Difference between revisions of "NFA to DFA"

Line 26: Line 26:
 
rankdir=LR;
 
rankdir=LR;
 
node [shape = none] "";
 
node [shape = none] "";
−
node [shape = doublecircle] q1, $q_12$;
+
node [shape = doublecircle] q1, q12;
 
node [shape = circle];
 
node [shape = circle];
  
 
""-> q0;
 
""-> q0;
 
q0->q1 [label="a"];
 
q0->q1 [label="a"];
−
q1->q0 [label="a"];
+
q0->q3 [label="b"];
−
q1->q1 [label="a"];
+
q1->q12 [label="a"];
−
q1 -> q0 [label="b"];
+
q1->q0 [label="b"];
 +
q12 -> q0 [label="b"];
 
}
 
}
 
</graphviz>
 
</graphviz>

Revision as of 00:43, 27 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=4146"

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