1 Theory of Automata & Formal Languages
Finite automata
Dec 14, 2014
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1
Theory of Automata &
Formal Languages
2
BOOKS
Theory of computer Science: K.L.P.Mishra &
N.Chandrasekharan
Intro to Automata theory, Formal languages and computation: Ullman,Hopcroft
Motwani
Elements of theory of computation Lewis & papadimitrou
3
Syllabus
Introduction
Deterministic and non deterministic Finite Automata, Regular Expression,Two way finite automata,Finite automata with output,properties of regular sets,pumping lemma, closure properties,Myhill nerode theorem
4
Context free Grammar: Derivation trees, Simplification forms
Pushdown automata: Def, Relationship between PDA and context free language,Properties, decision algorithms
Turing Machines: Turing machine model,Modification of turing machines,Church’s thesis,Undecidability,Recursive and recursively enumerable languages Post correspondence problems recursive functions
5
Chomsky Hierarchy: Regular grammars, unrestricted grammar, context sensitive language, relationship among languages
6
7
8
Finite AutomatonInput
String
•
Output
String
FiniteAutomaton
9
Finite Accepter
•
Input
“Accept” or“Reject”
String
FiniteAutomaton
Output
10
Transition Graph
•
initialstate
final state“accept”state
transition
Abba -Finite Accepter
0q 1q 2q 3q 4qa b b a
5q
a a bb
ba,
ba,
11
Initial Configuration
•
1q 2q 3q 4qa b b a
5q
a a bb
ba,
Input Stringa b b a
ba,
0q
12
Reading the Input
•
0q 1q 2q 3q 4qa b b a
5q
a a bb
ba,
a b b a
ba,
13
0q 1q 2q 3q 4qa b b a
5q
a a bb
ba,
a b b a
ba,
14
0q 1q 2q 3q 4qa b b a
5q
a a bb
ba,
a b b a
ba,
15
0q 1q 2q 3q 4qa b b a
5q
a a bb
ba,
a b b a
ba,
16
0q 1q 2q 3q 4qa b b a
Output: “accept”
5q
a a bb
ba,
a b b a
ba,
Input finished
17
Rejection
•
1q 2q 3q 4qa b b a
5q
a a bb
ba,
a b a
ba,
0q
18
•
0q 1q 2q 3q 4qa b b a
5q
a a bb
ba,
a b a
ba,
19
•
0q 1q 2q 3q 4qa b b a
5q
a a bb
ba,
a b a
ba,
20
•
0q 1q 2q 3q 4qa b b a
5q
a a bb
ba,
a b a
ba,
21
0q 1q 2q 3q 4qa b b a
5q
a a bb
ba,
Output:“reject”
a b a
ba,
Input finished
22
Another Example
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
25
a
b ba,
ba,
0q 1q 2q
a ba
26
a
b ba,
ba,
0q 1q 2q
a ba
Output: “accept”
Input finished
27
Rejection
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
30
a
b ba,
ba,
0q 1q 2q
ab b
31
a
b ba,
ba,
0q 1q 2q
ab b
Output: “reject”
Input finished
32
• Deterministic Finite Accepter (DFA)
FqQM ,,,, 0
Q
0q
F
: Finite set of states
: input alphabet
: transition function
: initial state is a member of Q
: set of final states
: Q X Q
33
Input Alphabet
0q 1q 2q 3q 4qa b b a
5q
a a bb
ba,
ba,
ba,
34
Set of States
Q
0q 1q 2q 3q 4qa b b a
5q
a a bb
ba,
543210 ,,,,, qqqqqqQ
ba,
35
Initial State
0q
1q 2q 3q 4qa b b a
