Improved schemes for visual secret sharing based on random grids

Abstract

Random grid (RG) is an alternative approach to realize a visual secret sharing (VSS) scheme. RG-based VSS has merits such as no pixel expansion and no tailor-made matrix requirement. Recently, many investigations on RG-based VSS are made. However, they need further improvements. In this paper, we obtain some improvements on RG-based VSS. Actually, two improved schemes are proposed, namely RG-based VSS for general access structure (GAS) with improved contrast and extended RG-based VSS with improved access structure. The first scheme can achieve better contrast than previous schemes. The second scheme reduces the chance of suspicion on secret image encryption by generating meaningful shares instead of noise-like shares in the first scheme, and improves the access structure from (k, k) to GAS while maintaining the property that the contrast of the recovered image is traded with that of share images by setting a certain parameter from small to large. Finally, theoretical analyses and experimental results are provided to demonstrate the effectiveness and advantages of the proposed schemes.

Appendices

Appendix A

Some basic symbols, concepts and definitions in this paper are provided as follows

Definition 1 (General access structure [1])

For VSS, let P = {1, 2, ⋯, n} be a set of participants and let 2^P denote the set of all subsets ofP. Let Γ_Qual ⊆ 2^P andΓ_Forb ⊆ 2^P, whereΓ_Qual ∩ Γ_Forb = ∅. Members of Γ_Qual are referred as qualified sets and Members of Γ_Forb are referred as forbidden sets. The pair (Γ_Qual, Γ_Forb) is called GAS.

Definition 2 (Minimal qualified set [1])

Let Γ₀ consist of all the minimal qualified sets as:

$$ {\varGamma}_0=\left\{Q\in {\varGamma}_{Qual}:{Q}^{\prime}\notin {\varGamma}_{Qual}\ \mathrm{for}\ \mathrm{all}\ {Q}^{\prime}\subset Q\right\} $$

(6)

Definition 3 (Strong access structure [1])

If Γ_Qual is monotone increasing and Γ_Forb is monotone decreasing, then the access structure is said to be strong, and Γ₀ is called a basis.

Definition 4 (Essential participant [1])

A participant p ∈ P is an essential participant if there exists a set X satisfies X ∪ {p} ∈ Γ_Qual but X ∉ Γ_Qual.

Definition 5 (Average light transmission [26])

For a certain pixel s in a secret image S whose size is M × N, the probability for s being transparent, say Prob(s = 0), is represented as the light transmission of s, say T(s). The light transmission of a white (resp. black) pixel is defined as T(s) = 1 (resp. T(s) = 0). Then the average light transmission of S is defined as

$$ T(S)=\frac{1}{M\times N}\times \sum \limits_{i=1}^M\sum \limits_{j=1}^NT\left(S\left[i,j\right]\right) $$

(7)

Definition 6 (RG-based VSS)

(Γ_Qual, Γ_Forb) is a strong access structure on a set of n participants and Γ₀ is defined as its basis. (Γ_Qual, Γ_Forb) constitutes a RG-based VSS if the following three conditions are met:

• 1) Each of n random grids R₁, ⋯, R_n is determined in a random way.

• 2) IfX = {i₁, i₂, ⋯, i_t} ∈ Γ_Qual, the stacking result of the t random grids, represented by ${R}_{i_1\otimes \cdots \otimes {i}_t}={R}_{i_1}\otimes \cdots \otimes {R}_{i_t}$, can recover the original secret image S:

$$ T\left({R}_{i_1\otimes \cdots \otimes {i}_t}\left[S(0)\right]\right)>T\left({R}_{i_1\otimes \cdots \otimes {i}_t}\left[S(1)\right]\right) $$

(8)

where S(0) (resp.S(1)) denote the area of all the white (resp. black) pixels in the secret image and ${R}_{i_1\otimes \cdots \otimes {i}_t}\left[S(0)\right]$ (resp. ${R}_{i_1\otimes \cdots \otimes {i}_t}\left[S(1)\right]$) denote the corresponding area of all the white (resp. black) pixels in the recovered secret image.

• 3) If X = {i₁, i₂, ⋯, i_t} ∈ Γ_Forb, the stacking result of the t random grids gives no clue about the original secret image S:

$$ T\left({R}_{i_1\otimes \cdots \otimes {i}_t}\left[S(0)\right]\right)=T\left({R}_{i_1\otimes \cdots \otimes {i}_t}\left[S(1)\right]\right) $$

(9)

The visual quality of the recovered secret image is measured by the contrast.

