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The pumping lemma is often used to prove that a particular language is non-regular: a proof by contradiction may consist of exhibiting a string (of the required length) in the language that lacks the property outlined in the pumping lemma. Example: The language = {:} over the alphabet = {,} can be shown to be non-regular as follows:
The pumping lemma for context-free languages (called just "the pumping lemma" for the rest of this article) describes a property that all context-free languages are guaranteed to have. The property is a property of all strings in the language that are of length at least p {\displaystyle p} , where p {\displaystyle p} is a constant—called the ...
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Ogden's lemma is often stated in the following form, which can be obtained by "forgetting about" the grammar, and concentrating on the language itself: If a language L is context-free, then there exists some number (where p may or may not be a pumping length) such that for any string s of length at least p in L and every way of "marking" p or more of the positions in s, s can be written as
To convert a grammar to Chomsky normal form, a sequence of simple transformations is applied in a certain order; this is described in most textbooks on automata theory. [4]: 87–94 [5] [6] [7] The presentation here follows Hopcroft, Ullman (1979), but is adapted to use the transformation names from Lange, Leiß (2009).
1.1 Pumping Lemma. 2 comments. 1.2 Analytic injective functions on the extended complex plane. 3 comments. 1.3 Lie Groups and Infinitesimal Actions. 4 comments.
Pumping lemma for context-free languages, the fact that all sufficiently long strings in such a language have a pair of substrings that can be repeated arbitrarily many times, usually used to prove that certain languages are not context-free; Pumping lemma for indexed languages; Pumping lemma for regular tree languages
The proof is essentially the same as the standard pumping lemma: use the pigeonhole principle to find copies of some nonterminal symbol in the longest path in the shortest derivation tree. Now we prove the first part of Parikh's theorem, making use of the above lemma.