Tight Bounds for Complementing Parity Automata
TTight Bounds for Complementing Parity Automata
Sven Schewe and Thomas VargheseDepartment of Computer Science, University of Liverpool
Abstract
We follow a connection between tight determinisation and complementation and establish acomplementation procedure from transition-labelled parity automata to transition-labelled nonde-terministic Büchi automata. We prove it to be tight up to an O ( n ) factor, where n is the size of thenondeterministic parity automaton. This factor does not depend on the number of priorities. The precise complexity of complementing ω -automata is an intriguing problem for two reasons: first,the quest for optimal algorithms is a much researched problem [1, 17, 12, 24, 15, 11, 25, 10, 8, 7, 6, 13,27, 29], and second, complementation is a valuable tool in formal verification (c.f., [9]), in particularwhen studying language inclusion problems of ω -regular languages. Complementation is also useful tocheck the correctness of translation techniques [27, 26]. The GOAL tool [26], for example, providessuch a test suite and incorporates recent algorithms [15, 25, 8, 13] for Büchi complementation.While devising optimal complementation algorithms for nondeterministic finite automata issimple—nondeterministic finite automata can be determinised using a simple subset construction, anddeterministic finite automata can be complemented by complementing the set of final states [14, 17]—devising optimal complementation algorithms for nondeterministic ω -automata is hard, because simplesubset constructions are not sufficient to determinise or complement them [11, 10].Given the hardness and importance of the problem, the complementation of ω -automata enjoyedmuch attention. The initial focus was on the complementation of Büchi automata with state-basedacceptance [1, 12, 24, 11, 15, 10, 25, 8, 7, 6, 27, 29, 18, 26], and it resulted in a continuous improvementof its upper and lower bounds.The first complementation algorithm dates back to the introduction of Büchi automata in 1962. Inhis seminal paper “On a decision method in restricted second order arithmetic” [1], Büchi developsa doubly exponential complementation procedure. While Büchi’s result shows that nondeterministicBüchi automata (and thus ω -regular expressions) are closed under complementation, complementing anautomaton with n states may, when using Büchi’s complementation procedure, result in an automatonwith O ( n ) states, while an Ω(2 n ) lower bound [17] is inherited from finite automata.In the late 80s, these bounds have been improved in a first sequence of results, starting with estab-lishing an EXPTIME upper bound [12, 24], which matches the EXPTIME lower bound [17] inheritedfrom finite automata. However, the early EXPTIME complementation techniques produce automatawith up to O ( n ) states [12, 24]; hence, these upper bounds were still exponential in the lower bounds.This situation changed in 1988, when Safra introduced his famous determinisation procedure fornondeterministic Büchi automata [15], resulting in an n O ( n ) bound for Büchi complementation, whileMichel [11] established a seemingly matching Ω( n !) lower bound in the same year. Together, theseresults imply that Büchi complementation is in n θ ( n ) , leaving again the impression of a tight bound.As pointed out by Vardi [27], this impression is misleading, because the O () notation hides an n θ ( n ) gap between both bounds. This gap has been narrowed down in 2001 to θ ( n ) by the introduction ofan alternative complementation technique that builds on level rankings and a cut-point construction [8].The complexity of the plain method is approximately (6 n ) n [8], leaving a (6 e ) n gap to Michel’s lowerbound [11]. 1 a r X i v : . [ c s . F L ] J un ubseqently, tight level rankings [6, 29] have been exploited by Friedgut, Kupferman, and Vardi [6]to improve the upper complexity bound to O (cid:0) (0 . n ) n (cid:1) , and by Yan [29] to improve the lower com-plexity bound to Ω (cid:0) (0 . n ) n (cid:1) . Schewe [18] has provided a matching upper bound, showing tightnessup to an O ( n ) factor.In recent works, more succinct acceptance mechanism have been studied, where the most importantones are parity and generalised Büchi automata, as they occur naturally in the translation of µ -calculiand LTL specifications, respectively. In [22], we gave tight bounds for the determinisation and comple-mentation of generalised Büchi automata. For Rabin, Streett, and parity automata, there has been muchprogress [4, 3, 2], in particular establishing an n θ ( n ) bound for parity complementation with state-basedacceptance, which has been a great improvement and pushed tightness of parity comple- mentation tothe level known from Büchi complementation since the late 80s [15, 11]. Contribution.
In this paper, we establish tight bounds for the complementation of parity automata withtransition-based acceptance. A generalisation of the ranking-based complementation procedures quotedabove to transition-based acceptance is straight forward, and the Safra-style determinisation proceduresfrom the literature [15, 16, 13, 19, 22] have a natural representation with an acceptance condition ontransitions. Their translation to state-based acceptance is by multiplying the acceptance from the lasttransition to the state space.A similar observation can be made for other automata transformations, like the removal of ε -transitions from translations of µ -calculi [28, 20] and the treatment of asynchronous systems [21], wherethe state-space grows by multiplication with the acceptance information (e.g., maximal priority on a fi-nite sequence of transitions), while it cannot grow in case of transition-based acceptance. Similarly, toolslike SPOT [5] offer more concise automata with