Definition 7 (Contrast [26])

ForX = {i₁, i₂, ⋯, i_t} ∈ Γ_Qual, the contrast of the recovered secret image ${R}_{i_1\otimes \cdots \otimes {i}_t}={R}_{i_1}\otimes \cdots \otimes {R}_{i_t}$ with respect to the original secret image S is

$$ {\alpha}_X=\frac{T\left({R}_{i_1\otimes \cdots \otimes {i}_t}\left[S(0)\right]\right)-T\left({R}_{i_1\otimes \cdots \otimes {i}_t}\left[S(1)\right]\right)}{1+T\left({R}_{i_1\otimes \cdots \otimes {i}_t}\left[S(1)\right]\right)} $$

(10)

Contrast determines how well human visual system can recognize the recovered secret image. A larger contrast is more easily to be accepted.

Definition 8 (Equivalence set)

A, B and C are three different nonempty sets. Let A ⊕ Bbe the XOR-ed result of all the corresponding elements of A and B. If A ⊕ B = B ⊕ C, then C is called an equivalence set with respect to A, denoted as C ≅ A.

Definition 9 (Equivalence qualified set)

Let D ∈ Γ_Qual be a qualified set. If there exists C ≅ A, where A ∈ 2^D and C ∉ 2^D, then D^′ is generated by replacing C for A with regard to the elements in D. D^′ is called an equivalence qualified set with respect to D, denoted as D^′ ∼ D.

Appendix B

Theoretical analyses on Algorithm 4 are provided as follows

Lemma 1

Given p random grids R₁(i, j), ⋯, R_p(i, j) assigned 0 or 1 randomly. We can derive$Prob\left({R}_{1\oplus \cdots \oplus p}\left(i,j\right)=0\right)=\frac{1}{2}$.

Proof: By using induction method,

• (1) when p = 2, we have

  $$ {\displaystyle \begin{array}{l}\kern0.5em Prob\left({R}_{1\otimes 2}\left(i,j\right)=0\right)\\ {}=\kern0.5em Prob\left({R}_1\left(i,j\right)=0\right)\times Prob\left({R}_2\left(i,j\right)=0\right)\\ {}\begin{array}{cc}& + Prob\left({R}_1\left(i,j\right)=1\right)\times Prob\left({R}_2\left(i,j\right)=1\right)\end{array}\\ {}\begin{array}{cc}=& \frac{1}{2}\end{array}\end{array}} $$

  (11)

• (2) Assume that the claim holds for p − 1, namely, $Prob\left({R}_{1\oplus \cdots \oplus p-1}\left(i,j\right)=0\right)=\frac{1}{2}$, and then we have

$$ {\displaystyle \begin{array}{l}\kern0.5em Prob\left({R}_{1\otimes p}\left(i,j\right)=0\right)\\ {}\begin{array}{cc}=& Prob\left({R}_{1\oplus \cdots \oplus p-1}\left(i,j\right)=0\right)\times Prob\left({R}_p\left(i,j\right)=0\right)\end{array}\\ {}\begin{array}{cc}& + Prob\left({R}_{1\oplus \cdots \oplus p-1}\left(i,j\right)=1\right)\times Prob\left({R}_p\left(i,j\right)=1\right)\end{array}\\ {}\begin{array}{cc}=& \frac{1}{2}\end{array}\end{array}} $$

(12)

Obviously, the claim also holds for p.

To sum up, Lemma 1 holds.

Lemma 2

Given p random grids R₁(i, j), ⋯, R_p(i, j) assigned 0 or 1 randomly. We can derive $Prob\left({R}_{1\otimes \cdots \otimes p}\left(i,j\right)=0\right)={\left(\frac{1}{2}\right)}^p$.

Proof: Since R₁(i, j), ⋯, R_p(i, j) are assigned 0 or 1 randomly, we obtain

$$ Prob\left({R}_1\left(i,j\right)=0\right)=\cdots = Prob\left({R}_p\left(i,j\right)=0\right)=\frac{1}{2} $$

(13)

Therefore, we have

$$ {\displaystyle \begin{array}{l}\kern0.5em Prob\left({R}_{1\otimes \cdots \otimes p}\left(i,j\right)=0\right)\\ {}\begin{array}{cc}=& Prob\left({R}_1\left(i,j\right)=0\right)\times \cdots \times Prob\left({R}_p\left(i,j\right)=0\right)\end{array}\\ {}{\begin{array}{cc}=& \left(\frac{1}{2}\right)\end{array}}^p\end{array}} $$

(14)

Lemma 3

For the secret pixelS(i, j), given p − 1 random grids R₁(i, j), ⋯, R_p − 1(i, j) assigned 0 or 1 randomly, and then assign the p-th random gridR_p(i, j) = S(i, j) ⊕ R₁(i, j) ⊕ ⋯ ⊕ R_p − 1(i, j). We have $Prob\left({R}_p\left(i,j\right)=0\right)=\frac{1}{2}$no matter S(i, j)is 0 or 1.

