├── An image encryption scheme based on Lorenz Hyperchaotic System and RSA algorithm-科研业绩成果材料附件.pdf ├── README.md ├── RSA ├── Copy_of_RSA.m ├── RSA.m ├── RSA_test.m ├── prime_produce.m └── rsa_code.m ├── encryption ├── BACIExpect.m ├── ENTROPYtest.m ├── GF257Table.m ├── GF257TableEx.m ├── House.tiff ├── ImCoef.m ├── ImCoef_test.m ├── Lake.tiff ├── Lena.bmp ├── LookUpGF257.m ├── LookUpGF257Ex.m ├── LookUpGF257Ex2.m ├── Male.tiff ├── NPCRUACI.m ├── NPCRUACIBACI.m ├── Peppers.tiff ├── TpDecrypt.m ├── TpEncrypt.m ├── Tree.tiff ├── UACIExpect.m ├── UACIExpect_test.m ├── gray21.512.tiff ├── hist_test.m ├── img_hist.m ├── lena256.bmp ├── lena512.bmp ├── lena512color.tiff ├── noise_attack.m ├── ruler.512.tiff └── variances.m ├── pictures ├── 5.1.09.tiff ├── 5.1.10.tiff ├── 5.1.11.tiff ├── 5.1.12.tiff ├── 5.1.13.tiff ├── 5.1.14.tiff ├── 5.2.08.tiff ├── 5.2.09.tiff ├── 5.3.01.tiff ├── 5.3.02.tiff ├── 7.1.01.tiff ├── 7.1.02.tiff ├── 7.1.03.tiff ├── 7.1.04.tiff ├── 7.1.05.tiff ├── 7.1.06.tiff ├── 7.1.07.tiff ├── 7.1.08.tiff ├── 7.1.09.tiff ├── 7.1.10.tiff ├── 7.2.01.tiff ├── boat.512.tiff ├── gray21.512.tiff └── ruler.512.tiff └── test_code ├── Copy_of_ImCoef_test.m ├── Copy_of_code4_25.m ├── Copy_of_crypt_test.m ├── code4_14.m ├── code4_16.m ├── code4_18.m ├── code4_19.m ├── code4_21.m ├── code4_22.m ├── code4_23.m ├── code4_24.m ├── code4_25.m ├── crypt_test.m └── crypt_test_color.m /An image encryption scheme based on Lorenz Hyperchaotic System and RSA algorithm-科研业绩成果材料附件.pdf: -------------------------------------------------------------------------------- https://raw.githubusercontent.com/linriguang/An-image-encryption-scheme-based-on-Lorenz-Hyperchaotic-System-and-RSA-algorithm/b546d489365c21d3973de7b5f96a760ab7c0a157/An image encryption scheme based on Lorenz Hyperchaotic System and RSA algorithm-科研业绩成果材料附件.pdf -------------------------------------------------------------------------------- /README.md: -------------------------------------------------------------------------------- 1 | # An-image-encryption-scheme-based-on-Lorenz-Hyperchaotic-System-and-RSA-algorithm 2 | This research proposes a new image encryption scheme based on Lorenz hyperchaotic system and Rivest-Shamir-Adleman (RSA) algorithm. Firstly, the initial values of the Lorenz hyperchaotic system are generated by RSA algorithm, and the keystream is produced iteratively. In order to change the position and gray value of the pixel, the image data are hidden by additive mode diffusion. Secondly, the diffusion image matrix is reshaped into a one-dimensional image matrix, which is confused without repetition to hide the image data again. Then, the finite field diffusion algorithm is executed to realize the third hiding of the image information. In order to diffuse the pixel information into the entire cipher image, the additive mode diffusion algorithm needs to be looped twice. Finally, the cipher image can be obtained. The experimental results prove that the image encryption scheme proposed in this research is effective, and has strong anti-attack and key sensitivity. Moreover, the security of this encryption scheme relies on the RSA algorithm, which has high security. 