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Which among the following cannot be accepted by a regular grammar?
- a)L is a set of numbers divisible by 2
- b)L is a set of binary complement
- c)L is a set of string with odd number of 0
- d)L is a set of 0n1n
Correct answer is option 'D'. Can you explain this answer?
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Which among the following cannot be accepted by a regular grammar?a)L ...
There exists no finite automata to accept the given language i.e. 0n1n. For other options, it is possible to make a dfa or nfa representing the language set.
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Which among the following cannot be accepted by a regular grammar?a)L ...
Regular Grammar and 0^n1^n
Regular grammars are a type of formal grammar used in formal language theory. They consist of a set of production rules that describe all possible strings in a given formal language. Regular grammars can generate regular languages, which are a type of formal language that can be recognized by a finite automaton.
0^n1^n
The language L = {0^n1^n | n ≥ 0} consists of all strings of the form 0^n1^n, where n is a non-negative integer. In other words, the number of 0s in the string is equal to the number of 1s in the string.
Why 0^n1^n cannot be accepted by a regular grammar?
- Regular grammars are limited in their ability to handle languages with nested structures or balancing properties, such as the language 0^n1^n.
- In the language 0^n1^n, each 0 must be followed by a corresponding 1, which requires some form of memory or counting mechanism to keep track of the number of 0s.
- Regular grammars do not have the ability to perform this kind of counting or matching of nested symbols, as they are equivalent in power to finite automata, which have limited memory capabilities.
- As a result, the language 0^n1^n cannot be described by a regular grammar, and more powerful formalisms such as context-free grammars or pushdown automata are required to generate or recognize this language.
In conclusion, the language 0^n1^n cannot be accepted by a regular grammar due to its nested structure and the counting requirements for matching 0s and 1s.
Regular grammars are a type of formal grammar used in formal language theory. They consist of a set of production rules that describe all possible strings in a given formal language. Regular grammars can generate regular languages, which are a type of formal language that can be recognized by a finite automaton.
0^n1^n
The language L = {0^n1^n | n ≥ 0} consists of all strings of the form 0^n1^n, where n is a non-negative integer. In other words, the number of 0s in the string is equal to the number of 1s in the string.
Why 0^n1^n cannot be accepted by a regular grammar?
- Regular grammars are limited in their ability to handle languages with nested structures or balancing properties, such as the language 0^n1^n.
- In the language 0^n1^n, each 0 must be followed by a corresponding 1, which requires some form of memory or counting mechanism to keep track of the number of 0s.
- Regular grammars do not have the ability to perform this kind of counting or matching of nested symbols, as they are equivalent in power to finite automata, which have limited memory capabilities.
- As a result, the language 0^n1^n cannot be described by a regular grammar, and more powerful formalisms such as context-free grammars or pushdown automata are required to generate or recognize this language.
In conclusion, the language 0^n1^n cannot be accepted by a regular grammar due to its nested structure and the counting requirements for matching 0s and 1s.
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Which among the following cannot be accepted by a regular grammar?a)L is a set of numbers divisible by 2b)L is a set of binary complementc)L is a set of string with odd number of 0d)L is a set of 0n1nCorrect answer is option 'D'. Can you explain this answer?
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