Arjun Suresh (talk | contribs) |
Arjun Suresh (talk | contribs) |
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Since, the strings of <math>L</math> can be enumerated it means <math>L</math> is recursively enumerable. That is we have a <math>TM</math> which accepts all strings in <math>L</math>. Now, to be recursive the <math>TM</math> should reject all strings not in <math>L</math>. Since, the strings of the language can be enumerated in lexicographic order, it's easy to do this. For any word <math>w</math>, if we see a word in the enumeration which is lexicographically higher than <math>w</math> but no <math>w</math>, it means <math>w</math> is not in the language. This makes <math>L</math> '''recursive'''. | Since, the strings of <math>L</math> can be enumerated it means <math>L</math> is recursively enumerable. That is we have a <math>TM</math> which accepts all strings in <math>L</math>. Now, to be recursive the <math>TM</math> should reject all strings not in <math>L</math>. Since, the strings of the language can be enumerated in lexicographic order, it's easy to do this. For any word <math>w</math>, if we see a word in the enumeration which is lexicographically higher than <math>w</math> but no <math>w</math>, it means <math>w</math> is not in the language. This makes <math>L</math> '''recursive'''. | ||
| − | Now, why | + | Now, why <math>L</math> need not be context free? Consider <math>L = \{a^nb^nc^n | n\ge 0\}</math>. The strings of this language can be enumerated in lexicographic order. But we know <math>L</math> is not context free as no <math>PDA</math> can accept <math>L</math>. |
If the strings of a language <math>L</math> can be effectively enumerated in lexicographic (i.e., alphabetic) order, which of the following statements is true?
(A) <math>L</math> is necessarily finite
(B) <math>L</math> is regular but not necessarily finite
(C) <math>L</math> is context free but not necessarily regular
(D) <math>L</math> is recursive but not necessarily context free
Since, the strings of <math>L</math> can be enumerated it means <math>L</math> is recursively enumerable. That is we have a <math>TM</math> which accepts all strings in <math>L</math>. Now, to be recursive the <math>TM</math> should reject all strings not in <math>L</math>. Since, the strings of the language can be enumerated in lexicographic order, it's easy to do this. For any word <math>w</math>, if we see a word in the enumeration which is lexicographically higher than <math>w</math> but no <math>w</math>, it means <math>w</math> is not in the language. This makes <math>L</math> recursive.
Now, why <math>L</math> need not be context free? Consider <math>L = \{a^nb^nc^n | n\ge 0\}</math>. The strings of this language can be enumerated in lexicographic order. But we know <math>L</math> is not context free as no <math>PDA</math> can accept <math>L</math>.
If the strings of a language <math>L</math> can be effectively enumerated in lexicographic (i.e., alphabetic) order, which of the following statements is true?
(A) <math>L</math> is necessarily finite
(B) <math>L</math> is regular but not necessarily finite
(C) <math>L</math> is context free but not necessarily regular
(D) <math>L</math> is recursive but not necessarily context free
Since, the strings of <math>L</math> can be enumerated it means <math>L</math> is recursively enumerable. That is we have a <math>TM</math> which accepts all strings in <math>L</math>. Now, to be recursive the <math>TM</math> should reject all strings not in <math>L</math>. Since, the strings of the language can be enumerated in lexicographic order, it's easy to do this. For any word <math>w</math>, if we see a word in the enumeration which is lexicographically higher than <math>w</math> but no <math>w</math>, it means <math>w</math> is not in the language. This makes <math>L</math> recursive.
Now, why <math>L</math> need not be context free? Consider <math>L = \{a^nb^nc^n | n\ge 0\}</math>. The strings of this language can be enumerated in lexicographic order. But we know <math>L</math> is not context free as no <math>PDA</math> can accept <math>L</math>.
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