3 | 具体的实验过程可以参考我的CSDN博客，超混沌Lorenz系统图像加密（MATLAB）：https://blog.csdn.net/qq_36439087/article/details/109964775?spm=1001.2014.3001.5502 4 | -------------------------------------------------------------------------------- /RSA/Copy_of_RSA.m: -------------------------------------------------------------------------------- 1 | % RSA加密和解密算法 2 | clear;clc%清除运行空间 3 | disp('RSA algorithm'); 4 | p = 3259;%大素数 最大的数是10000 5 | fprintf('\np=%d',p); 6 | q = 3821;%大素数 7 | fprintf('\nq=%d',q); 8 | n=p*q; %模数n 9 | fprintf('\nn=%d',n); 10 | phi=(p-1)*(q-1);%欧拉函数 11 | fprintf('\nphi(%d) is %d',n,phi); 12 | e = 1288367; 13 | d = 3385223; 14 | 15 | % %为了找到公钥e 16 | % while(cd~=1||val==0) 17 | % n1=randi(n,1,1);%生成随机数 18 | % e=randi([2 n1],1,1); 19 | % val=isprime(e);%确定哪些数组元素为质数 20 | % cd=gcd(e,phi);%G = gcd(A,B) 返回 A 和 B 的元素的最大公约数 21 | % end 22 | % 23 | % %vall 是否为1 d*e=1mod(phi(n)) d 为私钥 24 | % val1=0; 25 | % d=0; 26 | % while(val1~=1) 27 | % d=d+1; 28 | % val1=mod(d*e,phi); 29 | % end 30 | fprintf('\nd=%d',d); 31 | fprintf('\nPublic key is (%d,%d)',e,n); 32 | fprintf('\nPrivate key is (%d,%d)',d,n); 33 | m = [178333,38628,92873897,829809]; 34 | %m = [59806503,2349576,60074184,7488026]; 35 | m1=m; 36 | fprintf('\n'); 37 | disp('输入的明文大数M为： '); 38 | disp(m1); 39 | over=length(m1);%长度 40 | o=1; 41 | while(o<=over) 42 | m=m1(o);%文本字符串的数值 43 | diff=0; 44 | if(m>n) 45 | diff=m-n+1; 46 | end 47 | m=m-diff; 48 | 49 | qm=dec2bin(e); 50 | len=length(qm); 51 | c=1; 52 | xz=1; 53 | 54 | while(xz<=len) 55 | if(qm(xz)=='1') 56 | c=mod(mod((c^2),n)*m,n); 57 | elseif(qm(xz)=='0') 58 | c=(mod(c^2,n)); 59 | end 60 | xz=xz+1; 61 | end 62 | c1(o)=c; 63 | 64 | qm1=dec2bin(d); 65 | len1=length(qm1); 66 | nm=1; 67 | xy=1; 68 | 69 | while(xy<=len1) 70 | if(qm1(xy)=='1') 71 | nm=mod(mod((nm^2),n)*c,n); 72 | elseif(qm1(xy)=='0') 73 | nm=(mod(nm^2,n)); 74 | end 75 | xy=xy+1; 76 | end 77 | 78 | nm=nm+diff; 79 | nm1(o)=char(nm); 80 | o=o+1; 81 | end 82 | 83 | 84 | 85 | o=1; 86 | fprintf('\n密文C为：\n'); 87 | while(o<=over) 88 | fprintf('\t%d',c1(o)); 89 | o=o+1; 90 | end 91 | % o=1; 92 | % fprintf('\n明文M为： \n'); 93 | % while(o<=over) 94 | % fprintf('\t%d',nm1(o)); 95 | % o=o+1; 96 | % end 97 | 98 | %fprintf('\nThe decrypted message is: '); 99 | %disp(nm1); 100 | 101 | fprintf('\n'); 102 | 103 | Initial_value = sqrt(log(m1+c1)) -------------------------------------------------------------------------------- /RSA/RSA.m: -------------------------------------------------------------------------------- 1 | % RSA加密和解密算法 2 | clear;clc%清除运行空间 3 | disp('RSA algorithm'); 4 | p = prime_produce(10^4);%大素数 最大的数是10000 5 | fprintf('\np=%d',p); 6 | q = prime_produce(10^4);%大素数 7 | fprintf('\nq=%d',q); 8 | n=p*q; %模数n 9 | fprintf('\nn=%d',n); 10 | phi=(p-1)*(q-1);%欧拉函数 11 | fprintf('\nphi(%d) is %d',n,phi); 12 | val=0;%逻辑数组 13 | cd=0;%最大公约数，初值为0 14 | 15 | %为了找到公钥e 16 | while(cd~=1||val==0) 17 | n1=randi(n,1,1);%生成随机数 18 | e=randi([2 n1],1,1); 19 | val=isprime(e);%确定哪些数组元素为质数 20 | cd=gcd(e,phi);%G = gcd(A,B) 