NFA and regex€¦ · NFA and regex • the Boolean ... Can always convert to an equivalent DFA for...

clInformatics 1 School of Informatics, University of Edinburgh

NFA and regex

• the Boolean algebra of languages

• non-deterministic machines

• regular expressions

1

Informatics 1 School of Informatics, University of Edinburgh

The intersection of two regular languages is regular

2

Run both machines in parallel?

Build one machine that simulates two

machines running in parallel!

Keep track of the state of each

machine.

1

1

1

0 00 2

1

0

1

1 10

0

0

Informatics 1 School of Informatics, University of Edinburgh

The intersection of two regular languages is regular

3

1

0

0

1

1

1 10

0

0

1

0

0

1

0

10

20

100

11

0

0

1

11

21

01

1

|| =

0

0

1

2

10

mod 3

mod 2

0 1

3 2 3 2

0 0 0 0 1 11 0 2 0 0 12 0 1 0 2 10 1 0 0 1 11 1 2 0 0 12 1 1 0 2 1

1

Informatics 1 School of Informatics, University of Edinburgh

The intersection of two regular languages is regular

4

1

1 10

0

0|| =

1

0

0

1

0

4

2

10

11

0

0

1

1

5

3

1

0

mod 3

mod 2

mod 6

0 13 2 6 3 2 6

0 0 0 0 0 0 1 1 11 0 4 2 0 2 0 1 32 0 2 1 0 4 2 1 50 1 3 0 0 0 1 1 11 1 1 2 0 2 0 1 32 1 5 1 0 4 2 1 5

1

0

0

10

1

2

10

1

Informatics 1 School of Informatics, University of Edinburgh

The intersection of two regular languages is regular

5

1

1 10

0

0|| =

1

0

0

1

0

4

2

10

11

0

0

1

1

5

3

1

0

mod 3

mod 2

mod 6

0 13 2 6 3 2 6

0 0 0 0 0 0 1 1 11 1 1 2 0 2 0 1 32 0 2 1 0 4 2 1 50 1 3 0 0 0 1 1 11 0 4 2 0 2 0 1 32 1 5 1 0 4 2 1 5

1

0

0

10

1

2

10

1

Informatics 1 School of Informatics, University of Edinburgh

The regular languages A ⊆ Σ* form a Boolean Algebra

• Since they are closed under intersection and complement.

6

Informatics 1 School of Informatics, University of Edinburgh

Determinism

• Proof Add a new “black hole” state, ● For every pair (s, a) for which there is no state t with a transition T(s, a, t), add a transition T(s, a, ●). This includes a transition T(●, a, ●) for each a ∈ Σ . You cannot escape from the black hole. The black hole ● is not an accepting state.

This machine accepts the same language as the original.

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For this machine there is exactly one trace for each input

string

Can always convert to an equivalent DFA for which every state has exactly one transition leaving the state for each input symbol.

Informatics 1 School of Informatics, University of Edinburgh

Non DeterminismIn a non-deterministic machine (NFA), each state may have any number of transitions with the same input symbol, leaving to different successor states.

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0 1

1

1

0 0

2 0 1

0 0 0,1

1 2

2

Informatics 1 School of Informatics, University of Edinburgh

Non DeterminismIn a non-deterministic machine (NFA), each state may have any number of transitions with the same input symbol, leaving to different successor states.

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0 1

1

1

0 0

20 1

0 0 0,11 22

0,1 0,2 0,10,2 0 0,1

0 0,1

010

11

0

0,2

Informatics 1 School of Informatics, University of Edinburgh

Non DeterminismWe can simulate a non-deterministic machine using a deterministic machine – by keeping track of the set of states the NFA could possibly be in.

10

0 1

1

1

0 0

20 1

0 0 0,11 22

0,1 0,2 0,10,2 0 0,1

0 0,1

010

11

0

0,2

Informatics 1 School of Informatics, University of Edinburgh

Internal TransitionsWe sometimes add an internal transition ε to a non-deterministic machine (NFA)This is a state change that consumes no input.

11

0 1

10 0

2

0 1 ε0 0 1

1 2 0

2

ε

0 1

1

1

0 0

2

Informatics 1 School of Informatics, University of Edinburgh

Internal TransitionsWe sometimes add internal transitions – labelled ε – to a non-deterministic machine (NFA). This is a state change that consumes no input. It introduces non-determinism in the observed behaviour of the machine.

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0 1

10 0

2

0 1 ε0 0 11 2 02

ε

0ε* 1ε*0 0 1,01 22

NFA single start stateany number of accepting states

S

13

R

sequence RS

ε

14

SRε

alternation R|S

ε

ε

ε

15

SR

ε

ε

iteration R*

ε

16

R

εε

Informatics 1 School of Informatics, University of Edinburgh

Demohttp://ivanzuzak.info/noam/webapps/fsm_simulator/

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0 1

1

1

0 0

20 1

0 0 0,11 22

0,1 0,2 0,10,2 0 0,1

0 0,1

010

11

0

0,2

Informatics 1 School of Informatics, University of Edinburgh

Is there a regular expression for every FSM?

18

0 1

a b

2

c

Let Li be the language accepted if i is the accepting state

L0 = εL1 = L0 aL2 = L1 b | L0 c

L2 = L0 a b | ε cL2 = ε a b | ε cL2 = a b | c

0 1

a b

2

c

0 1

a b

2

c

L0

L1

L2

Informatics 1 School of Informatics, University of Edinburgh

Is there a regular expression for every FSM?

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L1 = L2 bL2 = L3 b | L1 aL3 = ε | L1 b

2

3

1 b

b

a

b

= ε | L2 b bL2 = (ε | L2 b b) b | L2 b a

= b | L2 b b b | L2 b a= b | L2 (b b b | b a)

Informatics 1 School of Informatics, University of Edinburgh

Is there a regular expression for every FSM?

20

2

3

1 b

b

a

b

L2 = b | L2 (b b b | b a)

L2 = b (b b b | b a)*

L1 = L2 b = b (b b b | b a)* b

L3 = ε | L2 b b = ε | b (b b b | b a)* bb

Informatics 1 School of Informatics, University of Edinburgh

Arden’s Lemma

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If R and S are regular expressions then the equation

X = R | X S has a solution X = R S*

If ε ∉ L(S) then this solution is unique. L2 = b | L2 (b b b | b a)L2 = b (b b b | b a)*