Linear Algebra and the C Language/a0ba

__NOTOC__ Install and compile this file in your working directory. /* ------------------------------------ */ /* Save as : c00a.c */ /* ------------------------------------ */ #include "v_a.h" /* ------------------------------------ */ /* ------------------------------------ */ #define RA R5 #define CA C5 #define Cb C1 /* ------------------------------------ */ /* ------------------------------------ */ int main(void) { double xy[6] ={ 1, -2, 2, -3, 3, 6 }; 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, +4, +1, -2, +1, +0, +4, +9, +2, -3, +1, +0, +9, +36, +3, +6, +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"); printf("invR*Q_T :"); mul_mR(invR,Q_T,invR_Q_T); p_mR(invR_Q_T,S10,P4,C6); 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 -2 +2 -3 +3 +6 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 +4.00 +1.00 -2.00 +1.00 +0.00 +4.00 +9.00 +2.00 -3.00 +1.00 +0.00 +9.00 +36.00 +3.00 +6.00 +1.00 +0.00 Press return to continue. Q : +0.1005 -0.4928 -0.5780 -0.3248 +0.5544 +0.0000 +0.1340 +0.0472 -0.9182 -0.3696 +0.1005 +0.0433 +0.7229 -0.2169 +0.6468 +0.4020 -0.7650 +0.3370 +0.0544 -0.3696 +0.9045 +0.3899 -0.1659 +0.0360 +0.0308 R : +9.9499 +36.5834 +3.6181 +4.0202 +1.4071 -0.0000 +7.4603 -0.3168 +4.5480 -0.3317 -0.0000 -0.0000 +0.8993 -3.4522 +0.8940 +0.0000 +0.0000 -0.0000 +0.4863 -0.1264 +0.0000 +0.0000 +0.0000 +0.0000 +0.3080 Press return to continue. Q_T : +1.0050e-01 +0.0000e+00 +1.0050e-01 +4.0202e-01 +9.0453e-01 -4.9284e-01 +1.3404e-01 +4.3327e-02 -7.6499e-01 +3.8994e-01 -5.7800e-01 +4.7225e-02 +7.2290e-01 +3.3703e-01 -1.6589e-01 -3.2485e-01 -9.1825e-01 -2.1687e-01 +5.4449e-02 +3.5992e-02 +5.5444e-01 -3.6962e-01 +6.4684e-01 -3.6962e-01 +3.0802e-02 invR : +1.0050e-01 -4.9284e-01 -5.7800e-01 -3.2485e-01 +5.5444e-01 +0.0000e+00 +1.3404e-01 +4.7225e-02 -9.1825e-01 -3.6962e-01 -0.0000e+00 +0.0000e+00 +1.1120e+00 +7.8932e+00 +1.2321e-02 -0.0000e+00 +0.0000e+00 -0.0000e+00 +2.0561e+00 +8.4398e-01 -0.0000e+00 +0.0000e+00 -0.0000e+00 -0.0000e+00 +3.2465e+00 Press return to continue. Solving this system yields a unique least squares solution, namely invR*Q_T : +1.0000 -0.0000 -0.0000 +0.0000 +0.0000 -0.0000 +1.0000 -0.0000 -0.0000 -0.0000 -3.2000 -7.2000 -0.9000 +0.8000 +0.1000 -0.2000 -2.2000 +0.1000 -0.2000 +0.1000 +1.8000 -1.2000 +2.1000 -1.2000 +0.1000 x = invR * Q_T * b : +1.00 +1.00 -10.40 -2.40 +0.60 The coefficients a, b, c, d of the curve are : +1.00x**2 +1.00y**2 -10.40x -2.40y +0.60 = 0 Press return to continue. Copy and paste in Octave: function xy = f (x,y) xy = +1.00*x^2 +1.00*y^2 -10.40*x -2.40*y +0.60; endfunction f (+1,-2) f (+2,-3) f (+3,+6)