juxtaposition of automata

juxtaposition of automata

Let

A

=

(

S

1

,

Σ

1

,

δ

1

,

I

1

,

F

1

)

and

B

=

(

S

2

,

Σ

2

,

δ

2

,

I

2

,

F

2

)

be two

automata

. We define the

juxtaposition

of

A

and

B

, written

A

⁢

B

, as the sextuple

(

S

,

Σ

,

δ

,

I

,

F

,

ϵ

)

, as follows:

1.

S

:=

S

1

∪

⋅

S

1

, where

∪

⋅

denotes

disjoint union

,

2.

Σ

:=

(

Σ

1

∪

Σ

2

)

∪

⋅

{

ϵ

}

,

3.

δ

:

S

×

Σ

→

P

⁢

(

S

)

given by

–

δ

⁢

(

s

,

ϵ

)

:=

I

2

if

s

∈

F

1

, and

δ

⁢

(

s

,

ϵ

)

:=

{

s

}

otherwise,

–

δ

|

(

S

1

×

Σ

1

)

:=

δ

1

,

–

δ

|

(

S

2

×

Σ

2

)

:=

δ

2

, and

–

δ

⁢

(

s

,

α

)

:=

∅

otherwise (where

α

≠

ϵ

).

4.

I

:=

I

1

,

5.

F

:=

F

2

.

Because

S

1

and

S

2

are considered as disjoint subsets of

S

,

I

∩

F

=

∅

. Also, from the

definition

above, we see that

A

⁢

B

is an automaton with

ϵ

-transitions (

http://planetmath.org/AutomatonWithEpsilonTransitions

).

The way

A

⁢

B

works is as follows: a word

c

=

a

⁢

b

, where

a

∈

Σ

1

*

and

b

∈

Σ

2

*

, is fed into

A

⁢

B

.

A

⁢

B

first reads

a

as if it were read by

A

, via transition function

δ

1

. If

a

is accepted by

A

, then one of its accepting states will be used as the initial state for

B

when it reads

b

. The word

c

is accepted by

A

⁢

B

when

b

is accepted by

B

.

Visually, the

state diagram

G

A

1

⁢

A

2

of

A

1

⁢

A

2

combines the state diagram

G

A

1

of

A

1

with the state diagram

G

A

2

of

A

2

by adding an edge from each final node of

A

1

to each of the start nodes of

A

2

with label

ϵ

(the

ϵ

-transition).

Proposition 1

.

L

⁢

(

A

⁢

B

)

=

L

⁢

(

A

)

⁢

L

⁢

(

B

)

Proof.

Suppose

c

=

a

⁢

b

is a words such that

a

∈

Σ

1

*

and

b

∈

Σ

2

*

. If

c

∈

L

⁢

(

A

⁢

B

)

, then

δ

⁢

(

q

,

a

⁢

ϵ

⁢

b

)

∩

F

≠

∅

for some

q

∈

I

=

I

1

. Since

δ

⁢

(

q

,

a

⁢

ϵ

⁢

b

)

∩

F

2

=

δ

⁢

(

q

,

a

⁢

ϵ

⁢

b

)

∩

F

≠

∅

and

b

∈

Σ

2

*

, we have, by the definition of

δ

, that

δ

⁢

(

q

,

a

⁢

ϵ

⁢

b

)

=

δ

⁢

(

δ

⁢

(

q

,

a

⁢

ϵ

)

,

b

)

=

δ

2

⁢

(

δ

⁢

(

q

,

a

⁢

ϵ

)

,

b

)

, which shows that

b

∈

L

⁢

(

B

)

and

δ

⁢

(

q

,

a

⁢

ϵ

)

∩

I

2

≠

∅

. But

δ

⁢

(

q

,

a

⁢

ϵ

)

=

δ

⁢

(

δ

⁢

(

q

,

a

)

,

ϵ

)

, by the definition of

δ

again, we also have

δ

⁢

(

q

,

a

)

∩

F

1

≠

∅

, which implies that

δ

⁢

(

q

,

a

)

=

δ

1

⁢

(

q

,

a

)

. As a result,

a

∈

L

⁢

(

A

)

.

Conversely

, if

a

∈

L

⁢

(

A

)

and

b

∈

L

⁢

(

B

)

, then for any

q

∈

I

=

I

1

,

δ

⁢

(

q

,

a

)

=

δ

1

⁢

(

q

,

a

)

, which has non-empty

intersection

with

F

1

. This means that

δ

⁢

(

q

,

a

⁢

ϵ

)

=

δ

⁢

(

δ

⁢

(

q

,

a

)

,

ϵ

)

=

I

2

, and finally

δ

⁢

(

q

,

a

⁢

ϵ

⁢

b

)

=

δ

⁢

(

δ

⁢

(

q

,

a

⁢

ϵ

)

,

b

)

=

δ

⁢

(

I

2

,

b

)

, which has non-empty intersection with

F

2

=

F

by

assumption

. This shows that

a

⁢

ϵ

⁢

b

∈

L

⁢

(

(

A

⁢

B

)

ϵ

)

, or

a

⁢

b

∈

L

⁢

(

A

⁢

B

)

.
∎

Title

juxtaposition of automata

Canonical

name

JuxtapositionOfAutomata

Date of creation

2013-03-22 18:03:51

Last modified on

2013-03-22 18:03:51

Owner

CWoo (3771)

Last modified by

CWoo (3771)

Numerical id

14

Author

CWoo (3771)

Entry type

Definition

Classification

msc 03D05

Classification

msc 68Q45

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