5q
a a bb
ba,
ba,
0q
36
Set of Final States
F
0q 1q 2q 3qa b b a
5q
a a bb
ba,
4qF
ba,
4q
37
Transition Function
0q 1q 2q 3q 4qa b b a
5q
a a bb
ba,
QQ :
ba,
38
10 , qaq
2q 3q 4qa b b a
5q
a a bb
ba,
ba,
0q 1q
39
50 , qbq
1q 2q 3q 4qa b b a
5q
a a bb
ba,
ba,
0q
40
0q 1q 2q 3q 4qa b b a
5q
a a bb
ba,
ba,
32 , qbq
41
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
2q 3q
4q 5q
ba,5q5q5q5q
42
Extended Transition Function
*
QQ *:*
0q 1q 2q 3q 4qa b b a
5q
a a bb
ba,
ba,
43
20 ,* qabq
3q 4qa b b a
5q
a a bb
ba,
ba,
0q 1q 2q
44
40 ,* qabbaq
0q 1q 2q 3q 4qa b b a
5q
a a bb
ba,
ba,
45
50 ,* qabbbaaq
1q 2q 3q 4qa b b a
5q
a a bb
ba,
ba,
0q
46
50 ,* qabbbaaq
1q 2q 3q 4qa b b a
5q
a a bb
ba,
ba,
0q
Observation: There is a walk from to with label
0qabbbaa
5q
47
Recursive Definition
•
)),,(*(,*
,*
awqwaq
0q 1q 2q 3q 4qa b b a
5q
a a bb
ba,
ba,
48
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
49
Languages Accepted by DFAs• Take DFA
• Definition:
– The language contains – all input strings accepted by
– = { strings that drive to a final state}
M
MLM
M ML
50
Example
•
0q 1q 2q 3q 4qa b b a
5q
a a bb
ba,
ba,
abbaML M
accept
51
Another Example
0q 1q 2q 3q 4qa b b a
5q
a a bb
ba,
ba,
abbaabML ,, M
acceptacceptaccept
•
52
Formally
• For a DFA Language accepted by :
FqQM ,,,, 0
M
FwqwML ,*:* 0
alphabet transitionfunction
initialstate
finalstates
53
Observation
• Language accepted by :
• Language rejected by :
FwqwML ,*:* 0M
FwqwML ,*:* 0M
54
More Examples•
a
b ba,
ba,
0q 1q 2q
}0:{ nbaML n
accept trap state or dead state
55
•
ML = { all strings with prefix }ab
a b
ba,
0q 1q 2q
accept
ba,3q
ab
56
•
ML = { all strings without substring }001
0 00 001
1
0
1
10
0 1,0
57
Regular Languages
• A language is regular if there is
• a DFA such that
• All regular languages form a language family–
LM MLL
58
Example
• The language
• is regular:
•
*,: bawawaL
a
b
ba,
a
b
ba
0q 2q 3q
4q
59
Non Deterministic Automata
60
Non Deterministic Finite Accepter
FqQM ,,,, 0
: 2QQ
61
1q 2q
3q
a
a
a
0q
}{aAlphabet =
Nondeterministic Finite Accepter (NFA)
62
1q 2q
3q
a
a
a
0q
Two choices
}{aAlphabet =
Nondeterministic Finite Accepter (NFA)
63
No transition
1q 2q
3q
a
a
a
0q
Two choices No transition
}{aAlphabet =
Nondeterministic Finite Accepter (NFA)
64
a a
0q
1q 2q
3q
a
a
First Choice
a
65
a a
0q
1q 2q
3q
a
a
a
First Choice
66
a a
0q
1q 2q
3q
a
a
First Choice
a
67
a a
0q
1q 2q
3q
a
a
a “accept”
First Choice
All input is consumed
68
a a
0q
1q 2q
3q
a
a
Second Choice
a
69
a a
0q
1q 2qa
a
Second Choice
a
3q
70
a a
0q
1q 2qa
a
a
3q
Second Choice
No transition:the automaton hangs
71
a a
0q
1q 2qa
a
a
3q
Second Choice
“reject”
Input cannot be consumed
72
An NFA accepts a string:
when there is a computation of the NFAthat accepts the string
