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

statistics-lab™ 为您的留学生涯保驾护航 在代写密码学Cryptography & Cryptanalysis方面已经树立了自己的口碑, 保证靠谱, 高质且原创的统计Statistics代写服务。我们的专家在代写密码学Cryptography & Cryptanalysis代写方面经验极为丰富，各种代写密码学Cryptography & Cryptanalysis相关的作业也就用不着说。

• Statistical Inference 统计推断
• Statistical Computing 统计计算
• Advanced Probability Theory 高等概率论
• Advanced Mathematical Statistics 高等数理统计学
• (Generalized) Linear Models 广义线性模型
• Statistical Machine Learning 统计机器学习
• Longitudinal Data Analysis 纵向数据分析
• Foundations of Data Science 数据科学基础

## 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 :

## 有限元方法代写

tatistics-lab作为专业的留学生服务机构，多年来已为美国、英国、加拿大、澳洲等留学热门地的学生提供专业的学术服务，包括但不限于Essay代写，Assignment代写，Dissertation代写，Report代写，小组作业代写，Proposal代写，Paper代写，Presentation代写，计算机作业代写，论文修改和润色，网课代做，exam代考等等。写作范围涵盖高中，本科，研究生等海外留学全阶段，辐射金融，经济学，会计学，审计学，管理学等全球99%专业科目。写作团队既有专业英语母语作者，也有海外名校硕博留学生，每位写作老师都拥有过硬的语言能力，专业的学科背景和学术写作经验。我们承诺100%原创，100%专业，100%准时，100%满意。

## MATLAB代写

MATLAB 是一种用于技术计算的高性能语言。它将计算、可视化和编程集成在一个易于使用的环境中，其中问题和解决方案以熟悉的数学符号表示。典型用途包括：数学和计算算法开发建模、仿真和原型制作数据分析、探索和可视化科学和工程图形应用程序开发，包括图形用户界面构建MATLAB 是一个交互式系统，其基本数据元素是一个不需要维度的数组。这使您可以解决许多技术计算问题，尤其是那些具有矩阵和向量公式的问题，而只需用 C 或 Fortran 等标量非交互式语言编写程序所需的时间的一小部分。MATLAB 名称代表矩阵实验室。MATLAB 最初的编写目的是提供对由 LINPACK 和 EISPACK 项目开发的矩阵软件的轻松访问，这两个项目共同代表了矩阵计算软件的最新技术。MATLAB 经过多年的发展，得到了许多用户的投入。在大学环境中，它是数学、工程和科学入门和高级课程的标准教学工具。在工业领域，MATLAB 是高效研究、开发和分析的首选工具。MATLAB 具有一系列称为工具箱的特定于应用程序的解决方案。对于大多数 MATLAB 用户来说非常重要，工具箱允许您学习应用专业技术。工具箱是 MATLAB 函数（M 文件）的综合集合，可扩展 MATLAB 环境以解决特定类别的问题。可用工具箱的领域包括信号处理、控制系统、神经网络、模糊逻辑、小波、仿真等。