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*'''Q:''' After converting from  NFA to DFA and  we get $2^n$ states. Now we apply minimization- how many states can we get after this?  
 
*'''Q:''' After converting from  NFA to DFA and  we get $2^n$ states. Now we apply minimization- how many states can we get after this?  
In the worst case this can remain $2^n$ - that is no minimization can be possible. We can see this in the following example  
+
A: In the worst case this can remain $2^n$ - that is no minimization can be possible. We can see this in the following example  
 
+
===NFA===
 
<graphviz caption='NFA'>
 
<graphviz caption='NFA'>
 
digraph NFA {
 
digraph NFA {
 
rankdir=LR;
 
rankdir=LR;
 
node [shape = none] "";
 
node [shape = none] "";
node [shape = doublecircle] q1;
+
node [shape = doublecircle] q2;
 
node [shape = circle];
 
node [shape = circle];
  
""-> q0;
+
""-> q1;
q0->q1 [label="a"];
 
q1->q0 [label="a"];
 
 
q1->q1 [label="a"];
 
q1->q1 [label="a"];
q1 -> q0 [label="b"];
+
q1->q2 [label="a"];
 +
q2 -> q2 [label="b"];
 
}
 
}
 
</graphviz>
 
</graphviz>
 +
 +
===DFA===
 
<graphviz caption='DFA'>
 
<graphviz caption='DFA'>
 
digraph DFA {
 
digraph DFA {
 
rankdir=LR;
 
rankdir=LR;
 
node [shape = none] "";
 
node [shape = none] "";
node [shape = doublecircle] q1, q12;
+
node [shape = doublecircle] q2, q12;
 
node [shape = circle];
 
node [shape = circle];
  
""-> q0;
+
""-> q1;
q0->q1 [label="a"];
 
q0->q3 [label="b"];
 
 
q1->q12 [label="a"];
 
q1->q12 [label="a"];
q1->q0 [label="b"];
+
q12->q12 [label="a"];
q12 -> q0 [label="b"];
+
q12->q2 [label="b"];
 +
q2->q2 [label="b"];
 +
q2 -> q3 [label="a"];
 +
q1 -> q3 [label="b"];
 
}
 
}
 
</graphviz>
 
</graphviz>

Latest revision as of 21:02, 28 July 2014

Conversion

  • Q: I know that $2^n$ are the maximum states in DFA for an $n$ state NFA. But what about the minimum no of states in a DFA for an $n$ state NFA?

A: I would say 1 because the $n$ states can be connected through $ε$ moves and hence all of these will be combines to a single start state in DFA.

  • Q: After converting from NFA to DFA and we get $2^n$ states. Now we apply minimization- how many states can we get after this?

A: In the worst case this can remain $2^n$ - that is no minimization can be possible. We can see this in the following example

NFA

NFA

DFA

DFA

Conversion

  • Q: I know that $2^n$ are the maximum states in DFA for an $n$ state NFA. But what about the minimum no of states in a DFA for an $n$ state NFA?

A: I would say 1 because the $n$ states can be connected through $ε$ moves and hence all of these will be combines to a single start state in DFA.

  • Q: After converting from NFA to DFA and we get $2^n$ states. Now we apply minimization- how many states can we get after this?

In the worst case this can remain $2^n$ - that is no minimization can be possible. We can see this in the following example

NFA

DFA