Motivation Alternating-time temporal logic plays a key role in - PowerPoint PPT Presentation

Probabilistic Alternating-Time µ -Calculus Fu Song 1 , Yedi Zhang 1 , Taolue Chen 2 , Yu Tang 3 , Zhiwu Xu 4 1 ShanghaiTech University, China 2 Birkbeck, University of London, UK 3 East China Normal University, Shanghai, China 4 Shenzhen University, Shenzhen, China January 7, 2019 1 / 53

Motivation Alternating-time temporal logic plays a key role in Stochastic Multi-Agent Systems reasoning about strategic abilities of agents Two fundamental problems: Model Checking 1 Satisfiability checking 2 2 / 53

Model Checking System input output Return True / False model checker input specification 3 / 53

Satisfiability Checking Return model Satisfiable input specification Solver Unsatisfiable Return UNSAT 4 / 53

System and Specification Stochastic Multi-agent Systems Stochastic Multi-agent Systems Systems PA 𝐔𝐌 A 𝐔𝐌 , 𝐁𝐔𝐌 ∗ 𝐐𝐃𝐔𝐌 , 𝐐𝐃𝐔𝐌 ∗ 𝐐𝐁𝐔𝐌 ∗ A 𝐍𝐃 5 / 53

Probabilistic Concurrent Game Structures (PCGS) M = ( Ag , Act , Q , Γ , δ, λ, q 0 ) � a 1 , b 2 � , 0 . 4 Ag := { 1 , 2 } q 3 Q := { q 0 , q 1 , q 2 , q 3 } Act := { a 1 , a 2 , b 1 , b 2 } � a 2 , b 2 � , 0 . 6 � a 1 , b 1 � , 0 . 9 Γ 1 ( q 0 ) := { a 1 } Γ 2 ( q 0 ) := { b 1 , b 2 } Γ 1 ( q 1 ) := { a 1 } Γ 2 ( q 1 ) := { b 1 , b 2 } q 0 � a 1 , b 2 � , 0 . 1 Γ 1 ( q 2 ) := { a 2 } Γ 2 ( q 2 ) := { b 1 , b 2 } Γ 1 ( q 3 ) := { a 1 , a 2 } Γ 2 ( q 3 ) := { b 2 } � a 1 , b 2 � , 0 . 3 � a 2 , b 2 � , 0 . 5 λ ( q 0 ) = λ ( q 2 ) = { p red } q 1 q 2 λ ( q 1 ) = λ ( q 3 ) = { p blue } � a 1 , b 1 � , 0 . 7 � a 2 , b 1 � , 0 . 5 A Probabilistic Boolean Network 6 / 53

Literature Stochastic Multi-agent Systems Stochastic Multi-agents Systems Systems PA 𝐔𝐌 A 𝐔𝐌 , 𝐁𝐔𝐌 ∗ 𝐐𝐁𝐔𝐌 ∗ A 𝐍𝐃 𝐐𝐃𝐔𝐌 , 𝐐𝐃𝐔𝐌 ∗ SAT Both Model SAT and Model Checking long-standing Checking open problem well studied 7 / 53

Goal Decidable logic for reasoning about strategic abilities in stochastic MAS: Satisfiability checking is useful Debugging specifications (Rozier and Vardi 2010) Social procedure or mechanism design (Pauly 2011) Assertion-based design (Foster, Krolnik, and Lacey 2004) 8 / 53

Alternating-Time µ -Calculus (AMC) Syntax φ ::= p | ¬ p | Z | φ ∨ φ | φ ∧ φ | µ Z .φ | ν Z .φ | �� A �� φ | [[ A ]] φ Semantics � p � ξ M = { q ∈ Q | p ∈ λ ( q ) } ; � ¬ p � ξ M = Q \ � p � ξ M ; � Z � ξ M = ξ ( Z ) ; � φ 1 ∧ φ 2 � ξ M = � φ 1 � ξ M ∩ � φ 2 � ξ M ; � φ 1 ∨ φ 2 � ξ M = � φ 1 � ξ M ∪ � φ 2 � ξ M ; M = � { Q ′ ⊆ Q | � φ � ξ [ Q ′ / Z ] � µ Z .φ � ξ ⊆ Q ′ } ; M M = � { Q ′ ⊆ Q | � φ � ξ [ Q ′ / Z ] � ν Z .φ � ξ ⊇ Q ′ } ; M υ A ,υ A � �� A �� φ � ξ , π 1 ∈ � φ � ξ M = { q ∈ Q | ∃ υ A ∈ Υ A , ∀ υ A ∈ Υ A , ∀ π ∈ O M } ; q υ A ,υ A � [[ A ]] φ � ξ , π 1 ∈ � φ � ξ M = { q ∈ Q | ∀ υ A ∈ Υ A , ∃ υ A ∈ Υ A , ∀ π ∈ O M } . q 9 / 53

