Search results
The first thing to note is that a regular language is a formal language, with some restrictions. A formal language is essentially a (possibly infinite) collection of strings. For example, the formal language Java is the collection of all possible Java files, which is a subset of the collection of all possible text files.
May 7, 2016 · A language is regular if and only if there is a regular expression for it. The "can be obtained from finite languages by applying the three operations union, concatenation, repetition a finite number of times" part is essentially a quick verbal definition of a regular expression. Usually, a regular expression (RE) is formally defined starting ...
Then, to accept the reverse language, all we need to do is reverse the direction of the transitions, change the start state to an accept state, and the accept state to the start state. Then we have a machine that is "backwards" compared to the original, and accepts the language LR. Share. Cite. Improve this answer.
Sep 13, 2016 · If a language is indeed regular that means there is an FA that accepts it. The complement of L is just the language of all words that are not in L. Thanks to Rick Decker for mentioning in the comments that this only works for FAs that are deterministic and to D.W for correcting the answer.
1. Pumping lemma can be used to disprove a language is not regular, Example: Let L = {a^mb^m | m ≥ 1} Then L is not regular. Proof: Assume L is regular. Let n be as in Pumping Lemma. Let w = a^nb^n. Let w = xyz be as in Pumping Lemma.
Jan 25, 2013 · Don't mix WXW R with WCW R, its X with + that makes language regular. Think by including X that is (a + b)* we can have finite choice for W that is a and b (finite is regular). Language WXW R can be say: if start with a ends with a and if start with b end with b. so correspondingly we need two final states. Q6 if W is a; Q5 if W is b; ITs DFA ...
My attempt on a proof: Yes, for contradiction assume that A is not regular. Then A2 = A ⋅ A. Since concatenation of two non-regular language is not regular A2 cannot be regular. This contradicts our assumption. So A is regular. So if A2 is regular then A is regular.
Nov 14, 2015 · The first of those questions asks about the intersection of a specific regular language with a specific non-regular language. The second is completely unrelated and I honestly don't know why you're even mentioning it. The third covers only intersection and my answer to it isn't particularly general. None of this justifies your shouty boldface ...
Feb 18, 2009 · Regular grammar is either right or left linear, whereas context free grammar is basically any combination of terminals and non-terminals. Hence you can see that regular grammar is a subset of context-free grammar. So for a palindrome for instance, is of the form, S->ABA. A->something. B->something.
6. In normal regular expression grammar, (a+b)* means zero or more of any sequence that start with a, then have zero or more a, then a b. This discounts things like baa (it doesn't start with a), abba, and a (there must be one exactlyb after each a group), so is not correct. (a*b*)* means zero or more of any sequence that contain zero or more a ...
