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Can a non-regular language have a regular expression?
Example 1 – All strings of length = 2 over {a, b}* i.e. L = {aa, ab, ba, bb} is regular. Given an expression of non-regular language, but the value of parameter is bounded by some constant, then the language is regular (means it has kind of finite comparison).
Can the intersection of two non-regular languages be regular?
(a) Union of two non-regular languages cannot be regular. Now L1 is regular (since regular languages are closed under complementation). Since, L1 is regular, hence its intersection with L i.e L1 ∩ L = L2 is regular (since regular languages are closed under intersection). Therefore, L2 is regular.
How can a language be regular?
A language is a regular language if there is a finite automaton that recognizes it. For example, this machine recognizes the language of strings that have an even number of zeroes since any string that has an even number of zeroes will go from the start state to an accepting state.
Is it possible for a language L to be regular while L ∗ is not regular?
Suppose L is any language over the alphabet Σ. If L is not regular then so is L+Σ, yet (L+Σ)∗=Σ∗ is regular. So you can see that L* is not always not regular.
What are non regular languages?
Definition: A language that cannot be defined by a regular expression is a nonregular language or an irregular language. 2.
How do you prove non regular language?
To prove that a language L is not regular, we use proof by contradiction.
What is the intersection of two non-regular languages?
2 Answers. Suppose that the two non-regular languages are distinct and have no strings in common. The intersection of these 2 languages will be the empty set, since no string exists in both languages. The empty set is a regular language, so this can happen sometimes.
Are non-regular languages closed under Kleene star?
The languages computed by this model are closed under union, concatenation, and star. A convenient syntax, Regular expressions, describe exactly the same languages that DFAs (and NFAs) recognize.
Is a * regular language?
Regular Expressions are used to denote regular languages. If a and b are regular expression, a + b is also a regular expression with language {a,b}. If a and b are regular expression, ab (concatenation of a and b) is also regular. If a is regular expression, a* (0 or more times a) is also regular.
Is L * a regular language?
However, the language L* is the language a*, which is regular. To see this, notice that since L contains the string a, the language L* contains all strings of the form an for any natural number n.
What makes a language a non-regular language?
Nevertheless, the point of establishing non-regular languages is not so much one of existence, but of illustrating that certain languages which are “computable” in some sense are not regular. What is it about this language which makes it non-regular? Imagine processing a string w ∈ L.
How to prove the non regularity of a language?
The proof of non-regularity of a language using the pumping lemma is a proof by contradiction. The goal is to assume that the language is regular and then derive strings which are not in the language, thereby contradicting the regularity assumption. This language is not regular :
How to write q in a non-regular language?
Consider the N+1 states {q0,q1,…,qN} related to the first N transitions. Here is a depiction: By the pigeonhole principle, at least two are equal: qi = qj, where 0 ≤ i < j ≤ N . Then write w = xyz where The substring y makes the transition from qi to qj, where i < j, and so |y| > 0 . Write q = qi = qj as the common value.
Why is L ( ababbababb ) a non-regular language?
I have found out that L ( ababbababb ) is non-regular due to “ababb” is a repetition that cannot be stored by the FSM due to its limited memory. -> based https://www.youtube.com/watch?v=WrzaPNj9OZ4 However, all finite languages are regular, isnt L (ababbababb) supposed to be regular then?