non deterministic automata
DESCRIPTION
Non Deterministic Automata. Nondeterministic Finite Accepter (NFA). Alphabet =. Nondeterministic Finite Accepter (NFA). Alphabet =. Two choices. Nondeterministic Finite Accepter (NFA). Alphabet =. Two choices. No transition. No transition. First Choice. First Choice. First Choice. - PowerPoint PPT PresentationTRANSCRIPT
1
Non Deterministic Automata
2
1q 2q
3q
a
a
a
0q
}{aAlphabet =
Nondeterministic Finite Accepter (NFA)
3
1q 2q
3q
a
a
a
0q
Two choices
}{aAlphabet =
Nondeterministic Finite Accepter (NFA)
4
No transition
1q 2q
3q
a
a
a
0q
Two choices No transition
}{aAlphabet =
Nondeterministic Finite Accepter (NFA)
5
a a
0q
1q 2q
3q
a
a
First Choice
a
6
a a
0q
1q 2q
3q
a
a
a
First Choice
7
a a
0q
1q 2q
3q
a
a
First Choice
a
8
a a
0q
1q 2q
3q
a
a
a “accept”
First Choice
9
a a
0q
1q 2q
3q
a
a
Second Choice
a
10
a a
0q
1q 2qa
a
Second Choice
a
3q
11
a a
0q
1q 2qa
a
a
3q
Second Choice
No transition:the automaton hangs
12
a a
0q
1q 2qa
a
a
3q
Second Choice
“reject”
13
Observation
An NFA accepts a stringifthere is a computation of the NFAthat accepts the string
14
Example
aa is accepted by the NFA:
0q
1q 2q
3q
a
a
a
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Lambda Transitions
1q 3qa0q
1q a
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a a
1q 3qa0q
2q a
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a a
1q 3qa0q
2q a
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a a
1q 3qa0q
2q a
(read head doesn’t move)
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a a
1q 3qa0q
2q a
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a a
1q 3qa0q
2q a
“accept”
String is acceptedaa
21
Language accepted: }{aaL
1q 3qa0q
2q a
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Another NFA Example
0q 1q 2qa b
3q
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a b
0q 1q 2qa b
3q
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0q 2qa b
3q
a b
1q
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a b
0q 1qa b
3q2q
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a b
0q 1qa b
3q2q
“accept”
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0qa b
a b
Another String
a b
1q 2q 3q
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0qa b
a b a b
1q 2q 3q
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0qa b
a b a b
1q 2q 3q
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0qa b
a b a b
1q 2q 3q
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0qa b
a b a b
1q 2q 3q
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0qa b
a b a b
1q 2q 3q
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0qa b
a b a b
1q 2q 3q
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a b a b
0qa b
1q 2q 3q
“accept”
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ab
ababababababL ...,,,
Language accepted
0q 1q 2qa b
3q
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Another NFA Example
0q 1q 2q0
11,0
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*10
...,101010,1010,10,
L
0q 1q 2q0
11,0
Language accepted
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Formal Definition of NFAs FqQM ,,,, 0
:Q
::0q
:F
Set of states, i.e. 210 ,, qqq
: Input aplhabet, i.e. ba,Transition function
Initial state
Final states
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10 1, qq
0
11,0
Transition Function
0q 1q 2q
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0q
0
11,0
},{)0,( 201 qqq
1q 2q
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0q
0
11,0
1q 2q
},{),( 200 qqq
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0q
0
11,0
1q 2q
)1,( 2q
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Extended Transition Function
*
0q
5q4q
3q2q1qa
aa
b
10 ,* qaq
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540 ,,* qqaaq
0q
5q4q
3q2q1qa
aa
b
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0320 ,,,* qqqabq
0q
5q4q
3q2q1qa
aa
b
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Formally
wqq ij ,*It holds
if and only if
there is a walk from towith label
iq jqw
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The Language of an NFA
0q
5q4q
3q2q1qa
aa
b
540 ,,* qqaaq
M
)(MLaa
50 ,qqF
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0q
5q4q
3q2q1qa
aa
b
0320 ,,,* qqqabq MLab
50 ,qqF
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0q
5q4q
3q2q1qa
aa
b
50 ,qqF
540 ,,* qqabaaq )(MLaaba
50
0q
5q4q
3q2q1qa
aa
b
50 ,qqF
10 ,* qabaq MLaba
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0q
5q4q
3q2q1qa
aa
b
aaababaaML *
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FormallyThe language accepted by NFA is:
where
and there is some
M
,...,, 321 wwwML
,...},{),(* 0 jim qqwq
Fqk (final state)
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0q kq
w
w
w
),(* 0 wq MLw
Fqk
iq
jq
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Equivalence of NFAs and DFAs
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Equivalence of Machines
For DFAs or NFAs:
Machine is equivalent to machine
if
1M 2M
21 MLML
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Example
0q 1q 2q
0
11,0
0q 1q 2q
0
11
0
1,0
NFA
DFA
*}10{1 ML
*}10{2 ML
1M
2M
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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
58
Equivalence of NFAs and DFAs
Question: NFAs = DFAs ?
