CS ASSIGNMENT 代写 : Secret Sharing Scheme For Image Encryption
A. Theorem
A primitive root of a number is the one whose powers generate all the integers from 0 to p-1. That is if " a" is a primitive root of the primitive number "p" then the number "a mod p" , "a2 mod p" ….. "ap-1 mod p" are distinct and consists of integers from I through p-1 in some permutation.
B. Block construction
Two block matrix of size 6 x 6 are constructed from the theorem, When p = 7 and a = 3
a1 = ( 3 2 6 4 5 1)
when p = 7 and a = 5
b1 = (5 4 6 2 3 1)
The first block matrix "A" , a1 values are taken as the first row elements. Then add the adjacent elements of the first row to get the second row. Then the same procedure is applied with the second row to get the third row. And finally this procedure is applied with the fifth row to get the sixth row. Next apply this procedure with b1 values to get the second block matrix "B".
C. Zone Construction
The zone matrix " M " of size 216 x 216 is computed by using the following formulas.
For , , k = 1 , 2 … r.
Here t = 6, r = 3 ,
D. Transformation Matrix
The Transformation matrix has the following form.
M (X1) M (X2) M (X3 )
M (Y1) M (Y2) M ( Y3)
T =
M (Z1) M (Z2) M (Z3)
Where
X1 = ( S * A) Mod 256
X2 = ( S * B) Mod 256
X3 = (( S+1) * A) Mod 256
Y1 = ( (S + 2) * A) Mod 256
Y2 = ( (S + 2) * B) Mod 256
Y3 = ( (S + 3) * A) Mod 256
Z1 = ( (S + 4) * A) Mod 256
Z2 = ( (S + 4) * B) Mod 256
Z3 = ( (S + 5) * A) Mod 256
M(X) ïƒ Zone of X
S ïƒ Secret
A , B ïƒ Block matrix 6 x 6
E. Encryption Process
The Transformation matrix is XOR with the image matrix to get the encrypted image.
Original Image
Generate random polynomial
Form transformation matrix
Calculate Shares from the polynomial.
XOR Transformation matrix with the Image
Shares are sending to the corresponding participants
Encrypted Image
Fig.2. Encryption process flow chart
IV. DECRYPTION PROCESS
A. Reconstruction
Each participant accepts the share from the sender for reconstructing the secret. The participant who is having the identification value ID1 accepts the share y(1), the participant having the identification value ID2 receives the share y(2) and so on. Only t shares are enough to reconstruct the secret. The combiner receives t shares y(1), y(2) , …y(t) and the polynomial is reconstructed by using Lagrange interpolation formula . The Lagrange interpolation formula is given below.
F(x) = (5)
Where x1 , x2 , … xt are the identification values. F(x) ïƒ The reconstructed polynomial . t ïƒ Threshold value
B. Decryption process
The receiver reconstructs the polynomial from the shares of the participants using Lagrange interpolation formula. The mean value is evaluated from the reconstructed polynomial. Next the cipher text is decrypted with the mean value to get the original text. The following flowchart explains this.
Shares
Generate polynomial using Lagrange interpolation
Generate Transformation Matrix
Encrypted Image
XOR Transformation matrix with the Encrypted Image
Original image
Fig.3. Decryption process
V. ENCRYPTION Algorithm
Take the Original Image.
Randomly generate the polynomial of degree (t-1)
Construct the Transformation Matrix
XOR Transformation matrix with Image matrix
The threshold value t , identification value of the participants ID (ID1, ID2, …, IDn ) and the Encrypted image are given in public.
Calculate the function values F(ID1),F(ID2), ..F(IDn).
Send F(IDi) to the participant having the ID value IDi , where i = 1 to n.
VI. Decryption algorithm
Reconstruct the polynomial from "t" shares using Lagrange interpolation.
Construct the Transformation Matrix
XOR Transformation matrix with Encrypted Image matrix
Get the Original Image
VII. Sample Images
Fig 4: Leena Image Fig 5 : Encrypted Leena
Fig6: Cameraman Fig7: Encrypted Cameraman
Fig8: Gold hill Fig 9: Encrypted Gold Hill
VIII. Analysis
A. Histogram Analysis
Histogram analysis is employed to illustrate its superior confusion and diffusion properties in the encrypted data. The histogram of the plain image cameraman is shown in figure 10 and the histogram of the encrypted image is shown in figure 11. Comparing the two histograms, we find that histogram of encrypted image is fairly uniform and is significantly different from that of the original image, and that the encrypted images transmitted do not provide any suspicion to the attacker, which can strongly resist statistical attacks.
Fig10: Histogram For Cameraman
Original Image
Fig11: Histogram for Cameraman
Encrypted Image
B. Encryption quality analysis
The quality of image encryption [12] may be determined as follows:
Let F and F' denote the original image and the encrypted image respectively each of size M * N pixels with L grey levels. F(x,y), F' (x,y) € {0, 1,….,L-1} are the grey levels of the images F and F' at position (x,y) ( 0 ≤ x ≥ M-1, 0 ≤ Y ≥ N-1 ). Let HL (F) denote the number of occurrences of each grey level L in the original image F. Similarly , HL ( F' ) denotes the number of occurrences of each grey level L in the encrypted image F'. The encryption quality represents the average number of changes to each grey level L and is expressed mathematically as
Encryption Quality =
The encryption Quality of this proposed algorithm is computed and is tabulated as follows.
Table 1: Encryption Quality Table
Image
Size
Encryption
Quality
Leena
256 x 256
149.1953
Cameraman
256 x 256
252.6563
Gold Hill
512 x 512
713.7500
C. Information Entropy
The entropy H of symbol S can be calculated using the following equation.
H(S) = ,
Where p(si) represents the probability of symbol si.
If the information entropy of encrypted image becomes larger, image distribution of gray scale will be more uniform. By calculation, the information entropy of Leena encrypted image is equal to 7.9972, which means that information leakage in the encrypted process is negligible and the encryption system is secure from the entropy attack.
IX. Conclusion
In the new algorithm a polynomial is generated randomly. It is almost impossible to extract the original image in the proposed method even if the algorithm is known. In this algorithm, the image is encrypted after a number of rounds, which makes the computation more complex. Compared with the encryption schemes[10] based on the secret sharing , the size of the shares is far smaller with the size of the image.
