Mathc complexes/06h
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c00a.c |
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/* ------------------------------------ */
/* Save as : c00a.c */
/* ------------------------------------ */
#include "w_a.h"
#include "d.h"
/* ------------------------------------ */
#define RA R5
#define CA C5
#define Cb C1
/* ------------------------------------ */
/* ------------------------------------ */
int main(void)
{
double xy[8] ={
1, 0,
2, 3,
3, 4,
4, 0 };
double ab[RA*(CA+Cb)]={
/* x**2 y**2 x y e = 0 */
+1, +0, +0, +0, +0, +1,
+1, +0, +1, +0, +1, +0,
+4, +9, +2, +3, +1, +0,
+9, +16, +3, +4, +1, +0,
+16, +0, +4, +0, +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**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, e, of the curve
ax**2 + by**2 + cx + dy + e = 0
that passes through these four points.
x y
+1 +0
+2 +3
+3 +4
+4 +0
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 +0.00 +1.00 +0.00 +1.00 +0.00
+4.00 +9.00 +2.00 +3.00 +1.00 +0.00
+9.00 +16.00 +3.00 +4.00 +1.00 +0.00
+16.00 +0.00 +4.00 +0.00 +1.00 +0.00
Press return to continue.
Q :
+0.0531 -0.0323 -0.2668 +0.1748 +0.9457
+0.0531 -0.0323 +0.7999 -0.5069 +0.3152
+0.2123 +0.4447 +0.4686 +0.7332 -0.0000
+0.4777 +0.7296 -0.2636 -0.4124 +0.0000
+0.8492 -0.5175 -0.0022 +0.0694 -0.0788
R :
+18.8414 +9.5534 +5.3074 +2.5476 +1.5922
+0.0000 +15.6759 +0.9758 +4.2525 +0.6244
-0.0000 -0.0000 +0.9375 +0.3514 +1.0027
-0.0000 -0.0000 +0.0000 +0.5499 -0.1167
+0.0000 +0.0000 +0.0000 +0.0000 +0.2364
Press return to continue.
Q_T :
+5.3074e-02 +5.3074e-02 +2.1230e-01 +4.7767e-01 +8.4919e-01
-3.2345e-02 -3.2345e-02 +4.4475e-01 +7.2957e-01 -5.1753e-01
-2.6681e-01 +7.9987e-01 +4.6856e-01 -2.6357e-01 -2.2010e-03
+1.7476e-01 -5.0692e-01 +7.3320e-01 -4.1242e-01 +6.9449e-02
+9.4573e-01 +3.1524e-01 -6.5153e-15 +5.3414e-15 -7.8811e-02
invR :
+5.3074e-02 -3.2345e-02 -2.6681e-01 +1.7476e-01 +9.4573e-01
+0.0000e+00 +6.3792e-02 -6.6396e-02 -4.5089e-01 -1.0946e-01
+0.0000e+00 +0.0000e+00 +1.0667e+00 -6.8168e-01 -4.8600e+00
-0.0000e+00 +0.0000e+00 -0.0000e+00 +1.8185e+00 +8.9757e-01
+0.0000e+00 -0.0000e+00 +0.0000e+00 -0.0000e+00 +4.2295e+00
Press return to continue.
Solving this system yields a unique
least squares solution, namely
x = invR * Q_T * b :
+1.00
-0.17
-5.00
+1.17
+4.00
The coefficients a, b, c, d, e, of the curve are :
+1.00x**2 -0.17y**2 -5.00x +1.17y +4.00 = 0
Press return to continue.