The document describes the process of minimizing a deterministic finite automaton (DFA) using the Myhill-Nerode theorem. It involves 4 steps: 1) creating a table of all state pairs, 2) marking pairs where one state is final and the other is not, 3) transitively marking additional pairs, 4) combining any remaining unmarked pairs into single states. An example is provided where the given 6-state DFA is minimized to a 3-state DFA using this process. Terminology used in DFA minimization is also defined.

1. Minimization of
DFA
-Sampath Kumar S,
AP/CSE, SECE

2. Minimization of DFA using Myhill-
Nerode Theorem:
Myphill-Nerode Theorem:
Step 1: Draw a table for all pairs of states (Qi, Qj) not
necessarily connected directly [All are unmarked initially].
Step 2: Consider every state pair (Qi, Qj) in the DFA where Qi ∈
F and Qj ∉ F or vice versa and mark them. [Here F is the set
of final states].
Step 3: Repeat this step until we cannot mark anymore states
−
If there is an unmarked pair (Qi, Qj), mark it if the pair
{δ(Qi, A), δ (Qj, A)} is marked for some input alphabet.
Step 4: Combine all the unmarked pair (Qi, Qj) and make them
a single state in the reduced DFA.
11/21/2017
Sampath Kumar S, AP/CSE, SECE
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3. Terminologies used:
 Final State: The state q1 is said to be final state if q1
F.
 Non-Final State: The state q1 is said to be non-final
state if q1 ∉ F.
 Equivalent State: The state q1 and q2 are equivalent
(q1 = q2 ) if both q1 and q2 are the final state or both
of them are non-final states. These states are said to
be equivalent state.
11/21/20173
Sampath Kumar S, AP/CSE, SECE

4. Problems to Discuss:
50. Find the minimum state DFA for the following
DFA.
11/21/2017
Sampath Kumar S, AP/CSE, SECE
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5. Problems no. 50:
Step 1: We draw a table for all pair of states.
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Sampath Kumar S, AP/CSE, SECE
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6. Problems no. 50:
Step 2 − We mark the state pairs (Qi, Qj) in the DFA
where Qi ∈ F and Qj ∉ F
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Sampath Kumar S, AP/CSE, SECE
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7. Problems no. 50:
Step 3 − We will try to mark the state pairs, with
green colored check mark, transitively. If we input
1 to state ‘a’ and ‘f’, it will go to state ‘c’ and ‘f’
respectively. (c, f) is already marked, hence we
will mark pair (a, f). Now, we input 1 to state ‘b’
and ‘f’; it will go to state ‘d’ and ‘f’ respectively. (d,
f) is already marked, hence we will mark pair (b, f).
11/21/2017
Sampath Kumar S, AP/CSE, SECE
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8. Problems no. 50:
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Sampath Kumar S, AP/CSE, SECE
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 Step 3

9. Problems no. 50:
 After step 3, we have got state combinations {a, b}
{c, d} {c, e} {d, e} that are unmarked.
 We can recombine {c, d} {c, e} {d, e} into {c, d, e}
 Hence we got two combined states as − {a, b} and
{c, d, e}
 So the final minimized DFA will contain three
states {f}, {a, b} and {c, d, e}
11/21/2017
Sampath Kumar S, AP/CSE, SECE
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10. Problems no. 50:
Final Diagram:
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Sampath Kumar S, AP/CSE, SECE
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11. Problems to Discuss:
51. Find the minimum state DFA for the following
DFA.
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Sampath Kumar S, AP/CSE, SECE
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12. 11/21/2017
Sampath Kumar S, AP/CSE, SECE
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13. நன்றி
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Sampath Kumar S, AP/CSE, SECE
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