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 9 recognizes token T2; and state 17 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, 10, 11, 13, 14, 16, 17 | 0 (T1) |
0 | a | 3, 12 | 2, 3, 4, 12, 17 | 1 (T1) |
0 | b | 6, 15 | 6, 7, 9, 14, 15, 16, 17 | 2 (T2) |
1 | a | 3 | 2, 3, 4 | 3 (T1) |
1 | b | - | - | - |
2 | a | 8 | 7, 8, 9 | 4 (T2) |
2 | b | 15 | 14, 15, 16, 17 | 5 (T3) |
3 | a | 3 | 2, 3, 4 | 3 (T1) |
3 | b | - | - | - |
4 | a | 8 | 7, 8, 9 | 4 (T2) |
4 | b | - | - | - |
5 | a | - | - | - |
5 | b | 15 | 14, 15, 16, 17 | 5 (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}).
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$ | 4 |
4 | bb$ | T2 (ba) |
0 | bb$ | 2 |
2 | b$ | 5 |
5 | $ | T3 (bb) |
The input string aababb is, after 9 steps, split into three tokens: T1 (corresponding to lexeme aa), T2 (ba), and T3 (bb).