Proof: ForS(i, j) = 0,R_p(i, j) = R₁(i, j) ⊕ ⋯ ⊕ R_p − 1(i, j). ForS(i, j) = 1, ${R}_p\left(i,j\right)=\overline{R_1\left(i,j\right)\oplus \cdots \oplus {R}_{p-1}\left(i,j\right)}$. By Lemma 1, we have

$$ Prob\left({R}_1\left(i,j\right)\oplus \cdots \oplus {R}_{p-1}\left(i,j\right)=0\right)=\frac{1}{2} $$

(15)

and

$$ Prob\left(\overline{R_1\left(i,j\right)\oplus \cdots \oplus {R}_{p-1}\left(i,j\right)}=0\right)=\frac{1}{2} $$

(16)

Thus, $Prob\left({R}_p\left(i,j\right)=0\right)=\frac{1}{2}$no matter S(i, j)is 0 or 1.

Lemma 4

Given a secret pixelS(i, j), the n random grids R₁(i, j), ⋯, R_n(i, j) generated by Steps 2–10 satisfy

$$ Prob\left({R}_1\left(i,j\right)=0\right)=\cdots = Prob\left({R}_n\left(i,j\right)=0\right)=\frac{1}{2} $$

(17)

Proof: By analyzing the generation of n random grids, the following three cases are considered.

• 1) Generate a random grid by assigning 0 or 1 randomly. For this case, each generated random grid satisfies Lemma 4.

• 2) Generate a random grid by XORing the secret pixelS(i, j)with some certain randomly generated random grids. For this case, each generated random grid satisfies Lemma 4 according to Lemma 3.

• 3) Generate a random grid by XORing some certain randomly generated random grids. For this case, each generated random grid satisfies Lemma 4according to Lemma 1.

To sum up, Lemma 4 holds.

Lemma 5

Given a secret pixel S(i, j), there are p random grids with $S\left(i,j\right)={R}_{i_1}\left(i,j\right)\oplus {R}_{i_2}\left(i,j\right)\oplus \cdots \oplus {R}_{i_p}\left(i,j\right)$. If S(i, j) = 1, we have $T\left({R}_{i_1\otimes \cdots \otimes {i}_p}\left[S\left(i,j\right)=1\right]\right)=0$.

Proof: Since $S\left(i,j\right)={R}_{i_1}\left(i,j\right)\oplus {R}_{i_2}\left(i,j\right)\oplus \cdots \oplus {R}_{i_p}\left(i,j\right)$, we have ${R}_{i_p}\left(i,j\right)={R}_{i_1}\left(i,j\right)\oplus {R}_{i_2}\left(i,j\right)\oplus \cdots \oplus S\left(i,j\right)$. If S(i, j) = 1, then ${R}_{i_p}\left(i,j\right)=\overline{R_{i_1}\left(i,j\right)\oplus {R}_{i_2}\left(i,j\right)\oplus \cdots \oplus {R}_{i_{p-1}}\left(i,j\right)}$. The stacking result of the p random grids is white only if all the p random grids are white. However, when the p − 1 random grids are white, ${R}_{i_p}\left(i,j\right)$ is black. The probability of the stacking result being white is zero. Therefore, we have$T\left({R}_{i_1\otimes \cdots \otimes {i}_p}\left[S\left(i,j\right)=1\right]\right)=0$.

Lemma 6

Given a secret pixel S(i, j), n random grids R₁(i, j), ⋯, R_n(i, j) and a collection of minimal qualified sets ${\varGamma}_0^{\prime }$, which are generated by Steps 2–10. For t random grids ${R}_{i_1}\left(i,j\right),\cdots, {R}_{i_t}\left(i,j\right)$, the following two conclusions can be obtained:

• 1) If there is at least one minimal qualified set$Q\subseteq {\varGamma}_0^{\prime }$satisfyingQ ⊆ {i₁, ⋯, i_t}, then we have

$$ T\left({R}_{i_1\otimes \cdots \otimes {i}_t}\left[S\left(i,j\right)=0\right]\right)={\left(\frac{1}{2}\right)}^{e_1} $$

(18)

and

$$ T\left({R}_{i_1\otimes \cdots \otimes {i}_t}\left[S\left(i,j\right)=1\right]\right)=0 $$

(19)

where e₁ denotes the number of randomly generated random grids for generating ${R}_{i_1}\left(i,j\right),\cdots, {R}_{i_t}\left(i,j\right)$ according to the proof of Lemma 4.

