 # On synchronization of partial automata

A goal of this paper is to introduce the new construction of an automaton with shortest synchronizing word of length O(d^n/d), where d ∈ℕ and n is the number of states for that automaton. Additionally we introduce new transformation from any synchronizable DFA or carefully synchronizable PFA of n states to carefully synchronizable PFA of d · n states with shortest synchronizing word of length Ω(d^n/d).

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## 1 Synchronization of partial automata

Partial finite automaton (PFA) is an ordered tuple where is a set of letters, is a set of states and is a transition function, not everywhere defined. For and we define inductively as and for where is the empty word and is defined. A word is called carefully synchronizing if there exists such that for every , and all transitions are defined. A PFA is called carefully synchronizing if it admits any carefully synchronizing word. Carefully synchronizing automaton is depicted on the Fig. 1 and its shortest carefully synchronizing word is what can be easily checked via power automaton construction analogous to the deterministic automata construction. The only difference is that we define only transitions we can define.

The concept of careful synchronization of PFA is a generalization of idea of synchronization for deterministic finite automata (DFA) with transition functions defined everywhere. The problem of estimating the value of

was considered first by Ito and Shikishima-Tsuji in [6-7] and later by Martyugin . Ito and Shikishima-Tsuji proved that and Martyugin improved the lower bound with the construction of automata of length . The best known upper bound for is due to .
Let . We define and . It can be easily verified from Fig 1. that the Černý Conjecture is not true for PFAs, since . We also recall following important facts.

###### Fact 1.

Let be a PFA and be its power automaton. Then is synchronizing if and only if for some state there exists a labelled path in from to . The shortest synchronizing word for corresponds to the shortest labelled path in as above.

###### Fact 2.

If automaton is carefully synchronizing then there exists such that transition under is defined for all states and such, that .

Now we are ready to give an example of a PFA with shortest carefully synchronizing word of length for all .

## 2 Automata with long shortest carefully synchronizing words

This section includes construction of an automaton for which the shortest carefully synchronizing word is of exponential length.
Let , , and . We understand as base representation of . Let , . We define automaton as follows:

Let , . we define partial transition function for as:

• ,

• ,

• ,

• ,

• ,

Let us remark some facts about the construction useful for further proofs.

###### Fact 3.

is not defined when and .

is not defined.

###### Fact 5.

is not defined when .

It is worth noticing that only transitions on letters and , join two states together and only letter is defined for all states.
Let and . We also define such, that:

• for corresponds to lower index of

• ,

In other words each corresponds to m-digit base representation of . For example if , then . Finally we define inductively word as:

Now we are ready to formulate first lemma of this section.

###### Lemma 1.

Let and be defined as above. For every there exists a path in and its transitions are labelled with consecutive letters of word .

###### Proof.

The result follows by induction on i.
Case is evident from the definition of since and for .
Now let us assume that the result holds for . From the induction hypothesis we know, that there exists a path whose transitions are labelled with consecutive letters of word . Let and notice that for every we have and on any of letters maps to itself, so for each there exist a path in (also labelled with letters of ). From the definition of we can see that for each it holds that . Using these two observations and the definition of we conclude that the lemma holds. ∎

###### Lemma 2.

Automaton is carefully synchronizing and its carefully synchronizing word is .

###### Proof.

We must show that . From the definition of we see that . From Lemma 1. we know that for every . Also from the definition of we see that for every . Joining those facts together we deduce that . ∎

###### Lemma 3.

Let be as in Lemma 2. Then and is the shortest carefully synchronizing word for automaton .

###### Proof.

First we will show that . It’s obvious that , so . We leave that identity as a simple exercise for a reader.

It can be easily verified that and is the only letter defined for all states. In order to prove minimality of it suffices to show that for each state in there is only one transition that leads to a state that has not been visited yet. All other transitions are either not defined or lead to states visited earlier. We must investigate two cases:

Case for some and
From the definition of , (which was visited) and from Fact 6. is not defined for any . From Lemma 2. it can be seen that for each there exists a letter which leads to an unvisited state . In order to show that there exists only one such letter let us assume, that , and . It is obvious from Fact 4. that for each such that the transition is not defined. If then it follows from Fact 5. that is not defined and the statement is true for that case.

Case for some and
From the definition of we see that . Furthermore when otherwise when , due to Fact 5, transitions are not defined. Notice that for . If , then is not defined. Moreover . Since there is only one letter leading to an unvisited state and transitions under other letters are either undefined or their result is already visited state of statement holds for that case.

Having that we know by induction that is minimal and that ends the proof. ∎

Following theorem is immediate from Lemma 4.

###### Theorem 1.

Let , . The shortest carefully synchronizing word for has length .

Using that theorem we can simply reproduce result obtained by Martyugin  as follows:

###### Corollary 1.

If then there exist a PFA with states and minimal carefully synchronizing word of length .

