On Decision Problems for Probabilistic Büchi Automata

Transcription

1 On Decision Prolems for Proailistic Büchi Automata Christel Baier 1, Nathalie Bertrand 2, Marcus Größer 1 1 TU Dresden, Germany 2 IRISA, INRIA Rennes, France Cachan 05 février 2008 Séminaire LSV 05/02/08, p.1

2 Introduction Proailistic Büchi Automata [Baier, Größer 05] PBA = NBA with proailities instead of non-determinism Séminaire LSV 05/02/08, p.2

3 Introduction Proailistic Büchi Automata [Baier, Größer 05] PBA = NBA with proailities instead of non-determinism a a a F L(A) = {w Σ ω ρ Runs(w) with ρ = F } Séminaire LSV 05/02/08, p.2

4 Introduction Proailistic Büchi Automata [Baier, Größer 05] PBA = NBA with proailities instead of non-determinism a, 1 2 a,1 a, 1 2 F,1 L(A) = {w Σ ω P A ({ρ Runs(w) ρ = F }) > 0} L NBA (A) = (a + ) a ω = L PBA (A) Séminaire LSV 05/02/08, p.2

5 Introduction Proailistic Büchi Automata [Baier, Größer 05] PBA = NBA with proailities instead of non-determinism c a, 1 2 a, 1 2 Séminaire LSV 05/02/08, p.3

6 Introduction Proailistic Büchi Automata [Baier, Größer 05] PBA = NBA with proailities instead of non-determinism c a, 1 2 a, 1 2 L NBA (A) = (a + ac) ω and L PBA (A) = Séminaire LSV 05/02/08, p.3

7 Introduction Expressiveness: PBA vs DBA PBA are strictly more expressive than DBA any DBA A can e turned into a PBA P Séminaire LSV 05/02/08, p.4

8 Introduction Expressiveness: PBA vs DBA PBA are strictly more expressive than DBA any DBA A can e turned into a PBA P there is a PBA whose language can t e recognized y a DBA a, 1 2 a a, 1 2 L(A) = (a + ) a ω Séminaire LSV 05/02/08, p.4

9 Introduction Expressiveness: PBA vs NBA PBA are strictly more expressive than NBA any NBA A can e turned into a PBA P Trick: replace A y an equivalent NBA deterministic in the limit Séminaire LSV 05/02/08, p.5

10 Introduction Expressiveness: PBA vs NBA PBA are strictly more expressive than NBA any NBA A can e turned into a PBA P Trick: replace A y an equivalent NBA deterministic in the limit there is a PBA whose language can t e recognized y an NBA a, 1 2 a a, 1 2 L(A) = {a k 1 a k 2 i (1 1 2k i ) > 0)} Séminaire LSV 05/02/08, p.5

11 Complementation Outline 1 Introduction 2 Complementation 3 Emptiness prolem Langage dependency on proailities Undecidaility of emptiness 4 Alternative semantics Expressivity Emptiness prolem 5 Conclusion Séminaire LSV 05/02/08, p.6

12 Complementation Complementation Theorem For each PBA P there exists a PBA P with P = O(exp( P ) such that L(P ) = Σ ω \ L(P). Moreover, P can e effectively constructed from P. Séminaire LSV 05/02/08, p.7

13 Complementation Complementation Theorem For each PBA P there exists a PBA P with P = O(exp( P ) such that L(P ) = Σ ω \ L(P). Moreover, P can e effectively constructed from P. Proof Scheme P PBA P R 0/1-PRA with L(P R ) = L(P) 0/1-PRA: Proailistic Rain Automaton s.t. all words have acceptance proaility in {0,1} Séminaire LSV 05/02/08, p.7

14 Complementation Complementation Theorem For each PBA P there exists a PBA P with P = O(exp( P ) such that L(P ) = Σ ω \ L(P). Moreover, P can e effectively constructed from P. Proof Scheme P PBA P R 0/1-PRA with L(P R ) = L(P) P S 0/1-PSA with L(P S ) = Σ ω \ L(P R ) 0/1-PSA: Proailistic Strett Automaton s.t. all words have acceptance proaility in {0,1} Séminaire LSV 05/02/08, p.7

15 Complementation Complementation Theorem For each PBA P there exists a PBA P with P = O(exp( P ) such that L(P ) = Σ ω \ L(P). Moreover, P can e effectively constructed from P. Proof Scheme P PBA P R 0/1-PRA with L(P R ) = L(P) P S 0/1-PSA with L(P S ) = Σ ω \ L(P R ) P PBA with L(P) = L(P S ) Séminaire LSV 05/02/08, p.7

