pumping-lemma

I saw this question , and was curious as to what the pumping lemma was ( Wikipedia didn't help much). I understand that it's basically a theoretical proof that must be true in order for a language to be in a certain class, but beyond that I don't really get it. Anyone care to try to explain it at a fairly granular level in a way understandable by non mathematicians/comp sci doctorates? Graphics Noob 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

I have the following problem: Languages L1 = {a^n * b^n : n>=0} and L2 = {b^n * a^n : n>=0} are context free languages so they are closed under the L1L2 so L={a^n * b^2n A^n : n>=0} must be context free too because it is generated by a closure property. I have to prove if this is true or not. So I checked the L language and I don’t think that it is context free then I also saw that L2 is just L1 reversed. Do I have to check if L1, L2 are deterministic? L1={a n b n : n>=0} and L2={b n a n : n>=0} are both context free. Since context-free languages are closed under concatenation, L3=L1L2 is also

Using pumping lemma, we can easily prove that the language L1 = {WcW^R|W ∈ {a,b}*} is not a regular language . (the alphabet is {a,b,c}; W^R represents the reverse string W) However, If we replace character c with "x"(x ∈ {a,b}+) , say, L2 = {WxW^R| x, W ∈ {a,b}^+} , then L2 is a regular language . Could you give me some ideas? Grijesh Chauhan If we replace character c with x where (x ∈ {a,b} + ), say, L2 = {WXW R | x, W ∈ {a,b} + }, then L2 is a regular language. Yes, L2 is Regular Language :). You can write regular expression for L2 too. Language L2 = {WXW R | x, W ∈ {a,b} + } means: string

So this is not about the pumping lemma and how it works, it's about a pre-condition. Everywhere in the net you can read, that regular languages must pass the pumping lemma, but noweher anybody talks about finite languages, which actually are a part of regular languages. So we might all aggree, that the following language is a finite language as well as it's a regular one, but it definitely does not pass the pumping lemma: L = {'abc', 'defghi'} Please, tell me if simply no one writes about it or why we're wrong - or even not. The reason that finite languages work with the pumping lemma is

I have a little confusion in checking whether the given language is regular or not using pumping lemma. Suppose we have to check whether: L. The language accepting even number of 0 's in regular or not? We know that it is regular because we can construct a DFA for L. But I want to prove this with pumping lemma. Now suppose, I take a String w= "0000" : Now will divide the string as x = 0 , y = 0 , and z = 00 . Now on applying pumping lemma for i = 2 , I will get the string "00000" , which is not present in my language so by pumping lemma its prove that the language is not regular. But it is