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.
標簽： monitoring sequential industria accurate
上傳時間： 2013-12-17
上傳用戶：familiarsmile
Linux.tar.gz //gz壓縮格式源代碼 linux0.0.1comment.zip //Linux 0.01核心源代碼的注釋 Linux0.01.zip //ZIP壓縮格式源代碼 Linux0.01雖然是Linux的第一個發行版本，但是卻基本具備了操作系統中最重要的組成部分，同時Linux0.01只有8500行左右的代碼，對操作系統的理解和實踐就非常簡單，輕松了。
上傳時間： 2014-01-20
上傳用戶：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.
標簽： generators contains Uniform randomc
上傳時間： 2014-12-01
上傳用戶：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.
標簽： transparent encryption and changes
上傳時間： 2013-12-27
上傳用戶：dongbaobao
ZLG7290例程*ZLG7290匯編例程*讀EEPROM并顯示例程* 16×2LCD模塊例程*128×64點陣LCD模塊例程* 直連KEY和LED例程 直連LED例程* lin模塊的原碼及例程。LIN總線例程 RS232例程(包括PC端和書上了串口例程) USB1.1例程(包括PC端)RS485例程 USB2.0例程(有3個，包括PC端) 基于ETHERNET的TCPIP例程自發自收例程 外中斷CAN例程USB2.0PC例程
上傳時間： 2014-12-22
上傳用戶：123456wh
嵌入式linux經典教程Embedded Linux Primer A Practical Real-World Approach (Original Edition) 由monta vista linux的開發者寫的.
標簽： Real-World Practical Embedded Approach
上傳時間： 2014-12-04
上傳用戶：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
標簽： responsibility Proceedings Interesting Algorithms
上傳時間： 2016-06-28
上傳用戶：wyc199288
一、問題的提出： 某廠根據計劃安排，擬將n臺相同的設備分配給m個車間，各車間獲得這種設備后，可以為國家提供盈利Ci j(i臺設備提供給j號車間將得到的利潤，1≤i≤n，1≤j≤m) 。問如何分配，才使國家得到最大的盈利L 二．算法的基本思想： 利用動態規劃算法的思想，設將i臺設備分配給j－1個車間，可以為國家得到最大利潤Li (j-1)(1≤i≤n，1≤j≤m)，那么將這i臺設備分配給j個車間，第j個車間只能被分配到0～i臺，所以我們只要算出當第j個車間分配到t(0<=t<=i)臺時提供的最大利潤Lt(j-1)+C(i-t)j，
標簽：
上傳時間： 2016-09-19
上傳用戶：希醬大魔王
ClustanGraphics聚類分析工具。提供了11種聚類算法。 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)
標簽： ClustanGraphics Complete Neighbor Linkage
上傳時間： 2014-01-02
上傳用戶：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.
標簽： the subsequence determine Instead
上傳時間： 2013-12-17
上傳用戶：evil
