Chapter 10 Equivalence and Inclusion Problem for Strongly Unambiguous Bu'Chi Automata

Equivalence and Inclusion Problem for Strongly

Unambiguous Buchi Automata

Nicolas Bousquet1 and Christof Loding2

1 ENS Chachan, [email protected]

2 RWTH Aachen, Informatik 7, 52056 Aachen, [email protected]

Abstract. We consider the inclusion and equivalence problem for un-ambiguous Buchi automata. We show that for a strong version of unam-biguity introduced by Carton and Michel these two problems are solvablein polynomial time. We generalize this to Buchi automata with a fixedfinite degree of ambiguity in the strong sense. We also discuss the prob-lems that arise when considering the decision problems for the standardnotion of ambiguity for Buchi automata.

1 Introduction

The model of unambiguous automata is located between deterministic and non-deterministic automata. An unambiguous automaton is a nondeterministic au-tomaton such that each input that is accepted has a unique accepting run. Theconcept of unambiguity also occurs in other areas of theoretical computer sci-ence, for example in complexity theory. The problems solvable in polynomialtime by unambiguous (nondeterministic) Turing machines are collected in thesubclass UP (Unambiguous Polynomial time) of NP [17].

There are two aspects in the study of unambiguous automata: expressivenessand computational complexity. Concerning expressiveness, because of the well-known equivalence between deterministic and nondeterministic finite automataover finite words and trees, unambiguous automata can recognize all the regularlanguages over these two domains. For automata over ω-words it is known thatdeterministic Buchi automata are strictly less expressive than nondeterministicones [10]. However, Arnold showed that all the ω-regular languages can be recog-nized by an unambiguous Buchi automaton [2]. For automata over ω-trees, theclass of unambiguous automata is not as expressive as the class of full nondeter-ministic tree automata (with standard acceptance conditions like parity, Rabinor Muller) [12,6].

An interesting subclass of the class of unambiguous Buchi automata is con-sidered by Carton and Michel [7]: Their definition requires that for each input(accepted or not) there is a unique run passing infinitely often through a finalstate (whether from an initial state or not). Thus, an infinite word is acceptedif the initial state is the first state of the path passing infinitely often through

A.-H. Dediu, H. Fernau, and C. Martın-Vide (Eds.): LATA 2010, LNCS 6031, pp. 118–129, 2010.c© Springer-Verlag Berlin Heidelberg 2010

Equivalence Problem for Strongly Unambiguous Buchi Automata 119

a final state. Non-acceptance means that the unique path of this form does notstart in an initial state. Carton and Michel show [7] that this restricted class ofBuchi automata suffices to capture the class of ω-regular languages. We considerin this paper a slight modification of their definition, and refer to these automataas strongly unambiguous.

The second interesting aspect for unambiguous models is the computationalcomplexity of algorithmic problems. We consider here the equivalence problem(as well as the inclusion problem which turns out to have the same complexity).It is well known that there is a gap in complexity for the equivalence problem be-tween deterministic and nondeterministic automata. The problem can be decidedin polynomial time over finite words [15] and finite trees for deterministic au-tomata, whereas the problem is PSPACE-complete over finite words (see Section10.6 of [1]1) and EXPTIME-complete over finite trees [8] for nondeterministicautomata.

As shown by Stearns and Hunt, the equivalence problem for unambiguous fi-nite automata over finite words is still polynomial [14] and Seidl showed the sameover finite trees [13]. In the present paper, we show that this result also holdsfor strongly unambiguous Buchi automata. To our knowledge, this identifies thefirst subclass of Buchi automata that is expressively complete for the ω-regularlanguages and at the same time allows a polynomial time equivalence test.

