Mathc complexes/06v
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c00d.c |
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/* ------------------------------------ */
/* Save as : c00d.c */
/* ------------------------------------ */
#include "w_a.h"
/* ------------------------------------ */
#define RA R5
#define CA C5
#define Cb C1
/* ------------------------------------ */
/* ------------------------------------ */
int main(void)
{
double xy[6] ={
10, 10,
-5, 1,
7, -10 };
double ab[RA*(CA+Cb)]={
/* x**2 y**2 x y = 0 */
+1, +0, +0, +0, +0, +1,
+0, +1, +0, +0, +0, +1,
+100, +100, +10, +10, +1, +0,
+25, +1, -5, +1, +1, +0,
+49, +100, +7, -10, +1, +0,
};
double **XY = ca_A_mRZ(xy,i_mZ(R3,C2));
double **Ab = ca_A_mRZ(ab,i_Abr_Ac_bc_mZ(RA,CA,Cb));
double **A = c_Ab_A_mZ(Ab,i_mZ(RA,CA));
double **b = c_Ab_b_mZ(Ab,i_mZ(RA,Cb));
double **Q = i_mZ(RA,CA);
double **R = i_mZ(CA,CA);
double **invR = i_mZ(CA,CA);
double **Q_T = i_mZ(CA,RA);
double **invR_Q_T = i_mZ(CA,RA);
double **x = i_mZ(CA,Cb); // x invR * Q_T * b
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_mRZ(XY,S5,P0,C6);
printf("\n");
printf(" Using the given XY, we obtain this matrix.\n");
printf(" (a = 1. This is my choice)\n\n");
printf(" x**2 y**2 x y c");
p_mRZ(Ab,S7,P2,C6);
stop();
clrscrn();
QR_mZ(A,Q,R);
printf(" Q :");
p_mRZ(Q,S10,P4,C6);
printf(" R :");
p_mRZ(R,S10,P4,C6);
stop();
clrscrn();
transpose_mZ(Q,Q_T);
printf(" Q_T :");
pE_mRZ(Q_T,S12,P4,C6);
inv_mZ(R,invR);
printf(" invR :");
pE_mRZ(invR,S12,P4,C6);
stop();
clrscrn();
printf(" Solving this system yields a unique\n"
" least squares solution, namely \n\n");
mul_mZ(invR,Q_T,invR_Q_T);
mul_mZ(invR_Q_T,b,x);
printf(" x = invR * Q_T * b :");
p_mRZ(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_mZ(XY);
f_mZ(A);
f_mZ(b);
f_mZ(Ab);
f_mZ(Q);
f_mZ(Q_T);
f_mZ(R);
f_mZ(invR);
f_mZ(invR_Q_T);
f_mZ(x);
return 0;
}
/* ------------------------------------ */
/* ------------------------------------ */
Exemple de sortie écran :
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
+10 +10
-5 +1
+7 -10
Using the given XY, we obtain this matrix.
(a = 1. This is my choice)
x**2 y**2 x y c
+1.00 +0.00 +0.00 +0.00 +0.00 +1.00
+0.00 +1.00 +0.00 +0.00 +0.00 +1.00
+100.00 +100.00 +10.00 +10.00 +1.00 +0.00
+25.00 +1.00 -5.00 +1.00 +1.00 +0.00
+49.00 +100.00 +7.00 -10.00 +1.00 +0.00
Press return to continue.
Q :
+0.0088 -0.0213 +0.0046 -0.6732 -0.7391
+0.0000 +0.0186 -0.0191 +0.7387 -0.6735
+0.8761 -0.2704 +0.3985 +0.0196 +0.0028
+0.2190 -0.5131 -0.8298 +0.0014 +0.0109
+0.4293 +0.8141 -0.3901 -0.0270 +0.0039
R :
+114.1359 +130.7652 +10.6715 +4.6874 +1.5245
-0.0000 +53.8745 +5.5600 -11.3588 +0.0306
-0.0000 -0.0000 +5.4043 +7.0561 -0.8214
+0.0000 +0.0000 +0.0000 +0.4678 -0.0060
+0.0000 +0.0000 +0.0000 -0.0000 +0.0176
Press return to continue.
Q_T :
+8.7615e-03 +0.0000e+00 +8.7615e-01 +2.1904e-01 +4.2931e-01
-2.1266e-02 +1.8562e-02 -2.7044e-01 -5.1309e-01 +8.1413e-01
+4.5779e-03 -1.9096e-02 +3.9854e-01 -8.2984e-01 -3.9005e-01
-6.7319e-01 +7.3872e-01 +1.9620e-02 +1.4100e-03 -2.7021e-02
-7.3910e-01 -6.7349e-01 +2.7660e-03 +1.0935e-02 +3.8595e-03
invR :
+8.7615e-03 -2.1266e-02 +4.5779e-03 -6.7319e-01 -7.3910e-01
-0.0000e+00 +1.8562e-02 -1.9096e-02 +7.3872e-01 -6.7349e-01
+0.0000e+00 +0.0000e+00 +1.8504e-01 -2.7909e+00 +7.7024e+00
-0.0000e+00 +0.0000e+00 -0.0000e+00 +2.1376e+00 +7.2929e-01
+0.0000e+00 +0.0000e+00 +0.0000e+00 +0.0000e+00 +5.6945e+01
Press return to continue.
Solving this system yields a unique
least squares solution, namely
x = invR * Q_T * b :
+1.00
+1.00
-11.07
-0.89
-80.44
The coefficients a, b, c, d, e, of the curve are :
+1.00x**2 +1.00y**2 -11.07x -0.89y -80.44 = 0
Press return to continue.