Linear Algebra and the C Language/a045

__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 FACTOR_E +1.E-0 /* ------------------------------------ */ int main(void) { double txy[8] ={ 1, 2, 2, -8, 3, -8, 4, -3 }; double tA[RA*CA]={ /* x**2 y**2 x y e */ +1, +0, +0, +0, +0, +1, +4, +1, +2, +1, +4, +64, +2, -8, +1, +9, +64, +3, -8, +1, +16, +9, +4, -3, +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" " %+.2f*x^2 %+.2f*y^2 %+.2f*x %+.2f*y %+.2f = 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 +2 +2 -8 +3 -8 +4 -3 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 +4.00 +1.00 +2.00 +1.00 +4.00 +64.00 +2.00 -8.00 +1.00 +9.00 +64.00 +3.00 -8.00 +1.00 +16.00 +9.00 +4.00 -3.00 +1.00 b : +1.00 +0.00 +0.00 +0.00 +0.00 Pinv = V * invS_T * U_T +1.0000e+00 +8.4318e-12 +5.5131e-11 -5.7896e-11 +1.5745e-11 +4.0000e-02 +2.0000e-02 -8.0000e-02 +1.0000e-01 -4.0000e-02 -5.0000e+00 -3.7921e-11 -1.0000e+00 +1.0000e+00 -7.7166e-11 +4.0000e-02 +2.2000e-01 -6.8000e-01 +7.0000e-01 -2.4000e-01 +3.7600e+00 +4.8000e-01 +2.6800e+00 -2.8000e+00 +6.4000e-01 Press return to continue. Solving this system yields a unique least squares solution, namely Pinv * b +1.0000 +0.0400 -5.0000 +0.0400 +3.7600 The coefficients a, b, c, d, e, of the curve are : +1.00*x^2 +0.04*y^2 -5.00*x +0.04*y +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)