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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

计算机代写|组合优化代写Combinatorial optimization代考|More Examples

Let us study more examples with divide-and-conquer technique and sorting algorithms.

Example 2.6.1 (Maximum Consecutive Subsequence Sum) Given a sequence of $n$ integers, find a consecutive subsequence with maximum sum.

Divide input sequence $S$ into two subsequence $S_1$ and $S_2$ such that $\left|S_1\right|=\lfloor n / 2\rfloor$ and $\left|S_2\right|=\lceil n / 2\rceil$. Let $\operatorname{Max} \operatorname{Sub}(S)$ denote the consecutive subsequence of $S$ with maximum sum. Then there are two cases.

Case 1. $\operatorname{Max} \operatorname{Sub}(S)$ is contained in either $S_1$ or $S_2$. In this case, $\operatorname{Max} \operatorname{Sub}(s)=$ $\operatorname{Max} \operatorname{Sub}\left(S_1\right)$ or $\operatorname{MaxSub}(s)=\operatorname{Max} \operatorname{Sub}\left(S_2\right)$.
Case 2. MaxSub(S) $\cap S_1 \neq \emptyset$ and $\operatorname{MaxSub}(S) \cap S_2 \neq \emptyset$. In this case, $\operatorname{Max} \operatorname{Sub}(S) \cap S_1$ is the tail subsequence with maximum sum. That is, suppose $S_1=\left{a_1, a_2, \ldots, a_p\right}$. Then among subsequences $\left{a_p\right}$, $\left{a_{p-1}, a_p\right}, \ldots,\left{a_1, \ldots, a_p\right}, \operatorname{MaxSub}(S) \cap S_1$ is the one with maximum sum. Therefore, it can be found in $O(n)$ time. Similarly, $\operatorname{Max} \operatorname{Sub}(S) \cap S_2$ is the head subsequence with maximum sum, which can be computed in $O(n)$ time.

Suppose MaxSub(S) can be computed in $T(n)$ time. Summarized from the above two cases, we obtain
$$T(n)=2 T(\lceil n / 2\rceil)+O(n) .$$
Therefore, $T(n)=O(n \log n)$.
Next, we present another algorithm running in $O(n)$ time.
Let $S_j$ be the maximum sum of a consecutive subsequence ending at the $j$ th integer $a_j$. Then, we have
\begin{aligned} S_1 &=a_1 \ S_{j+1} &= \begin{cases}S_j+a_{j+1} & \text { if } S_j>0, \ a_{j+1} & \text { if } S_j \leq 0 .\end{cases} \end{aligned}
This recursive formula gives a linear time algorithm to compute $S_j$ for all $1 \leq$ $j \leq n$. From them, find the maximum one, which is the solution for the maximum consecutive subsequence sum problem.

Example 2.6.2 (Closest Pair of Points) Given $n$ points in the Euclidean plane, find a pair of points to minimize the distance between them.

计算机代写|组合优化代写Combinatorial optimization代考|Dynamic Programming

Let us first study several examples and start from a simpler one.
Example 3.1.1 (Fibonacci Number) Fibonacci number $F_i$ for $i=0,1, \ldots$ is defined by
$$F_0=0, F_1=1, \text { and } F_i=F_{i-1}+F_{i-2} .$$
The computational process can be considered as a dynamic programming with self-reducibility structure as shown in Fig. 3.1.

Example $3.1 .2$ (Labeled Tree) Let $a_1, a_2, \ldots, a_n$ be a sequence of $n$ positive integers. A labeled tree for this sequence is a binary tree $T$ of $n$ leaves named $v_1, v_2, \ldots, v_n$ from left to right. We label $v_i$ by $a_i$ for all $i, 1 \leq i \leq n$. Let $D_i$ be the length of the path from $v_i$ to the root of $T$. The cost of $T$ is defined by
$$\operatorname{cost}(T)=\sum_{i=1}^n a_i D_i .$$
The problem is to construct a labeled tree $T$ to minimize the $\operatorname{cost} \operatorname{cost}(T)$ for a given sequence of positive integers $a_1, a_2, \ldots, a_n$.

Let $T(i, j)$ be the optimal labeled tree for subsequence $\left{a_i, a_{i+1}, \ldots, a_j\right}$ and $\operatorname{sum}(i, j)=a_i+a_{i+1}+\cdots+a_j$. Then
$$\cos t(T(i, j))=\min _{i \leq k<j}{\operatorname{cost}(T(i, k))+\cos t(T(k+1, j))}+\operatorname{sum}(i, j)$$
where
$$\operatorname{sum}(i, j)= \begin{cases}a_i & \text { if } i=j \ a_i+\operatorname{sum}(i+1, j) & \text { if } i<j .\end{cases}$$
As shown in Fig. 3.2, there are $1+2+\cdots+n=\frac{n(n+1)}{2}$ subproblems $T(i, j)$ in the table. From recursive formula, it can be seen that solution of each subproblem $T(i, j)$ can be computed in $O(n)$ time. Therefore, this dynamic programming runs totally in $O\left(n^3\right)$ time.

Actually, the running time of a dynamic programming is often estimated by the following formula:
running time $=$ (number of subproblems) $\times$ (computing time of recursion).

组合优化代考

计算机代写|组合优化代写Combinatorial optimization代考|More Examples

$\backslash$ left 的分隔符缺失或无法识别 是总和最大的那个。因此，可以在 $O(n)$ 时间。相似地， $\operatorname{Max} \operatorname{Sub}(S) \cap S_2$ 是总和最大的头子序列，可以计算 为 $O(n)$ 时间。

$$T(n)=2 T(\lceil n / 2\rceil)+O(n) .$$

$$S_1=a_1 S_{j+1} \quad=\left{S_j+a_{j+1} \quad \text { if } S_j>0, a_{j+1} \quad \text { if } S_j \leq 0 .\right.$$

计算机代写|组合优化代写Combinatorial optimization代考|Dynamic Programming

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

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

$$\cos t(T(i, j))=\min _{i \leq k<j} \operatorname{cost}(T(i, k))+\cos t(T(k+1, j))+\operatorname{sum}(i, j)$$

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

有限元方法代写

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MATLAB代写

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assignmentutor™您的专属作业导师
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