CN103825699B - To the Parallel Chaos encryption method storing file in cloud computing environment - Google Patents

Abstract

The encryption being data file to the Parallel Chaos encryption method of storage file in cloud computing environment provides a kind of lightweight encryption that can be parallel, belongs to information security field.Utilize cloud computing environment can parallel computation mechanism, such as MapReduce parallel computing, in conjunction with chaos encrypting method, by clear data is carried out block encryption, finally realize the encryption to whole file, and upload in the storage system of cloud computing environment.Method contains the calculating of encryption system initial value, encryption key sequence calculates and three steps of data encryption.In encryption system initializes, method uses three-dimensional Lorenz chaos system and Chen chaos system to generate the initial value for encryption and interference value；During calculating encryption key sequence, based on the initial value calculated and interference value, by Henon chaotic maps and the interative computation of 2 dimension Logistic chaotic maps, generate the key sequence for data encryption；In data encryption process, carry out computing by key sequence with plaintext, finally realize the encryption to clear data.

Description

To the Parallel Chaos encryption method storing file in cloud computing environment

Technical field

The present invention is directed to cloud computing environment needs the data file security problem of storage, it is proposed that one is applicable to cloud meter Calculating the parallel mixed chaos encipherment scheme of environment, its thinking is: the owner of data file will be stored in the literary composition in cloud environment Part is first encrypted, then uploads in the memory space rented, for user, it is achieved the confidentiality requirement to data.This Invention relates to chaos calculating, parallel computing and cloud computing technology, belongs to information security field.

Background technology

Big data security arts is the hot issue received much concern in recent years.On the one hand, data owner will be a large amount of Data be stored in cloud computing environment, for user access.But these mass data have a lot of sensitive data needs Ensure its confidentiality, such as the privacy information such as positional information, personal identification etc., how to realize data file in this kind of memory module Information encryption, it is achieved safely, effectively, simple data access control be cloud computing move towards actual application needs solution ask Topic.On the other hand, owing to file data quantity is big, need to build a kind of fast encryption scheme for big data, and at cloud meter Calculate in environment it is necessary to solve how can to utilize the cloud computing environment can the feature of parallel computation, it is achieved big data fast parallel Encryption.The present invention be aiming at file data confidentiality and can concurrency, it is proposed that a kind of cloud computing environment that can be applicable to Parallel mixed chaos encryption technology, the program can be effectively improved the enciphering rate of big data quantity.

Method contains the calculating of encryption system initial value, encryption key sequence calculates and three steps of data encryption.In encryption In system initialization, method uses three-dimensional Lorenz chaos system and Chen chaos system to generate the initial value for encryption and interference Value；During calculating encryption key sequence, based on the initial value calculated and interference value, by Henon chaotic maps and 2 dimensions The interative computation of Logistic chaotic maps, generates the key sequence for data encryption；In data encryption process, by close Key sequence carries out computing with plaintext, finally realizes the encryption to clear data.

Owing to using the computings such as block encryption, make the method for proposition can utilize the concurrent technique in cloud computing environment, real Existing encryption method can parallel computation.

Summary of the invention

Goal of the invention: it is an object of the invention to provide the Parallel Chaos encryption side to storage file in a kind of cloud computing environment Method, logical

Cross and combine the concurrent technique of cloud computing environment and Chaos Encryption Technology realizes a kind of parallel Chaotic Encryption, with Improve the encryption/decryption speed of big data quantity.

Technical scheme: the present invention for achieving the above object, adopts the following technical scheme that

Step one: data file piecemeal

Note data file a length of L the most in plain text, will be divided into n block in plain text, and the size of each piece is l_i, i≤n, and require l_i It is the integral multiple of 4 bytes, is designated as l_i=4N；Before so, n-1 block is isometric data block, and n-th piece is remainder data, it may be assumed that

l_i=4N (i ＜ n)

(1)

l_i=L-(n-1) 4N (i=n)

Step 2: application nonlinear chaotic system generates initiation sequence and interference sequence

For each piecemeal, at the beginning of being produced by three-dimensional continuous chaotic system Lorenz chaotic maps and Chen chaotic maps Value sequence and interference value sequence, method is as follows:

Lorenz chaotic maps equation is:

$\left\{\begin{array}{c}\frac{{\mathrm{dx}}_1}{\mathrm{dt}}=p(x_2-x_1)\\ \frac{{\mathrm{dx}}_2}{\mathrm{dt}}=-x_1{x}_3+r{x}_1-x_2\\ \frac{{\mathrm{dx}}_3}{\mathrm{dt}}=x_2{x}_1-t{x}_3\end{array}\right.---\left(2\right)$

Chen chaos system equation is:

$\left\{\begin{array}{c}\frac{{\mathrm{dx}}_4}{\mathrm{dt}}=a(x_5-x_4)\\ \frac{{\mathrm{dx}}_5}{\mathrm{dt}}=(c-a)x_4-x_4{x}_6+{\mathrm{cx}}_5\\ \frac{{\mathrm{dx}}_6}{\mathrm{dt}}=x_4{x}_5-b{x}_6\end{array}\right.---\left(3\right)$

Wherein χ=(x₁,x₂,x₃,x₄,x₅,x₆) it is chaos sequence to be generated, p, r, t and a, b, c are that system is to be given Parameter, utilizes 4 rank runge kutta methods to solve chaos system equation (2) and (3), and each iteration generates three random sequences respectively Value, i.e. 6 chaos sequence value χ=(x₁,x₂,x₃,x₄,x₅,x₆)；In order to be calculated this chaos sequence value, first obtain in plain text Side-play amount θ of block first character joint_i, the primary iteration number of times taking Lorenz system is N₀, the primary iteration number of times of Chen system is M₀, the iterations of note Lorenz system is ρ_L, the iterations of Chen system is ρ_C, then have:

${\mathrm{\ρ}}_L=\frac{{\mathrm{\θ}}_i}{N}+N_0,{\mathrm{\ρ}}_C=\frac{{\mathrm{\θ}}_i}{N}+M_0---\left(4\right)$

The corresponding chaos sequence of the most each piecemeal is

χ_ρ=(x₁(ρ_L),x₂(ρ_L),x₃(ρ_L),x₄(ρ_C),x₅(ρ_C),x₆(ρ_C)) (5)

Sequence in order to make two systems produce has coupling, the random sequence pair Lorenz system obtained produced With Chen system produce random sequence to carrying out XOR:

$\begin{array}{c}{x}_4\left({\mathrm{\ρ}}_C\right)=x_4\left({\mathrm{\ρ}}_C\right)\⊕x_1\left({\mathrm{\ρ}}_L\right)\\ x_5\left({\mathrm{\ρ}}_C\right)=x_5\left({\mathrm{\ρ}}_C\right)\⊕x_2\left({\mathrm{\ρ}}_L\right)\\ x_6\left({\mathrm{\ρ}}_C\right)=x_6\left({\mathrm{\ρ}}_C\right)\⊕x_3\left({\mathrm{\ρ}}_L\right)\end{array}---\left(6\right)$

Meter chaos sequence χ_ρFor first value sequence ${\mathrm{\χ}}_p^I=(x_1\left({\mathrm{\ρ}}_L\right),x_2\left({\mathrm{\ρ}}_L\right),x_3\left({\mathrm{\ρ}}_L\right),x_4\left({\mathrm{\ρ}}_C\right))$ And interference sequence ${\mathrm{\χ}}_{\mathrm{\ρ}}^d=(x_5\left({\mathrm{\ρ}}_C\right),x_6\left({\mathrm{\ρ}}_C\right));$

Step 3: calculate the chaos sequence of encryption key with first value sequence

By first value sequenceIn (x₁(ρ_L),x₂(ρ_L)) fractional part as H é non chaotic maps iteration initial value, (x₃(ρ_L),x₄(ρ_L)) fractional part produces the key sequence of encryption as the initial value of Logistic Map iteration, recurrence ω_ρ=(ω₁(ρ),ω₂(ρ),ω₃(ρ),ω₄(ρ)), wherein (ω₁(ρ),ω₂(ρ)) by the H é non system equation iteration of formula (7) Produce:

