Theoretical Aspects of Lexical Analysis/Exercise 12

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 = { b*|a, a*, b|a* }, input string = aababb

NFA

The following is the result of applying Thompson's algorithm. State 8 recognizes the first expression (token T1); state 12 recognizes token T2; and state 20 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]; 8 12 20
 node [shape=circle,fixedsize=true,width=0.2,fontsize=10];

 s -> 0

 0 -> 1 
 1 -> 2
 1 -> 4
 2 -> 3 [label="a",fontsize=10]
 3 -> 8
 4 -> 5
 4 -> 7
 5 -> 6 [label="b",fontsize=10]
 6 -> 5
 6 -> 7
 7 -> 8

 0 -> 9
 9 -> 10
 9 -> 12
 10 -> 11 [label="a",fontsize=10]
 11 -> 10
 11 -> 12

 0 -> 13
 13 -> 14
 13 -> 16
 14 -> 15 [label="b",fontsize=10]
 15 -> 20
 16 -> 17
 16 -> 19
 17 -> 18 [label="a",fontsize=10]
 18 -> 17
 18 -> 19
 19 -> 20
 fontsize=10

} </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, 5, 7, 8, 9, 10, 12, 13, 14, 16, 17, 19, 20	0 (T1)
0	a	3, 11, 18	3, 8, 10, 11, 12, 17, 18, 19, 20	1 (T1)
0	b	6, 15	5, 6, 7, 8, 15, 20	2 (T1)
1	a	11, 18	10, 11, 12, 17, 18, 19, 20	3 (T2)
1	b	-	-	-
2	a	-	-	-
2	b	6	5, 6, 7, 8	4 (T1)
3	a	11, 18	10, 11, 12, 17, 18, 19, 20	3 (T2)
3	b	-	-	-
4	a	-	-	-
4	b	6	5, 6, 7, 8	4 (T1)

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]; 0 1 2 3 4
 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]
 2 -> 4 [label="b",fontsize=10]
 3 -> 3 [label="a",fontsize=10]
 4 -> 4 [label="b",fontsize=10]
 fontsize=10

} </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}" -> "{}" [label="NF",fontsize=10]
 "{0, 1, 2, 3, 4}" -> "{0, 1, 2, 3, 4} " [label="  F",fontsize=10]
 "{0, 1, 2, 3, 4} " -> "{0, 1, 2, 4}" [label="  T1",fontsize=10]
 "{0, 1, 2, 3, 4} " -> "{3}" [label="  T2",fontsize=10]
 "{0, 1, 2, 4}" -> "{0}" //[label="  a",fontsize=10]
 "{0, 1, 2, 4}" -> "{1}" //[label="  a",fontsize=10]
 "{0, 1, 2, 4}" -> "{2, 4}" [label="  a",fontsize=10]
 fontsize=10
 //label="Minimization tree"

} </graph>

The tree expansion for non-splitting sets has been omitted for simplicity (transitions for super-state {2,4}).

Given the minimization tree, the final minimal DFA is as follows. Note that states 1 and 3 cannot be the same since they recognize different tokens.

<graph> digraph mindfa {

    { node [shape=circle style=invis] s }
 rankdir=LR; ratio=0.5
 node [shape=doublecircle,fixedsize=true,width=0.2,fontsize=10]; 0 1 24 3
 node [shape=circle,fixedsize=true,width=0.2,fontsize=10];
 s -> 0
 0 -> 1 [label="a",fontsize=10]
 1 -> 3 [label="a",fontsize=10]
 0 -> 24 [label="b",fontsize=10]
 24 -> 24 [label="b",fontsize=10]
 3 -> 3 [label="a",fontsize=10]
 fontsize=10

} </graph>

Input Analysis

In	Input	In+1 / Token
0	aababb$	1
1	ababb$	3
3	babb$	T2 (aa)
0	babb$	24
24	abb$	T1 (b)
0	abb$	1
1	bb$	T1 (a)
0	bb$	24
24	b$	24
24	$	T1 (bb)

The input string aababb is, after 10 steps, split into three tokens: T2 (corresponding to lexeme aa), T1 (b), T1 (a), and T1 (bb).