温度传感终端的在线手写数字识别系统设计

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"这篇研究论文描述了一种在温度传感终端上识别手写数字的系统。该系统接收手写数字作为一系列温度矩阵输入,经过简单的预处理后,利用可训练的级联相关神经网络(CCNN)进行分类。在分类之前,进行了混合特征提取以提高识别率。CCNN是在大约40位不同作者在温度传感器上书写的手写数字集上进行训练的,该集合包含800个数字。" 本文主要探讨了基于温度传感终端的在线手写数字识别技术,通过构建一种创新的系统来解决这一问题。系统的核心是将手写的数字转化为一系列的温度矩阵,这可能是通过某种特殊类型的传感器实现的,这种传感器能够捕捉到书写过程中的温度变化。手写数字的序列首先经过预处理步骤,这个步骤可能包括去除噪声、平滑线条等,以优化数据的质量,使其更适合后续的分析。 接下来,预处理后的数据被输入到一个可训练的级联相关神经网络(CCNN)中进行分类。CCNN是一种人工神经网络结构,特别适合于序列数据和模式识别任务。它通过级联多个小的、专门设计的网络模块来学习和识别复杂模式,每个模块都针对特定的特征或子任务进行训练。这种方法允许网络逐层学习,逐步提高识别精度,同时降低了过拟合的风险。 为了进一步提升识别效果,论文中提到了混合特征提取技术。这种技术通常结合了多种特征表示,如形状、纹理、方向和结构特征,以提供更全面的数字描述。通过提取这些多样化的特征,系统可以更好地理解和区分不同的手写数字,从而提高整体的识别率。 实验部分,研究者使用了一个包含800个手写数字的数据集,这些数字由大约40个不同的作者在温度传感终端上书写。使用这样的多作者数据集有助于增加模型的泛化能力,使其能适应各种不同的书写风格和习惯。训练过程中,CCNN对这些数据进行了学习,并根据反馈不断调整其权重和连接,以达到最佳的识别性能。 这篇研究论文介绍了一种新颖的、基于温度传感的在线手写数字识别方法,利用级联相关神经网络和混合特征提取技术,实现了高效且准确的识别效果。这项工作对于智能物联网设备、自动化办公系统以及无接触式交互界面等领域具有潜在的应用价值,为进一步优化和扩展手写识别技术提供了新的思路。

Description Consider the following 5 picture frames placed on an 9 x 8 array. ........ ........ ........ ........ .CCC.... EEEEEE.. ........ ........ ..BBBB.. .C.C.... E....E.. DDDDDD.. ........ ..B..B.. .C.C.... E....E.. D....D.. ........ ..B..B.. .CCC.... E....E.. D....D.. ....AAAA ..B..B.. ........ E....E.. D....D.. ....A..A ..BBBB.. ........ E....E.. DDDDDD.. ....A..A ........ ........ E....E.. ........ ....AAAA ........ ........ EEEEEE.. ........ ........ ........ ........ 1 2 3 4 5 Now place them on top of one another starting with 1 at the bottom and ending up with 5 on top. If any part of a frame covers another it hides that part of the frame below. Viewing the stack of 5 frames we see the following. .CCC.... ECBCBB.. DCBCDB.. DCCC.B.. D.B.ABAA D.BBBB.A DDDDAD.A E...AAAA EEEEEE.. In what order are the frames stacked from bottom to top? The answer is EDABC. Your problem is to determine the order in which the frames are stacked from bottom to top given a picture of the stacked frames. Here are the rules: 1. The width of the frame is always exactly 1 character and the sides are never shorter than 3 characters. 2. It is possible to see at least one part of each of the four sides of a frame. A corner shows two sides. 3. The frames will be lettered with capital letters, and no two frames will be assigned the same letter. Input Each input block contains the height, h (h<=30) on the first line and the width w (w<=30) on the second. A picture of the stacked frames is then given as h strings with w characters each. Your input may contain multiple blocks of the format described above, without any blank lines in between. All blocks in the input must be processed sequentially. Output Write the solution to the standard output. Give the letters of the frames in the order they were stacked from bottom to top. If there are multiple possibilities for an ordering, list all such possibilities in alphabetical order, each one on a separate line. There will always be at least one legal ordering for each input block. List the output for all blocks in the input sequentially, without any blank lines (not even between blocks). Sample Input 9 8 .CCC.... ECBCBB.. DCBCDB.. DCCC.B.. D.B.ABAA D.BBBB.A DDDDAD.A E...AAAA EEEEEE.. Sample Output EDABC

2023-06-06 上传
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The readPosInt method uses System.out.print (not println) to print its string argument on the screen (later when we use the readPosInt method, the string argument of the method will be a message telling the user to type some integer). Then the readPosInt method uses the input scanner object to read an integer from the user of the program. After reading the integer, the readPosInt method must also use the scanner’s nextLine method to read the single newline character that comes from the user pressing the Enter key on the keyboard after typing the integer (if you do not read this newline character using the nextLine method inside the readPosInt method, then the newline character will remain in the input stream, and, the next time you use the readLine method described above, the readLine method will just immediately read only the newline character from the input stream and return an empty string as result, without waiting for the user to type anything!) If the user types something which is not an integer, then the nextInt method of the scanner will throw an InputMismatchException. In that case the code of your readPosInt method must catch the exception, use System.out.println to print the error message "You must type an integer!" to the user (use System.out.println for this, not System.err.println, otherwise you might hit a bug in Eclipse...), use the scanner’s nextLine method to read (and ignore) the wrong input typed by the user of the program (if you do not do this, the wrong input typed by the user will remain in the input stream, and the next time you call the nextInt method again, you will get an InputMismatchException again!), and then do the whole thing again (including printing again the string argument of the readPosInt method) to try to read an integer again (hint: put the whole code of the method inside a while loop). After reading the integer and the newline character (which is just ignored), the readPosInt method tests the integer.写java文件

2023-05-25 上传

Write a program to 1.Setup a simulating backing store in memory. Read the data from pdata.bin to this backing store. 2.Initialize a page table for process p, set the frame number to be -1 for each page, indicating that the page is not loaded into memory yet. 3.Read logical addresses one by one from la.txt. 4.For each logical address, a)if its page has been loaded into physical memory, simply find the frame number in the page table, then generate physical address, find and print out the physical address and data inside this address. b)if the page is used for the first time, i.e., in page table, its frame number is -1,then the page that contains this address should be loaded into a free frame in physical memory (RAM). Then update the page table by adding the frame number to the right index in the page table. Then repeat 4a). Assumption: 1.Assume the file la.txt includes the sequence of generated addresses from CPU. 2.Use a part of memory as backing store that store data for a process. 3.The backing store size is 128 bytes 4.The size of process p is 128 bytes. 5.The contents of p is included in a file pdata.bin which is a binary file. 6.Use a part of memory as RAM. The size of physical memory is 256 bytes, from 0 to 255. All the physical memory is available, allocating starting from beginning in sequence. That is, allocate frame 0 first, then frame 1, then frame 2…. 7.The size of a frame is 32 bytes, i.e., 5 bits for the offset in a frame, total number of frames is 8. At beginning, no page table is available for process p.

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