Mathc complexes/06i
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c00b.c |
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/* ------------------------------------ */
/* Save as : c00b.c */
/* ------------------------------------ */
#include "w_a.h"
#include "d.h"
/* ------------------------------------ */
#define RA R5
#define CA C5
#define Cb C1
/* ------------------------------------ */
/* ------------------------------------ */
int main(void)
{
double xy[R4*(C2*C2)] ={
1, 1,
2, 4,
3, 9,
4, 16 };
double ab[RA*(CA+Cb)]={
/* x**2 y**2 x y e = +0 */
+1, +0, +0, +0, +0, +1,
+1, +1, +1, +1, +1, +0,
+4, +16, +2, +4, +1, +0,
+9, +81, +3, +9, +1, +0,
+16, +256, +4, +16, +1, +0
};
double **XY = ca_A_mRZ(xy,i_mZ(R4,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, 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_mRZ(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_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 = 0\n\n"
,x[R1][C1],x[R4][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, e, of the curve
ax**2 + by**2 + cx + dy + e = 0
that passes through these four points.
x y
+1 +1
+2 +4
+3 +9
+4 +16
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 +1.00 +1.00 +1.00 +1.00 +0.00
+4.00 +16.00 +2.00 +4.00 +1.00 +0.00
+9.00 +81.00 +3.00 +9.00 +1.00 +0.00
+16.00 +256.00 +4.00 +16.00 +1.00 +0.00
Press return to continue.
Q :
+0.0531 -0.1949 -0.6241 -0.7548 +0.0000
+0.0531 -0.1807 +0.6555 -0.4915 +0.5415
+0.2123 -0.5532 +0.3364 -0.1204 -0.7220
+0.4777 -0.6080 -0.2539 +0.4005 +0.4212
+0.8492 +0.5037 +0.0568 -0.1173 -0.0902
R :
+18.8414 +259.5343 +5.3074 +18.7884 +1.5922
+0.0000 +70.6822 -1.0960 +0.1949 -0.8381
-0.0000 -0.0000 +0.7936 +0.6241 +0.7948
+0.0000 +0.0000 +0.0000 +0.7548 -0.3287
+0.0000 +0.0000 +0.0000 +0.0000 +0.1504
Press return to continue.
Q_T :
+5.3074e-02 +5.3074e-02 +2.1230e-01 +4.7767e-01 +8.4919e-01
-1.9488e-01 -1.8073e-01 -5.5316e-01 -6.0796e-01 +5.0374e-01
-6.2405e-01 +6.5551e-01 +3.3642e-01 -2.5391e-01 +5.6750e-02
-7.5483e-01 -4.9155e-01 -1.2040e-01 +4.0047e-01 -1.1727e-01
+3.1185e-14 +5.4149e-01 -7.2199e-01 +4.2116e-01 -9.0249e-02
invR :
+5.3074e-02 -1.9488e-01 -6.2405e-01 -7.5483e-01 +0.0000e+00
+0.0000e+00 +1.4148e-02 +1.9537e-02 -1.9805e-02 -6.7686e-02
+0.0000e+00 -0.0000e+00 +1.2600e+00 -1.0417e+00 -8.9346e+00
-0.0000e+00 +0.0000e+00 -0.0000e+00 +1.3248e+00 +2.8955e+00
+0.0000e+00 +0.0000e+00 +0.0000e+00 -0.0000e+00 +6.6483e+00
Press return to continue.
Solving this system yields a unique
least squares solution, namely
x = invR * Q_T * b :
+1.00
-0.00
+0.00
-1.00
+0.00
The coefficients a, b, c, d, e, of the curve are :
+1.00x**2 -1.00y = 0
Press return to continue.