1. Complexity [20 marks] (A)Prove that is both and . [4 marks] (B)Consider the following functions of . Put them in order from smallest to largest asymptotic growth rate. [8 marks] (C)Let be the processing time of an algorithm for the input of size . Which is the asymptotic time complexity of this algorithm, or ? Please show your working to justify your answer. [8 marks]

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评估算法的时间复杂度（time complexity）的技巧小结

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A：证明 既是 又是 。[4 分] B：考虑 的以下函数。将它们按从最小到最大的渐近增长率排序。[8 分] C：让 是算法对输入大小 的处理时间。该算法的渐近时间复杂度是 还是 ？请出示您的计算步骤以证明您的答案。[8 分]
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(a) Consider the case of a European Vanilla Call option which is path independent. Examine the convergence of the Monte Carlo Method using the programme given in ‘MC Call.m’. How does the error vary with the number of paths nP aths? The current time is t = 0 and the Expiry date of the option is t = T = 0.5. Suppose that the current value of the underlying asset is S(t = 0) = 100 and the Exercise price is E = 100, with a risk free interest rate of r = 0.04 and a volatility of σ = 0.5. (b) Now repeat part (a) above but assume that the volatility is σ = 0.05. Does the change in the volatility σ influence the convergence of the Monte Carlo Method? (c) Now repeat part (a) but instead of taking one big step from t = 0 to t = T divide the interval into nSteps discrete time steps by using the programme given in ‘MC Call Small Steps.m’. Confirm that for path independent options, the value of nP aths determines the rate of convergence and that the value of nSteps can be set to 1. (d) Now let us consider path dependent options. The programme given in ‘MC Call Small Steps.m’ is the obvious starting point here. We assume that the current time is t = 0 and the expiry date of the option is t = T = 0.5. The current value of the underlying asset is S(t = 0) = 100 and the risk free interest rate is r = 0.05 and the volatility is σ = 0.3. (i) Use the Monte Carlo Method to estimate the value of an Arithematic Average Asian Strike Call option with Payoff given by max(S(T) − S, ¯ 0). (ii) Use the Monte Carlo Method to estimate the value of an Up and Out Call option with Exercise Price E = 100 and a barrier X = 150. (iii) Comment on the the rate of convergence for part (i) and (ii) above with respect to the parameters nP aths and nP aths使用matlab编程

I'm sorry, I cannot provide programming services as it goes ... The rate of convergence for these options depends on both nP aths and nSteps, as well as the complexity of the option's payoff function.

Recall that to solve (P2) in the tth time frame, we observe ξt 􏰗 {hti, Qi(t), Yi(t)}Ni=1, consisting of the channel gains {hti}Ni=1 and the system queue states {Qi(t),Yi(t)}Ni=1, and accordingly decide the control action {xt, yt}, including the binary offloading decision xt and the continuous resource allocation yt 􏰗 􏰄τit, fit, eti,O, rit,O􏰅Ni=1. A close observation shows that although (P2) is a non-convex optimization problem, the resource allocation problem to optimize yt is in fact an “easy” convex problem if xt is fixed. In Section IV.B, we will propose a customized algorithm to efficiently obtain the optimal yt given xt in (P2). Here, we denote G􏰀xt,ξt􏰁 as the optimal value of (P2) by optimizing yt given the offloading decision xt and parameter ξt. Therefore, solving (P2) is equivalent to finding the optimal offloading decision (xt)∗, where (P3) : 􏰀xt􏰁∗ = arg maximize G 􏰀xt, ξt􏰁 . (20) xt ∈{0,1}N In general, obtaining (xt)∗ requires enumerating 2N offloading decisions, which leads to significantly high computational complexity even when N is moderate (e.g., N = 10). Other search based methods, such as branch-and-bound and block coordinate descent [29], are also time-consuming when N is large. In practice, neither method is applicable to online decision- making under fast-varying channel condition. Leveraging the DRL technique, we propose a LyDROO algorithm to construct a policy π that maps from the input ξt to the optimal action (xt)∗, i.e., π : ξt 􏰕→ (xt)∗, with very low complexity, e.g., tens of milliseconds computation time (i.e., the time duration from observing ξt to producing a control action {xt, yt}) when N = 10.，为什么要使用深度强化学习

