润色：Previous studies have shown that the L1 norm is superior to the L2 norm in the processing of high-frequency detail features.

时间: 2024-06-07 22:07:55 浏览: 94
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L1—L2 optimization in signal and image processing

Revised: Previous research has demonstrated that the L1 norm performs better than the L2 norm when it comes to processing high-frequency detail features.
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vga.zip_All in the Timing_Quartus 12.1_dac vhdl_frequency contro

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BIFPN的全称是“双向特征金字塔网络”，是一种用于计算机视觉中物体检测的神经网络架构，2019年Tan等人发表的论文《BIFPN：带有重复的金字塔结构的高效多尺度融合》中首次提出。BIFPN是一种特征金字塔网络（FPN），...

Create a test class with three static methods shown below at minimum. public class Test{ public static void start(){ - Create an empty list of shapes and save the reference. - Pass the reference to create method that fills the list. - Pass the reference to display method that prints the list. //The following can be added here or into an additional method. - Remove a shape. - Display the list again. - Check the size of the list. - … } public static returnTypeOrVoid create(a reference to a list) { - Create shape objects using data from an input file. - Note: A shape can be a rectangle, a circle or a triangle. - Add the objects into the list. } public static returnTypeOrVoid display(a reference to a list) { - Display the objects of the list. … } }

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Remarks The provided hyperparameters are identical for CIFAR10, CIFAR100-20 and STL10. However, fine-tuning the hyperparameters can further improve the results. We list the most important hyperparameters of our method below: - Entropy weight: Can be adapted when the number of clusters changes. In general, try to avoid imbalanced clusters during training. - Confidence threshold: When every cluster contains a sufficiently large amount of confident samples, it can be beneficial to increase the threshold. This generally helps to decrease the noise. The ablation can be found in the paper. - Number of neighbors in SCAN: The dependency on this hyperparameter is rather small as shown in the paper.

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Find the tensile capacity P1 for the single-shear lap connection shown below Bolts are 3/4-in-diameter high-strength bolts..All structural steel is A36. Assume that the connection is A325SC.Assume a class A contact surface and a slip-critical connection. 12" xt" plate iedge distance (typ.) The properties of the steel are as follows: The tensile strength, Fu,ofA36 steel is 58 ksi, and the allowable shear stress F=17.0 ksi The cross-sectional area of the bolt Ab=0.44I in².

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There is a light sensor on your picoprobe development board. This question asks you to use this sensor to measure the ambient light and to control the three LEDs on the picoprobe board to make a night light. Configure your program to perform the following steps, after printing the message from Question 1. The steps should be repeated every 1 second. raw ADC result in the format shown below: Light Sensor: Raw: _____ (insert the actual value) (15 marks) (b) Convert the raw value to a voltage and print out the voltage that you calculated in the format displayed below. At least one decimal place is required. Voltage: _____ V (insert the actual value) (10 marks) (c) Use this voltage to control the three-colour LED to show a white colour, when the light sensor is covered by a piece of paper (e.g. when the room is getting dark). The white-coloured light will go off, as soon as the ambient light is high again, i.e. when the light sensor is exposed to normal room light. This is reminiscent of an automatic night light.

In the above code, we are dimming all three LEDs to create white light when the room is dark and turning them off when there is enough ambient light. The threshold voltage for detecting darkness is ...

The RGB images have three color channels (24-bit image), but the grayscale image had only a single grayscale image (8-bit image). In our previous experiment, we normalized the images of the pre- trained network by each color channel separately. This approach lost the information provided by the other two channels. Here, we used the same grayscale LDCT image for each channel to make it somewhat analogous to an RGB image for the CNN. Doing so engages all the weights and exploits all the learned knowledge when extracting features from the pretrained net- work. Since the images experimented with were smaller than the required input size of the pretrained CNN (224 × 224), we used bicubic interpolation for resizing. The dimension of a deep feature vector extracted for each image was 4096. The features were the output of the last fully connected layer (the full 2 layer as shown in Table 2) before the output layer in an ImageNet pretrained CNN. The architectures and para 解释