返回 A 和 B 的元素的最大公约数 21 | end 22 | 23 | %vall 是否为1 d*e=1mod(phi(n)) 24 | %d 为私钥 25 | val1=0; 26 | d=0; 27 | while(val1~=1) 28 | d=d+1; 29 | val1=mod(d*e,phi); 30 | end 31 | fprintf('\nd=%d',d); 32 | fprintf('\nPublic key is (%d,%d)',e,n); 33 | fprintf('\nPrivate key is (%d,%d)',d,n); 34 | m = [178334,38628,92873897,829809]; 35 | %m = [59806503,2349576,60074184,7488026]; 36 | m1=m; 37 | fprintf('\n'); 38 | disp('输入的明文大数M为： '); 39 | disp(m1); 40 | over=length(m1);%长度 41 | o=1; 42 | while(o<=over) 43 | m=m1(o);%文本字符串的数值 44 | diff=0; 45 | if(m>n) 46 | diff=m-n+1; 47 | end 48 | m=m-diff; 49 | 50 | qm=dec2bin(e); 51 | len=length(qm); 52 | c=1; 53 | xz=1; 54 | 55 | while(xz<=len) 56 | if(qm(xz)=='1') 57 | c=mod(mod((c^2),n)*m,n); 58 | elseif(qm(xz)=='0') 59 | c=(mod(c^2,n)); 60 | end 61 | xz=xz+1; 62 | end 63 | c1(o)=c; 64 | 65 | qm1=dec2bin(d); 66 | len1=length(qm1); 67 | nm=1; 68 | xy=1; 69 | 70 | while(xy<=len1) 71 | if(qm1(xy)=='1') 72 | nm=mod(mod((nm^2),n)*c,n); 73 | elseif(qm1(xy)=='0') 74 | nm=(mod(nm^2,n)); 75 | end 76 | xy=xy+1; 77 | end 78 | 79 | nm=nm+diff; 80 | nm1(o)=char(nm); 81 | o=o+1; 82 | end 83 | 84 | 85 | 86 | o=1; 87 | fprintf('\n密文C为：\n'); 88 | while(o<=over) 89 | fprintf('\t%d',c1(o)); 90 | o=o+1; 91 | end 92 | % o=1; 93 | % fprintf('\n明文M为： \n'); 94 | % while(o<=over) 95 | % fprintf('\t%d',nm1(o)); 96 | % o=o+1; 97 | % end 98 | 99 | %fprintf('\nThe decrypted message is: '); 100 | %disp(nm1); 101 | 102 | fprintf('\n'); 103 | 104 | Initial_value = sqrt(log(m1+c1)) -------------------------------------------------------------------------------- /RSA/RSA_test.m: -------------------------------------------------------------------------------- 1 | % RSA加密和解密算法 2 | clear;clc%清除运行空间 3 | disp('RSA algorithm'); 4 | p = 6899;%prime_produce(10^4);%大素数 5 | fprintf('\np=%d',p); 6 | q = 8329;%prime_produce(10^4);%大素数 7 | fprintf('\nq=%d',q); 8 | n=p*q; %模数n 9 | fprintf('\nn=%d',n); 10 | phi=(p-1)*(q-1);%欧拉函数 11 | fprintf('\nphi(%d) is %d',n,phi); 12 | val=0;%逻辑数组 13 | cd=0;%最大公约数，初值为0 14 | e = 6796771; 15 | d = 30904459; 16 | % %为了找到公钥e 17 | % while(cd~=1||val==0) 18 | % n1=randi(n,1,1);%生成随机数 19 | % e=randi([2 n1],1,1); 20 | % val=isprime(e);%确定哪些数组元素为质数 21 | % cd=gcd(e,phi);%G = gcd(A,B) 返回 A 和 B 的元素的最大公约数 22 | % end 23 | % 24 | % %vall 是否为1 d*e=1mod(phi(n)) 25 | % %d 为私钥 26 | % val1=0; 27 | % d=0; 28 | % while(val1~=1) 29 | % d=d+1; 30 | % val1=mod(d*e,phi); 31 | % end 32 | fprintf('\nd=%d',d); 33 | fprintf('\nPublic key is (%d,%d)',e,n); 34 | fprintf('\nPrivate key is (%d,%d)',d,n); 35 | %m = [2345,3456,4567,5678]; 36 | m = [45817714,24015420,36247848,32235019]; 37 | m1=m; 38 | fprintf('\n'); 39 | disp('输入的明文大数M为： '); 40 | disp(m1); 41 | over=length(m1);%长度 42 | o=1; 43 | while(o<=over) 44 | m=m1(o);%文本字符串的数值 45 | diff=0; 46 | if(m>n) 47 | diff=m-n+1; 48 | end 49 | m=m-diff; 50 | 51 | %qm=dec2bin(e); 52 | qm=dec2bin(d); 53 | len=length(qm); 54 | c=1; 55 | xz=1; 56 | 57 | while(xz<=len) 58 | if(qm(xz)=='1') 59 | c=mod(mod((c^2),n)*m,n); 60 | elseif(qm(xz)=='0') 61 | c=(mod(c^2,n)); 62 | end 63 | xz=xz+1; 64 | end 65 | c1(o)=c; 66 | 67 | %qm1=dec2bin(d); 68 | qm1=dec2bin(e); 69 | len1=length(qm1); 70 | nm=1; 71 | xy=1; 72 | 73 | while(xy<=len1) 74 | if(qm1(xy)=='1') 75 | nm=mod(mod((nm^2),n)*c,n); 76 | elseif(qm1(xy)=='0') 77 | nm=(mod(nm^2,n)); 78 | end 79 | xy=xy+1; 80 | end 81 | 82 | nm=nm+diff; 83 | nm1(o)=char(nm); 84 | o=o+1; 85 | end 86 | 87 | 88 | 89 | o=1; 90 | fprintf('\n密文C为：\n'); 91 | while(o<=over) 92 | fprintf('\t%d',c1(o)); 93 | o=o+1; 94 | end 95 | % o=1; 96 | % fprintf('\n明文M为： \n'); 97 | % while(o<=over) 98 | % fprintf('\t%d',nm1(o)); 99 | % o=o+1; 100 | % end 101 | 102 | %fprintf('\nThe decrypted message is: '); 103 | %disp(nm1); 104 | 105 | fprintf('\n'); -------------------------------------------------------------------------------- /RSA/prime_produce.m: -------------------------------------------------------------------------------- 1 | % 随机产生两个大素数 2 | function num = prime_produce(n) 3 | 4 | num = 0; 5 | val = 0; 6 | while(val == 0) 7 | prime1 = randi([1000 n],1,1); 8 | val=isprime(prime1);%确定哪些数组元素为质数 9 | end 10 | num = prime1; -------------------------------------------------------------------------------- /RSA/rsa_code.m: -------------------------------------------------------------------------------- https://raw.githubusercontent.com/linriguang/An-image-encryption-scheme-based-on-Lorenz-Hyperchaotic-System-and-RSA-algorithm/b546d489365c21d3973de7b5f96a760ab7c0a157/RSA/rsa_code.m -------------------------------------------------------------------------------- /encryption/BACIExpect.m: -------------------------------------------------------------------------------- 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-------------------------------------------------------------------------------- /encryption/NPCRUACI.m: -------------------------------------------------------------------------------- 1 | function nu = NPCRUACI(P1,P2) 2 | nu = zeros(1,2); 3 | P1 = double(P1);P2 = double(P2);[M,N] = size(P1); 4 | D = (P1~=P2);nu(1) = sum(sum(D))/(M * N) * 100; 5 | fprintf('NPCR = %8.4f% %. ',nu(1)); 6 | nu(2) = sum(sum(abs(P1 - P2)))/(255 * M *N) * 100; 7 | fprintf('UACI = %8.4f% % . \n',nu(2)); 8 | end 9 | -------------------------------------------------------------------------------- /encryption/NPCRUACIBACI.m: -------------------------------------------------------------------------------- 1 | % ����4-20 �Զ��庯��NPCPUACIBACI 2 | function nu = NPCRUACIBACI(P1,P2) 3 | nu = zeros(1,3); 4 | P1 = double(P1);P2 = double(P2);[M,N] = size(P1); 5 | D = (P1 ~= P2);nu(1) = sum(sum(D))/(M * N) * 100; % fprintf('NPCR = %8.4f% %. \n',nu(1)); 6 | nu(2) = sum(sum(abs(P1 - P2)))/(255 * M * N) * 100;% fprintf('UACI = %8.4f% % . \n',nu(2)); 7 | 8 | D = abs(P1 - P2);m = 0; 9 | for i = 1:M - 1 10 | for j = 1:N - 1 11 | d = D(i:i+1,j:j+1); 12 | m = m + (abs(d(1,1)-d(1,2))+abs(d(1,1)-d(2,1))+abs(d(1,1)-d(2,2))+abs(d(1,2)-d(2,1))+abs(d(1,2)-d(2,2))+abs(d(2,1)-d(2,2)))/6/255; 13 | end 14 | end 15 | nu(3) = m/((M - 1) * (N - 1)) * 100;% fprintf('BACI = %8.4f% % . \n',nu(3)); 16 | end 17 | 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A(i,j) 10 | tot_n(k + 1) = tot_n(k + 1) + 1; 11 | end 12 | end 13 | end 14 | end 15 | tot_n(1) = tot_n(1)/2;e_uaci = sum(tot_s.