An efficient XOR-based visual cryptography scheme with lossless reconfigurable algorithms

Abstract

This article proposes an efficient XOR-based visual cryptography scheme with lossless reconfigurable algorithms, which includes an encryption and a decryption schemes for gray-scale and color secret images. The encryption scheme aims to encrypt secret images by mapping them to several shared images that do not suffer any pixel expansion problem and do not reveal any information of the secret images by considering a random column selection algorithm based on the 0-mapping and 1-mapping matrices, where the two matrices are dynamically generated in the encryption scheme. Moreover, the proposed decryption scheme can reconstruct the secret images with only a series of XOR operations. The reconstructed images do not suffer any contrast distortion and pixel expansion problems comparing with their original secret images. Finally, the proposed two schemes are illustrated with the practical examples.

Introduction

A visual cryptography scheme (VCS) that belongs to one of the visual secret sharing schemes was first proposed by Naor and Shamir ¹ in 1994. In this scheme, a secret image can be printed on documents, photographs, and pictures, and it assumes that a secret image consists of a collection of only black and white pixels. The secret image can be directly revealed by stacking only $m$ of the $n$ shared images together if the secret image is encrypted to $n$ shared images. It means that the reconstruction processes of the revealed secret image can be automatically performed by human visual systems without any cryptography knowledge or computation. Therefore, the studies of VCSs have become more and more valuable. ²

In the aforementioned VCS, each pixel of the secret image is split into $m$ black and white subpixels, which means that the pixels of the reconstructed image are expanded comparing with the original image. ³ In recent years, a large number of studies aim to reduce the contrast loss rates and pixel expansion scales for reconstructed images. Hofmeister and Krause ⁴ proposed a ( $k$ , $n$ )-threshold VCS to compute the optimal contrast by resolving simple linear programming problems. In their proposed scheme, the optimal contrast is $1/4$ , $1/16$ , or $1/64$ if $k$ is $2$ , $3$ , or $4$ , respectively. Otherwise, the scheme that is proposed by Naor and Shamir ¹ has the ideal optimal contrast if $k=n$ . Viet and Kurosawa ⁵ proposed a VCS to generate a reconstructed image that almost has no contrast loss with simple reverse operations. The white area of the reconstructed image can be reconstructed completely. However, the size of reconstructed image is increased comparing with the original image. Cimato and DeSantis ⁶ proposed a VCS that can be reconstructed using white and black pixels completely. However, this scheme is only applicable to binary images and cannot be used to deal with gray-scale images. Wang and Song ⁷ introduced a reverse operation–based VCS that can encrypt the gray-scale images with the optimal contrast. Moreover, Hou ⁸ proposed a VCS for color images by adopting a color separation technology to separate a color image to three gray-scale images that are further converted into corresponding three binary images with a halftone technique, where the halftone technique is also extensively used in some other VCSs.^9–11 Finally, the binary images are encrypted to the shared images. Wang and Arce ¹² proposed an error diffusion algorithm to construct meaningful shared images, which aimed to as much as possible reduce errors during the encryption processes.

Pixel expansion problems of VCSs also should be considered during the encryption processes. ¹³ Each pixel of a secret image will be split into $m$ black and white subpixels that are used to generate the shared images. This process may suffer pixel expansion and contrast distortion problems comparing with the reconstructed image and the original image. Yang ¹⁴ proposed a probability-based VCS that uses the white pixel probability of a secret image as the contrast of the reconstructed image. Subsequently, this scheme is extended by Cimato et al. ¹⁵ Shyu ¹⁶ proposed a random grids-based VCS that can encrypt gray-scale and color images, where the size of the reconstructed image is same as the original image since the proposed scheme avoids the pixel expansion problems. However, this scheme is not suitable for any of the ( $k$ , $n$ )-threshold VCSs. Chen and Tsao ¹⁷ proposed a ( $2$ , $n$ )-threshold VCS and a ( $n$ , $n$ )-threshold VCS by extending the scheme that is proposed by Shyu. ¹⁶ Chen and Tsao ¹⁸ proposed a random grids-based ( $k$ , $n$ )-threshold VCS that can optimize the pixel expansion problems and construct basis matrices. Furthermore, more probability-based VCSs and random grids-based VCSs are proposed,^19–21 which aim to achieve the lossless reconstructed images.

An XOR-based visual cryptography system (XVCS) is developed to achieve more better contrast and resolution for the reconstructed images. In practical applications, although the XOR operations need the help of computers or other optical devices in decryption steps, but it provides a new way to optimize the pixel expansion and contrast distortion problems for the reconstructed images. ²² For a ( $2$ , $n$ )-threshold XVCS, the contrast of a reconstructed image is $0.5$ , and it does not suffer any pixel expansion problem. ²³ For a ( $n$ , $n$ )-threshold XVCS, the mentioned two problems can be avoided with some sufficient and necessary conditions. ²⁴ They proposed a region by region construction to achieve the preferable contrast with some small pixel expansion. Furthermore, their proposed scheme contains an exhaustive search method that is very hard and quite time-consuming. Some revised ( $k$ , $n$ )-threshold XVCSs are proposed to improve the pixel expansion problems by using linear coding, ²⁵ integer linear programming, ²⁶ and linear algebraic. ²⁴ However, these XVCSs include many complex computation processes to construct the basis matrices, and the pixel expansion problems do not be completely avoided. ²⁷

Inspiring from the aforementioned schemes, a ( $n$ , $n$ )-threshold XVCS with lossless reconfigurable algorithms is proposed in this article. It contains an encryption scheme and a decryption scheme. The former aims to encrypt secret images by mapping them to $n$ shared images. For a secret image, its each original bit will be mapped to $n$ shared bits that are randomly selected from the columns of a 0-mapping matrix and a 1-mapping matrix, where the two matrices are also dynamically generated. Therefore, the original bits are replaced by the shared bits that are used to construct the shared images. It is clear that the shared images do not reveal any information of the secret images. Moreover, the proposed decryption scheme can reconstruct the secret images with only a series of XOR operations. The reconstructed images do not suffer any contrast distortion problem and they also have the same size as their original secret images.

The main contributions of this article are concluded as follows:

1. An efficient XVCS with a lossless reconfigurable algorithm is proposed, which can encrypt gray-scale and color secret images because the two classes of secret images can be converted to the corresponding binary matrices and the core steps of the encryption and decryption schemes primarily consider the binary XOR operations that are a typical class of logical operations.

2. The shared images do not reveal any information of the secret images because each original bit of the secret images is randomly mapped to other $n$ shared bits. Moreover, they also do not suffer any contrast distortion and pixel expansion problems comparing with their original secret images.

3. The proposed encryption scheme can rapidly construct the shared images with a series of matrix column selection operations and the proposed decryption scheme can also rapidly reconstruct the secret images with only a series of XOR operations. The proposed schemes have an ideal computation complexity.

The rest of this article is organized as follows. Section “Basic definition” defines the basic concepts and parameters. The encryption and decryption schemes are presented in section “The encryption and decryption schemes.” In section “Examples,” the proposed schemes is illustrated with several practical examples, and section “Conclusion” concludes this article.

Basic definition

Let $A=[q_{\mathrm{ij}}{]}_{\theta \times \beta }$ be a secret image, where $q_{\mathrm{ij}}$ is a pixel of $A$ . Pixel $q_{\mathrm{ij}}$ is composed of three components $R$ , $G$ , and $B$ if $A$ is a color image, where $R$ , $G$ , and $B$ represent the red, green, and blue channels, respectively. Otherwise, pixel $q_{\mathrm{ij}}$ is a gray-scale pixel point if $A$ is a gray-scale image.

Definition 1. Let $A=[q_{\mathrm{ij}}{]}_{\theta \times \beta }$ be a secret image. Matrix $P=[p_{\mathrm{ik}}{]}_{\theta \times \xi }$ is called as the pixel matrix of $A$ such that (1) $P=A$ if $A$ is a gray-scale image, where $\xi =\beta$ . (2) $\forall i\in [0,\theta )$ , $\forall k\in [0,\xi )$ , and $\forall j\in [0,\beta )$ such that $p_{\mathrm{ik}}$ , $p_{i(k+1)}$ , and $p_{i(k+2)}$ are the red, green, and blue channels of $q_{\mathrm{ij}}$ , respectively, if $A$ is a color image, where $k=3\times j$ and $\xi =3\times \beta$ .

