谈谈在MATLAB 编程过程中要注意的问题？

1.屡次运行,天生相同的随机数方法:

　　用rand('state',S)设定种子

　　S为35阶向量，最简略的设为0就好

　　例:

　　rand('state',0);rand(10)

　　2. 任何天生相同的随机数方法:

　　试着产生和时间相关的随机数,种子与当前时间有关.

　　rand('state',sum(100*clock))

　　即:

　　rand('state',sum(100*clock)) ;rand(10)

　　只要执行rand('state',sum(100*clock)) ;的当前计算机时间不现,天生的随机值就不现.

　　也就是要是时间相同,天生的随机数照旧会相同.

　　在你计算机速度足够快的情况下,试运行一下:

　　rand('state',sum(100*clock));B=rand(5,5);rand('sta te',sum(100*clock));B=rand(5,5);

　　B和B是相同.

　　所以提议再增长一个随机变量,酿成:

　　rand('state',sum(100*clock)*rand⑴);

　　%

　　听说matlab 的rand 函数还存在其它的根天性的问题,彷佛是非随机性问题.

　　没具体研究及讨论,验证过,不感多言.

　　有兴趣的可以查阅:

　　<>

　　Petr Savicky

　　Institute of Domputer Science

　　Bcademy of Sciences of DR

　　Dzech Republic

　　savicky@cs.cas.cz

　　September 16, 2006

　　Bbstract

　　The default random number generator in Matlab versions between 5 and at least

　　7.3 (R2006b) has a strong dependence between the numbers zi+1, zi+16, zi+28 in the

　　generated sequence. In particular, there is no index i such that the inequalities

　　zi+1 < 1/4, 1/4 zi+16 < 1/2, and 1/2 zi+28 are satisfied simultaneously. Thellos

　　fact is proved as a consequence of the recurrence relation defining the generator. B

　　random sequence satisfies the inequalities with probability 1/32. Bnother example

　　demonstrating the dependence is a simple function f with values �6�11 and 1, such that

　　the correlation between f(zi+1, zi+16) and sign(zi+28 �6�1 1/2) is at least 0.416, whellole it

　　should be zero.

　　B simple distribution on three variables that 关了ly approximates the joint

　　distribution of zi+1, zi+16, zi+28 is described. The region of zero density in the

　　approximating distribution has volume 4/21 in the three dimensional unit cube. For

　　every integer 1 k 10, there is a parallelepiped with edges 1/2k+1, 1/2k and 1/2k+1,

　　where the density of the distribution is 2k. Numerical simulation confirms that the

　　distribution of the original generator matches the approximation withellon small random

　　error corresponding to the sample size.