Probabilistic Alternating-Time µ -Calculus (PAMC) Syntax φ ::= p | ¬ φ | Z | φ ∧ φ | φ ∨ φ | µ Z .φ | ν Z .φ | �� A �� ⊲⊳ k φ | [[ A ]] ⊲⊳ k φ Semantics � q ∈ Q | ∃ υ A ∈ Υ A , ∀ υ A ∈ Υ A : � � �� A �� ⊲⊳ k φ � ξ M = ; υ A ,υ A υ A ,υ A | π 1 ∈ � φ � ξ ( { π ∈ O M } ) ⊲⊳ k Pr q q � q ∈ Q | ∀ υ A ∈ Υ A , ∃ υ A ∈ Υ A : � � [[ A ]] ⊲⊳ k φ � ξ M = . υ A ,υ A υ A ,υ A | π 1 ∈ � φ � ξ Pr ( { π ∈ O M } ) ⊲⊳ k q q 10 / 53

Recall The PBN Example M = ( Ag , Act , Q , Γ , δ, λ, q 0 ) � a 1 , b 2 � , 0 . 4 Ag := { 1 , 2 } q 3 Q := { q 0 , q 1 , q 2 , q 3 } Act := { a 1 , a 2 , b 1 , b 2 } � a 2 , b 2 � , 0 . 6 � a 1 , b 1 � , 0 . 9 Γ 1 ( q 0 ) := { a 1 } Γ 2 ( q 0 ) := { b 1 , b 2 } Γ 1 ( q 1 ) := { a 1 } Γ 2 ( q 1 ) := { b 1 , b 2 } q 0 � a 1 , b 2 � , 0 . 1 Γ 1 ( q 2 ) := { a 2 } Γ 2 ( q 2 ) := { b 1 , b 2 } Γ 1 ( q 3 ) := { a 1 , a 2 } Γ 2 ( q 3 ) := { b 2 } � a 1 , b 2 � , 0 . 3 � a 2 , b 2 � , 0 . 5 λ ( q 0 ) = λ ( q 2 ) = { p red } q 1 q 2 λ ( q 1 ) = λ ( q 3 ) = { p blue } � a 1 , b 1 � , 0 . 7 � a 2 , b 1 � , 0 . 5 A Probabilistic Boolean Network 11 / 53

PBN with PAMC PBN Control inputs 𝒘 = (𝒘 𝟐 , 𝒘 𝟑 … 𝒘 𝒏 ) 𝝑 𝟏, 𝟐 𝒏 C- PBN Satisfies some properties? 12 / 53

PBN with PAMC PBN Control inputs 𝒘 = (𝒘 𝟐 , 𝒘 𝟑 … 𝒘 𝒏 ) 𝝑 𝟏, 𝟐 𝒏 C- PBN Satisfies some properties? For A ⊆ { n + 1 , ..., n + m } achievement property: µ Z . ( p ∨ �� A �� ≥ 0 . 8 Z ) maintenance property: ν Z . ( p ∧ �� A �� > 0 . 9 Z ) 13 / 53

Main Contributions Theorem PAMC subsumes AMC and P µ TL. PAMC and PATL/PATL ∗ are incomparable. 14 / 53

Main Contributions Theorem PAMC subsumes AMC and P µ TL. PAMC and PATL/PATL ∗ are incomparable. Theorem The model checking problem for PAMC over PCGSs is in UP ∩ co- UP and can be decided in O (( | φ | · |M| ) c · d ) time for some constant c, where d denotes the alternation depth of φ . 15 / 53

Main Contributions Theorem PAMC subsumes AMC and P µ TL. PAMC and PATL/PATL ∗ are incomparable. Theorem The model checking problem for PAMC over PCGSs is in UP ∩ co- UP and can be decided in O (( | φ | · |M| ) c · d ) time for some constant c, where d denotes the alternation depth of φ . Theorem The satisfiability problem for PAMC is EXPTIME -complete. Moreover, if φ is satisfiable, we can construct a model of φ in exponential size of | φ | s.t. the number of players is bounded by k + 1 , where k is the number of the players occurring in φ , the number of actions is bounded by | FL ∃ ( φ ) | + | FL ∀ ( φ ) | + 1 , the outdegree is bounded by | FL ∃ ( φ ) | + 2 . 16 / 53

PAMC: Expressiveness Proof sketch: 17 / 53

PAMC: Expressiveness Proof sketch: For every CGS M , AMC φ , � φ � M = � t ( φ ) � M . 1 t ( p ) = p , t ( ¬ p ) = ¬ p , t ( Z ) = Z t ( µ Z .φ ) = µ Z . t ( φ ) , t ( ν Z .φ ) = ν Z . t ( φ ) t ( �� A �� φ ) = �� A �� ≥ 1 t ( φ ) , t ([[ A ]] φ ) = [[ A ]] ≥ 1 t ( φ ) t ( φ 1 ◦ φ 2 ) = t ( φ 1 ) ◦ t ( φ 2 ) 18 / 53