transition-based acceptance mechanism as a translationfrom LTL. Using state-based acceptance in the automaton that we want to complement would also com-plicate the presentation of the complementation procedure. But first and foremost, using transition-basedacceptance provides cleaner results.This is the case because in state-based acceptance, the role of the states is overloaded. In finiteautomata over infinite structures, each state represents the class of tails of the word that can be acceptedfrom this state. In state-based acceptance, they have to account for the acceptance mechanism itself, too,while they are relieved from this burden in transition-based acceptance. In complementation techniquesbased on rankings, this results in a situation where states with certain properties, such as final states forBüchi automata, can only occur with some ranks, but not with all.As transition-based acceptance separates these concerns, the presentation becomes cleaner. Thenatural downside is that we lose the n O ( n ) bound [3] for parity complementation, as the number ofpriorities in a parity automaton with transition-based acceptance can grow arbitrarily. But in return, wedo get a clean and simple complementation procedure based on a data structure we call flattened nestedhistory trees (FNHTs), which is inspired by a generalisation of history trees [19] to multiple levels, onefor each even priority ≥ .In [22], we showed a connection between optimal determinisation and complementation for gener-alised Büchi automata, where we exploit the nondeterministic power of a Büchi automaton to devise atight complementation procedure. In this paper, we follow this connection between tight determinisa-tion [23] and complementation to devise a tight complementation construction from parity to nondeter-ministic Büchi automata.We show that any procedure that complements full parity automata with states Q and maximal pri-ority π has at least | fnht ( Q, π ) | / states, where fnht ( Q, π ) is the set of FNHTs for a given set Q ofstates and maximal priority π of the parity automaton that is to be complemented. Our complemen-tation construction uses a marker in addition for its acceptance mechanism. Essentially, it is used tomark some position of interest in an FNHT. It accounts for the O ( n ) gap between the upper and lowerbound. We show that, for π ≥ (and hence for Büchi automata upwards) the number of states of ourcomplementation construction is bounded by n + 1 times the lower bound.2 Preliminaries
We denote the non-negative integers by ω = { , , , , ... } . For a finite alphabet Σ , an infinite word α is an infinite sequence α α α · · · of letters from Σ . We sometimes interpret ω -words as functions α : i (cid:55)→ α i , and use Σ ω to denote the ω -words over Σ . ω -automata are finite automata that are interpreted over infinite words and recognise ω -regular lan-guages L ⊆ Σ ω . Nondeterministic parity automata are quintuples P = ( Q, Σ , I, T, pri : T → Π) , where Q is a finite set of states with a non-empty subset I ⊆ Q of initial states, Σ is a finite alphabet, T ⊆ Q × Σ × Q is a transition relation that maps states and input letters to sets of successor states,and pri is a priority function that maps transitions to a finite set Π ⊂ ω of non-negative integers.A run ρ of a nondeterministic parity automaton P on an input word α is an infinite sequence ρ : ω → Q of states of P , also denoted ρ = q q q · · · ∈ Q ω , such that the first symbol of ρ is an initialstate q ∈ I and, for all i ∈ ω , ( q i , α i , q i +1 ) ∈ T is a valid transition. For a run ρ on a word α , wedenote with ρ : i (cid:55)→ (cid:0) ρ ( i ) , α ( i ) , ρ ( i + 1) (cid:1) the transitions of ρ . Let infin ( ρ ) = { q ∈ Q | ∀ i ∈ ω ∃ j >i such that ρ ( j ) = q } denote the set of all states that occur infinitely often during the run ρ . Likewise,let infin ( ρ ) = { t ∈ T | ∀ i ∈ ω ∃ j > i such that ρ ( j ) = t } denote the set of all transitions that are takeninfinitely many times in ρ . Acceptance of a run is defined through the priority function pri . A run ρ of aparity automaton is accepting if lim sup n →∞ pri (cid:0) ρ ( n ) (cid:1) is even, that is, if the highest priority that occursinfinitely often in the transitions of ρ is even. A word α is accepted by a parity automaton P iff it has anaccepting run, and its language L ( P ) is the set of words it accepts.Parity automata with Π ⊆ { , } are called Büchi automata. Büchi automata are denoted B =( Q, Σ , I, T, F ) , where F ⊆ T are called the final or accepting transitions. A run is accepting if itcontains infinitely many accepting transitions. B is thus a rendering of the parity automaton, where pri : t (cid:55)→ if t ∈ F and pri : t (cid:55)→ if t / ∈ F .We assume w.l.o.g. that the set Π of priorities satisfies that min Π ∈ { , } . If this is not the case, wecan simply change pri accordingly to pri (cid:48) : t (cid:55)→ pri ( t ) − several times until this constraint is satisfied.We likewise assume that Π has no holes, that is, Π = { i ∈ ω | max Π ≥ i ≥ min Π } . If there isa hole h / ∈ Π with max Π > h > min Π , we can change pri to pri (cid:48) : t (cid:55)→ pri ( t ) if pri ( t ) < h and pri (cid:48) : t (cid:55)→ pri ( t ) − if pri ( t ) > h . Obviously, these changes do not affect the acceptance of any run, andapplying finitely many of these changes brings Π into this normal form.The different priorities have a natural order (cid:60) , where i (cid:31) j if i is even and j is odd; i is even and i > j ; or j is odd and i < j . For a non-empty set Π (cid:48) ⊆ Π of priorities, opt Π (cid:48) = { i ∈ Π (cid:48) | ∀ j ∈ Π (cid:48) . i (cid:60) j } denotes the optimal priority for acceptance.The complexity of a parity automaton P = ( Q, Σ , I, T, pri : T → Π) is measured by its size n = | Q | and its set of priorities Π . For