$\left\{\begin{array}{c}{x}_{n+1}=y_n+1-a{x}_n^2\\ y_{n+1}={\mathrm{bx}}_n\end{array}\right.---\left(7\right)$

(ω₃(ρ),ω₄(ρ)) produced by the Coupled Logistic chaotic map system equation iteration of formula (8):

$\left\{\begin{array}{c}{x}_{n+1}={\mathrm{\μ}}_1{x}_n(1-x_n)+{\mathrm{\γ}}_1{y}_n^2\\ y_{n+1}={\mathrm{\μ}}_2{y}_n(1-y_n)+{\mathrm{\γ}}_2(x_n^2+x_n{y}_n)\end{array}\right.---\left(8\right)$

Parameter value in mapping is: 2.75 ＜ μ₁＜ 3.4,2.7 ＜ μ₂＜ 3.45 and 0.15 ＜ γ₁＜ 0.21,0.13 ＜ γ₂＜ 0.15.

Step 4: reconstruct encryption key sequence

After obtaining encryption key sequence, due to its value produced can not with in plain text carry out xor operation, so need by Encryption key sequence is changed according to following formula:

Sequence in order to make two systems produce has coupling, carries out calculated as below:

$\begin{array}{c}{\mathrm{\ω}}_3\left(\mathrm{\ρ}\right)={\mathrm{\ω}}_1\left(\mathrm{\ρ}\right)\⊕{\mathrm{\ω}}_3\left(\mathrm{\ρ}\right)\\ {\mathrm{\ω}}_4\left(\mathrm{\ρ}\right)={\mathrm{\ω}}_2\left(\mathrm{\ρ}\right)\⊕{\mathrm{\ω}}_4\left(\mathrm{\ρ}\right)\end{array}---\left(10\right)$

Step 5: data encryption

After generating encryption key sequence, will carry out the encryption of clear text file, cryptographic operation is to take 4 successively in plaintext The ASCII character M that individual byte length data block is corresponding, carries out XOR (XOR) operation with encryption key sequence and obtains intermediate object program, so After carry out secondary xor operation with interference sequence again and obtain final ciphertext C:

$\left\{\begin{array}{c}{C}_{4k+1}=(M_{4k+1}\⊕{\mathrm{\ω}}_1(k+1))\⊕x_5(k+1)\\ C_{4k+2}=(M_{4k+2}\⊕{\mathrm{\ω}}_2(k+1))\⊕x_6(k+1)\\ C_{4k+3}=(M_{4k+3}\⊕{\mathrm{\ω}}_3(k+1))\⊕x_5(k+1)\\ C_{4k+4}=(M_{4k+4}\⊕{\mathrm{\ω}}_4(k+1))\⊕x_6(k+1)\end{array}\right.---\left(11\right)$

Wherein k represents the sequence number of data block, C_4k+1, C_4k+2, C_4k+3, C_4k+4Be respectively in data block corresponding to the 1st, 2,3 and The encrypted cipher text of 4 bytes.In the encryption method that method proposes, key space includes initial value χ (0)=(x of key sequence₁ (0),x₂(0),x₃(0),x₄(0),x₅(0),x₆(0)), the primary iteration times N of Lorenz and Chen system₀And M₀, and two Parameter p of individual chaos system, r, t and a, b, c.

Beneficial effect: the parallel mixed chaos encipherment scheme that the present invention designs solves in cloud computing actual application Two problems to be solved, one is data owner's confidentiality requirements to data file, by adding the packet of clear text file Close, upload and be stored in cloud environment, it is ensured that the sensitive information of data owner is protected；Two is to calculate for big data high-speed Requirement, uses block encryption etc. to see parallel computing technique, makes the ciphering process of data file may utilize in cloud computing environment Parallel computing, it is achieved the high-speed parallel of large data files encryption is calculated, finally meets big data in cloud computing environment Secure high-speed access.

Accompanying drawing explanation

Fig. 1 is the flow process of encipherment scheme.

