Difference between revisions of "Automata Doubts"

Revision as of 14:58, 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 for L' as follows:

For every state of D,

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

@@ Line 19: / Line 19: @@
  rankdir=LR;
  size="8,5"
- node [shape = doublecircle]; LR_0 LR_3 LR_4 LR_8;
+ node [shape = doublecircle]; LR_4;
  node [shape = circle];
- LR_0 -> LR_2 [ label = "SS(B)" ];
+ LR_0 -> LR_1 [ label = "a" ];
- LR_0 -> LR_1 [ label = "SS(S)" ];
+ LR_1 -> LR_2 [ label = "a" ];
- LR_1 -> LR_3 [ label = "S($end)" ];
+ LR_2 -> LR_3 [ label = "a" ];
- LR_2 -> LR_6 [ label = "SS(b)" ];
+ LR_3 -> LR_4 [ label = "a" ];
- LR_2 -> LR_5 [ label = "SS(a)" ];
+ LR_2 -> LR_5 [ label = "b" ];
- LR_2 -> LR_4 [ label = "S(A)" ];
+ LR_0-> LR_2 [ label = "b" ];
- LR_5 -> LR_7 [ label = "S(b)" ];
+ LR_1 -> LR_0 [ label = "b" ];
- LR_5 -> LR_5 [ label = "S(a)" ];
+ LR_2 -> LR_2 [ label = "b" ];
- LR_6 -> LR_6 [ label = "S(b)" ];
+ LR_3 -> LR_2 [ label = "b" ];
- LR_6 -> LR_5 [ label = "S(a)" ];
+ LR_4 -> LR_4 [ label = "b" ];
- LR_7 -> LR_8 [ label = "S(b)" ];
+ LR_4 -> LR_0 [ label = "a" ];
- LR_7 -> LR_5 [ label = "S(a)" ];
- LR_8 -> LR_6 [ label = "S(b)" ];
- LR_8 -> LR_5 [ label = "S(a)" ];
 }
 </graphviz>

Difference between revisions of "Automata Doubts"

Revision as of 14:58, 24 May 2014

Contents

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

Solution by Arjun Suresh

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

Solution by Arjun Suresh

Contents

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]

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

Solution by Arjun Suresh[edit]