The document discusses converting regular expressions to minimized deterministic finite automata (DFAs) in several steps: 1. Use Thompson's algorithm to convert the regular expression to a non-deterministic finite automaton (NFA). 2. Use the subset construction algorithm to convert the NFA to a DFA. 3. Use the tabulation method (also called Mark/Reduce procedure) to minimize the obtained DFA. Several examples are provided to demonstrate applying these steps to convert regular expressions like (a+b)*abb and abb(a+b)* to their minimized DFA representations. The subset construction algorithm and tabulation method for minimization are explained.

1. NFA to Minimized DFA

2. RE to DFA
• To convert RE to NFA use – Thompson’s algorithm
• To convert NFA to DFA use – Subset Construction algorithm
• To minimize the obtained DFA use – Tabulation method (also called
Mark/Reduce procedure)

3. Subset construction algorithm
Step 1: Find the λ-closure of all the states
Step 2: Let the λ-closure of the initial state be named as A
Step 3: Now find the transitions of each input symbol on this state A.
Then, find the λ-closure of the above set and name it as A if it appears
to be same otherwise give a new name as B.
Step 4: Repeat Step 3 for each of the new states added until there are
no new states in the queue.
Step 5: Convert the above representation to a DFA. The initial state of
the DFA will be A. Mark the final states of DFA.

4. Example 1
• Convert the RE (a+b)*abb to a minimized DFA
λ-closure(0) = {0, 1, 2, 4, 7} λ-closure(4) = {4} λ-closure(8) = {8}
λ-closure(1) = {1, 2, 4} λ-closure(5) = {1, 2, 4, 5, 6, 7} λ-closure(9) = {9}
λ-closure(2) = {2} λ-closure(6) = {1, 2, 4, 6, 7} λ-closure(10) = {10}
λ-closure(3) = {1, 2, 3, 4, 6, 7} λ-closure(7) = {7}

5. Example 1
λ-closure(0) = {0, 1, 2, 4, 7} – A
λ-closure(ẟ(A, a)) = λ-closure(3, 8) = {1, 2, 3, 4, 6, 7, 8} – B
λ-closure(ẟ(A, b)) = λ-closure(5) = {1, 2, 4, 5, 6, 7} – C
λ-closure(ẟ(B, a)) = λ-closure(3, 8) = {1, 2, 3, 4, 6, 7, 8} – B
λ-closure(ẟ(B, b)) = λ-closure(5, 9) = {1, 2, 4, 5, 6, 7, 9} – D
λ-closure(ẟ(C, a)) = λ-closure(3, 8) = {1, 2, 3, 4, 6, 7, 8} – B
λ-closure(ẟ(C, b)) = λ-closure(5) = {1, 2, 4, 5, 6, 7} – C
λ-closure(ẟ(D, a)) = λ-closure(3, 8) = {1, 2, 3, 4, 6, 7, 8} – B
λ-closure(ẟ(D, b)) = λ-closure(5, 10) = {1, 2, 4, 5, 6, 7, 10} – E
λ-closure(ẟ(E, a)) = λ-closure(3, 8) = {1, 2, 3, 4, 6, 7, 8} – B
λ-closure(ẟ(E, b)) = λ-closure(5) = {1, 2, 4, 5, 6, 7} – C

6. Example 1 – Final DFA
States a b
-> A B C
B B D
C B C
D B E
* E B C
A
B
C
D
E
E
a
a
a
a
a
b
b
b
b
b

7. DFA Minimization – Tabulation Method
States a b
-> A B C
B B D
C B C
D B E
* E B C
B
C
D
E
A B C D
Mark/Reduce Procedure
Pair – (AC)

8. Minimized DFA
States a b
-> A B C
B B D
C B C
D B E
* E B C
States a b
-> A B A
B B D
D B E
* E B A
A
B D
E
E
b b
b
b
a
a
a
a

9. Mark/Reduce Procedure

10. Example 2
• Find the minimal DFA for abb(a+b)*

11. Example 2
λ-closure(0) = {0} λ-closure(1) = {1} λ-closure(2) = {2}
λ-closure(3) = {3, 4, 5, 7, 10} λ-closure(4) = {4, 5, 7} λ-closure(5) = {5}
λ-closure(6) = {4, 5, 6, 7, 9, 10} λ-closure(7) = {7} λ-closure(8) = {4, 5, 7, 8, 9, 10}
λ-closure(9) = {4, 5, 7, 9, 10} λ-closure(10) = {10}

12. Example 2 – Subset Alg.
• λ-closure(0) = {0} - A
• λ-closure(ẟ(A, a)) = λ-closure(1) = {1} – B
• λ-closure(ẟ(A, b)) = φ
• λ-closure(ẟ(B, a)) = φ
• λ-closure(ẟ(B, b)) = λ-closure(2) = {2} – C
• λ-closure(ẟ(C, a)) = φ
• λ-closure(ẟ(C, b)) = λ-closure(3) = {3, 4, 5, 7, 10} – D