The polynomial time equivalence test over finite words from [14] uses a count-ing argument: The main idea is that, for unambiguous automata, the numberof accepting runs is equal to the number of accepted inputs. Stearns and Huntproved that it is sufficient to count the number of accepting paths of the givenunambiguous automata only up to a certain length and that this can be done inpolynomial time. The problem when trying to adapt such an approach to Buchiautomata is that runs of Buchi automata are infinite and one cannot simplycount the number of accepted words up to a certain length. However, it is possi-ble to restrict the problem of equivalence of regular languages of infinite wordsto ultimately periodic words [4] (see also [5]). A word is ultimately periodic if itis of the form u · vω where u and v are finite. It turns out that this restriction toultimately periodic words allows to adapt the counting argument to the case ofstrongly unambiguous Buchi automata. Instead of presenting a direct adaptionof the proof for finite words we show that the equivalence problem for stronglyunambiguous automata can be reduced in polynomial time to the equivalenceproblem for unambiguous automata on finite words.

This kind of reduction does not seem to work for unambiguous Buchi au-tomata. We show that deciding whether an unambiguous Buchi automaton (evena deterministic one) accepts some periodic word vω, where v is of a given lengthn, is NP-complete. Although this proof does not show that the equivalence prob-lem for unambiguous Buchi automata is difficult, it shows that different methodsare required.

1 In [1] the PSPACE-hardness of the non-universality problem for regular expressionsis shown. This can easily be turned into a PSPACE-hardness proof for the equivalenceproblem for nondeterministic finite automata.

120 N. Bousquet and C. Loding

The remainder of the paper is structured as follows. In the second section wegive some definitions and simple properties of Buchi automata. In Section 3 weshow how to reduce the equivalence problem for strongly unambiguous Buchiautomata to the case of unambiguous automata on finite words. In Section 4we extend these results to strongly k-ambiguous automata, a relaxed notion ofstrong unambiguity, where each word can have at most k final paths. In Section 5we show that deciding if a deterministic Buchi automaton accepts periodic wordsof a given length is NP-complete. We conclude in the last section.

2 Definitions and Background

For an alphabet Σ we denote as usual the set of finite words over Σ by Σ∗, theset of nonempty finite words by Σ+, and the set of infinite words by Σω. Thelength of a finite word u ∈ Σ∗ is denoted by |u|. For an infinite word α ∈ Σω wedenote the jth letter by α(j), i.e., α = α(0)α(1) · · · .

An infinite word of the form uvω = uvvv · · · for finite words u, v is calledultimately periodic.

We consider nondeterministic finite automata (NFA) on finite words of theform A = (Q, Σ, Qin, Δ, F ), where Q is a finite set of states, Σ is the inputalphabet, Qin ⊆ Q is the set of initial states, Δ ⊆ Q × Σ × Q is the transitionrelation, and F ⊆ Q is the set of final states. We use the standard terminologyfor NFAs (see e.g. [9]) and denote the language of words accepted by A by L(A).

A Buchi automaton A = (Q, Σ, Qin, Δ, F ) is of the same form as an NFA. Incontrast to NFAs, a Buchi automaton defines a language of infinite words. A pathfor the infinite word α ∈ Σω is an infinite sequence of states q0q1... such thatfor all j ∈ N, (qj−1, α(j), qj) ∈ Δ. A final path is a path that passes infinitelyoften through a final state. A path begins in q if q0 = q. If a final path for αbegins in some q0 ∈ Qin, then the word α is accepted by A. So a final pathis accepting if it starts with an initial state. If an accepting path for α exists,then A accepts α. The language L(A) is the set of infinite words α accepted byA. For an automaton A = (Q, Σ, Qin, Δ, F ) we denote by LA(q) the languageaccepted by (Q, Σ, {q}, Δ, F ). The class of languages that can be accepted byBuchi automata is called the class of ω-regular languages.

For a finite word u and two states q, q′ of A we write A : qu−→ q′ if one can

reach q′ from q on reading u, and we write A : qu−→F q′ if one can reach q′ from

q on reading u and by passing through a final state on the way.It is well known that the equivalence problem for NFAs, i.e., the question

whether two given NFAs accept the same language, is PSPACE-complete (see[1]). The same holds for the inclusion problem because equivalence can easilybe tested by checking for both inclusions. Furthermore, the lower bound on thecomplexity easily extends to Buchi automata.

