Transcript Finite Automata and Non Determinism

Finite Automata and
Non Determinism
http://cis.k.hosei.ac.jp/~yukita/
Definition 1.1: Finite Automaton
A finite automatonis a 5 - tupple(Q, ,  , q0 , F ), where
1. Q is a finiteset called the states,
2.  is a finiteset called thealphabet,
3. : Q    Q is thetransition function,
4. q0  Q is the start state, and
5. F  Q is the set of accept states.
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State Diagram for M1
1
0
q1
1
0
q2
q3
0, 1
3
Data Representation for M1
1. Q  {q1 , q2 , q3 }
2.   {0,1}
3. is described as
q1
q2
q3
0
q1
q3
q2
1
q2
q2
q2
,
4. q1 is thest art st ate,and
5. F  {q2 }.
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Task 01
DFA
1. Implement M1 with your favorite
programming language.
2. GUI
•
•
Two buttons for input 0 and 1
State chart with the current state highlighted
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Language of M1
T helanguage of machine M , writtenas L( M ),
is theset of all strings thatmachineM accepts.
We can say thatM recognizes A.
L( M 1 )  {w | w containsat least one1 and
an even number of 0s follow thelast 1}
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State Diagram for M5
0
q1
1
0, <RESET>
0
1
2
q0
2
q2
1, <RESET>
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Data Representation for M5
1. Q  {q0 , q1 , q2 }
2.   { RESET ,0,1,2}
3. is described as
R
0
1
2
q0
q0
q0
q1
q2
q1
q0
q1
q2
q0
q2
q0
q2
q0
q1
,
4. q0 is thestart state,and
5. F  {q0 }.
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Informal Description of M5
• M5 keeps a running count of the sum of
the numerical symbols it reads, modulo 3.
• Every time it receives the <RESET>
symbol it resets the count to 0.
• M5 accepts if the sum is 0, modulo 3.
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Definition 1.7: Regular Language
• A language is called a regular language if
some finite automaton recognizes it.
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Example 1.9: A finite automaton E2
• E2 recognizes the regular language of all
strings that contain the string 001 as a
substring.
• 0010, 1001, 001, and 1111110011110 are
all accepted,
• but 11 and 0000 are not.
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Find a set of states of E2
You
1. haven’t just seen any symbols of the
pattern,
2. have just seen a 0,
3. have just seen 00 or,
4. have just seen the entire pattern 001.
Assign the states q,q0,q00, and q001 to these
possibilities.
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Draw a State Diagram for E2
1
0, 1
0
0
q
q0
0
q00
1
q001
1
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Regular Operations on Languages
Let A and B be languages.
Un ion: A  B  {x | x  A or x  B}.
C on cate n at
ion: A  B  {xy | x  A and y  B}.
S tar: A  {x1 x2  xk | k  0 and each xi  A}.
*
T hefirst twooperationsare binary operations,
and thelast one a unary operation.
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Example 1.11
Let thealphabet be {a, b,  , z}.
Let A  {good, bad} and B  {boy, girl}. T hen,we have
A  B  {good, bad, boy, girl},
A  B  {goodboy,goodgirl, badboy, badgirl}, and
A*  { , good, bad, goodgood,goodbad, badbad,
goodgoodgood, good goodbad, goodbadgood,
goodbadbad, }.
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Theorem 1.12 Closedness for Union
T heclass of regular languages is closed under
theunion operation.
In other words, if A1 and A2 are regular languages,
so is A1  A2 .
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Proof of Theorem 1.12
Let M 1 recognizeA1 , where M 1  (Q1 , , 1 , q1 , F1 ),
and M 2 recognizeA2 , where M 2  (Q2 , ,  2 , q2 , F2 ).
Const ructM t o recognizeA1  A2 ,
where M  (Q, ,  , q, F ).
T hen,check thecorrectness of t heconstructon.
i
Proof of Th 1.12
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Construction of M
1. Q  Q1  Q2  {( r1 , r2 ) | r1  Q1 and r2  Q2 }.
2. Wecan assume thatM 1 and M 2 have thesame
alphabet. (Why?)
3. (r1 , r2 )  Q, a  ;  ((r1 , r2 ), a)  (1 (r1 , a ), 2 (r2 , a ))
4. q0  (q1 , q2 ).
5. F  ( F1  Q2 )  (Q1  F2 )  {( r1 , r2 ) | r1  F1 or r2  F2 }.
Proof of Th 1.12
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Correctness
You should check the following.
1. For any string recognized by M1 is
recognized by M.
2. For any string recognized by M2 is
recognized by M.
3. For any string recognized by M is
recognized by M1 or M2.
Proof of Th 1.12
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Theorem 1.13 Closedness for
concatenation
T heclass of regular languages is closed under concatenation
operation.In other words, if A1 and A2 are regular languages,
so is A1  A2 .
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Nondeterminism
• To prove Theorem 1.13, we need
nondeterminism.
• Nondeterminism is a generalization of
determinism. So, every deterministic
automaton is automatically a
nondeterministic automaton.
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Nondetermistic Finite Automata
• A nondeterministic finite automaton can be
different from a deterministic one in that
– for any input symbol, nondeterministic one
can transit to more than one states.
– epsilon transition
• NFA and DFA stand for nondeterministic
finite automaton and deterministic finite
automaton, respectively.
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NFA N1
0,1
0,1
q1
1
q2
0,
1
q3
q4
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Parallel world and NFA
...
...
accept
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Example 1.14
NFA N2
0,1
q1
1
q2
0,1
q3
0,1
q4
Let language A consist of all strings over {0,1} containing a 1 in the third
position from the end. N2 recognizes A.
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A DFA equivalent to N2
0
q000
0
0
1
q100
0
q010
0
q110
1
1
0
0
1
0
1
q001
q101
1
1
q011
1
q111
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Example 1.15
NFA N3
0
0


