Minimization of DFA

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.
11/21/2017
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
11/21/2017
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:
11/21/2017
Sampath Kumar S, AP/CSE, SECE
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11. Problems to Discuss:
51. Find the minimum state DFA for the following
DFA.
11/21/2017
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
11/21/2017
Sampath Kumar S, AP/CSE, SECE
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