Complexity of infinite tree languages

Transcription

1 Complexity of infinite tree languages when automata meet topology Damian Niwiński University of Warsaw joint work with André Arnold, Szczepan Hummel, and Henryk Michalewski Liverpool, October

2 Example 1 Is there a path that meets and infinitely often, but only finitely often? 2

3 Example 1 Is there a path that meets and infinitely often, but only finitely often? 3

4 Example 2 Is there a strategy for against to achieve the same condition? 4

5 Example 2 Is there a strategy for against to achieve the same condition? 5

6 General method we opponent we we opp opp opp Strategies are (possibly infinite) trees. Construct an automaton accepting the winning strategies, and test it for non-emptiness. 6

7 Infinite computations Büchi (1960) and Rabin (1969) considered infinite computations of finite automata in their proofs of decidability of the theories S1S and S2S. D. Muller(1960) used similar concepts to analyse asynchronous digital circuits. Since 1980s, computer scientists study infinite computations in context of verification of computing systems (reactive, concurrent, open,... ). Non-termination is an expected behaviour. Mathematicians have been playing infinite games since the 1930s: Banach Mazur, later Gale Stewart (1953),... 7

8 Automata can be classified along several axes working on infinite words or trees, in deterministic, non-deterministic, or alternating mode, with a certain acceptance condition and, specifically, a Rabin Mostowski index. All these features relate to the complexity of the non-emptiness problem. Sets of infinite words or trees can be also classified by hierarchies of set-theoretical topology (Borel, projective, Wadge). Topological hardness is often behind the hierarchy results. 8

9 Classical definability theory Σ 1 2 Π Borel, Baire, Lebesgues Lusin, Suslin Σ 1 1 Π Tarski, Kuratowski Σ Mostowski, Kleene 0 1 Π 0 1 Σ 0 0 Π 0 0 9

10 ,,Easy can be hard. We will see regular sets of infinite trees beyond the Borel hierarchy, inseparable pair of regular sets. 10

11 For R ω k ({0, 1} ω ) l, let 0 R = { m, α : ( n) R(m, n, α)} 1 R = { m, α : ( β) R(m, α, β)} Arithmetical hierarchy Analytical hierarchy Σ 0 0 = recursive relations Σ 1 0 = arithmetical relations Π 0 n = {R : R Σ 0 n} Π 1 n = {R : R Σ 0 n} Σ 0 n+1 = { 0 R : R Π 0 n} Σ 1 n+1 = { 1 R : R Π 1 n} 0 n = Σ 0 n Π 0 n 1 n = Σ 1 n Π 1 n 11

12 Arithmetical hierarchy Σ 0 0 = recursive relations Π 0 n = {R : R Σ 0 n} Σ 0 n+1 = { 0 R : R Π 0 n} Analytical hierarchy Σ 1 0 = arithmetical relations Π 1 n = {R : R Σ 0 n} Σ 1 n+1 = { 1 R : R Π 1 n} 0 n = Σ 0 n Π 0 n Relativized (boldface) hierarchies 1 n = Σ 1 n Π 1 n For β {0, 1} ω, let R[β] = { m, α : R(m, α, β)}. Σ i n = {R[β] : R Σ i n, β {0, 1} ω } i n = Σ i n Π i n Π i n = {R[β] : R Π i n, β {0, 1} ω } i {0, 1} Σ 0 1 = open Π 0 1 = closed 1 1 = Borel 12

13 Büchi automata on infinite words where Tr Q Σ Q, F Q. A = Σ, Q, q I, Tr,F a a,b b a b a a ((a + b) b) ω (a + b) a ω The second one cannot be recognized by a deterministic automaton. a b a a b a a a b a a a a b a a b a... 13

14 So (a + b) a ω cannot be recognized by a deterministic automaton. But this also follows by a topological argument! We assume the Cantor topology on X ω, induced by the metric d(u, v) = 2 min{m : u m v m } If A is deterministic then the mapping Σ ω u run(u) Q ω satisfies L(A) u run(u) (Q F ) ω, hence continuously reduces L(A) to (Q F ) ω. 14

15 If A is deterministic then the mapping Σ ω u run(u) Q ω continuously reduces L(A) to (Q F ) ω. But (a + b) a ω Σ 0 2 Π 0 2, (Q F ) ω Π 0 2. Hence (a + b) a ω cannot be recognized by a deterministic automaton. 15

16 But (a + b) aω Σ02 Π02, (Q F )ω Π02. Hence (a + b) aω cannot be recognized by a deterministic automaton. photo P.Jahr 16?

