The combinatorial core of the OVSF code assignment problem that arises in UMTS is to assign some nodes of a complete binary tree of height h (the code tree) to n simultaneous connections, such that no two assigned nodes (codes) are on the same root-to-leaf path. Each connection requires a code on a specified level. The code can change over time as long as it is still on the same level. We consider the one-step code assignment problem: Given an assignment, move the minimum number of codes to serve a new request. Minn and Siu proposed the so-called DCAalgorithm to solve the problem optimally. We show that DCA does not always return an optimal solution, and that the problem is NP-hard. We give an exact nO(h)-time algorithm, and a polynomial time greedy algorithm that achieves approximation ratio Θ(h). Finally, we consider the online code assignment problem for which we derive several results
標簽： combinatorial assignment problem arises
上傳時間： 2014-01-19
上傳用戶：BIBI
A Flash Player with ActionScript support. Write in C and C++. It have two part, one is Player and another is ActionScript Library. Runs on Linux and *BSD. It s very rough now, it have long way to go.
標簽： Player ActionScript and support
上傳時間： 2013-12-24
上傳用戶：時代電子小智
這個是一個vb編的GPS程序的源代碼，呵呵希望對大家會有所幫助！-this a vb series of the GPS program s source code, as long as you will want to help!
上傳時間： 2014-12-22
上傳用戶：啊颯颯大師的
A "code-what"? Unless you have spent some time working in the area of reverse engineering, chances are you have not heard of the term "codecave" before. If you have heard of it, you might not have read a clear definition of it or quite understand what it is or why it is useful. I have even asked seasoned assembly programmers about the term before and most of them had not heard of it. If it is new to you, do not worry, you are not the only one. It is a term that is scarcely used and is only useful in a reverse engineering context. Furthermore, is it "codecave" or "code cave"? I am not quite sure, but I will try my best to refer to it consistently as a "codecave". A space may sneak in there from time to time
標簽： engineering code-what chances reverse
上傳時間： 2014-01-17
上傳用戶：hn891122
1. 8623L平臺，其并行flash已經被燒入uClinux系統，并有IP地址. 2. Windows客戶機,使用tera term通過串口,把kernel燒到并行flash上,用命令boot rom啟動uClinux并且登陸. 3. Linux服務器，主要作用是編譯軟件,有nfs服務，nfs服務目錄是/nfsdir,文件包括有8623L的驅動程序,應用程序和音視頻文件,可以被8620L使用. em8623的開發包
標簽： 8623L
上傳時間： 2014-07-25
上傳用戶：youth25
We obtained the energy transport velocity distribution for a three dimensional ideal cloak explicitly. Near the operation frequency, the energy transport velocity has rather peculiar distribution. The velocity along a line joining the origin of the cloak is a constant, while the velocity approaches zero at the inner boundary of the cloak. A ray pointing right into the origin of the cloak will experience abrupt changes of velocities when it impinges on the inner surface of the cloak. This peculiar distribution causes long time delays for beams passing through the ideal cloak within a geometric optics description.
標簽： distribution dimensional transport obtained
上傳時間： 2013-12-19
上傳用戶：zhliu007
The existence of numerous imaging modalities makes it possible to present different data present in different modalities together thus forming multimodal images. Component images forming multimodal images should be aligned, or registered so that all the data, coming from the different modalities, are displayed in proper locations. The term image registration is most commonly used to denote the process of alignment of images , that is of transforming them to the common coordinate system. This is done by optimizing a similarity measure between the two images. A widely used measure is Mutual Information (MI). This method requires estimating joint histogram of the two images. Experiments are presented that demonstrate the approach. The technique is intensity-based rather than feature-based. As a comparative assessment the performance based on normalized mutual information and cross correlation as metric have also been presented.
標簽： present modalities existence different
上傳時間： 2017-04-03
上傳用戶：qunquan
Maya Calendar During his last sabbatical, professor M. A. Ya made a surprising discovery about the old Maya calendar. From an old knotted message, professor discovered that the Maya civilization used a 365 day long year, called Haab, which had 19 months. Each of the first 18 months was 20 days long, and the names of the months were pop, no, zip, zotz, tzec, xul, yoxkin, mol, chen, yax, zac, ceh, mac, kankin, muan, pax, koyab, cumhu. Instead of having names, the days of the months were denoted by numbers starting from 0 to 19. The last month of Haab was called uayet and had 5 days denoted by numbers 0, 1, 2, 3, 4. The Maya believed that this month was unlucky, the court of justice was not in session, the trade stopped, people did not even sweep the floor.
標簽： A. M. sabbatical surprising
上傳時間： 2014-01-05
上傳用戶：libenshu01
Watermarking schemes evaluation Abstract鈥擠igital watermarking has been presented as a solution to copy protection of multimedia objects and dozens of schemes and algorithms have been proposed. Two main problems seriously darken the future of this technology though. Firstly, the large number of attacks and weaknesses which appear as fast as new algorithms are proposed, emphasizes the limits of this technology and in particu-lar the fact that it may not match users expectations. Secondly, the requirements, tools and methodologies to assess the current technologies are almost non-existent. The lack of benchmarking of current algorithms is bla-tant. This confuses rights holders as well as software and hardware manufacturers and prevents them from using the solution appropriate to their needs. Indeed basing long-lived protection schemes on badly tested watermarking technology does not make sense.
標簽： Watermarking watermarking evaluation presented
上傳時間： 2013-12-04
上傳用戶：thinode
Generate Possion Dis. step1:Generate a random number between [0,1] step2:Let u=F(x)=1-[(1/e)x] step3:Slove x=1/F(u) step4:Repeat Step1~Step3 by using different u,you can get x1,x2,x3,...,xn step5:If the first packet was generated at time [0], than the second packet will be generated at time [0+x1],The third packet will be generated at time [0+x1+x2], and so on …. Random-number generation 1.static method random from class Math -Returns doubles in the range 0.0 <= x < 1.0 2.class Random from package java.util -Can produce pseudorandom boolean, byte, float, double, int, long and Gaussian values -Is seeded with the current time of day to generate different sequences of numbers each time the program executes
標簽： Generate Possion between random
上傳時間： 2017-05-25
上傳用戶：bibirnovis
