🚀 go-pugleaf

Zapomniałem dodać link do pracy skąd zaczerpnąłem pomysł na taki program  :
http://arxiv.org/abs/math/0505036


hth


Adam


On 15.04.2012 15:13, Adam Majewski wrote:
> Witam.
>
> Napisałem program w c ( kod poniżej). Oblicza ile potrzeba kroków (
> iteracji tu LastIteration) aby punkt z1 uciekł z koła o promieniu ER,
> czyli sprawdza abs(z1)<ER
>
>
>
>
> Program działa, ale dla n=27 wpada w nieskończoną pętlę.
> Przypuszczam, że wtedy program nie rozróżnia z1 i zp :
> bo wtedy distance(z1,zp)<Calculated Machine epsilon
>
>
>
> gdzie :
> distance(z1,zp) = 1.490116e-08
> Calculated Machine epsilon = 1.19209E-07 ( używając programu z wikipedii )
> http://en.wikipedia.org/wiki/Machine_epsilon
> punkt jest liczbą zespoloną)
> punkt zp jest punktem stałym = nie ucieka z koła o promieniu ER
>
>
> Jak to się ma do :
>
> DBL_EPSILON = 2.220446e-16 = 0.00000000000000022204460492503130808473 (
> z float.h )
> ?
>
>
>
> Pozdrawiam
>
> Adam
>
>
>
>
>
>
>
>
>
>
>
>
> ============== Uruchomienie==========================
>
> gcc -lm -Wall i.c
> ./a.out
>
>
> =============wyniki ====================
>
>
> n= 1 distance(z1,zp) = 5.000000e-01 LastIteration = 3
> n= 2 distance(z1,zp) = 2.500000e-01 LastIteration = 5
> n= 3 distance(z1,zp) = 1.250000e-01 LastIteration = 10
> n= 4 distance(z1,zp) = 6.250000e-02 LastIteration = 19
> n= 5 distance(z1,zp) = 3.125000e-02 LastIteration = 35
> n= 6 distance(z1,zp) = 1.562500e-02 LastIteration = 68
> n= 7 distance(z1,zp) = 7.812500e-03 LastIteration = 133
> n= 8 distance(z1,zp) = 3.906250e-03 LastIteration = 261
> n= 9 distance(z1,zp) = 1.953125e-03 LastIteration = 518
> n= 10 distance(z1,zp) = 9.765625e-04 LastIteration = 1031
> n= 11 distance(z1,zp) = 4.882812e-04 LastIteration = 2055
> n= 12 distance(z1,zp) = 2.441406e-04 LastIteration = 4104
> n= 13 distance(z1,zp) = 1.220703e-04 LastIteration = 8201
> n= 14 distance(z1,zp) = 6.103516e-05 LastIteration = 16394
> n= 15 distance(z1,zp) = 3.051758e-05 LastIteration = 32778
> n= 16 distance(z1,zp) = 1.525879e-05 LastIteration = 65547
> n= 17 distance(z1,zp) = 7.629395e-06 LastIteration = 131084
> n= 18 distance(z1,zp) = 3.814697e-06 LastIteration = 262156
> n= 19 distance(z1,zp) = 1.907349e-06 LastIteration = 524301
> n= 20 distance(z1,zp) = 9.536743e-07 LastIteration = 1048590
> n= 21 distance(z1,zp) = 4.768372e-07 LastIteration = 2097167
> n= 22 distance(z1,zp) = 2.384186e-07 LastIteration = 4194322
> n= 23 distance(z1,zp) = 1.192093e-07 LastIteration = 8388675
> n= 24 distance(z1,zp) = 5.960464e-08 LastIteration = 16778821
> n= 25 distance(z1,zp) = 2.980232e-08 LastIteration = 33604781
> n= 26 distance(z1,zp) = 1.490116e-08 LastIteration = 68563774
>
>
> =================== kod ==================
>
>
> #include <stdio.h>
> #include <math.h> // pow
> #include <float.h> // DBL_EPSILON
>
>
>
> double GiveLastIteration(double Zx, double Zy, double Cx, double Cy, int
> IterationMax, int ER2)
> {
> double Zx2, Zy2; /* Zx2=Zx*Zx; Zy2=Zy*Zy */
> double i=0.0;
> Zx2=Zx*Zx;
> Zy2=Zy*Zy;
> while (Zx2+Zy2<ER2) /* ER2=ER*ER */
> {
> Zy=2*Zx*Zy + Cy;
> Zx=Zx2-Zy2 + Cx;
> Zx2=Zx*Zx;
> Zy2=Zy*Zy;
> i+=1.0;
> //printf("i = %f Zx = %e = %40.38f \n", i,Zx, Zx);
> }
> return i;
> }
>
>
> int main()
>
> {
> double distance;
>
> int n;
> // parabolic fixed point zp = zpx = zpy*I = 0.5
> double zpx = 0.5;
> double zpy = 0.0;
> // z1 = z1x + z1y*I = point of exterior of Julia set but near parabolic
> // fixed point zp
> double z1x;
> double z1y = zpy;
> double cx= 0.25;
> double cy= 0.0;
> // Escape Radius ; it defines target set = { z: abs(z)>ER}
> // all points z in the target set are escaping to infinity
> double ER=2.0;
> double ER2;
>
> double LastIteration;
>
>
> //n = 1000;
> ER2= ER*ER;
>
>
>
> for (n =1; n<101; n++)
> {
> distance = pow(2.0,-n);
> printf("n= %d distance(z1,zp) = %e = %40.38f ",n, distance, distance);
> z1x = zpx + distance;
> LastIteration = GiveLastIteration(z1x,z1y, cx,cy, n+n,ER2 );
> printf("LastIteration = %.0f \n", LastIteration);
> }
>
> printf("DBL_EPSILON = %e = %40.38f ",DBL_EPSILON , DBL_EPSILON );
>
>
> return 0;
> }