🚀 go-pugleaf

http://www.fractalforums.com/programming/numerical-problem-with-bailout-test/






























n 16.04.2012 18:26, Adam Majewski wrote:
> 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;
>> }
>