Search results
Results from the WOW.Com Content Network
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 ...
Proof (1) If L {\displaystyle L} is regular, construct a minimal DFA to accept it. Clearly, if x , y {\displaystyle x,y} end up in the same state after running through the DFA, then x ∼ L y {\displaystyle x\sim _{L}y} , thus the number of equivalence classes of ∼ L {\displaystyle \sim _{L}} is at most the number of DFA states, which must be ...
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.
The simplest example: S → aSb S → ab. This grammar generates the language {:}, which is not regular (according to the pumping lemma for regular languages). The special character ε stands for the empty string. By changing the above grammar to S → aSb S → ε
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
Hi Jochen Burghardt, the example of a non-regular language given to satisfy the non-generalised Pumping Lemma does not satisfy it. For example the word abc can be pumped down to bc which is not in the given language. This is a deep flaw in the example that cannot be easily fixed - as it relies on m ≥ 1.
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