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411 समाचार
Public archive edition · Attributed source reporting with original dates · AI drafts are excluded

411 REFERENCE · Wikibooks

Linear Algebra and the C Language/a0jp

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__NOTOC__ Install and compile this file in your working directory. /* ------------------------------------ */ /* Save as : c00d.c */ /* ------------------------------------ */ #include "v_a.h" /* ------------------------------------ */ #define RA R5 #define CA C5 /* ------------------------------------ */ int main(void) { double xy[6] ={ 10, 10, -5, 1, 7, -10 }; double tA[RA*CA]={ /* x**2 y**2 x y e */ +1, +0, +0, +0, +0, +0, +1, +0, +0, +0, +100, +100, +10, +10, +1, +25, +1, -5, +1, +1, +49, +100, +7, -10, +1, }; double tb[RA*C1]={ /* = 0 */ +1, +1, +0, +0, +0, }; double **XY = ca_A_mR(xy,i_mR(R3,C2)); double **A = ca_A_mR(tA,i_mR(RA,CA)); double **b = ca_A_mR(tb,i_mR(RA,C1)); double **Inv = i_mR(CA,RA); double **Invb = i_mR(CA,C1); clrscrn(); printf("\n"); printf(" Find the coefficients a, b, c, d, of a circle \n\n"); printf(" ax**2 + ay**2 + bx + cy + d = 0 \n\n"); printf(" that passes through these three XY. \n\n"); printf(" x y"); p_mR(XY,S5,P0,C6); stop(); clrscrn(); printf(" Using the given XY, 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(" Inv :"); invgj_mR(A,Inv); pE_mR(Inv,S12,P4,C10); stop(); clrscrn(); printf(" Inv :"); p_mR(Inv,S10,P4,C10); printf(" x = Inv * b "); mul_mR(Inv,b,Invb); p_mR(Invb,S10,P4,C10); printf(" The coefficients a, b, c, d, e, of the curve are : \n\n" " %+.2fx**2 %+.2fy**2 %+.2fx %+.2fy %+.2f = 0\n\n" ,Invb[R1][C1],Invb[R2][C1],Invb[R3][C1], Invb[R4][C1],Invb[R5][C1]); stop(); f_mR(XY); f_mR(A); f_mR(b); f_mR(Inv); f_mR(Invb); return 0; } /* ------------------------------------ */ /* ------------------------------------ */ Screen output example: Find the coefficients a, b, c, d, of a circle ax**2 + ay**2 + bx + cy + d = 0 that passes through these three XY. x y +10 +10 -5 +1 +7 -10 Press return to continue. Using the given XY, we obtain this matrix. (a = 1. This is my choice) A : +1.00 +0.00 +0.00 +0.00 +0.00 +0.00 +1.00 +0.00 +0.00 +0.00 +100.00 +100.00 +10.00 +10.00 +1.00 +25.00 +1.00 -5.00 +1.00 +1.00 +49.00 +100.00 +7.00 -10.00 +1.00 b : +1.00 +1.00 +0.00 +0.00 +0.00 Inv : +1.0000e+00 +1.1102e-16 +3.4694e-18 -3.6863e-18 +8.6736e-19 -4.4409e-16 +1.0000e+00 +0.0000e+00 +4.3368e-19 +0.0000e+00 -3.8132e+00 -7.2527e+00 +4.0293e-02 -7.3260e-02 +3.2967e-02 -1.9780e+00 +1.0879e+00 +4.3956e-02 +1.0989e-02 -5.4945e-02 -4.2088e+01 -3.8352e+01 +1.5751e-01 +6.2271e-01 +2.1978e-01 Press return to continue. Inv : +1.0000 +0.0000 +0.0000 -0.0000 +0.0000 -0.0000 +1.0000 +0.0000 +0.0000 +0.0000 -3.8132 -7.2527 +0.0403 -0.0733 +0.0330 -1.9780 +1.0879 +0.0440 +0.0110 -0.0549 -42.0879 -38.3516 +0.1575 +0.6227 +0.2198 x = Inv * b +1.0000 +1.0000 -11.0659 -0.8901 -80.4396 The coefficients a, b, c, d, e, of the curve are : +1.00x**2 +1.00y**2 -11.07x -0.89y -80.44 = 0 Press return to continue. Copy and paste in Octave: function xy = f (x,y) xy = +1.000000000*x^2 +1.000000000*y^2 -11.065934068*x -0.890109889*y -80.439560439; endfunction f (+10,+10) f (-5,+1) f (+7,-10)