###### Proof.

We construct automaton with and denote rest of states as . Now we can add a letter to the automaton, say , and add a transition over that letter to , resulting with , such that it acts like identity on and . ∎

## 3 Further improvements

Define a relation on the set of states for an automaton and a given letter such that if, and only if . It is obvious that, for any , is an equivalence relation on the set of states. We also define -transversal as such that each equivalence class has at most one representative in . Let be equivalence classes of on . Finally we say that letter is -preserving with respect to if for any -transversal , such that lower index of corresponds to -th equivalence class, such that lower index of corresponds to -th equivalence class. It can be easily seen that letter in automaton defines on the , resulting with partition of it on pairwise disjunctive sets, and letters for are -preserving. In Section 2. we defined the automaton that first creates on and then traverses some of transversals of that relation. Specifically, after applying letter on we reducing number of equivalence classes possible to traverse by one. It is natural question to ask if we can traverse more transversals than we have shown in Section 2. We give universal construction that can be used to improve the lower bound obtained by Martyugin.
Main idea is to immediately reduce to equivalence classes and then treat those classes as states of some synchronizable DFA or carefully synchronizable PFA. First we define construction sufficient to construct carefully synchronizing automaton with long shortest carefully synchronizing word for any given synchronizable DFA or carefully synchronizable PFA and next we apply that construction to Černý automata in order to give an upper bound for shortest carefully synchronizing word for such created automaton.
Let be a finite automaton. Let and . We define PFA as follows:

Let , . we define partial transition function for as:

• ,

• ,

• ,

• ,

• , for all such that

###### Fact 6.

Let be defined as above on . Letters are -preserving.

Before moving further we prove following lemma.

###### Lemma 4.

Let be a power automaton for automaton . Let . Let also and . Then the shortest path from to in is of length .

###### Proof.

Since showing that path of desired length exists is similar to the proof of Lemma 1. and the word which traverses all sets on is analogous to the word in the proof of Lemma 1. we omit that part of proof and focus on proving that at any point on there exist only one transition to state not visited before in order to show minimality of .
Notice that we can treat any on as -ary representation of just like in Lemma 3, the only difference is that we omit ”empty spots” in . Another similarity is that after is directly after on . Assume that , and . It is obvious from Fact 4. that for each such that the transition is not defined. If we must investigate two cases. If then from Fact 5. is not defined. Else notice from definition of that transition zeros some of positions younger than and maps older positions and position to itself, so the result of that transition is such that which was visited earlier since all numbers between and must be on . None of letters is defined. That concludes the proof. ∎

Having that construction we may prove main theorem of that section.

###### Theorem 2.

Let . is a synchronizing DFA(carefully synchronizing PFA) if, and only if is carefully synchronizing.

###### Proof.

Let be a power automaton for automaton and be a power automaton for automaton . Fix . First we prove right implication. Since is (carefully) synchronizing there exist a (carefully) synchronizing word . We now construct carefully synchronizing word for . Let . Since only letter is defined for all states we append letter to . Notice that . From Lemma 1. We know that there exist of length such that . Notice that for each or is undefined. Let . It is easy to observe that if such that , then . Any such is subset of , so from Lemma 4. there exist word (of length ), such that . From that it is easy to notice that word carefully synchronizes .
In order to prove left implication suppose, for the sake of contradiction, that there exist non-synchronizable DFA (non-carefully synchronizable PFA) such that its is carefully synchronizable. Any (carefully) synchronizing word for must start with letter , which defines on . Notice that for any -transversal , is not defined for and, due to Fact 7, there is no letter that can change traversed equivalence classes. That leads to contradiction, since must be (carefully) synchronizing so to be carefully synchronizing. ∎

###### Corollary 2.

If is a synchronizing DFA(carefully synchronizing PFA) then the shortest carefully synchronizing word for is

###### Proof.

Since and it labels the shortest path from to we conclude corollary holds. ∎

Now we are ready to bound shortest carefully synchronizing word for . Let be a DFA such that , and is defined as follows:

• for

Despite the shortest synchronizing word for is of length , we find another synchronizing word, more appropriate word to bound .

###### Lemma 5.

If is even, then word synchronizes , otherwise synchronizes .

###### Proof.

We prove for even

since proof for odd

is similar. Let be a power automaton for automaton . Denote and . It is easy to check by induction on that if , then . So . Now consider action of on set . We will proof by induction on that if , then . If then, from definition of , . So it is easily seen that , and . Assume that lemma holds for every , then . Similarly as in case, applying word results with so the statement holds. That implies . Notice that and that ends proof. ∎

Having that we prove following theorem.

###### Theorem 3.

If is even, then . Otherwise .

###### Proof.

Let be even. We construct for a given automaton a reset word of desired length. From Lemma 5 we know that , so we know that there exist carefully synchronizing word for of form . Denote and . It is obvious that . Now we calculate . It is easy to notice that after applying letter number of equivalence classes traversed by reduces by one, so from Lemma 4. . Consider . From the proof of Lemma 5 we can deduce that we reduce number of equivalence classes by one after two letters in . Thus, from Lemma 4 we obtain . Because we do not traverse all equivalence classes when and (see proof of Lemma 5), we substract . After simple calculations we obtain . Similar analysis of word when is odd results with . ∎

We are now able to formulate following corollary.

###### Corollary 3.

Let , . Then and .

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