Question

Find the final DFA by performing the DFA minimization process. A is initial state and final states are B, C, D, E, F, G.

Answer 1

The final minimized DFA is given by the transition table in the solution.

Answer 2

To perform DFA minimization, we will use the table-filling algorithm (also known as the Myhill-Nerode theorem based approach).

1. Complete the DFA:

The given DFA has undefined transitions: δ(A, 1) and δ(F, 1). To make the DFA complete, we introduce a new non-final trap state, T. All undefined transitions will go to T, and from T, all transitions will loop back to T.

• Original states: Q = {A, B, C, D, E, F, G}
• Initial state: A
• Final states: F_orig = {B, C, D, E, F, G}
• Non-final states: NF_orig = {A}
• New set of states: Q' = Q ∪ {T} = {A, B, C, D, E, F, G, T}
• New Final States: F_new = F_orig = {B, C, D, E, F, G}
• New Non-Final States: NF_new = NF_orig ∪ {T} = {A, T}

The complete transition table is:

Q/Σ	0	1
A	B	T
B	C	C
C	D	E
D	C	C
E	B	G
F	E	T
G	B	G
T	T	T

2. Initial Partition (P₀):

Separate final states from non-final states.

P₀ = { {A, T}, {B, C, D, E, F, G} }

Let P₀_NF = {A, T} and P₀_F = {B, C, D, E, F, G}.

3. First Refinement (P₁):

• For P₀_NF = {A, T}:

  • δ(A, 0) = B ∈ P₀_F
  • δ(A, 1) = T ∈ P₀_NF
  • δ(T, 0) = T ∈ P₀_NF
  • δ(T, 1) = T ∈ P₀_NF

  A and T are distinguishable because δ(A, 0) leads to a state in P₀_F while δ(T, 0) leads to a state in P₀_NF. So, P₀_NF splits into {A} and {T}.

• For P₀_F = {B, C, D, E, F, G}:

  We check which states transition to P₀_NF for any input.

  • F: δ(F, 1) = T ∈ P₀_NF.

  All other states in P₀_F (B, C, D, E, G) transition only to states within P₀_F for both inputs. So, P₀_F splits into {F} and {B, C, D, E, G}.

Thus, P₁ = { {A}, {T}, {F}, {B, C, D, E, G} }

Let P₁_1 = {A}, P₁_2 = {T}, P₁_3 = {F}, P₁_4 = {B, C, D, E, G}.

4. Second Refinement (P₂):

• P₁_1, P₁_2, P₁_3 are singletons, so they cannot be split further.

• For P₁_4 = {B, C, D, E, G}:

  We check pairs of states within this partition based on which partition in P₁ their next states fall into.

  • Compare B and D:

    • δ(B, 0) = C ∈ P₁_4, δ(D, 0) = C ∈ P₁_4
    • δ(B, 1) = C ∈ P₁_4, δ(D, 1) = C ∈ P₁_4

    Since B and D transition to states in the same partitions for both inputs, they are equivalent. Group: {B, D}.

  • Compare E and G:

    • δ(E, 0) = B ∈ P₁_4, δ(G, 0) = B ∈ P₁_4
    • δ(E, 1) = G ∈ P₁_4, δ(G, 1) = G ∈ P₁_4

    Since E and G transition to states in the same partitions for both inputs, they are equivalent. Group: {E, G}.

  • State C is left alone for now.

  Now, check if {B, D}, {E, G}, and {C} are distinguishable from each other considering their transitions to these new sub-partitions.

  • B,D: δ({B,D}, 0) = C ∈ {C}, δ({B,D}, 1) = C ∈ {C}
  • C: δ(C, 0) = D ∈ {B,D}, δ(C, 1) = E ∈ {E,G}
  • E,G: δ({E,G}, 0) = B ∈ {B,D}, δ({E,G}, 1) = G ∈ {E,G}

  All these groups are distinguishable. For example, δ({B,D},0) goes to {C}, but δ(C,0) goes to {B,D}.

Thus, P₁_4 splits into {B, D}, {E, G}, and {C}.

So, P₂ = { {A}, {T}, {F}, {B, D}, {E, G}, {C} }

5. Third Refinement (P₃):

Let the partitions in P₂ be:

P₂_1 = {A} P₂_2 = {T} P₂_3 = {F} P₂_4 = {B, D} P₂_5 = {E, G} P₂_6 = {C}

• All single-state partitions ({A}, {T}, {F}, {C}) cannot be split further.

• For P₂_4 = {B, D}:

  • δ(B, 0) = C ∈ P₂_6, δ(D, 0) = C ∈ P₂_6
  • δ(B, 1) = C ∈ P₂_6, δ(D, 1) = C ∈ P₂_6

  Since B and D transition to states in the same partitions (P₂_6) for both inputs, {B, D} remains a single partition.

• For P₂_5 = {E, G}:

  • δ(E, 0) = B ∈ P₂_4, δ(G, 0) = B ∈ P₂_4
  • δ(E, 1) = G ∈ P₂_5, δ(G, 1) = G ∈ P₂_5

  Since E and G transition to states in the same partitions (P₂_4 and P₂_5 respectively) for both inputs, {E, G} remains a single partition.

Since P₃ = P₂, the algorithm terminates.

6. Construct the Minimized DFA:

Assign new names to the equivalent classes:

• S_A = {A} (Initial State, Non-Final)
• S_T = {T} (Non-Final)
• S_F = {F} (Final State)
• S_BD = {B, D} (Final State)
• S_EG = {E, G} (Final State)
• S_C = {C} (Final State)

The initial state of the minimized DFA is S_A (since A is the original initial state). The final states of the minimized DFA are S_F, S_BD, S_EG, S_C (since F, B, D, E, G, C are original final states).

Minimized DFA Transition Table:

State/input	0	1
->S_A	S_BD	S_T
S_T	S_T	S_T
S_F*	S_EG	S_T
S_BD*	S_C	S_C
S_EG*	S_BD	S_EG
S_C*	S_BD	S_EG

Where * denotes a final state.

The final DFA is represented by the transition table above.

Explanation of the solution:

1. Completion: The initial DFA had undefined transitions, which were handled by adding a trap state T (non-final).
2. Initial Partition: States were grouped into non-final ({A, T}) and final ({B, C, D, E, F, G}).
3. Iterative Refinement: Partitions were iteratively split by identifying distinguishable states. States p and q are distinguishable if their transitions on some input a lead to states in different partitions from the previous step.
   • In P₀, A and T were separated. F was separated from other final states because its 1-transition went to the non-final trap state.
   • In P₁, within {B, C, D, E, G}, B and D were found to be equivalent, and E and G were found to be equivalent, leading to further splits.
4. Termination: The process stopped when no more partitions could be split.
5. Final DFA Construction: Each resulting partition forms a new state in the minimized DFA. The initial state is the partition containing the original initial state, and final states are partitions containing any original final states.

Answer:

The final minimized DFA is given by the following transition table:

State/input	0	1
->S_A	S_BD	S_T
S_T	S_T	S_T
S_F*	S_EG	S_T
S_BD*	S_C	S_C
S_EG*	S_BD	S_EG
S_C*	S_BD	S_EG

Where:

• S_A represents the original state {A} (initial, non-final)
• S_T represents the trap state {T} (non-final)
• S_F represents the original state {F} (final)
• S_BD represents the equivalent states {B, D} (final)
• S_EG represents the equivalent states {E, G} (final)
• S_C represents the original state {C} (final)