在上下文无关的语言中抽取引理 [英] Pumping Lemma in context-free languages
问题描述
A = {0^a 1^b 2^c | a < b < c}
我需要证明A不是上下文无关的。我猜我必须为此使用Pumping Lemma,但是如何使用?
I need to show that A is not context-free. I'm guessing I have to use the Pumping Lemma for this, but how?
推荐答案
目标是证明对于任何长度大于等于最小抽运长度的字符串,无法抽运该字符串。也就是说,如果将其拆分为子字符串 uvxyz
,则该字符串是由于复制(或删除副本) v
和 y
仍使用语言 A
。
The goal is to prove that for any string with length >= a minimum pumping length, the string cannot be pumped. That is, if you split it into substrings uvxyz
, the string that results from making copies (or removing copies) of v
and y
are still in language A
.
请注意,您只需要显示该语言中的一个字符串就不能被泵送(只要满足最小泵送长度p)即可。
Note that you only have to show that one string in the language cannot be pumped (as long as it meets the minimum pumping length p)
请考虑该语言及其相关性到 A
:
Consider this language and how it relates to A
:
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