Theoretical Aspects of Lexical Analysis/Exercise 6: Difference between revisions
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{{CollapsedCode|Graphical representation of the DFA| | {{CollapsedCode|Graphical representation of the DFA| | ||
< | <dot-hack> | ||
digraph dfa { | digraph dfa { | ||
{ node [shape=circle style=invis] s } | { node [shape=circle style=invis] s } | ||
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4 -> 5 [label="a",fontsize=10] | 4 -> 5 [label="a",fontsize=10] | ||
fontsize=10 | fontsize=10 | ||
} | } | ||
</ | </dot-hack> | ||
}} | }} | ||
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Note that before considering transition behavior, states are split according to the token they recognize. | Note that before considering transition behavior, states are split according to the token they recognize. | ||
< | <dot-hack> | ||
digraph mintree { | digraph mintree { | ||
node [shape=none,fixedsize=true,width=0.3,fontsize=10] | node [shape=none,fixedsize=true,width=0.3,fontsize=10] | ||
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"{1, 2}" -> "{2}" [label=" a",fontsize=10] | "{1, 2}" -> "{2}" [label=" a",fontsize=10] | ||
fontsize=10 | fontsize=10 | ||
} | } | ||
</ | </dot-hack> | ||
The tree expansion for sets {0, 4} and {1, 2} was only tested for "a" (sufficient). | The tree expansion for sets {0, 4} and {1, 2} was only tested for "a" (sufficient). | ||
Revision as of 10:13, 12 February 2019
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.
- G = { aa, aaaa, a|b}, input string = aaabaaaaa
NFA
The following is the result of applying Thompson's algorithm.
| NFA built by Thompson's algorithm |
|---|
| {{{2}}} |
DFA
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 |
|---|
| {{{2}}} |
| DFA minimization tree |
|---|
| {{{2}}} |
Input Analysis
| 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).