TWO-WAY FINITE AUTOMATA &

PEBBLE AUTOMATA

Written by Liat Peterfreund 1

AUTOMATA FINITE WAY-WOT

A two way deterministic finite automata (2DFA)

is a quintuple 0, , , ,M Q q F where:

0, , ,Q q F are as before

: { , }Q Q L R

o If ( , ) ( , )q a p L then in state q ,

scanning the input symbol a , the 2DFA enters state

p and moves its head left one square.

o If ( , ) ( , )q a p R then in state q ,

scanning the input symbol a , the 2DFA enters state

p and moves its head right one square. 2

:OMMENTSC

o In describing the behavior of one-way finite automata,

we extend to *Q . This notion is insufficient for 2DFA

because the head may move left.

o Instead we introduce the notion of instantaneous

description ( ID ) of 2DFA which describes the input string,

current state and current position if the input head.

o The relation M on ID 's such that 1 2MI I if and only if M

can go from 1I to 2I in one move.

o An ID of M is a sting in * *Q . 3

o The ID wqx whereas *,w x and q Q is

intended to represent :

1) wx is the input string

2) q is the current state

3) the input head is scanning the first symbol of x

(If x then the input head has moved off the

right hand end of the input).

4

o We define the relation M

or just if M is

understood, by

1) 1 2 1 1 2 1 1... ... ... ...i i n i i i na a a qa a a a a a pa a

whenever ( , ) ( , )iq a p R

2) 1 2 2 1 1 2 2 1... ... ... ...i i i n i i i na a a a qa a a a a pa a a

whenever ( , ) ( , )iq a p L and 1i (The

condition 1i prevents any action in the event

that the tape head would move off the left hand

end of the tape).

o Note that no move is possible if 1i n (the tape

head has moved off the right hand end).

5

Let * be the reflexive and transitive closure of .

*

1 kI I whenever 1 2 ... kI I I for some

2 1,..., kI I we

define *

0( ) | ,L M w q w wp p F . That is w is

accepted by M if, starting in state 0q with w on the input

tape and the head at the left end of w , M eventually enters

a final state at the same time it falls off the right hand of the

input tape. 6

Example:

Consider a 2DFA M that behaves as follows: starting in

state 0q , M repeats a cycle of moves wherein the tape head

moves right until two ones have been encountered then left

until encountering a zero at which point state 0q is reentered

and the cycle repeated. More precisely M has three states

all of which are final. is given as follows:

0 1

0q 0 ,q R 1,q R

1q 1,q R 2 ,q L

2q 0 ,q R 2 ,q L

7

Consider the input 101001:

0 1

1

2

0

1

1

1

2

101001 1 01001

10 1001

1 01001

10 1001

101 001

1010 01

10100 1

1010 01

q q

q

q

q

q

q

q

q

0

1

10100 1

101001

q

q

than 101001 ( )L M

8

Crossing sequences

The list of states below each boundary between squares is

termed a crossing sequence.

1 0 1 0 0 1

0q

1q 1q

2q

0q

1q 1q

1q

2q

0q

1q

9

Observations:

o If a 2DFA accepts its input, no crossing sequence

may have a repeated state with the head moving in

the same direction. Otherwise the 2DFA being

deterministic would be in a loop.

o The first time a boundary is crossed, the head must

be moving right.

o Subsequent crossings must be in alternate directions:

odd-numbered elements of the crossing

sequence represents right moves

even-numbered elements of the crossing

sequence represents left moves

10

o If the input is accepted then all crossing sequences

are of odd length

o A crossing sequence 1,..., kq q is said to be valid if :

it is of odd length

no two odd-numbered and no two even-

numbered elements are identical

o The number of valid crossing sequences is finite (a

2DFA with s states can have valid crossing sequence

of length at most 2s-1). 11

Theorem:

2DFA and DFA (deterministic finite automata) have the

same expressive power.

12

Proof:

o We construct a NFA (non-deterministic finite automata)

that will simulate the 2DFA M and whose states are

the valid crossing sequences of M .

o In order to construct the transition function we first

examine the relationship between adjacent crossing

sequences.

