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  • What is the Pumping Lemma in Laymans terms? - Stack Overflow
    The pumping lemma is a simple proof to show that a language is not regular, meaning that a Finite State Machine cannot be built for it The canonical example is the language (a^n)(b^n) This is the simple language which is just any number of a s, followed by the same number of b s So the strings ab aabb aaabbb aaaabbbb etc are in the language
  • Clarification on Regular Language pumping lemma - Stack Overflow
    The pumping lemma says that: if a language L is regular: then there is some positive integer p such that if ω ∈ L and | ω | ≥ p (| ω | is the length of ω) then ω can be pumped (see below) Now, I assert that any finite language satisfies this constraint, without defining what it means to "be pumped" A finite language – that is, a language which only includes a finite number of
  • What exactly is a pumping lemma and how do you do one?
    The pumping lemma (for regular languages; there's a more complicated one for context free languages) is a result about regular languages that says "if L is a regular language, then there's an integer p so that every word in L that's at least as long as p can be divided into three parts x, y, and z so that xz ∈ L, xyz ∈ L, xyyz ∈ L, xyyyz
  • Using Pumping Lemma to prove non regularity of a language
    To answer your questions: The idea is correct You want to use the Pumping Lemma for Regular Languages, and if you can prove that applying the Pumping Lemma to a word of a given language results in a word that is not in the language then you have shown that that language cannot be regular The Pumping Lemma is often used and useful in that sense
  • O que é o Pumping Lemma (ou Lema do Bombeamento) ? E como aplicá-lo?
    Estava lendo pelo HOPCROFT e tive dificuldade em aplicar o lema do bombeamento de maneira formal aos exercícios para provar que uma linguagem não é regular Neste caso, me refiro ao Lema do Bombea
  • How do we choose a good string for the pumping lemma?
    How to Use the Pumping Lemma for Regular Languages When you are given a language L and are using the pumping lemma to prove that it is not regular, do this: : :
  • A detail on the Pumping Lemma for regular languages
    Yes, that's how the pumping lemma works It's only useful for proving languages to not be regular Satisfying the pumping lemma is only a necessary but not a sufficient condition for a language being regular (Nota bene: Likewise for context-free languages and the respective pumping lemma there)
  • To make sure: Pumping lemma for infinite regular languages only?
    The reason that finite languages work with the pumping lemma is because you can make the pumping length longer than the longest word in the language The pumping lemma, as stated on Wikipedia (I don't have my theory of computation book with me) is the following: Let L be a regular language Then there exists an integer p ≥ 1 depending only on L such that every string w in L of length at
  • computational complexity - Understanding the Pumping Lemma . . .
    I have been having an extremely hard time proving a language is irregular using the pumping lemma I looked and dozens of examples and spent hours on this one topic, and I am still not able to wrap
  • pumping lemma - why is {a^nb^n} context-free? - Stack Overflow
    Using pumping lemma you can proof that certain language is not regular (or not context free) but con't proof that language is regular or context free (depends which lemma you applies) Pumping lemma is Sufficient but not necessary condition





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