### 数学代写|组合优化代写Combinatorial optimization代考|MTH5107

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• 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

## 数学代写|组合优化代写Combinatorial optimization代考|Optimal and Approximation Solutions

Let us show an optimality condition for the minimum spanning tree.
Theorem 1.2.1 (Path Optimality) A spanning tree $T^*$ is a minimum spanning tree if and only if it satisfies the following condition:

Path Optimality Condition For every edge $(u, v)$ not in $T^$, there exists a path $p$ in $T^$, connecting $u$ and $v$, and moreover, $c(u, v) \geq c(x, y)$ for every edge $(x, y)$ in path $p$.

Proof Suppose, for contradiction, that $c(u, v)<c(x, y)$ for some edge $(x, y)$ in the path $p$. Then $T^{\prime}=\left(T^* \backslash(x, y)\right) \cup(u, v)$ is a spanning tree with cost less than $c\left(T^\right)$, contradicting the minimality of $T^$.

Conversely, suppose that $T^$ satisfies the path optimality condition. Let $T^{\prime}$ be a minimum spanning tree such that among all minimum spanning tree, $T^{\prime}$ is the one with the most edges in common with $T^$. Suppose, for contradiction, that $T^{\prime} \neq T^$. We claim that there exists an edge $(u, v) \in T^$ such that the path in $T^{\prime}$ between $u$ and $v$ contains an edge $(x, y)$ with length $c(x, y) \geq c(u, v)$. If this claim is true, then $\left(T^{\prime} \backslash(x, y)\right) \cup(u, v)$ is still a minimum spanning tree, contradicting the definition of $T^{\prime}$.

Now, we show the claim by contradiction. Suppose the claim is not true. Consider an edge $\left(u_1, v_1\right) \in T^* \backslash T^{\prime}$. the path in $T^{\prime}$ connecting $u_1$ and $v_1$ must contain an edge $\left(x_1, y_1\right)$ not in $T^$. Since the claim is not true, we have $c\left(u_1, v_1\right)$ connecting $x_1$ and $y_1$, which must contain an edge $\left(u_2, v_2\right) \notin$ $T^{\prime}$. Since $T^$ satisfies the path optimality condition, we have $c\left(x_1, y_1\right) \leq c\left(u_2, v_2\right)$. Hence, $c\left(u_1, v_1\right)$ such that $c\left(u_1, v_2\right)<c\left(u_2, v_2\right)<c\left(u_3, v_3\right)<\cdots$, contradicting the finiteness of $T^*$.

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

The most important measure of quality for algorithms is the running time. However, for the same algorithm, it may take different times when we run it in different computers. To give a uniform standard, we have to get an agreement that runs algorithms in a theoretical computer model. This model is the multi-tape Turing machine which has been accepted by a very large population. Based on the Turing machine, the theory of computational complexity has been built up. We will touch this part of theory in Chap. 8.

But, we will use RAM model to evaluate the running time for algorithms throughout this book except Chap. 8. In RAM model, assume that each line of pseudocode requires a constant time. For example, the running time of insertion sort is calculated in Fig. 1.6.

Actually, RAM model and Turing machine model are closely related. The running time estimated based on these two models is considered to be close enough. However, they are sometimes different in estimation of running time. For example, the following is a piece of pseudocode.
for $i=1$ to $n$
do assign First $(i) \leftarrow i$
end-for
According to RAM model, the running time of this piece is $O(n)$. However, based on the Turing machine, the running time of this piece is $O(n \log n)$ because the assigned value has to be represented by a string with $O(\log n)$ symbols.

Theoretically, a constant factor is often ignored. For example, we usually say that the running time of insertion sort is $O\left(n^2\right)$ instead of giving the specific quadratic function with respect to $n$. Here $f(n)=O(g(n))$ means that there exist constants $c>0$ and $n_0>0$ such that
$$f(n) \leq c \cdot g(n) \text { for } n \geq n_0$$

# 组合优化代写

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

$$F_0=F_1=1, F_i=F_{i-2}+F_{i-1} \text { for } i \geq 2$$

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

$$F_0=0, F_1=1 \text {, and } F_i=F_{i-1}+F_{i-2} .$$

$$\operatorname{cost}(T)=\sum_{i=1}^n a_i D_i .$$

$\operatorname{sum}(i, j)=a_i+a_{i+1}+\cdots+a_j$. 然后
$$\operatorname{cost}(T(i, j))=\min _{i \leq k<j} \cos t(T(i, k))+\cos t(T(k+1, j))+\operatorname{sum}(i, j)$$

$$\operatorname{sum}(i, j)=\left{a_i \quad \text { if } i=j a_i+\operatorname{sum}(i+1, j) \quad \text { if } i<j .\right.$$

running time $=($ 子问题的数量 $) \times($ 递归的计算时间 $)$ 。

## 有限元方法代写

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## MATLAB代写

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