How to encode images using chaotic mapping?

Hi everyone, I want to encode images using chaotic mappings. I divided it into 3 steps: bit permutation, pixel permutation, block permutation. I use logistic mapping as the key to encrypt. I hope everyone will guide me to write this program. I tried the program but not. Thanks very much.

4 comentarios

Geoff Hayes
Geoff Hayes el 18 de Mayo de 2018
Huong - can you show us what you have tried so far?
Huong Hoang
Huong Hoang el 19 de Mayo de 2018
Yes. But I think it is totally wrong. I bit permutation by converting the binary pixel R, G, B in a pixel image. But I did not know how to do it, because I study by myself so it took a lot of time. Can you help me show code? Very grateful to you.
Image Analyst
Image Analyst el 19 de Mayo de 2018
Probably no one has such a program, nor would be inclined to read and understand a paper and then code it up. Good luck though. If you have a smaller, more specific/target question though, feel free to ask.
Hello Image Analyst, this is my code. I want in a pixel, each bit of the image plane R, G, B will be permutation using Logistic map: x (n + 1) = r * x (n) * (1-x (n)). Can you see and fix me?
function key = creatkey(m);
a = imread('E:\_temp_matlab_R2018a_win64\bin\y.jpg');
m=length(a);
r=3.7
x(1)=0.4
for i=2:m
x(i)= r*x(i-1)*(1-x(i-1));
end
disp(x)
[heigth,width,~]=zeros(0);
a=24*width*heigth
binary=zeros(1,n)
k=1
for i=1:heigth
for j=1:width
bin = c(i,j,1);
bin1 = dec2bin(binary);
for i=1:8
if bin1(i)=='1'
binary(k)=1
k=k+1
end
end
end
for j=1:width
bin = c(i,j,2);
bin1 = dec2bin(binary);
for i=1:8
if bin1(i)=='1'
binary(k)=1
k=k+1
end
end
end
for j=1:width
bin = c(i,j,3);
bin1 = dec2bin(binary);
for i=1:8
if bin1(i)=='1'
binary(k)=1
k=k+1
end
end
end

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Respuestas (2)

Chandu Sree
Chandu Sree el 10 de Feb. de 2020
function key = creatkey(m);
a = imread('E:\_temp_matlab_R2018a_win64\bin\y.jpg');
m=length(a);
r=3.7
x(1)=0.4
for i=2:m
x(i)= r*x(i-1)*(1-x(i-1));
end
disp(x)
[heigth,width,~]=zeros(0);
a=24*width*heigth
binary=zeros(1,n)
k=1
for i=1:heigth
for j=1:width
bin = c(i,j,1);
bin1 = dec2bin(binary);
for i=1:8
if bin1(i)=='1'
binary(k)=1
k=k+1
end
end
end
for j=1:width
bin = c(i,j,2);
bin1 = dec2bin(binary);
for i=1:8
if bin1(i)=='1'
binary(k)=1
k=k+1
end
end
end
for j=1:width
bin = c(i,j,3);
bin1 = dec2bin(binary);
for i=1:8
if bin1(i)=='1'
binary(k)=1
k=k+1
end
end
end
Chandu Sree
Chandu Sree el 10 de Feb. de 2020
function key = creatkey(m);
a = imread('E:\_temp_matlab_R2018a_win64\bin\y.jpg');
m=length(a);
r=3.7
x(1)=0.4
for i=2:m
x(i)= r*x(i-1)*(1-x(i-1));
end
disp(x)
[heigth,width,~]=zeros(0);
a=24*width*heigth
binary=zeros(1,n)
k=1
for i=1:heigth
for j=1:width
bin = c(i,j,1);
bin1 = dec2bin(binary);
for i=1:8
if bin1(i)=='1'
binary(k)=1
k=k+1
end
end
end
for j=1:width
bin = c(i,j,2);
bin1 = dec2bin(binary);
for i=1:8
if bin1(i)=='1'
binary(k)=1
k=k+1
end
end
end
for j=1:width
bin = c(i,j,3);
bin1 = dec2bin(binary);
for i=1:8
if bin1(i)=='1'
binary(k)=1
k=k+1
end
end
end

2 comentarios

Amani Theramban
Amani Theramban el 28 de Mzo. de 2020
any one here to do the Novel Medical Image Encryption Scheme Based
on Chaos and DNA Encoding project??
Rik
Rik el 18 de Abr. de 2020
It is unclear to me what this comment has to with this thread or this answer in particular.

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el 18 de Mayo de 2018

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Rik
el 18 de Abr. de 2020

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