• 2) If there is no minimal qualified set $Q\subseteq {\varGamma}_0^{\prime }$ satisfying Q ⊆ {i₁, ⋯, i_t}, then we have

$$ T\left({R}_{i_1\otimes \cdots \otimes {i}_t}\left[S\left(i,j\right)=0\right]\right)=T\left({R}_{i_1\otimes \cdots \otimes {i}_t}\left[S\left(i,j\right)=1\right]\right)={\left(\frac{1}{2}\right)}^{e_2} $$

(20)

where e₂ denotes the number of randomly generated random grids for generating ${R}_{i_1}\left(i,j\right),\cdots, {R}_{i_t}\left(i,j\right)$ according to the proof of Lemma 4.

Proof: For S(i, j) = 0, since ${R}_{i_1}\left(i,j\right),\cdots, {R}_{i_t}\left(i,j\right)$ are generated by random grids comprising stochastic 0 or 1, the probability of ${R}_{i_1}\left(i,j\right)=0,\cdots, {R}_{i_t}\left(i,j\right)=0$ is equal to the case when all the randomly generated random grids are white pixels. Therefore, for the first condition, we have

$$ T\left({R}_{i_1\otimes \cdots \otimes {i}_t}\left[S\left(i,j\right)=0\right]\right)={\left(\frac{1}{2}\right)}^{e_1} $$

(21)

and for the second condition, we have

$$ T\left({R}_{i_1\otimes \cdots \otimes {i}_t}\left[S\left(i,j\right)=0\right]\right)={\left(\frac{1}{2}\right)}^{e_2} $$

(22)

For S(i, j) = 1, two cases are considered:

• 1) If there is at least one minimal qualified set$Q\subseteq {\varGamma}_0^{\prime }$satisfyingQ ⊆ {i₁, ⋯, i_t}, assume Q = {j₁, j₂, ⋯, j_p} and according to Steps 6–7 of Algorithm 4, we have

$$ S\left(i,j\right)={R}_{j_1}\left(i,j\right)\oplus {R}_{j_2}\left(i,j\right)\oplus \cdots \oplus {R}_{j_p}\left(i,j\right) $$

(23)

By Lemma 5, we obtain $T\left({R}_{j_1\otimes \cdots \otimes {j}_p}\left[S\left(i,j\right)=1\right]\right)=0$. Since {j₁, j₂, ⋯, j_p} ⊆ {i₁, ⋯, i_t}, we have

$$ T\left({R}_{i_1\otimes \cdots \otimes {i}_t}\left[S\left(i,j\right)=1\right]\right)=0 $$

(24)

• 2) If there is no minimal qualified set $Q\subseteq {\varGamma}_0^{\prime }$ satisfying Q ⊆ {i₁, ⋯, i_t}, then the stacking result of the t random grids is white only if all the t random grids are white. The probability of ${R}_{i_1}\left(i,j\right)=0,\cdots, {R}_{i_t}\left(i,j\right)=0$ is equal to the case when all the e₂ randomly generated random grids are white. Therefore, $T\left({R}_{i_1\otimes \cdots \otimes {i}_t}\left[S\left(i,j\right)=1\right]\right)={\left(\frac{1}{2}\right)}^{e_2}$.

To sum up, this lemma holds.

Theorem 1

Given a secret imageSand a strong access structure(Γ_Qual, Γ_Forb), the proposed Algorithm 4 is a valid construction for RG-based VSS, which meets the following three conditions:

• 1) Each of n shares R₁, ⋯, R_n is determined in a random way:

$$ Prob\left({R}_t=0\right)= Prob\left({R}_t=1\right)=\frac{1}{2},t=1,2,\cdots, n $$

(25)

• 2) IfX = {i₁, i₂, ⋯, i_t} ∈ Γ_Qual, the stacking result of the t shares can decode the original secret image S:

$$ T\left({R}_{i_1\otimes \cdots \otimes {i}_t}\left[S(0)\right]\right)>T\left({R}_{i_1\otimes \cdots \otimes {i}_t}\left[S(1)\right]\right) $$

(26)

• 3) IfX = {i₁, i₂, ⋯, i_t} ∈ Γ_Forb, the stacking result of the t shares gives no clue about the original secret image S:

$$ T\left({R}_{i_1\otimes \cdots \otimes {i}_t}\left[S(0)\right]\right)=T\left({R}_{i_1\otimes \cdots \otimes {i}_t}\left[S(1)\right]\right) $$

(27)

Proof: By Lemma 4, we can conclude that the first condition of this theorem holds immediately.

