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==={{Template:Author|Arjun Suresh|{{arjunweb}} }}===
 
==={{Template:Author|Arjun Suresh|{{arjunweb}} }}===
  
We have a DFA for L, let it be D and have n states. Now we can make a NFA for L' as follows:
+
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:
 
+
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:
For every state of D,  
+
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 
 
<graphviz>
 
<graphviz>
 
digraph finite_state_machine {
 
digraph finite_state_machine {
 
rankdir=LR;
 
rankdir=LR;
 
size="8,5"
 
size="8,5"
node [shape = doublecircle]; LR_3;
+
node [shape = doublecircle]; 3;
 
node [shape = circle];
 
node [shape = circle];
LR_0 -> LR_1 [ label = "a" ];
+
0 -> 1 [ label = "a" ];
LR_1 -> LR_2 [ label = "a" ];
+
1 -> 2 [ label = "a" ];
LR_2 -> LR_3 [ label = "a" ];
+
2 -> 3 [ label = "a" ];
LR_3 -> LR_2 [ label = "a" ];
+
3 -> 2 [ label = "a" ];
LR_0-> LR_2 [ label = "b" ];
+
0-> 2 [ label = "b" ];
LR_1 -> LR_0 [ label = "b" ];
+
1 -> 0 [ label = "b" ];
LR_2 -> LR_2 [ label = "b" ];
+
2 -> 2 [ label = "b" ];
LR_3 -> LR_2 [ label = "b" ];
+
3 -> 2 [ label = "b" ];
 
}
 
}
 
</graphviz>
 
</graphviz>
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node [shape = doublecircle]; LR_0;
 
node [shape = doublecircle]; LR_0;
 
node [shape = circle];
 
node [shape = circle];
LR_1 -> LR_0 [ label = "a" ];
+
0 -> 1 [ label = "a" ];
 +
      1 -> 0 [ label = "b" ];
 
 
 
}
 
}

Revision as of 15:12, 24 May 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 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.

2. CYCLE(L) ={xy | yx is in L ,L is regular }

Is this statement true or false ?

Solution by Arjun Suresh

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: 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: 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

This is a graph with borders and nodes. Maybe there is an Imagemap used so the nodes may be linking to some Pages.

h

This is a graph with borders and nodes. Maybe there is an Imagemap used so the nodes may be linking to some Pages.



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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 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.

2. CYCLE(L) ={xy | yx is in L ,L is regular }[edit]

Is this statement true or false ?

Solution by Arjun Suresh[edit]

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: 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: 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

This is a graph with borders and nodes. Maybe there is an Imagemap used so the nodes may be linking to some Pages.

h[edit]

This is a graph with borders and nodes. Maybe there is an Imagemap used so the nodes may be linking to some Pages.



blog comments powered by Disqus