数学代写|密码学作业代写Cryptography & Cryptanalysis代考|Basic Visual Cryptography Algorithms

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数学代写|密码学作业代写Cryptography & Cryptanalysis代考|Deterministic Visual Cryptography

A simple example is shown in Fig. 2.1. When sharing a white pixel $s=0$, one randomly chooses one row from the second and the third rows in Fig. 2.1, and distributes the second column to Share 1 and the third column to Share 2. The final stacking result is a $2 \times 2$ block with two black pixels, as shown in the last column. Similarly, while sharing a black pixel $s=1$, one randomly chooses one row from the fourth and the fifth rows of the table, and distributes the second column to Share 1 and the third column to Share 2 . The final stacking result is a $2 \times 2$ block with four black pixels, as shown in the last column. So, one secret pixel is expanded to a block having four pixels, or equivalently, one secret pixel is represented by four sub-pixels.

If we look at one share, no matter whether a white pixel is shared or a black pixel is shared, we always see roughly equal number of the following two types of blocks on each share:
$$\mathbf{A}{1}=\left[\begin{array}{ll} 0 & 1 \ 1 & 0 \end{array}\right], \quad \mathbf{A}{2}=\left[\begin{array}{ll} 1 & 0 \ 0 & 1 \end{array}\right]$$
Thus, by checking only one share, it is impossible to tell if a black pixel is shared or a white pixel is shared. This is the security consideration in visual cryptography. But

if we check the target image, the block corresponding to white secret pixel has two black pixels, while the block corresponding to the black secret pixel has four black pixels. Thus, one can distinguish between the black region and the white region of the secret image by checking the target image. This is the contrast consideration in visual cryptography.

When designing a deterministic VC algorithm, the blocks corresponding to a secret pixel are usually vectorized and organized into a matrix. For example, the two blocks in (2.5) are represented as a matrix
$$\mathbf{C}=\left[\begin{array}{c} \operatorname{vec}\left(\mathbf{A}{1}\right)^{T} \ \operatorname{vec}\left(\mathbf{A}{2}\right)^{T} \end{array}\right]$$
where each block $\mathbf{A}{i}, i=1,2$, is stretched to a column vector by function vec $\left(\mathbf{A}{i}\right)$, transposed, and filled into one row of $\mathbf{C}$. So, each row of $\mathbf{C}$ should be distributed to one share and then reshaped to appropriate shape. For the design of VC algorithm, the reshaping is not a central issue. So, it is usually omitted when defining the $V C$.
For the example in Fig. 2.1, we have the following matrices:
\begin{aligned} &\mathbf{C}{0}=\left{\mathbf{C}{1}^{0}, \mathbf{C}{2}^{0}\right}=\left{\left[\begin{array}{llll} 0 & 1 & 1 & 0 \ 0 & 1 & 1 & 0 \end{array}\right],\left[\begin{array}{llll} 1 & 0 & 0 & 1 \ 1 & 0 & 0 & 1 \end{array}\right]\right} \ &\mathbf{C}{1}=\left{\mathbf{C}{1}^{1}, \mathbf{C}{2}^{1}\right}=\left{\left[\begin{array}{llll} 0 & 1 & 1 & 0 \ 1 & 0 & 0 & 1 \end{array}\right],\left[\begin{array}{llll} 1 & 0 & 0 & 1 \ 0 & 1 & 1 & 0 \end{array}\right]\right} \end{aligned}

数学代写|密码学作业代写Cryptography & Cryptanalysis代考|Definition

Let $\mathcal{H}(\mathbf{v})$ be the Hamming weight of a Boolean vector $\mathbf{v}$, i.e., the number of $1 \mathrm{~s}$ in $\mathbf{v}$. Based on the above explanation, the $(k, n)$-threshold VC can be formally defined by the following definition [13]:

Definition 2.1 $(k, n)$-threshold VC: $\mathrm{A}(k, n)$-threshold VC consists of two collections of $n \times m$ matrices $\mathrm{C}{0}$ and $\mathrm{C}{1}$. To share a white (black resp.) pixel, one randomly chooses one matrix from $\mathrm{C}{0}\left(\mathrm{C}{1}\right.$ resp.). Each row of the chosen matrix is distributed to one share. The following two conditions should be satisfied:

1. Contrast condition: For any $\mathbf{C} \in \mathbf{C}{0}$, let $\mathbf{C}=\left[\mathbf{c}{1}^{T}, \ldots, \mathbf{c}{n}^{T}\right]^{T}$, where $\mathbf{c}{i}$ is the $i$-th row of matrix $C$. Then, the stacking of any $k$ rows or more satisfies $\mathcal{H}\left(\vee_{j=1}^{k} \mathbf{c}{i{j}}\right) \leq d-\alpha$, where $i_{1}, \ldots, i_{k}$ are $k$ randomly chosen row indices. For any $\mathbf{C} \in \mathrm{C}{1}, \mathcal{H}\left(\mathrm{V}{j=1}^{k} \mathbf{c}{i{j}}\right) \geq d$.
2. Security condition: For any $q<k$ randomly chosen shares $\left{i_{1}, \ldots, i_{q}\right} \in$ ${1,2, \ldots, n}$, the two collections of matrices $\hat{C}{0}$ and $\hat{C}{1}$, obtained by restricting each matrix to the rows $i_{1}, \ldots, i_{q}$, should be indistinguishable. Namely, $\hat{\mathrm{C}}{0}$ and $\hat{C}{1}$ should contain the same matrices with the same frequencies.

The parameter $m$ is pixel expansion rate and parameter a gives the minimum difference in Hamming weight between the stacking result for a black secret pixel and the stacking result for a white secret pixel.

数学代写|密码学作业代写Cryptography & Cryptanalysis代考|Constructions

The construction of the two sets of matrices $C_{0}$ and $C_{1}$ can be based on two basis matrices $\mathbf{B}{0}$ and $\mathbf{B}{1}$. After designing $\mathbf{B}{0}\left(\mathbf{B}{1}\right.$ resp.), $\mathbf{C}{0}$ may include all column permutations of $\mathbf{B}{0} \mathrm{~ ( B ⿱}$ corresponding $\mathrm{C}_{0}$.

One optimal construction for $(n, n)$-threshold scheme was proposed by Naor and Shamir [13]. It starts with a set having $n$ components, which is called a ground set. Let the set be $\mathrm{W}=\left{e_{1}, \ldots, e_{n}\right}$, which has $2^{n}$ subsets. Among these subsets, let $U_{1}, \ldots, U_{2^{w-1}}$ be the $2^{n-1}$ subsets with even cardinality, and let $V_{1}, \ldots, V_{2^{w-1}}$ be the $2^{n-1}$ subsets with odd cardinality. Then we obtain two basis matrices $\mathbf{B}{0}$ and $\mathbf{B}{1}$ as follows:
\begin{aligned} &B_{0}[i, j]= \begin{cases}1, & \text { if } e_{i} \in \mathrm{U}{j} \ 0, & \text { otherwise. }\end{cases} \ &B{1}[i, j]= \begin{cases}1, & \text { if } e_{i} \in \mathrm{V}{j} \ 0, & \text { otherwise. }\end{cases} \end{aligned} Then the set of matrices $C{0}\left(C_{1}\right.$ resp.) is obtained by all possible column permutations of $\mathbf{B}{0}$ ( $\mathbf{B}{1}$ resp.).