a given size n and set of priorities Π , there is an automaton that recognisesa hardest language. This automaton is referred to as the full automaton P Π n = ( Q, Σ , I, T, pri : T → Π) ,with | Q | = n , I = Q , Σ = Q × Q → Π , T = { q, σ, q (cid:48) ) | σ ( q, q (cid:48) ) (cid:54) = ∅ , and pri ( q, σ, q (cid:48) ) = opt σ ( q, q (cid:48) ) .Note that partial functions from Q × Q to Π would work as well as the alphabet. The larger alphabetis chosen for technical convenience in the proofs. Any other language recognised by a nondeterministicparity automaton P with n states and priorities Π can essentially be obtained by a language restrictionvia alphabet restriction from P Π n . The construction described in this section draws from two main sources of inspiration. One source isthe introduction of efficient techniques for the determinisation of parity automata in [23]. The nestedhistory trees used there have been our inspiration for the flattened nested history trees that form the coredata structure in the complementation from Subsection 3.2 and are the backbone of the lower boundproof from Subsection 3.4. The second source of inspiration is the connection [22] between the efficientdeterminisation based on history trees [19] for Büchi automata and generalised Büchi automata [22] andtheir level ranking based complementation [8, 6, 18, 22].3he intuition for the complementation is to use the nondeterministic power of a Büchi automaton toreduce the size of the data stored for determinisation. As usual, this nondeterministic power is intuitivelyused to guess a point in time, where all nodes of the nested history trees from parity determinisation[23], which are eventually always stable, are henceforth stable. Alongside, the set of stable nodes canbe guessed.Like in the construction for Büchi automata, the structure can then be flattened, preserving the‘nicking order’, the order in which older nodes and descendants take preference in taking states of thenondeterministic parity automaton that is determinised. The complement automaton runs in two phases:a first phase before this guessed point in time, and a second phase after this point, where the run startsin such a flattened tree.In the first subsection, we introduce flattened nested history trees as our main data structure. Whilewe take inspiration from nested history trees [23], the construction is self-contained. In the second sub-section, we show that Büchi automata recognising the complement language of the full nondeterministicparity automaton P Π n need to be large by showing disjointness properties of accepting runs for a largeclass of words, one for each full flattened nested history tree introduced in Subsection 3.1. The definitionof this language is also instructive in how the data structure is exploited.We extend our data structure by markers, resulting in marked flattened trees , which are then used asthe main part of the state space of the natural complementation construction introduced in Subsection3.2. We show correctness of our complementation construction in Subsection 3.3 and tightness up to an O ( n ) factor in Subsection 3.4.Note that all our constructions assume max Π ≥ , and therefore do not cover the less expressiveCoBüchi automata. Flattened nested history trees (FNHTs) are the main data structure used in our complementation al-gorithm. For a given parity automaton P = ( Q, Σ , I, T, pri : T → Π) , an FNHT over the set Q of states, maximal priority π m = max Π and maximal even priority π e = opt Π , is a tuple ( T , l s : T → Q , l l : T → N , l p : T → Q , l r : T → Q ) , where T (an ordered, labelled tree)is a non-empty, finite, and prefix closed subset of finite sequences of natural numbers and a special sym-bol s (for stepchild ), ω ∪{ s } , that satisfies the constraints given below. We call a node v s ∈ T a stepchild of v , and refer to all other nodes vc with c ∈ ω as the natural children of v . nc ( v ) = { vc | c ∈ ω and vc ∈ T } is the set of natural children of v . The root is a stepchild.The constraints an FNHT quintuple has to satisfy are as follows: • Stepchildren have only natural children, and natural children only stepchildren. • Only natural children and, when the highest priority π is odd, the root may be leafs. • T is order closed: for all c, c (cid:48) ∈ ω with c < c (cid:48) , vc (cid:48) ∈ T implies vc ∈ T . • For all v ∈ T , l s ( v ) (cid:54) = ∅ . • If v is a stepchild, then l p ( v ) = ∅ . • If v is a stepchild, then l s ( v ) = l r ( v ) ∪ (cid:83) v (cid:48) ∈ nc ( v ) l s ( v (cid:48) ) .The sets l s ( v (cid:48) ) and l s ( v (cid:48)(cid:48) ) are disjoint for all v (cid:48) , v (cid:48)(cid:48) ∈ nc ( v ) with v (cid:48) (cid:54) = v (cid:48)(cid:48) , and l r ( v ) is disjointwith (cid:83) v (cid:48) ∈ nc ( v ) l s ( v (cid:48) ) . • If v is a natural child, then l p ( v ) (cid:54) = ∅ , l s ( v ) = l p ( v ) ∪ l r ( v ) , and l p ( v ) ∩ l r ( v ) = ∅ . • If a natural child v is not a leaf, then l s ( v s ) = l p ( v ) . • l l ( ε ) = π e and, for all v ∈ T , l l ( v ) ≥ . 