Detailed description of the invention

Enforcement to technical scheme below is described in further detail:

Step one: data file piecemeal

If the file that length of the plaintext is 15 bytes, i.e. L=15.Being divided into 4 pieces, the greatest length of each piece is 4 bytes, Then there is N=1.First 3 pieces is isometric 4 byte data blocks, and the 4th piece is 3 bytes of remainder data, it may be assumed that

l_i=4N=4 (i=1,2,3)

(12)

l_i=L-(n-1) 4N=15-(4-1) 4=3 (i=4)

Step 2: application nonlinear chaotic system generates initiation sequence and interference sequence

For each piecemeal, at the beginning of being produced by three-dimensional continuous chaotic system Lorenz chaotic maps and Chen chaotic maps Value sequence and interference value sequence.Method is as follows:

Lorenz chaotic maps equation is:

$\left\{\begin{array}{c}\frac{{\mathrm{dx}}_1}{\mathrm{dt}}=p(x_2-x_1)\\ \frac{{\mathrm{dx}}_2}{\mathrm{dt}}=-x_1{x}_3+r{x}_1-x_2\\ \frac{{\mathrm{dx}}_3}{\mathrm{dt}}=x_2{x}_1-t{x}_3\end{array}\right.---\left(13\right)$

Wherein p=10, r=28, t=8/3.Chen chaos system equation is:

$\left\{\begin{array}{c}\frac{{\mathrm{dx}}_4}{\mathrm{dt}}=a(x_5-x_4)\\ \frac{{\mathrm{dx}}_5}{\mathrm{dt}}=(c-a)x_4-x_4{x}_6+{\mathrm{cx}}_5\\ \frac{{\mathrm{dx}}_6}{\mathrm{dt}}=x_4{x}_5-b{x}_6\end{array}\right.---\left(14\right)$

Wherein χ=(x₁,x₂,x₃,x₄,x₅,x₆) it is chaos sequence to be generated, a=35, b=3, c=28.Utilize 4 rank dragons Ge Kutafa solves chaos system equation (13) and (14), and each iteration generates three stochastic ordering train values, i.e. 6 chaos sequences respectively Train value χ=(x₁,x₂,x₃,x₄,x₅,x₆).In order to be calculated this chaos sequence value, according to the division of data block, can calculate Go out side-play amount θ of each Plaintext block first character joint₁,θ₂,θ₃,θ₄, divide according to data block and have θ₁=0, θ₂=4, θ₃=8, θ₄= 12.The primary iteration number of times taking Lorenz system is N₀The primary iteration number of times of=2000, Chen system is M₀=3000.Then two The iterations of system is respectively as follows:

$\begin{array}{cc}{\mathrm{\ρ}}_L\left(1\right)=\frac{0}{4}+2000=2000& {\mathrm{\ρ}}_C\left(1\right)=\frac{0}{4}+3000=3000\\ {\mathrm{\ρ}}_L\left(2\right)=\frac{4}{4}+2000=2001& {\mathrm{\ρ}}_C\left(2\right)=\frac{4}{4}+3000=3001\\ {\mathrm{\ρ}}_L\left(3\right)=\frac{8}{4}+2000=2002& {\mathrm{\ρ}}_C\left(3\right)=\frac{8}{4}+3000=3002\\ {\mathrm{\ρ}}_L\left(4\right)=\frac{12}{4}+2000=2003& {\mathrm{\ρ}}_C\left(4\right)=\frac{12}{4}+3000=3003\end{array}---\left(15\right)$

Taking initial value is x₁(0)=1,x₂(0)=2,x₃(0)=3,x₄(0)=4,x₅(0)=5,x₆(0)=6, then it is calculated each The corresponding chaos sequence of piecemeal is:

χ_ρ=(x₁(ρ_L),x₂(ρ_L),x₃(ρ_L),x₄(ρ_C),x₅(ρ_C),x₆(ρ_C))

χ_ρ(1)=(x₁(2000),x₂(2000),x₃(2000),x₄(3000),x₅(3000),x₆(3000))

=(17.875805137193584,23.577775139444757,35.317757483789414,

6.717742409639023,7.835244360075882,10.201089575432276)

χ_ρ(2)=(x₁(2001),x₂(2001),x₃(2001),x₄(3001),x₅(3001),x₆(3001))