Example 1: Let $A=\left[\begin{array}{cc}{q}_{00}\hfill & q_{01}\hfill \\ \multicolumn{1}{c}{q_{10}}& q_{11}\hfill \end{array}\right]$ be a secret image. Then, the pixel matrix of $A$ is

$$P=\left[\begin{array}{cc}{p}_{00}\hfill & p_{01}\hfill \\ \multicolumn{1}{c}{p_{10}}& p_{11}\hfill \end{array}\right]=\left[\begin{array}{cc}{q}_{00}\hfill & q_{01}\hfill \\ \multicolumn{1}{c}{q_{10}}& q_{11}\hfill \end{array}\right]$$

if $A$ is a gray-scale image. Otherwise,

$$P=\left[\begin{array}{cccccc}{p}_{00}\hfill & p_{01}\hfill & p_{02}\hfill & p_{03}\hfill & p_{04}\hfill & p_{05}\hfill \\ \multicolumn{1}{c}{p_{10}}& p_{11}\hfill & p_{12}\hfill & p_{13}\hfill & p_{14}\hfill & p_{15}\hfill \end{array}\right]$$

if $A$ is a color image, where items $p_{00}$ , $p_{01}$ , and $p_{02}$ are the $R$ , $G$ , and $B$ components of pixel $q_{00}$ , respectively. Items $p_{03}$ , $p_{04}$ , and $p_{05}$ are the $R$ , $G$ , and $B$ components of pixel $q_{01}$ , respectively. Similarly, it can obtain the items $p_{10}$  –  $p_{15}$ . It is shown that the secret image $A$ and its pixel matrix $P$ have the same rows. However, the pixel matrix $P$ has the three times as many columns as $A$ . This means that the pixels of $A$ are expanded three times in $P$ if $A$ is a color image. Of course, the pixels do not be expand if $A$ is a gray-scale image.

Let $p\in [0,255]$ be a pixel of an image. Then, $p$ can be transformed into a 8-bit binary number, denoted as

$$b_0{b}_1{b}_2{b}_3{b}_4{b}_5{b}_6{b}_7=\mathrm{Binary}(p)$$

(1)

and the inverse function is denoted as

$$p=\mathrm{Binar}{y}^{-1}(b_0{b}_1{b}_2{b}_3{b}_4{b}_5{b}_6{b}_7)$$

(2)

where $\forall b_i\in \{0,1\}$ and $i\in [0,7]$ . For example, we have $00010101=\mathrm{Binary}(21)$ and $21=\mathrm{Binar}{y}^{-1}(00010101)$ .

Definition 2. Let $P=[p_{\mathrm{ik}}{]}_{\theta \times \xi }$ be a pixel matrix. Matrix $B=[b_{\mathrm{ir}}{]}_{\theta \times \omega }$ is called as the binary matrix of $P$ if $\forall i\in [0,\theta )$ and $\forall k\in [0,\xi )$ such that $\omega =8\times \xi$ and

$$b_{\mathrm{ir}},b_{i(r+1)},b_{i(r+2)},\cdots ,b_{i(r+7)}=\mathrm{Binary}(p_{\mathrm{ik}})$$

where $r=8\times k$ and $r\in [0,\omega )$ .

Example 2: Let $P=\left[\begin{array}{cc}{p}_{11}\hfill & p_{12}\hfill \\ \multicolumn{1}{c}{p_{21}}& p_{22}\hfill \end{array}\right]=\left[\begin{array}{cc}21\hfill & 182\hfill \\ \multicolumn{1}{c}{3}& 255\hfill \end{array}\right]$ be a pixel matrix. The binary matrix of $P$ is

$$\begin{array}{c}B=\left[\begin{array}{cc}{b}_{00}{b}_{01}{b}_{02}\dots b_{07}\hfill & b_{08}{b}_{09}{b}_{0(10)}\dots b_{0(15)}\hfill \\ \multicolumn{1}{c}{b_{10}{b}_{11}{b}_{12}\dots b_{17}}& b_{18}{b}_{19}{b}_{1(10)}\dots b_{1(15)}\hfill \end{array}\right]\hfill \\ =\left[\begin{array}{cc}00010101\hfill & 10110110\hfill \\ \multicolumn{1}{c}{00000011}& 11111111\hfill \end{array}\right]\hfill \end{array}$$

In a VCS, it is inadvisable that directly share the original bits of a secret image with the participants since it will reveal the information of the secret images. A better idea is to construct some transparent random bits to replace the original bits. This means that the original bits should be mapped to other irrelevant bits that are shared for participants. Therefore, a mapping matrix is defined as follows.

Definition 3. Let $n$ be the number of participants. Matrix $M=[e_{\mathrm{xy}}{]}_{n\times 2^n}$ is called as a mapping matrix if $\forall e_{\mathrm{xy}}\in M$ such that $e_{\mathrm{xy}}\in \{0,1\}$ and $\forall y\in [0,2^n-1]$ such that

$$e_{0y}{e}_{1y}{e}_{2y}\cdots e_{(n-1)y}=\mathrm{Binary}(y).$$

Example 3: Assume that there are three participants, that is, $n=3$ . The mapping matrix $M$ is

$$\begin{array}{c}\begin{array}{cccccccc}0\hfill & 1\hfill & 2\hfill & 3\hfill & 4\hfill & 5\hfill & 6\hfill & 7\hfill \end{array}\hfill \\ M=[e_{\mathrm{xy}}{]}_{3\times 8}=\left(\begin{array}{cccccccc}0\hfill & 0\hfill & 0\hfill & 0\hfill & 1\hfill & 1\hfill & 1\hfill & 1\hfill \\ \multicolumn{1}{c}{0}& 0\hfill & 1\hfill & 1\hfill & 0\hfill & 0\hfill & 1\hfill & 1\hfill \\ \multicolumn{1}{c}{0}& 1\hfill & 0\hfill & 1\hfill & 0\hfill & 1\hfill & 0\hfill & 1\hfill \end{array}\right)\hfill \end{array}$$

where $000=\mathrm{Binary}(0)$ , $001=\mathrm{Binary}(1)$ , $010=\mathrm{Binary}(2)$ , $\dots$ , and $111=\mathrm{Binary}(7)$ .

Since each original bit of a secret image should be mapped to other irrelevant bits, it can randomly select one of the columns from the mapping matrix $M$ to as the mapped bits. Moreover, the value of each original bit of secret images is ‘0’ or ‘1.’ It is necessary to consider which one column is selected to as the mapped bits for the original bit ‘0’ and which one column is selected for the original bit ‘1.’ Considering the parity-check of XOR operations on a series of bits, their result is zero (resp. one) if these bits contain an even (resp. odd) number of ‘1.’ Therefore, the mapping matrix $M$ should be divided into two submatrices according to the number of “1” in its each column.

Definition 4. Let $n$ be the number of participants, $M=[e_{\mathrm{xy}}{]}_{n\times 2^n}$ be a mapping matrix, $V=(1,1,1,\dots ,1{)}_n$ be a vector, $M^0=[e_{\mathrm{xy}}^0{]}_{n\times 2^{n-1}}$ and $M^1=[e_{\mathrm{xy}}^1{]}_{n\times 2^{n-1}}$ be two matrices such that $M^0\cap M^1=\varnothing$ , $M^0\cup M^1=M$ , and $\forall y\in [0,2^n-1]$ such that

$$e_{(*y)}\in \{\begin{array}{c}{M}^0\mathrm{if}(V\times e_{(*y)})\mathrm{mod}2=0\hfill \\ \multicolumn{1}{c}{M^1\mathrm{if}(V\times e_{(*y)})\mathrm{mod}2\ne 0}\end{array}$$

where $e_{(*y)}\in M$ represents the $y$ th column of $M$ . Then, $M^0$ is called as a 0-mapping matrix and $M^1$ is called as a 1-mapping matrix.

This means that a mapping matrix $M$ is divided into a 0-mapping matrix $M^0$ and a 1-mapping matrix $M^1$ , where each column of $M^0$ contains an even number of “1” and each column of $M^1$ contains an odd number of “1.”

