Arjun Suresh (talk | contribs) |
Arjun Suresh (talk | contribs) |
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} | } | ||
</graphviz> | </graphviz> | ||
− | + | ==h== | |
+ | <graphviz> | ||
+ | digraph finite_state_machine { | ||
+ | rankdir=LR; | ||
+ | size="8,5" | ||
+ | node [shape = doublecircle]; LR_0; | ||
+ | node [shape = circle]; | ||
+ | LR_1 -> LR_0 [ label = "a" ]; | ||
+ | |||
+ | } | ||
+ | </graphviz> | ||
{{Template:FBD}} | {{Template:FBD}} | ||
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,