Infinite Synchronizing Words for Probabilistic Automata (Erratum)
Abstract
In [DMS11b], we introduced the weakly synchronizing languages for probabilistic automata. In this report, we show that the emptiness problem of weakly synchronizing languages for probabilistic automata is undecidable. This implies that the decidability result of [DMS11b, tech, DMS11] for the emptiness problem of weakly synchronizing language is incorrect.
1 Definitions.
We present the main notations and definitions. We refer to [DMS11b] for detailed preliminaries.
A probability distribution over a finite set is a function such that . The support of is the set . We denote by the set of all probability distributions over . Given a finite alphabet , we denote by the set of all finite words and by the set of all infinite words over . The length of a finite word is denoted by .
Probabilistic Automata.
A probabilistic automaton (PA) consists of a finite set of states, an initial probability distribution , a finite alphabet , and a probabilistic transition function . In a state , the probability to go to a state after reading a letter is . We define , and for sets and , let . The outcome of on an infinite word is the infinite sequence of probability distributions such that is the initial distribution, and for all and ,
The norm of a probability distribution over is .
Weakly Synchronizing Language for PAs.
An infinite word is said to be weakly synchronizing for PA , if
We denote by the set of all weakly synchronizing words, named weakly synchronizing language, of . Given a PA , the emptiness problem of weakly synchronizing language asks whether .
2 The emptiness problem of weakly synchronizing languages for PAs is undecidable.
Theorem 2.2 states that the emptiness problem of weakly synchronizing languages for PAs is undecidable. To show that, we present a reduction from the value 1 problem for PAs which is undecidable [GO10], to our problem.
The value 1 problem for PAs.
Let be a PA with a single initial state where and a set of accepting states . The value 1 problem considers finite words: the computation of on the word is the sequence where and for all . The acceptance probability of by is given by . The value of , denoted , is the supremum acceptance probability . Given a PA , the value 1 problem asks whether . It is equivalent to check if there are some words accepted by with probability arbitrarily close to 1.
Theorem 2.1
The value 1 problem for probabilistic automata is undecidable [GO10].
Undecidability result.
Theorem 2.2
The emptiness problem of weakly synchronizing languages for probabilistic automata is undecidable.
Proof
We present a proof using a reduction from the value 1 problem for PAs. Given a PA equipped with the single initial state and accepting states , we construct another PA , such that iff .
First, from the PA we construct another PA such that has value 1 iff does. The state space is extended, by adding two new states and (). The alphabet where . The initial state is in common, but only is an accepting state (). The transition function globally remains unchanged; only the following transitions are added in : for all , and for all . In addition, for all . Figure 2 illustrates the definition of from . By construction, since is the only accepting state and is only reachable with , and . We see that if with , otherwise . Thus, has value 1 iff has value 1.
Then from the PA , we construct a PA such that , the weakly synchronizing language of is not empty, iff has value 1. For each state except , a twin state is added to the state space. Thus, . The alphabet where . The initial distribution is . The probabilistic transitions function is defined as follows.

for all states and all .

for all states .

for all states and all .

In addition, for all .
Figure 3 shows the construction. For convenience, a twin and are drawn in an oval; Figure 2 illustrates the transitions between two pairs of twin states where each pair is replaced with an oval. Intuitively, the PA mimics the behavior of where each state (except the accepting state ) shares the probability to be in with the twin , in all steps . Moreover, each “resets” .
Let us shortly formalize two important and intuitive properties of resulting from the construction.
Property : Let be an infinite word containing the symbol such that where and . Let be the outcome of on the word ; and be the outcome on . Since with uniform distribution, on inputting the letter , the automaton is reset to the initial distribution , and one may “forget” the prefix . Formally, for all and ,
Property : For all words , the computations of and on give
 a)

for all ,
 b)

.
Let us recall that . Property can easily be proved since and no labeled transition with reaches . We prove Property by induction on the length of .
Base (): Let be the empty word . By construction, we know the initial distribution for is and for is .
Induction: Assume that the statement holds for all words with . Let where and . By definition, for . By induction hypothesis, since for , and since , we conclude . Similarly, we obtain .
Now, we show that iff the PA has value 1.
First, we assume that has value 1. So, for all , there exists a finite word such that . Let be such that for all . We claim that the infinite word is a weakly synchronizing word for . Let be the outcome of on . For , let be the prefix of which ends with . Let denote the length of , then The last of the prefix is located at (for ). Thus, which is the initial distribution of . By construction, . Hence, and . It implies that . Hence is a weakly synchronizing word for , thus .
Now let us assume that . So, by definition there exists an infinite word such that . We claim that contains infinitely many . Towards contradiction, assume that there exists such that for all , .
Let be the outcome of on . There are two cases:

There exists such that . Since and , we have for all , a contradiction with the fact that is weakly synchronizing.

For all we have . Let be one position after the last in , or if there is no . For all , let be the subword starting at with length ; and let be the computation of on the finite word . By Properties and , for all and all and . This gives , a contradiction.
Now, we claim that contains infinitely many too. Towards contradiction, assume that there exists such that for all , . Let be the sequence of all positions in after where () in increasing order (). So, and for all and all . Let be the finite subword of between the positions and where for some . We see that for some . Let be the computation of on . By Properties and , for all and all , and also . This gives for all , a contradiction with the fact that is weakly synchronizing.
We showed that contains infinitely many and . Since is weakly synchronizing, for all there exists where (In fact, since is weakly synchronizing, for all for all there exists where , but we do not need here). For fixed , let be such that . Let be one position after the last before (i.e., ), or if there is no before . Let be the finite subword of starting from position to the position . By Properties and , for all . Therefore, since , we have and . Hence has value 1.