Linear Algebra and the C Language/a04b
Install and compile this file in your working directory.
/* ------------------------------------ */
/* Save as: c00a.c */
/* ------------------------------------ */
#include "v_a.h"
/* ------------------------------------ */
#define RA R5
#define CA C5
/* ------------------------------------ */
#define FACTOR_E +1.E-2
/* ------------------------------------ */
int main(void)
{
double txy[6] ={
1, -2,
2, -3,
3, 6 };
double tA[RA*CA]={
/* x**2 y**2 x y e */
+1.00, +0.00, +0.00, +0.00, +0.00,
+0.00, +1.00, +0.00, +0.00, +0.00,
+1.00, +4.00, +1.00, -2.00, +1.00,
+4.00, +9.00, +2.00, -3.00, +1.00,
+9.00, +36.00, +3.00, +6.00, +1.00
};
double tb[RA*C1]={
/* = 0 */
+1.00,
+1.00,
+0.00,
+0.00,
+0.00
};
double **xy = ca_A_mR(txy, i_mR(R3,C2));
double **A = ca_A_mR(tA, i_mR(RA,CA));
double **b = ca_A_mR(tb, i_mR(RA,C1));
double **Pinv = Pinv_Rn_mR(A, i_mR(CA,RA),FACTOR_E);
double **Pinvb = mul_mR(Pinv,b, i_mR(CA,C1));
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);
stop();
clrscrn();
printf(" Using the given xy, 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 ");
pE_mR(Pinv, S12,P4,C10);
stop();
clrscrn();
printf(" Pinv = V invS_T U_T ");
p_mR(Pinv, S10,P4,C10);
printf(" Pinv b ");
p_mR(Pinvb, S10,P4,C10);
printf(" The coefficients a, b, c, d, e, of the curve are: \n\n"
" %+.2f*x^2 %+.2f*y^2 %+.2f*x %+.2f*y %+.2f = 0\n\n"
,Pinvb[R1][C1],Pinvb[R2][C1],Pinvb[R3][C1],
Pinvb[R4][C1],Pinvb[R5][C1]);
stop();
f_mR(xy);
f_mR(A);
f_mR(b);
f_mR(Pinv);
f_mR(Pinvb);
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 -2
+2 -3
+3 +6
Press return to continue.
Using the given XY, we obtain this matrix.
(a = 1. This is my choice)
A :
+1.00 +0.00 +0.00 +0.00 +0.00
+0.00 +1.00 +0.00 +0.00 +0.00
+1.00 +4.00 +1.00 -2.00 +1.00
+4.00 +9.00 +2.00 -3.00 +1.00
+9.00 +36.00 +3.00 +6.00 +1.00
b :
+1.00
+1.00
+0.00
+0.00
+0.00
Pinv = V * invS_T * U_T
+1.0000e+00 +5.4042e-11 +9.1213e-12 -6.6832e-12 -7.8143e-13
-1.7705e-11 +1.0000e+00 -1.1338e-11 +7.5043e-12 +1.2205e-13
-3.2000e+00 -7.2000e+00 -9.0000e-01 +8.0000e-01 +1.0000e-01
-2.0000e-01 -2.2000e+00 +1.0000e-01 -2.0000e-01 +1.0000e-01
+1.8000e+00 -1.2000e+00 +2.1000e+00 -1.2000e+00 +1.0000e-01
Press return to continue.
Pinv = V * invS_T * U_T
+1.0000 +0.0000 +0.0000 -0.0000 -0.0000
-0.0000 +1.0000 -0.0000 +0.0000 +0.0000
-3.2000 -7.2000 -0.9000 +0.8000 +0.1000
-0.2000 -2.2000 +0.1000 -0.2000 +0.1000
+1.8000 -1.2000 +2.1000 -1.2000 +0.1000
Pinv * b
+1.0000
+1.0000
-10.4000
-2.4000
+0.6000
The coefficients a, b, c, d, e, of the curve are :
+1.00*x^2 +1.00*y^2 -10.40*x -2.40*y +0.60 = 0
Press return to continue.
Copy and paste in Octave:
function xy = f (x,y)
xy = +1.00*x^2 +1.00*y^2 -10.40*x -2.40*y +0.60;
endfunction
f (+1,-2)
f (+2,-3)
f (+3,+6)