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.
| NFA built by Thompson's algorithm |
|---|
|
State 3 recognizes the first expression (token T1); state 8 recognizes token T2; and state 14 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, 4, 9, 10, 12 | 0 |
| 0 | a | 2, 5, 11 | 2, 5, 11, 14 | 1 (T3) |
| 0 | b | 13 | 13, 14 | 2 (T3) |
| 1 | a | 3, 6 | 3, 6 | 3 (T1) |
| 1 | b | - | - | - |
| 2 | a | - | - | - |
| 2 | b | - | - | - |
| 3 | a | 7 | 7 | 4 |
| 3 | b | - | - | - |
| 4 | a | 8 | 8 | 5 (T2) |
| 4 | b | - | - | - |
| 5 | a | - | - | - |
| 5 | b | - | - | - |
| Graphical representation of the DFA |
|---|
| DFA minimization tree |
|---|
|
Note that before considering transition behavior, states are split according to the token they recognize. The tree expansion for sets {0, 4} and {1, 2} was only tested for "a" (sufficient). Given the minimization tree, the final minimal DFA is exactly the same as the original DFA (all leaf sets are singular). |
| In | Input | In+1 / Token |
|---|---|---|
| 0 | aaabaaaaa$ | 1 |
| 1 | aabaaaaa$ | 3 |
| 3 | abaaaaa$ | 4 |
| 4 | baaaaa$ | error (backtracking) |
| 3 | abaaaaa$ | T1 (aa) |
| 0 | abaaaaa$ | 1 |
| 1 | baaaaa$ | T3 (a) |
| 0 | baaaaa$ | 2 |
| 2 | aaaaa$ | T3 (b) |
| 0 | aaaaa$ | 1 |
| 1 | aaaa$ | 3 |
| 3 | aaa$ | 4 |
| 4 | aa$ | 5 |
| 5 | a$ | T2 (aaaa) |
| 0 | a$ | 1 |
| 1 | $ | T3 (a) |
The input string aaabaaaaa is, after 16 steps, split into three tokens: T1 (corresponding to lexeme aa), T3 (a), T3 (b), T2 (aaaa), and T3 (a).