13

o Suppose we are given an isolated tape square holding the

symbol a and are also given valid crossing sequences

1 2, ,..., kq q q and 1 2, ,..., lp p p at the left and the right

boundaries of the square, respectively.

Note that there may be no input strings that can

be attached to the left and right of symbol a to

actually produce these two crossing sequences

14

o We can test two sequences for local compatibility as

follows:

If the tape head moves left from the square

holding a in state iq , restart the automaton on

the square holding a in state 1iq

If the tape head moves right from the square

holding a in state ip , restart the automaton on

the square holding a in state 1ip

15

We take 1 2, ,..., kq q q to appear at the left boundary of a and

1 2, ,..., lp p p at the right boundary of a then,

1 2, ,..., kq q q right-match

1 2, ,..., lp p p on a if these

sequences are consistent, assuming we initially reach

a in state 1q moving right

1 2, ,..., kq q q left-match 1 2, ,..., lp p p on a if these

sequences are consistent, assuming we initially reach

a in state 1p moving left 16

We define right-matching and left matching pairs of crossing

sequences recursively in (1) through (5):

(1) The null sequence left- and right-matches the null

sequence. (if we never reach the square holding a ,

then it is consistent that the boundaries on neither

side should be crossed).

17

(2) If 3,..., kq q right-matches 1 2, ,..., lp p p

and

1 2( , ) ( , )q a q L then 1 2, ,..., kq q q

right-matches

1 2, ,..., lp p p (if the first crossing of the left boundary

is in state 1q and the head immediately moves left in

state 2q, then if we follow these two crossings by any

consistent behavior starting from another crossing of

the left boundary, we obtain a consistent pair of

sequences with first crossing moving right i.e. a right

matched pair).

(3) If 2 ,..., kq q left-matches

2 ,..., lp p and

1 1( , ) ( , )q a p R then 1 2, ,..., kq q q right-matches

1 2, ,..., lp p p .

18

(4) If 1,..., kq q left-matches

3,..., lp p and

1 2( , ) ( , )p a p R then 1,..., kq q left-matches

1 2, ,..., lp p p .

(5) If 2 ,..., kq q right matches 2 ,..., lp p and

1 1( , ) ( , )p a q L then 1,..., kq q left-matches

1 2, ,..., lp p p .

19

The construction:

Let 0, , , ,M Q q F be a 2DFA . We contruct a NFA

0' ', , ', ', 'M Q q F which accepts ( )L M such that:

1) 'Q consists of all valid crossing sequences for M .

2) 0 'q is the crossing sequence consisting of

0q alone.

3) 'F is the set of all crossing sequences of length one

consisting of a state in F .

4) | is a valid crossing sequence

'( , )that is right matched by on input

d dc a

c a

20

The intuitive idea is that 'M puts together pieces of

computations of M as it scans the input string. This is

done by guessing successive crossing sequences. If 'M

has guessed that c is the crossing sequence at a

boundary, and a is the next input symbol then 'M can

guess any valid crossing sequence that c right matches

on input a .

If the guessed computation results in M moving off the

right end of the input in an accepting state, then 'M

accepts. 21

Claim:

( ) ( ')L M L M

Proof:

( ) ( ')L M L M :

Let ( )w L M . Look at the crossing sequences generated by

an accepting computation of M on w . Each crossing

sequence right-matches the one at the next boundary, so 'M

can guess the proper crossing sequence (among its other

guesses) and accept i.e. ( ')w L M .

22

( ') ( )L M L M :

Let ( ')w L M .

Consider the crossing sequence 0 ,..., nc c of M

corresponding the states of 'M as 'M scans

1,..., nw a a .

For each 0 i n ic right matches 1ic on ia

We can construct an accepting computation of M on

input w by determining when the head reverses

direction. 23

We prove by induction on i that 'M on reading 1... ia a

can enter state 1,...,i kc q q only if the following two

conditions hold:

1) M started in state 0q on

1... ia a will first move right

from position i in state 1q

2) for 2,4,...j , if M is started at position i in state

jq , M will eventually move right from position i in

state 1jq (this implies k must be odd). 24

Basis: 0i then 0 0c q . (1) is satisfied since M

begins its computation by "moving right from position

0" in state 0q . (2) holds vacuously.

