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
∴ T_a = {2}	∴ T_b = {5}	T_c = ∅
∴ ε − closure (T_a) = ε − closure (2) = {2} = B	∴ ε − closure (T_b) = 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
∴ T_a = {3}	∴ T_b = {}	T_c = {}
∴ ε − closure (3) = = {3, 12} = E	∴ ε − closure (T_b) = { } = ∅ = D	∴ ε − closure (T_c) = ∅ = D

For State C

For input symbol a	For input symbol b	For input symbol c
∴ T_a = {}	∴ T_b = {7}	T_c = {10}
∴ ε − closure (T_a) = ∅ = D	∴ ε − closure (7) = { 7, 8, 6, 9, 11, 12} = F	∴ ε − closure (10) = {10, 9, 11, 12} = G

For State E

∴ T_a = ∅	∴ T_b = ∅	T_c = ∅
∴ ε − closure (T_a) = ∅ = D	∴ ε − closure (T_b) = ф = D	∴ ε − closure (Tc) = ф = D

For State F

∴ T_a = ∅	∴ T_b = {7}	T_c = {10}
∴ ε − closure (T_a) = ∅ = D	∴ ε − closure (T_b) = = ε − closure (7) = {7,8,6,9,12} = F	∴ ε − closure (10) = {10, 9, 11, 12} = G

For State G

∴ T_a = ∅	∴ T_b = {7}	T_c = {10}
∴ ε − closure (ф) = ф = D	∴ ε − closure (ф) = ф = D	∴ ε − closure (10) = G

For State D

D = ∅

T_a = T_b = T_c = ∅

ε − closure (T_a) = ε − closure (T_b) = ε − closure (T_c) = ф = D

Transition Table and Diagram for DFA will be −

(Initial State)

Ginni

Updated on: 29-Oct-2021

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Construct NFA for the following language and convert it into DFA using the algorithm - L = (aa+ (bb*)c*)