Bill的小茶坊

巴赫的《无伴奏大提琴组曲》在电脑里放了很久，最近拿出来听了一下，结果就被电到了，太喜欢了，非常赞。

 

没想到大提琴能营造出如此瑰丽的音乐色彩，在巴赫缜密严谨的谱曲下，6首组曲像广袤的大海一样让听者完全沉醉了。

 

强烈推荐一记。

最后一周，红宝已经差不多都背出来了，昨天小模考了一下，一般般，类反错1、2个，填空2个，阅读3个，数学很Happy，这样。

其他的course project也做的差不多了，dlx终于知道怎么写vhdl了，小小地写了一下，理论上这个pipeline是可以跑起来了，下周去做做实验~OO的project也开始有点领悟了，其实如果直接做的话1天不到就能做完，但是我总是想设计得更好，结果就一直没写，总想有人在旁边reassure我一下

红宝开始过最后认真的一遍，这次要把反义词一起背进去，之后就快速地过红宝吧，也没时间慢慢来了。

TODOlist上N多事情没完成，心里一直觉得不踏实，好多好多的project还有考试。这个学期犯了几个错误，唉。

感觉这个学期精力太分散，要做的事情太多，结果一件都没做好。

稀疏的光线透过窗帘照进来，照在昏黄的壁橱上，希望就这么一点么。。。。。

The two examples of most importance are pumping lemma for regular languages and pumping lemma for context-free grammar(CFG).

 

Pumping Lemma for regular languages:

Let L be a regular language. Then there exists an integer p which is no less than 1, depending only on L, such that every string w in L of length at least p, which is often called the "pumping length", can be written as w = xyz, satisfying the following conditions:
a) |y| > 0
b) |xy| <= p
c) for all i >= 0, x(y^i)z belongs to L

Proof. For any regular language L, there exists an FA that accepts it, and we assume such FA has minimal states, denoting p as the number of states. For any given string w in L, whose length is at least p, there is a path from start state A_0 to final state A_p-1. Since such path length is at least p and we have p states, according to pigeonhole principle, there exists two states, denoted as A_i and A_j, which are the same, i.e. A_i = A_j. Thus, this FA has three parts. The first one is from A_0 to A_i-1, which is denoted as x, the second one is from A_i to A_j, which is a cycle and is denoted as y, the last part is the rest. Note that for all i >=0, x(y^i)z can be accepted by this FA. Thus, x(y^i)z belongs to L.

 

Pumping Lemma for CFG:

Let L be a context-free language, then there exists some integer p > 0 such that for any string s in L with |s| >= p can be written as s = uvxyz satisfying the following conditions:
a) |vxy| <= p
b) |vy| >= 1
c) for all i >=0, u(v^i)x(y^i)z belongs to L

 

今天写程序的时候抛出了这个异常，怎么调都不对

后来查了API才明白，是对Iterator的理解不够啊~

 

常用的Iterator使用方法是这样的（假设这是一个Hashset<T>的迭代器）：

while (iterator.hasNext()) {

    T t = iterator.next();

}

 

如果在while中对Hashset进行了增加或删除，则会抛出这个异常，原因是Hashset中所做的修改不会通知它的迭代器，而在迭代器中有一个状态，一旦它检查到状态和它预计的不一致时，为了安全考虑就抛出了这样的异常

 

正确的使用方法应该是用Iterator自己的remove()，但是对于插入似乎没有好的方法

 

暂时没有查到很好的解决方法，我只能自己搞了一个ArrayList<T>来模拟集合的行为。