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Write me a java project named pontoon game, which needs to meet the following requirements:Pontoon is a card game played with three decks of playing cards (i.e. cards of 2-10, Jack, Queen, King, and Ace from the four suits of Hearts, Diamonds, Spades, and Clubs). There are two players: the user of the computer, and the computer itself. The object of the game is to beat the other player by either getting as close to a score of 21 as possible (without exceeding 21) or by having 5 cards and a score of 21 or less. Each card has a value: numeric cards have their numeric value, 'face' cards have value 10, and aces have value 11. The game is played as follows, Firstly the user is told about the game and invited to play. If they opt to play then four cards are dealt. The first card goes to the user, the second to the computer, the third to the user, and the fourth to the computer. The computer's card remains hidden, all others are shown. The user can then indicate they wish to have an additional card ("hit"). If they are this card is also displayed. Additional cards may be taken until either the user's score exceeds 21 (they "bust"), or they choose not to take another card ("sit"/"stand"). Once the user has finished taking cards the computer will take cards. The computer will always take a card if their total is less than 16 (and the user's score doesn't exceed 21).

Sure, here is an implementation of the Pontoon game in Java: import java.util.ArrayList; import java.util.Collections; import java.util.List; import java.util.Scanner; public class PontoonGame {...

下面是一道java编程题In the well-known story of "The tortoise and the hare", the hare was too confident and took a break in the middle of the competition, which led to the tortoise winning the competition. A multi-threaded program is required to simulate the race between the tortoise and the hare. The speed of the tortoise is 1 meter per 1500 milliseconds, and the speed of the hare is 5 meters per 500 milliseconds. When the hare reached the 700th meter, it chose to rest for 10,000 milliseconds. As a result, the tortoise wins the race. The output result of the program running is shown in the figure.

Sure, here is a possible implementation of the program in Java: java public class TortoiseAndHareRace { private static final int RACE_LENGTH = 1000; private static final int TORTOISE_SPEED = 1; ...

将用sort函数数组X中元素S字母按照a,..,z的顺序排列 matlabS = 'Scientists look for universal patterns of complex systems because such invariant features may help to unveil the principles of system organization Complex network studies can not only provide a unique viewpoint of nature and society but also reveal ubiquitous patterns, e.g. small world and scale free, characteristic of various complex systems. However, ecological studies have shown that binary food webs, which depict trophic interactions in ecosystems, refuse to become part of the small world and scale free networks family. Although some common features, including "two degrees separation", which means the very small average distance, are shared among food webs, other meaningful attributes such as degree distribution and clustering coefficient change with the size and complexity (connectance)of the network.';

However, ecological studies have shown that binary food webs, which depict trophic interactions in ecosystems, refuse to become part of the small world and scale free networks family. Although some ...

7 errors have detailed information that is not shown. Use 'stats.errorDetails: true' resp. '--stats-error-details' to show it. webpack 5.83.1 compiled with 11 errors in 1777 ms

这个错误提示是 webpack 在编译过程中出现了 11 个错误，但是并没有给出详细的错误信息，建议你可以加上 --stats-error-details 参数重新编译，以显示详细的错误信息。 你可以在命令行中输入以下命令： ...

翻译一下To validate the effectiveness of modality-aware representation learning (MARL) and adaptive graph learning (AGL), we replace MARL with MLP and direct concatenation respectively, and replace AGL with the construction method adopted in popGCN [11] and kNN graph GkNN using RBF kernel. Table IV shows the ablation study results on applying different modules in our models. Specifically, the performance of the constructed graph GP opGCN of popGCN is the worst, especially on the ABIDE dataset, indicating that handconstructed graph is indeed not a desirable choice. Furthermore, both “MARL+GpopGCN ” and “MARL+GkNN ” worsen the performance of MARL, which means that an inappropriate metric can have negative impacts. Besides, AGL achieves a favorable performance despite the absence of MARL, which again validates the effectiveness of adaptive graph learning. More importantly, it can be observed that the combination of MARL and AGL achieves a performance that far exceeds the other combinations. In addition, we also compare the performance of AGL and transformer [35], which is a popular architecture, as shown in Table V. IPT is designed as an interpatient transformer into which the patients’ representations are fed to obtain the results. It can be seen that the performance of the two methods is very close to each other.

