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

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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|Algorithms with Self-Reducibility

There exist a large number of algorithms in which the problem is reduced to several subproblems, each of which is the same problem on a smaller-size input. Such a problem is said to have the self-reducibility, and the algorithm is said to be with self-reducibility.

For example, consider sorting problem again. Suppose input contains $n$ numbers. We may divide these $n$ numbers into two subproblems. One subproblem is the sorting problem on $\lfloor n / 2\rfloor$ numbers, and the other subproblem is the sorting problem on $\lceil n / 2\rceil$ numbers. After completely sorting each subproblem, combine two sorted sequences into one. This idea will result in a sorting algorithm, called the merge sort. The pseudocode of this algorithm is shown in Algorithm 1.

The main body calls a procedure. This procedure contains two self-calls, which means that the merge sort is a recursive algorithm, that is, the divide will continue until each subproblem has input of single number. Then this procedure employs another procedure (Merge) to combine solutions of subproblems with smaller

$$T(n)=2 \cdot T(\lceil n / 2\rceil)+c \cdot n$$
By induction, we can prove that
$$t(n) \leq T(n) \text { for all } n \geq 1$$
For base step, $t(1)=0=T(1)$. For induction step,
\begin{aligned} t(n) & \leq 2 \cdot t(\lceil n / 2\rceil)+c \cdot n \ & \leq 2 \cdot T(\lceil n / 2\rceil)+c \cdot n \quad \text { (by induction hypothesis) } \ & =T(n) . \end{aligned}
Now, let us discuss how to solve recursive equation about $T(n)$. Usually, we use two stages. In the first stage, we consider special numbers $n=2^k$ and employ the recursive tree to find $T\left(2^k\right)$ (Fig. $\left.2.2\right)$, that is,
\begin{aligned} T\left(2^k\right) & =2 \cdot T\left(2^{k-1}\right)+c \cdot 2^k \ & =2 \cdot\left(2 \cdot T\left(2^{k-2}\right)+c \cdot 2^{k-1}\right)+c \cdot 2^k \ & =\ldots \ & =2^k T(1)+k c \cdot 2^k \end{aligned}

## 数学代写|组合优化代写Combinatorial optimization代考|Rectilinear Minimum Spanning Tree

Consider two points $A=\left(x_1, y_1\right)$ and $B=\left(x_2, y_2\right)$ in the plane. The rectilinear distance of $A$ and $B$ is defined by
$$d(A, B)=\left|x_1-x_2\right|+\left|y_1-y_2\right| .$$
The rectilinear plane is the plane with the rectilinear distance, denoted by $L_1$-plane. In this section, we study the following problem.

Problem 2.2.1 (Rectilinear Minimum Spanning Tree) Given $n$ points in the rectilinear plane, compute the minimum spanning tree on those $n$ given points.
In Chap. 1, we already present Kruskal algorithm which can compute a minimum spanning tree within $O(m \log n)$ time. In this section, we will improve this result by showing that the rectilinear minimum spanning tree can be computed in $O(n \log n)$ time. To do so, we first study an interesting problem as follows.

Problem 2.2.2 (All Northeast Nearest Neighbors) Consider a set $P$ of $n$ points in the rectilinear plane. For each $A=\left(x_A, y_A\right) \in P$, another point $B=\left(x_B, y_B\right) \in P$ is said to lie in northeast (NE) area of $A$ if $x_A \leq x_B$ and $y_A \leq y_B$, but $A \neq B$. Furthermore, $B$ is the NE nearest neighbor of $A$ if $B$ has the shortest distance from A among all points lying in the NE area of A. This problem is required to compute the NE nearest neighbor for every point in $P$. (The NE nearest neighbor of a point A is “none” if no given point lies in the northeast area of A.)

Let us design a divide-and-conquer algorithm to solve this problem. For simplicity of description, assume all $n$ points have distinct $x$-coordinates and distinct $y$-coordinates. Now, we bisect $n$ points by a vertical line $L$. Let $P_l$ be the set of points lying on the left side of $L$ and $P_r$ the set of points lying on the right side of $L$. Suppose we already solve the all NE nearest neighbors problem on input point sets $P_l$ and $P_r$, respectively. Let us discuss how to combine solutions for two subproblems into a solution for all NE nearest neighbors on $P$.

For point $A$ in $P_r$, the NE nearest neighbor in $P_r$ is also the NE nearest neighbor in $P$. However, for point $A$ in $P_l$, the NE nearest neighbor in $P_l$ may not be the NE nearest neighbor in $P$. Actually, let $B_1$ denote the NE nearest neighbor of $A$ in $P_l$ and $B_r$ the NE nearest neighbor of $A$ for $B_2$ in $P_r$. Then, if $d\left(A, B_1\right) \leq d\left(A, B_2\right)$, then the NE nearest neighbor of $A$ in $P$ is $B_1$; otherwise, it is $B_2$. Therefore, to complete the combination task, it is sufficient to compute the NE nearest neighbors in $P_r$ for all points in $P_l$. We will show that this computation takes $O(n)$ time. To do so, let us first show a lemma.

# 组合优化代写

## 数学代写|组合优化代写Combinatorial optimization代考|Algorithms with Self-Reducibility

$$T(n)=2 \cdot T(\lceil n / 2\rceil)+c \cdot n$$

$$t(n) \leq T(n) \text { for all } n \geq 1$$

$$t(n) \leq 2 \cdot t(\lceil n / 2\rceil)+c \cdot n \quad \leq 2 \cdot T(\lceil n / 2\rceil)+c \cdot n \quad \text { (by induction hypothesis) }=T(n) .$$

$$T\left(2^k\right)=2 \cdot T\left(2^{k-1}\right)+c \cdot 2^k \quad=2 \cdot\left(2 \cdot T\left(2^{k-2}\right)+c \cdot 2^{k-1}\right)+c \cdot 2^k=\ldots \quad=2^k T(1)$$

## 数学代写|组合优化代写Combinatorial optimization代考|Rectilinear Minimum Spanning Tree

$$d(A, B)=\left|x_1-x_2\right|+\left|y_1-y_2\right|$$

## 有限元方法代写

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## MATLAB代写

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