0
0
0
Let language A consist of all strings 0k , where k is a multiple of 2 or 3. N3
recognizes A.
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A DFA equivalent to N3
0, 1
q-1
1
1
1
1
q0
0
q１
0
1
0
q２
1
q3
0
q4
0
q5
0
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Example 1.16 NFA N4
q1
a
b

a
q2
a,b
q3
N4 accepts , a, baba, and baa. N4 does not accept b, nor babba.
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Definition 1.17: NFA
A nondeterministic finite automatonis a 5 - tupple(Q, ,  , q0 , F ), where
1. Q is a finiteset called thestates,
2.  is a finiteset called thealphabet,
3. : Q    2Q is thetransition function,
4. q0  Q is thestart state, and
5. F  Q is theset of accept states.
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Example 1.18 NFA N1
0,1
1
1. Q  {q1 , q2 , q3 , q4 },
0,
q1
2.   {0,1}
0
q1
0,1
1
{q1} {q1 , q2 }
q2
1
q3
q4


3.  is given as q2 {q3 }
q3 

{q4 }
{q3 }

q4 {q4 }
{q4 }

4. q1 is thestart stat e.
5. F  {q4 }.
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In what situation is Non
Determinism relevant?
• Von Neumann machines are deterministic.
• However, there are many cases where
machine specification is all we need.
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Theorem 1.19
• Every nondeterministic finite automaton
has an equivalent deterministic finite
automaton.
– Def. The two machines are equivalent is they
recognize the same language.
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Proof of Th. 1.19
Let N  (Q, ,  , q0 , F ) be theNFA recognizing some language A.
Let us assume first thatN has no  arrows.
ConstructM  (Q, ,  , q0 , F ) such that
1. Q  2Q ,
2.  ( R, a)    (r , a),
rR
3. q0  {q0 }, and
4. F   {R  Q | R  F  }.
Proof of Th 1.19
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Incorporate  arrows
E ( R)  {q | q can be reached from R by traveling along zero or
more  arrows.}
We modify   as follows.
 ( R, a)  {q  Q | q  E ( (r , a)) for some r  R}.
We modify q0 as follows.
q0  E ({q0 }).
We omit the correctness proof.
Proof of Th 1.19
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Corollary 1.20
• A language is regular if and only if some
nondeterministic finite automaton
recognizes it.
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Example 1.21 NFA N4 to DFA
1
b
a
2

a,b
a
3
Given N 4  {{1,2,3}, {a, b},  ,1,{3}}, we want toconstruct
an equivalent DFA D. T heD' s stateset may be takesas
2{1, 2,3}  {,{1}, {2}, {3}, {1,2}, {1,3}, {2,3}, {1,2,3}}.
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Task 02
Parallel World
1. Write a program that simulates N4.
2. GUI
– Three buttons for input 0, 1, and epsion.
– State chart that reflect the branching of the
world.
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Start and Accept states
T hestart stateis E ({1})  {1,3}.
T heset of acceptstatesis {{3}, {1,3}, {2,3}, {1,2,3}}.
T hestatediagram is given as follows.
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The state diagram of D
a,b
a
f
b
{1}
{2}
{1,2}
a,b
b
b
b
a
a
a
{3}
a
a
{1,3}
{2,3}
{1,2,3}
b
b
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Theorem 1.22 The class of regular languages is
closed under the union operation.
N
N1


N2
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Proof of Th. 1.22
Let N1  (Q1 , , 1 , q1 , F1 ) recognizeA1 , and
N 2  (Q2 , ,  2 , q2 , F2 ) recognizeA2 .
Construct N  {Q, ,  , q0 , F ) to recognizeA1  A2 .
1. Q  {q0 }  Q1  Q2
2. T hestate q0 is thestart stateof N.
3. F  F1  F2
1 (q, a )
 (q, a )
 2
4.  (q, a )  
{q1 , q2 }

for q  Q1
for q  Q2
for q  q0 and a  
for q  q0 and a   .
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Theorem 1.23 The class of regular languages is
closed under the concatenation operation.
N2
N1

N

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Proof of Th. 1.23
Let N1  (Q1 , , 1 , q1 , F1 ) recognizeA1 , and
N 2  (Q2 , ,  2 , q2 , F2 ) recognizeA2 .
Construct N  {Q, ,  , q1 , F2 ) to recognizeA1  A2 .
1. Q  Q1  Q2
2. T hestart state q1 is thesame as thatof N1.
3. T heset of acceptstatesis thesame as thatof N 2
1 (q, a )
 (q, a )
 1
4.  (q, a )  
1 (q, a )  {q2 }
 2 (q, a )
for q  Q1 and q  F1
for q  F1 and a  
for q  F1 and a  
for q  Q2 .
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Theorem 1.24 The class of regular languages is
closed under the star operation.
N
N1



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Proof of Th. 1.24
Let N1  (Q1 , , 1 , q1 , F1 ) recognizeA1.
Construct N  {Q, ,  , q0 , F ) to recognizeA .
*
1
1. Q  {q0 }  Q1
2. T hestart state q0 is thenew start state.
3. F  {q0 }  F1
1 (q, a )
 (q, a )
 1
4.  (q, a )  1 (q, a )  {q1}
{q }
 1

for q  Q1 and q  F1
for q  F1 and a  
for q  F1 and a  
for q  q0 and a  
for q  q0 and a   .
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