Linear Algebra and the C Language/a0bg

__NOTOC__ Install and compile this file in your working directory. /* ------------------------------------ */ /* Save as : c00c.c */ /* ------------------------------------ */ #include "v_a.h" /* ------------------------------------ */ /* ------------------------------------ */ #define RA R5 #define CA C5 #define Cb C1 /* ------------------------------------ */ /* ------------------------------------ */ int main(void) { double xy[8] ={ 1, 4, 2, 5, 3, -7, 4, 5 }; double ab[RA*(CA+Cb)]={ /* x**2 y**2 x y e = 0 */ +1, +0, +0, +0, +0, +1, +1, +16, +1, +4, +1, +0, +4, +25, +2, +5, +1, +0, +9, +49, +3, -7, +1, +0, +16, +25, +4, +5, +1, +0 }; 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); 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, 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_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, e, of the curve ax**2 + by**2 + cx + dy + e = 0 that passes through these four points. x y +1 +4 +2 +5 +3 -7 +4 +5 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 +16.00 +1.00 +4.00 +1.00 +0.00 +4.00 +25.00 +2.00 +5.00 +1.00 +0.00 +9.00 +49.00 +3.00 -7.00 +1.00 +0.00 +16.00 +25.00 +4.00 +5.00 +1.00 +0.00 Press return to continue. Q : +0.0531 -0.0740 -0.3021 +0.8975 -0.3081 +0.0531 +0.3652 +0.4419 +0.4171 +0.7033 +0.2123 +0.3903 +0.6371 +0.0179 -0.6296 +0.4777 +0.6791 -0.5380 -0.1419 +0.0335 +0.8492 -0.4977 +0.1346 -0.0069 +0.1139 R : +18.8414 +50.7923 +5.3074 +2.1761 +1.5922 +0.0000 +36.4300 +1.1919 -3.8300 +0.9368 -0.0000 -0.0000 +0.6405 +9.3923 +0.6756 +0.0000 +0.0000 +0.0000 +2.7168 +0.2863 -0.0000 -0.0000 -0.0000 +0.0000 +0.2210 Press return to continue. Q_T : +5.3074e-02 +5.3074e-02 +2.1230e-01 +4.7767e-01 +8.4919e-01 -7.3999e-02 +3.6520e-01 +3.9025e-01 +6.7906e-01 -4.9773e-01 -3.0208e-01 +4.4185e-01 +6.3712e-01 -5.3802e-01 +1.3462e-01 +8.9751e-01 +4.1711e-01 +1.7931e-02 -1.4185e-01 -6.8557e-03 -3.0810e-01 +7.0327e-01 -6.2960e-01 +3.3489e-02 +1.1386e-01 invR : +5.3074e-02 -7.3999e-02 -3.0208e-01 +8.9751e-01 -3.0810e-01 +0.0000e+00 +2.7450e-02 -5.1082e-02 +2.1530e-01 -2.3912e-01 +0.0000e+00 -0.0000e+00 +1.5613e+00 -5.3975e+00 +2.2203e+00 +0.0000e+00 -0.0000e+00 +0.0000e+00 +3.6808e-01 -4.7684e-01 -0.0000e+00 +0.0000e+00 -0.0000e+00 +0.0000e+00 +4.5243e+00 Press return to continue. Solving this system yields a unique least squares solution, namely x = invR * Q_T * b : +1.00 +0.28 -6.00 +0.48 -1.39 The coefficients a, b, c, d, e, of the curve are : +1.00x**2 +0.28y**2 -6.00x +0.48y -1.39 = 0 Press return to continue. Copy and paste in Octave: function xy = f (x,y) xy = +1.000000000*x^2 +0.280303030*y^2 -6.000000000*x +0.477272727*y -1.393939394; endfunction f (+1,+4) f (+2,+5) f (+3,-7) f (+4,+5)