25

Assume the hypothesis is true for 1i . Suppose that

'M on reading 1... ia a can enter state

1,...,i lc p p from state 1 1,...,i kc q q .

o Since k and l are odd, and 1ic right

matches ic on ia there must exist an odd j

such that in state jq on input ia , M moves

right. Let 1j be the smallest such j . By

definition of "right matches" it follows that

1 1 1( , ) ( , )jq a p R . This proves (1).

o We will prove (2) by induction:

Basis: vacuously

Inductive step: by the definition of

"right-matches" (rule (3))

1 1[ ,..., ]j kq q left-matches 2[ ,..., ]lp p

26

Now if 1( , ) ( , )j i jp a p R for all

even j , then (2) follows immediately.

In the case that for some smallest

even 2j ,

2( , ) ( , )j ip a q L , then by

definition of "left-matches" (rule (5))

q must be 1 1jq

and 1 2[ ,..., ]j kq q

right matches 2 1[ ,..., ]j lp p

By induction hypothesis, (2) holds for

1 2[ ,..., ]j kq q and 2 1[ ,..., ]j lp p 27

With the induction hypothesis for all i established, the

fact that [ ]nc p for some p F implies that M

accepts 1... na a .

28

Reminder - Example:

Consider a 2DFA M that behaves as follows: starting in

state 0q , M repeats a cycle of moves wherein the tape head

moves right until two ones have been encountered then left

until encountering a zero at which point state 0q is reentered

and the cycle repeated. More precisely M has three states

all of which are final. is given as follows:

0 1

0q 0 ,q R 1,q R

1q 1,q R 2 ,q L

2q 0 ,q R 2 ,q L

29

Consider the construction of NFA 'M equivalent to the

2DFA M :

Since 2q is only entered on a left move and 1q and 3q are

only entered on right moves, all even numbered

components of valid crossing sequences must be 2q .

Since a valid crossing sequence must be of odd length,

and no two odd numbered states can be the same, nor

can two even-numbered states be the same, there are

only four crossing sequences of interest:

0 1 0 2 1 1 2 0[ ],[ ],[ , , ],[ , , ]q q q q q q q q . 30

Their right matches:

Valid crossing sequences

Right matches on 0

Right matches on 1

0[ ]q 0[ ]q 1[ ]q

1[ ]q 1 1 2 0[ ],[ , , ]q q q q -

0 2 1[ , , ]q q q - -

1 2 0[ , , ]q q q - 1[ ]q

Thus the automaton is:

0 0

1

1 0

0[ ]q

1[ ]q

1 2 0[ , , ]q q q

31

Pebble Automata

32

Preface:

We consider strings over infinite alphabet D .

A -stringD is a finite sequence 1... nd d where 0n

and 1 : ii n d D .

dom( ) 1,2,...,w w .

For dom( )i w we write ( )wval i for id .

A language is a set of -stringsD .

33

We delimit input strings by two special symbols ,

for the left and for the right of the string, neither of

which are in D .

Therefore the automata work on extended strings of

the form w u where *u D .

The positions of , are 0 and 1w respectively

We write ( )dom w for the extended set

0,..., 1w of positions

We likewise extend the function wval by defining

(0)wval and ( 1)wval w

34

Pebble automata are finite state machines equipped

with a finite number of pebbles.

The use of pebbles is restricted by a stack discipline

(pebble 1i can only be placed when pebble i is

present on the string and pebble i can only be lifted

when pebble 1i is not present on the string).

The highest numbered pebble present on the string

acts as the head of the automaton.

35

A transition depends on:

o The current state

o The pebbles placed on the current position of

the head

o Equality type of the data values under the

placed pebbles.

