The gray-scale encryption of image encryption: the principle and steps of a round of encryption algorithm based on the key × decryption key ≡ 1 mod gray level

Algorithm basis

  Given number $p$ and number$G$ , consider the congruence
$$px\equiv 1\mathrm{m}\mathrm{o}\mathrm{d}G$$
当$\gcd \left(p,G\right)|1$即$\gcd \left(p,G\right)=1$ time congruence is solvable, Remember the solution as
$$x\equiv qmodG$$
$\forall a\in \mathbb{Z}$，记$pa\mathrm{m}\mathrm{o}\mathrm{d}G=b$，即有
$$pa\equiv bmodG$$
则有
$$qb\equiv q\left(pa\right)=\left(pq\right)a\equiv amodG$$

Modular gray level encryption algorithm

  Given a picture size $M\times$ImageII of$N$ (unit: pixel)$I$ , determine its gray level number$G$ (usually$2$ ofnnPower of$n$ , for example$2$ of$The\; 8th$ power is 256).

  Key : $p\in Z$ , satisfy$\gcd \left(p,G\right)=1$

  解钥：$q\in Z$ , by solving$pq\equiv 1\mathrm{m}\mathrm{o}\mathrm{d}G$ find out.

Encryption process

1. Put the plain text $I\; is$ encrypted as ciphertext$I_1$，$I_1$Is also a size $M\times N$ , the gray level isGGThe image of$G$ , and the gray value of each pixel$I_1\left(i,j\right)$为
   $$I_1\left(i,j\right)=I\left(i,j\right)\times p\mathrm{m}\mathrm{o}\mathrm{d}G,i=0,1,\cdots ,M-1,j=0,1,\cdots ,N-1$$

2. Decrypt ciphertext $I_1$Get plaintext $I$ , the gray value at each pixel$I\left(i,j\right)$为
   $$I\left(i,j\right)=I_1\left(i,j\right)\times q\mathrm{m}\mathrm{o}\mathrm{d}G,i=0,1,\cdots ,M-1,j=0,1,\cdots ,N-1$$

  It can be seen from the above steps that this is an encryption in the image space domain, an encryption that only changes the gray value of the pixel , and the decrypted image is no different from the original image . Since the key and the decryption key are generally not equal, they are also a kind of public key encryption .

Matlab code (with gray level of $2^8=256$ images as an example)

I = imread('待读入的图像路径+名称');  % 读入图像
G = 256;  % 确定灰度级，默认图像是8位的，即灰度级是2^8=256
p = 5;  % 确定密钥

% 检验密钥p与灰度级G的最大公因数是否为1
if gcd(p, G) ~= 1
    error('密钥p与灰度级G的最大公因数不是1，不存在解钥');
end

% 展示原图像
figure, imshow(I);

% 将图像I进行加密得到密文I1, 并可视化
I1 = mod( double(I) * p, G );
figure, imshow( uint8(I1) );

% 利用简单的循环迭代找出解钥q
% 因为gcd(p, 256) = 1, 因此p一定是奇数。
% 又由于pq mod256 = 1, 因此q也一定是奇数。
q = 1;
while mod( p*q, G ) ~= 1
    q = q + 2;
end

% 将密文I1解密得到I2, 并可视化
I2 = mod( I1 * q, G );
figure, imshow( uint8(I2) );

% 比较原图像I 与 解密得到的I2 是否一致（即该算法是否有损）
% 使用isequal(), 该函数返回真值当且仅当两个矩阵相同索引处的值均相等
if isequal( double(I), I2 )
    disp('解密得到的图像与原图像一致');
else
    disp('解密得到的图像与原图像不一致');
end