Difference between revisions of "Automata Doubts"

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Given <math>L</math> is $RE$. So there is a $TM$, which accepts and halts for all words in <math>L</math>.  
 
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>.
 
Now, if <math>L'</math> is $RE$, then there is a $TM$, which accepts and halts for all words not in <math>L</math>.
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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.
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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 it is from $L$, the first $TM$ will halt and we say it belongs to $L$. If it is 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.
 
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=='''2.''' CYCLE(L) ={xy | yx is in L,L is regular } ==
 
=='''2.''' CYCLE(L) ={xy | yx is in L,L is regular } ==

Revision as of 12:51, 14 June 2014

1. If <math>L</math> and <math>L'</math> are both recursively enumerable, then <math>L</math> is recursive. Why?

Solution by Arjun Suresh

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 it is from $L$, the first $TM$ will halt and we say it belongs to $L$. If it is 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.



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1. If <math>L</math> and <math>L'</math> are both recursively enumerable, then <math>L</math> is recursive. Why?[edit]

Solution by Arjun Suresh[edit]

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 it is from $L$, the first $TM$ will halt and we say it belongs to $L$. If it is 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.



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