Image processing using chaotic map

The chaotic tent map is given by:

xi+1 = f (xi,μ)

(f (xi,μ)= μxi, if xi < 0.5

μ(1−xi), otherwise;) ……………… Eq.1

where xi ∈[ 0,1], for i ≥ 0. This map transforms an interval [0, 1] onto itself and contains only one control parameter μ, respectively, where μ ∈[ 0,2].

Steps:

(1) Read plain-images (original image) (Ma×b), get size of M, e.g., use [a, b, c] to save the size of M, a = 256, b = 256, c = 3, let N = a∗b∗c, and initiate control parameter (μ) of the chaotic tent map.

(2) Input the secret (encryption) key x0 into the algorithm. Iterate the chaotic tent map N times using equation 1, and obtain the key array x(n), the size of x(n) is N;

(3) Encrypt each element of matrix (Ma×b) using the key array x(n), that is, mix the original image M(a×b) components with the key array x(n);

(4) The resulting image of step (3) is the ciphered image.

Coding:

M=imread('cameraman.tif');
[a,b] = size(M);
N=a*b;
u=1.999;
x(1)=0.5;
for i=1:N
        if x(i) < 0.5 %and condition
            x(i+1)=u*x(i); %is it right
        else
            if x(i) >= 0.5 %and this condition 
            x(i+1)=u*(1-x(i)); %is it right
            end
        end
    end
r=xor(x(i+1),M);

I wrote this code by following the steps mentioned above(research paper attached). I also attached the required output but i am getting a full white or full black image. Can someone help me identify the problem.

[Karbala'a Unvi. Science]

Dear Sir,

will you send me the final code to look at it and find how did you get the result..

Thanks in advance

Your

Dr.Zeyad

Answer 1

[Walter Roberson]

Your final r is a scalar double. You will not get much from it.

Your x will be a vector double. You would need to reshape it and uint8 it.

[Nazish]

Thank you.. stay blessed

[Shankar R]

Wts wrong in above code.. ? I am not getting output?? Can you explain?

[Walter Roberson]

r = bitxor(im2uint8(reshape(x(2:end),a,b)),M);