=(18.17831668337063,22.534873609894326,37.37817682181235,

6.782891166690998,7.8362000710961714,10.394920549118666)

χ_ρ(3)=(x₁(2002),x₂(2002),x₃(2002),x₄(3002),x₅(3002),x₆(3002))

=(18.395763937117206,21.26573786112585,39.3236292734919,

6.844042435179011,7.828307249610151,10.589251029269274)

χ_ρ(4)=(x₁(2003),x₂(2003),x₃(2003),x₄(3003),x₅(3003),x₆(3003))

=(18.546007636272776,18.12250541456983,42.723826295121185,

6.953205127666364,7.785230347792439,10.977112780831336)

Step 3: calculate the chaos sequence of encryption key with first value sequence

By chaos sequence χ_ρIt is divided into just value sequence ${\mathrm{\χ}}_p^I=(x_1\left({\mathrm{\ρ}}_L\right),x_2\left({\mathrm{\ρ}}_L\right),x_3\left({\mathrm{\ρ}}_L\right),x_4\left({\mathrm{\ρ}}_C\right)),$ And interference sequenceBy first value sequenceFractional part will reflect as H é non chaotic maps and 2D-Logistic The initial value penetrated, carries out recurrence and produces ciphering sequence ω_ρ=(ω₁(ρ),ω₂(ρ),ω₃(ρ),ω₄(ρ)), wherein (ω₁(ρ), ω₂(ρ)) produced by the H é non system iterative equation of formula (16):

$\left\{\begin{array}{c}{x}_{n+1}=y_n+1-a{x}_n^2\\ y_{n+1}={\mathrm{bx}}_n\end{array}\right.---\left(16\right)$

Parameter in formula (16) is a=1.4, b=0.3.Sequential value (ω₃(ρ),ω₄(ρ)) by the two dimension of formula (17) Logistic chaotic mapping system iterative equation produces:

$\left\{\begin{array}{c}{x}_{n+1}={\mathrm{\μ}}_1{x}_n(1-x_n)+{\mathrm{\γ}}_1{y}_n^2\\ y_{n+1}={\mathrm{\μ}}_2{y}_n(1-y_n)+{\mathrm{\γ}}_2(x_n^2+x_n{y}_n)\end{array}\right.---\left(17\right)$

Wherein parameter is μ₁=2.93, μ₂=3.17, γ₁=0.197, γ₂=0.139.Produce the corresponding encryption with 4 piecemeals Key sequence ω_ρ=(ω₁(ρ),ω₂(ρ),ω₃(ρ),ω₄(ρ)):

ω (1)=(0.503926645776215,0.26274154115807524,0.7310172809052533, 0.5911631451546245)

ω(2)=(1.490358034498705,0.053495005011188465,0.635768102381588, 0.610385893368047)

ω(3)=(1.0464571296343332,0.1187291811351617,0.5208502734446377, 0.6470227289937563)

ω(4)=(1.407740930215608,0.15611128793982337,0.39125888333654657, 0.6995879558829032)

Step 4: reconstruct encryption key sequence

After obtaining encryption key sequence, due to its value produced can not with in plain text carry out xor operation, so need by Encryption key sequence is changed according to formula (18):

Sequence in order to make two systems produce has coupling, calculates:

$\begin{array}{c}{\mathrm{\ω}}_3\left(\mathrm{\ρ}\right)={\mathrm{\ω}}_1\left(\mathrm{\ρ}\right)\⊕{\mathrm{\ω}}_3\left(\mathrm{\ρ}\right)\\ {\mathrm{\ω}}_4\left(\mathrm{\ρ}\right)={\mathrm{\ω}}_2\left(\mathrm{\ρ}\right)\⊕{\mathrm{\ω}}_4\left(\mathrm{\ρ}\right)\end{array}---\left(19\right)$

The chaos encryption sequence obtained through conversion is:

ω_i(1)=(56,114,83,76)x_i(1)=(54,39)

ω_i(2)=(7,100,101,63)x_i(2)=(31,100)

ω_i(3)=(54,123,25,47)x_i(3)=(126,70)

ω_i(4)=(7,35,39,18)x_i(4)=(13,124)