In [14] unambiguous NFAs are considered and it is shown that the equiva-lence and inclusion problem for these automata are solvable in polynomial time.Unambiguous automata are nondeterministic automata in which for each wordthere is at most one accepting path.


Our aim is to see to what extent these results can be lifted to Buchi automata.We introduce two notions of unambiguity, the standard one and a stronger notionintroduced in [7].

Definition 1. A Buchi automaton A is called unambiguous if every infiniteword has at most one accepting path in A, and it is called strongly unambiguousif every infinite word has at most one final path.

Clearly, if A is strongly unambiguous, then for each infinite word α there is atmost one state q such that α ∈ LA(q). We state this observation as a remark forlater reference.

Remark 1. Let A be a strongly unambiguous Buchi automaton and α be aninfinite word. If α ∈ LA(q1) and α ∈ LA(q2) then q1 = q2.

Consider, for example, the automata shown in Figure 1. Both automata acceptthe language over {a, b} consisting of all words that contain infinitely many b.The automaton on the left-hand side is deterministic (where deterministic au-tomata as usual only have a single initial state and for each state and letter atmost one outgoing transition) and therefore unambiguous, but it is not stronglyunambiguous: the word bω is accepted from both states q0 and q1. The automa-ton on the right-hand side is strongly unambiguous. It accepts the same languageas the deterministic automaton, but from state p0 all accepted words start witha, and from p1 all accepted words start with b.

q0 q1

a

b

b

a

p0 p1

a

a

b

b

Fig. 1. Example for a deterministic Buchi automaton (left-hand side) and a stronglyunambiguous Buchi automaton (right-hand side) for the same language

Note that each strongly unambiguous automaton is unambiguous becauseeach accepting path is also a final path. It has been shown in [2] that each ω-regular language can be accepted by an unambiguous Buchi automaton. Theclass of strongly unambiguous automata has been introduced in [7]2 and it hasbeen shown that this class is expressively complete for the ω-regular languages.

Theorem 1 ([7]). Every ω-regular language can be recognized by a stronglyunambiguous Buchi automaton.

2 The definition in [7] is even more restrictive: It is required that each word has exactlyone final path. This allows an easy complementation by complementing the set ofinitial states. We have chosen the more relaxed notion because the polynomial timeequivalence test also works in this setting. Further note that in [7] theses automataare simply called unambiguous and not strongly unambiguous.


This expressive completeness makes strongly unambiguous Buchi automata aninteresting class. It is also worth noting that strongly unambiguous Buchi au-tomata naturally occur in the translation from linear temporal logic formulasinto Buchi automata. In the standard approach for this translation the Buchiautomaton guesses valuations of all subformulas of the given formula and ver-ifies that the guesses are correct (see [3]). An input word is accepted from theunique state that evaluates all subformulas correctly. Hence the automaton thatis constructed in this standard way is strongly unambiguous.

Before we turn to the decision problems for strongly unambiguous Buchi au-tomata, we compare them to deterministic automata. Note that deterministicBuchi automata do not capture the full class of ω-regular languages, but usingextended acceptance conditions like the Muller acceptance condition, determin-istic automata become expressively complete [11] (see also [16]).

The example from Figure 1 already shows that deterministic Buchi automataneed not to be strongly unambiguous. In fact, there is no deterministic Buchi au-tomaton that is strongly unambiguous and equivalent to the one fromFigure 1: Assume A is a deterministic and strongly unambiguous Buchi au-tomaton accepting all words containing infinitely many b. Then the word bω isaccepted from the unique initial state. Then bω is not accepted from any otherstate by Remark 1. Thus, the initial state is final and has a b-loop. The only wayto accept abω would be to also have an a-loop on the initial state. This wouldmean that A accepts all ω-words over {a, b}. Note that we only used the factthat a deterministic automaton has only a single initial state. So this exampleshows that a set of initial states is necessary for strongly unambiguous Buchiautomata, in general.

