资源列表
CMA
- 恒模算法仿真，该算法是Bussgang算法的特例，是一种工程应用非常广的算法。-constant modulus algorithm simulation, the algorithm is a special case Bussgang algorithm, Application is a very broad algorithm.
ADAPTIVEGA
- ADPTIVE GA是改进遗传算法程序,提供了各种交叉算子,变异算子,具有强大的计算功能-ADPTIVE GA is improved genetic algorithms and procedures, and providing a variety of cross-operator, mutation operator, with powerful computing capabilities
Isomap
- ISOMAP算法，包括dfun.m dijk.m isomap.m l2_distance.m等等等等，一种流形学习算法（很好用，本人编译运行并仿真实现过）-ISOMAP algorithm, including dfun.m dijk.m isomap.m l2_distance.m etc., a learning algorithm Manifold (useful, I compile and run off Simulation)
tangxin
- 贪心算法求磁盘最优存储问题,能够解决此问题.不失为一种好办法.-greedy algorithm for optimal disk storage, to solve this problem. It is a good approach.
0240302314jiangwei13852910295
- 游程长度编码非常简单，编码、解码速度快，应用广泛。本文主要介绍了游程长度编码的原理和实现技术，对游程长度编码技术做了较为全面地研究。包括游程压缩模型、数据压缩、解压缩过程，并给出了流程图和相应的程序。 关键词：编码；解码；游程长度编码；压缩；解压缩。 -run-length coding is very simple, encoding, decoding speed, widespread use. This paper describes a run-length coding
yimaqi
- 银行管理系统，数据结构，C/C++，树，下载后解压才可以用-banking management systems, data structures, C / C, trees, and the downloading of decompression before they can use
TSPprogrammatlab
- 已知n个城市之间的相互距离，现有一个推销员必须遍访这n个城市，并且每个城市只能访问一次，最后又必须返回出发城市。如何安排他对这些城市的访问次序，可使其旅行路线的总长度最短？-known cities between the distance from the existing one must salesman n visited this city, every city and can only be visited once, finally starting to return to t
matlaborder
- matlab命令大全电子书，matlab入门必备-Matlab order Daquan e-books, Matlab entry required
gonge
- 本人自编的共轭梯度法。为c源码，已在TC下运行通过。-I directed the conjugate gradient method. C source code for, the TC has run through.
hunanface_orientation
- 人脸的检测与定位算法，采用降维算法实现：）-Face Detection and localization algorithm, using dimensionality reduction algorithm :)
hand_recognization
- 手写数字识别之模板匹配法，VC＋＋6.0编写的：）-handwritten figures identifiable template matching method, VC + +6.0 prepared :)
signal1hugo
- 自己的一个实验,验证频率的采样定理.即采样点数要足够大.-own an experiment to test frequency of the sampling theorem. That sampling points should be powerful enough.