Step 5: data encryption

After generating encryption key sequence, the encryption of clear text file will be carried out.If the ASC of the plaintext of 15 bytes II yard M is respectively (97,98,99,100,101,102,103,104,105,106,107,108,109,110,111), and cryptographic operation is In plaintext, take 4 ASCII character M corresponding to byte length data block successively, carry out XOR (XOR) operation with encryption key sequence Obtain intermediate object program, carry out secondary xor operation with interference sequence the most again and obtain final ciphertext C:

$\left\{\begin{array}{c}{C}_{4k+1}=(M_{4k+1}\⊕{\mathrm{\ω}}_1(k+1))\⊕x_5(k+1)\\ C_{4k+2}=(M_{4k+2}\⊕{\mathrm{\ω}}_2(k+1))\⊕x_6(k+1)\\ C_{4k+3}=(M_{4k+3}\⊕{\mathrm{\ω}}_3(k+1))\⊕x_5(k+1)\\ C_{4k+4}=(M_{4k+4}\⊕{\mathrm{\ω}}_4(k+1))\⊕x_6(k+1)\end{array}\right.---\left(20\right)$

Wherein k represents the sequence number of data block, C_4k+1, C_4k+2, C_4k+3, C_4k+4Be respectively in data block corresponding to the 1st, 2,3 and The encrypted cipher text of 4 bytes.Concrete encrypted result is:

$\left\{\begin{array}{c}{C}_1=(M_1\⊕{\mathrm{\ω}}_1\left(1\right))\⊕x_5\left(1\right)=97\⊕56\⊕54=111\\ C_2=(M_2\⊕{\mathrm{\ω}}_2\left(1\right))\⊕x_6\left(1\right)=98\⊕114\⊕54=38\\ C_3=(M_3\⊕{\mathrm{\ω}}_3\left(1\right))\⊕x_5\left(1\right)=99\⊕83\⊕39=23\\ C_4=(M_4\⊕{\mathrm{\ω}}_4\left(1\right))\⊕x_6\left(1\right)=100\⊕76\⊕39=15\end{array}\right.$

$\left\{\begin{array}{c}{C}_5=(M_5\⊕{\mathrm{\ω}}_1\left(2\right))\⊕x_5\left(2\right)=101\⊕7\⊕31=125\\ C_6=(M_6\⊕{\mathrm{\ω}}_2\left(2\right))\⊕x_6\left(2\right)=102\⊕100\⊕31=29\\ C_7=(M_7\⊕{\mathrm{\ω}}_3\left(2\right))\⊕x_5\left(2\right)2=103\⊕101\⊕100=102\\ C_8=(M_8\⊕{\mathrm{\ω}}_4\left(2\right))\⊕x_6\left(2\right)=104\⊕63\⊕100=51\end{array}\right.$

$\left\{\begin{array}{c}{C}_9=(M_9\⊕{\mathrm{\ω}}_1\left(3\right))\⊕x_5\left(3\right)=105\⊕54\⊕126=33\\ C_{10}=(M_{10}\⊕{\mathrm{\ω}}_2\left(3\right))\⊕x_6\left(3\right)=106\⊕123\⊕126=111\\ C_{11}=(M_{11}\⊕{\mathrm{\ω}}_3\left(3\right))\⊕x_5\left(3\right)=107\⊕25\⊕70=52\\ C_{12}=(M_{12}\⊕{\mathrm{\ω}}_4\left(3\right))\⊕x_6\left(3\right)=108\⊕47\⊕70=5\end{array}\right.$

$\left\{\begin{array}{c}{C}_{13}=(M_{13}\⊕{\mathrm{\ω}}_1\left(4\right))\⊕x_5\left(4\right)=109\⊕7\⊕13=103\\ C_{14}=(M_{14}\⊕{\mathrm{\ω}}_2\left(4\right))\⊕x_6\left(4\right)=110\⊕35\⊕13=64\\ C_{15}=(M_{15}\⊕{\mathrm{\ω}}_3\left(4\right))\⊕x_5\left(4\right)=111\⊕39\⊕124=52\end{array}\right.$

It is M=(97,98,99,100,101,102,103,104,105,106,107,108,109,110,111) i.e. in plain text, Ciphertext is C=(111,38,23,15,125,29,102,51,33,111,52,5,103,64,52).

