Linear Algebra and the C Language/a042

__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 FACTOR_E +1.E-0 /* ------------------------------------ */ int main(void) { double txy[8] ={ 1, 0, 2, 3, 3, 4, 4, 0 }; double tA[RA*CA]={ /* x**2 y**2 x y e */ +1.00, +0.00, +0.00, +0.00, +0.00, +1.00, +0.00, +1.00, +0.00, +1.00, +4.00, +9.00, +2.00, +3.00, +1.00, +9.00, +16.00, +3.00, +4.00, +1.00, +16.00, +0.00, +4.00, +0.00, +1.00 }; double tb[RA*C1]={ /* = 0 */ +1.00, +0.00, +0.00, +0.00, +0.00, }; 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 +0 +2 +3 +3 +4 +4 +0 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 +0.00 +1.00 +0.00 +1.00 +4.00 +9.00 +2.00 +3.00 +1.00 +9.00 +16.00 +3.00 +4.00 +1.00 +16.00 +0.00 +4.00 +0.00 +1.00 b : +1.00 +0.00 +0.00 +0.00 +0.00 Pinv = V * invS_T * U_T +1.0000e+00 +4.2688e-14 -5.9459e-14 +3.4212e-14 -6.0507e-15 -1.6667e-01 +1.3889e-01 -3.3333e-01 +2.5000e-01 -5.5556e-02 -5.0000e+00 -3.3333e-01 +2.7142e-13 -1.5601e-13 +3.3333e-01 +1.1667e+00 -6.3889e-01 +1.3333e+00 -7.5000e-01 +5.5556e-02 +4.0000e+00 +1.3333e+00 -1.3500e-13 +7.7716e-14 -3.3333e-01 Press return to continue. Solving this system yields a unique least squares solution, namely Pinv * b +1.0000 -0.1667 -5.0000 +1.1667 +4.0000 The coefficients a, b, c, d, e, of the curve are : +1.000000000*x^2 -0.166666667*y^2 -5.000000000*x +1.166666667*y +4.000000000 = 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)