Example 4: For the mapping matrix $M$ in Example 3, the $1$ th column is $e_{(*0)}=(0,0,0{)}^T$ . We have $e_{(*0)}\in M^0$ since $(V\times e_{(*0)})\mathrm{mod}2=((1,1,1)\times (0,0,0{)}^T)\mathrm{mod}2=0$ . The $2$ th column is $e_{(*1)}\in M^1$ since $(V\times e_{(*1)})\mathrm{mod}2=((1,1,1)\times (0,0,1{)}^T)\mathrm{mod}2=1$ . We have $e_{(*2)}\in M^1$ , $e_{(*3)}\in M^0$ , $e_{(*4)}\in M^1$ , $e_{(*5)}\in M^0$ , $e_{(*6)}\in M^0$ , and $e_{(*7)}\in M^1$ . Then, $M^0$ and $M^1$ are

$$\begin{array}{c}\begin{array}{cccc}0\hfill & 3\hfill & 5\hfill & 6\hfill \end{array}\hfill \\ M^0=\left(\begin{array}{cccc}0\hfill & 0\hfill & 1\hfill & 1\hfill \\ \multicolumn{1}{c}{0}& 1\hfill & 0\hfill & 1\hfill \\ \multicolumn{1}{c}{0}& 1\hfill & 1\hfill & 0\hfill \end{array}\right)\mathrm{and}\hfill \\ \begin{array}{c}\begin{array}{cccc}1\hfill & 2\hfill & 4\hfill & 7\hfill \end{array}\hfill \\ M^1=\left(\begin{array}{cccc}0\hfill & 0\hfill & 1\hfill & 1\hfill \\ \multicolumn{1}{c}{0}& 1\hfill & 0\hfill & 1\hfill \\ \multicolumn{1}{c}{1}& 0\hfill & 0\hfill & 1\hfill \end{array}\right)\hfill \end{array}\hfill \end{array}$$

Definition 5. Let $n$ be the number of participants, $M^0=[e_{\mathrm{xy}}^0{]}_{n\times 2^{n-1}}$ and $M^1=[e_{\mathrm{xy}}^1{]}_{n\times 2^{n-1}}$ be the 0-mapping and 1-mapping matrices, respectively, and $b_{\mathrm{ir}}$ be an original bit of a secret image. The $y$ th column of $M^0$ (resp. $M^1$ ), denoted as $e_{*y}^0$ (resp. $e_{*y}^1$ ), is called as the 0-mapping(resp. 1-mapping) vector of $b_{\mathrm{ir}}$ if $e_{*y}^0$ (resp. $e_{*y}^1$ ) is selected to replace the original element $b_{\mathrm{ir}}$ to be shared for the $n$ participants.

Example 5: In Example 2, we know that $b_{00}=0$ and $b_{17}=1$ are two original bits in the binary matrix $B$ . If the secret image $A$ will be shared for three participants, we can construct a 0-mapping matrix $M^0$ and a 1-mapping matrix $M^1$ , as shown in Example 4. Therefore, the $1$ th column of $M^0$ , denoted as $e_{*1}^0=(0,1,1{)}^T$ , is called as the 0-mapping vector for the original bit $b_{00}$ if it is selected to replace $b_{00}$ to be shared for the three participants, where the first participant obtains “0” and the other two participants both obtain “1.” Similarly, the $3$ th column of $M^1$ , denoted as $e_{*3}^1=(1,1,1{)}^T$ , is called as the 1-mapping vector for the original bit $b_{17}$ if it is selected to replace $b_{17}$ to be shared for the three participants, where they all obtain “1.”

For a secret image, its original bits can be shared and reconstructed according to the column items of 0-mapping and 1-mapping matrices. Assume that the value of an original bit is ’0.’ It can be mapped by selecting one of the columns from the 0-mapping matrix. Certainly, the original bit can be reconstructed by a series of XOR operations for all bits of the selected column. Similarly, an original bit can also be mapped and reconstructed by one of the columns in a 1-mapping matrix if its value is “1.” To further enhance the secrecy of the secret image, the column selections in 0-mapping and 1-mapping matrices should consider their randomness.

Definition 6. Let $n$ be the number of participants, $M^0$  = $[e_{\mathrm{xy}}^0{]}_{n\times 2^{n-1}}$ and $M^1$  = $[e_{\mathrm{xy}}^1{]}_{n\times 2^{n-1}}$ be the 0-mapping and 1-mapping matrices, respectively, $V$  = ( $v_0$ , $v_1$ , $\dots$ , $v_{{(2}^{n-1}-1)})$ be a vector such that $\sum _{\tau =0}^{{(2}^{n-1}-1)}(v_{\tau })=1$ and $\forall \tau \in [0,2^{n-1})$ , $v_{\tau }\in [0,1]$ . Vector $V$ is called as a probability vector of matrices $M^0$ and $M^1$ if $\forall \tau \in [0,2^{n-1})$ such that $v_{\tau }$ is the probability of selecting $e_{*\tau }^0$ (resp. $e_{*\tau }^1$ ) to be the 0-mapping (resp. 1-mapping) vector of an original bit “0” (resp. “1”).

Example 6: If a secret image will be shared for three participants, we can construct a 0-mapping matrix $M^0$ and a 1-mapping matrix $M^1$ , as shown in Example 4. Let $V=(0.2,0.4,0.1,0.3)$ be a probability vector of $M^0$ and $M^1$ . The probabilities of selecting columns $(0,0,0{)}^T$ , $(0,1,1{)}^T$ , $(1,0,1{)}^T$ , and $(1,1,0{)}^T$ from $M^0$ are 0.2, 0.4, 0.1, and 0.3, respectively. In the same way, the probabilities of selecting columns $(0,0,1{)}^T$ , $(0,1,0{)}^T$ , $(1,0,0{)}^T$ , and $(1,1,1{)}^T$ from $M^1$ are same with the former.

In the actual applications of Example 6, it is difficult to precisely generate a random number $\varphi$ such that $\varphi \in \{0.2,0.4,0.1,0.3\}$ for each bit of the secret image. An ideal method is to construct a probability interval $({\gamma }_i,{\gamma }_j]$ for each column selection from $M^0$ or $M^1$ according to the probability vector $V$ . The corresponding column will be selected from $M^0$ or $M^1$ if ${\gamma }_i<\varphi \le {\gamma }_j$ . Therefore, a probability interval vector is defined as follows.

Definition 7. Let $n$ be the number of participants and $V$  = ( $v_0$ , $v_1$ , $\dots$ , $v_{{(2}^{n-1}-1)})$ be a probability vector. Vector $\mathrm{\Gamma }$  = ( ${\gamma }_0$ , ${\gamma }_1$ , $\dots$ , ${\gamma }_{2^{n-1}})$ is called as the probability interval vector if ${\gamma }_0=0$ and $\forall \rho \in \{1,\dots ,2^{n-1}\}$ such that ${\gamma }_{\rho }=\sum _{k=0}^{\rho -1}(v_k)$ .

Example 7: In Example 6, we know that $V$  = (0.2, 0.4, 0.1, 0.3) is a probability vector and $n=3$ . We have ${\gamma }_0=0$ , ${\gamma }_1=v_0=0.2$ , ${\gamma }_2=v_0+v_1=0.6$ , $\dots$ , and ${\gamma }_4=v_0+v_1+v_2+v_3=1$ . Therefore, the probability interval vector is $\mathrm{\Gamma }=(0,0.2,0.6,0.7,1)$ . The probability intervals of selecting columns $(0,0,0{)}^T$ , $(0,1,1{)}^T$ , $(1,0,1{)}^T$ , and $(1,1,0{)}^T$ from $M^0$ to be the 0-mapping vectors are $(0,0.2]$ , $(0.2,0.6]$ , $(0.6,0.7]$ , and $(0.7,1]$ , respectively. It is similar for the column selections of $M^1$ .

The encryption and decryption schemes

Let $n$ be the number of participants and $V=(v_0$ , $v_1$ , $\dots$ , $v_{{(2}^{n-1}-1)})$ be a probability vector. For each original bit of a secret image, the probability of selecting its 0-mapping vector or 1-mapping vector should be equal in order to ensure the fairness in the whole encryption processes. Then,

$$\{\begin{array}{c}{v}_0=v_1=\cdots =v_{{(2}^{n-1}-1)}\hfill \\ \multicolumn{1}{c}{v_0+v_1+\cdots +v_{{(2}^{n-1}-1)}=1}\end{array}$$

(3)

Therefore, $\forall \tau \in [0,2^{n-1}-1]$ , we have

$$v_{\tau }=\frac{1}{2^{n-1}}$$

(4)

The encryption scheme

In the encryption processes, Algorithms 1 and 2 are proposed to encrypt a secret image to its shared images for $n$ participants. Figure 1 shows the whole framework of the proposed encryption scheme.

Figure 1.

The whole framework of the proposed encryption scheme.

Algorithm 1.

Computing the intermediate variables according to the secret image and the number of participants.

Input: A secret image $A=[q_{\mathrm{ij}}{]}_{\theta \times \beta }$ ;

The number of participants, denoted as $n$ .

