Table of Contents
Can NFA accept non-regular languages?
A nondeterministic finite automaton (NFA), or nondeterministic finite-state machine, does not need to obey these restrictions. Using the subset construction algorithm, each NFA can be translated to an equivalent DFA; i.e., a DFA recognizing the same formal language. Like DFAs, NFAs only recognize regular languages.
Can a DFA accept a non-regular language?
The DFA will either accept a string not in the language or reject a string in the language, which it shouldn’t be able to do. Can’t place all these strings into different states; there are only finitely many states! Theorem: The language L = { anbn | n ∈ ℕ } is not regular.
What should be done in order to prove a language is non-regular?
(*) for any string x ∈ L where |x| ≥ n, there are strings u, v, w such that (i) x = uvw, (ii) v = ǫ, (iii) |uv| ≤ n, (iv) uvkw ∈ L for all k ∈ N. To prove that a language L is not regular, we use proof by contradiction.
Can a regular expression generate a non-regular language?
So, yes, all regular expressions define regular languages, and every regular language is the set of strings that match some regular expression.
Is there a language accepted by NFA but not by DFA?
Deterministic Finite Automata are strictly weaker class than Non-deterministic Finite Automata (NFAs), i.e., there exists a language that is accepted by an NFA but is not accepted by any DFA. It is known that context-free languages are not closed under intersection.
What language does the following NFA accept?
The language accepted by an NFA < Q, , q0, , A > is the set of strings that are accepted by the NFA. Some of the strings accepted by the NFA given above are , a, ab, aaa, abbbb etc. and the language it accepts is a*( ab + a + ba )(bb)* .
Are non regular languages countable?
The existence of non-regular languages is guaranteed by the fact that the regular languages of any alphabet are countable, and we know that the set of all subsets of strings is not countable.
What makes a language non regular?
Definition: A language that cannot be defined by a regular expression is a nonregular language or an irregular language.
Are non regular languages Context free?
A context-free language is a language generated by a context-free grammar. Here is an example of a language that is not regular (proof here) but is context-free: { a n b n ∣ n ≥ 0 } \{a^nb^n | n \geq 0\} {anbn∣n≥0}. This is the language of all strings that have an equal number of a’s and b’s.
How can you prove a language is not regular using closure properties?
Once we have some languages that we can prove are not regular, such as anbn, we can use the closure properties of regular languages to show that other languages are also not regular. L = {w : w contains an equal number of a’s and b’s } a*b* is regular. So, if L is regular, then L1 = L ∩ a*b* is regular.
How do you prove a language is not context free?
An expression that doesn’t form a pattern on which linear comparison could be carried out using stack is not context free language. Example 1 – L = { a^m b^n^2 } is not context free. Example 2 – L = { a^n b^2^n } is not context free. Example 3 – L = { a^n^2 } is not context free.
Can a regular language be infinite?
Now, when a language is infinite (it has an infinite number of strings), it might be any type of a language (regular or not) in the Chomsky hierarchy.
Is empty language regular?
Formal definition The empty language Ø is a regular language. For each a ∈ Σ (a belongs to Σ), the singleton language {a} is a regular language. If A is a regular language, A* (Kleene star) is a regular language. Due to this, the empty string language {ε} is also regular.
Is 0 * a regular language?
Yes, Language {an an | n >= 0} is a regular language. To proof that certain language is regular, you can draw its dfa/regular expression.
Is a regular expression a regular language?
Regular expressions are used to denote regular languages. They can represent regular languages and operations on them succinctly. The set of regular expressions over an alphabet is defined recursively as below. Any element of that set is a regular expression.
What is the relationship between NFA accepted languages and DFA languages?
What is the relation between NFA-accepted languages and DFA accepted languages? Explanation: The no of languages accepted by NFA and DFA is equal.
Can a NFA recognize the WW language?
Answer: The NFA M below recognizes the language C = { w ∈ Σ∗ | w ends with 00 }, where Σ = {0, 1}. Note that M′ accepts the string 100 ∈ C = { w | w does not end with 00 }, so M′ does not recognize the language C. Since every NFA has an equivalent DFA (Theorem 1.39), there is a DFA D such that L(D) = L(M) = C.
Are all formal languages regular?
Note that all finite languages are regular, but not all regular languages are finite; our double-0 language contains an infinite number of words ( 007 , 008 , but also 004242 and 0012345 ), but can be tested with constant memory: To test whether a word belongs in it, check whether the first symbol is 0 , and whether Jul 16, 2011.
How do you create a NFA language?
It is easy to construct an NFA than DFA for a given regular language. The finite automata are called NFA when there exist many paths for specific input from the current state to the next state. Every NFA is not DFA, but each NFA can be translated into DFA.Example 1: Q = {q0, q1, q2} ∑ = {0, 1} q0 = {q0} F = {q2}.
Which of the technique can be used to prove that a language is non regular Ardens theorem pumping lemma Ogden S Lemma none of the mentioned?
Which of the technique can be used to prove that a language is non regular? Explanation: We use the powerful technique called Pumping Lemma, for showing certain languages not to be regular. We use Ardens theorem to find out a regular expression out of a finite automaton.