Claims (1)

1. to the Parallel Chaos encryption method storing file in a cloud computing environment, it is characterised in that the method includes following step Rapid:

Step one: data file piecemeal

Note data file a length of L the most in plain text, will be divided into n block in plain text, and the size of each piece is l_i, i≤n, and require l_iIt is 4 The integral multiple of byte, is designated as l_i=4N；Before so, n-1 block is isometric data block, and n-th piece is remainder data, it may be assumed that

$\begin{array}{cc}{l}_i=4N& (i<n)\\ l_i=L-(n-1)\&CenterDot;4N& (i=n)\end{array}---\left(1\right)$

Step 2: application nonlinear chaotic system generates initiation sequence and interference sequence

For each piecemeal, produce initial value sequence by three-dimensional continuous chaotic system Lorenz chaotic maps and Chen chaotic maps Row and interference value sequence, method is as follows:

Lorenz chaotic maps equation is:

$\left\{\begin{array}{c}\frac{{\mathrm{dx}}_1}{dt}=p(x_2-x_1)\\ \frac{{\mathrm{dx}}_2}{dt}=-x_1{x}_3+{\mathrm{rx}}_1-x_2\\ \frac{{\mathrm{dx}}_3}{dt}=x_2{x}_1-{\mathrm{tx}}_3\end{array}\right.---\left(2\right)$

Chen chaos system equation is:

$\left\{\begin{array}{c}\frac{{\mathrm{dx}}_4}{dt}=a(x_5-x_4)\\ \frac{{\mathrm{dx}}_5}{dt}=(c-a)x_4-x_4{x}_6+{\mathrm{cx}}_5\\ \frac{{\mathrm{dx}}_6}{dt}=x_4{x}_5-{\mathrm{bx}}_6\end{array}\right.---\left(3\right)$

Wherein χ=(x₁,x₂,x₃,x₄,x₅,x₆) it is chaos sequence to be generated, p, r, t and a, b, c are the ginseng that system is to be given Number, utilizes 4 rank runge kutta methods to solve chaos system equation (2) and (3), and each iteration generates three stochastic ordering train values respectively, I.e. 6 chaos sequence value χ=(x₁,x₂,x₃,x₄,x₅,x₆)；In order to be calculated this chaos sequence value, first obtain Plaintext block Side-play amount θ of first character joint_i, the primary iteration number of times taking Lorenz system is N₀, the primary iteration number of times of Chen system is M₀, the iterations of note Lorenz system is ρ_L, the iterations of Chen system is ρ_C, then have:

$\begin{array}{cc}{\mathrm{\&rho;}}_L=\frac{{\mathrm{\&theta;}}_i}{N}+N_0& {\mathrm{\&rho;}}_C=\frac{{\mathrm{\&theta;}}_i}{N}+M_0\end{array}---\left(4\right)$ The corresponding chaos sequence of the most each piecemeal is

χ_ρ=(x₁(ρ_L),x₂(ρ_L),x₃(ρ_L),x₄(ρ_C),x₅(ρ_C),x₆(ρ_C)) (5)

Sequence in order to make two systems produce has coupling, the random sequence that the Lorenz system obtained is produced to The random sequence that Chen system produces is to carrying out XOR:

$\begin{array}{c}{x}_4\left({\mathrm{\&rho;}}_C\right)=x_4\left({\mathrm{\&rho;}}_C\right)\&CirclePlus;x_1\left({\mathrm{\&rho;}}_L\right)\\ x_5\left({\mathrm{\&rho;}}_C\right)=x_5\left({\mathrm{\&rho;}}_C\right)\&CirclePlus;x_2\left({\mathrm{\&rho;}}_L\right)\\ x_6\left({\mathrm{\&rho;}}_C\right)=x_6\left({\mathrm{\&rho;}}_C\right)\&CirclePlus;x_3\left({\mathrm{\&rho;}}_L\right)\end{array}---\left(6\right)$