*tot_n)/sum(tot_n)/255; 16 | fprintf('UACI = %10.4f% % \n',e_uaci * 100); 17 | end 18 | 19 | -------------------------------------------------------------------------------- /encryption/UACIExpect_test.m: -------------------------------------------------------------------------------- https://raw.githubusercontent.com/linriguang/An-image-encryption-scheme-based-on-Lorenz-Hyperchaotic-System-and-RSA-algorithm/b546d489365c21d3973de7b5f96a760ab7c0a157/encryption/UACIExpect_test.m -------------------------------------------------------------------------------- /encryption/gray21.512.tiff: -------------------------------------------------------------------------------- https://raw.githubusercontent.com/linriguang/An-image-encryption-scheme-based-on-Lorenz-Hyperchaotic-System-and-RSA-algorithm/b546d489365c21d3973de7b5f96a760ab7c0a157/encryption/gray21.512.tiff 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-------------------------------------------------------------------------------- /encryption/noise_attack.m: -------------------------------------------------------------------------------- 1 | % 加密和解密测试 2 | 3 | %噪声攻击测试 4 | 5 | clear;clc;close all; 6 | %K = [4.0349 3.7979 4.2980 3.9282]; %读入混沌系统的初始值 7 | K = [3.5739 3.7979 4.2980 3.9282]; 8 | img = imread('ruler.512.tiff');%读入图像 9 | %img = imread('Baboon.tiff'); 10 | %img = imread('Peppers.tiff'); 11 | %img = imread('boat.tiff'); 12 | %img = imread('Clock.tiff'); 13 | %img = imread('House.tiff'); 14 | %img = imread('Male.tiff'); 15 | %img = imread('Tree.tiff'); 16 | 17 | %P = rgb2gray(img); %彩色图像要转化为灰度图像 18 | %figure(1);imshow(P); 19 | P = img; 20 | 21 | tic;C = TpEncrypt(P,K);toc; 22 | C = uint8(C); 23 | figure(1); 24 | g = imnoise(C,'gaussian',0.01);%添加高斯噪声 25 | g = uint8(g); 26 | imshow(g);%加密计时，显示加了噪声的密文图像 27 | 28 | tic;P1 = TpDecrypt(g,K);toc;figure(2);imshow(P1);%解密计时，显示解密后的明文图像 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imhist(C2(:),256); 18 | %计算方差 19 | var_y = 0; 20 | for i = 1:256 21 | for j = 1:256 22 | var_y = var_y + (y2(i)-y3(j))^2/2; 23 | end 24 | end 25 | format long 26 | var_z = var_y/(256*256)%方差值 27 | 28 | 29 | 30 | 31 | 32 | 33 | 34 | 35 | 36 | -------------------------------------------------------------------------------- /pictures/5.1.09.tiff: -------------------------------------------------------------------------------- https://raw.githubusercontent.com/linriguang/An-image-encryption-scheme-based-on-Lorenz-Hyperchaotic-System-and-RSA-algorithm/b546d489365c21d3973de7b5f96a760ab7c0a157/pictures/5.1.09.tiff -------------------------------------------------------------------------------- /pictures/5.1.10.tiff: -------------------------------------------------------------------------------- https://raw.githubusercontent.com/linriguang/An-image-encryption-scheme-based-on-Lorenz-Hyperchaotic-System-and-RSA-algorithm/b546d489365c21d3973de7b5f96a760ab7c0a157/pictures/5.1.10.tiff 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