For example, Let $n=3$ and $m=4$, then we use the ground set $\mathrm{W}=\left{e_{1}, e_{2}, e_{3}\right}$. So, we have the following subsets:
\begin{aligned} &\mathrm{U}{1}=\left{e{1}, e_{2}\right}, \quad \mathrm{U}{2}=\left{e{1}, e_{3}\right}, \quad \mathrm{U}{3}=\left{e{2}, e_{3}\right}, \quad \mathrm{U}{4}=\emptyset \ &\mathrm{V}{1}=\left{e_{1}\right}, \quad \mathrm{V}{2}=\left{e{2}\right}, \quad \mathrm{V}{3}=\left{e{3}\right}, \quad \mathrm{V}{4}=\left{e{1}, e_{2}, e_{3}\right} \end{aligned}
From (2.10), we get two basis matrices:
$$\mathbf{B}{0}=\left[\begin{array}{llll} 1 & 1 & 0 & 0 \ 1 & 0 & 1 & 0 \ 0 & 1 & 1 & 0 \end{array}\right], \quad \mathbf{B}{1}=\left[\begin{array}{llll} 1 & 0 & 0 & 1 \ 0 & 1 & 0 & 1 \ 0 & 0 & 1 & 1 \end{array}\right]$$
The set of matrices $\mathrm{C}{0}\left(\mathrm{C}{1}\right.$ resp.) contains $2^{n-1} !=4 !=24$ matrices that are all possible column permutations of the matrix $\mathbf{B}{0}\left(\mathbf{B}{1}\right.$ resp.).

Naor and Shamir proved that this scheme is optimal, in that it has the minimum pixel expansion $m$ and maximum contrast $\alpha$ for a given $n$ :

Theorem $2.1$ ([13]) For any ( $n, n)$-threshold deterministic VC, the contrast is upper bounded by $1 / 2^{n-1}$, and the pixel expansion $m$ is lower bounded by $2^{n-1}$.

Proof for security and contrast condition can be found in [13], which is thus omitted here. We also direct the reader to $[13]$ for more general constructions such as $(k, n)$-threshold scheme.
An experimental result using the scheme in Fig. 2.1 is shown in Fig. 2.2.

数学代写|密码学作业代写Cryptography & Cryptanalysis代考|Deterministic Visual Cryptography

C=[一个东西⁡(一种1)吨 一个东西⁡(一种2)吨]

\begin{对齐} &\mathbf{C}{0}=\left{\mathbf{C}{1}^{0}, \mathbf{C}{2}^{0}\right}=\left{ \left[\begin{array}{llll}0&1&1&0\0&1&1&0 \end{array}\right],\left[\begin{array}{llll}1&0&0 0&1\1&0&0&1\end{array}\right]\right}\ &\mathbf{C}{1}=\left{\mathbf{C}{1}^{1},\mathbf{C}{2}^{1}\right}=\left{\left[\begin {array}{llll}0&1&1&0\1&0&0&1\end{array}\right],\left[\begin{array}{llll}1&0&0&1\0&1&1&0\end{array}\right]\right}\end{aligned}\begin{对齐} &\mathbf{C}{0}=\left{\mathbf{C}{1}^{0}, \mathbf{C}{2}^{0}\right}=\left{ \left[\begin{array}{llll}0&1&1&0\0&1&1&0 \end{array}\right],\left[\begin{array}{llll}1&0&0 0&1\1&0&0&1\end{array}\right]\right}\ &\mathbf{C}{1}=\left{\mathbf{C}{1}^{1},\mathbf{C}{2}^{1}\right}=\left{\left[\begin {array}{llll}0&1&1&0\1&0&0&1\end{array}\right],\left[\begin{array}{llll}1&0&0&1\0&1&1&0\end{array}\right]\right}\end{aligned}

数学代写|密码学作业代写Cryptography & Cryptanalysis代考|Definition

1. 对比条件：任意C∈C0， 让C=[C1吨,…,Cn吨]吨， 在哪里C一世是个一世- 矩阵的第 行C. 然后，任意堆叠ķ行或更多满足H(∨j=1ķC一世j)≤d−一种， 在哪里一世1,…,一世ķ是ķ随机选择的行索引。对于任何C∈C1,H(在j=1ķC一世j)≥d.
2. 安全条件：任意q<ķ随机选择的股票\left{i_{1}, \ldots, i_{q}\right} \in\left{i_{1}, \ldots, i_{q}\right} \in 1,2,…,n, 两个矩阵集合C^0和C^1，通过将每个矩阵限制为行获得一世1,…,一世q, 应该是无法区分的。即，C^0和C^1应该包含具有相同频率的相同矩阵。

数学代写|密码学作业代写Cryptography & Cryptanalysis代考|Constructions

Naor 和 Shamir 证明了该方案是最优的，因为它具有最小的像素扩展米和最大对比度一种对于给定的n :

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