Linear Algebra and the C Language/a043
Install and compile this file in your working directory.
/* ------------------------------------ */
/* Save as : c00b.c */
/* ------------------------------------ */
#include "v_a.h"
/* ------------------------------------ */
/* ------------------------------------ */
#define RA R5
#define CA C5
/* ------------------------------------ */
#define FACTOR_E +1.E-0
/* ------------------------------------ */
int main(void)
{
double xy[8] ={
1, -8,
2, 2,
3, 1,
4, 2 };
double tA[RA*CA]={
/* x**2 y**2 x y e */
+1, +0, +0, +0, +0,
+1, +64, +1, -8, +1,
+4, +4, +2, +2, +1,
+9, +1, +3, +1, +1,
+16, +4, +4, +2, +1,
};
double tb[RA*C1]={
/* = 0 */
+1,
+0,
+0,
+0,
+0
};
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 -8
+2 +2
+3 +1
+4 +2
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 +64.00 +1.00 -8.00 +1.00
+4.00 +4.00 +2.00 +2.00 +1.00
+9.00 +1.00 +3.00 +1.00 +1.00
+16.00 +4.00 +4.00 +2.00 +1.00
b :
+1.00
+0.00
+0.00
+0.00
+0.00
Pinv = V * invS_T * U_T
+1.0000e+00 +1.1983e-15 -2.5951e-15 +4.9571e-14 -2.9560e-14
-1.4444e-01 +1.1111e-02 +3.8889e-02 -1.1111e-01 +6.1111e-02
-6.0000e+00 -1.0143e-14 -5.0000e-01 -4.2810e-13 +5.0000e-01
-5.6667e-01 -3.3333e-02 +3.8333e-01 -6.6667e-01 +3.1667e-01
+9.7111e+00 +2.2222e-02 +1.0778e+00 +1.7778e+00 -1.8778e+00
Press return to continue.
Solving this system yields a unique
least squares solution, namely
Pinv * b
+1.0000
-0.1444
-6.0000
-0.5667
+9.7111
The coefficients a, b, c, d, e, of the curve are :
+1.000000000*x^2 -0.144444444*y^2 -6.000000000*x -0.566666667*y +9.711111111 = 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)