The next example shows that strongly unambiguous automata can be ex-ponentially more succinct than deterministic ones. We formulate the followingremark for deterministic Muller automata, because the Muller condition is themost general one of the standard acceptance conditions that are usually con-sidered: A Muller condition is specified by a family F of state sets. A run isaccepting if the set of states that appear infinitely often in this run is a memberof F .

Remark 2. There is a family (Ln)n≥1 of ω-languages over the alphabet {a, b}such that each Ln can be accepted by a strongly unambiguous Buchi automatonwith n+2 states, and each deterministic Muller automaton for Ln needs at least2n states.

Proof. We use the standard syntactic right-congruence for ω-languages L, de-fined by u ∼L v iff uα ∈ L ⇔ vα ∈ L for all α ∈ Σω. As for automata on finitewords (see [9]), one can show that each Muller automaton for L needs at leastas many states as there are classes of ∼L.

The language Ln = Σ∗aΣn−1abω can be recognized by a strongly unambigu-ous Buchi automaton of size n + 2 as shown in Figure 2. The number of ∼Ln

classes is at least 2n, so a deterministic Muller automaton which recognizes Ln

has at least 2n states. ��


q0 p1 p2 · · · pn q1

a, b

a a, b a, b a, b a

b

Fig. 2. A strongly unambiguous Buchi automaton for L = Σ∗aΣn−1abω

Finally, we would like to mention that there are also deterministic Buchi automataexponentially smaller than strongly unambiguous ones. This is in contrast to un-ambiguous automata, since each deterministic automaton is unambiguous.

Remark 3. There is a family (Ln)n≥1 of ω-languages over the alphabet {a, b}such that each Ln can be accepted by a deterministic Buchi automaton withn + 1 states, and each strongly unambiguous Buchi automaton for Ln needs atleast 2n−1 states.

Proof. Let Ln be the language of all words in {a, b}n in which the nth letteris a. Using n + 1 states a deterministic automaton can check this property.

Let An be a strongly unambiguous automaton for Ln. Assume that An hasless than 2n−1 states. Then there are two different words w1, w2 ∈ {a, b}n−1 oflength n − 1 such that w1abω and w2abω are accepted by An from the samestate q. Since w1 and w2 are different, we can assume w.l.o.g. that w1 = waw′

1

and w2 = wbw′2 for some words w, w′

1, w′2. Let m = |w| and v be a word of length

n−m−1. Then vwaw′1b

ω is in Ln. The corresponding accepting run must reach qafter having read v because otherwise there would be two different final pathsfor waw′

1bω. Hence, there is also an accepting run for vwbw′

2bω: the one that

moves to q on reading v, and then accepting wbw′2b

ω from q. Since wbw′2b

ω /∈ Ln

we get a contradiction and thus An has at least 2n−1 states. ��

3 Equivalence for Strongly Unambiguous BuchiAutomata

In this section we prove that the equivalence problem for strongly unambigu-ous Buchi automata can be solved in polynomial time. For the case of finitewords, the proof of Stearns and Hunt [14] uses a counting argument: For anunambiguous automaton over finite words the number of accepted words of acertain length is the same as the number of accepting paths of this length. Theidea is to decide if L(A1) ∩ L(A2) = L(Ai) for i = 1, 2 by simply counting thenumber of accepting paths. Since L(A1) ∩ L(A2) ⊆ L(Ai), the equality holds iffor each n there is the same number of accepting paths in the automaton forL(A1) ∩ L(A2) and in Ai. The key argument is then that a comparison of thenumber of accepting paths is sufficient up to a certain bound of the length.

For general infinite words it is impossible to count final paths in a reasonableway. However, there are infinite words that can be represented in a finite way: theultimately periodic words introduced in the previous section. This class of words


is particularly interesting because it can be used to characterize equivalence andinclusion of ω-regular languages. This easily follows from the closure propertiesof the class of ω-regular languages and the fact that each non-empty ω-regularlanguage contains an ultimately periodic word.

Theorem 2 ([4]). Let A1 and A2 be two Buchi automata.