16 Complementation Complementation Theorem For each PBA P there exists a PBA P with P = O(exp( P ) such that L(P ) = Σ ω \ L(P). Moreover, P can e effectively constructed from P. Proof Scheme P PBA P R 0/1-PRA with L(P R ) = L(P) P S 0/1-PSA with L(P S ) = Σ ω \ L(P R ) P PBA with L(P) = L(P S ) Difficult step: PBA equivalent 0/1-PRA Séminaire LSV 05/02/08, p.7

17 Complementation Complementation: First step in details From P uild an equivalent 0/1-PRA. Construction idea: Organize the infinite computation tree into a finite-state automaton y merging runs meeting at some point. Séminaire LSV 05/02/08, p.8

18 Complementation Complementation: First step in details From P uild an equivalent 0/1-PRA. Construction idea: Organize the infinite computation tree into a finite-state automaton y merging runs meeting at some point. States: tuples p 1, ξ 1,, p k, ξ k, R p i Q pairwise distinct, ξ i {0, 1} and R Q. R-component: usual powerset construction p i state witnessing sample runs ξ i it indicating whether the last step is a proper P-transition Séminaire LSV 05/02/08, p.8

19 Complementation Complementation: First step in details From P uild an equivalent 0/1-PRA. Construction idea: Organize the infinite computation tree into a finite-state automaton y merging runs meeting at some point. States: tuples p 1, ξ 1,, p k, ξ k, R p i Q pairwise distinct, ξ i {0, 1} and R Q. R-component: usual powerset construction p i state witnessing sample runs ξ i it indicating whether the last step is a proper P-transition Rain condition: for some index j, the j-th run visits F infinitely often and from some point on the attached it is 0. Séminaire LSV 05/02/08, p.8

20 Complementation Complementation: First step in details (2) Possile a-successors of p = p 1, ξ 1,, p k, ξ k, R : q = q 1, ζ 1,, q k, ζ k, q k+1, ζ k+1 q m, ζ m, S 1. q i δ(p i,a) for 1 i k 2. {q k+1,,q m } = (δ(r,a) F) \ {q 1,,q k } 3. ζ 1 = = ζ k = 0 and ζ k+1 = = ζ m = 1 4. S = δ(r,a) Séminaire LSV 05/02/08, p.9

21 Complementation Complementation: First step in details (2) Possile a-successors of p = p 1, ξ 1,, p k, ξ k, R : q = q 1, ζ 1,, q k, ζ k, q k+1, ζ k+1 q m, ζ m, S 1. q i δ(p i,a) for 1 i k 2. {q k+1,,q m } = (δ(r,a) F) \ {q 1,,q k } 3. ζ 1 = = ζ k = 0 and ζ k+1 = = ζ m = 1 4. S = δ(r,a) P PRA (ρ) > 0 P PBA (ρ) > 0 P PRA (ρ) = 1 Séminaire LSV 05/02/08, p.9

22 Complementation Complementation: First step in details (2) Possile a-successors of p = p 1, ξ 1,, p k, ξ k, R : q = q 1, ζ 1,, q k, ζ k, q k+1, ζ k+1 q m, ζ m, S 1. q i δ(p i,a) for 1 i k 2. {q k+1,,q m } = (δ(r,a) F) \ {q 1,,q k } 3. ζ 1 = = ζ k = 0 and ζ k+1 = = ζ m = 1 4. S = δ(r,a) P PRA (ρ) > 0 P PBA (ρ) > 0 P PRA (ρ) = 1 Example a a q r s L(P) = (a + ) a ω Séminaire LSV 05/02/08, p.9

23 Complementation Complementation: First step in details (3) q, 0, {q} Séminaire LSV 05/02/08, p.10

24 Complementation Complementation: First step in details (3) q, 0, {q} q, 0, r, 1, {q, r} r, 0, {q, r} Séminaire LSV 05/02/08, p.10

25 Complementation Complementation: First step in details (3) q, 0, {q} q, 0, r, 1, {q, r} q,0, s,0, {q, s} q, 0, r, 0, {q, r} r, 0, {q, r} Séminaire LSV 05/02/08, p.10

26 Complementation Complementation: First step in details (3) a q, 0, {q} q, 0, r, 1, {q, r} q,0, s,0, {q, s} q, 0, r, 0, {q, r} q, 0, s, 0, r, 1, {q, r, s} r, 0, {q, r} s, 0, {q, s} r, 0, s, 0, {q, r, s} a a a s,0, r, 1, {q,r, s} s,0, r, 0, {q,r, s} q, 0, s, 0, r, 0, {q, r, s} a Séminaire LSV 05/02/08, p.10

