Theoretical Aspects of Lexical Analysis/Exercise 4

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Use Thompson's algorithm to build the NFA for the following regular expression. Build the corresponding DFA and minimize it.

  • (a|b)*abb(a|b)*

NFA

The following is the result of applying Thompson's algorithm.

<graph> digraph nfa {

    { node [shape=circle style=invis] s }
 rankdir=LR; ratio=0.5
 node [shape=doublecircle,fixedsize=true,width=0.2,fontsize=10]; 17
 node [shape=circle,fixedsize=true,width=0.2,fontsize=10];
 s -> 0
 0 -> 1 
 1 -> 2 
 1 -> 4
 2 -> 3 [label="a",fontsize=10]
 4 -> 5 [label="b",fontsize=10]
 3 -> 6
 5 -> 6
 6 -> 1
 6 -> 7
 0 -> 7
 7 -> 8 [label="a",fontsize=10]
 8 -> 9 [label="b",fontsize=10]
 9 -> 10 [label="b",fontsize=10]
 10 -> 11 
 11 -> 12 
 11 -> 14
 12 -> 13 [label="a",fontsize=10]
 14 -> 15 [label="b",fontsize=10]
 13 -> 16
 15 -> 16
 16 -> 11
 16 -> 17
 10 -> 17


 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, 2, 4, 7 0
0 a 3, 8 1, 2, 3, 4, 6, 7, 8 1
0 b 5 1, 2, 4, 5, 6, 7 2
1 a 3, 8 1, 2, 3, 4, 6, 7, 8 1
1 b 5, 9 1, 2, 4, 5, 6, 7, 9 3
2 a 3, 8 1, 2, 3, 4, 6, 7, 8 1
2 b 5 1, 2, 4, 5, 6, 7 2
3 a 3, 8 1, 2, 3, 4, 6, 7, 8 1
3 b 5, 10 1, 2, 4, 5, 6, 7, 10, 11, 12, 14, 17 4
4 a 3, 8, 13 1, 2, 3, 4, 6, 7, 8, 11, 12, 13, 14, 16, 17 5
4 b 5, 15 1, 2, 4, 5, 6, 7, 11, 12, 14, 15, 16, 17 6
5 a 3, 8, 13 1, 2, 3, 4, 6, 7, 8, 11, 12, 13, 14, 16, 17 5
5 b 5, 9, 15 1, 2, 4, 5, 6, 7, 9, 11, 12, 14, 15, 16, 17 7
6 a 3, 8, 13 1, 2, 3, 4, 6, 7, 8, 11, 12, 13, 14, 16, 17 5
6 b 5, 15 1, 2, 4, 5, 6, 7, 11, 12, 14, 15, 16, 17 6
7 a 3, 8, 13 1, 2, 3, 4, 6, 7, 8, 11, 12, 13, 14, 16, 17 5
7 b 5, 10, 15 1, 2, 4, 5, 6, 7, 10, 11, 12, 14, 15, 16, 17 8
8 a 3, 8, 13 1, 2, 3, 4, 6, 7, 8, 11, 12, 13, 14, 16, 17 5
8 b 5, 15 1, 2, 4, 5, 6, 7, 11, 12, 14, 15, 16, 17 6


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]; 4 5 6 7 8
 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 -> 1  [label="a",fontsize=10]
 1 -> 3  [label="b",fontsize=10]
 2 -> 1 [label="a",fontsize=10]
 2 -> 2 [label="b",fontsize=10]
 3 -> 1 [label="a",fontsize=10]
 3 -> 4 [label="b",fontsize=10]
 4 -> 5 [label="a",fontsize=10]
 4 -> 6 [label="b",fontsize=10]
 5 -> 5 [label="a",fontsize=10]
 5 -> 7 [label="b",fontsize=10]
 6 -> 5 [label="a",fontsize=10]
 6 -> 6 [label="b",fontsize=10]
 7 -> 5 [label="a",fontsize=10]
 7 -> 8 [label="b",fontsize=10]
 8 -> 5 [label="a",fontsize=10]
 8 -> 6 [label="b",fontsize=10]
 fontsize=10
 //label="DFA for (a|b)*abb(a|b)*"

} </graph>

Given the minimization tree to the right, the final minimal DFA is: <graph> digraph dfamin {

    { node [shape=circle style=invis] s }
 rankdir=LR; ratio=0.5
 node [shape=doublecircle,fixedsize=true,width=0.4,fontsize=10]; 456
 node [shape=circle,fixedsize=true,width=0.3,fontsize=10];
 s -> 02
 02 -> 1 [label="a",fontsize=10]
 02 -> 02 [label="b",fontsize=10]
 1 -> 1  [label="a",fontsize=10]
 1 -> 3  [label="b",fontsize=10]
 3 -> 1 [label="a",fontsize=10]
 3 -> 456 [label="b",fontsize=10]
 456 -> 456 [label="a",fontsize=10]
 456 -> 456 [label="b",fontsize=10]
 fontsize=10
 //label="DFA for (a|b)*abb(a|b)*"

} </graph>

The minimization tree is as follows.

<graph> digraph mintree {

 node [shape=none,fixedsize=true,width=0.3,fontsize=10]
 "{0, 1, 2, 3, 4, 5, 6}" -> "{0, 1, 2, 3}" [label="NF",fontsize=10]
 "{0, 1, 2, 3, 4, 5, 6}" -> "{4, 5, 6}" [label="  F",fontsize=10]
 //"{0, 1, 2, 3}" -> "{0, 1, 2, 3} " [label="  a",fontsize=10]
 "{0, 1, 2, 3}" ->  "{0, 1, 2}"
 "{0, 1, 2, 3}" -> "{3} " [label="  b",fontsize=10]
 "{0, 1, 2}" -> "{0, 2} "
 "{0, 1, 2}" -> "{1} " [label="  b",fontsize=10]
 fontsize=10
 //label="Minimization tree"

} </graph>

The tree expansion for non-splitting sets has been omitted for simplicity ("a" transition for non-final states and the {0, 2} "a" and "b" transitions. The final states are all indistinguishable, regarding either "a" or "b" transitions.