1. A1 and A2 accept the same language if and only if they accept the sameultimately periodic words.

2. L(A1) ⊆ L(A2) if and only if the ultimately periodic words recognized by A1

are in L(A2).

Using this fact one can indeed adapt the counting argument from [14] to count ul-timately periodic words accepted by Buchi automata. However, instead of adapt-ing the proof of [14] we can also give a reduction of the equivalence problem forstrongly unambiguous Buchi automata to the equivalence problem for unam-biguous automata over finite words. The idea of reducing decision problems forBuchi automata to automata over finite words has already been used in [5].The main difference here is that the reduction is computable in polynomial timewhen starting from a strongly unambiguous automaton. Before we present thereduction we state a simple but important property of strongly unambiguousBuchi automata that makes our reduction work.

Lemma 1. Let A = (Q, Σ, Qin, Δ, F ) be a strongly unambiguous Buchi automa-ton. An ultimately periodic word uvω is accepted by A iff there are states q0 ∈ Qin

and q ∈ Q such that A : q0u−→ q

v−→F q.

Proof. Obviously, if A : q0u−→ q

v−→F q for states q0 ∈ Qin and q ∈ Q, then uvω

is accepted. Now suppose that uvω is accepted by A. Then there is an acceptingpath ρ = q0q1.... on uvω that starts in some q0 ∈ Qin. Let q = q|u| be thestate reached in ρ after reading u, and let q′ = q|u|+|v| be the state reachedafter reading uv. Then vω is in LA(q) and in LA(q′) and thus, by Remark 1 wehave q = q′. Furthermore, that path A : q

v−→ q must pass through a final statebecause otherwise there would be another accepting path for vω that starts fromq. By prefixing this accepting path with the v-loop from q to q we would obtainmore than one accepting path for vω starting in q, contradicting the strongunambiguity of A. ��For the reduction to unambiguous automata on finite words we now build froma strongly unambiguous Buchi automaton an unambiguous automaton on finitewords that accepts precisely the words of the form u#v such that uvω is acceptedby A. By Theorem 2 two strongly unambiguous Buchi automata are equivalentiff the resulting finite automata are.

Let A = (Q, Σ, Qin, Δ, F ) be a strongly unambiguous Buchi automaton. Thefinite automaton we are constructing simulates A on the first part u of theinput u#v. When reading # it stores the current state q and continues readingthe input. It accepts if it reaches q again after having read v, and if it haspassed through a final state on the way. Since the automaton accepts codings


of ultimately periodic words, we call it Aup, where the subscript abbreviatesultimately periodic.

Formally, Aup = (Q′, Σ ∪ {#}, Qin, Δ′, F ′) is defined as follows:

– Q′ = Q ∪ (Q × Q × {0, 1}).– Δ′ contains

• all transitions from Δ,• all transitions of the form (q, #, (q, q, 0)) with q ∈ Q, and• all transitions of the form ((q, p, i), a, (q, p′, i′)), where (p, a, p′) ∈ Δ, and

i′ ={

1 if p′ ∈ F,i if p′ /∈ F.

– F ′ = {(q, q, 1) | q ∈ Q}.Lemma 2. The automaton Aup over finite words is unambiguous and acceptsthe language L(Aup) = {u#v ∈ Σ∗#Σ+ | uvω ∈ L(A)}.Proof. It is clear from the construction that Aup only accepts words of the formu#v ∈ Σ∗#Σ+.

Furthermore, one easily sees that Aup accepts precisely those u#v such thatthere are states q0 ∈ Qin and q ∈ Q with A : q0

u−→ q and A : qv−→F q. Lemma 1

allows us to conclude that L(Aup) = {u#v ∈ Σ∗#Σ+ | uvω ∈ L(A)}.The automaton Aup simulates the automaton A when reading its input. In

particular, if there are two different paths for accepting u#v, then there are alsotwo different paths in A accepting uvω. Thus, since A is strongly unambiguous,Aup is unambiguous. ��Since inclusion and equivalence for unambiguous automata on finite words aredecidable in polynomial time [14], we obtain the following result for stronglyunambiguous Buchi automata.

