Some recursively unsolvable problems relating to isolated ... - LaBRI

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to the "cl.assical" theory: however, ~he first problem was solved (in a negative ... strings from right to left. ... To prove the recursive unsolvability of this problem it is possible ... whether 1/2 is isolated for p, contrary to the statement of th. 3.1.
SOME R E C U R S ! V E L Y U N S O L V A B L E P R O B L E M S R E L A T I N G TO ISOLATED C U T P 0 1 N T S IN PROBABILISTIC AUTOMATA A. Bertoni~

G. Mauri,

M. Torelli

Istituto di C i b e r n e t i c a U n i v e r s i t A di Milano - Italy

I. I n t r o d u c t i o n In the present work we examine the d e c i d a b i l i t y of some questions c o n c e r n i n g the existence of cutpoints probabilistic

automaton

isolated with respect to a given

(PA). As is well known,

the q u e s t i o n of the e-

x i s t e n c e of an i s o l a t e d cutpoint~ b e s i d e s an intrinsic interest, has a considerable

importance

in c o n n e c t i o n with R a b i n ' s theorem

ensures that a cutpoint event T(~,~)

is regular if ~

[i], w h i c h

is an isolated cut-

point for the PA ~. Rabin

[2] m e n t i o n s

PA ~ w i t h

rational

the f o l l o w i n g tw9 open problems:

for every given

t r a n s i t i o n probabilities,

l) decide w h e t h e r every given

(rational)

number ~

is i s o l a t e d with

respect to ~; 2) decide w h e t h e r ~ h a s

any i s o l a t e d cutpoints.

These p r o b l e m s have b e e n studied from the point of view of the const__ructive theory of PA ([8], ~ ] ) ,

coming to c o n c l u s i o n that w i t h i n the

frame of such theory it is u n d e c i d a b l e w h e t h e r every given PA has an isolated cutpoint.

On the other hand,

M e t r a and Smilgais

[~

an a l g o r i t h m w h i c h decides on the existence of a e u t p o i n t respect to a PA h a v i n g only two states and rational

discovered isolated with

transition probabi-

lities. The results of the c o n s t r u c t i v e to the "cl.assical" theory: however, negative

sense)

theory cannot be carried over d i r e c t l y ~he first p r o b l e m was solved

in ~6J, m a k i n g use of some ideas taken from

Here we want to summarize and extend in part the results in that the first p r o b l e m is s e m i d e c i d a b l e of events,

(in a

[7~ and

[~ .

[6] , Showing

if r e f e r r e d to a certain class

and then p r o v i n g the u n s o l v a b i l i t y of the second ~ r o b l e m .

88

2. Definitions Let

~

{~,~'"~M}

be a finite

the free m o n o i d generated by [, of symbols

in ~,

including

alphabet:

in which

we shall denote by , where

a) ~ is a stochastic vector

l~n,

b) A ( ~ ) is, for every

, a stochastic matrix nxn,

row in the matrix

~eZ

i. e . ~ = ( ~ , ..... K~)',

is a stochastic

vector;

e) ~ is a vector nxl with components

0 or i.

We shall

indicate by

~T the transpose

of ~,

From now on, we shall consider numbers

~= ~ = i

~ ~0~

that is every

i. e. the vector

and elements

Ixn.

of matrices

and

vectors which all are rational numbers. Def.

2. A probabilisti c event

p : [~--~,i] as follows:

given by pK(x)=~A(x)~, A(A)=I,

Def.

[0,I~ denotes

where A(x)

is reeursively

defined

the identity matrix;

A(x~)=A~x)A(~ and

(PE) generated by the PA uq~is the function

) for every ~ e Z

the interval

of rational

; numbers

r, 0 4 r 4 i.

3. A rational number A is said to be an isolated cutpoint with res-

pect to ~ i f f

there exists an

E>0 such that for every x E ["

Ip(x)-ala ~. Def.

4. A PA is said to be quasidefinite

number k(~)

iff for any

such that for every x ~ [ * with l(x)) k(6)

ial distributions

~ and ~

, ]PE (x)-p9 (x) I 4 6

6~0

there

is a

and any two init-

89

3. U n s o l v a b i l i t y of the first n r o b l e m In this section we shall resume the results given in [6J, r e f e r r i n g to that w o r k for details,

the m a i n difference b e i n g that here we use,

i n s t e a d o~ the longest prefix common to two strings, in order to a t t a i n such results more directly.

the longest suffix,

This introduces into

p r o o f s only m i n o r changes, which s u b s t a n t i a l l y come down to reading strings from right to left. Let Z and K be two finite a l p h a b e t s and c o n s i d e r two h o m o m o r p h i s m s ~

~:

~*,.jA

>

~



e s p o n d e n c e p r o b l e m consists

.

The well k n o w ~ Post's corr-

in deciding,

w h e t h e r there exists x ~ ~ * s u c h

for any two given h o m o m o r p h i s m s ,

that ~ ( x ) = ~ ( x ) .

It is also well known

that such a p r o b l e m is r e c u r s i v e l y unsolvable. The existence

of a solution for a given instance of Post's c o r r e s p o n d -

ence p r o b l e m o b v i o u s l y implies that ~ ( J ~

~(x) o~x))=

o~

;

we want to examine now the d e c i d a b i l i t y of the f o l l o w i n g question: the o a r d i n a l i t y of the set

~'

x~Z~

for any two given h o m o m o r p h i s m s

~(xjo~(x) ~,

is

finite or infinite

j ~

To prove the r e c u r s i v e u n s o l v a b i l i t y of this p r o b l e m it is possible to p r o c e e d in a way similar to the one f o l l o w e d to prove the u n s o l v a b i l ity of Post's c o r r e s p o n d e n c e p r o b l e m

(see for instance

[i0]), e s t a b l i s h -

ing a c o r r e s p o n d e n c e b e t w e e n the sequence of c o m p u t a t i o n of every Tuming m a c h i n e with a given input9 ~{(x)o~(x)

common to the two h o m o m o r p h i s m s

ular T u r i n g machine. { U

, ~)o

possible

and the longest

If it were possible

@.(×)~

a s s o c i a t e d with that partic-

to determine w h e t h e r

is finite or infinite,

it w o u l d then be

to decide w h e t h e r every given Turing machine

latter p r o b l e m is n o t o r i o u s l y unsolvable, is u n s o l v a b l e Now,

(for every x e ~ * ) suffix

stops.

