The need for accurate monitoring and analysis of sequential data arises in many scientic, industrial and nancial problems. Although the Kalman lter is effective in the linear-Gaussian case, new methods of dealing with sequential data are required with non-standard models. Recently, there has been renewed interest in simulation-based techniques. The basic idea behind these techniques is that the current state of knowledge is encapsulated in a representative sample from the appropriate posterior distribution. As time goes on, the sample evolves and adapts recursively in accordance with newly acquired data. We give a critical review of recent developments, by reference to oil well monitoring, ion channel monitoring and tracking problems, and propose some alternative algorithms that avoid the weaknesses of the current methods.
标签： monitoring sequential industria accurate
上传时间： 2013-12-17
上传用户：familiarsmile
Linux.tar.gz //gz压缩格式源代码 linux0.0.1comment.zip //Linux 0.01核心源代码的注释 Linux0.01.zip //ZIP压缩格式源代码 Linux0.01虽然是Linux的第一个发行版本，但是却基本具备了操作系统中最重要的组成部分，同时Linux0.01只有8500行左右的代码，对操作系统的理解和实践就非常简单，轻松了。
上传时间： 2014-01-20
上传用户：lunshaomo
Uniform random number generators by Agner Fog, 2001 - 2007 randomc.zip contains a C++ class library of uniform random number generators of good quality. The random number generators found in standard libraries are often of a poor quality, insufficient for large Monte Carlo calculations. This C++ implementation provides random number generators of a much better quality: Better randomness, higher resolution, and longer cycle lengths. The same random number generators are available as libraries coded in assembly language for higher speed. These libraries can be linked into projects coded in other programming languages under Windows, Linux, BSD, etc. The library files are available in the archive asmlib.zip. Non-uniform random number generators are provided in stocc.zip.
标签： generators contains Uniform randomc
上传时间： 2014-12-01
上传用户：royzhangsz
Fast and transparent file system and swap encryption package for linux. No source code changes to linux kernel. Works with 2.6, 2.4, 2.2 and 2.0 kernels.
标签： transparent encryption and changes
上传时间： 2013-12-27
上传用户：dongbaobao
ZLG7290例程*ZLG7290汇编例程*读EEPROM并显示例程* 16×2LCD模块例程*128×64点阵LCD模块例程* 直连KEY和LED例程 直连LED例程* lin模块的原码及例程。LIN总线例程 RS232例程(包括PC端和书上了串口例程) USB1.1例程(包括PC端)RS485例程 USB2.0例程(有3个，包括PC端) 基于ETHERNET的TCPIP例程自发自收例程 外中断CAN例程USB2.0PC例程
上传时间： 2014-12-22
上传用户：123456wh
嵌入式linux经典教程Embedded Linux Primer A Practical Real-World Approach (Original Edition) 由monta vista linux的开发者写的.
标签： Real-World Practical Embedded Approach
上传时间： 2014-12-04
上传用户：wfl_yy
Proceedings of Practice of Interesting Algorithms 2007 The editor assumes no responsibility for the accuracy, completeness or usefulness of the information disclosed in this volume. Unauthorized use might infringe on privately owned patents of publication right. Please contact the individual authors for permission to reprint or otherwise use information from their papers. First edition 2007 Publication Planned by Prof. Wenxin Li Edited by Yili Zhao All rights reserved by Artificial Intelligence Laboratory, Peking University June 26, 2007
标签： responsibility Proceedings Interesting Algorithms
上传时间： 2016-06-28
上传用户：wyc199288
一、问题的提出： 某厂根据计划安排，拟将n台相同的设备分配给m个车间，各车间获得这种设备后，可以为国家提供盈利Ci j(i台设备提供给j号车间将得到的利润，1≤i≤n，1≤j≤m) 。问如何分配，才使国家得到最大的盈利L 二．算法的基本思想： 利用动态规划算法的思想，设将i台设备分配给j－1个车间，可以为国家得到最大利润Li (j-1)(1≤i≤n，1≤j≤m)，那么将这i台设备分配给j个车间，第j个车间只能被分配到0～i台，所以我们只要算出当第j个车间分配到t(0<=t<=i)台时提供的最大利润Lt(j-1)+C(i-t)j，
标签：
上传时间： 2016-09-19
上传用户：希酱大魔王
ClustanGraphics聚类分析工具。提供了11种聚类算法。 Single Linkage (or Minimum Method, Nearest Neighbor) Complete Linkage (or Maximum Method, Furthest Neighbor) Average Linkage (UPGMA) Weighted Average Linkage (WPGMA) Mean Proximity Centroid (UPGMC) Median (WPGMC) Increase in Sum of Squares (Ward s Method) Sum of Squares Flexible (ß space distortion parameter) Density (or k-linkage, density-seeking mode analysis)
标签： ClustanGraphics Complete Neighbor Linkage
上传时间： 2014-01-02
上传用户：003030
Instead of finding the longest common subsequence, let us try to determine the length of the LCS. Then tracking back to find the LCS. Consider a1a2…am and b1b2…bn. Case 1: am=bn. The LCS must contain am, we have to find the LCS of a1a2…am-1 and b1b2…bn-1. Case 2: am≠bn. Wehave to find the LCS of a1a2…am-1 and b1b2…bn, and a1a2…am and b b b b1b2…bn-1 Let A = a1 a2 … am and B = b1 b2 … bn Let Li j denote the length of the longest i,g g common subsequence of a1 a2 … ai and b1 b2 … bj. Li,j = Li-1,j-1 + 1 if ai=bj max{ L L } a≠b i-1,j, i,j-1 if ai≠j L0,0 = L0,j = Li,0 = 0 for 1≤i≤m, 1≤j≤n.
标签： the subsequence determine Instead
上传时间： 2013-12-17
上传用户：evil
