# 数学代写|组合优化代写Combinatorial optimization代考|CSC205

#### Doug I. Jones

Lorem ipsum dolor sit amet, cons the all tetur adiscing elit

couryes-lab™ 为您的留学生涯保驾护航 在代写组合优化Combinatorial optimization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。我们的专家在代写组合优化Combinatorial optimization代写方面经验极为丰富，各种代写组合优化Combinatorial optimization相关的作业也就用不着说。

• Statistical Inference 统计推断
• Statistical Computing 统计计算
• (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
couryes™为您提供可以保分的包课服务

## 数学代写|组合优化代写Combinatorial optimization代考|The Integer Hull of a Polyhedron

As linear programs, integer programs can be infeasible or unbounded. It is not easy to decide whether $P_I=\emptyset$ for a polyhedron $P$. But if an integer program is feasible we can decide whether it is bounded by simply considering the LP relaxation.

Proposition 5.2. Let $P={x: A x \leq b}$ be some rational polyhedron whose integer hull is nonempty, and let $c$ be some vector (not necessarily rational). Then $\max {c x: x \in P}$ is bounded if and only if $\max \left{c x: x \in P_I\right}$ is bounded.

Proof: Suppose max ${c x: x \in P}$ is unbounded. Then Corollary $3.28$ says that the system $y A=c, y \geq 0$ has no solution. By Corollary $3.26$ there is a vector $z$ with $c z<0$ and $A z \geq 0$. Then the LP $\min {c z: A z \geq 0,-\mathbb{1} \leq z \leq \mathbb{1}}$ is feasible. Let $z^$ be an optimum basic solution of this LP. $z^$ is rational as it is a vertex of a rational polytope. Multiply $z^*$ by a suitable natural number to obtain an integral vector $w$ with $A w \geq 0$ and $c w<0$. Let $v \in P_I$ be some integral vector. Then $v-k w \in P_I$ for all $k \in \mathbb{N}$, and thus $\max \left{c x: x \in P_I\right}$ is unbounded.
The other direction is trivial.
Definition 5.3. Let $A$ be an integral matrix. A subdeterminant of $A$ is $\operatorname{det} B$ for some square submatrix $B$ of $A$ (defined by arbitrary row and column indices). We write $\Xi(A)$ for the maximum absolute value of the subdeterminants of $A$.

Lemma 5.4. Let $C={x: A x \leq 0}$ be a polyhedral cone, where $A$ is an integral matrix. Then $C$ is generated by a finite set of integral vectors, each having components with absolute value at most $\Xi(A)$.

Proof: By Lemma $3.12, C$ is generated by some of the vectors $y_1, \ldots, y_t$, such that for each $i, y_i$ is the solution to a system $M y=b^{\prime}$ where $M$ consists of $n$ linearly independent rows of $\left(\begin{array}{c}A \ I\end{array}\right)$ and $b^{\prime}=\pm e_j$ for some unit vector $e_j$. Set $z_i:=$ $|\operatorname{det} M| y_i$. By Cramer’s rule, $z_i$ is integral with $\left|z_i\right|_{\infty} \leq \Xi(A)$. Since this holds for each $i$, the set $\left{z_1, \ldots, z_t\right}$ has the required properties.

## 数学代写|组合优化代写Combinatorial optimization代考|Unimodular Transformations

In this section we shall prove two lemmas for later use. A square matrix is called unimodular if it is integral and has determinant 1 or $-1$. Three types of unimodular matrices will be of particular interest: For $n \in \mathbb{N}, p \in{1, \ldots, n}$ and $q \in{1, \ldots, n} \backslash{p}$ consider the matrices $\left(a_{i j}\right){i, j \in{1, \ldots, n}}$ defined in one of the following ways: $$a{i j}=\left{\begin{array}{ll} 1 & \text { if } i=j \neq p \ -1 & \text { if } i=j=p \ 0 & \text { otherwise } \end{array} \quad a_{i j}= \begin{cases}1 & \text { if } i=j \notin{p, q} \ 1 & \text { if }{i, j}={p, q} \ 0 & \text { otherwise }\end{cases}\right.$$
$$a_{i j}= \begin{cases}1 & \text { if } i=j \ -1 & \text { if }(i, j)=(p, q) \ 0 & \text { otherwise }\end{cases}$$
These matrices are evidently unimodular. If $U$ is one of the above matrices, then replacing an arbitrary matrix $A$ (with $n$ columns) by $A U$ is equivalent to applying one of the following elementary column operations to $A$ :