Given a secret pixelS(i, j), a collection of minimal qualified sets ${\varGamma}_0^{\prime }$ are generated by Steps 2–10. Let k =  ∣ Γ₀∣ denote the number of minimal qualified sets in Γ₀ and ${k}^{\prime }=\mid {\varGamma}_0^{\prime}\mid$ (k^′ ≤ k) denote the number of minimal qualified sets in${\varGamma}_0^{\prime }$. Assume that the elements in X = {i₁, i₂, ⋯, i_t}can form d minimal qualified sets. The probability for the case that at least one minimal qualified set $Q\subseteq {\varGamma}_0^{\prime }$ satisfies Q ⊆ {i₁, ⋯, i_t} is $1-\frac{\left(\begin{array}{l}k-{k}^{\prime}\\ {}d\end{array}\right)}{\left(\begin{array}{l}k\\ {}d\end{array}\right)}$ and the probability for no minimal qualified set $Q\subseteq {\varGamma}_0^{\prime }$ satisfying Q ⊆ {i₁, ⋯, i_t} is $\frac{\left(\begin{array}{l}k-{k}^{\prime}\\ {}d\end{array}\right)}{\left(\begin{array}{l}k\\ {}d\end{array}\right)}$. Therefore, the light transmissions of the stacking result are

$$ T\left({R}_{i_1\otimes \cdots \otimes {i}_t}\left[S\left(i,j\right)=0\right]\right)=\left(1-\frac{\left(\begin{array}{l}k-{k}^{\prime}\\ {}d\end{array}\right)}{\left(\begin{array}{l}k\\ {}d\end{array}\right)}\right){\left(\frac{1}{2}\right)}^{e_1}+\frac{\left(\begin{array}{l}k-{k}^{\prime}\\ {}d\end{array}\right)}{\left(\begin{array}{l}k\\ {}d\end{array}\right)}{\left(\frac{1}{2}\right)}^{e_2} $$

(28)

and

$$ T\left({R}_{i_1\otimes \cdots \otimes {i}_t}\left[S\left(i,j\right)=1\right]\right)=\frac{\left(\begin{array}{l}k-{k}^{\prime}\\ {}d\end{array}\right)}{\left(\begin{array}{l}k\\ {}d\end{array}\right)}{\left(\frac{1}{2}\right)}^{e_2} $$

(29)

Then

$$ {\displaystyle \begin{array}{l}\kern0.5em T\left({R}_{i_1\otimes \cdots \otimes {i}_t}\left[S\left(i,j\right)=0\right]\right)-T\left({R}_{i_1\otimes \cdots \otimes {i}_t}\left[S\left(i,j\right)=1\right]\right)\\ {}\begin{array}{cc}=& \left(1-\frac{\left(\begin{array}{l}k-{k}^{\prime}\\ {}d\end{array}\right)}{\left(\begin{array}{l}k\\ {}d\end{array}\right)}\right){\left(\frac{1}{2}\right)}^{e_1}\end{array}\end{array}} $$

(30)

If X = {i₁, i₂, ⋯, i_t} ∈ Γ_Qual, d ≥ 1 and we have

$$ T\left({R}_{i_1\otimes \cdots \otimes {i}_t}\left[S\left(i,j\right)=0\right]\right)>T\left({R}_{i_1\otimes \cdots \otimes {i}_t}\left[S\left(i,j\right)=1\right]\right) $$

(31)

If X = {i₁, i₂, ⋯, i_t} ∈ Γ_Forb, d = 0 and we have

$$ T\left({R}_{i_1\otimes \cdots \otimes {i}_t}\left[S\left(i,j\right)=0\right]\right)=T\left({R}_{i_1\otimes \cdots \otimes {i}_t}\left[S\left(i,j\right)=1\right]\right) $$

(32)

According to Definition 5, we conclude the second and third conditions of this theorem.

Theorem 2

The RG-based VSS for GAS presented in [30] is the special case of the proposed RG-based VSS for GAS with contrast $\frac{2d}{\left({2}^t+1\right)k-d}$.