Output: A binary matrix $B=[b_{\mathrm{ir}}{]}_{\theta \times \omega }$ ;

A 0-mapping matrix $M^0=[e_{\mathrm{xy}}^0{]}_{n\times 2^{n-1}}$ ;

A 1-mapping matrix $M^1=[e_{\mathrm{xy}}^1{]}_{n\times 2^{n-1}}$ ;

A probability interval vector $\mathrm{\Gamma }=({\gamma }_0,{\gamma }_1,\cdots$ , ${\gamma }_{2^{n-1}})$ ;

A random probability matrix $\Phi =[{\varphi }_{\mathrm{ir}}{]}_{\theta \times \omega }$ .

1: Getting the pixel matrix $P=[p_{\mathrm{ik}}{]}_{\theta \times \xi }$ according to the secret image $A$ (Definition 1);

2: Computing the binary matrix $B$  = $[b_{\mathrm{ir}}{]}_{\theta \times \omega }$ according to pixel matrix $P$ (Definition 2), where $\omega =8\times \xi$ ;3: Getting a mapping matrix $M=[e_{\mathrm{xy}}{]}_{n\times 2^n}$ according to the number of participants $n$ (Definition 3);

4: Computing the 0-mapping and 1-mapping matrices $M^0$  = $[e_{\mathrm{xy}}^0{]}_{n\times 2^{n-1}}$ and $M^1$  = $[e_{\mathrm{xy}}^1{]}_{n\times 2^{n-1}}$ according to the mapping matrix $M$  = $[e_{\mathrm{xy}}{]}_{n\times 2^n}$ (Definition 4);

5: According to the number of participants $n$ , getting a probability vector $V=(v_0,v_1,\cdots ,v_{{(2}^{n-1}-1)})$ by using the methods of Definition 6;

6: Computing the probability interval vector $\mathrm{\Gamma }=({\gamma }_0,$ ${\gamma }_1,\cdots ,{\gamma }_{2^{n-1}})$ according to $V$ (Definition 7);

7: Randomly generating a random probability matrix $\Phi =[{\varphi }_{\mathrm{ir}}{]}_{\theta \times \omega }$ for binary matrix $B$ , where $\forall {\varphi }_{\mathrm{ir}}\in \Phi$ such that ${\varphi }_{\mathrm{ir}}$ is a random value and $0<{\varphi }_{\mathrm{ir}}\le 1$ ;

8: return $B$ , $M^0$ , $M^1$ , $\mathrm{\Gamma }$ , and $\Phi$ ;

Algorithm 2.

Generating $n$ shared images $A_0^r$ , $A_1^r$ , $\dots$ , $A_{n-1}^r$ for the $n$ participants.

Input: A binary matrix $B=[b_{\mathrm{ir}}{]}_{\theta \times \omega }$ ;

A 0-mapping matrix $M^0=[e_{\mathrm{xy}}^0{]}_{n\times 2^{n-1}}$ ;

A 1-mapping matrix $M^1=[e_{\mathrm{xy}}^1{]}_{n\times 2^{n-1}}$ ;

A probability interval vector $\mathrm{\Gamma }=({\gamma }_0,{\gamma }_1,\cdots$ , ${\gamma }_{2^{n-1}})$ ;

A random probability matrix $\Phi =[{\varphi }_{\mathrm{ir}}{]}_{\theta \times \omega }$ .

Output: The $n$ shared images $A_0^r$ , $A_1^r$ , $\cdots$ , $A_{n-1}^r$ .

1: Let $B_0^r$ , $B_1^r$ , $\cdots$ , and $B_{n-1}^r$ be $n$ binary matrices with the size of $\theta \times \omega$ and their all bits be ‘0’;

2: For( $i=0$ ; $i\le \theta$ ; $i++$ )

3: For( $r=0$ ; $r\le \omega$ ; $r++$ )

4: If( $b_{\mathrm{ir}}$ == 0) then

5: $\alpha =0$ ;

6: Else

7: $\alpha =1$ ;

8: End if

9: For ( $x=0$ ; $x<n$ ; $x++$ )

10: $B_x^r[\mathrm{ir}]=e_{\mathrm{xj}}^{\alpha }$ , where $j$ $s.t.$ ${\varphi }_{\mathrm{ir}}\in ({\gamma }_{j-1},{\gamma }_j]$ ;

11: End for

12: End for

13: End for

14: Converting the $n$ binary matrices $B_0^r-B_{n-1}^r$ to $n$ pixel matrices $P_0^r$ - $P_{n-1}^r$ according to the reverse processes of definition 2;

15: Computing the shared images $A_0^r$ , $A_1^r$ , $\cdots$ , $A_{n-1}^r$ according to the $n$ pixel matrices $P_0^r$ - $P_{n-1}^r$ with the reverse processes of definition 1;

16: return $A_0^r$ , $A_1^r$ , $\cdots$ , and $A_{n-1}^r$ ;

In Algorithm 1, several intermediate variables that are used to generate the shared images are computed according to the input secret image $A$ and the number of participants $n$ , such as the binary matrix $B$ , the 0-mapping and 1-mapping matrices $M^0$ and $M^1$ , a probability interval vector $\mathrm{\Gamma }$ , and a random probability matrix $\mathrm{\Phi }$ that are the inputs of Algorithm 2.

In Steps 1 and 2 in Algorithm 1, the pixel values of the input secret image $A$ are converted into binary matrix according to definitions 1 and 2, since the proposed scheme is based on the XOR operations. The binary matrix $B$ of the secret image $A$ can be directly obtained by equation (1) if $A$ is a gray-scale image. If $A$ is a color image, it should be first converted to a pixel matrix $P$ and continually converted into the binary matrix $B$ that is three times as the width as the case of the gray-scale image.

In Steps 3 and 4, the 0-mapping and 1-mapping matrices $M^0$ and $M^1$ are computed, where $n$ is the number of participants. For each original bit $b_{\mathrm{ir}}\in B$ , it is inadvisable that directly sharing $b_{\mathrm{ir}}$ to the $n$ participants. An ideal scheme is that the original bit $b_{\mathrm{ir}}$ is mapped to other $n$ bits in which each bit is used to replace the original bit $b_{\mathrm{ir}}$ and is shared to a corresponding participant. Conversely, the $n$ shared bits of the $n$ participants can be used to reconstruct the original bit $b_{\mathrm{ir}}$ by a series of XOR operations. Step 3 is the mapping processes from the original bit $b_{\mathrm{ir}}$ to the $n$ shared bits. We first generate a series of numbers, that is, 0, 1, 2, $\cdots$ , $2^n-1$ , and each number can be converted to a $n$ -bit binary number. Therefore, a mapping matrix $M$ is constructed, where each original bit $b_{\mathrm{ir}}$ can be mapped to one of the columns of the mapping matrix $M$ . Considering the reconstruction problems of the original bit $b_{\mathrm{ir}}$ , the column selections from the mapping matrix $M$ should consider the fact that the original bit $b_{\mathrm{ir}}$ has two values $b_{\mathrm{ir}}=0$ or $b_{\mathrm{ir}}=1$ . Therefore, the mapping matrix $M$ should be divided into two parts. Step 4 realizes the dividing processes. According to Definition 4, a column of the mapping matrix $M$ is also the column of the 0-mapping matrix $M^0$ (resp. 1-mapping matrix $M^1$ ) if the bits of this column contain an even (resp. odd) number of “1.” This means that the mapped $n$ bits can be randomly selected from the 0-mapping matrix $M^0$ if $b_{\mathrm{ir}}=0$ . Otherwise, the mapped $n$ bits can be randomly selected from the 1-mapping matrix $M^1$ .