4 If v s ∈ T , then l l ( v s ) = l l ( v ) − , and if vc ∈ T for c ∈ ω , then l l ( vc ) = l l ( v ) .The elements in l s ( v ) are called the states, l p ( v ) the pure states, and l r ( v ) the recurrent states of anode v , and l l ( v ) is called its level. Note that the level follows a simple pattern: the root is labelledwith the maximal even priority, l l ( ε ) = π e , the level of natural children is the same as the level of theirparents, and the level of a stepchild v s of a node v is two less than the level of v . For a given maximaleven priority π e , the level is therefore redundant information that can be reconstructed from the nodeand π i . For a given set Q and maximal priority π , fnht ( Q, π ) denotes the flattened nested history treesover Q . An FNHT is called full if the states l s ( ε ) = Q of the root is the full set Q .To include an acceptance mechanism, we enrich FNHTs to marked flattened tress (MFTs), whichadditionally contain a marker v m and a marking set Q m , such that • either v m = ( v, r ) with v ∈ T is used to mark that we follow a breakpoint construction on therecurrent states, in this case l r ( v ) ⊇ Q m (cid:54) = ∅ , • or v m = ( v, p ) such that v is a leaf in T is used to mark that we follow a breakpoint constructionon the pure states of a leaf v , in this case l p ( v ) ⊇ Q m (cid:54) = ∅ .The marker is used to mark a property to be checked. For markers v m = ( v, r ) , the property is thata particular node would not spawn stable children in a nested history tree [23]. As usual in Safra likeconstructions, this is checked with a breakpoint, where a breakpoint is reached when all children of anode spawned prior to the last breakpoint die. For markers v m = ( v, p ) , the property is that all runsthat are henceforth trapped in the pure nodes of v must eventually encounter a priority l l ( v ) − . Thispriority is then dominating, and implies rejection as an odd priority. We check these properties roundrobin for all nodes in T , skipping over nodes, where the respective sets l r ( v ) or l p ( v ) are empty, as thebreakpoint there is trivially reached immediately.For a given FNHT ( T , l s , l l , l p , l r ) , next ( v m ) is a mapping from a marker v m to a marker/markingset pair ( v, r ) , l r ( v ) or ( v, p ) , l p ( v ) . The new marker is the first marker after v m in some round robinorder such that the set l r ( v ) or l p ( v ) , resp., is non-empty.If ( T , l s , l l , l p , l r ) is an FNHT and v m and Q m satisfy the constraints for markers and marking setsfrom above, then ( T , l s , l l , l p , l r ; v m , Q m ) is a marked flattened tree. For a given set Q and priorities Π with maximal priority π = max Π , mft ( Q, π ) denotes the marked flattened trees over Q . A marking iscalled full if either v m = ( v, r ) and Q m = l r ( v ) , or v m = ( v, p ) and Q m = l p ( v ) . For a given nondeterministic parity automaton P = ( Q, Σ , I, T, pri : T → Π) with maximal even prior-ity π e > , we construct a nondeterministic Büchi automaton C = ( Q (cid:48) , Σ , { I } , T (cid:48) , F ) that recognises thecomplement language of P as follows. First we set Q (cid:48) = Q ∪ Q with Q = 2 Q and Q = mft ( Q, π ) ,and T (cid:48) = T ∪ T t ∪ T , where • T ⊆ Q × Σ × Q are transitions in an initial part Q of the states of C , • T t ⊆ Q × Σ × Q are transfer transitions that can be taken only once in a run, and • T ⊆ Q × Σ × Q , are transitions in a final part Q of the states of C ,where T and T are deterministic. We first define a transition function δ for the subset construction andfunctions δ i for all priorities i ∈ Π , and then the sets T , T t , and T . • δ : ( S, σ ) (cid:55)→ { q ∈ Q | ∃ s ∈ S. ( s, σ, q ) ∈ T } , • δ i : ( S, σ ) (cid:55)→ (cid:8) q ∈ Q | ∃ s ∈ S. ( s, σ, q ) ∈ T and pri (cid:0)(cid:0) s, σ, q ) (cid:1) (cid:60) i (cid:9) , • T = (cid:8) ( S, σ, S (cid:48) ) ∈ Q × Σ × Q | S (cid:48) = δ ( S, σ ) (cid:9) ,where only transitions ( ∅ , σ, ∅ ) are accepting.5 T t = (cid:8)(cid:0) S, σ, ( T , l s , l l , l p , l r ; v m , Q m ) (cid:1) ∈ Q × Σ × Q | l s ( ε ) = δ ( S, σ ) (cid:9) and we havethat ( T , l s , l l , l p , l r ; v m , Q m ) is a marked flattened tree. • T = (cid:8)(cid:0) ( T , l s , l l , l p , l r ; v m , Q m ) , σ, s (cid:1) ∈ Q × Σ × Q | – if v is a stepchild, then l (cid:48)(cid:48) s ( v ) = δ l l ( v )+1 (cid:0) l s ( v ) , σ (cid:1) – if v is a natural child, then l (cid:48)(cid:48) s ( v ) = δ l l ( v ) − (cid:0) l s ( v ) , σ (cid:1) – if v is a natural child, then l (cid:48)(cid:48) r ( v ) = δ l l ( v ) − (cid:0) l r ( v ) , σ (cid:1) ∪ δ l l ( v ) (cid:0) l s ( v ) , σ (cid:1) , – starting at the root, we then define inductively: ∗ l (cid:48) s ( ε ) = l (cid:48)(cid:48) s ( ε ) , ∗ if vc is a natural child, then l (cid:48) s ( vc ) = (cid:0) l (cid:48)(cid:48) s ( vc ) ∩ l (cid:48) s ( v ) (cid:1) (cid:114) (cid:83) c (cid:48) Let ρ = S S . . . be an accepting run of C on α that stays in Q . Thus, there is an i ∈ ω suchthat, for all j ≥ i , S j = ∅ . But if we consider any run ρ (cid:48) = q q q . . . of P on α , then it is easy to showby induction that q k ∈ S k holds for all k ∈ ω , which contradicts S i = ∅ ; that is, in this case P has norun on α .Let us now assume that ρ = S S . . . S i s i +1 s i +2 . . . is an accepting run of C on α , where ( S i , α i , s i +1 ) ∈ T t is the transfer transition taken. (Recall that runs of C must either stay in Q orcontain exactly one transfer transition.)Let us assume for contradiction that P has an accepting run ρ (cid:48) = q q q . . . with even dominatingpriority e = lim sup j →∞ pri (cid:0) ( q j , α j , q j +1 ) (cid:1) . Let, for all j > i , s j = ( T , l js , l l , l jp , l jr ; v jm , Q jm (cid:1) and S j = l js ( ε ) . It is again easy to show by induction that q j ∈ S j for all j ∈ ω . Let now v j ∈ T be thelongest node with l jl ( v j ) ≥ e and q j ∈ l js ( v j ) . Note that such a node exists, as q j ∈ S j = l js ( ε ) holds.We now distinguish the two cases that the v j do and do not stabilise eventually. First case. Assume that there are an i (cid:48) > i and a v ∈ T such that, for all j ≥ i (cid:48) , v j = v . We choose i (cid:48) big enough that pri ( q j − , α j − , q j ) (cid:31) e + 1 holds for all j ≥ i (cid:48) . If v is a stepchild , then q j ∈ l jr ( v ) for all j ≥ i (cid:48) . Using the assumption that ρ is accepting, thereis an i (cid:48)(cid:48) > i (cid:48) such that ( s i (cid:48)(cid:48) − , α i (cid:48)(cid:48) − , s i (cid:48)(cid:48) ) is accepting, and v i (cid:48)(cid:48) m = ( v, r ) . (Note that q i (cid:48)(cid:48) ∈ l i (cid:48)(cid:48) r ( v ) implies l i (cid:48)(cid:48) r ( v ) (cid:54) = ∅ .) But then we have q i (cid:48)(cid:48) ∈ Q i (cid:48)(cid:48) m = l i (cid:48)(cid:48) r ( v ) , and an inductive argument provides ( s j , α j , s j +1 ) / ∈ F and q j ∈ Q jm for all j ≥ i (cid:48)(cid:48) . This contradicts that ρ is accepting. If v is a natural child , then we distinguish three cases. The first one is that there is a j (cid:48) ≥ i (cid:48) suchthat q j (cid:48) ∈ l j (cid:48) r ( v ) . Then we can show by induction that q j ∈ l jr ( v ) for all j ≥ j (cid:48) and follow the sameargument as for stepchildren, using i (cid:48)(cid:48) > j (cid:48) .The second is that q j ∈ l jp ( v ) holds for all j ≥ i (cid:48) . There are now again a few sub-cases thateach lead to contradiction. The first is that l l ( v ) = e . But in this case, we can choose a j > i (cid:48) with6 ri (cid:0) ( q j , α j , q j +1 ) (cid:1) = e and get q j +1 ∈ l j +1 r ( v ) (contradiction). The second is that l l ( v ) > e and v isnot a leaf. But in that case, l l ( v s ) ≥ e holds and q j ∈ l jp ( v ) implies q j ∈ l jp ( v s ) , which contradicts themaximality of v . Finally, if l l ( v ) > e and v is a leaf of T , we get a similar argument as for stepchildren:Using the assumption that ρ is accepting, there is an i (cid:48)(cid:48) > i (cid:48) such that ( s i (cid:48)(cid:48) − , α i (cid:48)(cid:48) − , s i (cid:48)(cid:48) ) is accepting,and v i (cid:48)(cid:48) m = ( v, p ) . (Note that q i (cid:48)(cid:48) ∈ l i (cid:48)(cid:48) p ( v ) implies l i (cid:48)(cid:48) p ( v ) (cid:54) = ∅ .) But then we have q i (cid:48)(cid:48) ∈ Q i (cid:48)(cid:48) m = l i (cid:48)(cid:48) p ( v ) , andan inductive argument provides ( s j , α j , s j +1 ) / ∈ F and q j ∈ Q jm for all j ≥ i (cid:48)(cid:48) . This contradicts that ρ is accepting. Second case. Assume that the v j do not stabilise. Let v be the longest sequence such that v is an initialsequence of almost all v j , and let i (cid:48) > i be an index such that v is an initial sequence of v j for all j ≥ i (cid:48) .Note that q j is in l s ( v (cid:48) j ) for all ancestors v (cid:48) j of v j .First, we assume for contradiction that there is a j > i (cid:48) with pri (cid:0) ( q j , α j , q j +1 ) (cid:1) = e (cid:48) (cid:31) l l ( v ) (notethat the ‘better than’ relation implies that e (cid:48) > l l ( v ) is even). Then we select a maximal ancestor v (cid:48) of v with l l ( v (cid:48) ) = e (cid:48) ; note that such an ancestor is a natural child, as a stepchild has only natural children,and all of them have the same level.As v (cid:48) is an ancestor of v j and v j +1 , q j ∈ l js ( v (cid:48) ) and q j +1 ∈ l j +1 s ( v (cid:48) ) hold, and by the transition rulesthus imply q j +1 ∈ l j +1 r ( v (cid:48) ) , which contradicts q j +1 ∈ l j +1 s ( v j +1 ) . (Note that l l ( v (cid:48) ) > l l ( v ) ≥ l l ( v j +1 ) holds.)Second, we show that pri (cid:0) ( q j , α j , q j +1 ) (cid:1) (cid:52) l l ( v ) + 1 holds infinitely many times. For this, we firstnote that the non-stability of the sequence of v j -s implies that at least one of the following three eventshappen for infinitely many j > i (cid:48) .1. v is a stepchild, q j ∈ l js ( vc ) for some child vc of v , but, for all children vc (cid:48) of v , q j +1 / ∈ l j +1 s ( vc (cid:48) ) ,2. v is a stepchild, q j ∈ l js ( vc ) for some child vc of v , and q j +1 ∈ l j +1 s ( vc (cid:48) ) for some older sibling vc (cid:48) of vc , that is, for c (cid:48) > c , or3. v is a natural child, q j / ∈ l js ( v s ) , but q j +1 ∈ l j +1 s ( v s ) .Note that this is just the counter position to “ v j stabilises or v is not maximal”. In all three cases, thedefinition of T requires that pri (cid:0) ( q j , α j , q j +1 ) (cid:1) (cid:52) l l ( v ) + 1 .As the first observation implies that there may only be finitely many transitions with even priority > l l ( v ) and the second observation implies that there are infinitely many transitions in ρ (cid:48) with oddpriority > l l ( v ) , they together imply that lim sup j →∞ pri (cid:0) ( q j , α j , q j +1 ) (cid:1) is odd, which leads to the finalcontradiction. (cid:3) Lemma 3.2 If P has an accepting run on α , then C rejects α . Proof. Let ρ = q q q . . . be an accepting run of P on α with even dominating priority e =lim sup j →∞ pri (cid:0) ( q j , α j , q j +1 ) (cid:1) .Let us first assume for contradiction that C has an accepting run ρ (cid:48) = S S . . . which is entirely in Q . It is then easy to show by induction that q i ∈ S i holds for all i ∈ ω , such that no transition of ( S i , α i , S i +1 ) is accepting.Let us now assume for contradiction that C has an accepting run ρ (cid:48) = S S . . . S i s i +1 s i +2 . . . , where ( S i , α i , s i +1 ) ∈ T t is the transfer transition taken. (Recall that runs of C must either stay in Q or containexactly one transfer transition.) Let further s j = ( T , l js , l l , l jp , l jr ; v jm , Q jm (cid:1) and S j = l js ( ε ) for all j > i .It is easy to show by induction that, for all j ∈ ω , q j ∈ S j holds. We choose an i ε > i such that, forall k ≥ i ε , pri (cid:0) ( q k − , α k − , q k ) (cid:1) ≤ e holds.Let us now look at the nodes v ∈ T , such that q j ∈ l js ( v ) , where j ≥ i ε . Construction basis. We have already shown q j ∈ S j = l js ( ε ) for all j > i , and thus in particular forall j ≥ i ε . 