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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