17 Parity automata where rank : Q {0, 1,..., k}. A = Σ, Q, q I, Tr, rank A run is accepting if lim sup i rank(q i ) is even. a 0 b a 1 b (a + b) a ω The Rabin-Mostowski index of a parity automaton A is We can assume min rank {0, 1}. (min rank, max rank) 17

18 The McNaughton Theorem (1966) A nondeterministic Büchi automaton can be simulated by a deterministic parity automaton of some index (i, k). The minimal index (i, k) may be arbitrarily high (Wagner 1979, Kaminski 1985). Again, it can be inferred by a topological argument. Let M i,k = {u {i,..., k} ω : lim sup l u l is even} 18

19 M 1,4 M 1,3 M 1,2 M 0,3 M 0,2 M 0,1 M 1,1 M 0,0 No continuous reduction down the hierarchy. 19

20 Wadge game G(A, B) Spoiler a 0 Σ Duplicator b 0 Σ a 1 b 1 Here A, B Σ ω (Σ finite). a 2 b 2. a 12 b 12. Duplicator wins if a 0 a 1 a 2... A b 0 b 1 b 2 B. a 13 wait Fact a 14 wait Duplicator has a winning strategy iff there is a a 15 b 13 continuousf : Σ ω Σ ω s.t. A = f 1 (B),.. in symbols, A w B. 20

21 Spoiler s strategy, e.g., in G(M 0,5, M 1,6 ) Spoiler Duplicator Note i If a deterministic automaton of index (1, 6), i 1. accepted M 0,5 there would be a continuous. reduction of M 0,5 to M 1,6 wait u rank run(u). 0 Contradiction!.. 21

22 Parity tree automata A = Σ, Q, q I, Tr, rank where Tr Q Σ Q Q, rank : Q {0, 1,...,k}. σ 22

23 Parity tree automata cont d A run of A on a tree t : {l, r} Σ is a tree ρ : {l, r} Q, such that, ρ(w), t(w), ρ(wl), ρ(wr) Tr, for each w dom (ρ) ρ(w), t(w) ρ(wl) ρ(wr) The run is accepting if, for each path P = p 0 p 1... {l, r} ω, lim sup k rank(ρ(p 0 p 1... p k ) is even. 23

24 Example a a/b, a a b a/b, b b rank(a) = 0 rank(b) = 1 recognizes the set of trees where, on each branch, b appears only finitely often. The complement can be recognized by a skip a/b, a a/b, a skip a b skip a/b, b a/b, b skip b skip skip, a/b rank(a) = 1 skip rank(b) = rank(skip) = 2 24

25 Example cont d a a/b, a a b a/b, b b rank(a) = 0 rank(b) = 1 This set cannot be recognized by a Büchi automaton (i.e., of index ( 1,2)), Rabin

26 Rabin s proof Again, a topological argument could be used instead, as this set is Π 1 1 complete, while the Büchi automata can recognize only Σ 1 1 sets. 26

27 Rabin s example generalizes to T i,k = {t {i,..., k} {l,r} : each branch is in M i,k } (1, 3) (0, 2) (1, 2) (0, 1) (1, 1) (0, 0) These sets witness the strictness of the non-deterministic index hierarchy (N 1986). A topological argument is not helpful here, as all the sets T i,k are Π 1 1 complete, hence Wadge equivalent (except for T 0,0, T 1,1, T 1,2 ). 27

28 But topology comes back in the proof of the strictness of the alternating index hierarchy (Bradfield 1996, 1998, another proof by Arnold 1998). 28

29 Parity games V V positions of Eve positions of Adam (disjoint) V V possible moves (with V = V V ) p 1 V rank : Q ω initial position the ranking function. An infinite play v 0 v 1 v 2... is won by Eve iff lim sup n rank(v n ) is even. Parity games enjoy positional determinacy (Emerson and Jutla 1991, Mostowski 1991). 29

30 Game tree languages Alphabet : {, } {i,..., k}. Eve : Adam :, j, j, j, j Eve wins an infinite play (x 0, i 0 ), (x 1, i 1 ), (x 2, i 2 ),... (x l {, }) iff lim sup l i l is even. The set W i,k consists of all trees such that Eve has a winning strategy. 30

31 Example, 1, 0, 1, 0, 1, 2, 1, 0 31

32 Alternating parity tree automata where Q = Q Q, Tr Q Σ {1, 2, ε} Q, rank : Q {0, 1,...,k}. A = Σ, Q, Q, Q, q I, Tr, rank An input tree t is accepted by A iff Eve has a winning strategy in the parity game Q {1, 2}, Q {1, 2}, (q 0, ε), Mov = {((p, v), (q, vd)): v dom(t), (p, t(v), d, q) Tr} rank(q, v) = rank(q). 32

