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'''1.''' If <math>L</math> and <math>L'</math> are both recursively enumerable, then <math>L</math> is recursive. Why?
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=='''1.''' If $L$ and $L'$ are both recursively enumerable, then $L$ is recursive. Why?==
  
===Solution===
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==={{Template:Author|Arjun Suresh|{{arjunweb}} }}===
  
  
Given <math>L</math> is $RE$. So there is a $TM$, which accepts and halts for all words in <math>L</math>.  
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Given $L$ is $RE$. So there is a $TM$, which accepts and halts for all words in $L$.  
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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Now, if $L'$ is $RE$, then there is a $TM$, which accepts and halts for all words not in $L$.
So, if a word, from <math>L</math> or not from <math>L</math>,is given, 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 $L$ or not from $L$), 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 $L$. Thus, $L$ becomes recursive.
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<!--
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=='''2.''' CYCLE(L) ={xy | yx is in L,L is regular } ==
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Is this statement true or false ?
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==={{Template:Author|Arjun Suresh|{{arjunweb}} }}===
  
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We have a DFA for L, let it be D and have n states. Now we can make a NFA N for L' as follows:
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Our start state will be the final state of D. For every transition on a symbol s from state x to y in D, we will have a transition from y to x. Further we add the following modification to the DFA:
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For every state of N, we append the part of D till that state (all reachable transitions reaching that state). For example, for the start state in N(the final state in D), we'll append the whole D as it is. (This would mean every string accepted by D will be accepted by N also). For the below DFA, the appending is shown for state 1 
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<graphviz>
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digraph finite_state_machine {
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rankdir=LR;
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size="8,5"
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node [shape = doublecircle]; 3;
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node [shape = circle];
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0 -> 1 [ label = "a" ];
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1 -> 2 [ label = "a" ];
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2 -> 3 [ label = "a" ];
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3 -> 2 [ label = "a" ];
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0-> 2 [ label = "b" ];
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1 -> 0 [ label = "b" ];
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2 -> 2 [ label = "b" ];
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3 -> 2 [ label = "b" ];
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}
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</graphviz>
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<graphviz>
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digraph finite_state_machine2 {
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rankdir=LR;
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size="8,5"
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node [shape = doublecircle]; 0,1,2,3;
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node [shape = circle];
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0 -> 1 [ label = "a" ];
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      1 -> 0 [ label = "b" ];
 +
        1 -> 0 [ label = "a" ];
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2 -> 1 [ label = "a" ];
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3 -> 2 [ label = "a" ];
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2 -> 3 [ label = "a" ];
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2-> 0 [ label = "b" ];
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0 -> 1 [ label = "b" ];
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2 -> 2 [ label = "b" ];
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2 -> 3 [ label = "b" ];
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}
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</graphviz>
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-->
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{{Template:FBD}}
  
  
{{Template:FB}}
 
  
 
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[[Category: Theory of Computation]]
 
 
 
 
<disqus/>
 
 
 
[[Category: Automata Theory]]
 

Latest revision as of 00:30, 7 May 2015

1. If $L$ and $L'$ are both recursively enumerable, then $L$ is recursive. Why?

Solution by Arjun Suresh

Given $L$ is $RE$. So there is a $TM$, which accepts and halts for all words in $L$. Now, if $L'$ is $RE$, then there is a $TM$, which accepts and halts for all words not in $L$. So, if a word is given (either from $L$ or not from $L$), 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 $L$. Thus, $L$ 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?

Solution[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, from <math>L</math> or not from <math>L</math>,is given, 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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