Linear Algebra and the C Language/a04x
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)