Difference between revisions of "NFA to DFA"

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[http://condor.depaul.edu/glancast/444class/docs/nfa2dfa.html Conversion]
 
[http://condor.depaul.edu/glancast/444class/docs/nfa2dfa.html 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?
+
*'''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.  
 
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.  
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q1 -> q0 [label="b"];
 
q1 -> q0 [label="b"];
 
}
 
}
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+
</graphviz>
 
<graphviz caption='DFA'>
 
<graphviz caption='DFA'>
 
digraph DFA {
 
digraph DFA {
 
rankdir=LR;
 
rankdir=LR;
 
node [shape = none] "";
 
node [shape = none] "";
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node [shape = doublecircle] q1, q12;
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node [shape = doublecircle] q1;
 
node [shape = circle];
 
node [shape = circle];
  
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q1 -> q0 [label="b"];
 
q1 -> q0 [label="b"];
 
}
 
}
 +
</graphviz>

Revision as of 22:54, 26 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?

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
Retrieved from "https://gatecse.in/w/index.php?title=NFA_to_DFA&oldid=4142"

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

<graphviz caption='NFA'> digraph NFA { rankdir=LR; node [shape = none] ""; node [shape = doublecircle] q1; node [shape = circle];

""-> q0; q0->q1 [label="a"]; q1->q0 [label="a"]; q1->q1 [label="a"]; q1 -> q0 [label="b"]; }

<graphviz caption='DFA'> digraph DFA { rankdir=LR; node [shape = none] ""; node [shape = doublecircle] q1, q12; node [shape = circle];

""-> q0; q0->q1 [label="a"]; q1->q0 [label="a"]; q1->q1 [label="a"]; q1 -> q0 [label="b"]; }