33 Topological argument: reducibility (Arnold & N, 2008). The sets W i,k form a strict hierarchy w.r.t. the Wadge W 1,3 W 1,2 W 0,2 W 0,1 W 1,1 W 0,0 = (1, 3) (1, 2) (0, 2) (0, 1) (1, 1) (0, 0) This gives an alternative proof of the strictness of the alternating index hierarchy (Bradfield 1998). 33

34 The importance of the alternating index hierarchy follows from its relation to the µ-calculus. (1, 3) (1, 2) (0, 2) (0, 1) (µνµ) (νµ) (νµν) (µν) (1, 1) (0, 0) (µ) (ν) 34

35 An idea of the proof: Given an alternating automaton A of index (i, k), an input tree t induces a full tree of the game G(A, t). The mapping t G(A, t) is continuous and satisfies L(A) t G(A, t) W i,k hence it reduces L(A) to W i,k. Therefore any automaton recognizing W i,k must have an index at least (i, k). 35

36 Sketch of proof that W i,k w W i,k : Up to renaming, W i,k W i,k By Banach Fixed-Point Theorem, there is no contracting reduction of L to L x fix L f(x fix ) L x fix L Main Lemma If f reduces W i,k to some L then there is a mapping h : {i,..., k} {l,r} {i,..., k} {l,r} (padding), such that h reduces W i,k to itself, f h is contracting. About h: For W 0,k, it stretches the original tree completing by the nodes labeled by (, 0). For W 1,k, by (, 1). 36

37 Separation problem Given disjoint sets A and B, find a simple set C, such that A C B 37

38 We are interested in separation of sets of trees photo M.Bojańczyk 38

39 Lusin separation theorem. Any two disjoint Σ 1 1 sets are separable by a Borel set. There exist two disjoint Π 1 1 sets not separable by a any Borel set. We show that these sets can be chosen as regular sets of trees. Let W 0,1 be obtained from W 0,1 by interchanging and 0 1. That is, in W 0,1 Adam has a strategy to force that there is only finitely many 0 s. Note that W 0,1 is a copy of W 0,1 included in W 0,1. Fact (Hummel, Michalewski, N. 2009) W 0,1 and W 0,1 are not separable by any Borel set. 39

40 Consequently, W 0,1 and W 0,1 weak monadic second-order logic (WS2S) are not separable by any set definable in of index simultaneously (1, 2) and (0, 1) (Büchi and co-büchi) (Rabin 1970) recognized by weak alternating automaton (Muller, Schupp, Saoudi 1986) definable in the alternation-free µ-calculus (Arnold & N. 1991). In contrast, any two disjoint sets of trees recognizable by automata of index (1, 2) (Büchi) are separable by a set from this class (Rabin 1970). 40

41 Lemma For any Borel set B {0, 1} ω, there is a continuous reduction such that u B f(u) W 0,1 u B f(u) W 0,1 f : {0, 1} ω T {0,1} {, } If there were a Borel set C s.t. W 0,1 C W 0,1, we would have u B f(u) C u B f(u) C Hence any Borel set would be a continuous inverse image of C, which is impossible, since the Borel hierarchy is strict. 41

42 42

43 Proof of the lemma. For a clopen set B it is enough to take B x t 1 W 0,1 B x t 2 W 0,1 Suppose B = B n and we have suitable reductions F Bn. The case B = B n follows from symmetry. 43

44 What about the next levels of the hierarchy? (1, 3) (1, 2) (1, 1) (0, 2) (0, 1) (0, 0) Here k = (1, k) (0, k 1). 44

45 (1, 3) (1, 2) (1, 1) (0, 2) (0, 1) (0, 0) Conjecture Separation property in the class (i, n): holds for n even, fails for n odd. 45

46 In terms of the µ-calculus (µνµ) (µν) (µ) (νµν) (νµ) (ν) Conjecture Separation property holds for ν... classes, fails for µ... classes. 46

47 Evidence For n odd, one can define a copy W i,n W i,n by renaming, and this is impossible for n even. For example, W 1,3 : Conjecture: W i,n and W i,n (for n odd) are inseparable by a set. Result: W 1,3 is complete in the class of Σ1 1-inductive sets. 47

48 Related work: Santocanale and Arnold 2005 showed that the characterization of 2 in terms of the alternation free µ-calculus does not extend to n 3. Σ µ 3 Comp (Σ µ 3 Πµ 3 ) Π µ 3 Σ µ 3 Πµ 3 Σ µ 2 Comp (Σ µ 2 Πµ 2 ) Comp (Σ µ 1 Πµ 1 ) Π µ 2 48

49 Conclusion Topological complexity often, but not always, underlines the automata-theoretic complexity. Topological arguments appear to work better for deterministic or alternating, rather than for non-deterministic automata (?). The concept of inseparability sheds some light on the fine structure of finite-state recognizable sets of trees. Parity games seem to be the core concept of the theory. 49