Formal Languages Finite Automata
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DESCRIPTION
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Fall 2004 Costas Busch - RPI
Formal Languages
Finite Automata
2
Type-0 : Unrestricted;recursively enumerated; phrase structured; turing m/c
Type-1: CSL;LBAType-2: CFL;PDAType-3:RL:FA
Chomsky Hierarchy
3
Finite Automaton
Input
“Accept” or“Reject”
String
FiniteAutomaton
Output
4
Transition Graph
initialstate
accepting state
statetransition
0q 1q 2q 3q 4qa b b a
5q
a a bb
ba,
ba,
5
Initial Configuration
1q 2q 3q 4qa b b a
5q
a a bb
ba,
Input Stringa b b a
ba,
0q
6
Reading the Input
0q 1q 2q 3q 4qa b b a
5q
a a bb
ba,
a b b a
ba,
7
0q 1q 2q 3q 4qa b b a
5q
a a bb
ba,
a b b a
ba,
8
0q 1q 2q 3q 4qa b b a
5q
a a bb
ba,
a b b a
ba,
9
0q 1q 2q 3q 4qa b b a
5q
a a bb
ba,
a b b a
ba,
10
0q 1q 2q 3q 4qa b b a
accept
5q
a a bb
ba,
a b b a
ba,
Input finished
11
Rejection
1q 2q 3q 4qa b b a
5q
a a bb
ba,
a b a
ba,
0q
12
0q 1q 2q 3q 4qa b b a
5q
a a bb
ba,
a b a
ba,
13
0q 1q 2q 3q 4qa b b a
5q
a a bb
ba,
a b a
ba,
14
0q 1q 2q 3q 4qa b b a
5q
a a bb
ba,
a b a
ba,
15
0q 1q 2q 3q 4qa b b a
5q
a a bb
ba,
reject
a b a
ba,
Input finished
16
Acceptance or Rejection?
1q 2q 3q 4qa b b a
5q
a a bb
ba,
ba,
0q
17
Initial State
1q 2q 3q 4qa b b a
5q
a a bb
ba,
ba,
0q
18
1q 2q 3q 4qa b b a
5q
a a bb
ba,
ba,
0q
reject
Rejection
19
Language?
a
b ba,
ba,
0q 1q 2q
20
Another Example
a
b ba,
ba,
0q 1q 2q
a ba
21
a
b ba,
ba,
0q 1q 2q
a ba
22
a
b ba,
ba,
0q 1q 2q
a ba
23
a
b ba,
ba,
0q 1q 2q
a ba
24
a
b ba,
ba,
0q 1q 2q
a ba
accept
Input finished
25
Rejection Example
a
b ba,
ba,
0q 1q 2q
ab b
26
a
b ba,
ba,
0q 1q 2q
ab b
27
a
b ba,
ba,
0q 1q 2q
ab b
28
a
b ba,
ba,
0q 1q 2q
ab b
29
a
b ba,
ba,
0q 1q 2q
ab b
reject
Input finished
30
Languages Accepted by FAsFA
Definition:The language contains all input strings accepted by
= { strings that bring to an accepting state}
M
MLM
M ML
31
Example: L(M) = ?
0q 1q 2q 3q 4qa b b a
5q
a a bb
ba,
ba,
M
accept
32
Example
0q 1q 2q 3q 4qa b b a
5q
a a bb
ba,
ba,
M
accept
33
Example: L(M) = ?