The transition relation specifies:

o Change of state

o Movement of the head

o Possibly whether the head pebble is removed

or a new pebble is placed 36

Definition:

A nondeterministic two-way k-pebble automaton A over D

(infinite alphabet) is a tuple 0( , , , )Q q F T where:

o Q is a finite set of states

o 0q Q is the initial state

o F Q is a set of finite states

37

o T is a finite set of transitions of the form

where

o is of the form ( , , , , )i s P V q or ( , , , )i P V q

where 1,...,i k , ,s D ,

, 1,..., 1P V i

o is of the form ( , )q d with q Q and

stay,left,right,place-new-pebble,

lift-current-pebbled

38

Given a string w , a configuration of A on w is of the form

, ,i q where 1,...,i k , q Q and

:{1,..., } dom ( )i w .

We call pebble assignment and i the depth of

configuration (and of pebble assignment).

Initial configuration: 0 0 0: 1, ,q where 0(1) 0 .

Accepting configuration: , ,i q where F . 39

A transition ( , , , , )i s P V q applies to a configuration

, ,j q if:

1) ,i j p q

2) | ( ( )) ( ( ))w wV l i val l val i

3) | ( ) ( )P l i l i

4) ( ( ))wval i s

A transition ( , , , )i P V q applies to a configuration

, ,j q if 1-3 holds and no transition

( , , , , )i s P V q applies to . 40

We define the transition relation as follows:

[ , , ] [ ', ', ']i q i q iff there is a transition ( , )p d that

applies to such that 'q p and : '( ) ( )j i j j and:

If stayd , then 'i i and '( ) ( )i i

If leftd , then 'i i and '( ) ( ) 1i i

If rightd , then 'i i and '( ) ( ) 1i i

If place-new-pebbled , then ' 1i i and

'( ) '( 1) ( )i i i

If lift-current-pebbled , then ' 1i i

41

An automaton is deterministic if in each configuration at most

one transition applies.

If there are no left transitions then the automaton is one-

way.

The language of the automaton: A string w is accepted by A

if *

0 for some accepting configuration .

42

Example 1:

Let 1... | 0, , :n i jL d d n i j i j d d .

We define a 2-pebble-automata A accepting this language.

0( , , , )A Q q F T is defined as follows:

1 2, , , accQ q q q q

{ }accF q

T consists of the following transitions:

(1) 1 1(1, , , ) ( , right)q q

(2) 1(1, , , ) ( ,place-new-pebble)q q

(3) 2(2,{1},{1}, ) ( , right)q q

(4) 2 2(2, , , ) ( , right)q q

(5) 2(2, ,{1}, ) ( ,stay)accq q

43

Intuitive explanation:

(1-2) A stays in state 1q while moving to the right or places a

pebble

(3) After the placement of the pebble A moves one position

to the right

(4-5) A continues moving to the right and when it reads a

symbol equal to the symbol under the first pebble, it moves

to the final state 44

Example 2:

Let *$ | , $ [ ] [ ]L u v u v D D u v whereas

[ ],[ ]u v are the sets of all the letters of ,u v respectively.

We define a 2-pebble-automata A accepting this language

that operates as follows:

o The first pebble passes through all of the positions of

the word u (until it encounters the sign $ ) and for

each such position it checks whether there is a

position in v with the same letter.

o The above check is done by placing the second pebble

and iterating through all of the positions in v - if

there is such a position it lifts the pebble and

continues with the next position in u , if there is not it

rejects.

o When the first pebble reaches the $ it moves to an

accepting state.

45

Formally, 0( , , , )A Q q F T whereas

0 , , , , ,u uv v rej accQ q q q q q q , accF q and T contains

the following transitions:

0( 1)(1, , , ) ( , right)

( 1)(1,$, , , ) ( ,stay)

( 1)(1, , , ) ( , place-new-pebble)

( 2)(2,{1},{1}, ) ( , right)

( 2)(2,$, , , ) ( , right)

( 2)(2, , , ) ( , right)

( 2)(2, ,{1}, ) ( , r

u

u acc

u uv

uv uv

uv v

uv uv

uv uv

a q q

b q q

c q q

a q q

b q q

c q q

d q q

0

ight)