27 Complementation Complementation: First step in details (3) a q, 0, {q} q, 0, r, 1, {q, r} q,0, s,0, {q, s} q, 0, r, 0, {q, r} q, 0, s, 0, r, 1, {q, r, s} r, 0, {q, r} s, 0, {q, s} r, 0, s, 0, {q, r, s} a a a s,0, r, 1, {q,r, s} s,0, r, 0, {q,r, s} q, 0, s, 0, r, 0, {q, r, s} P PR (aa ω ) > 0 a Séminaire LSV 05/02/08, p.10

28 Complementation Complementation: First step in details (3) a q, 0, {q} q, 0, r, 1, {q, r} q,0, s,0, {q, s} q, 0, r, 0, {q, r} q, 0, s, 0, r, 1, {q, r, s} r, 0, {q, r} s, 0, {q, s} r, 0, s, 0, {q, r, s} a a a s,0, r, 1, {q,r, s} s,0, r, 0, {q,r, s} q, 0, s, 0, r, 0, {q, r, s} P PR (aa ω ) = 1 a Séminaire LSV 05/02/08, p.10

29 Emptiness prolem Outline 1 Introduction 2 Complementation 3 Emptiness prolem Langage dependency on proailities Undecidaility of emptiness 4 Alternative semantics Expressivity Emptiness prolem 5 Conclusion Séminaire LSV 05/02/08, p.11

30 Emptiness prolem A weird example a, λ a a, 1 λ P λ : q r L(P λ ) = {a k 1 a k 2 i (1 λk i) > 0)} Séminaire LSV 05/02/08, p.12

31 Emptiness prolem A weird example a, λ a a, 1 λ P λ : q r L(P λ ) = {a k 1 a k 2 i (1 λk i) > 0)} Lemma For 0 < λ < 1 2 < µ < 1, L(P λ) L(P µ ). Hint w = a k 1 a k 2 with for all m, 2 m elements of (k i ) set to m. w L(P λ ) \ L(P µ ) Séminaire LSV 05/02/08, p.12

32 Emptiness prolem An undecidale prolem for PFA The emptiness prolem is undecidale for Proailistic Finite Automata (as well as some variants). Séminaire LSV 05/02/08, p.13

33 Emptiness prolem An undecidale prolem for PFA The emptiness prolem is undecidale for Proailistic Finite Automata (as well as some variants). Undecidaility result for PFA [MHC03] The following prolem is undecidale: Given 0 < ε < 1 and P a PFA such that either w P P (w) > 1 ε or w P P (w) ε tell which is the case. Séminaire LSV 05/02/08, p.13

34 Emptiness prolem Emptiness prolem for PBA Theorem The emptiness prolem is undecidale for PBA. Séminaire LSV 05/02/08, p.14

35 Emptiness prolem Emptiness prolem for PBA Theorem The emptiness prolem is undecidale for PBA. Proof Sketch Reduction of the modified emptiness prolem for PFA Séminaire LSV 05/02/08, p.14

36 Emptiness prolem Emptiness prolem for PBA Theorem The emptiness prolem is undecidale for PBA. Proof Sketch Reduction of the modified emptiness prolem for PFA { w P R (w) ε or R PFA with w P R (w) > 1 ε P 1 and P 2 PBA s.t. L >ε (R) = L(P 1 ) L(P 2 ) = Séminaire LSV 05/02/08, p.14

37 Emptiness prolem Proof in more details: P 1 a, λ a q a, 1 λ r L(P λ ) = {a k1 a k2 i (1 λk i ) > 0)} Séminaire LSV 05/02/08, p.15

38 Emptiness prolem Proof in more details: P 1 a, λ a a, 1 λ L(P λ ) = {a k1 a k2 q i (1 λk i r ) > 0)} P 1 : s 0 R q, 1, 1 s 0, 1 p p f $, 1 R r, 1 f $, 1 F From s 0 in R q, reading w# leads to R r with proaility P R (w). a w λ 1 P R (w) $$ $, 1 $, 1 Séminaire LSV 05/02/08, p.15

39 Emptiness prolem Proof in more details: P 1 a, λ a a, 1 λ L(P λ ) = {a k1 a k2 q i (1 λk i r ) > 0)} P 1 : s 0 R q, 1, 1 s 0, 1 p p f $, 1 R r, 1 f $, 1 F From s 0 in R q, reading w# leads to R r with proaility P R (w). a w λ 1 P R (w) $$ j $, 1 $, 1 L(P 1 ) = {w1 1# w1 k 1 $$w1 2# w2 k 2 $$ ( 1 ( k j 1 i=1 (1 P R(w j i )))) > 0} Séminaire LSV 05/02/08, p.15

