wondering how to convert this non-deterministic FSA
to a deterministic one, specifically the tables.
I have got this so far but am stuck:
How do i assign BHMQ? For state one, won't there be four digits? (C, I, N, R)?
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make a new state {B,H,M,Q} as S, then S(0) = B(0) + H(0) + M(0) + Q(0) and then do it for S(1). And continue adding states till no new state is reached – Kishan Kumar Apr 30 '16 at 11:21
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http://web.cecs.pdx.edu/~harry/compilers/slides/LexicalPart3.pdf this pdf may help. Check this once – Kishan Kumar Apr 30 '16 at 11:25
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I am confused @KishanKumar, how would that look on the table? – Emolk Apr 30 '16 at 11:26
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the states in the newly constructed dfa should not necessarily be same as the in nfa. You could consider state BHMQ as a single state and use it – Kishan Kumar Apr 30 '16 at 11:28
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And in table write the new state in the place of BHMQ, and make a new row for that state Only. Make row for tables of only the states which are derived by the start state in DFA – Kishan Kumar Apr 30 '16 at 11:33
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@KishanKumar might not have removed the nulls correctly, any pdfs on how to do that properly? – Emolk Apr 30 '16 at 11:46
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http://stackoverflow.com/questions/31120670/converting-epsilon-nfa-to-nfa see this for e-transition – Kishan Kumar Apr 30 '16 at 11:49
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@KishanKumar Is there a way to do this purely based off the tables? – Emolk Apr 30 '16 at 12:09
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yeah, you can do it. By only using tables. – Kishan Kumar Apr 30 '16 at 14:39
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The concept you are looking for is the so-called power set construction. From the initial automaton, the state set is converted by taking its power set, which is the set of all of its subsets, to be the the state set of the resulting automaton; the final states are the ones which contain at least one of the initial final states.
The resulting automaton is equivalent to the initial one in the sense that the accepted languages are equal. However, the number of states may grow exponentially.
Codor
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