( 2)(2, ,{1}, ) ( , lift-current-pebble)

( 2)(2, , , , ) ( ,stay)

( 2)(2, , , ) ( , right)

v

v rej

v v

e q q

f q q

g q q

46

Intuitive explanation:

( 1)a is used to skip a sign

( 1),( 1)b c are used to iterate all over u 's elements

( 2) ( 2)a d skip the end of u and the $

( 2) ( 2)e f iterates through v and compares the sign under

the current pebble with the sign under pebble 1

47

Example 3:

We define a two-way nondeterministic pebble

automaton with three pebbles that accepts all words

w where 2w and there exists some position i in

( )dom w such that the set of symbols occurring at

positions smaller than i is disjoint from the set of

symbols occurring at position larger than or equal to

i .

For example the automaton accepts abb and

rejects abab 48

The automaton will work as follows:

o The first pebble is used nondeterministically

to guess i

o The second pebble is used to step through

position to the left of i one at a time, from

1i to

o For each such position of the second pebble,

the third pebble is used to check that all

symbols at positions i are distinct from the

symbol under the second pebble.

o The input is excepted if the second pebble

can reach the leftmost symbol

49

Formal definition: , , ,oA Q q F T where

0 7,...,Q q q , 5{ }F q and T is the set of

transitions:

50

0 0

0 1

1 2

( 1) (1, , , , ) ( , right) } skip

(b1) (1, , , ) ( , right) } 1

(c1) (1, , , , ) ( ,stay) } reached the e

a q q

q q i

q q

1 1

1 3

3 4

nd without guessing

(d1) (1, , , ) ( , right) guess

(e1) (1, , , ) ( , place-new-pebble)

( 2) (2,{1},{1}, ) ( , left) } move to pos

i

q qi

q q

a q q

3 4

4 5

ition 1

( 2) (2, , , ) ( , left) } move one position left (after lifting pebble 3)

( 2) (2, , , , ) ( ,stay) } finished reading

i

b q q

c q q

4 6

6 6

6 6

6 6

positions 1

( 2) (2, , , ) ( , place-new-pebble) } "save" position

( 3) (3,{2},{2}, ) ( , right)

(b3) (3, , ,q ) ( , right)

( 3) (3, ,{2}, ) ( , right)

( 3) (3,{1}

i

d q q

a q q

q

c q q

d

6 7

7 7

7 3

7 7

move right to position 1

,{1}, ) ( , right)

(e3) (3, ,{1},q ) ( , right)

( 3) (3, , , , ) ( , lift-pebble) scan positions >

( 3) (3, , , ) ( , right)

i

q q

q

f q q i

g q q

51

For example, an accepting run on abb will be:

52

In the above definition, new pebbles are placed at the

position if the most recent pebble. An alternative would

be to place new pebbles at the beginning of the string.

While the choice makes no difference in the two-way

case, it is significant in the one way case. We refer to the

model as defined above as weak pebble automata and to

the latter as strong pebble automata. Strong pebble

automata are formally defined by setting '( 1) 0i in

the place-new-pebble case of the definition of the

transition relation. 53

54

Our next result shows that the variants of strong pebble

automata, one-way and two-way, collapse.

Theorem:

Two-way weak 2-pebble automata and two-way strong 2-

pebble automata have the same expressive power.

55

Proof:

Let , , ,oA Q q F T be a two-way strong 2-pebble

automaton. We want to construct a two-way weak 2-pebble

automaton that will simulate A 's behavior.

Idea of construction: For every state in q Q we will add a

new state 'q and transition that such that the automaton head

will move left until it encounters the left delimiter and then

move to the appropriate state.

56

Let , , ,oA Q q F T be a two-way weak 2-pebble

automaton. We construct a two-way 2-pebble strong

automaton that will simulate A 's behavior. The only difference

between the automata is in placing a new pebble.

Idea of construction: For every state in q Q we will add a

new state 'q and transition that such that the automaton head

will move right until it encounters the first pebble and then

move to the appropriate state.

