Mathc matrices/c23n
Application ou QR décomposition
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c00o.c |
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/* ------------------------------------ */
/* Save as : c00o.c */
/* ------------------------------------ */
#include "v_a.h"
/* ------------------------------------ */
/* ------------------------------------ */
#define RA R5
#define CA C5
#define Cb C1
/* ------------------------------------ */
/* ------------------------------------ */
void fun(void)
{
double xy[8] ={
1, 1,
2, 4,
3, 9,
4, 16 };
double ab[RA*(CA+Cb)]={
/* 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
};
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 = 0\n\n"
,x[R1][C1],x[R4][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);
}
/* ------------------------------------ */
int main(void)
{
fun();
return 0;
}
/* ------------------------------------ */
/* ------------------------------------ */
Trouver les coefficients a, b, c, d, e du conique,
ax**2 + by**2 + cx + dy + e = 0
qui passe par ces quatre points.
(x[1],y[1]) (x[2],y[2]) (x[3],y[3]) (x[4],y[4])
En utilisant les quatre points nous obtenons la matrice.
(a)x**2 (b)y**2 (c)x (d)y (e) = 0
x[1]**2 y[1]**2 x[1] y[1] 1 0
x[2]**2 y[2]**2 x[2] y[2] 1 0
x[3]**2 y[3]**2 x[3] y[3] 1 0
x[4]**2 y[4]**2 x[4] y[4] 1 0
Ce système à quatre lignes et cinq inconnus (a, b, c, d, e).
C'est un système homogène, il a donc une infinité de solution.
Pour trouver une solution j'ai choisi de poser que a = 1.
Nous avons donc cinq lignes et cinq inconnus.
1 0 0 0 0 1
x[1]**2 y[1]**2 x[1] y[1] 1 0
x[2]**2 y[2]**2 x[2] y[2] 1 0
x[3]**2 y[3]**2 x[3] y[3] 1 0
x[4]**2 y[4]**2 x[4] y[4] 1 0
Il suffit maintenant de résoudre le système.
On utilise la QR décomposition. Dans l'exemple suivant on utilise un autre algorithme sur le même exemple. 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
+3.1923e-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.