0q 1q 2q 3q 4qa b b a
5q
a a bb
ba,
ba,
M
acceptacceptaccept
34
Example
0q 1q 2q 3q 4qa b b a
5q
a a bb
ba,
ba,
abbaabML ,, M
acceptacceptaccept
35
Example: L(M) = ?
a
b ba,
ba,
0q 1q 2q
accept trap state
36
Example
a
b ba,
ba,
0q 1q 2q
}0:{ nbaML n
accept trap state
37
Formal Definition Finite Automaton (FA)
FqQM ,,,, 0
Q
0q
F
: set of states
: input alphabet
: transition function
: initial state
: set of accepting states
38
Input Alphabet
0q 1q 2q 3q 4qa b b a
5q
a a bb
ba,
ba,
ba,
39
Set of States
Q
0q 1q 2q 3q 4qa b b a
5q
a a bb
ba,
543210 ,,,,, qqqqqqQ
ba,
40
Initial State
0q
1q 2q 3q 4qa b b a
5q
a a bb
ba,
ba,
0q
41
Set of Accepting States
F
0q 1q 2q 3qa b b a
5q
a a bb
ba,
4qF
ba,
4q
42
Transition Function
0q 1q 2q 3q 4qa b b a
5q
a a bb
ba,
QQ :
ba,
43
10 , qaq
2q 3q 4qa b b a
5q
a a bb
ba,
ba,
0q 1q
44
50 , qbq
1q 2q 3q 4qa b b a
5q
a a bb
ba,
ba,
0q
45
0q 1q 2q 3q 4qa b b a
5q
a a bb
ba,
ba,
32 , qbq
46
Transition Function
0q 1q 2q 3q 4qa b b a
5q
a a bb
ba,
a b
0q
1q
2q
3q
4q
5q
1q 5q
5q 2q
5q 3q
4q 5q
ba,5q5q5q5q
47
Extended Transition Function
*
QQ *:*
0q 1q 2q 3q 4qa b b a
5q
a a bb
ba,
ba,
48
20 ,* qabq
3q 4qa b b a
5q
a a bb
ba,
ba,
0q 1q 2q
49
40 ,* qabbaq
0q 1q 2q 3q 4qa b b a
5q
a a bb
ba,
ba,
50
50 ,* qabbbaaq
1q 2q 3q 4qa b b a
5q
a a bb
ba,
ba,
0q
51
qwq ,*
Observation: if there is a walk from to with label then
q qw
q qw
q qkw 21
1 2 k
52
50 ,* qabbbaaq
1q 2q 3q 4qa b b a
5q
a a bb
ba,
ba,
0q
Example: There is a walk from to with label
0qabbbaa
5q
53
Recursive Definition )),,(*(,*
,*
wqwq
q qw1q
qwq
),(
,*
1
1
1
,*
),(,*
qwq
qwq
)),,(*(,* wqwq
54
0q 1q 2q 3q 4qa b b a
5q
a a bb
ba,
ba,
2
1
0
0
0
0
,
,,
,,,*
),,(*
,*
q
bq
baq
baq
baq
abq
55
Language Accepted by FAs
For a FA
Language accepted by :
FqQM ,,,, 0
M
FwqwML ,*:* 0
0q qw Fq
56
ObservationLanguage rejected by :
FwqwML ,*:* 0
M
0q qw Fq
57
L(M) ? ab
a b
ba,
0q 1q 2q
accept
ba,3q
ab
58
Example ML = { all strings with prefix }ab
a b
ba,
0q 1q 2q
accept
ba,3q
ab
59
0 00 001
1
0
1
10
0 1,0
L(M)?
60
Example ML = { all strings without
substring }001
0 00 001
1
0
1
10
0 1,0
61
a
b
ba,
a
b
ba
0q 2q 3q
4q
L(M) ?
62
Example
*,:)( bawawaML
a
b
ba,
a
b
ba
0q 2q 3q
4q
63
Regular LanguagesDefinition:A language is regular if there is FA such that
Observation:All languages accepted by FAs form the family of regular languages
LM MLL
64
abba abbaab,,}0:{ nban
{ all strings with prefix }ab
{ all strings without substring }001
Examples of regular languages:
There exist automata that accept theseLanguages (see previous slides).
*,: bawawa
65
There exist languages which are not Regular:
}0:{ nbaL nn
There is no FA that accepts such a language
Example:
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