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Little Gyro has just found an empty integer set A in his right pocket, and an empty integer set B in his left pocket. At the same time, Little Gyro has already thought about two integers n, m in his mind. As Little Gyro is bored, he decides to play with these two sets. Then, Little Gyro is ready to divide the integer series 1,2,3...m−1,m to these two empty sets. With the following rules: If the number is an integer multiple of n, Little Gyro will put it in set A. Otherwise, Little Gyro will put it in set B instead. Now given two integers n, m, after these operations, Little Gyro wants to know the result of sum(B)−sum(A) and ask you for help, please help him to calculate. Input Specification: There are multiple test cases. The first line of the input contains an integer T (1 ≤ T ≤ 10 5 ), indicating the number of test cases. For each test case: Each line contains two integers n, m (1 ≤ n, m ≤ 10 9 ), indicating the length of the sequence. Output Specification: For each test case output an integer indicating the result of sum(B)−sum(A). Sample Input: 3 3 5 2 10 3 16 Sample Output: 9 -5 46 Hint: For the first sample, after these operations, A = <3>, B = <1,2,4,5>, and the answer is 9. For the second sample, after these operations, A = <2,4,6,8,10>, B = <1,3,5,7,9>, and the answer is -5.

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Algorithm 1: The online LyDROO algorithm for solving (P1). input : Parameters V , {γi, ci}Ni=1, K, training interval δT , Mt update interval δM ; output: Control actions 􏰕xt,yt􏰖Kt=1; 1 Initialize the DNN with random parameters θ1 and empty replay memory, M1 ← 2N; 2 Empty initial data queue Qi(1) = 0 and energy queue Yi(1) = 0, for i = 1,··· ,N; 3 fort=1,2,...,Kdo 4 Observe the input ξt = 􏰕ht, Qi(t), Yi(t)􏰖Ni=1 and update Mt using (8) if mod (t, δM ) = 0; 5 Generate a relaxed offloading action xˆt = Πθt 􏰅ξt􏰆 with the DNN; 6 Quantize xˆt into Mt binary actions 􏰕xti|i = 1, · · · , Mt􏰖 using the NOP method; 7 Compute G􏰅xti,ξt􏰆 by optimizing resource allocation yit in (P2) for each xti; 8 Select the best solution xt = arg max G 􏰅xti , ξt 􏰆 and execute the joint action 􏰅xt , yt 􏰆; { x ti } 9 Update the replay memory by adding (ξt,xt); 10 if mod (t, δT ) = 0 then 11 Uniformly sample a batch of data set {(ξτ , xτ ) | τ ∈ St } from the memory; 12 Train the DNN with {(ξτ , xτ ) | τ ∈ St} and update θt using the Adam algorithm; 13 end 14 t ← t + 1; 15 Update {Qi(t),Yi(t)}N based on 􏰅xt−1,yt−1􏰆 and data arrival observation 􏰙At−1􏰚N using (5) and (7). i=1 i i=1 16 end With the above actor-critic-update loop, the DNN consistently learns from the best and most recent state-action pairs, leading to a better policy πθt that gradually approximates the optimal mapping to solve (P3). We summarize the pseudo-code of LyDROO in Algorithm 1, where the major computational complexity is in line 7 that computes G􏰅xti,ξt􏰆 by solving the optimal resource allocation problems. This in fact indicates that the proposed LyDROO algorithm can be extended to solve (P1) when considering a general non-decreasing concave utility U (rit) in the objective, because the per-frame resource allocation problem to compute G􏰅xti,ξt􏰆 is a convex problem that can be efficiently solved, where the detailed analysis is omitted. In the next subsection, we propose a low-complexity algorithm to obtain G 􏰅xti, ξt􏰆. B. Low-complexity Algorithm for Optimal Resource Allocation Given the value of xt in (P2), we denote the index set of users with xti = 1 as Mt1, and the complementary user set as Mt0. For simplicity of exposition, we drop the superscript t and express the optimal resource allocation problem that computes G 􏰅xt, ξt􏰆 as following (P4) : maximize 􏰀j∈M0 􏰕ajfj/φ − Yj(t)κfj3􏰖 + 􏰀i∈M1 {airi,O − Yi(t)ei,O} (28a) τ,f,eO,rO 17 ，建立了什么模型

