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Total number of DFA possible with 2 states q0 → start and non-final, q1 → final
over Ʃ = {a,b} is
(a) 16                 (b) 32
(c) 48                 (d) 64
in Theory of Computation by (169 points) | 179 views

2 Answers

+3 votes
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On every state it has two inputs , so each input can go to one of the two states. So for every state its arrows (->) have four possible combinations. Similarly on second state it also has four combinations.of arrows emerging from it. So i think its 16.
by (377 points)
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I think if we have.

N is the number of states

S is the alphabet symbols

Then total dfa possible is (choice for initial state) * (choice for final state) *(choice for transition function)



= 16
by (67 points)
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