In order to facilitate the random column selections in step 4, a probability interval vector $\mathrm{\Gamma }=({\gamma }_0,{\gamma }_1,\cdots ,{\gamma }_{2^{n-1}})$ is proposed in Steps 5 and 6. For all original bits in $B$ , the probabilities of their column selections should be in step 4 according to equations (3) and (4). We have the probability vector is

$$\begin{array}{c}V=(v_0,v_1,\dots ,v_{{(2}^{n-1}-1)})\hfill \\ =(\frac{1}{2^{n-1}},\frac{1}{2^{n-1}},\dots ,\frac{1}{2^{n-1}})\hfill \end{array}$$

Therefore, the probability interval vector is

$$\begin{array}{c}\mathrm{\Gamma }=({\gamma }_0,{\gamma }_1,\dots ,{\gamma }_{2^{n-1}})\hfill \\ =(0,\frac{1}{2^{n-1}},\frac{2}{2^{n-1}},\dots ,\frac{2^{n-1}-1}{2^{n-1}},1)\hfill \end{array}$$

For the binary matrix $B=[b_{\mathrm{ir}}{]}_{\theta \times \omega }$ of secret image $A$ , $\forall b_{\mathrm{ir}}\in B$ , bit $b_{\mathrm{ir}}$ will be mapped to $n$ shared bits by randomly selecting a column from the 0-mapping or 1-mapping matrixes according to the probability interval vector $\mathrm{\Gamma }$ and a random value ${\varphi }_{\mathrm{ir}}$ ( $0\le {\varphi }_{\mathrm{ir}}\le 1$ ). In step 7, a corresponding random probability matrix $\mathrm{\Phi }=[{\varphi }_{\mathrm{ir}}{]}_{\theta \times \omega }$ is randomly generated for the binary matrix $B=[b_{\mathrm{ir}}{]}_{\theta \times \omega }$ , where it is a one-to-one correspondence between ${\varphi }_{\mathrm{ir}}$ and $b_{\mathrm{ir}}$ and $0<{\varphi }_{\mathrm{ir}}\le 1$ . This means that, $\forall b_{\mathrm{ir}}\in B$ , there must be one and only one ${\varphi }_{\mathrm{ir}}\in \mathrm{\Phi }$ such that the $n$ mapped bits of $b_{\mathrm{ir}}$ are uniquely determined by ${\varphi }_{\mathrm{ir}}$ .

In Algorithm 2, the $n$ shared images $A_0^r$ , $A_1^r$ , $\cdots$ , and $A_{n-1}^r$ are computed according to the outputs of Algorithm 1, which have a same size as the secret image $A$ , that is, $\theta \times \beta$ .

In step 1, the $n$ temporary binary matrices $B_0^r$ , $B_1^r$ , $\cdots$ , and $B_{n-1}^r$ are created and their all bits are initialized to “0.” They are used to save the mapping bits that are randomly selected from the 0-mapping matrix $M^0$ or 1-mapping matrix $M^1$ .

In Steps 2 to 13, the shared bits are randomly selected from the 0-mapping matrix $M^0$ and the 1-mapping matrix $M^1$ . They are stored to the $n$ temporary binary matrices $B_0^r$ , $B_1^r$ , $\cdots$ , and $B_{n-1}^r$ . In Steps 4 to 8, the value of a temporary variable $\alpha$ is first determined according to the value of original bit $b_{\mathrm{ir}}$ . In steps 9 to 11, the $j$ th column are selected to as the shared bits from the 0-mapping or 1-mapping matrices according to the temporary variable $\alpha$ since $j$ such that ${\varphi }_{\mathrm{ir}}\in ({\gamma }_{j-1},{\gamma }_j]$ , where ${\varphi }_{\mathrm{ir}}\in \mathrm{\Phi }$ is a random value and ${\gamma }_{j-1},{\gamma }_j\in \mathrm{\Gamma }$ .

In Steps 14 and 15, the $n$ temporary binary matrices $B_0^r$ , $B_1^r$ , $\cdots$ , and $B_{n-1}^r$ are converted to $n$ pixel matrices $P_0^r$  –  $P_{n-1}^r$ according to the reverse processes of definition 2 and the pixel matrices $P_0^r$  –  $P_{n-1}^r$ are continuously converted to $n$ shared images $A_0^r$ , $A_1^r$ , $\cdots$ , and $A_{n-1}^r$ according to the reverse processes of definition 2. Therefore, the $n$ shared images $A_0^r$ , $A_1^r$ , $\cdots$ , $A_{n-1}^r$ are encrypted and shared for $n$ participants.

The decryption scheme

After the $n$ shared images $A_0^r$ , $A_1^r$ , $\cdots$ , and $A_{n-1}^r$ are collected from the $n$ participants, the secret image $A$ can be reconstructed by Algorithm 3.

Algorithm 3.

The secret image $A$ can be reconstructed according to $n$ shared images $A_0^r$ , $A_1^r$ , $\dots$ , and $A_{n-1}^r$ .

Input: The $n$ shared images $A_0^r$ , $A_1^r$ , $\cdots$ , and $A_{n-1}^r$ .

Output: The secret image $A=[q_{\mathrm{ij}}{]}_{\theta \times \beta }$ .

1: Converting the $n$ shared images $A_0^r$ - $A_{n-1}^r$ to $n$ pixel matrices $P_0^r$ - $P_{n-1}^r$ (definition 1);

2: Getting the $n$ binary matrices $B_0^r$ - $B_{n-1}^r$ according to the $n$ pixel matrices $P_0^r$ - $P_{n-1}^r$ (definition 2);

3: Let $B=[b_{\mathrm{ir}}{]}_{\theta \times \omega }$ be a binary matrix and their all bits be initialized to ’0’;

4: For( $i=0$ ; $i\le \theta$ ; $i++$ )

5: For( $r=0$ ; $r\le \omega$ ; $r++$ )

6: $b_{\mathrm{ir}}=b_{0[\mathrm{ir}]}^r\oplus b_{1[\mathrm{ir}]}^r\oplus \cdots \oplus b_{n-1[\mathrm{ir}]}^r$ , where $b_{k[\mathrm{ir}]}^r\in B_k^r$ and $k\in \{0,1,\cdots ,n-1\}$ ;

7: End for

8: End for

9: Converting the binary matrix $B$ to a pixel matrix $P$ by the reverse processes of definition 2;

10: Getting the secret image $A$ according to the pixel matrix $P$ by the reverse processes of definition 1;

11: return $A$ ;

In Steps 1 and 2, the $n$ shared images $A_0^r$ , $A_1^r$ , $\dots$ , and $A_{n-1}^r$ are converted to $n$ pixel matrices $P_0^r$  –  $P_{n-1}^r$ according to definition 1 and the pixel values of the $n$ pixel matrices are continuously converted to binary numbers, denoted as $B_0^r$  –  $B_{n-1}^r$ , according to definition 2 since the proposed decryption scheme is based on the XOR operations.

In Step 3, a binary matrix $B=[b_{\mathrm{ir}}{]}_{\theta \times \omega }$ is created and their all bits are initialized to “0.” It is used to save the binary bits that are reconstructed according to the corresponding bits of the $n$ shared images $A_0^r$ , $A_1^r$ , $\dots$ , and $A_{n-1}^r$ .

In Steps 4–8, $\forall b_{\mathrm{ir}}\in B$ , $b_{\mathrm{ir}}$ is reconstructed by a series of XOR operations among the binary bits $B_{0[\mathrm{ir}]}^r$ , $B_{1[\mathrm{ir}]}^r$ , $\dots$ , and $B_{n-1[\mathrm{ir}]}^r$ that are one of the columns in the 0-mapping matrix $M^0$ or 1-mapping matrix $M^1$ and are randomly selected to as the mapping bits in the steps 9–11 of Algorithm 2. According to Steps 4–8 of Algorithm 2, the group of $B_{0[\mathrm{ir}]}^r$ , $B_{1[\mathrm{ir}]}^r$ , $\dots$ , and $B_{n-1[\mathrm{ir}]}^r$ is selected from the columns of the 0-mapping matrix if the original bit is “0,” that is, $b_{\mathrm{ir}}=0$ . Otherwise, it is selected from the columns of the 1-mapping matrix $M^1$ . In other words, according to Definition 4, the selected $n$ binary bits $B_{0[\mathrm{ir}]}^r$ , $B_{1[\mathrm{ir}]}^r$ , $\dots$ , and $B_{n-1[\mathrm{ir}]}^r$ should contain an even number of “1” if they are selected from the columns of 0-mapping matrix $M^0$ . Otherwise, they should contain an odd number of “1.” In step 6 of Algorithm 3, we have $b_{\mathrm{ir}}=b_{0[\mathrm{ir}]}^r\oplus b_1^r[\mathrm{ir}]\oplus \cdots \oplus b_{n-1}^r[\mathrm{ir}]$ , where $b_{k[\mathrm{ir}]}^r\in B_k^r$ and $k\in \{0,1,\dots ,n-1\}$ . Therefore, we have $b_{\mathrm{ir}}=0$ if the $n$ binary bits $b_{0[\mathrm{ir}]}^r$ , $b_{1[\mathrm{ir}]}^r$ , $\cdots$ , and $b_{n-1[\mathrm{ir}]}^r$ contain an even number of “1.” Otherwise, we have $b_{\mathrm{ir}}=1$ .

In Steps 9 and 10, the binary matrix $B$ is converted to a pixel matrix $P$ according to the reverse processes of Definition 2 and the pixel values of the pixel matrix $P$ are continuously converted to binary numbers, denoted as $A$ , according to the reverse processes of definition 2.

