Linear Algebra and the C Language/a0bh

__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[8] ={ 1, 2, 2, -8, 3, -8, 4, -3 }; double ab[RA*(CA+Cb)]={ /* x**2 y**2 x y e = 0 */ +1, +0, +0, +0, +0, +1, +1, +4, +1, +2, +1, +0, +4, +64, +2, -8, +1, +0, +9, +64, +3, -8, +1, +0, +16, +9, +4, -3, +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 +2 +2 -8 +3 -8 +4 -3 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 +4.00 +1.00 +2.00 +1.00 +0.00 +4.00 +64.00 +2.00 -8.00 +1.00 +0.00 +9.00 +64.00 +3.00 -8.00 +1.00 +0.00 +16.00 +9.00 +4.00 -3.00 +1.00 +0.00 Press return to continue. Q : +0.0531 -0.0369 -0.3184 +0.6481 +0.6888 +0.0531 +0.0166 +0.9276 +0.3589 +0.0879 +0.2123 +0.7087 +0.1022 -0.4485 +0.4909 +0.4777 +0.5240 -0.1642 +0.4552 -0.5129 +0.8492 -0.4707 +0.0287 -0.2069 +0.1172 R : +18.8414 +52.0130 +5.3074 -7.9612 +1.5922 +0.0000 +74.7238 +1.1234 -8.4165 +0.7786 -0.0000 -0.0000 +0.7543 +2.2650 +0.8943 +0.0000 -0.0000 +0.0000 +1.2851 +0.1587 +0.0000 +0.0000 +0.0000 +0.0000 +0.1832 Press return to continue. Q_T : +5.3074e-02 +5.3074e-02 +2.1230e-01 +4.7767e-01 +8.4919e-01 -3.6944e-02 +1.6587e-02 +7.0871e-01 +5.2400e-01 -4.7065e-01 -3.1842e-01 +9.2757e-01 +1.0219e-01 -1.6420e-01 +2.8745e-02 +6.4808e-01 +3.5888e-01 -4.4855e-01 +4.5520e-01 -2.0685e-01 +6.8878e-01 +8.7929e-02 +4.9094e-01 -5.1292e-01 +1.1724e-01 invR : +5.3074e-02 -3.6944e-02 -3.1842e-01 +6.4808e-01 +6.8878e-01 -0.0000e+00 +1.3383e-02 -1.9930e-02 +1.2278e-01 -6.5947e-02 +0.0000e+00 -0.0000e+00 +1.3257e+00 -2.3367e+00 -4.4478e+00 -0.0000e+00 +0.0000e+00 -0.0000e+00 +7.7818e-01 -6.7412e-01 -0.0000e+00 +0.0000e+00 -0.0000e+00 +0.0000e+00 +5.4589e+00 Press return to continue. Solving this system yields a unique least squares solution, namely x = invR * Q_T * b : +1.00 +0.04 -5.00 +0.04 +3.76 The coefficients a, b, c, d, e, of the curve are : +1.00x**2 +0.04y**2 -5.00x +0.04y +3.76 = 0 Press return to continue. Copy and paste in Octave: function xy = f (x,y) xy = +1.00*x^2 +0.04*y^2 -5.00*x +0.04*y +3.76; endfunction f (+1,+2) f (+2,-8) f (+3,-8) f (+4,-3)