57

Theorem:

Two-way strong 2-pebble automata and one-way strong 2-

pebble automata have the same expressive power.

Proof:

A one-way strong 2-pebble automaton is a particular case of

two-way strong automaton.

58

Claim: For each two-way non-deterministic pebble automata

A there is a strong one-way deterministic pebble automata that accepts the same language.

Proof: Assumptions:

(1) A lifts pebble only in the right delimiter (2) A accepts its input only in configurations [1, q, θ],

i.e., with only one pebble.

59

By virtually adding two steps we view an accepting computation as consisting of: (1) a first step in which the first pebble is placed at the left delimiter, i.e., position 0. (2) a computation in which always at least one pebble is present. (3) a final step in which the only remaining pebble is removed.

60

Writing [0, , ]p for a (virtual) configuration without pebble,

to determine whether A accepts, one has to find out whether *

0[0, , ] [0, , ]q q for some final state q.

The latter can be done by recursively solving subproblems of

the form *[ , , ] [ , ', ]ii q i q

where the subscript > i indicates that only subcomputations are considered in which, at every step, more than i pebbles are present.

61

More formally, we show the following claim by induction on i (starting from i = k).

Claim: For each {0,...,k}i and each finite set of states R,

there is a strong one-way deterministic pebble automata Bi

(with k pebbles) such that, whenever Bi starts from a

configuration[i,p, ] , where p R , the next configuration of

depth i of Bi is [i, (p, S), ] , where *S {(q,q ) | [i,q, ] [i, q , ]}.iQ Q

The set of states of Bi includes R and 2R RR .

62

The theorem indeed follows from the claim: Let i = 0 and let

0R {p } (the intended initial state of 0B ). We obtain an

automaton which ends up in a state 0(p , S) where *

0S {(q,q ) | [0,q, ] [0,q , ]}Q Q

The set of final states of 0B consists of all states 0(p , S)

where S contains a pair 0(q , q) with q F .

63

For i k the proof of the claim is trivial, as there are no

configurations of depth k . Hence, kB can compute (p, S)

by a stay-transition.

Therefore, let i k and suppose the claim holds for all j i

64

Let the input string w of length n be fixed. Hence iB

works on the string w with positions from

dom ( ).w

For 1l n , let l denote the ( 1)i -pebble

assignment that coincides with in the first i pebbles

and for which ( 1)l i l .

( )lS is the set of pairs ( , )q q of states such that

there is a computation starting at [ 1, , ]li q and

reaching [ 1, ', ]li q which only includes

configurations [ , , ]j q that fulfill 1j i or (

1j i and ( 1)i l ). Intuitively, this says that

pebble 1i is not allowed to move to the right of

position l .

65

We write ( )lS for the set of pairs ( , )q q of states

for which [ 1, , ]li q can be reached from 1[ 1, , ]i q by a subcomputation satisfying the same

property.

The set ( )lS can be computed as follows:

Let ( )lR be the set of pairs ( , )q q for which one

of the following holds:

(a) There exist 1 2,p p such that 1

1[ 1, , ] [ 1, , ]l li q i p , 1

1 2( , ) ( )lp p S and 1

2[ 1, , ] [ 1, , ]l li p i q .

(b) *

1[ 1, , ] [ 1, , ]l l

ii q i q

(c) [ 1, , ] [ 1, , ]l li q i q

66

( )lS is the transitive closure of ( )lR .

They are computed simultaneously.

The information needed for (a) can be computed in one

left-to-right pass of pebble 1i .

By induction we can assume a subautomaton 1iB that

computes, for each position l , the part of ( )lR

contributed by condition (b).

(c) and the computation of the transitive closure do not require any pebble movements.

During the same pass, the automaton can compute, for

each position l , the set ( )lS . The computation of

( )lS makes use of the sets ( )mS , m l .

67

From 1( )nS , 1( )nS and the transition relation

of A one can deduce, during a lift-pebble step, the set

S as in the claim. Note that 1n is the position of the

right delimiter and recall that A lifts its pebbles only at that position.

This completes the proof of the claim and of the

theorem.