Yes, a regular language can be infinite.
Yes, it is true that if a language is undecidable, then it must be infinite.
No, it is not necessarily true that if language A is regular and language B reduces to A, then language B is also regular.
The complement of a regular language is regular because regular languages are closed under complementation. This means that if a language is regular, its complement is also regular.
The complement of a regular language is the set of all strings that are not in the original language. In terms of regular expressions, the complement of a regular language can be represented by negating the regular expression that defines the original language.
No, it is not possible to show that the language recognized by an infinite pushdown automaton is decidable.
Regular languages are not closed under infinite union because while the union of a finite number of regular languages results in a regular language, an infinite union can produce a language that is not regular. For example, the set of languages {a^n | n ≥ 0} for n = 0, 1, 2, ... represents an infinite union of regular languages, but the resulting language {a^n | n ≥ 0} is not regular, as it cannot be recognized by any finite automaton. This is due to the limitations of finite state machines, which cannot handle the potentially unbounded complexity of infinite unions.
Yes, it is true that if a language is undecidable, then it must be infinite.
no
it is not regular language .it is high level language
There are an infinite number of regular polygons.
No, it is not necessarily true that if language A is regular and language B reduces to A, then language B is also regular.
The complement of a regular language is regular because regular languages are closed under complementation. This means that if a language is regular, its complement is also regular.
The complement of a regular language is the set of all strings that are not in the original language. In terms of regular expressions, the complement of a regular language can be represented by negating the regular expression that defines the original language.
No, it is not possible to show that the language recognized by an infinite pushdown automaton is decidable.
The reverse of a regular language is regular because for every string in the original language, there exists a corresponding string in the reversed language that is also regular. This is because regular languages are closed under the operation of reversal, meaning that if a language is regular, its reverse will also be regular.
infinite.
infinitagon