7 onstruction step. If, for some stepchild v ∈ T with l l ( v ) ≥ e and some i v ≥ i ε , it holds for all j ≥ i v that q j ∈ l js ( v ) , then the following holds for all j ≥ i v : if v (cid:48) ∈ nc ( v ) is a natural child of v and q j ∈ l js ( v (cid:48) ) , then either q j +1 ∈ l j +1 s ( v (cid:48) ) , or there is a younger sibling v (cid:48)(cid:48) of v (cid:48) in T such that q j +1 ∈ l j +1 s ( v (cid:48)(cid:48) ) .As transitions to younger siblings can only occur finitely often without intermediate transitions toolder siblings, we have one of the following two cases:1. for all j ≥ i v , q j ∈ l js ( v ) , but for every natural child v (cid:48) of v , q j / ∈ l js ( v (cid:48) ) , or2. there is a natural child v (cid:48) of v and an index i v (cid:48) ≥ i v such that, for all j ≥ i v (cid:48) , q j ∈ l js ( v (cid:48) ) .As v is a stepchild, the first case implies that q j ∈ l jr ( v ) for all j ≥ i v . However, using the assumptionthat ρ (cid:48) is accepting, there is an i (cid:48) v > i v such that ( s i (cid:48) v − , α i (cid:48) v − , s i (cid:48) v ) is accepting, and v i (cid:48) v m = ( v, r ) , as themarker is circulating in a round robin fashion. (Note that q i (cid:48) v ∈ l i (cid:48) v r ( v ) implies l i (cid:48) v r ( v ) (cid:54) = ∅ .) But then wehave q i (cid:48) v ∈ Q i (cid:48) v m = l i (cid:48) v r ( v ) , and an inductive argument provides ( s j , α j , s j +1 ) / ∈ F and q j ∈ Q jm for all j ≥ i (cid:48) v .In the second case, we continue with v (cid:48) and the index i v (cid:48) .If, for some natural child v ∈ T with l l ( v ) > e and some i v ≥ i ε , it holds for all j ≥ i v that q j ∈ l js ( v ) , then one of the following holds.1. There is an i (cid:48) v ≥ i v such that q i (cid:48) v ∈ l i (cid:48) v r ( v ) .2. For all j ≥ i v , q j ∈ l jp ( v ) .In the first case, it is easy to show by induction that q j ∈ l jr ( v ) holds for all j ≥ i v (cid:48) . We can then againuse the assumption that ρ (cid:48) is accepting. Consequently, there is an i (cid:48)(cid:48) v > i (cid:48) v such that ( s i (cid:48)(cid:48) v − , α i (cid:48)(cid:48) v − , s i (cid:48)(cid:48) v ) isaccepting, and v i (cid:48)(cid:48) v m = ( v, r ) , as the marker is circulating in a round robin fashion. (Note that q i (cid:48)(cid:48) v ∈ l i (cid:48)(cid:48) v r ( v ) implies l i (cid:48)(cid:48) v r ( v ) (cid:54) = ∅ .) But then we have again q i (cid:48)(cid:48) v ∈ Q i (cid:48)(cid:48) v m = l i (cid:48)(cid:48) v r ( v ) , and an inductive argument provides ( s j , α j , s j +1 ) / ∈ F and q j ∈ Q jm for all j ≥ i (cid:48)(cid:48) v .In the second case, if v is not a leaf, then it holds for all j ≥ i v s = i v that q j ∈ l js ( v s ) , and we cancontinue with v s . If v is a leaf, we again use the assumption that ρ (cid:48) is accepting. Consequently, there isan i (cid:48) v > i v such that ( s i (cid:48) v − , α i (cid:48) v − , s i (cid:48) v ) is accepting, and v i (cid:48) v m = ( v, p ) , as the marker is circulating in around robin fashion. (Note that v is a leaf and that q i (cid:48) v ∈ l i (cid:48) v p ( v ) implies l i (cid:48) v p ( v ) (cid:54) = ∅ .) But then we have q i (cid:48) v ∈ Q i (cid:48) v m = l i (cid:48) v p ( v ) , and an inductive argument provides ( s j , α j , s j +1 ) / ∈ F and q j ∈ Q jm for all j ≥ i (cid:48) v .If, for some natural child v ∈ T with l l ( v ) = e and some i v ≥ i ε , it holds for all j ≥ i v that q j ∈ l js ( v ) , then there is, by the definition of e , a j > i v with pri ( q j − , α j , q j ) = e . But then q j − ∈ l j − s ( v ) and q j ∈ l js ( v ) imply q j ∈ l jr ( v ) . It is then easy to establish by induction that q j (cid:48) ∈ l j (cid:48) r ( v ) for all j (cid:48) ≥ j . We can then again use the assumption that ρ (cid:48) is accepting. Consequently, there is a j (cid:48) > j such that ( s j (cid:48) − , α j (cid:48) − , s j (cid:48) ) is accepting, and v j (cid:48) m = ( v, r ) , as the marker is circulating in a round robinfashion. (Note that q j (cid:48) ∈ l j (cid:48) r ( v ) implies l j (cid:48) r ( v ) (cid:54) = ∅ .) But then we have again q j (cid:48) ∈ Q j (cid:48) m = l j (cid:48) r ( v ) , and aninductive argument provides ( s k , α k , s k +1 ) / ∈ F and q k ∈ Q km for all k ≥ j (cid:48) . Contradiction. As the level is reduced by two every second step, one of the arguments that contradictthe assumption that ρ (cid:48) is accepting is reached in at most π e steps. (cid:3) Corollary 3.3 C recognises the complement language of P . (cid:3) .4 Lower bound and tightness In order to establish a lower bound, we use a sub-language of the full automaton P Π n , and show that anautomaton that recognises it must have at least as many states as there are full FNHTs in fnht ( Q, π ) for n = | Q | and π = max Π .To show this, we define two letters for each full FNHT t = ( T , l s , l l , l p , l r ) ∈ fnht ( Q, π ) . β t : Q × Q → Π is the letter where: • if v is a stepchild and p, q ∈ l s ( v ) , then l l ( v )+1 ∈ β t ( p, q ) (provided l l ( v )+1 ∈ Π ), • if v is a stepchild, p ∈ l r ( v ) , and q ∈ l s ( vc ) for some c ∈ ω , then l l ( v ) ∈ β t ( p, q ) , • if v is a stepchild, c, c (cid:48) ∈ ω , c < c (cid:48) , vc (cid:48) ∈ T , p ∈ l s ( vc (cid:48) ) , and q ∈ l s ( vc ) , then l l ( v ) ∈ β t ( p, q ) , • if v is a natural child, p ∈ l p ( v ) , and q ∈ l r ( v ) then l l ( v ) ∈ β t ( p, q ) . • if v is a natural child and p, q ∈ l r ( v ) , then l l ( v ) − ∈ β t ( p, q ) , and • if v is a natural child and p, q ∈ l p ( v ) , then l l ( v ) − ∈ β t ( p, q ) . γ t : Q × Q → Π is the letter where i ∈ γ t ( p, q ) if i ∈ β t ( p, q ) and additionally: • if v is a natural child, l l ( v ) − ∈ Π , and p, q ∈ l r ( v ) , then l l ( v ) − ∈ γ t ( p, q ) , • if v is a stepchild and p, q ∈ l r ( v ) , then l l ( v ) ∈ γ t ( p, q ) , and • if v is a natural child, l l ( v ) − ∈ Π , and p, q ∈ l p ( v ) , then l l ( v ) − ∈ γ t ( p, q ) .For a high integer h > | fnht ( Q, π ) | , we now define the ω -word α t = ( β t γ th − ) ω , which consists ofinfinitely many sequences of length h that start with a letter β t and continue with h − repetitions ofthe letter γ t , for each full FNHT t ∈ fnht ( Q, π ) .We first observe that α t is rejected by P Π n . Lemma 3.4 α t / ∈ L ( P Π n ) . Proof. By Lemma 