40 Emptiness prolem Proof in more details: P 2 Σ,1 Σ, 1, ε, 1 ε p 0 p 1, 1 P 2 : $$, 1 Σ, 1 Σ, 1 $$, 1 F p 2 From p 0, reading v (Σ {#}) leads to p 1 with proaility 1 (1 ε) v i # Séminaire LSV 05/02/08, p.16

41 Emptiness prolem Proof in more details: P 2 Σ,1 Σ, 1, ε, 1 ε p 0 p 1, 1 P 2 : $$, 1 Σ, 1 Σ, 1 $$, 1 F p 2 From p 0, reading v (Σ {#}) leads to p 1 with proaility 1 (1 ε) v i # L(P 2 ) = {v 1 $$v 2 $$ v i (Σ #) and i (1 (1 ε) v i # ) = 0} Séminaire LSV 05/02/08, p.16

42 Emptiness prolem Proof in more details: P 2 Σ,1 Σ, 1, ε, 1 ε p 0 p 1, 1 P 2 : $$, 1 Σ, 1 Σ, 1 $$, 1 F p 2 From p 0, reading v (Σ {#}) leads to p 1 with proaility 1 (1 ε) v i # L(P 2 ) = {w 1 1 # w1 k 1 $$w 2 1 # w2 k 2 $$ i (1 (1 ε)k i 1 ) = 0} Séminaire LSV 05/02/08, p.16

43 Emptiness prolem Proof conclusion L(P 1) = {w1# w 1 k 1 1 $$w1# w 2 k 2 2 $$ Q j 1 `Q )) k j 1 i=1 (1 PR(wj i > 0} L(P 2) = {w1# w 1 k 1 1 $$w1# w 2 k 2 2 $$ Q i (1 (1 ε)k i 1 ) = 0} Séminaire LSV 05/02/08, p.17

44 Emptiness prolem Proof conclusion L(P 1) = {w1# w 1 k 1 1 $$w1# w 2 k 2 2 $$ Q j 1 `Q )) k j 1 i=1 (1 PR(wj i > 0} L(P 2) = {w1# w 1 k 1 1 $$w1# w 2 k 2 2 $$ Q i (1 (1 ε)k i 1 ) = 0} If w, P R (w) ε w, w L(P 2 ) w / L(P 1 ) = L(P 1 ) L(P 2 ) = Séminaire LSV 05/02/08, p.17

45 Emptiness prolem Proof conclusion L(P 1) = {w1# w 1 k 1 1 $$w1# w 2 k 2 2 $$ Q j 1 `Q )) k j 1 i=1 (1 PR(wj i > 0} L(P 2) = {w1# w 1 k 1 1 $$w1# w 2 k 2 2 $$ Q i (1 (1 ε)k i 1 ) = 0} If w, P R (w) ε w, w L(P 2 ) w / L(P 1 ) = L(P 1 ) L(P 2 ) = If w, P R (w) > 1 ε Let w = (w#) k1 w$$(w#) k2 w$$ P P1 ( w) > j (1 εk j 1 ) and P P2 ( w) = i (1 (1 ε)k i 1 ) = L(P 1 ) L(P 2 ) Séminaire LSV 05/02/08, p.17

46 Emptiness prolem Proof conclusion L(P 1) = {w1# w 1 k 1 1 $$w1# w 2 k 2 2 $$ Q j 1 `Q )) k j 1 i=1 (1 PR(wj i > 0} L(P 2) = {w1# w 1 k 1 1 $$w1# w 2 k 2 2 $$ Q i (1 (1 ε)k i 1 ) = 0} If w, P R (w) ε w, w L(P 2 ) w / L(P 1 ) = L(P 1 ) L(P 2 ) = If w, P R (w) > 1 ε Let w = (w#) k1 w$$(w#) k2 w$$ P P1 ( w) > j (1 εk j 1 ) and P P2 ( w) = i (1 (1 ε)k i 1 ) = L(P 1 ) L(P 2 ) L >ε (R) = L(P 1 ) L(P 2 ) = Séminaire LSV 05/02/08, p.17

47 Emptiness prolem PBA-related Consequences Immediate consequences of the undecidaility result. Corollary The following prolems are undecidale. Given P 1 and P 2 PBA L(P 1 ) = Σ ω? L(P 1 ) = L(P 2 )? L(P 1 ) L(P 2 )? Séminaire LSV 05/02/08, p.18