Theorem 3. The inclusion and equivalence problem for strongly unambiguousBuchi automata are decidable in polynomial time.

Proof. Given two strongly unambiguous Buchi automata A and B we trans-form them into the unambiguous automata Aup and Bup. For these we can testinclusion or equivalence in polynomial time. By Lemma 2 and Theorem 2 thecorresponding result is correct for A and B. ��

4 Extension to Strongly k-Ambiguous Automata

An automaton over finite words is called k-ambiguous if each accepted word hasat most k accepting runs. In [14] it is shown that equivalence and inclusion of k-ambiguous automata can be solved in polynomial time (for a fixed k). Followingthis idea, we extend the result from the previous section to strongly k-ambiguousBuchi automata, as defined below.

Definition 2. A Buchi automaton A is called k-ambiguous if each infinite wordhas at most k accepting paths in A, and is called strongly k-ambiguous if eachinfinite word has at most k final paths in A.


The accepting paths for ultimately periodic words are not as constrained instrongly k-ambiguous automata as they are in strongly unambiguous automata,but there is a similar characterization. The difference is that the loop on theperiodic part can be longer, but it can consist of at most k repetitions of theperiodic pattern.

Lemma 3. Let A = (Q, Σ, Qin, Δ, F ) be a strongly k-ambiguous Buchi automa-ton. An ultimately periodic word uvω is accepted by A iff there are states q0 ∈ Qin

and q1, . . . , qk+1 ∈ Q such that

A : q0u−→ q1

v−→ q2v−→ · · · v−→ qk

v−→ qk+1 ,

qk+1 = qs for some 1 ≤ s ≤ k, and qiv−→F qi+1 for some s ≤ i ≤ k.

Proof. The proof is similar to the proof of Lemma 1. Assume that uvω is acceptedby A, let ρ be an accepting path, and let q0, . . . , qk+1 be such that ρ starts asfollows:

q0u−→ q1

v−→ q2v−→ · · · v−→ qk+1.

Then vω is in LA(qi) for all i ∈ {1, . . . , k + 1}. Since A is strongly k-ambiguous,there must be some i �= j such that qi = qj . Now assume that qk+1 �= qs for alls ∈ {1, . . . , k}. Then we get arbitrarily many different accepting paths of A onuvω by repeating the loop between qi and qj an arbitrary number of times beforecontinuing the path towards qk+1. Hence, there must exist some 1 ≤ s ≤ k withqk+1 = qs.

It remains to show that there is some i ∈ {s, . . . , k} such that qiv−→F qi+1.

Assume the contrary. Then we can again produce an arbitrary number of ac-cepting paths for uvω by repeating the loop qs

v−→ · · · qkv−→ qs (that does not

contain a final state) before continuing the path with the accepting part afterqk+1 in ρ. All these paths are different because we always increase the part thatdoes not contain a final state before continuing with the part of ρ after qk+1

that contains infinitely many accepting states. This proves one direction of theclaim. The other direction is obvious. ��We now construct an automaton Ak

up that has the same property as Aup fromthe previous section. This automaton, after having read u and when reading #,guesses the states q2, . . . , qk and then verifies that the properties from Lemma 3are satisfied.

Formally Akup = (Q′, Σ ∪ {#}, Qin, Δ

′, F ′) is constructed as follows:

– Q′ = Q ∪ ((Q × Q)k × {0, . . . , k}).– Δ′ contains all transitions from Δ, all transitions of the form

(q1, #, [(q1, q1), . . . , (qk, qk), 0])

and all transitions of the form (written with an arrow for better readability)

[(q1, p1), . . . , (qk, pk), i] a−→ [(q1, p′1), . . . , (qk, p′k), i′]


where (pj , a, p′j) ∈ Δ for all j, and

i′ = max({i} ∪ {j | p′j ∈ F}).– F ′ = {[(q1, q2), (q2, q3) . . . , (qk, qs), i] | 1 ≤ s ≤ k and s ≤ i ≤ k}.

