chaotic sequence generation from chen's hyperchaotic system

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Abirami
Abirami 2014 年 8 月 15 日
コメント済み: mosarrat jahan 2016 年 9 月 23 日
Hello
I need to generate two chaotic sequences based on chen's hyperchaotic system.It has to be generated from the following four formulas
X=ay-x;
Y=-xz+dx+cy-q;
Y=xy-bz;
Q=x+k;
where a,b,c,d,x,y,z,q are all initialised as follows. I need only X and Y where
X=[x1,x2,...x4n]
Y=[y1,y2,...y4n]
a=36 ;
b=3 ;
c=28 ;
d=16 ;
k=0.2 ;
x=0.3 ;
y=-0.4 ;
z=1.2 ;
q=1 ;
n=256 ;
I tried the following code but i'm not able to get it properly.
clc
clear all
close all
w=imread('C:\Users\Abzz\Desktop\lena.png');
[m n]=size(w)
a=36;
b=3;
c=28;
d=16;
k=0.2;
x(1)=0.3;
y(1)=-0.4;
z(1)=1.2;
q(1)=1;
for i=1:1:4(n)
x(i+1)=(a*(y(i)-x(i)));
y(i+1)=-(x(i)*z(i))+(d*x(i))+(c*y(i))-q(i);
z(i+1)=(x(i)*y(i))-(b*z(i));
q(i+1)=x(i)+k;
end
disp(x);
disp(y);
pls help. thanks in advance.
  3 件のコメント
Abirami
Abirami 2014 年 8 月 16 日
編集済み: Abirami 2014 年 8 月 16 日
sir i used the wrong variable a...i corrected my code by changine the variable....im planning to scramble the image with the index values obtained from the two chaotic sequences X and Y...
mosarrat jahan
mosarrat jahan 2016 年 9 月 23 日
how to generate chaotic sequence using logistic map of n*n image

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回答 (1 件)

Geoff Hayes
Geoff Hayes 2014 年 8 月 16 日
Abirami - you should provide a reference to a paper or document that describes Chen's Hyperchaotic System with respect to chaotic sequence generation. Or is this something new? From http://www.ijest.info/docs/IJEST11-03-05-262.pdf, there is some text that describes the hyperchaotic Chen dynamics as
xp = a*(y x) + v
yp = d*x x*u + c*y
up = x*y b*u
vp = y*u + r*v
where x,y,u,v are the states of the system and a,b,c,d,r are constant, positive parameters of the system. This is very similar to your lines of code of
x(i+1) = (a*(y(i)-x(i)));
y(i+1) = -(x(i)*z(i))+(d*x(i))+(c*y(i))-q(i);
z(i+1) = (x(i)*y(i))-(b*z(i));
q(i+1) = x(i)+k;
where z and q are u and v respectively from the previous set of equations. Note that there are a couple of differences between the first and last equations in the two sets. As the first set is also similar to that found from http://www.ncnsd.org/proceedings/proceeding09/Paper/49.pdf, how did you decide on how to define your equations?
As for the code that generates the sequences,
for i=1:1:4(n)
x(i+1)=(a*(y(i)-x(i)));
y(i+1)=-(x(i)*z(i))+(d*x(i))+(c*y(i))-q(i);
z(i+1)=(x(i)*y(i))-(b*z(i));
q(i+1)=x(i)+k;
end
Your for loop iterates from 1 to 4(n)...which is just 4 as the (n) seems to be ignored (or rather, just displayed). Why have you chosen this? Are you trying to do this for just four iterations so that a chaotic sequence of length four is generated? Is your idea to generate a chaotic sequence for each pixel as
w=imread('C:\Users\Abzz\Desktop\lena.png');
[m n]=size(w);
% define a,b,c,d
for r=1:m
for s=1:n
% generate new pixel coordinate for img(r,s) using
% four iterations of the chaotic sequence
for k=1:4
% generate sequence
end
% last (fourth iteration) value from x and y sequences defines the
% scrambled pixel coordinate for w(r,s)
end
end
Is that the idea?
  3 件のコメント
Geoff Hayes
Geoff Hayes 2014 年 8 月 16 日
Abirami - I can't download the pdf (as there is a fee attached to it) so you will have to outline it in more detail. If you need 1024 iterations, then you can just change your code from
for i=1:1:4(n)
% etc.
end
to
for i=1:1:4*n
% etc.
end
Now your X and Y will each have 1024 elements. What is the next step?
Abirami
Abirami 2014 年 11 月 2 日
編集済み: Abirami 2014 年 11 月 2 日
https://drive.google.com/file/d/0BzotgT5a_8mfTHVsd1FXbGFDeVk/view?usp=sharing
sir please do view this link to get an idea about my paper. i still have a doubt about the number of elements and how the sequences should be used. pls help.thanks in advance.

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