### 数学代写|组合优化代写Combinatorial optimization代考|ORIE6334

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

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