Arjun Suresh (talk | contribs) |
Arjun Suresh (talk | contribs) |
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LR_3 -> LR_2 [ label = "b" ]; | LR_3 -> LR_2 [ label = "b" ]; | ||
} | } | ||
| + | </graphviz> | ||
| + | <graphviz> | ||
digraph finite_state_machine { | digraph finite_state_machine { | ||
rankdir=LR; | rankdir=LR; | ||
Given <math>L</math> is $RE$. So there is a $TM$, which accepts and halts for all words in <math>L</math>. Now, if <math>L'</math> is $RE$, then there is a $TM$, which accepts and halts for all words not in <math>L</math>. So, if a word is given (either from <math>L</math> or not from <math>L</math>), give it to both those $TM$s. If its from $L$, the first $TM$ will halt and we say it belongs to $L$. If its not from $L$, the second one will halt and we say it doesn't belong to <math>L</math>. Thus, <math>L</math> becomes recursive.
Is this statement true or false ?
We have a DFA for L, let it be D and have n states. Now we can make a NFA for L' as follows:
For every state of D,
Given <math>L</math> is $RE$. So there is a $TM$, which accepts and halts for all words in <math>L</math>. Now, if <math>L'</math> is $RE$, then there is a $TM$, which accepts and halts for all words not in <math>L</math>. So, if a word is given (either from <math>L</math> or not from <math>L</math>), give it to both those $TM$s. If its from $L$, the first $TM$ will halt and we say it belongs to $L$. If its not from $L$, the second one will halt and we say it doesn't belong to <math>L</math>. Thus, <math>L</math> becomes recursive.
Is this statement true or false ?
We have a DFA for L, let it be D and have n states. Now we can make a NFA for L' as follows:
For every state of D,
GATECSE