•All the input is consumed and the automaton is in a final state
73
An NFA rejects a string:
when there is no computation of the NFAthat accepts the string
• All the input is consumed and the automaton is in a non final state
• The input cannot be consumed
74
Example
aa is accepted by the NFA:
0q
1q 2q
3q
a
a
a
“accept”
0q
1q 2qa
a
a
3q “reject”
because this computationaccepts aa
75
a
0q
1q 2q
3q
a
a
Rejection example
a
76
a
0q
1q 2q
3q
a
a
a
First Choice
77
a
0q
1q 2q
3q
a
a
a
First Choice
“reject”
78
Second Choice
a
0q
1q 2q
3q
a
a
a
79
Second Choice
a
0q
1q 2qa
a
a
3q
80
Second Choice
a
0q
1q 2qa
a
a
3q “reject”
81
Examplea is rejected by the NFA:
0q
1q 2qa
a
a
3q “reject”
0q
1q 2qa
a
a
3q
“reject”
All possible computations lead to rejection
82
Rejection example
a a
0q
1q 2q
3q
a
a
a
a
83
a a
0q
1q 2q
3q
a
a
a
First Choice
a
84
a a
0q
1q 2q
3q
a
a
First Choice
a
a
No transition:the automaton hangs
85
a a
0q
1q 2q
3q
a
a
a “reject”
First Choice
a
Input cannot be consumed
86
a a
0q
1q 2q
3q
a
a
Second Choice
a
a
87
a a
0q
1q 2qa
a
Second Choice
a
3q
a
88
a a
0q
1q 2qa
a
a
3q
Second Choice
No transition:the automaton hangs
a
89
a a
0q
1q 2qa
a
a
3q
Second Choice
“reject”
a
Input cannot be consumed
90
aaais rejected by the NFA:
0q
1q 2q
3q
a
a
a
“reject”
0q
1q 2qa
a
a
3q “reject”
All possible computations lead to rejection
91
1q 2q
3q
a
a
a
0q
Language accepted: }{aaL
92
Lambda →Transitions
1q 3qa0q
2q a
93
a a
1q 3qa0q
2q a
94
a a
1q 3qa0q
2q a
95
a a
1q 3qa0q
2q a
(read head doesn’t move)
96
a a
1q 3qa0q
2q a
97
a a
1q 3qa0q
2q a
“accept”
String is acceptedaa
all input is consumed
98
a a
1q 3qa0q
2q a
Rejection Example
a
99
a a
1q 3qa0q
2q a
a
100
a a
1q 3qa0q
2q a
(read head doesn’t move)
a
101
a a
1q 3qa0q
2q a
a
No transition:the automaton hangs
102
a a
1q 3qa0q
2q a
“reject”
String is rejectedaaa
a
Input cannot be consumed
103
Language accepted: }{aaL
1q 3qa0q
2q a
104
Another NFA Example
0q 1q 2qa b
3q
105
a b
0q 1q 2qa b
3q
106
0q 2qa b
3q
a b
1q
107
a b
0q 1qa b
3q2q
108
a b
0q 1qa b
3q2q
“accept”
109
0qa b
a b
Another String
a b
1q 2q 3q
110
0qa b
a b a b
1q 2q 3q
111
0qa b
a b a b
1q 2q 3q
112
0qa b
a b a b
1q 2q 3q
113
0qa b
a b a b
1q 2q 3q
114
0qa b
a b a b
1q 2q 3q
115
0qa b
a b a b
1q 2q 3q
116
a b a b
0qa b
1q 2q 3q
“accept”
117
ab
ababababababL ...,,,
Language accepted
0q 1q 2qa b
3q
118
Another NFA Example
0q 1q 2q0
11,0
119
{ }{ } *10=
...,101010,1010,10,λ=)(ML
0q 1q 2q0
11,0
Language accepted
120
Remarks:
•The symbol never appears on the input tape
0q2M
0q1M
{}=)M(L 1 }λ{=)M(L 2
•Extreme automata:
121
0q
2q
1qa
0q 1qa
}{=)( 1 aML
2M1M
}{=)( 2 aML
NFA DFA
•NFAs are interesting because we can express languages easier than DFAs
ba,b
a,b
122
Formal Definition of NFAs