• multiply a column by $-1$;
• exchange two columns;
• subtract one column from another column.
A series of the above operations is called a unimodular transformation. Obviously the product of unimodular matrices is unimodular. It can be shown that a matrix is unimodular if and only if it arises from an identity matrix by a unimodular transformation (equivalently, it is the product of matrices of the above three types); see Exercise 6. Here we do not need this fact.

Proposition 5.9. The inverse of a unimodular matrix is also unimodular For each unimodular matrix $U$ the mappings $x \mapsto U x$ and $x \mapsto x U$ are bijections on $\mathbb{Z}^n$.
Proof: Let $U$ be a unimodular matrix. By Cramer’s rule the inverse of a unimodular matrix is integral. Since $(\operatorname{det} U)\left(\operatorname{det} U^{-1}\right)=\operatorname{det}\left(U U^{-1}\right)=\operatorname{det} I=1, U^{-1}$ is also unimodular. The second statement follows directly from this.

Lemma 5.10. For each rational matrix $A$ whose rows are linearly independent there exists a unimodular matrix $U$ such that $A U$ has the form $(B 0)$, where $B$ is a nonsingular square matrix.
Proof: Suppose we have found a unimodular matrix $U$ such that
$$A U=\left(\begin{array}{ll} B & 0 \ C & D \end{array}\right)$$
for some nonsingular square matrix $B$. (Initially $U=I, D=A$, and the parts $B$, $C$ and 0 have no entries.)

# 组合优化代考

## 数学代写|组合优化代写组合优化代考|The Integer Hull of a Polyhedron

. The Integer Hull of a Polyhedron

## 数学代写|组合优化代写combinatoroptimization代考|Unimodular transforms

. 在本节中，我们将证明两个引理供以后使用。如果一个方阵是积分且行列式为1或$-1$，则称为一模方阵。有三种类型的单模矩阵是特别有趣的:对于$n \in \mathbb{N}, p \in{1, \ldots, n}$和$q \in{1, \ldots, n} \backslash{p}$，考虑以下列方式之一定义的矩阵$\left(a_{i j}\right){i, j \in{1, \ldots, n}}$: $$a{i j}=\left{\begin{array}{ll} 1 & \text { if } i=j \neq p \ -1 & \text { if } i=j=p \ 0 & \text { otherwise } \end{array} \quad a_{i j}= \begin{cases}1 & \text { if } i=j \notin{p, q} \ 1 & \text { if }{i, j}={p, q} \ 0 & \text { otherwise }\end{cases}\right.$$
$$a_{i j}= \begin{cases}1 & \text { if } i=j \ -1 & \text { if }(i, j)=(p, q) \ 0 & \text { otherwise }\end{cases}$$

• 将某列乘以 $-1$
• 交换两列;
• 从另一列减去一列。上面的一系列操作称为单模变换。显然，单模矩阵的乘积是单模的。可以证明，一个矩阵当且仅当它由单位矩阵通过一模变换产生(等价地，它是上述三种类型的矩阵的乘积)，它是一模的;见练习6。

$$A U=\left(\begin{array}{ll} B & 0 \ C & D \end{array}\right)$$
。(最初是$U=I, D=A$, $B$, $C$和0部分没有条目。)

## 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。

Days
Hours
Minutes
Seconds

# 15% OFF

## On All Tickets

Don’t hesitate and buy tickets today – All tickets are at a special price until 15.08.2021. Hope to see you there :)