Theoretical Aspects of Lexical Analysis/Exercise 16

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, c }. Indicate the number of processing steps for the given input string.

• G = { bc*, a*|c, a|b* }, input string = aaabcacc

NFA

The following is the result of applying Thompson's algorithm. State 5 recognizes the first expression (token T1); state 13 recognizes token T2; and state 21 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]; 5 13 21
 node [shape=circle,fixedsize=true,width=0.2,fontsize=10];

 s -> 0

 0 -> 1 
 1 -> 2 [label="b",fontsize=10]
 2 -> 3
 2 -> 5
 3 -> 4 [label="c",fontsize=10]
 4 -> 3
 4 -> 5

 0 -> 6 
 6 -> 7
 6 -> 11
 7 -> 8
 7 -> 10
 8 -> 9 [label="a",fontsize=10]
 9 -> 8
 9 -> 10
 11 -> 12 [label="c",fontsize=10]
 12 -> 13
 10 -> 13

 0 -> 14
 14 -> 15
 14 -> 17
 15 -> 16 [label="a",fontsize=10]
 17 -> 18
 17 -> 20
 18 -> 19 [label="b",fontsize=10]
 19 -> 18
 19 -> 20
 20 -> 21
 16 -> 21

 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, 6, 7, 8, 10, 11, 13, 14, 15, 17, 18, 20, 21	0 (T2)
0	a	9, 16	8, 9, 10, 13, 16, 21	1 (T2)
0	b	2, 19	2, 3, 5, 18, 19, 20, 21	2 (T1)
0	c	12	12, 13	3 (T2)
1	a	9	8, 9, 10, 13	4 (T2)
1	b	-	-	-
1	c	-	-	-
2	a	-	-	-
2	b	19	18, 19, 20, 21	5 (T3)
2	c	4	3, 4, 5	6 (T1)
3	a	-	-	-
3	b	-	-	-
3	c	-	-	-
4	a	9	8, 9, 10, 13	4 (T2)
4	b	-	-	-
4	c	-	-	-
5	a	-	-	-
5	b	19	18, 19, 20, 21	5 (T3)
5	c	-	-	-
6	a	-	-	-
6	b	-	-	-
6	c	4	3, 4, 5	6 (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 5 6
 node [shape=circle,fixedsize=true,width=0.2,fontsize=10];
 s -> 0
 0 -> 1 [label="a",fontsize=10]
 0 -> 2 [label="b",fontsize=10]
 0 -> 3 [label="c",fontsize=10]
 1 -> 4 [label="a",fontsize=10]
 2 -> 5 [label="b",fontsize=10]
 2 -> 6 [label="c",fontsize=10]
 4 -> 4 [label="a",fontsize=10]
 5 -> 5 [label="b",fontsize=10]
 6 -> 6 [label="c",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, 5, 6}" -> "{}       " [label="NF",fontsize=10]
 "{0, 1, 2, 3, 4, 5, 6}" -> "{0, 1, 2, 3, 4, 5, 6} " [label="  F",fontsize=10]
 "{0, 1, 2, 3, 4, 5, 6} " -> "{2, 6}" [label="  T1",fontsize=10]
 "{0, 1, 2, 3, 4, 5, 6} " -> "{0, 1, 3, 4}" [label="  T2",fontsize=10]
 "{0, 1, 2, 3, 4, 5, 6} " -> "{5}" [label="  T3",fontsize=10]
 "{0, 1, 3, 4}" -> "{0, 1, 4}" [label="  a",fontsize=10]
 "{0, 1, 3, 4}" -> "{3}" //[label="  a",fontsize=10]
 "{2, 6}" -> "{2}" [label="  b",fontsize=10]
 "{2, 6}" -> "{6}" //[label="  b",fontsize=10]
 "{0, 1, 4}" -> "{0}" [label="  b",fontsize=10]
 "{0, 1, 4}" -> "{1, 4}" //[label="  b",fontsize=10]
 "{1, 4}" -> "{1, 4} " [label="  a, b, c",fontsize=10]
 fontsize=10
 //label="Minimization tree"

} </graph>

Given the minimization tree, the final minimal DFA is 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 14 2 3 5 6
 node [shape=circle,fixedsize=true,width=0.2,fontsize=10];
 s -> 0
 0 -> 14 [label="a",fontsize=10]
 0 -> 2 [label="b",fontsize=10]
 0 -> 3 [label="c",fontsize=10]
 14 -> 14 [label="a",fontsize=10]
 2 -> 5 [label="b",fontsize=10]
 2 -> 6 [label="c",fontsize=10]
 5 -> 5 [label="b",fontsize=10]
 6 -> 6 [label="c",fontsize=10]
 fontsize=10

} </graph>

Input Analysis

In	Input	In+1 / Token
0	aaabcacc$	14
14	aabcacc$	14
14	abcacc$	14
14	bcacc$	T2 (aaa)
0	bcacc$	2
2	cacc$	6
6	acc$	T1 (bc)
0	acc$	14
14	cc$	T2 (a)
0	cc$	3
3	c$	T2 (c)
0	c$	3
3	$	T2 (c)

The input string aaabcacc is, after 13 steps, split into five tokens: T2 (corresponding to lexeme aaa), T1 (bc), T2 (a), T2 (c), T2 (c).