为了验证模态感知表示学习（MARL）和自适应图学习（AGL）的有效性，我们分别使用MLP和直接串联来代替MARL，使用popGCN的构建方法和使用RBF内核的kNN图GkNN来代替AGL，并在我们的模型中应用不同的模块，表IV显示了...

Figure 7: You will use the inverse compositional image alignment to track 4 frames of images. Given a template and a set of consecutive images, you will (1) initialize the affine transform using the feature based alignment and then (2) track over frames using the inverse compositional image alignment. function [A_cell] = TrackMultiFrames(template, image_cell) Input: template is gray-scale template. image_cell is a cell structure that contains a set of consecutive image frames, i.e., image_cell{i} is the i th frame. Output: A_cell is the set of affine transforms from the template to each frame of image, i.e., A_cell{i} is the affine transform from the template to the i th image. Description: You will apply the inverse compositional image alignment sequentially to track over frames as shown in Figure 7. Note that the template image needs to be updated at every frame, i.e., template←WarpImage(I, inv(A), size(template))翻译

% Warp template image using inverse of previous affine transform if i > 1 template = warp_image(template, inv(A_cell{i-1}), size(template)); end % Compute inverse compositional image alignment ...

用C++写You are given two integers x and k . Grasshopper starts in a point 0 on an OX axis. In one move, it can jump some integer distance, that is not divisible by k , to the left or to the right. What's the smallest number of moves it takes the grasshopper to reach point x ? What are these moves? If there are multiple answers, print any of them. Input The first line contains a single integer t (1≤t≤1000 ) — the number of testcases. The only line of each testcase contains two integers x and k (1≤x≤100 ; 2≤k≤100 ) — the endpoint and the constraint on the jumps, respectively. Output For each testcase, in the first line, print a single integer n — the smallest number of moves it takes the grasshopper to reach point x . In the second line, print n integers, each of them not divisible by k . A positive integer would mean jumping to the right, a negative integer would mean jumping to the left. The endpoint after the jumps should be exactly x . Each jump distance should be from −109 to 109 . In can be shown that, for any solution with the smallest number of jumps, there exists a solution with the same number of jumps such that each jump is from −109 to 109 . It can be shown that the answer always exists under the given constraints. If there are multiple answers, print any of them.

以下是本题的 C++ 代码实现： c++ #include #include using namespace std; int main() ...接着根据上文中提到的算法思想进行判断，如果 $x$ 能被 $k$ 整除，则至少需要跳两步，输出 $k$ 和 $x$；...

A. Encoding Network of PFSPNet The encoding network is divided into three parts. In the part I, RNN is adopted to model the processing time pij of job i on all machines, which can be converted into a fixed dimensional vector pi. In the part II, the number of machines m is integrated into the vector pi through the fully connected layer, and the fixed dimensional vector p˜i is output. In the part III, p˜i is fed into the convolution layer to improve the expression ability of the network, and the final output η p= [ η p1, η p2,..., η pn] is obtained. Fig. 2 illustrates the encoding network. In the part I, the modelling process for pij is described as follows, where WB, hij , h0 are k-dimensional vectors, h0, U, W, b and WB are the network parameters, and f() is the mapping from RNN input to hidden layer output. The main steps of the part I are shown as follows. Step 1: Input pij to the embedding layer and then obtain the output yij = WB pij ; Step 2: Input yi1 and h0 to the RNN and then obtain the hidden layer output hi1 = f(yi1,h0; U,W, b). Let p1 = h1m ; Step 3: Input yij and hi,j−1, j = 2, 3 ··· , m into RNN in turn, and then obtain the hidden layer output hij = f(yij ,hi,j−1; U,W, b), j = 2, 3 ··· , m. Let pi = him . In the part II, the number of machines m and the vector pi are integrated by the fully connected layer. The details are described as follows. WB and h˜i are d-dimensional vectors, WB W and ˜b are network parameters, and g() denotes the mapping from the input to the output of full connection layer. Step 1: Input the number of machines m to the embedding layer, and the output m = WB m is obtained。Step 2: Input m and pi to the fully connected layer and then obtain the output hi = g([m, pi];W, b); Step 3: Let pi = Relu(hi). In the part III, pi, i = 1, 2,...,n are input into onedimensional convolution layer. The final output vector η pi, i = 1, 2, ··· , n are obtained after the output of convolutional layer goes through the Relu layer.首先逐行仔细的分析此过程，其次怎么使用pytorch用EncoderNetwork类完全实现这个过程的所有功能和步骤

