5. Minimal DFAs

States 2 and 5 are indistinguishable.

Looking at the DFA we can see that states 2 and 5 both go to state 6 on “1” and to 3 on “0”.

States 1 and 2 are indistinguishable.

Looking at the DFA we can see that states 1 and 2 both go to state 2 on “a” and to 5 on “b”.

States 4 and 5 are distinguishable because we can accept “1” from 4, but not from 5.

States 0 and 5 are distinguishable because from 0 we can accept word “b”, but not from 5.

Initially marked pairs of (final, non-final) states:

• (1, 6)
• (2, 6)
• (3, 6)
• (4, 6)
• (5, 6)

Predecessors for (6, 1):

• 0: -
• 1: -

Predecessors for (6, 2):

• 0: -
• 1: (2, 1), (5, 1) ⇒ mark them

Predecessors for (6, 3):

• 0: -
• 1: -

Predecessors for (6, 4):

• 0: -
• 1: -

Predecessors for (6, 5):

• 0: -
• 1: (2, 3), (5, 3), (2, 4), (5, 4) ⇒ mark them

Predecessors for (2, 1):

• 0: -
• 1: -

Predecessors for (5, 1):

• 0: -
• 1: -

Predecessors for (3, 2):

• 0: -
• 1: -

Predecessors for (4, 2):

• 0: -
• 1: -

Predecessors for (5, 3):

• 0: -
• 1: -

Predecessors for (5, 4):

• 0: -
• 1: -

The partitions of undistinguishable states are given by the missing marks:

• (5, 2)
• (3, 1), (4, 3), (4, 1)

⇒ The partitions are: {2, 5} and {1, 3, 4}

Initially marked pairs of (final, non-final) states:

• (0, 1)
• (0, 2)
• (0, 4)
• (3, 1)
• (3, 2)
• (3, 4)
• (5, 1)
• (5, 2)
• (5, 4)

Predecessors for (5, 4):

• a: -
• b: (3, 4), (3, 1), (3, 2), (3, 5) ⇒ mark them

Predecessors for (5, 1):

• a: -
• b: (5, 0) ⇒ mark it

Predecessors for (3, 1):

• a: -
• b: -

Predecessors for (3, 2):

• a: (0, 1), (0, 2), (0, 4) ⇒ already marked
• b: -

Predecessors for (5, 2):

• a: (5, 1), (5, 2), (5, 4) ⇒ already marked
• b: -

Predecessors for (3, 4):

• a: -
• b: -

Predecessors for (0, 1):

• a: -
• b: -

Predecessors for (0, 2):

• a: (3, 1), (3, 2), (3, 4) ⇒ already marked
• b: -

Predecessors for (0, 4):

• a: -
• b: -

Predecessors for (0, 5):

• a: (3, 5) ⇒ mark it
• b: -

Predecessors for (3, 5):

• a: (0, 5) ⇒ already marked
• b: -

The partitions of undistinguishable states are given by the missing marks:

• (0, 3)
• (4, 1), (4, 2), (2, 1)

⇒ The partitions are: {0, 3} and {1, 2, 4}

• assume the states are labeled as integers starting with 0, and that 0 is the unique initial state of both DFAs (this can be easily obtained from any previous transformation).
• give an ordering to the symbols of the alphabet. Write $ \delta$ as a string by the following rules:
  • we sort each transition $ \delta(q,c)=q'$ , by $ q$ first and by $ c$ , second. We append each transition given this ordering (smaller states first, then for each state, output the order of the $ \delta(q,c)$ transitions according to $ c$ .
• two minimal DFAs accept the same language iff:
  • they have the same string representation of $ \delta$ and
  • have the same final states.

Another more elegant approach is to use state equivalence. We can treat two (not necessarily minimal) DFAs as one and the same, and compute the (in)distinguishability table over all states, in a single go. Finally, the DFAs accept the same language iff their initial states are indistinguishable.

1. The first way will be presented in the next seminar. Using closure properties we will prove both the following properties:

• L(E1) ∩ L(E2) = ∅
• L(complement(E1)) ∩ L(complement(E2)) = ∅

2. Check that the initial state of minDfa(E1) is indistinguishable from the initial state of minDfa(E2).

We will solve 2, but we will assume we have already computed minDfa for the 2 regular expressions and got the following DFAs as result:

Obs: We can tell they are the same by looking at them but the computer needs an algorithm to check this equivalence.

Initially marked pairs of (final, non-final) states:

• (0, 1)
• (0, 2)
• (0, 4)
• (0, 5)
• (3, 1)
• (3, 2)
• (3, 4)
• (3, 5)

Predecessors for (0, 1):

• a: -
• b: -

Predecessors for (0, 2):

• a: -
• b: (1, 2) ⇒ mark it

Predecessors for (0, 4):

• a: -
• b: -

Predecessors for (0, 5):

• a: -
• b: (1, 5) ⇒ mark it

Predecessors for (2, 1):

• a: (1, 2), (0, 2) ⇒ already marked
• b: -

Predecessors for (3, 1):

• a: -
• b: -

Predecessors for (5, 1):

• a: (3, 2), (4, 2) ⇒ mark them
• b: -

Predecessors for (3, 2):

• a: -
• b: (4, 2) ⇒ already marked

Predecessors for (4, 2):

• a: (5, 1), (5, 0) ⇒ already marked
• b: -

Predecessors for (4, 3):

• a: -
• b: -

Predecessors for (5, 3):

• a: -
• b: (5, 4) ⇒ mark it

Predecessors for (5, 4):

• a: (3, 5), (4, 5) ⇒ already marked
• b: -

So, we get 3 sets of indistinguishable states: {0, 3}, {1, 4} and {2, 5}.

Because states 0 and 3, which are initial states in the 2 DFAs, we can conclude that the 2 DFAs accept the same language.

• reverse transitions
• the initial state becomes final in Reverse(A)
• the final state becomes initial in Reverse(A), but here we have more final states:
  • normally, to compute Reverse(A) we should add a new initial state in Reverse(A); the final states in A become non-final in Reverse(A); then add $\varepsilon$ - transitions from the new initial state of Reverse(A) to the former final states of A
  • for Brzozowski's minimization algorithm however, we will not add this new initial state and, instead, we will start the subset construction in Step 2 from the set of final states that were in A, that now become “multiple” initial states in Reverse(A)
• we can also not include the sink state in the reversal of the DFA
• the set of initial states is: {1, 2, 3, 4, 5}

• rename the subsets as follows: {1,2,3,4,5} → 1 and {1,3,4} → 2

• we can ignore the sink state
• as previously, we remember a set of initial states: {1,2}

• this is minDfa(A)