RE2FA(After Lesson 13)

8/9/2019 RE2FA(After Lesson 13)

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RERE Finite AutomataFinite AutomataRERE Finite AutomataFinite Automata

Can we build a finiteautomaton for every regular

expression? Yes, build FA inductively

based on the definition ofRegular Expression


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NFANFANFANFA

Nondeterministic FiniteAutomaton (NFA)

Can have multipletransitions for one input

in a given state Can have I - moves


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Epsilon MovesEpsilon MovesEpsilon MovesEpsilon Moves moves

machine can move from stateA to state B without consuminginput

I

A B


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NFANFANFANFAoperation of the automaton is notcompletely defined by input

1

1

0

On input 11, automaton could bein either state

A B C


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Execution of FAExecution of FAExecution of FAExecution of FA

A NFA can choose

Whether to make I-moves.

Which of multiple

transitions to take for a

single input.


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Acceptance ofNFAAcceptance ofNFAAcceptance ofNFAAcceptance ofNFA

NFA can get into multiple states Rule: NFA accepts if it can get

in a final state

1

1

0

A B C

0


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DFA and NFADFA and NFADFA and NFADFA and NFA

Deterministic Finite Automata

(DFA)

One transition per input perstate.

No I - moves


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Execution of FAExecution of FAExecution of FAExecution of FA

A DFA

can take only one path

through the state graph.

Completely determined by

input.


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NFA vs DFANFA vs DFANFA vs DFANFA vs DFA

NFAs and DFAs recognize

the same set of languages

(regular languages)

DFAs are easierto

implement table driven.


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For a given language, the

NFA can be simpler than

the DFA.

DFA can be exponentially

largerthan NFA.


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NFAs are the key to

automating RE DFA

construction.


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RERE NFA ConstructionNFA ConstructionRERE NFA ConstructionNFA Construction

hompsons construction

( 1968)

Build an NFA for each REterm.

Combine NFAs withI-moves.


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Key idea:

NFA pattern for each

symbol and each operator.

Join them with I-moves in

precedence order.


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I

s0

as1

NFA fora

s0

as1

NFA forab

s3

bs4


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s0

a s1

NFA fora


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s0

a s1

NFA fora

s3

bs4NFA forb


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s0

a s1

NFA fora

s0

as1

s3

bs4

s3

bs4NFA forb


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I

s0

a s1

NFA fora

s0

as1

NFA forab

s3

bs4

s3

bs4NFA forb


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I

s0 s5

s1

as2

NFA fora | b

s3

bs4

I

I

I


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s1

as2

NFA fora


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s1

as2

s3

bs4

NFA fora and b


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I

s0 s5

s1

as2

NFA fora | b

s3

bs4

I

I

I


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Is0 s4s1 a s2

NFA fora*

I

I

I


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s1 a s2

NFA fora


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Is0 s4s1 a s2

NFA fora*

I

I

I


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Example REExample RE NFANFAExample REExample RE NFANFA

NFA fora ( b|c )* I

s3

s9

s4 s5

s6 s7

s8s0 s1 s2a II I

I

I I

Ib

c

I


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building NFA fora ( b|c )*

s0 s1a


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NFA fora, b and c

s4 s5

s6 s7

s0 s1a

b

c


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NFA fora and b|c

s3

s4 s5

s6 s7

s8s0 s1a

I

I I

Ib

c


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NFA fora and ( b|c )*I

s3

s9

s4 s5

s6 s7

s8s0 s1 s2a II

I

I I

Ib

c

I


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NFA fora ( b|c )* I

s3

s9

s4 s5

s6 s7

s8s0 s1 s2a II I

I

I I

Ib

c

I


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NFANFA DFA ConstructionDFA ConstructionNFANFA DFA ConstructionDFA Construction

he algorithm is called subsetconstruction.

In the transition table of anNFA, each entry is a set ofstates.

In DFA, each entry is a singlestate


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he general idea behind

NFA-to-DFA construction is

that each DFA state

corresponds to a set of

NFA states.


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he DFA uses its state to

keep track of all possible

states the NFA can be in

after reading each input

symbol.


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We will use the followingoperations.

I-closure(T):set ofNFA states reachable

from some NFA states

in Ton I-transitions alone.


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move(T,a):

set ofNFA states to which

there is a transition on inputa from some NFA state s in

set of states T.


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Before it sees the first input

symbol, NFA can be in

any of the state in the setI-closure(s

0), where s

0is the

start state of the NFA.


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Suppose that exactly the

states in set Tare

reachable from s0 on agiven sequence of input

symbols.


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Let a be the next input

symbol.

On seeing a, the NFA can

move to any of the states in

the set move(T,a).


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Let a be the next input

symbol.

On seeing a, the NFA can

move to any of the states in

the set move(T,a).


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When we allow for

I-transitions, NFA can be in

any of the states inI-closure(move(T,a))

after seeing a.