PAMC: Expressiveness Proof sketch: For every CGS M , AMC φ , � φ � M = � t ( φ ) � M . 1 t ( p ) = p , t ( ¬ p ) = ¬ p , t ( Z ) = Z t ( µ Z .φ ) = µ Z . t ( φ ) , t ( ν Z .φ ) = ν Z . t ( φ ) t ( �� A �� φ ) = �� A �� ≥ 1 t ( φ ) , t ([[ A ]] φ ) = [[ A ]] ≥ 1 t ( φ ) t ( φ 1 ◦ φ 2 ) = t ( φ 1 ) ◦ t ( φ 2 ) For every MC M , P µ TL φ , � φ � M = � t µ ( φ ) � M . 2 t µ ( p ) = p , t µ ( ¬ p ) = ¬ p , t µ ( Z ) = Z t µ ( µ Z .φ ) = µ Z . t µ ( φ ) , t µ ( ν Z .φ ) = ν Z . t µ ( φ ) t µ ([ X φ ] ⊲⊳ k ) = ��∅�� ⊲⊳ k t µ ( φ ) t µ ( φ 1 ◦ φ 2 ) = t µ ( φ 1 ) ◦ t µ ( φ 2 ) P µ TL and PCTL are incomparable (Liu et al. 2015). 3 19 / 53

PAMC: Expressiveness Proof sketch: For every CGS M , AMC φ , � φ � M = � t ( φ ) � M . 1 t ( p ) = p , t ( ¬ p ) = ¬ p , t ( Z ) = Z t ( µ Z .φ ) = µ Z . t ( φ ) , t ( ν Z .φ ) = ν Z . t ( φ ) t ( �� A �� φ ) = �� A �� ≥ 1 t ( φ ) , t ([[ A ]] φ ) = [[ A ]] ≥ 1 t ( φ ) t ( φ 1 ◦ φ 2 ) = t ( φ 1 ) ◦ t ( φ 2 ) For every MC M , P µ TL φ , � φ � M = � t µ ( φ ) � M . 2 t µ ( p ) = p , t µ ( ¬ p ) = ¬ p , t µ ( Z ) = Z t µ ( µ Z .φ ) = µ Z . t µ ( φ ) , t µ ( ν Z .φ ) = ν Z . t µ ( φ ) t µ ([ X φ ] ⊲⊳ k ) = ��∅�� ⊲⊳ k t µ ( φ ) t µ ( φ 1 ◦ φ 2 ) = t µ ( φ 1 ) ◦ t µ ( φ 2 ) P µ TL and PCTL are incomparable (Liu et al. 2015). 3 AMC is strictly more expressive than ATL ∗ . 4 20 / 53

PAMC: Model Checking Idea : a recursive procedure (adapted from AMC Model Checking) 21 / 53

PAMC: Model Checking Idea : a recursive procedure (adapted from AMC Model Checking) Challenge : �� A �� ⊲⊳ k φ , [[ A ]] ⊲⊳ k φ 22 / 53

PAMC: Model Checking Idea : a recursive procedure (adapted from AMC Model Checking) Challenge : �� A �� ⊲⊳ k φ , [[ A ]] ⊲⊳ k φ Solution : take ⊲⊳ ∈ { >, ≥} and �� A �� ⊲⊳ k φ for example:   q ∈ � �� A �� ⊲⊳ k φ � ξ iff  �  �  δ ( q , d , q ′ )   sup  inf Pr f , g ( d ) ·  ⊲⊳ k        g ∈ F q   f ∈ F q d ∈ D q ′ ∈ � φ � ξ A A 23 / 53

PAMC: Model Checking Idea : a recursive procedure (adapted from AMC Model Checking) Challenge : �� A �� ⊲⊳ k φ , [[ A ]] ⊲⊳ k φ Solution : take ⊲⊳ ∈ { >, ≥} and �� A �� ⊲⊳ k φ for example:   q ∈ � �� A �� ⊲⊳ k φ � ξ iff  �  �  δ ( q , d , q ′ )    sup inf Pr f , g ( d ) ·  ⊲⊳ k        g ∈ F q   f ∈ F q d ∈ D q ′ ∈ � φ � ξ A A Lemma For any PAMC formula φ , PCGS M and valuation ξ , � φ � ξ M = eval ( φ, ξ ) . Moreover, the procedure eval runs in O (( | φ | · |M| ) c · dep ( φ ) ) time for some constant c. 24 / 53

PAMC: Satisfiability Checking Solving SAT Two-player Checking Parity Game of 𝓗 𝝔 𝝔 φ is satisfiable iff Player-0 has a winning strategy in G φ . 25 / 53