Linear Algebra and the C Language/a044

__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 FACTOR_E +1.E-0 /* ------------------------------------ */ int main(void) { double txy[8] ={ 1, 4, 2, 5, 3, -7, 4, 5 }; double tA[RA*CA]={ /* x**2 y**2 x y e */ +1, +0, +0, +0, +0, +1, +16, +1, +4, +1, +4, +25, +2, +5, +1, +9, +49, +3, -7, +1, +16, +25, +4, +5, +1, }; double tb[RA*C1]={ /* = 0 */ +1, +0, +0, +0, +0 }; double **xy = ca_A_mR(txy, i_mR(R4,C2)); double **A = ca_A_mR(tA, i_mR(RA,CA)); double **b = ca_A_mR(tb, i_mR(RA,C1)); double **Pinv = Pinv_Rn_mR(A, i_mR(CA,RA),FACTOR_E); double **Pinvb = mul_mR(Pinv,b, i_mR(CA,C1)); 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, S10,P0,C6); stop(); clrscrn(); printf(" Using the given points, we obtain this matrix.\n"); printf(" (a = 1. This is my choice)\n\n"); printf(" A:"); p_mR(A, S10,P2,C7); printf(" b:"); p_mR(b, S10,P2,C7); printf(" Pinv = V invS_T U_T "); pE_mR(Pinv, S12,P4,C10); stop(); clrscrn(); printf(" Solving this system yields a unique\n" " least squares solution, namely \n\n"); printf(" Pinv b "); p_mR(Pinvb, S10,P4,C10); printf(" The coefficients a, b, c, d, e, of the curve are: \n\n" " %+.9f*x^2 %+.9f*y^2 %+.9f*x %+.9f*y %+.9f = 0\n\n" ,Pinvb[R1][C1],Pinvb[R2][C1],Pinvb[R3][C1], Pinvb[R4][C1],Pinvb[R5][C1]); stop(); f_mR(xy); f_mR(b); f_mR(A); f_mR(Pinv); f_mR(Pinvb); 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 Press return to continue. Using the given points, we obtain this matrix. (a = 1. This is my choice) A : +1.00 +0.00 +0.00 +0.00 +0.00 +1.00 +16.00 +1.00 +4.00 +1.00 +4.00 +25.00 +2.00 +5.00 +1.00 +9.00 +49.00 +3.00 -7.00 +1.00 +16.00 +25.00 +4.00 +5.00 +1.00 b : +1.00 +0.00 +0.00 +0.00 +0.00 Pinv = V * invS_T * U_T +1.0000e+00 +9.8734e-11 -8.1187e-11 +4.7280e-12 +8.7404e-12 +2.8030e-01 -9.0909e-02 +1.3258e-01 +7.5758e-03 -4.9242e-02 -6.0000e+00 -6.1187e-10 -5.0000e-01 -2.9365e-11 +5.0000e-01 +4.7727e-01 -1.8182e-01 +3.0682e-01 -6.8182e-02 -5.6818e-02 -1.3939e+00 +3.1818e+00 -2.8485e+00 +1.5152e-01 +5.1515e-01 Press return to continue. Solving this system yields a unique least squares solution, namely Pinv * b +1.0000 +0.2803 -6.0000 +0.4773 -1.3939 The coefficients a, b, c, d, e, of the curve are : +1.000000000*x^2 +0.280303030*y^2 -6.000000000*x +0.477272727*y -1.393939394 = 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)