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28 Int'l Conf. Foundations of Computer Science | FCS'11 | A 4 out of n Secret Sharing Scheme in Visual Cryptography without Expansion Ying-Yu Chen Department of Computer Science and Information Engineering, National Chi Nan University Puli, Nantou 54561, Taiwan, R.O.C. e-mail: s99321514@ncnu.edu.tw AbstractVisual cryptography (VC, for short) encrypts the secret image into n shares (transparency). We cannot see any information from any one share, and decrypt the original image by stacking all of the shares. Now, we extend it to the k out of n secret sharing scheme. (k, n) threshold secret sharing scheme encrypts as the same way and decrypts the original image by stacking at least k shares. If one stacks less than k shares, he (or she) cannot recognize the secret image. In this paper, we construct a new scheme for (4, n) threshold secret sharing encrypt in VC by using a method of combination and the size of the share is as small as the original image. That is, there is no expanded need while some of the previous scheme need. Keywords-visual cryptography; secret sharing scheme; security; share. I. INTRODUCTION Visual cryptography (VC, for short) and the (k, n)-threshold secret sharing scheme were proposed by Naor and Shamir in 1995 [9]. Visual cryptography means the secret image is turned into n shares that combined with black and white pixel, and the decrypting by stacking the shares together to reveal the secret image. So we can decrypt the secret image by human’s eye without using computer. The (k, n)-threshold secret sharing scheme means a dealer sends a share to each of the n users. In the condition of k is small or equal than n, the fewer than k users stack their shares together, they cannot see any information from the image. But at least k users stack their shares together, they will find out the secret from the image. In most of the VC scheme [2, 4, 7, 8, 9, 10, 12], the pixel of each share will expand. The more value of n is, the more value of the expansion will be. However, the size of the share is larger than the original image. In 1995, Naor and Shamir [9] proposed some (k, n)-threshold secret sharing in VC for three kinds of condition. First, some of their schemes are the efficient solutions for (2, n) and (3, n). Second, they propose a general k out of k scheme. _________________________________ * Corresponding author. Justie Su-Tzu Juan * Department of Computer Science and Information Engineering, National Chi Nan University Puli, Nantou 54561, Taiwan, R.O.C. e-mail: jsjuan@ncnu.edu.tw Third, they propose a general k out of n scheme when k is small or equal than n. But the pixel of each share will expand in their later two methods. For convenience, the third method is called NS scheme in this paper. In 2008, Fang et al. propose a new algorithm (called FLL scheme in this paper) [6]. They solve the problem in expansion. In FLL scheme, the authors use Hilbert-curve [1] and two queues to present a VC scheme. The shares they generate are as small as the input image S, so the pixel of each share won’t expand. But since Fang et al. use Naor and Shamir’s scheme to design their scheme, the more number of the expansion in Naor and Shamir’s scheme is, the image we decrypted will be more unclear in FLL scheme. Another subject has been considered these years, progressive visual secret sharing (PVSS, for shout) scheme [3, 5]. In 2011, Hou et al. propose a new algorithm for (2, n) threshold PVSS scheme [3]. In above researches, no matter the shares will be expansion or not, it cannot reveal the secret image by stacking less than k shares and it can reveal the secret image by stacking at least k shares for k ≥ 4. Because the expansion in the most of (4, n)-threshold secret sharing scheme in VC is quite large and the image we decrypted will be not clear in FLL scheme for (4, n)-threshold secret sharing scheme in VC. Hence, we propose a new scheme to improve it. We use the theory of combination to construct the scheme. Actually, our scheme is also a (4, n) threshold PVSS scheme. The detail of our scheme is presented on next section. Some experiment results are given in section , and the conclusion is stated in section . II. THE PROPOSED SCHEME We will show our (4, n)-threshold secret sharing scheme in VC as follows. For convenience, let
