Linear Algebra and the C Language/a04x

__NOTOC__ Install and compile this file in your working directory. /* ------------------------------------ */ /* Save as: c00b.c */ /* ------------------------------------ */ #include "v_a.h" /* ------------------------------------ */ /* ------------------------------------ */ #define RA R5 #define CA C5 #define Cb C1 /* ------------------------------------ */ /* ------------------------------------ */ int main(void) { double txy[8] ={ 1, 0, 2, 3, 3, 4, 4, 0 }; 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, +1.00, +0.00, +1.00, +0.00, +1.00, +0.00, +4.00, +9.00, +2.00, +3.00, +1.00, +0.00, +9.00, +16.00, +3.00, +4.00, +1.00, +0.00, +16.00, +0.00, +4.00, +0.00, +1.00, +0.00, }; double **xy = ca_A_mR(txy, i_mR(R4,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, 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, S5,P0,C6); printf(" Using the given points, we obtain this matrix.\n"); printf(" (a = 1. This is my choice)\n\n"); printf(" Ab:\n"); printf(" x**2 y**2 x y e = 0 "); 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(" x = inv(A_TA)A_T b:"); p_mR(x, S10,P4,C7); stop(); clrscrn(); printf(" 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" " %+.9f*x^2 %+.9f*y^2 %+.9f*x %+.9f*y %+.9f = 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, e of the conic, ax**2 + by**2 + cx + dy + e = 0 which crosses these four points (x[1],y[1]) (x[2],y[2]) (x[3],y[3]) (x[4],y[4]) Using the four points we obtain the matrix. (a)x**2 (b)y**2 (c)x (d)y (e) = 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 x[4]**2 y[4]**2 x[4] y[4] 1 0 This system has four lines and five unknowns (a, b, c, d, e). 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. 1 0 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 x[4]**2 y[4]**2 x[4] y[4] 1 0 All that remains is to solve the system. 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 Using the given points, we obtain this matrix. (a = 1. This is my choice) Ab : x**2 y**2 x y e = 0 +1.00 +0.00 +0.00 +0.00 +0.00 +1.00 +1.00 +0.00 +1.00 +0.00 +1.00 +0.00 +4.00 +9.00 +2.00 +3.00 +1.00 +0.00 +9.00 +16.00 +3.00 +4.00 +1.00 +0.00 +16.00 +0.00 +4.00 +0.00 +1.00 +0.00 Press return to continue. A_T : +1.00 +1.00 +4.00 +9.00 +16.00 +0.00 +0.00 +9.00 +16.00 +0.00 +0.00 +1.00 +2.00 +3.00 +4.00 +0.00 +0.00 +3.00 +4.00 +0.00 +0.00 +1.00 +1.00 +1.00 +1.00 A_TA : +355.00 +180.00 +100.00 +48.00 +30.00 +180.00 +337.00 +66.00 +91.00 +25.00 +100.00 +66.00 +30.00 +18.00 +10.00 +48.00 +91.00 +18.00 +25.00 +7.00 +30.00 +25.00 +10.00 +7.00 +4.00 Press return to continue. inv(A_TA) : +1.0000 -0.1667 -5.0000 +1.1667 +4.0000 -0.1667 +0.2238 +0.7685 -0.9182 -0.4630 -5.0000 +0.7685 +25.2222 -5.6019 -20.5556 +1.1667 -0.9182 -5.6019 +4.1127 +3.7963 +4.0000 -0.4630 -20.5556 +3.7963 +17.8889 inv(A_TA)*A_T : +1.0000 +0.0000 -0.0000 -0.0000 -0.0000 -0.1667 +0.1389 -0.3333 +0.2500 -0.0556 -5.0000 -0.3333 +0.0000 +0.0000 +0.3333 +1.1667 -0.6389 +1.3333 -0.7500 +0.0556 +4.0000 +1.3333 -0.0000 -0.0000 -0.3333 x = inv(A_TA)*A_T*b : +1.0000 -0.1667 -5.0000 +1.1667 +4.0000 Press return to continue. x = inv(A_TA)*A_T*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)