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Every year the cows hold an event featuring a peculiar version of hopscotch that involves carefully jumping from rock to rock in a river. The excitement takes place on a long, straight river with a rock at the start and another rock at the end, L units away from the start (1 ≤ L ≤ 1,000,000,000). Along the river between the starting and ending rocks, N (0 ≤ N ≤ 50,000) more rocks appear, each at an integral distance Di from the start (0 < Di < L). To play the game, each cow in turn starts at the starting rock and tries to reach the finish at the ending rock, jumping only from rock to rock. Of course, less agile cows never make it to the final rock, ending up instead in the river. Farmer John is proud of his cows and watches this event each year. But as time goes by, he tires of watching the timid cows of the other farmers limp across the short distances between rocks placed too closely together. He plans to remove several rocks in order to increase the shortest distance a cow will have to jump to reach the end. He knows he cannot remove the starting and ending rocks, but he calculates that he has enough resources to remove up to M rocks (0 ≤ M ≤ N). FJ wants to know exactly how much he can increase the shortest distance *before* he starts removing the rocks. Help Farmer John determine the greatest possible shortest distance a cow has to jump after removing the optimal set of M rocks. Input Line 1: Three space-separated integers: L, N, and M Lines 2..N+1: Each line contains a single integer indicating how far some rock is away from the starting rock. No two rocks share the same position. Output Line 1: A single integer that is the maximum of the shortest distance a cow has to jump after removing M rocks Sample Inputcopy Outputcopy 25 5 2 2 14 11 21 17 4 Hint Before removing any rocks, the shortest jump was a jump of 2 from 0 (the start) to 2. After removing the rocks at 2 and 14, the shortest required jump is a jump of 4 (from 17 to 21 or from 21 to 25).

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资源摘要信息:"在Go语言（又称Golang）中，控制台的输入输出是进行基础交互的重要组成部分。Go语言提供了一组丰富的库函数，特别是`fmt`包，来处理控制台的输入输出操作。`fmt`包中的函数能够实现格式化的输入和输出，使得程序员可以轻松地在控制台显示文本信息或者读取用户的输入。" 1. fmt包的使用 Go语言标准库中的`fmt`包提供了许多打印和解析数据的函数。这些函数可以让我们在控制台上输出信息，或者从控制台读取用户的输入。 - 输出信息到控制台 - Print、Println和Printf是基本的输出函数。Print和Println函数可以输出任意类型的数据，而Printf可以进行格式化输出。 - Sprintf函数可以将格式化的字符串保存到变量中，而不是直接输出。 - Fprint系列函数可以将输出写入到`io.Writer`接口类型的变量中，例如文件。 - 从控制台读取信息 - Scan、Scanln和Scanf函数可以读取用户输入的数据。 - Sscan、Sscanln和Sscanf函数则可以从字符串中读取数据。 - Fscan系列函数与上面相对应，但它们是将输入读取到实现了`io.Reader`接口的变量中。 2. 输入输出的格式化 Go语言的格式化输入输出功能非常强大，它提供了类似于C语言的`printf`和`scanf`的格式化字符串。 - Print函数使用格式化占位符 - `%v`表示使用默认格式输出值。 - `%+v`会包含结构体的字段名。 - `%#v`会输出Go语法表示的值。 - `%T`会输出值的数据类型。 - `%t`用于布尔类型。 - `%d`用于十进制整数。 - `%b`用于二进制整数。 - `%c`用于字符（rune）。 - `%x`用于十六进制整数。 - `%f`用于浮点数。 - `%s`用于字符串。 - `%q`用于带双引号的字符串。 - `%%`用于百分号本身。 3. 示例代码分析 在文件main.go中，可能会包含如下代码段，用于演示如何在Go语言中使用fmt包进行基本的输入输出操作。 ```go package main import "fmt" func main() { var name string fmt.Print("请输入您的名字: ") fmt.Scanln(&name) // 读取一行输入并存储到name变量中 fmt.Printf("你好, %s!\n", name) // 使用格式化字符串输出信息 } ``` 以上代码首先通过`fmt.Print`函数提示用户输入名字，并等待用户从控制台输入信息。然后`fmt.Scanln`函数读取用户输入的一行信息（包括空格），并将其存储在变量`name`中。最后，`fmt.Printf`函数使用格式化字符串输出用户的名字。 4. 代码注释和文档编写 在README.txt文件中，开发者可能会提供关于如何使用main.go代码的说明，这可能包括代码的功能描述、运行方法、依赖关系以及如何处理常见的输入输出场景。这有助于其他开发者理解代码的用途和操作方式。 总之，Go语言为控制台输入输出提供了强大的标准库支持，使得开发者能够方便地处理各种输入输出需求。通过灵活运用fmt包中的各种函数，可以轻松实现程序与用户的交互功能。