Can we convert a non-deterministic finite automata into a deterministic finite Automata?

Compiler Design Programming Languages Computer Programming

Yes, we can convert a NFA into DFA. For every NFA there exists an equivalent DFA. The equivalence is defined in terms of languages acceptance. Since NFA is nothing but a finite automata in which zero, one or more transitions on an input symbols are permitted. It can always construct finite automata which will simulate all moves of DFA on a particular input symbol in parallel, then get a finite automata in which there will be exactly one transition on every input symbol. Here, corresponding to a NFA there exist a DFA. To construct DFA equivalent to NFA, it should be remembered that states of DFA are a collection of states of NFA.

Algorithm NFA – to – DFA

Input − NFA with set of states N = {n₀,n₁……n_n}, with start state n₀.

Output − DFA, with set of states D′={d₀,d₁,d₂……d_n}, with start state d₀.

d₀=ε−closure (n₀)

D′={d₀}

set d₀ unmarked

while there is an unmarked state d in D′. {
   set d marked {
      For each input symbol 'a' {
         Let T be a set of states in NFA to which there is a transition on 'a' from some state ni in d d′=ε−closure (T).
         If d′ is not already present in D′ {
            D′=D′∪{d′}
            Add transition d→d′,labeled 'a'
            set d′ unmarked
         }
      }
   }
}

Example − Design Lexical Analyzer for the following LEX Program.

AUXILIARY DEFINITION S
letter = A|B|C|…….|Z
digit = 0 |1|2|………|9
TRANSLATION RULES
begin    {return 1}
end      {return 2}
If       {return 3}
then     {return 4}
else     {return 5}
letter  (letter+digit)* {value=Install ( );return 6}
digit + {value=Install ( );return 7}
<      {value=1;return 8}
<=     {value=2;return 8}
=      {value=3;return 8}
< >    {value=4;return 8}
>      {value=5;return 8}
>=     {value=6;return 8}

Solution