3.3, it suffices to show that the complement automaton C of P Π n , asdefined in Section 3.2 accepts α t . The language is constructed such that C has a run ρ = Q ( t ; v m , Q m )( t ; v m , Q m )( t ; v m , Q m ) . . . , such that the transition (cid:0) ( t ; v im , Q im ) , α ti , ( t ; v i +1 m , Q i +1 m ) (cid:1) isaccepting for i > if i mod h = 0 . (cid:3) Let B be some automaton with states S that recognises the complement language of P Π n . We nowfix an accepting run ρ t = s s s . . . for each word α t and define the set A t of states in an ‘acceptingcycle’ as A t = (cid:8) s ∈ S | ∃ i, j, k ∈ ω with ≤ j < k ≤ h such that s = s ih + j = s ih + k (cid:9) holds, anddefine the interesting states I t = A t ∩ infin ( ρ t ) . Lemma 3.5 For t (cid:54) = t (cid:48) , I t and I t (cid:48) are disjoint ( I t ∩ I t (cid:48) = ∅ ).Proof idea. The proof idea is to assume that a state s ∈ I t ∩ I t (cid:48) , and use it to construct a word from α t and α t (cid:48) and an accepting run of B on the resulting word from ρ t and ρ t (cid:48) , and then show that it is alsoaccepted by P Π n . Proof. Let us assume for contradiction that s ∈ I t ∩ I t (cid:48) for t = ( T , l s , l l , l p , l r ) (cid:54) = t (cid:48) = ( T (cid:48) , l (cid:48) s , l (cid:48) l , l (cid:48) p , l (cid:48) r ) .Noting that we can change the role of t and t (cid:48) , we fix two positions i and i (cid:48) in the run ρ t of α t suchthat s = s i = s i (cid:48) , and there is a j ∈ ω such that jh < i < i (cid:48) ≤ j ( h + 1) , and two positions j and j (cid:48) in ρ t (cid:48) = s (cid:48) s (cid:48) s (cid:48) . . . such that j < j (cid:48) , s = s (cid:48) j = s (cid:48) j (cid:48) and there is a k ∈ ω with j ≤ k < j (cid:48) such that ( s (cid:48) k , α t (cid:48) k , s (cid:48) k +1 ) is an accepting transition of B . Note that the definition of I t provides the first and thedefinition of I t (cid:48) the latter. 9or the finite words β = α t α t . . . α ti − , γ = s s . . . s i − , β = γ ti (cid:48) − , γ = s i s i +1 . . . s i (cid:48) − , β = α t (cid:48) j α t (cid:48) j +1 . . . α t (cid:48) j (cid:48) − , and γ = s j s j +1 . . . s j (cid:48) − , ρ t (cid:48) t = γ ( γ γ ) ω is an accepting run of the inputword α t (cid:48) t = β ( β β ) ω = α α α . . . .We now show that α t (cid:48) t or α tt (cid:48) is accepted by P Π n .We start with the degenerated case that T = { ε } is the FNHT where the root is a leaf, and thus π = max Π odd. (The case T (cid:48) = { ε } is similar.) We select a q ∈ l (cid:48) s (0) , and consider the run ρ = q ω of P Π n on α t (cid:48) t . By construction, pri ( q, α k , q ) ≤ opt Π = l l ( ε ) holds for all k ≥ i . Moreover, α k = γ t holdsfor infinitely many k ∈ ω . (In particular, it holds if k ≥ i and ( k − i ) mod ( i (cid:48) − i + j (cid:48) − j ) < i (cid:48) − i .)For all of these transitions, pri ( q, α k , q ) = opt Π = l l ( ε ) holds, such that lim sup n →∞ (cid:0) ρ ( i ) (cid:1) = opt Π iseven.Starting with the level λ = opt Π of the root and the whole trees T and T (cid:48) , we now run through thefollowing construction.We firstly look at the case that there is some difference in the label of some natural child v ∈T ∩ T (cid:48) on the level λ . If there is an oldest child v ∈ T ∩ T (cid:48) with l s ( v ) (cid:54) = l (cid:48) s ( v ) , we assume w.l.o.g.that there is a q ∈ l s ( v ) (cid:114) l (cid:48) s ( v ) . Then there are two sub-cases, first that there is a q (cid:48) ∈ l s ( v ) ∩ l (cid:48) s ( v ) ,and second that l s ( v ) ∩ l (cid:48) s ( v ) = ∅ . In the latter case we choose a q (cid:48) ∈ l (cid:48) s ( v ) . In both sub-cases, the run ρ = q i (cid:48) ( q (cid:48) j (cid:48) − j q i (cid:48) − i ) ω = q q q . . . of P Π n on α t (cid:48) t satisfies pri ( q k , α k , q k +1 ) (cid:60) λ − for all k ∈ ω , and pri ( q k , α k , q k +1 ) (cid:60) λ when q k = q and q k +1 = q (cid:48) . (Note that in this case α k ∈ { β t (cid:48) , γ t (cid:48) } holds.)Otherwise l s ( v ) = l (cid:48) s ( v ) holds for all natural children v ∈ T ∩T (cid:48) on level λ , and there is a v ∈ T ∩T (cid:48) on level λ such that l r ( v ) (cid:54) = l (cid:48) r ( v ) . We assume w.l.o.g. that there is a q ∈ l r ( v ) (cid:114) l (cid:48) r ( v ) . We choose a q (cid:48) ∈ l p ( v ) . (Note that q (cid:54) = q (cid:48) ∈ l s ( v ) = l (cid:48) s ( v ) .) Then the run ρ = q i (cid:48) ( q (cid:48) j (cid:48) − j q i (cid:48) − i ) ω = q q q . . . of P Π n on α t (cid:48) t satisfies pri ( q k , α k , q k +1 ) (cid:60) λ − for all k ∈ ω , and pri ( q k , α k , q k +1 ) (cid:60) λ when q k = q and q k +1 = q (cid:48) . (Note that in this case α k = γ t holds.)We secondly look at the case where l s ( v ) = l (cid:48) s ( v ) and l r ( v ) = l (cid:48) r ( v ) holds for all natural children v ∈ T ∩ T on level π e , but there is a natural child v on level λ in the symmetrical difference of T and T (cid:48) . Let us assume w.l.o.g. that v ∈ T (cid:48) . Let q ∈ l (cid:48) s ( v ) and let v be the child of v (cid:48) . This immediatelyimplies that q ∈ l r ( v ) . Thus, the run ρ = q ω of P Π n on α t (cid:48) t satisfies pri ( q, α k , q ) (cid:60) λ − for all k > i ,and pri ( q, α k , q ) (cid:60) λ whenever α k = γ t , which happens infinitely often.We finally look at the case where the nodes of T and T (cid:48) on level λ are the same, and where l s ( v ) = l (cid:48) s ( v ) and l r ( v ) = l (cid:48) r ( v ) hold for all nodes v of T on level λ , but there is a node v on level λ whichis a leaf in T but not in T (cid:48) . (The case “leaf in T (cid:48) but not in T ” is entirely symmetric.) Thus, v s isa node in T (cid:48) , and we select a q ∈ l s ( v s . If we now consider the run ρ = q ω of P Π n on α t (cid:48) t , then pri ( q, α k , q ) (cid:60) λ − holds for all k > i . At the same time pri ( q, α k , q ) (cid:60) λ − holds whenever α k = γ t , which happens infinitely often.If neither of these cases holds, then there must be a natural child v on level λ such that v s ∈ T ∩ T (cid:48) and l s ( v s ) = l p ( v ) = l (cid:48) p ( v ) = l (cid:48) s ( v s ) , such that t and t (cid:48) differ on the descendants of v . We thencontinue the construction by reducing λ to λ − and intersecting T and T (cid:48) with the descendants of v in t and t (cid:48) , respectively, and restrict the co-domain of the labelling functions of t and t (cid:48) accordingly. Thisconstruction will lead to a difference in at most . · opt Π steps. (cid:3) Theorem 3.6 B has at least as many states as fnht ( Q, max Π) contains full FNHTs. Proof. We prove the claim with a case distinction. The first case is that I t (cid:54) = ∅ holds for all fullFNHT t ∈ fnht ( Q, max Π) . Lemma 3.5 shows that the sets of interesting states are pairwise disjoint fordifferent trees t (cid:54) = t (cid:48) , such that, as none of them is empty, B has at least as many states as fnht ( Q, max Π) contains full FNHTs.The second case is there is a full FNHT t ∈ fnht ( Q, max Π) such that I t = ∅ . By Lemma 3.4, each ρ t = s s s . . . is an accepting run. Let now i ∈ ω be an index, such that, for all j ≥ i , s j ∈ infin ( ρ t ) ,and k ≥ i an integer with k mod h = 0 . I t = ∅ implies that s k + j (cid:54) = s k + j (cid:48) for all j, j (cid:48) with ≤ j < j (cid:48) ≤ h . Then B , and even infin ( ρ t ) , has at least h − different states, and the claim follows with h > | fnht ( Q, max Π) | . (cid:3) 10o show tightness, we proceed in three steps. In a first step, we provide an injection from MFTswith non-full marking to MFTs with full marking.Next, we argue that the majority of FNHTs is full. Taking into account that there are at most | Q | different markers makes it simple to infer that the states of our complementation construction dividedby the lower bound from Theorem 3.6 is in O ( n ) . Lemma 3.7 There is an injection from MFTs with non-full marking to MFTs with full marking in mft ( Q, π ) . Proof. For non-trivial trees T (cid:54) = {∅} , we can simply map an MFT ( T , l s , l l , l p , l r ; v m , Q m ) • for v m = ( v, p ) to the MFT (cid:0) T (cid:48) , l (cid:48) s , l (cid:48) l , l (cid:48) p , l (cid:48) r ; v m , l (cid:48) p ( v ) (cid:1) and • for v m = ( v, r ) to the MFT (cid:0) T (cid:48) , l (cid:48) s , l (cid:48) l , l (cid:48) p , l (cid:48) r ; v m , l (cid:48) r ( v ) (cid:1) , where T (cid:48) differs from T only in that it has a fresh node v , which is the youngest sibling of v m . l (cid:48) s , l (cid:48) p , l (cid:48) r differfrom l s , l p , l r only in v and v (where v is only in the pre-image of l (cid:48) s , l (cid:48) l , l (cid:48) p , l (cid:48) r ). We set l (cid:48) s ( v ) = l (cid:48) p ( v ) = Q m and, consequently, l (cid:48) r ( v ) = ∅ . We also set l (cid:48) s ( v ) = l s ( v ) (cid:114) Q m .For v m = ( v, p ) , we set l (cid:48) r ( v ) = l r ( v ) and l (cid:48) p ( v ) = l p ( v ) (cid:114) Q m . Note that, by the definition ofmarkers, v is a leaf, and l (cid:48) p ( v ) is non-empty because the marking in ( T , l s , l l , l p , l r ; v m , Q m ) is not full.For v m = ( v, r ) , we set l (cid:48) r ( v ) = l r ( v ) (cid:114) Q m and l (cid:48) p ( v ) = l p ( v ) . Note that l (cid:48) r ( v ) is non-empty becausethe marking in ( T , l s , l l , l p , l r ; v m , Q m ) is not full.It is easy to see that the resulting MFT is well formed in both cases. What remains is the corner caseof T = { ε } . ( T , l s , l l , l p , l r ; ( ε, r ) , Q m ) and map it to ( T (cid:48) , l (cid:48) s , l (cid:48) l , l (cid:48) p , l (cid:48) r ; ( ε, r ) , Q m ) for T (cid:48) = { ε, } and l (cid:48) s ( ε ) = l s ( ε ) , l (cid:48) s ( ε ) = Q m , l (cid:48) p ( ε ) = l (cid:48) p (0) = ∅ , and consequently l (cid:48) s (0) = l (cid:48) r (0) = l s ( ε ) (cid:114) Q m . (Note that thelatter is non-empty because the marking in ( T , l s , l l , l p , l r ; ( ε, r ) , Q m ) is not full.) This is again a wellformed MFT with full marking.It is easy to see that the resulting function is injective. (cid:3) In Lemma 3.7, we have shown that the majority of MFTs have a full marking. Next we will see thatthe majority of FNHTs is full. (Note that neither mapping is surjective.) Lemma 3.8 There is an injection from non-full to full FNHTs in fnht ( Q, π ) . Proof. To obtain such an injection, it suffices to map a non-full FNHT ( T , l (cid:48) s , l (cid:48) l , l (cid:48) p , l (cid:48) r ) to the FNHT ( T (cid:48) , l (cid:48) s , l (cid:48) l , l (cid:48) p , l (cid:48) r ) where T (cid:48) differs from T only in that it has a fresh youngest child v of the root. l (cid:48) s agrees with l s on every node of T except for the root ε , and l (cid:48) p , l (cid:48) r agree with l p , l r on every nodeof T . We set l (cid:48) s ( ε ) = Q , l (cid:48) s ( v ) = l (cid:48) p ( v ) = Q (cid:114) l s ( ε ) , and l (cid:48) r ( v ) = ∅ .It is obvious that the resulting FNHT ( T (cid:48) , l (cid:48) s , l (cid:48) l , l (cid:48) p , l (cid:48) r ) is full and well formed, and it is also obviousthat the mapping is injective. (cid:3) Theorem 3.9 The complementation construction from this section is tight up to a factor of n +1 , where n = | Q | is the number of states of the complemented parity automaton. Proof. For the number of MFTs, Lemma 3.7 shows that they are at most twice the number of MFTswith full marking. Note that the marker ( v m , p ) can only refer to leafs where l p ( v m ) is non-emptyand markers ( v m , r ) can only refer to nodes where l r ( v m ) is non-empty. It is easy to see that all setsdescribed in this way are pairwise disjoint. This implies that there are at most | Q | such markers. 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