01 +4.7767e-01 +8.4919e-01
-3.6944e-02 +1.6587e-02 +7.0871e-01 +5.2400e-01 -4.7065e-01
-3.1842e-01 +9.2757e-01 +1.0219e-01 -1.6420e-01 +2.8745e-02
+6.4808e-01 +3.5888e-01 -4.4855e-01 +4.5520e-01 -2.0685e-01
+6.8878e-01 +8.7929e-02 +4.9094e-01 -5.1292e-01 +1.1724e-01
invR :
+5.3074e-02 -3.6944e-02 -3.1842e-01 +6.4808e-01 +6.8878e-01
-0.0000e+00 +1.3383e-02 -1.9930e-02 +1.2278e-01 -6.5947e-02
+0.0000e+00 -0.0000e+00 +1.3257e+00 -2.3367e+00 -4.4478e+00
-0.0000e+00 +0.0000e+00 -0.0000e+00 +7.7818e-01 -6.7412e-01
-0.0000e+00 +0.0000e+00 -0.0000e+00 +0.0000e+00 +5.4589e+00
Press return to continue.
Solving this system yields a unique
least squares solution, namely
x = invR * Q_T * b :
+1.00
+0.04
-5.00
+0.04
+3.76
The coefficients a, b, c, d, e, of the curve are :
+1.00x**2 +0.04y**2 -5.00x +0.04y +3.76 = 0
Press return to continue.
Copy and paste in Octave:
function xy = f (x,y)
xy = +1.00*x^2 +0.04*y^2 -5.00*x +0.04*y +3.76;
endfunction
f (+1,+2)
f (+2,-8)
f (+3,-8)
f (+4,-3)