Meter chaos sequence χ_ρFor first value sequenceAnd interference sequence

Step 3: calculate the chaos sequence of encryption key with first value sequence

By first value sequenceIn (x₁(ρ_L),x₂(ρ_L)) fractional part is as the initial value of H é non chaotic maps iteration, (x₃ (ρ_L),x₄(ρ_C)) fractional part produces the key sequence of encryption as the initial value of Coupled Logistic chaotic map iteration, recurrence Row ω_ρ=(ω₁(ρ),ω₂(ρ),ω₃(ρ),ω₄(ρ)), wherein (ω₁(ρ),ω₂(ρ)) by the H é non system equation of formula (7) Iteration produces:

$\left\{\begin{array}{c}{x}_{n+1}=y_n+1-a{x}_n^2\\ y_{n+1}={\mathrm{bx}}_n\end{array}\right.---\left(7\right)$

(ω₃(ρ),ω₄(ρ)) produced by the Coupled Logistic chaotic map system equation iteration of formula (8):

$\left\{\begin{array}{c}{x}_{n+1}={\mathrm{\&mu;}}_1{x}_n(1-x_n)+{\mathrm{\&gamma;}}_1{y}_n^2\\ y_{n+1}={\mathrm{\&mu;}}_2{y}_n(1-y_n)+{\mathrm{\&gamma;}}_2(x_n^2+x_n{y}_n)\end{array}\right.---\left(8\right)$

Parameter value in mapping is: 2.75 ＜ μ₁＜ 3.4,2.7 ＜ μ₂＜ 3.45 and 0.15 ＜ γ₁＜ 0.21,0.13 ＜ γ₂＜ 0.15；

Step 4: reconstruct encryption key sequence

After obtaining encryption key sequence, owing to its value produced can not carry out xor operation with plaintext, so needing to encrypt Key sequence is changed according to following formula:

Sequence in order to make two systems produce has coupling, carries out calculated as below:

$\begin{array}{c}{\mathrm{\&omega;}}_3\left(\mathrm{\&rho;}\right)={\mathrm{\&omega;}}_1\left(\mathrm{\&rho;}\right)\&CirclePlus;{\mathrm{\&omega;}}_3\left(\mathrm{\&rho;}\right)\\ {\mathrm{\&omega;}}_4\left(\mathrm{\&rho;}\right)={\mathrm{\&omega;}}_2\left(\mathrm{\&rho;}\right)\&CirclePlus;{\mathrm{\&omega;}}_4\left(\mathrm{\&rho;}\right)\end{array}---\left(10\right)$

Step 5: data encryption

After generating encryption key sequence, will carry out the encryption of clear text file, cryptographic operation is to take 4 words successively in plaintext The ASCII character M that joint length data block is corresponding, carries out XOR (XOR) operation with encryption key sequence and obtains intermediate object program, the most again Carry out secondary xor operation with interference sequence and obtain final ciphertext C:

$\left\{\begin{array}{c}{C}_{4k+1}=(M_{4k+1}\&CircleTimes;{\mathrm{\&omega;}}_1(k+1\left)\right)\&CirclePlus;x_5(k+1)\\ C_{4k+2}=(M_{4k+2}\&CircleTimes;{\mathrm{\&omega;}}_2(k+1\left)\right)\&CirclePlus;x_6(k+1)\\ C_{4k+3}=(M_{4k+3}\&CircleTimes;{\mathrm{\&omega;}}_3(k+1\left)\right)\&CirclePlus;x_5(k+1)\\ C_{4k+4}=(M_{4k+4}\&CircleTimes;{\mathrm{\&omega;}}_4(k+1\left)\right)\&CirclePlus;x_6(k+1)\end{array}\right.---\left(11\right)$

Wherein k represents the sequence number of data block, C_4k+1, C_4k+2, C_4k+3, C_4k+4It is respectively in data block corresponding to the 1st, 2,3 and 4 words The encrypted cipher text of joint；In the encryption method that method proposes, key space includes initial value χ (0)=(x of key sequence₁(0), x₂(0),x₃(0),x₄(0),x₅(0),x₆(0)), the primary iteration times N of Lorenz and Chen system₀And M₀, and two mix Parameter p of ignorant system, r, t and a, b, c.