Proof: For Algorithm 4, if k^′ = 1, we have e₁ = t − 1 and e₂ = t. Thus, by Theorem 1 we have

$$ {\displaystyle \begin{array}{l}T\left({R}_{i_1\otimes \cdots \otimes {i}_t}\left[S\left(i,j\right)=0\right]\right)=\left(1-\frac{\left(\begin{array}{l}k-1\\ {}d\end{array}\right)}{\left(\begin{array}{l}k\\ {}d\end{array}\right)}\right){\left(\frac{1}{2}\right)}^{t-1}+\frac{\left(\begin{array}{l}k-1\\ {}d\end{array}\right)}{\left(\begin{array}{l}k\\ {}d\end{array}\right)}{\left(\frac{1}{2}\right)}^t\\ {}\begin{array}{cccc}\begin{array}{cc}& \end{array}& & & \end{array}\kern1.50em =\frac{d}{k}{\left(\frac{1}{2}\right)}^{t-1}+\left(1-\frac{d}{k}\right){\left(\frac{1}{2}\right)}^t\end{array}} $$

(33)

and

$$ T\left({R}_{i_1\otimes \cdots \otimes {i}_t}\left[S\left(i,j\right)=1\right]\right)=\frac{\left(\begin{array}{l}k-1\\ {}d\end{array}\right)}{\left(\begin{array}{l}k\\ {}d\end{array}\right)}{\left(\frac{1}{2}\right)}^t=\left(1-\frac{d}{k}\right){\left(\frac{1}{2}\right)}^t $$

(34)

By Definition 5 and Definition 7, the contrast is

$$ {\displaystyle \begin{array}{l}\kern0.5em \frac{T\left({R}_{i_1\otimes \cdots \otimes {i}_t}\left[S(0)\right]\right)-T\left({R}_{i_1\otimes \cdots \otimes {i}_t}\left[S(1)\right]\right)}{1+T\left({R}_{i_1\otimes \cdots \otimes {i}_t}\left[S(1)\right]\right)}\\ {}\begin{array}{cc}=& \frac{\frac{d}{k}{\left(\frac{1}{2}\right)}^{t-1}}{1+\left(1-\frac{d}{k}\right){\left(\frac{1}{2}\right)}^t}\end{array}\\ {}\begin{array}{cc}=& \frac{2d}{\left({2}^t+1\right)k-d}\end{array}\end{array}} $$

(35)

Herein, Algorithm 4 is reduced to the scheme [30].

Appendix C

Theoretical analyses on Algorithm 5 are provided as follows

Lemma 7

Given a secret pixel S(i, j) and n share pixels ES₁(i, j), ⋯, ES_n(i, j), any of the n random grids ${R}_1^{\prime}\left(i,j\right),\cdots, {R}_n^{\prime}\left(i,j\right)$ generated by Algorithm 5 satisfies the following conditions:

• 1) For any share pixel ES_t(i, j) = 0, the average light transmission of the corresponding random grid ${R}_t^{\prime}\left(i,j\right)$ is

$$ T\left({R}_t^{\prime}\left[E{S}_t\left(i,j\right)=0\right]\right)=\frac{1}{2}\beta +\left(\frac{1}{2}-\delta \right)\left(1-\beta \right) $$

(36)

where t = 1, 2, ⋯, n and $\delta \le \frac{1}{8}$.

• 2) For any share pixelES_t(i, j) = 1, the average light transmission of the corresponding random grid ${R}_t^{\prime}\left(i,j\right)$ is

$$ T\left({R}_t^{\prime}\left[E{S}_t\left(i,j\right)=1\right]\right)=\frac{1}{2}\beta $$

(37)

Proof: When d = 0, the random grid ${R}_t^{\prime}\left(i,j\right)$ is generated by Step 5 of Algorithm 5 and by Lemma 4 we have

$$ T\left({R}_t^{\prime}\left[E{S}_t\left(i,j\right)=0\right]\right)=T\left({R}_t^{\prime}\left[E{S}_t\left(i,j\right)=1\right]\right)=\frac{1}{2} $$

(38)

When d = 1, the random grid ${R}_t^{\prime}\left(i,j\right)$ is generated by Step 6 of Algorithm 5 and we obtain $T\left({R}_t^{\prime}\left[E{S}_t\left(i,j\right)=1\right]\right)=0$ immediately. For ES_t(i, j) = 0, the random grid ${R}_t^{\prime}\left(i,j\right)$ is generated randomly and then assigned black with a probability δ since that the code “randomly choose one from them and assign it black” occurs only when all ${R}_{i_1}^{\prime}\left(i,j\right),\cdots, {R}_{i_t}^{\prime}\left(i,j\right)$ are all white. For t = 2, $\delta =\frac{1}{2}\times {\left(\frac{1}{2}\right)}^2=\frac{1}{2}$ is the worst situation. For other cases, the value of δ decreases exponentially as the value of t increases and we can obtain $\delta \ll \frac{1}{8}$. Since Step 5 occurs with the probability β and Step 6 occurs with the probability 1 − β, we can conclude the above two conditions of this lemma.

