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If the strings of a language L can be effectively enumerated in lexicographic (i.e., alphabetic) order, which of the following statements is true ?
- a)L is necessarily finite
- b)L is regular but not necessarily finite
- c)L is context free but not necessarily regular
- d)L is recursive but not necessarily context free
Correct answer is option 'D'. Can you explain this answer?
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Verified Answer
If the strings of a language L can be effectively enumerated in lexico...
The strings of a language L can be effectively enumerated means a Turing machine exists for language L which will enumerate all valid strings of the language.
If the string is in lexicographic order then TM will accept the string and halt in the final state.
But, if the string is not lexicographic order then TM will reject the string and halt in non-final state.
Thus, L is recursive language.
We can not construct PDA for language L. So, the given language is not context free.
Thus, option (D) is correct.
Most Upvoted Answer
If the strings of a language L can be effectively enumerated in lexico...
Explanation:
To understand why option D is true, let's go through the other options one by one.
a) L is necessarily finite:
If the language L can be effectively enumerated in lexicographic order, it means that all the strings in L can be listed in alphabetic order. However, this does not imply that the language is finite. It is possible to have infinite languages that can be lexicographically ordered. For example, the language L = {a, aa, aaa, aaaa, ...} is infinite but can still be enumerated in lexicographic order.
b) L is regular but not necessarily finite:
If the language L can be effectively enumerated in lexicographic order, it means that there exists an algorithm to list all the strings in L. However, this does not imply that L is regular. There are regular languages that cannot be effectively enumerated, such as the language L = {a^n b^n | n >= 0}. This language is not lexicographically enumerable because the strings in L are not in lexicographic order.
c) L is context-free but not necessarily regular:
Similar to option b, the fact that L can be effectively enumerated in lexicographic order does not imply that L is context-free. There are context-free languages that cannot be effectively enumerated, such as the language L = {ww^R | w is a string of 0s and 1s}. This language is not lexicographically enumerable because the strings in L are not in lexicographic order.
d) L is recursive but not necessarily context-free:
This option is true. If the language L can be effectively enumerated in lexicographic order, it means that there exists an algorithm to list all the strings in L. This algorithm can be implemented using a Turing machine, which makes L a recursive language. Recursive languages are more powerful than context-free languages, so L can be recursive but not necessarily context-free.
Therefore, the correct answer is option d) L is recursive but not necessarily context-free.
To understand why option D is true, let's go through the other options one by one.
a) L is necessarily finite:
If the language L can be effectively enumerated in lexicographic order, it means that all the strings in L can be listed in alphabetic order. However, this does not imply that the language is finite. It is possible to have infinite languages that can be lexicographically ordered. For example, the language L = {a, aa, aaa, aaaa, ...} is infinite but can still be enumerated in lexicographic order.
b) L is regular but not necessarily finite:
If the language L can be effectively enumerated in lexicographic order, it means that there exists an algorithm to list all the strings in L. However, this does not imply that L is regular. There are regular languages that cannot be effectively enumerated, such as the language L = {a^n b^n | n >= 0}. This language is not lexicographically enumerable because the strings in L are not in lexicographic order.
c) L is context-free but not necessarily regular:
Similar to option b, the fact that L can be effectively enumerated in lexicographic order does not imply that L is context-free. There are context-free languages that cannot be effectively enumerated, such as the language L = {ww^R | w is a string of 0s and 1s}. This language is not lexicographically enumerable because the strings in L are not in lexicographic order.
d) L is recursive but not necessarily context-free:
This option is true. If the language L can be effectively enumerated in lexicographic order, it means that there exists an algorithm to list all the strings in L. This algorithm can be implemented using a Turing machine, which makes L a recursive language. Recursive languages are more powerful than context-free languages, so L can be recursive but not necessarily context-free.
Therefore, the correct answer is option d) L is recursive but not necessarily context-free.
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If the strings of a language L can be effectively enumerated in lexicographic (i.e., alphabetic) order, which of the following statements is true ?a)L is necessarily finiteb)L is regular but not necessarily finitec)L is context free but not necessarily regulard)L is recursive but not necessarily context freeCorrect answer is option 'D'. Can you explain this answer?
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