this is apart from a code for image encryption using chaotic map, please help me to know what is the wrong with this code
3 次查看(过去 30 天)
显示 更早的评论
x(1)=0.2350;
y(1)=0.3500;
z(1)=0.7350;
a(1)=0.0125;
b(1)=0.0157;
l(1)=3.7700;
image_height=256;
for i=1:1:70000
x(i+1)=l*x(i)*(1-x(i))+b*y(i)*y(i)*x(i)+a*z(i)*z(i)*z(i);
y(i+1)=l*y(i)*(1-y(i))+b*z(i)*z(i)*y(i)+a*x(i)*x(i)*x(i);
z(i+1)=l*z(i)*(1-z(i))+b*x(i)*x(i)*z(i)+a*y(i)*y(i);
end
x=ceil(mod((x*100000),image_height));
y=ceil(mod((y*100000),image_height));
z=ceil(mod((z*100000),image_height));
original=imread('peppers.png');
rgb=rgb2gray(original);
[row,col]=size(rgb);
n=500;
p=600;
q=700;
for j=1:1:row
k(j)=x(j+n);
l(j)=y(j+p);
end
% for i=1:1:row
% k(i)=x(i+n);
% end
% for j=1:1:col
% l(j)=y(j+p);
% end
for j=1:1:col*row
m(j)=z(j+q);
end
please i want to know what's the error in the last command line " m(j)=z(j+q);"
when i run i got error in this line , please help me
7 个评论
Rik
2021-5-4
@DGM That is actually what allows this discussion. If this code wasn't fundamentally broken it would be a question about implementing cryptography, which is classified as a weapon in US law (yes, seriously). Discussing it on a website that can be reached from outside of the US is considered as a weapons export, which requires permission being granted prior to publishing.
It is a stupid system that makes no sense in this case, but since the Mathworks website is governed by US law, in general discussions about cryptography aren't allowed. (although there is hardly any enforcement of this rule, for more context, see this thread)
DGM
2021-5-4
Yeah, I figured I wouldn't bring it up. The ridiculousness of reality needs no aid to stay in memory.
回答(0 个)
另请参阅
类别
在 Help Center 和 File Exchange 中查找有关 Encryption / Cryptography 的更多信息
Community Treasure Hunt
Find the treasures in MATLAB Central and discover how the community can help you!
Start Hunting!