Finally, the secret image $A=[q_{\mathrm{ij}}{]}_{\theta \times \beta }$ can be reconstructed according to the $n$ shared images $A_0^r$ , $A_1^r$ , $\cdots$ , and $A_{n-1}^r$ using Algorithm 3. It is shown that the reconstructed image $A$ has the same size as the original secret image. Moreover, the reconstructed image $A$ can remain the same quality comparing with the original secret image. Furthermore, all shared images do not reveal any information of the original secret image.

Examples

In this section, a brief example is presented to illustrate the proposed schemes and a series of experiments is given to demonstrate the feasibility of the proposed schemes.

A brief example

Figure 2(a) shows a secret image $A$ that will be encrypted and shared to three participants, that is, $n=3$ . According to the step 1 of Algorithms 1, the pixel matrix $P$ of $A$ is

$$P=[p_{\mathrm{ik}}{]}_{\theta \times \xi }=[p_{\mathrm{ij}}{]}_{5\times 5}=\left[\begin{array}{ccccc}0\hfill & 0\hfill & 0\hfill & 0\hfill & 0\hfill \\ \multicolumn{1}{c}{0}& 255\hfill & 255\hfill & 255\hfill & 0\hfill \\ \multicolumn{1}{c}{0}& 0\hfill & 0\hfill & 0\hfill & 0\hfill \\ \multicolumn{1}{c}{255}& 255\hfill & 255\hfill & 255\hfill & 255\hfill \\ \multicolumn{1}{c}{0}& 0\hfill & 0\hfill & 0\hfill & 0\hfill \end{array}\right]$$

In Algorithms 1–3, both the pixel matrices $P$ and $P^r$ are identical with the secret image $A$ since $A$ is agray-scale image. According to the step 2 of Algorithms 1, the binary matrix $B$ of the pixel matrices $P$ is

$$\begin{array}{c}B=[b_{\mathrm{ir}}{]}_{5\times 40}=\hfill \\ \left[\begin{array}{ccccc}00000000\hfill & 00000000\hfill & 00000000\hfill & 00000000\hfill & 00000000\hfill \\ \multicolumn{1}{c}{00000000}& 11111111\hfill & 11111111\hfill & 11111111\hfill & 00000000\hfill \\ \multicolumn{1}{c}{00000000}& 00000000\hfill & 00000000\hfill & 00000000\hfill & 00000000\hfill \\ \multicolumn{1}{c}{11111111}& 11111111\hfill & 11111111\hfill & 11111111\hfill & 11111111\hfill \\ \multicolumn{1}{c}{00000000}& 00000000\hfill & 00000000\hfill & 00000000\hfill & 00000000\hfill \end{array}\right]\hfill \end{array}$$

We know $n=3$ . According to the Steps 3 and 4 of Algorithms 1, the mapping matrix $M$ is constructed, as shown in Example 3. The 0-mapping matrix $M^0$ and 1-mapping matrix $M^1$ are obtained, as shown in Example 4.

Figure 2.

(a) A secret image $A$ ; (b)–(d) the shared images $A_0^r$ , $A_1^r$ , and $A_2^r$ , respectively; and (e) the reconstructed image.

According to equations (3) and (4), $\forall b_{\mathrm{ir}}\in B$ , the probabilities of column selections for $b_{\mathrm{ir}}$ from the 0-mapping matrix $M^0$ or 1-mapping matrix $M^1$ are equal to $1/12^{n-1}{2}^{n-1}$ , where $n=3$ . According to the Steps 5 and 6 of Algorithms 1, the probability vector $V$ is

$$\begin{array}{c}V=(v_0,v_1,\cdots ,v_{{(2}^{n-1}-1)})\hfill \\ =(\frac{1}{2^{n-1}},\frac{1}{2^{n-1}},\cdots ,\frac{1}{2^{n-1}})\hfill \\ =(0.25,0.25,0.25,0.25)\hfill \end{array}$$

and the probability interval vector $\mathrm{\Gamma }$ is

$$\begin{array}{c}\mathrm{\Gamma }=({\gamma }_0,{\gamma }_1,\cdots ,{\gamma }_{2^{n-1}})\hfill \\ =(0,v_0,v_0+v_1,v_0+v_1+v_2,v_0+v_1+v_2+v_3)\hfill \\ =(0,0.25,0.5,0.75,1).\hfill \end{array}$$

Considering Step 7 of Algorithms 1, a temporary random matrix ${\mathrm{\Phi }}^{\prime }$ that has the same size $(5\times 40)$ with the binary matrix $B$ is randomly generated, that is,

$${\mathrm{\Phi }}^{\prime }=[{\varphi }_{\mathrm{ir}}^{\prime }{]}_{5\times 40}=\left[\begin{array}{ccccc}52347812\hfill & 75141237\hfill & 65364217\hfill & 98651234\hfill & 67547681\hfill \\ \multicolumn{1}{c}{12678297}& 23498764\hfill & 32579812\hfill & 99531467\hfill & 27468913\hfill \\ \multicolumn{1}{c}{14785348}& 32764589\hfill & 76579876\hfill & 17643257\hfill & 53214789\hfill \\ \multicolumn{1}{c}{97651258}& 56417961\hfill & 23455687\hfill & 74352691\hfill & 42356797\hfill \\ \multicolumn{1}{c}{54971645}& 76976512\hfill & 47921523\hfill & 56123975\hfill & 37215689\hfill \end{array}\right]$$

where ${\varphi }_{\mathrm{ir}}^{\prime }\in \mathbb{Z}$ and ${\varphi }_{\mathrm{ir}}^{\prime }\in [0,9]$ . Therefore, the random probability matrix $\mathrm{\Phi }=[{\varphi }_{\mathrm{ir}}{]}_{5\times 40}$ can be computed with $\forall i\in [1,5]$ and $\forall r\in [1,40]$ such that ${\varphi }_{\mathrm{ir}}=({\varphi }_{\mathrm{ir}}^{\prime }+1)/10$ .

Let $B_0^r$  =  $[b_{0[\mathrm{ir}]}^r{]}_{\theta \times \omega }$ , $B_1^r$ = $[b_{1[\mathrm{ir}]}^r{]}_{\theta \times \omega }$ , and $B_2^r$ = $[b_{2[\mathrm{ir}]}^r{]}_{\theta \times \omega }$ be three binary matrices such that

$$B_0^r=B_1^r=B_2^r=\left[\begin{array}{ccccc}00000000\hfill & 00000000\hfill & 00000000\hfill & 00000000\hfill & 00000000\hfill \\ \multicolumn{1}{c}{00000000}& 00000000\hfill & 00000000\hfill & 00000000\hfill & 00000000\hfill \\ \multicolumn{1}{c}{00000000}& 00000000\hfill & 00000000\hfill & 00000000\hfill & 00000000\hfill \\ \multicolumn{1}{c}{00000000}& 00000000\hfill & 00000000\hfill & 00000000\hfill & 00000000\hfill \\ \multicolumn{1}{c}{00000000}& 00000000\hfill & 00000000\hfill & 00000000\hfill & 00000000\hfill \end{array}\right]$$

According to Steps 2 to 13 in Algorithm 2, $\forall i\in [0,4]$ and $\forall r\in [0,39]$ , if

• $i=0$ and $r=0$ , we have $b_{[0,0]}=0$ and $\alpha =0$ . Since ${\varphi }_{[0,0]}={\varphi }_{[0,0]}^{\prime }/10=5/10=0.5\in (0.25,0.5]$ in $\mathrm{\Gamma }$ , we have $j=1$ and the 1th column of $M^0$ is selected to as the mapping bits for $b_{[0,0]}$ , that is, $(0,1,1{)}^T$ . Therefore,

  $b_{0[0,0]}^r=0$ , $b_{1[0,0]}^r=1$ , and $b_{2[0,0]}^r=1$ .

• $i=0$ and $r=1$ , we have $b_{[0,1]}=0$ and $\alpha =0$ . Since ${\varphi }_{[0,1]}={\varphi }_{[0,1]}^{\prime }/10=2/10=0.2\in (0,0.25]$ in $\mathrm{\Gamma }$ , we have $j=0$ and the 0th column of $M^0$ is selected to as the mapping bits for $b_{[0,1]}$ , that is, $(0,0,0{)}^T$ . Therefore,

  $b_{0[0,1]}^r=0$ , $b_{1[0,1]}^r=0$ , and $b_{2[0,1]}^r=0$ .