48 Emptiness prolem PBA-related Consequences Immediate consequences of the undecidaility result. Corollary The following prolems are undecidale. Given P 1 and P 2 PBA L(P 1 ) = Σ ω? L(P 1 ) = L(P 2 )? L(P 1 ) L(P 2 )? Verification against PBA specifications The following prolems are undecidale. Given a transition system T and a PBA P is there a path in T whose trace is in L(P)? do the traces of all paths in T elong to L(P)? Séminaire LSV 05/02/08, p.18

49 Emptiness prolem Consequences for POMDP Partially Oservale MDP A POMDP (M, ) consists of an MDP M equipped with an equivalence relation over states of M. Séminaire LSV 05/02/08, p.19

50 Emptiness prolem Consequences for POMDP Partially Oservale MDP A POMDP (M, ) consists of an MDP M equipped with an equivalence relation over states of M. Undecidaility results The following prolems are undecidale Given (M, ) and F set of states of M, is there an oservation-ased U such that P U ( F) > 0. Given (M, ) and F set of states of M, is there an oservation-ased U such that P U ( F) = 1. Séminaire LSV 05/02/08, p.19

51 Emptiness prolem Consequences for POMDP Partially Oservale MDP A POMDP (M, ) consists of an MDP M equipped with an equivalence relation over states of M. Undecidaility results The following prolems are undecidale Given (M, ) and F set of states of M, is there an oservation-ased U such that P U ( F) > 0. Given (M, ) and F set of states of M, is there an oservation-ased U such that P U ( F) = 1. First undecidaility results in qualitative verification of POMDP. Séminaire LSV 05/02/08, p.19

52 Alternative semantics Outline 1 Introduction 2 Complementation 3 Emptiness prolem Langage dependency on proailities Undecidaility of emptiness 4 Alternative semantics Expressivity Emptiness prolem 5 Conclusion Séminaire LSV 05/02/08, p.20

53 Alternative semantics Almost-sure semantics for PBA Alternative semantics L(A) = {w Σ ω P A ({ρ Runs(w) ρ = F })= 1} Séminaire LSV 05/02/08, p.21

54 Alternative semantics Almost-sure semantics for PBA Alternative semantics L(A) = {w Σ ω P A ({ρ Runs(w) ρ = F })= 1} Expressivity almost-sure PBA are strictly less expressive than PBA almost-sure PBA and ω-regular languages are incomparale almost-sure PBA are not closed under complementation Séminaire LSV 05/02/08, p.21

55 Alternative semantics Recap: expressivity (a + ) a ω PBA PBA =1 DBA NBA {a k 1 a k 2 i (1 λk i) = 0} {a k 1 a k 2 i (1 λk i) > 0} Séminaire LSV 05/02/08, p.22

56 Alternative semantics Emptiness prolem and related results Decidaility result for POMDP Almost-sure reachaility in POMDP is decidale (EXPTIME). Séminaire LSV 05/02/08, p.23

57 Alternative semantics Emptiness prolem and related results Decidaility result for POMDP Almost-sure reachaility in POMDP is decidale (EXPTIME). Corollary The emptiness prolem is decidale for almost-sure PBA. Proof Sketch for PBA almost-sure reachaility and almost-sure repeated reachaility are interreducile PBA are a special instance of POMDP Séminaire LSV 05/02/08, p.23

58 Conclusion Outline 1 Introduction 2 Complementation 3 Emptiness prolem Langage dependency on proailities Undecidaility of emptiness 4 Alternative semantics Expressivity Emptiness prolem 5 Conclusion Séminaire LSV 05/02/08, p.24

59 Conclusion Conclusion Results concerning PBA complementation operator emptiness (and related prolems) undecidale for PBA expressivity of almost-sure PBA emptiness decidale for almost-sure PBA Séminaire LSV 05/02/08, p.25

60 Conclusion Conclusion Results concerning PBA complementation operator emptiness (and related prolems) undecidale for PBA expressivity of almost-sure PBA emptiness decidale for almost-sure PBA Results concerning POMDP positive repeated reachaility undecidale for POMDP almost-sure reachaility decidale for POMDP Séminaire LSV 05/02/08, p.25

61 Conclusion Conclusion Results concerning PBA complementation operator emptiness (and related prolems) undecidale for PBA expressivity of almost-sure PBA emptiness decidale for almost-sure PBA Results concerning POMDP positive repeated reachaility undecidale for POMDP almost-sure reachaility decidale for POMDP Open questions emptiness prolem for PBA with small alphaet efficient transformation from LTL to PBA Séminaire LSV 05/02/08, p.25

62 Conclusion Thank you for your attention! Questions? Séminaire LSV 05/02/08, p.26