Linear Algebra and the C Language/a04y

__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 Cb C1 /* ------------------------------------ */ /* ------------------------------------ */ int main(void) { double txy[6] ={ 1, -2, 2, -3, 3, 6 }; double ab[RA*(CA+Cb)]={ /* x**2 y**2 x y e = 0 */ +1.00, +0.00, +0.00, +0.00, +0.00, +1.00, +0.00, +1.00, +0.00, +0.00, +0.00, +1.00, +1.00, +4.00, +1.00, -2.00, +1.00, +0.00, +4.00, +9.00, +2.00, -3.00, +1.00, +0.00, +9.00, +36.00, +3.00, +6.00, +1.00, +0.00, }; double **xy = ca_A_mR(txy, i_mR(R3,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 **A_T = transpose_mR(A, i_mR(CA,RA)); double **A_TA = mul_mR(A_T,A, i_mR(CA,CA)); double **invA_TA = inv_mR(A_TA, i_mR(CA,CA)); double **invA_TAA_T = mul_mR(invA_TA,A_T, i_mR(CA,RA)); double **x = mul_mR(invA_TAA_T,b, i_mR(CA,Cb)); /* x = inv(A_TA)A_T b */ 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); printf("\n"); printf(" Using the given xy, we obtain this matrix.\n"); printf(" (a = 1. This is my choice)\n\n"); printf(" x**2 y**2 x y "); p_mR(Ab, S7,P2,C6); stop(); clrscrn(); printf(" A_T:"); p_mR(A_T, S10,P2,C7); printf(" A_TA:"); p_mR(A_TA, S10,P2,C7); stop(); clrscrn(); printf(" inv(A_TA):"); p_mR(invA_TA, S10,P4,C7); printf(" inv(A_TA)A_T:"); p_mR(invA_TAA_T, S10,P4,C7); printf("\n x = inv(A_TA)A_T b:"); p_mR(x, S10,P4,C7); stop(); clrscrn(); printf("\n x = inv(A_TA)A_T b:"); p_mR(x, S10,P4,C7); 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(A_T); f_mR(A_TA); f_mR(invA_TA); f_mR(invA_TAA_T); f_mR(x); return 0; } /* ------------------------------------ */ /* ------------------------------------ */ Presentation : Find the coefficients a, b, c, d of the circle, ax**2 + ay**2 + bx + cy + d = 0 which crosses these three points (x[1],y[1]) (x[2],y[2]) (x[3],y[3]) Using the three points we obtain the matrix. (a)x**2 (a)y**2 (b)x (c)y (d) = 0 x[1]**2 y[1]**2 x[1] y[1] 1 0 x[2]**2 y[2]**2 x[2] y[2] 1 0 x[3]**2 y[3]**2 x[3] y[3] 1 0 This system has three lines and four unknowns (a, b, c, d). It is a homogeneous system, so it has an infinite number of solutions. To find a solution, I chose to set a = 1. We therefore have five rows and five unknowns. (a)x**2 (a)y**2 x y 1 0 0 0 0 1 0 1 0 0 0 1 x[1]**2 y[1]**2 x[1] y[1] 1 0 x[2]**2 y[2]**2 x[2] y[2] 1 0 x[3]**2 y[3]**2 x[3] y[3] 1 0 All that remains is to solve the system. 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 +1 -2 +2 -3 +3 +6 Using the given XY, we obtain this matrix. (a = 1. This is my choice) x**2 y**2 x y +1.00 +0.00 +0.00 +0.00 +0.00 +1.00 +0.00 +1.00 +0.00 +0.00 +0.00 +1.00 +1.00 +4.00 +1.00 -2.00 +1.00 +0.00 +4.00 +9.00 +2.00 -3.00 +1.00 +0.00 +9.00 +36.00 +3.00 +6.00 +1.00 +0.00 Press return to continue. A_T : +1.00 +0.00 +1.00 +4.00 +9.00 +0.00 +1.00 +4.00 +9.00 +36.00 +0.00 +0.00 +1.00 +2.00 +3.00 +0.00 +0.00 -2.00 -3.00 +6.00 +0.00 +0.00 +1.00 +1.00 +1.00 A_TA : +99.00 +364.00 +36.00 +40.00 +14.00 +364.00 +1394.00 +130.00 +181.00 +49.00 +36.00 +130.00 +14.00 +10.00 +6.00 +40.00 +181.00 +10.00 +49.00 +1.00 +14.00 +49.00 +6.00 +1.00 +3.00 Press return to continue. inv(A_TA) : +1.0000 -0.0000 -3.2000 -0.2000 +1.8000 -0.0000 +1.0000 -7.2000 -2.2000 -1.2000 -3.2000 -7.2000 +63.5400 +16.2400 +0.0400 -0.2000 -2.2000 +16.2400 +4.9400 +2.7400 +1.8000 -1.2000 +0.0400 +2.7400 +10.5400 inv(A_TA)*A_T : +1.0000 +0.0000 -0.0000 -0.0000 -0.0000 +0.0000 +1.0000 -0.0000 -0.0000 -0.0000 -3.2000 -7.2000 -0.9000 +0.8000 +0.1000 -0.2000 -2.2000 +0.1000 -0.2000 +0.1000 +1.8000 -1.2000 +2.1000 -1.2000 +0.1000 x = inv(A_TA)*A_T*b : +1.0000 +1.0000 -10.4000 -2.4000 +0.6000 Press return to continue. x = inv(A_TA)*A_T*b : +1.0000 +1.0000 -10.4000 -2.4000 +0.6000 The coefficients a, b, c, d, e, of the curve are : +1.00x**2 +1.00y**2 -10.40x -2.40y +0.60 = 0 Press return to continue. Copy and paste in Octave: function xy = f (x,y) xy = +1.00*x^2 +1.00*y^2 -10.40*x -2.40*y +0.60; endfunction f (+1,-2) f (+2,-3) f (+3,+6)