Linear Algebra and the C Language/a0bf

__NOTOC__ Install and compile this file in your working directory. /* ------------------------------------ */ /* Save as : c00b.c */ /* ------------------------------------ */ #include "v_a.h" /* ------------------------------------ */ /* ------------------------------------ */ #define RA R5 #define CA C5 #define Cb C1 /* ------------------------------------ */ /* ------------------------------------ */ int main(void) { double xy[8] ={ 1, -8, 2, 2, 3, 1, 4, 2 }; double ab[RA*(CA+Cb)]={ /* x**2 y**2 x y e = 0 */ +1, +0, +0, +0, +0, +1, +1, +64, +1, -8, +1, +0, +4, +4, +2, +2, +1, +0, +9, +1, +3, +1, +1, +0, +16, +4, +4, +2, +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 -8 +2 +2 +3 +1 +4 +2 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 +64.00 +1.00 -8.00 +1.00 +0.00 +4.00 +4.00 +2.00 +2.00 +1.00 +0.00 +9.00 +1.00 +3.00 +1.00 +1.00 +0.00 +16.00 +4.00 +4.00 +2.00 +1.00 +0.00 Press return to continue. Q : +0.0531 -0.0068 -0.2458 +0.1164 +0.9608 +0.0531 +0.9973 -0.0277 -0.0429 +0.0022 +0.2123 +0.0357 +0.7522 +0.6136 +0.1066 +0.4777 -0.0452 +0.4432 -0.7365 +0.1759 +0.8492 -0.0454 -0.4203 +0.2563 -0.1858 R : +18.8414 +8.1204 +5.3074 +2.1761 +1.5922 +0.0000 +63.7421 +0.7515 -8.0429 +0.9424 -0.0000 -0.0000 +1.1253 +1.3284 +0.7475 +0.0000 +0.0000 +0.0000 +1.3462 +0.0905 +0.0000 +0.0000 +0.0000 -0.0000 +0.0989 Press return to continue. Q_T : +5.3074e-02 +5.3074e-02 +2.1230e-01 +4.7767e-01 +8.4919e-01 -6.7614e-03 +9.9728e-01 +3.5707e-02 -4.5164e-02 -4.5430e-02 -2.4581e-01 -2.7670e-02 +7.5217e-01 +4.4320e-01 -4.2025e-01 +1.1638e-01 -4.2860e-02 +6.1361e-01 -7.3649e-01 +2.5628e-01 +9.6082e-01 +2.1987e-03 +1.0664e-01 +1.7589e-01 -1.8579e-01 invR : +5.3074e-02 -6.7614e-03 -2.4581e-01 +1.1638e-01 +9.6082e-01 +0.0000e+00 +1.5688e-02 -1.0477e-02 +1.0407e-01 -1.6551e-01 +0.0000e+00 +0.0000e+00 +8.8867e-01 -8.7694e-01 -5.9111e+00 -0.0000e+00 -0.0000e+00 -0.0000e+00 +7.4285e-01 -6.7975e-01 -0.0000e+00 -0.0000e+00 -0.0000e+00 +0.0000e+00 +1.0107e+01 Press return to continue. Solving this system yields a unique least squares solution, namely x = invR * Q_T * b : +1.00 -0.14 -6.00 -0.57 +9.71 The coefficients a, b, c, d, e, of the curve are : +1.00x**2 -0.14y**2 -6.00x -0.57y +9.71 = 0 Press return to continue. Copy and paste in Octave: function xy = f (x,y) xy = +1.000000000*x^2 -0.144444444*y^2 -6.000000000*x -0.566666667*y +9.711111111; endfunction f (1, -8) f (2, 2) f (3, 1) f (4, 2)