h0 = torch.zeros(1, x.size(0), self.rnn.hidden_size).to(x.device) out, _ = self.rnn(y, h0) p = out[:, -1, :] # Part II: Fully Connected Layer m = m.view(-1, 1) h = self.fc(torch.cat((p, m), dim...

用C语言解决下列问题：Kirill wants to weave the very beautiful blanket consisting of n×m of the same size square patches of some colors. He matched some non-negative integer to each color. Thus, in our problem, the blanket can be considered a B matrix of size n×m consisting of non-negative integers. Kirill considers that the blanket is very beautiful, if for each submatrix A of size 4×4 of the matrix B is true: A11⊕A12⊕A21⊕A22=A33⊕A34⊕A43⊕A44, A13⊕A14⊕A23⊕A24=A31⊕A32⊕A41⊕A42, where ⊕ means bitwise exclusive OR Kirill asks you to help her weave a very beautiful blanket, and as colorful as possible! He gives you two integers n and m . Your task is to generate a matrix B of size n×m , which corresponds to a very beautiful blanket and in which the number of different numbers maximized. Input The first line of input data contains one integer number t (1≤t≤1000 ) — the number of test cases. The single line of each test case contains two integers n and m (4≤n,m≤200) — the size of matrix B . It is guaranteed that the sum of n⋅m does not exceed 2⋅105 . Output For each test case, in first line output one integer cnt (1≤cnt≤n⋅m) — the maximum number of different numbers in the matrix. Then output the matrix B (0≤Bij<263) of size n×m . If there are several correct matrices, it is allowed to output any one. It can be shown that if there exists a matrix with an optimal number of distinct numbers, then there exists among suitable matrices such a B that (0≤Bij<263) .

我可以回答这个问题。这是一个矩阵问题，可以使用C语言解决。具体来说，可以使用循环嵌套来生成矩阵B，同时使用位运算符来实现异或操作。对于每个4×4的子矩阵A，可以使用条件判断来判断是否满足美丽的条件，如果...

Using an ACL connection, a Central can establish one or more isochronous connections that use the isochronous physical channel. An isochronous connection is used to transfer isochronous data between the Central and a Peripheral by using a logical transport, which is referred to as a Connected Isochronous Stream (CIS). A CIS consists of CIS events that occur at regular intervals (designated ISO_Interval). Every CIS event consists of one or more subevents. In each subevent, the Central transmits once and the Peripheral responds. If the Central and Peripheral have completed transferring the scheduled isochronous data in a CIS event, all remaining subevents in that event will have no radio transmissions and the event is closed. Each subevent uses a PHY channel which is determined by using the channel selection algorithm. The PHY channel that is used for a subevent is marked as ISO Ch(eventcount, subeventcount), as shown in Figure 1.4.

通过ACL连接，中央设备可以建立一个或多个使用同步物理通道的同步连接。同步连接用于通过逻辑传输（称为连接的同步流，CIS）在中央设备和外围设备之间传输同步数据。CIS由定期发生的CIS事件组成（指定为ISO_Interval...

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