带有连接的常规语言
常规语言在以下操作下是封闭的：
init(L) = 字符串 w 的集合，使得对于某些 x，wx 在 L 中。
编辑： x 可以是任何字符串、字符或空字符串 我怎样才能证明这一点？
the regular languages are closed under the operation:
init(L) = the set of the strings w such that for some x, wx is in L.
EDIT :
x can be any string, a character or empty string
How can I prove that ?
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好吧，第一次看错问题，现在明白了。这仍然是微不足道的。查看自动化，您搜索的是自动化的一部分，分为两个状态集 S1 和 S2，因此它们之间只有一个转换（如果它从 S1->S2 S1 当然包含起始节点，而 S2 包含结束节点）。这样的情况总是存在的（空语言除外），如果没有这样的节点可以添加一个，所以w只是一个包含空词的集合，这当然也是正则的（空语言情况也是如此）。
OK, misread the quesion on the first time, now I get it. It still trivial. Looking at the automate what you searching is a partion of the automate into two state sets S1 and S2, so that just one transition is between them (and if its from S1->S2 S1 contains of course the start node, and S2 the end node). Such exist always (exception empty language), in case there is no such node you can add one, so w is just a set containing the empty word, which is of course also regular (as well as the empty language case).
除非我误会了，否则答案是你不能。因为这不是真的。
首先，让我们考虑语言
L = {aa, bb, cc}和字母表{a, b, c}因此，
init(L) = {a ，b，c}。但是，init(L)中的每个元素都不在L中。编辑：如果我们连接空字符，则 init(L) = {a, b, c, aa, bb, cc}。这仍然不等于
L。Unless I'm misunderstanding, the answer is that you can't. Because it's not true.
First, let's consider the language
L = {aa, bb, cc}and alphabet{a, b, c}So,
init(L) = {a, b, c}. However, each of the elemnts ininit(L)are not inL.Edit: If we are concatenating empty characters, then
init(L) = {a, b, c, aa, bb, cc}. Which is still not equal toL.一种语言是正规的，当且仅当有一个有限状态自动机可以识别它。因此，假设 L 是一种常规语言，而 A 是一个能够识别它的自动机。现在，如果有一组可能的转换从 A 开始并以“接受”状态结束，则称 A 的状态是“好”。定义一个新的自动机 A'，其中所有到“好”状态的转换都被直接转换到接受状态所取代。那么A'识别的语言正是init(L)。
A language is regular iff there's a finite-state automaton that recognizes it. So, suppose L is a regular language and let A be an automaton that recognizes it. Now, say that a state of A is "good" if there's some set of possible transitions starting there and ending in the "accept" state. Define a new automaton A' in which all transitions to "good" states are replaced by direct transitions to the accept state. Then the language recognized by A' is exactly init(L).
我认为这是一个新的DFA B，它使A（原始DFA）的所有状态可以达到A的最终状态为B的最终状态。
I think it's a new DFA B, that makes all the state of A(the original DFA) that can reach to the final states of A the final state of B.