Int'l Conf. Foundations of Computer Science | FCS'11 | 29 C n! ⎧ , if 0 ≤ j ≤ n; ( n j)! j! ⎨ ⎩ 0 , otherwise. n − j = Definition 1. An n × m 0-1 matrix M(n, j) is called totally symmetric if each column has the same weight, say j, and m equal to the number of C and every column vectors are difference to each others, where the weight of a column vector means the sum of each entry in this column vector. Definition 2. Given an n × m1 matrix A and an n × m2 matrix B, we define: 1. [A||B]be an n × (m1 + m2) matrix that obtained by concatenating A and B; 2. [a × A||b × B], for any two positive integer a and b, be an n × (a × m1 + b × m2) matrix that obtained by concatenating A for a times and B for b times. Lemma 1. Give an n × m totally symmetric matrix A = M(n, j), For i = 1, 2, …, n, let fi(A) represent the Hamming weight of the row vector that is the result of applying “or” operation for any i rows in A. Then fi(A) n n − i = fi(M(n, j)) = C − C . j j Proof. The number of column in M(n, j) is equal to . When we choose any i rows, because if any one column vector has all zeros in these i rows, the result entry for applying “or” operation for these row vectors still will be zero. In the other way, if any one column vector has all zeros in these i rows, there must has j ones show on the other n − i rows. Hence, the number of those kind column vectors is n j n i C j − . So fi(M(n, j)) equals to the number of all columns subtract the number of columns that has all zeros in these i rows (and has j ones in the other n − i rows). Hence, fi(M(n, j)) = C n j C − . n − i j We use an example to demonstrate Lemma 1: Example 1. Let A = M(4, 2) = 1 1 1 0 0 0 1 0 0 1 0 0 1 1 1 0 0 . 1 0 0 1 0 1 1 Then we have f1(A) = 3, f2(A) = 5, f3(A) = 6, and f4(A) = 6. Lemma 2. fi([A||B]) = fi(A) + fi(B) for any two totally symmetric matrices A and B. Proof. A matrix [A||B] is obtained by concatenating matrices A and B. Let A denote an n × m1 matrix, B denote an n × m2 matrix, and A = M(n, jA), B = M(n, jB). Then fi([A||B]) = fi([M(n, jA)||M(n, jB)]). Because do Hamming weight for the resulting row vector of applying “or” operation for some i row of [A||B] is equal to that [A||B] be divided into two parts A, B and do the same thing to these two parts, then add those two results together. According to the above reason, fi([M(n, jA)||M(n, jB)]) = fi(M(n, jA)) + fi (M(n, jB)). That is, fi([A||B]) = fi(A) + fi(B). Let [A1||A2||…||Ak] = […[[A1||A2]||A3]||…||Ak] for any matrices A1, A2, …, Ak, with the size of Ai is n × mi for i = 1, 2, …, k. We have the following corollary. Corollary 1. For any k totally symmetric matrices A1, A2, …, Ak, where Ai = M(n, ji) for some 0 ≤ ji ≤ n for any 1 ≤ i ≤ k. Let [A1||A2||…||Ak] = B. Then fi(B) = fi(A1) + fi(A2) + … + fi(Ak). Definition 3. The light transmission rate ℑ(S) = w / p = 1 − (b / p), where w means the number of the white pixel in the image S, p means the number of the all pixel in the image S, b means the number of the black pixel in the image S. Definition 4. Given totally symmetric matrices A1, A2, …, Ak. For any 1 ≤ i ≤ k, Ai = M(n, ji) for some 0 ≤ ji ≤ n. Let [A1||A2||…||Ak] = B and B is an n × m matrix. Define ℑ(B, k) = 1 – (fk(B) / m), where k ≤ n. To construct the (4, n) secret sharing scheme, we will construct two matrices, C0 and C1, which will be used for constructing n shares later. Since we will choose any one column vector of C0 (or C1, respectively) randomly when construct the pixels of n shares according to a white pixel of secret image S (or a black pixel of secret image S, respectively), we have to follow two conditions: − ℑ(C0, k) = ℑ(C1, k) for 1 ≤ k ≤ 3. − ℑ(C0, k) > ℑ(C1, k) for k ≥ 4. The first rule ensures C0 and C1 have the same light transmission rate when k is between one and three, so it won’t see any information when we stack less than 4 shares. The second ensure C0 and C1 have the different light transmission rate when k is larger than three, and the light transmission rate for the white pixel of secret image S is greater than that for the black pixel of secret image S, so it can reveal the
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