  $\cdots ⋮$

• $i=1$ and $r=8$ , we have $b_{[1,8]}=1$ and $\alpha =1$ . Since ${\varphi }_{[1,8]}={\varphi }_{[1,8]}^{\prime }/10=2/10=0.2\in (0,0.25]$ in $\mathrm{\Gamma }$ , we have $j=0$ and the 0th column of $M^1$ is selected to as the mapping bits for $b_{[1,8]}$ , that is, $(0,0,1{)}^T$ . Therefore,

  $b_{0[1,8]}^r=0$ , $b_{1[1,8]}^r=0$ , and $b_{2[1,8]}^r=1$ .

• $i=1$ and $r=9$ , we have $b_{[1,9]}=1$ and $\alpha =1$ . Since ${\varphi }_{[1,9]}={\varphi }_{[1,9]}^{\prime }/10=3/10=0.3\in (0.25,0.5]$ in $\mathrm{\Gamma }$ , we have $j=1$ and the 1th column of $M^1$ is selected to as the mapping bits for $b_{[1,9]}$ , that is, $(0,1,0{)}^T$ . Therefore,

  $b_{0[1,9]}^r=0$ , $b_{1[1,9]}^r=1$ , and $b_{2[1,9]}^r=0$ .

  $\cdots ⋮$

• $i=4$ and $r=39$ , we have $b_{[4,39]}=0$ and $\alpha =0$ . Since ${\varphi }_{[4,39]}={\varphi }_{[4,39]}^{\prime }/10=9/10=0.9\in (0.75,1]$ in $\mathrm{\Gamma }$ , we have $j=3$ and the 3th column of $M^0$ is selected to as the mapping bits for $b_{[4,39]}$ , that is, $(1,1,0{)}^T$ . Therefore,

  $b_{0[4,39]}^r=1$ , $b_{1[4,39]}^r=1$ , and $b_{2[4,39]}^r=0$ .

Therefore, the binary matrices $B_0^r$ , $B_1^r$ , and $B_2^r$ are

$$B_0^r=\left[\begin{array}{ccccc}00001100\hfill & 10000001\hfill & 10010001\hfill & 11100000\hfill & 11001110\hfill \\ \multicolumn{1}{c}{00111011}& 00011110\hfill & 00011100\hfill & 11000011\hfill & 01011100\hfill \\ \multicolumn{1}{c}{00110001}& 00110011\hfill & 11011111\hfill & 01100001\hfill & 00000111\hfill \\ \multicolumn{1}{c}{11100001}& 01001110\hfill & 00000111\hfill & 10000110\hfill & 00001111\hfill \\ \multicolumn{1}{c}{00110100}& 11111000\hfill & 01100000\hfill & 00000110\hfill & 01000011\hfill \end{array}\right]$$

$$B_1^r=\left[\begin{array}{ccccc}10110100\hfill & 01010010\hfill & 01101000\hfill & 11010011\hfill & 00110010\hfill \\ \multicolumn{1}{c}{00001010}& 01111001\hfill & 10101100\hfill & 11110100\hfill & 00101101\hfill \\ \multicolumn{1}{c}{01011111}& 10001111\hfill & 00101100\hfill & 00011010\hfill & 11001011\hfill \\ \multicolumn{1}{c}{10010011}& 10100100\hfill & 01111010\hfill & 01110010\hfill & 10110010\hfill \\ \multicolumn{1}{c}{11100011}& 00100100\hfill & 10100101\hfill & 11001101\hfill & 10001111\hfill \end{array}\right]$$

$$B_2^r=\left[\begin{array}{ccccc}10111000\hfill & 11010011\hfill & 11111001\hfill & 00110011\hfill & 11111100\hfill \\ \multicolumn{1}{c}{00110001}& 10011000\hfill & 01001111\hfill & 11001000\hfill & 10001110\hfill \\ \multicolumn{1}{c}{01101110}& 10111100\hfill & 11110011\hfill & 01111011\hfill & 11001100\hfill \\ \multicolumn{1}{c}{10001101}& 00010101\hfill & 10000010\hfill & 00001011\hfill & 01000010\hfill \\ \multicolumn{1}{c}{11010111}& 11011100\hfill & 11000101\hfill & 11001011\hfill & 11001100\hfill \end{array}\right]$$

By converting the binary matrices $B_0^r$ , $B_1^r$ , and $B_2^r$ to the pixel matrices $P_0^r$ , $P_1^r$ , and $P_2^r$ , we have

$$P_0^r=\left[\begin{array}{ccccc}12\hfill & 129\hfill & 145\hfill & 224\hfill & 206\hfill \\ \multicolumn{1}{c}{59}& 30\hfill & 56\hfill & 195\hfill & 92\hfill \\ \multicolumn{1}{c}{49}& 51\hfill & 223\hfill & 97\hfill & 7\hfill \\ \multicolumn{1}{c}{225}& 78\hfill & 7\hfill & 134\hfill & 15\hfill \\ \multicolumn{1}{c}{52}& 248\hfill & 96\hfill & 6\hfill & 67\hfill \\ \multicolumn{1}{c}{}& \hfill & \hfill & \hfill & \hfill \end{array}\right]$$

$$P_1^r=\left[\begin{array}{ccccc}180\hfill & 82\hfill & 104\hfill & 211\hfill & 50\hfill \\ \multicolumn{1}{c}{10}& 121\hfill & 184\hfill & 244\hfill & 45\hfill \\ \multicolumn{1}{c}{95}& 143\hfill & 44\hfill & 26\hfill & 203\hfill \\ \multicolumn{1}{c}{147}& 164\hfill & 122\hfill & 114\hfill & 178\hfill \\ \multicolumn{1}{c}{227}& 36\hfill & 165\hfill & 205\hfill & 143\hfill \\ \multicolumn{1}{c}{}& \hfill & \hfill & \hfill & \hfill \end{array}\right]$$

$$P_2^r=\left[\begin{array}{ccccc}184\hfill & 211\hfill & 249\hfill & 51\hfill & 252\hfill \\ \multicolumn{1}{c}{49}& 152\hfill & 79\hfill & 200\hfill & 142\hfill \\ \multicolumn{1}{c}{110}& 188\hfill & 243\hfill & 123\hfill & 204\hfill \\ \multicolumn{1}{c}{141}& 21\hfill & 130\hfill & 11\hfill & 66\hfill \\ \multicolumn{1}{c}{207}& 220\hfill & 197\hfill & 203\hfill & 204\hfill \\ \multicolumn{1}{c}{}& \hfill & \hfill & \hfill & \hfill \end{array}\right]$$

The three shared images $A_0^r$ , $A_1^r$ , and $A_2^r$ are same as the three pixel matrices $P_0^r$ , $P_1^r$ , and $P_2^r$ since the secret image $A$ is a gray-scale image. Therefore, we have $A_0^r=P_0^r$ , $A_1^r=P_1^r$ , and $A_2^r=P_2^r$ , as shown in Figure 2(b)–(d), respectively. It is obvious that each shared image does not reveal any information of the original secret image.

After the shared images $A_0^r$ , $A_1^r$ , and $A_2^r$ are collected from the three participants, the secret image $A$ can be reconstructed by Algorithm 3. First, the shared images $A_0^r$ , $A_1^r$ , and $A_2^r$ should be converted to the binary matrices $B_0^r$ , $B_1^r$ , and $B_2^r$ that are presented above. Let $B=[b_{\mathrm{ir}}{]}_{5\times 40}$ be a binary matrix. According to Steps 4–8 of Algorithm 3, $\forall i\in [0,4]$ and $\forall r\in [0,39]$ , if

• $i=0$ and $r=0$ , we have

  $b_{[0,0]}=b_{0[0,0]}^r\oplus b_{1[0,0]}^r\oplus b_{2[0,0]}^r=0\oplus 1\oplus 1=0$ .

• $i=0$ and $r=1$ , we have

  $b_{[0,1]}=b_{0[0,1]}^r\oplus b_{1[0,1]}^r\oplus b_{2[0,1]}^r=0\oplus 0\oplus 0=0$ .

  $\cdots ⋮$

• $i=1$ and $r=8$ , we have

  $b_{[1,8]}=b_{0[1,8]}^r\oplus b_{1[1,8]}^r\oplus b_{2[1,8]}^r=0\oplus 0\oplus 1=1$ .

• $i=1$ and $r=9$ , we have

  $b_{[1,9]}=b_{0[1,9]}^r\oplus b_{1[1,9]}^r\oplus b_{2[1,9]}^r=0\oplus 1\oplus 0=1$ .