Lemma 8

Given a secret pixel S(i, j), an access structure (Γ_Qual, Γ_Forb) and n share pixelsES₁(i, j), ⋯, ES_n(i, j). The n random grids ${R}_1^{\prime}\left(i,j\right),\cdots, {R}_n^{\prime}\left(i,j\right)$ generated by Algorithm 5 satisfy the following conditions:

• 1) If X = {i₁, i₂, ⋯, i_t} ∈ Γ_Qual and β > 0, the stacking result of the t random grids can recover the secret pixelS(i, j):

$$ T\left({R}_{i_1\otimes \cdots \otimes {i}_t}^{\prime}\left[S\left(i,j\right)=0\right]\right)>T\left({R}_{i_1\otimes \cdots \otimes {i}_t}^{\prime}\left[S\left(i,j\right)=1\right]\right) $$

(39)

• 2) If X = {i₁, i₂, ⋯, i_t} ∈ Γ_Forb, the stacking result of the t random grids gives no clue about the secret pixel S(i, j):

$$ T\left({R}_{i_1\otimes \cdots \otimes {i}_t}^{\prime}\left[S\left(i,j\right)=0\right]\right)=T\left({R}_{i_1\otimes \cdots \otimes {i}_t}^{\prime}\left[S\left(i,j\right)=1\right]\right) $$

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Proof: When d = 0, the n random grids ${R}_1^{\prime}\left(i,j\right),\cdots, {R}_n^{\prime}\left(i,j\right)$are generated by Step 5 and by Theorem 1 we have

$$ T\left({R}_{i_1\otimes \cdots \otimes {i}_t}^{\prime}\left[S\left(i,j\right)=0\right]\right)=\left(1-\frac{\left(\begin{array}{l}k-{k}^{\prime}\\ {}d\end{array}\right)}{\left(\begin{array}{l}k\\ {}d\end{array}\right)}\right){\left(\frac{1}{2}\right)}^{e_1}+\frac{\left(\begin{array}{l}k-{k}^{\prime}\\ {}d\end{array}\right)}{\left(\begin{array}{l}k\\ {}d\end{array}\right)}{\left(\frac{1}{2}\right)}^{e_2} $$

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and

$$ T\left({R}_{i_1\otimes \cdots \otimes {i}_t}^{\prime}\left[S\left(i,j\right)=1\right]\right)=\frac{\left(\begin{array}{l}k-{k}^{\prime}\\ {}d\end{array}\right)}{\left(\begin{array}{l}k\\ {}d\end{array}\right)}{\left(\frac{1}{2}\right)}^{e_2} $$

(42)

When d = 1, the n random grids ${R}_1^{\prime}\left(i,j\right),\cdots, {R}_n^{\prime}\left(i,j\right)$ are generated by Step 6, which is independent of the secret pixel. Thus, the light transmission of the stacking result is fixed to the same value no matter what the secret pixel is, namely, $T\left({R}_{i_1\otimes \cdots \otimes {i}_t}^{\prime}\left[S\left(i,j\right)=0\right]\right)=T\left({R}_{i_1\otimes \cdots \otimes {i}_t}^{\prime}\left[S\left(i,j\right)=1\right]\right)=\theta$.where 0 ≤ θ ≤ 1.

Due to that Step 5 occurs with the probability β and Step 6 occurs with the probability1 − β, we have

$$ {\displaystyle \begin{array}{l}\kern0.5em T\left({R}_{i_1\otimes \cdots \otimes {i}_t}^{\prime}\left[S\left(i,j\right)=0\right]\right)\\ {}\begin{array}{cc}=& \beta \left(\left(1-\frac{\left(\begin{array}{l}k-{k}^{\prime}\\ {}d\end{array}\right)}{\left(\begin{array}{l}k\\ {}d\end{array}\right)}\right){\left(\frac{1}{2}\right)}^{e_1}+\frac{\left(\begin{array}{l}k-{k}^{\prime}\\ {}d\end{array}\right)}{\left(\begin{array}{l}k\\ {}d\end{array}\right)}{\left(\frac{1}{2}\right)}^{e_2}\right)\end{array}+\left(1-\beta \right)\theta \end{array}} $$