FqQM ,,,, 0
:Q
::0q
:F
Set of states, i.e. 210 ,, qqq
: Input alphabet, i.e. ba,
Transition function
Initial state
Final states
123
10 1, qq
0
11,0
Transition Function
0q 1q 2q
124
0q
0
11,0
},{)0,( 201 qqq
1q 2q
125
0q
0
11,0
1q 2q
},{),( 200 qqq
126
0q
0
11,0
1q 2q
)1,( 2q
127
Extended Transition Function
•
*
0q
5q4q
3q2q1qa
aa
b
10 ,* qaq
128
540 ,,* qqaaq
0q
5q4q
3q2q1qa
aa
b
129
0320 ,,,* qqqabq
0q
5q4q
3q2q1qa
aa
b
130
Formally
wqq ij ,*
It holdsif and only if
there is a walk from towith label
iq jqw
131
The Language of an NFA
•
0q
5q4q
3q2q1qa
aa
b
540 ,,* qqaaq
M
)(MLaa
50 ,qqF
132
0q
5q4q
3q2q1qa
aa
b
0320 ,,,* qqqabq MLab
50 ,qqF
133
•
0q
5q4q
3q2q1qa
aa
b
50 ,qqF
540 ,,* qqabaaq )(MLaaba
134
0q
5q4q
3q2q1qa
aa
b
50 ,qqF
10 ,* qabaq MLaba
135
0q
5q4q
3q2q1qa
aa
b
aaababaaML *
136
Formally
• The language accepted by NFA is:
• where
• and there is some
M
,...,, 321 wwwML
,...},{),(* 0 jim qqwq
Fqk (final state)
137
•
0q kq
w
w
w
),(* 0 wq MLw
Fqk
iq
jq
138
Equivalence of NFAs and DFAs
139
Equivalence of Machines
• For DFAs or NFAs:
• Machine is equivalent to machine
• if
•
1M 2M
21 MLML
140
• 0q 1q 2q
0
11,0
0q 1q 2q
0
11
0
1,0
NFA
DFA
*}10{1 ML
*}10{2 ML
1M
2M
141
• Since
• machines and are equivalent
*1021 MLML
1M 2M
0q 1q 2q
0
11,0
0q 1q 2q
0
11
0
1,0
DFA
NFA 1M
2M
142
Equivalence of NFAs and DFAs
Question: NFAs = DFAs ?
Same power?Accept the same languages?
143
Equivalence of NFAs and DFAs
Question: NFAs = DFAs ?
Same power?Accept the same languages?
YES!
144
We will prove:
Languages acceptedby NFAs
Languages acceptedby DFAs
NFAs and DFAs have the same computation power
145
Languages acceptedby NFAs
Languages acceptedby DFAs
Step 1
Proof: Every DFA is trivially an NFA
A language accepted by a DFAis also accepted by an NFA
146
Languages acceptedby NFAs
Languages acceptedby DFAs
Step 2
Proof: Any NFA can be converted to anequivalent DFA
A language accepted by an NFAis also accepted by a DFA
147
NFA to DFA
•
a
b
a
0q 1q 2q
NFA
DFA
0q
M
M
148
•
a
b
a
0q 1q 2q
NFA
DFA
0q 21,qqa
M
M
149
•
a
b
a
0q 1q 2q
NFA
DFA
0q 21,qqa
b
M
M
150
•
a
b
a
0q 1q 2q
NFA
DFA
0q 21,qqa
b
a
M
M
151
•
a
b
a
0q 1q 2q
NFA
DFA
0q 21,qqa
b
ab
M
M
152
•
a
b
a
0q 1q 2q
NFA
DFA
0q 21,qqa
b
ab
ba,
M
M
153
•
a
b
a
0q 1q 2q
NFA
DFA
0q 21,qqa
b
ab
ba,
M
M
)(MLML
154
NFA to DFA: Remarks
• We are given an NFA
• We want to convert it to an equivalent DFA
M
M
)(MLML
155
• If the NFA has states
• the DFA has states in the power set
•
,...,, 210 qqq
,....,,,,,,, 7432110 qqqqqqq
156
Procedure NFA to DFA
•
• 1. Initial state of NFA:
•
• Initial state of DFA:
0q
0q
157
•
a
b
a
0q 1q 2q
NFA
DFA
0q
M
M
158
Procedure NFA to DFA
• 2. For every DFA’s state
• Compute in the NFA
•
• Add transition to DFA
},...,,{ mji qqq
...
