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.

Task 4.1 The following recursive algorithm performs a binary search in an ordered array. Test the algorithm by searching the value 105 from the array: A = [11, 23, 31, 47, 52, 68, 71, 89, 94, 105, 112, 126, 133, 148] Use recursion tree method to find out the complexity class of the algorithm, when the length of the array is the size of the input. The “r = ë(p + q)/2û” on line 7 means floor function. For example, ë11/2û = 5. Input: Array A[1,..,n], n >= 1, with elements in increasing order A[1] <= A[2] <= … <= A[n]. Numbers 1<=p<=q<=n. Number x that is searched in A[p,..,q]. Output: The index of x in the array or -1 if x is not present in A[p,..,q]. SEARCH(A,p,q,x) 1. if p==q 2. if A[p]==x 3. return p 4. else 5. return -1 6. else 7. r = ë(p + q)/2û 8. if x<=A[r] 9. return SEARCH(A,p,r,x) 10. else 11. return SEARCH(A,r+1,q,x)

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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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We also assessed the computational efficiency of our subsampling method. The computations were carried out using R (R Core Team, 2021) on a desktop computer with 64GB memory. We restricted the calculations to access one CPU core and recorded the average CPU time from ten repetitions. Table 5 reports the results for Case I, where the subsample size is r = 1000. The computational speed of the OSP estimator is much faster than that of the full data estimator. Particularly, the computational burden of the full data method gets heavier as the increase of full data sample size. The computation time of the full data estimator increases in a polynomial order of n which agrees with the theoretical time complexity of O(n2). This full data time complexity is different from linear models (Wang et al., 2019) or logistic regression models (Wang et al., 2018), for which the time complexity is linear in sample size, i.e., O(n). In other words, subsampling is more desirable in Cox’s regression because it reduce the computational cost more significantly. The UNIF estimator is faster to compute than the OSP estimator, because it does not need the step of calculating the sampling probabilities, but it has a lower estimation efficiency as we have seen in previous results

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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 ，算法建立的模型是什么

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It looks like you are adding an Embedding layer to your model using Keras. This layer will be used to convert your input text data into a dense vector representation that can be processed by your model. Here's what each of the arguments you've provided to the Embedding layer mean: input_dim: This specifies the size of the vocabulary, or the number of unique words in your input data. It should be set to MAX_WORDS_NUM+1, which ensures that all words in your data are assigned a unique index. output_dim: This specifies the size of the embedding vector for each word. Typically, this is set to a value between 50 and 300, depending on the size of your dataset and complexity of the task. input_length: This specifies the length of each input sequence, or the number of words in each document. It should be set to MAX_SEQUENCE_LEN, which ensures that all input sequences have the same length and can be processed by your model. By adding the Embedding layer to your model, you are essentially creating a lookup table that maps each word in your vocabulary to a dense vector representation. This allows your model to learn meaningful relationships between words and make predictions based on the context of the input data.翻译

它应该设置为MAX_WORDS_NUM+1，这样确保所有单词都被分配了一个唯一的索引。 - output_dim: 这个参数指定了每个单词的嵌入向量的大小。通常，这个值设置为50到300之间的一个值，取决于数据集的大小和任务的复杂性。...
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