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
Use this resource to teach yourself Visual C# .NET version 2003, start developing Microsoft .NET–connected applications—one step at a time, master language fundamentals at your own pace and use the learn-by-doing exercises to dig in and code!
This book uses the Python language to teach pro
-
gramming concepts and problem
-solving skills, without assuming any previous program- ming experience. With easy-to-understand examples, pseudocode, flowcharts, and other
tools, the student learns how to design the logic of programs and then implement those
programs using Python. This book is ideal for an introductory programming course or a
programming logic and design course using Python as the language.
As with all the boolts in the
Starting Out With
series, the hallmark of this text is its clear,
friendly, and easy
-to-understand writing. In addition, it is rich in example programs that
are concise and practical. The programs in this book include short examples that highlight
specific programming topics, as well as more involved examples that focus on problem
solving. Each chapter provides one or more case studies that provide step
-by-step analysis
of a specific problem and shows the student how to solve it.
A new type of cloak is discussed: one that gives all cloaked objects the appearance of a flat conducting
sheet. It has the advantage that none of the parameters of the cloak is singular and can in fact be made
isotropic. It makes broadband cloaking in the optical frequencies one step closer.
The main MIPS processor of SMP8630 comes with a JTAG interface, allowing:
access to caches and data bus (DRAM) with a bandwidth of about 200kbit/s
examining the processor state whatever the execution mode (monice)
connecting to monice using mdi-server and using a gdb client on the processor to step and break
accurately whatever the execution mode
running semi-hosted applications
fl ash write tool
memory testing (MT command)
real-time traces: has not been built in CPU (Config3_TL=0) and only supported by MajicPLUS probes
(maybe built into emulator?)
The following Matlab code converts a Matrix into it a diagonal and off-diagonal component and performs up to 100 iterations of the Jacobi method or until 蔚step < 1e-5
Login Manager V3.0(LM3.0) is an authentication system which can integrate with any existing website that meets the requirements. LM3.0 provides a gatekeeper where user must be authorized before entering the membership secured areas.
Features:
1. Flexibility
LM3.0 allows administrator to integrate it with the current unprotected website. This is especially useful if major changes are going to be painful. With LM3.0, you re just one step towards getting the security you needed most.
2. Speed
LM3.0 uses PHP and MySQL which enables fast data transactions.
迷宮算法（maze）
/* Maze * Starting point is m[0][0], need to find a path go to m[9][9]. 0 means OK,
* 1 means cannot go there, boundary is 0 and 9, cannot go beyond boundary.
* Each step can be made horizontally or vertically for one more grid (diagonal
* jump is not allowed).
* Your program should print a series of grid coordinates that start from m[0][0]
* and go to m[9][9]
* Hint: No need to find the shortest path, only need to find one path that gets
* you to desitination.
*/