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Subset ConstructionSubset ConstructionSubset ConstructionSubset Construction

Algorithm:

Input:

NFA N with state set S,

alphabet 7, start state

s0, final states F


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Output:DFA D with state set S,

alphabet 7, start statess0 = I-closure(s0), final

states F, transitiontable: S x 7 S


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Subset Construction ExampleSubset Construction ExampleSubset Construction ExampleSubset Construction Example

NFA for(a | b )*abb

0I

I

I I

I

a

b

I

I

1

2 3

4 5

6 8 9

I

7

a b b

10


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he start state of equivalentDFA is I-closure(0), which isA = {0,1,2,4,7}

I

0I

I

I I

I

a

bI

1

2 3

4 5

6 8 9I

7a b b

10

I


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A = {0,1,2,4,7}, these areexactly the states reachablefrom state 0 via I-transition.

I

0I

I

I I

I

a

bI

1

2 3

4 5

6 8 9I

7a b b

10

I


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he input symbol alphabethere is {a,b}.

I

0I

I

I I

I

a

b

I

1

2 3

4 5

6 8 9I

7a b b

10

I


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he algorithm tells us to markA and then compute

I-closure(move(A,a))

I

0I

I

I I

I

a

b

I

1

2 3

4 5

6 8 9I

7a b b

10

I


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move(A,a)), is the set of statesofNFA that have transition ona from members ofA.

I

0I

I

I I

I

a

bI

1

2 3

4 5

6 8 9I

7a b b

10

I


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Only 2 and 7 have suchtransition, to 3 and 8.

I

0I

I

I I

I

a

b

I

1

23

4 5

6 8 9I

7a b b

10

I


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So,

I

-closure(move(A

,a)) =I-closure({3,8}) ={1,2,3,4,6,7,8}

I

0I

I

I I

I

a

b

I

1

2 3

4 5

6 8 9I

7a b b

10

I


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Let B = {1,2,3,4,6,7,8}. hus Dtran[A,a] = B

I

0I

I

I I

I

a

b

I

1

2 3

4 5

6 8 9I

7a b b

10

I


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For input b, among states in A,only 4 has transition on b to 5

I

0I

I

I I

I

a

b

I

1

2 3

4 5

6 8 9I

7a b b

10

I


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C = I-closure({5})= {1,2,4,5,6,7}

hus, Dtran[A,b] = C

I

0I

I

I I

I

a

bI

1

2 3

4 5

6 8 9I

7a b b

10

I


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We continue this process withthe unmarked sets B and C

I

0I

I

I I

I

a

bI

1

2 3

4 5

6 8 9I

7a b b

10

I


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i.e., I-closure(move(B,a)),I-closure(move(B,b)),

I

0I

I

I I

I

a

bI

1

2 3

4 5

6 8 9I

7a b b

10

I


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I-closure(move(C,a)) andI-closure(move(C,b))

0I

I

I I

I

a

bI

1

2 3

4 5

6 8 9I

7a b b

10

I


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Until all sets and states ofDFAare marked.

I

0I

I

I I

I

a

b

I

1

2 3

4 5

6 8 9I

7a b b

10

I


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Eventually, the 5 sets are:A={0,1,2,4,7}

B={1,2,3,4,6,7,8}C={1,2,4,5,6,7}

D={1,2,4,5,6,7,9}E={1,2,4,5,6,7,10}



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A is start state

A={0,1,2,4,7} D={1,2,4,5,6,7,9}B={1,2,3,4,6,7,8} E={1,2,4,5,6,7,10}

C={1,2,4,5,6,7}

I

0I

I

I I

I

a

b

I

1

2 3

4 5

6 8 9I

7a b b

10

I


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E is accepting state

A={0,1,2,4,7} D={1,2,4,5,6,7,9}B={1,2,3,4,6,7,8} E={1,2,4,5,6,7,10}

C={1,2,4,5,6,7}

0I

I

I I

I

a

b

I

1

2 3

4 5

6 8 9I

7a b b

10

I


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Resulting DFAResulting DFAResulting DFAResulting DFA

DFA for(a | b )*abb

A

a

b a

b bEB D

C

a

b

ba

a


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Resulting DFAResulting DFAResulting DFAResulting DFA

A

a

b a

b bEB D

C

a

b

ba

a

a a aa b b


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Final Transition TableFinal Transition TableFinal Transition TableFinal Transition Table

StateInput symbola b

A B C

B B D

C B C

D B E

E B C


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DFA MinimizationDFA MinimizationDFA MinimizationDFA Minimization

DFA for(a | b )*abb

A

a

b a

b bEB D

C

a

b

ba

a


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DFA MinimizationDFA MinimizationDFA MinimizationDFA Minimization

MinimizedDFA for(a | b )*abb

A,C

ab

b bEB D

a

b

a

a

RE2FA(After Lesson 13)

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