### 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|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 $N E$ 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 $P$.

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

Consider a sequence of $n$ distinct integers which are stored in an array $A[1 . . n]$. An element $A[i]$ is a local maximum if $A[i-1]A[i+1]$ for $1S[i+1]$ for $i=1$, and $A[i-1]<A[i]$ for $i=n$. The sequence $A[1 . . n]$ is said to be bitonic if it contains exactly one local maximum, which is actually the global maximum one. Consider the following problem.

Problem 2.3.1 (Maximum Element in Bitonic Sequence) Given a sequence $A[1 . . n]$ of $n$ distinct integers, find the maximum element.
The problem can be solved by the following lemma.
Lemma 2.3.2 Assume $1 \leq iA[j]$, then $A[1 . . j-1]$ must contain a local maximum.
Proof First, assume $A[i]A[j+1]$. In this case, if none of $A[j], A[j-1], \ldots, A[i-1]$ is a local maximum, then $A[j]<A[j-1]<\cdots<A[i]$, contradicting to $A[i]<A[j]$
Similarly, we can show the second statement.
For $n \geq 4$, we can choose $i$ and $j$ such that $1 \leq i<j \leq n, i \geq n / 3$, and $n-j+1 \geq n / 3$. With such $i$ and $j$, for each comparison, the sequence can be cut off at least one third. Therefore, the maximum element can be found within $O(\log n)$ comparisons.

Next, we consider a situation that $A[i]=f(i)$, that is, $A[i]$ has to be obtained through evaluation of a function $f(i)$. Therefore, we want to find the maximum element with the minimum number of evaluations. In this situation, $i$ and $j$ will be selected based on a rule with Fibonacci number $F_i$ defined as follows:
$$F_0=F_1=1, F_i=F_{i-2}+F_{i-1} \text { for } i \geq 2$$

# 组合优化代写

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

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

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

$A[1 . n n]$ 如果它恰好包含一个局部最大值，而实际上是全局最大值，则称它是双调的。考虑以 下问题。

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

## 有限元方法代写

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## MATLAB代写

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