,,*
,,*
aq
aq
j
i
},...,,{ mji qqq
},...,,{},,...,,{ mjimji qqqaqqq
159
•
a
b
a
0q 1q 2q
NFA
0q 21,qqa
DFA
},{),(* 210 qqaq
210 ,, qqaq
M
M
},{),(* 210 qqaq
160
Procedure NFA to DFA
• Repeat Step 2 for all letters in alphabet,
• until
• no more transitions can be added.
161
•
a
b
a
0q 1q 2q
NFA
DFA
0q 21,qqa
b
ab
ba,
M
M
162
Procedure NFA to DFA
• 3. For any DFA state
• If some is a final state in the NFA
• Then, • is a final state in the DFA•
},...,,{ mji qqq
jq
},...,,{ mji qqq
163
•
a
b
a
0q 1q 2q
NFA
DFA
0q 21,qqa
b
ab
ba,
Fq 1
Fqq 21,
M
M
164
Theorem
• Take NFA M
Apply procedure to obtain DFA M
Then and are equivalent :M M
MLML
165
FinallyWe have proven
Languages acceptedby NFAs
Languages acceptedby DFAs
166
Languages acceptedby NFAs
Languages acceptedby DFAs
We have proven
Regular Languages
167
Languages acceptedby NFAs
Languages acceptedby DFAs
We have proven
Regular LanguagesRegular Languages
168
Languages acceptedby NFAs
Languages acceptedby DFAs
We have proven
Regular LanguagesRegular Languages
Thus, NFAs accept the regular languages
169
Single Final State for NFAs and DFAs
170
Observation
• Any Finite Automaton (NFA or DFA)
• can be converted to an equivalent NFA
• with a single final state
171
a
b
b
aNFA
Equivalent NFA
a
b
b
a
Example
172
NFAIn General
Equivalent NFA
Singlefinal state
173
Extreme Case
NFA without final state
Add a final stateWithout transitions
174
Some Properties of Regular Languages
175
1L2L
21LLConcatenation:
*1LStar:
21 LL Union:
Are regularLanguages
For regular languages and we will prove that:
properties
176
We Say:Regular languages are closed under
21LLConcatenation:
*1LStar:
21 LL Union:
177
Properties
1L 2L
21LLConcatenation:
*1LStar:
21 LL Union:
Are regularLanguages
For regular languages and we will prove that:
178
1LRegular language
11 LML
1M
Single final state
NFA 2M
2L
Single final state
22 LML
Regular language
NFA
179
Example
}{1 baL na
b 1M
baL 2ab
2M
180
Union
• NFA for 1M
2M
21 LL
181
• a
b
ab
}{1 baL n
}{2 baL
}{}{21 babaLL n NFA for
182
Concatenation
• NFA for
21LL
1M 2M
183
Example
•
• NFA for
a
b ab
}{1 baL n}{2 baL
}{}}{{21 bbaababaLL nn
184
Star Operation• NFA for *1L
1M
*1L
185
Example
• NFA for
*}{*1 baL n
a
b
}{1 baL n
N>= 0
186
Procedure: NFA to DFA
1 Create a graph with vertex {q0}.Identify this vertex as initial vertex of DFA.
2 Repeat the following steps until no more edges are missing.Take any vertex {qi,qj,….qk} of G that has no outgoing edge for some symbol a of the alphabet.Compute (qi, a), (qj, a)…. (qk, a)Then form the union of all these yielding the set {ql, qm, …qn}.Create a vertex for G labeled {ql, qm,…qn} if it does not already exist.Add to G an edge from {qi, qj,…qk} to {ql,qm…qn} and label it with a.
3 Every state of G whose label contains any qf of F is identified as a final vertex of DFA.
4 If NFA accepts then vertex {q0} in G is also made a final vertex.
**
*
*
187
q0 q2q10,1 0,1
10
188
q0
{q0,q1}
q1
q0,q1,q2q1,q2
q2
0
01
1
0
0,1
0,1
1
1
0,1
Equivalent DFA
Start
0
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