Lemma 4. The automaton Akup over finite words is k-ambiguous and accepts

the language L(Akup) = {u#v ∈ Σ∗#Σ+ | uvω ∈ L(A)}.

Proof. The proof is based on Lemma 3 and is similar to the proof of Lemma 2.��

As in the previous section we conclude:

Theorem 4. For a fixed k, the inclusion and equivalence problem for stronglyk-ambiguous Buchi automata can be decided in polynomial time.

5 Periodic Words in Deterministic Automata

In this section we show that deciding for a given deterministic Buchi automa-ton whether it accepts a periodic word with a period of a given length is NP-complete. Since deterministic automata are special cases of unambiguous au-tomata, this shows that it is unlikely that the counting techniques that workfor unambiguous automata on finite words and for strongly unambiguous Buchiautomata can be transferred to unambiguous Buchi automata.

More formally, consider for a class C of Buchi automata the following decisionproblem Periodic(C):

Given: A Buchi automaton A from the class C and a natural number n.Question: Does there exist a word v with |v| ≤ n such that A accepts vω?

Proposition 1. The problem Periodic(C) is NP-complete for the class C ofdeterministic Buchi automata.

Proof. Membership in NP can easily be verified: If n is bigger than the numberof states of A, then A accepts a periodic word of length n iff the initial state ofA is contained in a loop with a final state. This can be checked in polynomialtime by standard graph algorithms. If n is smaller than the number of states ofA, then we can guess a word v of length at most n and verify in polynomial timewhether vω is accepted by A.

For the NP-hardness we give a reduction from the satisfiability problem ofBoolean formulas in conjunctive normal form (CNF). Let ϕ = C1 ∧ · · · ∧ Ck besuch a formula over n variables x1, . . . , xn consisting of clauses C1, . . . , Ck. Atruth assignment of the variables can naturally be encoded by a word of lengthn over the alphabet {0, 1}: Position i of the word is 0 if xi is false, and 1 if xi istrue.

The deterministic Buchi automaton that we construct works over the alphabet{0, 1, #} and only accepts words from ({0, 1}n#)ω : It basically consists of a loop


of length (n + 1) · k that reads words v1#v2# · · · vk# where each vi is from{0, 1}n. Each vi encodes an assignment of the n variables as described above.The automaton A checks whether the assignment coded by vi satisfies the clauseCi: Assume that A reads letter j of vi. If this letter is 0 and ¬xj is contained inCi or the letter is 1 and xj is contained in Ci, then A sets a bit indicating thatthe current clause is satisfied. If it reaches the end of vi without the bit beingset it rejects by moving to a sink state. Otherwise it reads # and proceeds tovi+1. After having processed k such words, A loops back to the initial state. Allstates except the sink state are accepting.

If this automaton accepts a periodic word with period of length at most n+1,then it must be of the form (v#)ω , where v ∈ {0, 1}n satisfies all the clausesof ϕ. This shows that ϕ is satisfiable iff the constructed automaton accepts aperiodic word with period of length at most n + 1. ��This is of course not a proof for the hardness of equivalence for unambiguousBuchi automata. It only shows that the techniques that have been used so farfor obtaining polynomial time equivalence tests are unlikely to work for the caseof unambiguous Buchi automata.

6 Conclusion

The class of strongly unambiguous Buchi automata is the first known class ofBuchi automata as expressive as nondeterministic Buchi automata for whichthe inclusion and equivalence problem can be decided in polynomial time. How-ever, this class is quite difficult to understand because strongly unambiguousBuchi automata are co-deterministic [7] and we usually think in a deterministicway. In addition, there are deterministic Buchi automata exponentially smallerthan strongly unambiguous ones, which is impossible for unambiguous Buchiautomata, because every deterministic Buchi automaton is unambiguous. There-fore, it would be interesting to settle the complexity of the equivalence problemfor unambiguous Buchi automata.

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Chapter 10 Equivalence and Inclusion Problem for Strongly Unambiguous Bu'Chi Automata

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