Since the

~t follows that our p r o b l e m

too.

let Km{0,1 ..... k-l} be a finite set of symbols to be i n t e r p r e t e d m

as digits,

be given by r ( x l . . . X m ) = =I x .3k J _ l ~ m

and let r : K ~ - - ~ l ]

w h e r e x . e K, [ ~ j ~ m . $ A n y given h o m o m o r p h i s m ~:

< ~ , ," A >

i n t e r p r e t e d as a f u n c t i o n ~er~: Moreover,

+

~--~,

~

m a y then be

I] .

such a f u n c t i o n is a PE: in fact the 2-state PA

90

~.~, ~

< ~

(~o)

~-~(~)-~(%)

, A(~)~

where k ( ~ ) = k - I ( ~ ( ~ ) ) ,

generates

~(%)~%

the probabilistic

, ~

~

event ~(x),

,

(3.0

as is

easily verified. Given two h o m o m o r p h i s m s (note the exclusion fractional prove

k-adie

~

of k-l),

and ~z~

mapping

and denoted by ~ a n d

representations

r~

and r ~ ,

that number 0 is isolated with respect

From the foregoing Th. 3.i.

result we derive

It is undecidable

~into

whether

{O,i ..... k-2)

~ respectively

the

it is not difficult to

~Z~ (~)-~(~))

the two following

to iff

theorems.

1/2 is isolated with respect

to

every given 4-state PA. Proof. ed PEs, too,

Let

~4 a n d ~ b e

two h o m o m o r p h i s m s

and

generated by 2-state automata according

as well as the convex combination

a/~ ~

~

and

~z their associat-

to (3.I):

I-@2 is a PE Z4e ~ ( ~ - ~ ) = p j

~ ~(~-~=

which can be generated by a 4-state PA. It is obvious

that 1/2 is isol-

ated with respect

to

undecidable, Th. 3.2.

to p iff O is isolated with respect

which

Let ~ be a rational number,

Let 0 < A < 1 / 2

O¢ A < I: it is undeeidable

isolated

whether

and p be the PE defined in th. 3.I:

for 2lp, since

12~p-AI=2Alp-i/21it

1/2 is isolated for p, contrary

Moreover, a parallel

if 1/2 < ~ < i ,

whether

to every given PA. 2Ap is a PE

(which can be generated by a 5-state PA) and if it were decidable ~is

is

as we showed above. D

A is isolated with respect Proof.

~-~,

l-2(l-l)p=q

whether

would be decidable

to the statement

is a PE, and since

of th. 3.1. Iq-II=2(I-A)Ip-I/21

argument h o l d s . ~

As one can see,

the question

of the u n d e c i d a b i l i t y

for the extreme

values ~=O and A=I remains open.

4. Partial

decidability

of the first problem for a certain class of PEs

Let be a quasidefinite

PA: pZ(x)=IIA(x)~,

p~(zx)=~A(z)A(x)~=~(z)A(x)~,

if we write ~(z)=~A(z):

a stochastic

f>O

vector,

with l(x)~ k([)

for any

and every z ~ [~

there is a k([)

then and since ~(z)

is

such that for every x

91

Informally, sible"

to what can be p r e f i x e d

We note page

~he PE generated by a quasidefinite

that the PA ~ ? o f

178 of

[93 ) and

Th. 4.1. k(6)

to a given string of sufficient

(3.1)

is quasidefinite

~zx)=k-l(~(x))~(z)+~(x),

We can state the following

(p being

ated with respect

(see exercise

is recursively

nonincreasing

l(x)> k(£)

the PE generated b y e ) ,

to~"

length. 4,

theorem:

such that for every x,z ~ Z ~ i f

Ip(zx)-p#x)I$6

insen-

so that I~(zx)-~(x)li~i+ of;

¢(x))

i/2

i/2

:

z ~S.

We can conclude that the value 1/2+~/2,

for instance,

b) Suppose 0 is not isolated for ~1-@2:

in that case there exists a

sequence IXn] of words in Z + such that (5.1),

is isolated.

lim p(x )=I/2. But n --,o~ n

because of

(5.2) and (5.3), if

z=c(x c) i~ (0C) k~ (x c) i2 (0c) k~ ...(x c)i~(0e) k~ n n n with ij,kj~O,

i.~ j g m ,

then ~(z)=fP(Xn)(y) , where

y=l i~ 0 k~ 1 ~ 0 k~ ...lifO k~ and lim

n --> o ~

@(z)= n lim - ~

fP(Xn ) (y)=fl/2(Y)=0. ~.

Since the set of rational numbers whose representation

is of type

0.~ is dense in [0,1] , we conclude that there cannot exist an isolated cutpoint with respect to ~. Th. 5.3. The problem of deciding whether every given PA with rational probability

transitions has an isolated cutpoint is recursively unsol-

94

vable.

Proof.

Suppose the problem is solvable for the a b o v e - m e n t i o n e d P A ~ :

it w o u l d then be possible to ~ i - ~ 2 ,

to decide w h e t h e r 0 is isolated with respect

c o n t r a r y to what we prove d in §3.

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