every ϵ-automaton is equivalent to an automaton

every

ϵ

-automaton is equivalent to an automaton

In this entry, we show that an

automaton

with

ϵ

-transitions (

http://planetmath.org/EpsilonTransition

) is no more power than one without. Having

ϵ

-transitions is purely a matter of convenience.

Proposition 1

.

Every

ϵ

-automaton (

http://planetmath.org/EpsilonAutomaton

) is

equivalent

to an automaton.

For the proof, we use the following setup (see the parent entry for more detail):

•

E

=

(

S

,

Σ

,

δ

,

I

,

F

,

ϵ

)

is an

ϵ

-automaton, and

E

ϵ

is the automaton associated with

E

,

•

h

:

(

Σ

∪

{

ϵ

}

)

*

→

Σ

*

is the

homomorphism

that erases

ϵ

(it takes

ϵ

to the

empty word

, also denoted by

ϵ

). From the parent entry,

L

⁢

(

E

)

:=

h

⁢

(

L

⁢

(

E

ϵ

)

)

.

Proof.

Define a

function

δ

1

:

S

×

Σ

→

P

⁢

(

S

)

, as follows: for each pair

(

s

,

a

)

∈

S

×

Σ

, let

δ

1

⁢

(

s

,

a

)

=

⋃

{

δ

⁢

(

s

,

u

)

∣

h

⁢

(

u

)

=

a

}

.

In other words,

δ

1

⁢

(

s

,

a

)

is the set of all states

reachable

from

s

by words of the form

ϵ

m

⁢

a

⁢

ϵ

n

. As usual, we extend

δ

1

so its domain is

S

×

Σ

*

. By abuse of notation, we use

δ

1

again for this

extension

. First, we set

δ

1

⁢

(

s

,

ϵ

)

:=

{

s

}

. Then we inductively define

δ

1

⁢

(

s

,

u

⁢

a

)

=

δ

1

⁢

(

δ

1

⁢

(

s

,

u

)

,

a

)

. Using

induction

,

δ

1

⁢

(

s

,

u

⁢

a

)

=

δ

1

⁢

(

δ

1

⁢

(

s

,

u

)

,

a

)

=

δ

1

⁢

(

⋃

h

⁢

(

v

)

=

u

δ

⁢

(

s

,

v

)

,

a

)

=

⋃

h

⁢

(

v

)

=

u

δ

1

⁢

(

δ

⁢

(

s

,

v

)

,

a

)

=

⋃

h

⁢

(

v

)

=

u

⋃

t

∈

δ

⁢

(

s

,

v

)

δ

1

⁢

(

t

,

a

)

=

⋃

h

⁢

(

v

)

=

u

⋃

t

∈

δ

⁢

(

s

,

v

)

⋃

h

⁢

(

w

)

=

a

δ

⁢

(

t

,

w

)

=

⋃

h

⁢

(

v

)

=

u

⋃

h

⁢

(

w

)

=

a

⋃

t

∈

δ

⁢

(

s

,

v

)

δ

⁢

(

t

,

w

)

=

⋃

h

⁢

(

v

)

=

u

⋃

h

⁢

(

w

)

=

a

δ

⁢

(

δ

⁢

(

s

,

v

)

,

w

)

=

⋃

h

⁢

(

v

)

=

u

⋃

h

⁢

(

w

)

=

a

δ

⁢

(

s

,

v

⁢

w

)

=

⋃

{

δ

⁢

(

s

,

v

⁢

w

)

∣

h

⁢

(

v

)

=

u

⁢

and

⁢

h

⁢

(

w

)

=

a

}

=

⋃

{

δ

⁢

(

s

,

x

)

∣

h

⁢

(

x

)

=

u

⁢

a

}

So for any non-empty word

u

, we have the following

equation

:

δ

1

⁢

(

s

,

u

)

=

⋃

{

δ

⁢

(

s

,

v

)

∣

h

⁢

(

v

)

=

u

}

.

(1)

In other words, if

u

=

a

1

⁢

a

2

⁢

⋯

⁢

a

n

, then

δ

1

⁢

(

s

,

u

)

is the set of all states reachable from

s

by words of the form

ϵ

i

0

⁢

a

1

⁢

ϵ

i

1

⁢

a

2

⁢

ϵ

i

2

⁢

⋯

⁢

ϵ

i

n

-

1

⁢

a

n

⁢

ϵ

i

n

.

(2)

Now, define

A

to be the automaton

(

S

,

Σ

,

δ

1

,

I

,

F

)

. Then, from equation (1) above, a word

u

=

a

1

⁢

a

2

⁢

⋯

⁢

a

n

is accepted by

A

iff some word

v

of the form (2) is accepted by

E

ϵ

iff

u

=

h

⁢

(

v

)

is accepted by

E

, proving the

proposition

.
∎

Remark

. Another approach is to use the

concept

of

ϵ

-

closure

(

http://planetmath.org/EpsilonClosure

). The proof is very similar to the one given above, and the resulting equivalent automaton is a DFA.

Title

every

ϵ

-automaton is equivalent to an automaton

Canonical name

EveryepsilonautomatonIsEquivalentToAnAutomaton

Date of creation

2013-03-22 19:02:03

Last modified on

2013-03-22 19:02:03

Owner

CWoo (3771)

Last modified by

CWoo (3771)

Numerical id

8

Author

CWoo (3771)

Entry type

Definition

Classification

msc 03D05

Classification

msc 68Q45

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