chaotic sequence generation from chen's hyperchaotic system

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.

[Geoff Hayes]

Abirami - why do you read in the image data into the a variable as

 a=imread('C:\Users\Abzz\Desktop\lena.png');

only to replace this a couple of lines later with

 a=36;

What are you attempting to do with the image? Since you have its dimensions m and n (you are assuming that the input image is 2D grayscale ) are you going to manipulate or scramble or encrypt the pixels?

[Abirami]

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]

how to generate chaotic sequence using logistic map of n*n image

Answer 1

[Geoff Hayes]

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?

[Abirami]

i had used the equations from the following paper...and n=256 here...so basically i need 1024 iterations..

http://ac.els-cdn.com/S0030402612009126/1-s2.0-S0030402612009126-main.pdf?_tid=c172834a-1eae-11e4-a37e-00000aacb362&acdnat=1407469844_1ef0993683899635fc568b808692fb1d

i need to get only the vectors X and Y...i have defined X and Y earlier....pls help...thanks in advance....

[Geoff Hayes]

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]

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.