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Construct NFA for the following language and convert it into DFA using the algorithm - L = (aa+ (bb*)c*)
Solution
NFA for the above language will be −
Conversion from NFA to DFA −
ε − closure(0) = {0, 1, 4} = A
For State A
| For input symbol a | For input symbol b | For input symbol c |
|---|---|---|
| ∴ Ta = {2} | ∴ Tb = {5} | Tc = ∅ |
| ∴ ε − closure (Ta) = ε − closure (2) = {2} = B |
∴ ε − closure (Tb) = ε − closure (5) = {5, 6, 8, 9, 11, 12} = C |
∴ ε − closure (∅) = ∅ = D |
For State B
| For input symbol a | For input symbol b | For input symbol c |
|---|---|---|
| ∴ Ta = {3} | ∴ Tb = {} | Tc = {} |
| ∴ ε − closure (3) = = {3, 12} = E |
∴ ε − closure (Tb) = { } = ∅ = D |
∴ ε − closure (Tc) = ∅ = D |
For State C
| For input symbol a | For input symbol b | For input symbol c |
|---|---|---|
| ∴ Ta = {} | ∴ Tb = {7} | Tc = {10} |
| ∴ ε − closure (Ta) = ∅ = D |
∴ ε − closure (7) = { 7, 8, 6, 9, 11, 12} = F |
∴ ε − closure (10) = {10, 9, 11, 12} = G |
For State E
| ∴ Ta = ∅ | ∴ Tb = ∅ | Tc = ∅ |
| ∴ ε − closure (Ta) = ∅ = D |
∴ ε − closure (Tb) = ? = D |
∴ ε − closure (Tc) = ? = D |
For State F
| ∴ Ta = ∅ | ∴ Tb = {7} | Tc = {10} |
| ∴ ε − closure (Ta) = ∅ = D |
∴ ε − closure (Tb) = = ε − closure (7) = {7,8,6,9,12} = F |
∴ ε − closure (10) = {10, 9, 11, 12} = G |
For State G
| ∴ Ta = ∅ | ∴ Tb = {7} | Tc = {10} |
| ∴ ε − closure (?) = ? = D |
∴ ε − closure (?) = ? = D |
∴ ε − closure (10) = G |
For State D
D = ∅
Ta = Tb = Tc = ∅
ε − closure (Ta) = ε − closure (Tb) = ε − closure (Tc) = ? = D
Transition Table and Diagram for DFA will be −
(Initial State)
| a | B | c | Here | |
| A | B | C | D | A = {0,1,4} |
| B | E | D | D | B = {2} |
| C | D | F | G | C = {5,6,8,9,11,12} |
| D | D | D | D | D = ∅ |
| E | D | D | D | E = {3, 12} |
| F | D | F | G | F = {7,8,6,9,11,12} |
| G | D | D | G | G = {10,9,11,12} |
Updated on: 2021-10-29T08:28:42+05:30
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