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and

$$ T\left({R}_{i_1\otimes \cdots \otimes {i}_t}^{\prime}\left[S\left(i,j\right)=1\right]\right)=\beta \left(\frac{\left(\begin{array}{l}k-{k}^{\prime}\\ {}d\end{array}\right)}{\left(\begin{array}{l}k\\ {}d\end{array}\right)}{\left(\frac{1}{2}\right)}^{e_2}\right)+\left(1-\beta \right)\theta $$

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Therefore, we obtain

$$ {\displaystyle \begin{array}{l}\kern0.5em T\left({R}_{i_1\otimes \cdots \otimes {i}_t}^{\prime}\left[S\left(i,j\right)=0\right]\right)-T\left({R}_{i_1\otimes \cdots \otimes {i}_t}^{\prime}\left[S\left(i,j\right)=1\right]\right)\\ {}\begin{array}{cc}=& \beta \left(1-\frac{\left(\begin{array}{l}k-{k}^{\prime}\\ {}d\end{array}\right)}{\left(\begin{array}{l}k\\ {}d\end{array}\right)}\right){\left(\frac{1}{2}\right)}^{e_1}\end{array}\end{array}} $$

(45)

If X = {i₁, i₂, ⋯, i_t} ∈ Γ_Qual and β > 0, by Theorem 1 we have

$$ T\left({R}_{i_1\otimes \cdots \otimes {i}_t}^{\prime}\left[S\left(i,j\right)=0\right]\right)>T\left({R}_{i_1\otimes \cdots \otimes {i}_t}^{\prime}\left[S\left(i,j\right)=1\right]\right) $$

(46)

If X = {i₁, i₂, ⋯, i_t} ∈ Γ_Forb, by Theorem 1 we have

$$ T\left({R}_{i_1\otimes \cdots \otimes {i}_t}^{\prime}\left[S\left(i,j\right)=0\right]\right)=T\left({R}_{i_1\otimes \cdots \otimes {i}_t}^{\prime}\left[S\left(i,j\right)=1\right]\right) $$

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Theorem 3

Given a secret imageSand a strong access structure(Γ_Qual, Γ_Forb), Algorithm 5 is a valid construction for extended RG-based VSS, which meets the following three conditions:

• (1) If β < 1, each of n random grids ${R}_1^{\prime },\cdots, {R}_n^{\prime }$ can reveal the corresponding share image no matter what the secret image is:

$$ T\left({R}_t^{\prime}\left[E{S}_t(0)\right]\right)>T\left({R}_t^{\prime}\left[E{S}_t(1)\right]\right),t=1,2,\cdots, n $$

(48)

• (2) If X = {i₁, i₂, ⋯, i_t} ∈ Γ_Qual and β > 0, the stacking result of the t random grids can recover the secret image S:

$$ T\left({R}_{i_1\otimes \cdots \otimes {i}_t}^{\prime}\left[S(0)\right]\right)>T\left({R}_{i_1\otimes \cdots \otimes {i}_t}^{\prime}\left[S(1)\right]\right) $$

(49)

• (3) If X = {i₁, i₂, ⋯, i_t} ∈ Γ_Forb, the stacking result of the t random grids gives no clue about the secret image S:

$$ T\left({R}_{i_1\otimes \cdots \otimes {i}_t}^{\prime}\left[S(0)\right]\right)=T\left({R}_{i_1\otimes \cdots \otimes {i}_t}^{\prime}\left[S(1)\right]\right) $$

(50)

Proof: By Lemma 7, we have

$$ T\left({R}_t^{\prime}\left[E{S}_t\left(i,j\right)=0\right]\right)-T\left({R}_t^{\prime}\left[E{S}_t\left(i,j\right)=1\right]\right)=\left(\frac{1}{2}-\delta \right)\left(1-\beta \right) $$

(51)

If β < 1, $T\left({R}_t^{\prime}\left[E{S}_t\left(i,j\right)=0\right]\right)>T\left({R}_t^{\prime}\left[E{S}_t\left(i,j\right)=1\right]\right)$ since $\delta \le \frac{1}{8}$. According to Definition 5, we have

$$ T\left({R}_t^{\prime}\left[E{S}_t(0)\right]\right)>T\left({R}_t^{\prime}\left[E{S}_t(1)\right]\right) $$

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By Lemma 8 and Definition 5, the second and third conditions of this theorem can be concluded.