  $\cdots ⋮$

• $i=4$ and $r=39$ , we have

  $b_{[4,39]}=b_{0[4,39]}^r\oplus b_{1[4,39]}^r\oplus b_{2[4,39]}^r=1\oplus 1\oplus 0=0$ .

We have

$$B=[b_{\mathrm{ir}}{]}_{5\times 40}=\left[\begin{array}{ccccc}00000000\hfill & 00000000\hfill & 00000000\hfill & 00000000\hfill & 00000000\hfill \\ \multicolumn{1}{c}{00000000}& 11111111\hfill & 11111111\hfill & 11111111\hfill & 00000000\hfill \\ \multicolumn{1}{c}{00000000}& 00000000\hfill & 00000000\hfill & 00000000\hfill & 00000000\hfill \\ \multicolumn{1}{c}{11111111}& 11111111\hfill & 11111111\hfill & 11111111\hfill & 11111111\hfill \\ \multicolumn{1}{c}{00000000}& 00000000\hfill & 00000000\hfill & 00000000\hfill & 00000000\hfill \end{array}\right]$$

Therefore, the reconstructed image is

$$\left[\begin{array}{ccccc}0\hfill & 0\hfill & 0\hfill & 0\hfill & 0\hfill \\ \multicolumn{1}{c}{0}& 255\hfill & 255\hfill & 255\hfill & 0\hfill \\ \multicolumn{1}{c}{0}& 0\hfill & 0\hfill & 0\hfill & 0\hfill \\ \multicolumn{1}{c}{255}& 255\hfill & 255\hfill & 255\hfill & 255\hfill \\ \multicolumn{1}{c}{0}& 0\hfill & 0\hfill & 0\hfill & 0\hfill \end{array}\right]=A$$

Figure 2(e) shows the reconstructed image $A$ . It is shown that the reconstructed image has the same size as the original secret image. Moreover, the reconstructed image remains the same quality comparing with the original secret image.

The practical examples

To further demonstrate the feasibility of the proposed encryption and decryption schemes, a series of practical examples are given including some classical gray-scale and color images. The secret images are tested with an Qt-based tool that is developed with the language C++ and Qt development platform ²⁸ in our laboratory according to the proposed schemes. Figures 3–7(a) show a part of secret images, where Figures 3(a) and 4(a) are gray-scale images and Figures 5–7(a) are color images. Figures 3–6(a) will be encrypted and shared to four participants, that is, $n=4$ . Figure 7(a) will be encrypted and shared to seven participants, that is, $n=7$ .

Figure 3.

(a) A gray-scale secret image $A$ ; (b)–(e) are the shared images $A_0^r$ , $A_1^r$ , $A_2^r$ , and $A_3^r$ , respectively; and (f) is the reconstructed image.

Figure 4.

(a) A gray-scale secret image $B$ ; (b)–(e) are the shared images $B_0^r$ , $B_1^r$ , $B_2^r$ , and $B_3^r$ , respectively; and (f) is the reconstructed image.

Figure 5.

(a) A color secret image $C$ ; (b)–(e) are the shared images $C_0^r$ , $C_1^r$ , $C_2^r$ , and $C_3^r$ , respectively; and (f) is the reconstructed image.

Figure 6.

(a) A color secret image $D$ ; (b)–(e) are the shared images $D_0^r$ , $D_1^r$ , $D_2^r$ , and $D_3^r$ , respectively; and (f) is the reconstructed image.

Figure 7.

(a) A color secret image $E$ ; (b)–(h) are the shared images $E_0^r$  –  $E_6^r$ , respectively; and (i) is the reconstructed image.

In Figures 3–6, each secret image can be encrypted into four shared images $S_0$ , $S_1$ , $S_2$ , and $S_3$ according to Algorithms 1 and 2. Figures 3–6(b)–(e) show the four shared images that have the same size as the secret images, respectively. In Figure 7, the secret image $E$ is encrypted into seven shared images $S_0$  –  $S_6$ according to Algorithms 1 and 2. It is obvious that the shared images do not reveal any information of the corresponding secret images because they are meaningless. After each shared images are collected from all the participants, the corresponding secret images can be reconstructed by Algorithm 3. It is shown that the reconstructed images have the same size as the original secret images, which also remain the same contrasts comparing with the original secret images, respectively.

The comparison

Table 1 shows a comparison between the schemes that are proposed by Shen et al., ²⁴ Shyu, ²⁶ and Guo et al., ²⁷ and this article, which mainly consider the pixel expansion, contrast, and computation complexity problems. The size of the reconstructed image is equal to that one of secret image if its pixel expansion is zero. Otherwise, the size of the reconstructed image relative to the secret image will gradually increase as the increasing of its pixel expansion. In Table 1, the schemes that are proposed by Shen et al., ²⁴ Shyu, ²⁶ and Guo et al. ²⁷ suffer the pixel expansion problems since their pixel expansions are six, eight, and seven, respectively. However, our proposed scheme do not suffer any pixel expansion problem since its pixel expansion is zero. The main reason is that the matrices of the secret image, shared images, and reconstructed image all maintain the same size.

Table 1.

The comparison.

Schemes	Pixelexpansions	Contrasts	Computationcomplexities
Shen et al. 24	6	$1/8$	$O(n^3)$
Shyu 26	8	$1/8$	$O(n^2)$
Guo et al. 27	7	$1/2$	$O(n^3)$
This article	0	$1$	$O(n^2)$

Moreover, the reconstructed image has the same quality as the secret image if its contrast is one. Otherwise, the quality of the reconstructed image relative to the secret image will gradually decrease as the decreasing of its contrast. In Table 1, the qualities of the reconstructed images that are reconstructed by Guo et al., ²⁷ Shyu, ²⁶ and Shen et al. ²⁴ gradually deteriorate since their contrasts gradually decrease. Fortunately, our proposed scheme do not suffer any contrast problem since its contrast is one. The main reason is that each bit of the secret image can be correctly reconstructed for the reconstructed image with the reversibility of XOR operations. Therefore, it can be concluded that our proposed scheme do not suffer any pixel expansion and contrast distortion problems.

Furthermore, the main steps of Algorithm 1 can be divided into two categories. One is to compute the binary matrix of a secret image. The another one is to computer the 0-mapping and 1-mapping matrixes for the binary matrix. Let $n$ be the number of participants and $A=[q_{\mathrm{ij}}{]}_{\theta \times \beta }$ be a secret image, where $n<<\theta$ and $n<<\beta$ . The binary matrix of $A$ is $B=[b_{\mathrm{ir}}{]}_{\theta \times 8\beta }$ and the 0-mapping and 1-mapping matrixes are $M^0=[e_{\mathrm{xy}}^0{]}_{n\times 2^{n-1}}$ and $M^1=[e_{\mathrm{xy}}^1{]}_{n\times 2^{n-1}}$ , respectively. Therefore, the computation complexity of Algorithm 1 is $O(n^2)$ since $n<<\theta$ and $n<<\beta$ . Algorithm 2 is the encryption scheme that is used to generate $n$ shared images for the $n$ participants. It contains $\theta \times 8\beta$ bits and each bit will be randomly selected the shared bits from the columns of the 0-mapping or 1-mapping matrixes. Therefore, the computation complexity of Algorithm 2 is also $O(n^2)$ . Algorithm 3 is the decryption scheme that is used to reconstruct the secret image $A$ based on the $n$ shared images. Each shared image contains $\theta \times 8\beta$ bits since it has the same size with the secret image $A$ . In Algorithm 3, all bits of the $n$ shared images will be sequentially and simultaneously extracted to reconstruct the original bits with a series of XOR operations. Similarly, the computation complexity of Algorithm 3 is $O(n^2)$ . Comparing with the existing schemes, the proposed schemes have an ideal computation complexity.

Conclusion

In this article, an efficient ( $n$ , $n$ )-threshold XVCS with lossless reconfigurable algorithms is proposed. It contains an encryption scheme and a decryption scheme. The former aims to encrypt gray-scale and color secret images by mapping a secret image to $n$ shared images. The extensive experimental results show that the shared images do not suffer any pixel expansion problem but also do not reveal any information of the secret images. The main reason is that each original bit of the secret images will be mapped to other $n$ bits by randomly selecting a column from the 0-mapping and 1-mapping matrices that are dynamically generated in the proposed encryption scheme. In the latter scheme, the reconstructed images have the same size and quality as the original secret images. In the future, a ( $k$ , $n$ )-threshold visual cryptography scheme will be proposed by expanding the 0-mapping and 1-mapping matrices based on their partial reconfigurability. Furthermore, some other performance analyses will be considered, such as the security of the proposed methods.