RK密钥图像简单加密和恢复.rar_RK秘钥_Rk恢复密钥_图像密钥_异或加密_异或运算

%yeohyong %2018.04.02 %龙格库塔加密和恢复 clc; clear all; %清除所有变量 data = imread('yang2.jpg'); %读取图像，从默认路径读取图像到data，原图是彩色图像 [m,n,r] = size(data); %获取data图像矩阵的尺寸， if r ~= 1 %当r>1时是彩色图像，r=1是灰度图像 data1 = rgb2gray(data); %彩色图像转成灰度图像 else data1 = data; %data是灰度图像直接赋值給data1矩阵 end figure(1); %图1 imshow(data1); %显示灰度图像data1 title('灰度图像'); %*********************龙格库塔算法****************************** %函数句柄定义 f1 = @(x, y, z, t)(10*(y-x)); f2 = @(x, y, z, t)(28 * x - x* z+ y); f3 = @(x, y, z, t)(x* y - 8 /3 *z); %定义4个1行100000列的矩阵，元素全部是0 x = zeros(1, 1000); %事先知道m和n相等，则需要保存n*n个像素点 y = zeros(1, 1000); z = zeros(1, 1000); t = zeros(1, 1000); %初始值 h = 0.0005; %步长 t(1) = 0.001; x(1) = 0.01; y(1) = 0.02; z(1) = 0.05; for i = 2 : 1000 %共n*n+1-2+1=n*n个点 k11 = f1(x(i - 1), y(i - 1), z(i - 1), t(i-1)); k21 = f2(x(i - 1), y(i - 1), z(i - 1), t(i-1)); k31 = f3(x(i - 1), y(i - 1), z(i - 1), t(i-1)); k12 = f1(x(i - 1) + 0.5 * h * k11, y(i - 1) + 0.5 * h * k21, z(i - 1) + 0.5 * h * k31,t(i-1)+0.5*h); k22 = f2(x(i - 1) + 0.5 * h * k11, y(i - 1) + 0.5 * h * k21, z(i - 1) + 0.5 * h * k31,t(i-1)+0.5*h); k32 = f3(x(i - 1) + 0.5 * h * k11, y(i - 1) + 0.5 * h * k21, z(i - 1) + 0.5 * h * k31,t(i-1)+0.5*h); k13 = f1(x(i - 1) + 0.5 * h * k12, y(i - 1) + 0.5 * h * k22, z(i - 1) + 0.5 * h * k32,t(i-1)+0.5*h); k23 = f2(x(i - 1) + 0.5 * h * k12, y(i - 1) + 0.5 * h * k22, z(i - 1) + 0.5 * h * k32,t(i-1)+0.5*h); k33 = f3(x(i - 1) + 0.5 * h * k12, y(i - 1) + 0.5 * h * k22, z(i - 1) + 0.5 * h * k32,t(i-1)+0.5*h); k14 = f1(x(i - 1) + 0.5 * h * k13, y(i - 1) + 0.5 * h * k23, z(i - 1) + 0.5 * h * k33,t(i-1)+0.5*h); k24 = f2(x(i - 1) + 0.5 * h * k13, y(i - 1) + 0.5 * h * k23, z(i - 1) + 0.5 * h * k33,t(i-1)+0.5*h); k34 = f3(x(i - 1) + 0.5 * h * k13, y(i - 1) + 0.5 * h * k23, z(i - 1) + 0.5 * h * k33,t(i-1)+0.5*h); t(i) = t(i - 1) + h; x(i) = x(i - 1) + h*(1/6*k11 +1/3*k12+1/3*k13+1/6*k14); y(i) = y(i - 1) + h*(1/6*k21 +1/3*k22+1/3*k23+1/6*k24); z(i) = z(i - 1) + h*(1/6*k31 +1/3*k32+1/3*k33+1/6*k34); end %从上面算法中得到z(i),i=2、3、4.......1000，由已知得到z(1) = 0.05,所以i 取到1~1000 %计算z(n)的模 % a = z.*z; % b = sum(a); % c = sqrt(b); for i = 1 : n for j = 1 : n temp = data1(i, j); % 提去(i, j)处的像素值，相当于取出矩阵data1中的元素 % temp1 = floor(double(temp)^0.5); %floor()取下整数 %temp1 = mod(floor(double(temp) ^ 2 * rand(1)), 256); % rand(1)随机加密 %key = mod(floor(256 ^ 3 * rand(1)), 256); %key = mod(floor(c * (256 ^ 3)) , 256); key = mod(floor(sqrt(sum(z.*z)) * (256 ^ 3)) , 256); temp1 = bitxor(temp, key);% key必须是整数 data1(i, j) = temp1; temp = bitxor(temp1, key); data2(i, j) = temp; end end % end figure(2); imshow(uint8(data1)); title('R-K加密'); imwrite(data2,'D:\Databank_Yeoh\Matlab_databank\Matlab_Path_Yeoh\yangyong3.jpg'); figure(3); imshow(uint8(data2)); title('R-K加密恢复后的图像'); %imtool(uint8(data3)); %可以显示全图