Same power?Accept the same languages?
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Equivalence of NFAs and DFAs
Question: NFAs = DFAs ?
Same power?Accept the same languages?
YES!
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We will prove:
Languages acceptedby NFAs
Languages acceptedby DFAs
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We will prove:
Languages acceptedby NFAs
Languages acceptedby DFAs
NFAs and DFAs have the same computation power
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Languages acceptedby NFAs
Languages acceptedby DFAs
Step 1
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Languages acceptedby NFAs
Languages acceptedby DFAs
Step 1
Proof: Every DFA is also an NFA
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Languages acceptedby NFAs
Languages acceptedby DFAs
Step 1
Proof: Every DFA is also an NFA
A language accepted by a DFAis also accepted by an NFA
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Languages acceptedby NFAs
Languages acceptedby DFAs
Step 2
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Languages acceptedby NFAs
Languages acceptedby DFAs
Step 2
Proof: Any NFA can be converted to anequivalent DFA
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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
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NFA to DFA
a
b
a
0q 1q 2q
NFA
DFA
0q
69
NFA to DFA
a
b
a
0q 1q 2q
NFA
DFA
0q 21,qqa
70
NFA to DFA
a
b
a
0q 1q 2q
NFA
DFA
0q 21,qqa
b
71
NFA to DFA
a
b
a
0q 1q 2q
NFA
DFA
0q 21,qqa
b
a
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NFA to DFA
a
b
a
0q 1q 2q
NFA
DFA
0q 21,qqa
b
ab
73
NFA to DFA
a
b
a
0q 1q 2q
NFA
DFA
0q 21,qqa
b
ab
ba,
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NFA to DFA
a
b
a
0q 1q 2q
NFA
DFA
0q 21,qqa
b
ab
ba,
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NFA to DFA: Remarks
We are given an NFA
We want to convert it to an equivalent DFA
With
M
M
)(MLML
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If the NFA has states
the DFA has states in the powerset
,...,, 210 qqq
,....,,,,,,, 7432110 qqqqqqq
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Procedure NFA to DFA
1. Initial state of NFA:
Initial state of DFA:
0q
0q
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Example
a
b
a
0q 1q 2q
NFA
DFA
0q
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Procedure NFA to DFA
2. For every DFA’s state
Compute in the NFA
Add transition
},...,,{ mji qqq
...
,,*
,,*
aq
aq
j
i
},...,,{ mji qqq
},...,,{},,...,,{ mjimji qqqaqqq
80
Exampe
a
b
a
0q 1q 2q
NFA
0q 21,qqa
DFA
},{),(* 210 qqaq
210 ,, qqaq
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Procedure NFA to DFA
Repeat Step 2 for all letters in alphabet,
untilno more transitions can be added.
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Example
a
b
a
0q 1q 2q
NFA
DFA
0q 21,qqa
b
ab
ba,
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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
84
Example
a
b
a
0q 1q 2q
NFA
DFA
0q 21,qqa
b
ab
ba,
Fq 1
Fqq 21,
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Theorem Take NFA M
Apply procedure to obtain DFA M
Then and are equivalent :M M
MLML
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Finally
We have proven
Languages acceptedby NFAs
Languages acceptedby DFAs
87
Languages acceptedby NFAs
Languages acceptedby DFAs
We have proven
Regular Languages
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Languages acceptedby NFAs
Languages acceptedby DFAs
We have proven
Regular LanguagesRegular Languages