Linear Algebra and the C Language/a0be

__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[8] ={ 1, 0, 2, 3, 3, 4, 4, 0 }; double ab[RA*(CA+Cb)]={ /* x**2 y**2 x y e = 0 */ +1, +0, +0, +0, +0, +1, +1, +0, +1, +0, +1, +0, +4, +9, +2, +3, +1, +0, +9, +16, +3, +4, +1, +0, +16, +0, +4, +0, +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 +0 +2 +3 +3 +4 +4 +0 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 +0.00 +1.00 +0.00 +1.00 +0.00 +4.00 +9.00 +2.00 +3.00 +1.00 +0.00 +9.00 +16.00 +3.00 +4.00 +1.00 +0.00 +16.00 +0.00 +4.00 +0.00 +1.00 +0.00 Press return to continue. Q : +0.0531 -0.0323 -0.2668 +0.1748 +0.9457 +0.0531 -0.0323 +0.7999 -0.5069 +0.3152 +0.2123 +0.4447 +0.4686 +0.7332 -0.0000 +0.4777 +0.7296 -0.2636 -0.4124 +0.0000 +0.8492 -0.5175 -0.0022 +0.0694 -0.0788 R : +18.8414 +9.5534 +5.3074 +2.5476 +1.5922 +0.0000 +15.6759 +0.9758 +4.2525 +0.6244 -0.0000 -0.0000 +0.9375 +0.3514 +1.0027 +0.0000 -0.0000 +0.0000 +0.5499 -0.1167 +0.0000 +0.0000 +0.0000 +0.0000 +0.2364 Press return to continue. Q_T : +5.3074e-02 +5.3074e-02 +2.1230e-01 +4.7767e-01 +8.4919e-01 -3.2345e-02 -3.2345e-02 +4.4475e-01 +7.2957e-01 -5.1753e-01 -2.6681e-01 +7.9987e-01 +4.6856e-01 -2.6357e-01 -2.2010e-03 +1.7476e-01 -5.0692e-01 +7.3320e-01 -4.1242e-01 +6.9449e-02 +9.4573e-01 +3.1524e-01 -8.6284e-15 +7.0729e-15 -7.8811e-02 invR : +5.3074e-02 -3.2345e-02 -2.6681e-01 +1.7476e-01 +9.4573e-01 +0.0000e+00 +6.3792e-02 -6.6396e-02 -4.5089e-01 -1.0946e-01 +0.0000e+00 +0.0000e+00 +1.0667e+00 -6.8168e-01 -4.8600e+00 -0.0000e+00 -0.0000e+00 -0.0000e+00 +1.8185e+00 +8.9757e-01 -0.0000e+00 -0.0000e+00 -0.0000e+00 -0.0000e+00 +4.2295e+00 Press return to continue. Solving this system yields a unique least squares solution, namely x = invR * Q_T * b : +1.00 -0.17 -5.00 +1.17 +4.00 The coefficients a, b, c, d, e, of the curve are : +1.00x**2 -0.17y**2 -5.00x +1.17y +4.00 = 0 Press return to continue. Copy and paste in Octave: function xy = f (x,y) xy = +1.000000000*x^2 -0.166666667*y^2 -5.000000000*x +1.166666667*y +4.000000000; endfunction f (+1,+0) f (+2,+3) f (+3,+4) f (+4,+0)