Compute the non-deterministic finite automaton (NFA) by using Thompson's algorithm. Compute the minimal deterministic finite automaton (DFA).
The alphabet is Σ = { a, b }. Indicate the number of processing steps for the given input string.
The following is the result of applying Thompson's algorithm. State 4 recognizes the first expression (token T1); state 12 recognizes token T2; and state 20 recognizes token T3.
Determination table for the above NFA:
| In | α∈Σ | move(In, α) | ε-closure(move(In, α)) | In+1 = ε-closure(move(In, α)) |
|---|---|---|---|---|
| - | - | 0 | 0, 1, 2, 4, 5, 6, 7, 9, 10, 12, 13, 14, 16, 17, 19, 20 | 0 (T1) |
| 0 | a | 3, 8, 15 | 2, 3, 4, 7, 8, 9, 12, 15, 20 | 1 (T1) |
| 0 | b | 11, 18 | 11, 12, 17, 18, 19, 20 | 2 (T2) |
| 1 | a | 3, 8 | 2, 3, 4, 7, 8, 9, 12 | 3 (T1) |
| 1 | b | - | - | - |
| 2 | a | - | - | - |
| 2 | b | 18 | 17, 18, 19, 20 | 4 (T3) |
| 3 | a | 3, 8 | 2, 3, 4, 7, 8, 9, 12 | 3 (T1) |
| 3 | b | - | - | - |
| 4 | a | - | - | - |
| 4 | b | 18 | 17, 18, 19, 20 | 4 (T3) |
Graphically, the DFA is represented as follows:
The minimization tree is as follows. Note that before considering transition behavior, states are split according to the token they recognize.
The tree expansion for non-splitting sets has been omitted for simplicity ("a" transitions for super-state {0, 1, 3}, and "a" and "b" transitions for super-state {1,3} -- note that these transitions would not split state {1,3}).
Given the minimization tree, the final minimal DFA is as follows. Note that states 2 and 4 cannot be the same since they recognize different tokens.
| In | Input | In+1 / Token |
|---|---|---|
| 0 | aababb$ | 13 |
| 13 | ababb$ | 13 |
| 13 | babb$ | T1 (aa) |
| 0 | babb$ | 2 |
| 2 | abb$ | T2 (b) |
| 0 | abb$ | 13 |
| 13 | bb$ | T1 (a) |
| 0 | bb$ | 2 |
| 2 | b$ | 4 |
| 4 | $ | T3 (bb) |
The input string aababb is, after 10 steps, split into three tokens: T1 (corresponding to lexeme aa), T2 (b), T1 (a), and T3 (bb).