Linear Algebra and the C Language/a0bb

__NOTOC__ Install and compile this file in your working directory. /* ------------------------------------ */ /* Save as : c00b.c */ /* ------------------------------------ */ #include "v_a.h" /* ------------------------------------ */ /* ------------------------------------ */ #define RA R5 #define CA C5 #define Cb C1 /* ------------------------------------ */ /* ------------------------------------ */ int main(void) { double xy[6] ={ 1, -1, 2, -9, 3, -8 }; double ab[RA*(CA+Cb)]={ /* x**2 y**2 x y = 0 */ +1, +0, +0, +0, +0, +1, +0, +1, +0, +0, +0, +1, +1, +1, +1, -1, +1, +0, +4, +81, +2, -9, +1, +0, +9, +64, +3, -8, +1, +0, }; double **XY = ca_A_mR(xy,i_mR(R3,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 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); 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 = 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); invgj_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 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_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); 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 -1 +2 -9 +3 -8 Using the given points, we obtain this matrix. (a = 1. This is my choice) Ab : x**2 y**2 x y = 0 +1.00 +0.00 +0.00 +0.00 +0.00 +1.00 +0.00 +1.00 +0.00 +0.00 +0.00 +1.00 +1.00 +1.00 +1.00 -1.00 +1.00 +0.00 +4.00 +81.00 +2.00 -9.00 +1.00 +0.00 +9.00 +64.00 +3.00 -8.00 +1.00 +0.00 Press return to continue. Q : +0.1005 -0.1835 -0.3374 -0.6212 +0.6757 +0.0000 +0.0202 -0.0143 +0.7368 +0.6757 +0.1005 -0.1634 +0.9344 -0.1904 +0.2323 +0.4020 +0.8993 +0.0633 -0.1202 +0.1056 +0.9045 -0.3612 -0.0944 +0.1436 -0.1478 R : +9.9499 +90.5539 +3.6181 -10.9549 +1.4071 +0.0000 +49.5882 +0.5518 -5.0413 +0.3748 -0.0000 -0.0000 +0.7776 -0.7482 +0.9032 -0.0000 -0.0000 -0.0000 +0.1234 -0.1670 -0.0000 -0.0000 -0.0000 +0.0000 +0.1900 Press return to continue. Q_T : +1.0050e-01 +0.0000e+00 +1.0050e-01 +4.0202e-01 +9.0453e-01 -1.8353e-01 +2.0166e-02 -1.6337e-01 +8.9933e-01 -3.6116e-01 -3.3742e-01 -1.4312e-02 +9.3436e-01 +6.3264e-02 -9.4444e-02 -6.2115e-01 +7.3679e-01 -1.9038e-01 -1.2020e-01 +1.4359e-01 +6.7567e-01 +6.7567e-01 +2.3226e-01 +1.0557e-01 -1.4780e-01 invR : +1.0050e-01 -1.8353e-01 -3.3742e-01 -6.2115e-01 +6.7567e-01 -0.0000e+00 +2.0166e-02 -1.4312e-02 +7.3679e-01 +6.7567e-01 -0.0000e+00 +0.0000e+00 +1.2861e+00 +7.7945e+00 +7.3667e-01 -0.0000e+00 +0.0000e+00 -0.0000e+00 +8.1005e+00 +7.1180e+00 -0.0000e+00 +0.0000e+00 -0.0000e+00 -0.0000e+00 +5.2623e+00 Press return to continue. Solving this system yields a unique least squares solution, namely x = invR * Q_T * b : +1.00 +1.00 +1.44 +10.56 +7.11 The coefficients a, b, c, d of the curve are : +1.00x**2 +1.00y**2 +1.44x +10.56y +7.11 = 0 Press return to continue. Copy and paste in Octave: function xy = f (x,y) xy = +0.999999988*x^2 +1.000000219*y^2 +1.444445879*x +10.555557922*y +7.111111809; endfunction f (+1,-1) f (+2,-9) f (+3,-8)