Theoretical Aspects of Lexical Analysis/Exercise 6
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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. State 3 recognizes the first expression (token T1); state 8 recognizes token T2; and state 14 recognizes token T3.
<graph> digraph nfa {
{ node [shape=circle style=invis] s } rankdir=LR; ratio=0.5 node [shape=doublecircle,fixedsize=true,width=0.2,fontsize=10]; 3 8 14 node [shape=circle,fixedsize=true,width=0.2,fontsize=10];
s -> 0
0 -> 1 1 -> 2 [label="a",fontsize=10] 2 -> 3 [label="a",fontsize=10]
0 -> 4 4 -> 5 [label="a",fontsize=10] 5 -> 6 [label="a",fontsize=10] 6 -> 7 [label="a",fontsize=10] 7 -> 8 [label="a",fontsize=10]
0 -> 9 9 -> 10 9 -> 12 10 -> 11 [label="a",fontsize=10] 12 -> 13 [label="b",fontsize=10] 11 -> 14 13 -> 14 fontsize=10 //label="NFA for (a|b)*abb(a|b)*"
} </graph>
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 | - | - | - |
Graphically, the DFA is represented as follows:
<graph> digraph dfa {
{ node [shape=circle style=invis] s } rankdir=LR; ratio=0.5 node [shape=doublecircle,fixedsize=true,width=0.2,fontsize=10]; 1 2 3 5 node [shape=circle,fixedsize=true,width=0.2,fontsize=10]; s -> 0 0 -> 1 [label="a",fontsize=10] 0 -> 2 [label="b",fontsize=10] 1 -> 3 [label="a",fontsize=10] 3 -> 4 [label="a",fontsize=10] 4 -> 5 [label="a",fontsize=10] fontsize=10 //label="DFA for (a|b)*abb(a|b)*"
} </graph>
The minimization tree is as follows. Note that before considering transition behavior, states are split according to the token they recognize.
<graph> digraph mintree {
node [shape=none,fixedsize=true,width=0.3,fontsize=10] "{0, 1, 2, 3, 4, 5}" -> "{0, 4}" [label=" NF",fontsize=10] "{0, 1, 2, 3, 4, 5}" -> "{1, 2, 3, 5}" [label=" F",fontsize=10] "{1, 2, 3, 5}" -> "{3}" [label=" T1",fontsize=10] "{1, 2, 3, 5}" -> "{5}" [label=" T2",fontsize=10] "{1, 2, 3, 5}" -> "{1, 2}" [label=" T3",fontsize=10] "{0, 4}" -> "{0}" //[label=" T3",fontsize=10] "{0, 4}" -> "{4}" [label=" a",fontsize=10] "{1, 2}" -> "{1}" //[label=" T3",fontsize=10] "{1, 2}" -> "{2}" [label=" a",fontsize=10] fontsize=10 //label="Minimization tree"
} </graph>
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).
Input Analysis
In | Input | In+1 / Token |
---|---|---|
0 | aaabaaaaa$ | |
$ | ||
$ | ||
$ | ||
$ | ||
$ | ||
$ | ||
$ | ||
$ |