Arjun Suresh (talk | contribs) (Created page with "[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$$ state NFA. But what about the minimum...") |
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[http://condor.depaul.edu/glancast/444class/docs/nfa2dfa.html Conversion] | [http://condor.depaul.edu/glancast/444class/docs/nfa2dfa.html Conversion] | ||
| − | Q: I know that $2^n$ are the maximum states in DFA for an n$$ 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. | ||
| − | 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 | 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'> | <graphviz caption='NFA'> | ||
| + | digraph example2 { | ||
| + | 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 example2 { | digraph example2 { | ||
rankdir=LR; | rankdir=LR; | ||
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.
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 example2 { 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 example2 { 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"]; }
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.
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 example2 { 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 example2 { 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"]; }
GATECSE