Linear Algebra and the C Language/a0bd

__NOTOC__ Install and compile this file in your working directory. /* ------------------------------------ */ /* Save as : c00d.c */ /* ------------------------------------ */ #include "v_a.h" /* ------------------------------------ */ /* ------------------------------------ */ #define RA R5 #define CA C5 #define Cb C1 /* ------------------------------------ */ /* ------------------------------------ */ int main(void) { double xy[6] ={ 10, 10, -5, 1, 7, -10 }; double ab[RA*(CA+Cb)]={ /* x**2 y**2 x y = 0 */ +1, +0, +0, +0, +0, +1, +0, +1, +0, +0, +0, +1, +100, +100, +10, +10, +1, +0, +25, +1, -5, +1, +1, +0, +49, +100, +7, -10, +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 +10 +10 -5 +1 +7 -10 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 +100.00 +100.00 +10.00 +10.00 +1.00 +0.00 +25.00 +1.00 -5.00 +1.00 +1.00 +0.00 +49.00 +100.00 +7.00 -10.00 +1.00 +0.00 Press return to continue. Q : +0.0088 -0.0213 +0.0046 -0.6732 -0.7391 +0.0000 +0.0186 -0.0191 +0.7387 -0.6735 +0.8761 -0.2704 +0.3985 +0.0196 +0.0028 +0.2190 -0.5131 -0.8298 +0.0014 +0.0109 +0.4293 +0.8141 -0.3901 -0.0270 +0.0039 R : +114.1359 +130.7652 +10.6715 +4.6874 +1.5245 -0.0000 +53.8745 +5.5600 -11.3588 +0.0306 -0.0000 -0.0000 +5.4043 +7.0561 -0.8214 -0.0000 -0.0000 -0.0000 +0.4678 -0.0060 -0.0000 +0.0000 +0.0000 +0.0000 +0.0176 Press return to continue. Q_T : +8.7615e-03 +0.0000e+00 +8.7615e-01 +2.1904e-01 +4.2931e-01 -2.1266e-02 +1.8562e-02 -2.7044e-01 -5.1309e-01 +8.1413e-01 +4.5779e-03 -1.9096e-02 +3.9854e-01 -8.2984e-01 -3.9005e-01 -6.7319e-01 +7.3872e-01 +1.9620e-02 +1.4100e-03 -2.7021e-02 -7.3910e-01 -6.7349e-01 +2.7660e-03 +1.0935e-02 +3.8595e-03 invR : +8.7615e-03 -2.1266e-02 +4.5779e-03 -6.7319e-01 -7.3910e-01 -0.0000e+00 +1.8562e-02 -1.9096e-02 +7.3872e-01 -6.7349e-01 -0.0000e+00 +0.0000e+00 +1.8504e-01 -2.7909e+00 +7.7024e+00 -0.0000e+00 +0.0000e+00 -0.0000e+00 +2.1376e+00 +7.2929e-01 -0.0000e+00 +0.0000e+00 -0.0000e+00 +0.0000e+00 +5.6945e+01 Press return to continue. Solving this system yields a unique least squares solution, namely x = invR * Q_T * b : +1.00 +1.00 -11.07 -0.89 -80.44 The coefficients a, b, c, d of the curve are : +1.00x**2 +1.00y**2 -11.07x -0.89y -80.44 = 0 Press return to continue. Copy and paste in Octave: function xy = f (x,y) xy = +1.000000000*x^2 +1.000000000*y^2 -11.065934068*x -0.890109889*y -80.439560439; endfunction f (+10,+10) f (-5,+1) f (+7,-10)