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"{0, 1, 3}" -> "{0}" | "{0, 1, 3}" -> "{0}" | ||
"{0, 1, 3}" -> "{1,3}" [label=" b",fontsize=10] | "{0, 1, 3}" -> "{1,3}" [label=" b",fontsize=10] | ||
| + | "{1,3}" -> "{1,3} " [label=" a,b",fontsize=10] | ||
"{2, 4}" -> "{2}" | "{2, 4}" -> "{2}" | ||
"{2, 4}" -> "{4}" [label=" b",fontsize=10] | "{2, 4}" -> "{4}" [label=" b",fontsize=10] | ||
| Line 174: | Line 175: | ||
} | } | ||
</dot-hack> | </dot-hack> | ||
| − | |||
| − | |||
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. | 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. | ||
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.
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).