13. Example 2
• λ-closure(ẟ(D, a)) = λ-closure(6) = {4, 5, 6, 7, 9, 10} – E
• λ-closure(ẟ(D, b)) = λ-closure(8) = {4, 5, 7, 8, 9, 10} – F
• λ-closure(ẟ(E, a)) = λ-closure(6) = {4, 5, 6, 7, 9, 10} – E
• λ-closure(ẟ(E, b)) = λ-closure(8) = {4, 5, 7, 8, 9, 10} – F
• λ-closure(ẟ(F, a)) = λ-closure(6) = {4, 5, 6, 7, 9, 10} – E
• λ-closure(ẟ(F, b)) = λ-closure(8) = {4, 5, 7, 8, 9, 10} – F

14. Example 2 - Minimization
States a b
-> A B φ
B φ C
C φ D
* D E F
* E E F
* F E F
B
C
D
E
F
A B C D E
Pairs
{{DE}{DF}}
{EF} => {DEF}

15. Example 2 – Minimized DFA
States a b
-> A B φ
B φ C
C φ D
* D D D
A B C D
D
T
a
b
a
b
a
b
a, b
a, b

16. Example 3
• Convert to minimized DFA
• λ-closure(q0) = {q0}
• λ-closure(q1) = {q1}
• λ-closure(q2) = {q2}
q0 q1
q1 q2
0
0, 1 0, 1
1

17. Example 3
• λ-closure(q0) = {q0} – A
• λ-closure(ẟ(A, 0)) = λ-closure(q0,q1} = {q0,q1} – B
• λ-closure(ẟ(A, 1)) = λ-closure(q1} = {q1} – C
• λ-closure(ẟ(B, 0)) = λ-closure(q0,q1,q2} = {q0,q1,q2} – D
• λ-closure(ẟ(B, 1)) = λ-closure(q1,q2} = {q1,q2} – E
• λ-closure(ẟ(C, 0)) = λ-closure(q2} = {q2} – F
• λ-closure(ẟ(C, 1)) = λ-closure(q2} = {q2} – F
• λ-closure(ẟ(D, 0)) = λ-closure(q0,q1,q2} = {q0,q1,q2} – D
• λ-closure(ẟ(D, 1)) = λ-closure(q1,q2} = {q1,q2} - E

18. Example 3
• λ-closure(ẟ(E, 0)) = λ-closure(q2} = {q2} – F
• λ-closure(ẟ(E, 1)) = λ-closure(q2} = {q2} – F
• λ-closure(ẟ(F, 0)) = φ
• λ-closure(ẟ(F, 1)) = λ-closure(q2} = {q2} - F
States 0 1
-> A B C
*B D E
*C F F
*D D E
*E F F
F φ F

19. Example 3
B
C
D
E
F
A B C D E
States 0 1
-> A B C
*B D E
*C F F
*D D E
*E F F
F φ F
Pairs:
(BD), (CE)

20. Example 3 – Minimized DFA
• Let (BD) be X
• Let (CE) be Y
States 0 1
-> A B C
*B D E
*C F F
*D D E
*E F F
F φ F
States 0 1
-> A X Y
*X X Y
*Y F F
F φ F
A
X
Y
F
X
Y
1
0
1
0
0, 1
0, 1
0

21. Example 4
• Find the minimized DFA for the given NFA
• λ-closure(q0) = {q0, q1}
• λ-closure(q1) = {q1}
• λ-closure(q2) = {q2}
q0 q1 q2
q1
0
0, λ
1
0, 1
1

22. Example 4
• λ-closure(q0) = {q0, q1} - A
• λ-closure(ẟ(A, 0)) = λ-closure(q0, q1, q2) = {q0, q1, q2} – B
• λ-closure(ẟ(A, 1)) = λ-closure(q1, q2) = {q1, q2} – C
• λ-closure(ẟ(B, 0)) = λ-closure(q0, q1, q2) = {q0, q1, q2} – B
• λ-closure(ẟ(B, 1)) = λ-closure(q1, q2) = {q1, q2} – C
• λ-closure(ẟ(C, 0)) = λ-closure(q0, q2) = {q0, q1, q2} – B
• λ-closure(ẟ(C, 1)) = λ-closure(q1, q2) = {q1, q2} - C

23. Example 4
B
C
A B
States 0 1
->* A B C
*B B C
*C B C
All pairs are
indistinguishable
So, {ABC} is a single
state
X
X
0, 1

24. Example 5 - Minimize the given DFA
• First draw the DFA and check if all
the states are reachable from start
state
• The sets are {q1, q2, q3} and {q2,
q3}.
• So, {q1, q2, q3} can be combined
into one state
States 0 1
-> q0 q1 q3
q1 q2 q4
q2 q1 q4
q3 q2 q4
*q4 q4 q4
q1
q2
q3
q4
q0 q1 q2 q3
States 0 1
-> q0 X X
X X q4
*q4 q4 q4

25. Exercises – NFA to Minimized DFA
1.
2.
A B X
C
a
b c
λ λ
A B C
B
0 1
0,1 0,1

26. Exercises – NFA to Minimized DFA
3.
4. RE ab*a(a+b)
5. RE (0+1)+(0+1)*
6. 𝐿 = 𝑎𝑛𝑏𝑚 ∶ n ≥ 1, m ≥ 2
7. 𝐿 = 𝑎𝑛b: n ≥ 1} ⋃ {𝑏𝑛a ∶ n ≥ 1
A B X
C
a
b a, b
a, λ b, λ
b, λ

27. Exercises – Minimize the DFA
8. 9.
States 0 1
-> A B F
B G C
*C A C
D C G
E H F
F C G
G G E
H G C

28. Exercises
10.