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

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Probabilistic Alternating-Time µ-Calculus

Fu Song1, Yedi Zhang1, Taolue Chen2, Yu Tang3, Zhiwu Xu4

1ShanghaiTech University, China 2Birkbeck, University of London, UK 3East China Normal University, Shanghai, China 4Shenzhen University, Shenzhen, China

January 7, 2019

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Motivation

Alternating-time temporal logic plays a key role in Stochastic Multi-Agent Systems reasoning about strategic abilities of agents Two fundamental problems:

1

Model Checking

2

Satisfiability checking

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Model Checking

System model checker Return True / False specification

input input

utput

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Satisfiability Checking

Return model specification Solver Return UNSAT

Satisfiable Unsatisfiable input 4 / 53

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System and Specification

Stochastic Systems Multi-agent Systems 𝐐𝐃𝐔𝐌 , 𝐐𝐃𝐔𝐌∗ A𝐔𝐌 , 𝐁𝐔𝐌∗ A𝐍𝐃 PA𝐔𝐌 𝐐𝐁𝐔𝐌∗ Stochastic Multi-agent Systems

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Probabilistic Concurrent Game Structures (PCGS)

M = (Ag, Act, Q, Γ, δ, λ, q0) q3 q0 q1 q2 a1, b1, 0.9 a1, b2, 0.1 a1, b1, 0.7 a1, b2, 0.3 a2, b2, 0.5 a2, b1, 0.5 a2, b2, 0.6 a1, b2, 0.4

A Probabilistic Boolean Network

Ag := {1, 2} Q := {q0, q1, q2, q3} Act := {a1, a2, b1, b2} Γ1(q0) := {a1} Γ2(q0) := {b1, b2} Γ1(q1) := {a1} Γ2(q1) := {b1, b2} Γ1(q2) := {a2} Γ2(q2) := {b1, b2} Γ1(q3) := {a1, a2} Γ2(q3) := {b2} λ(q0) = λ(q2) = {pred} λ(q1) = λ(q3) = {pblue}

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Literature

Stochastic Systems Multi-agents Systems 𝐐𝐃𝐔𝐌 , 𝐐𝐃𝐔𝐌∗ A𝐔𝐌 , 𝐁𝐔𝐌∗ A𝐍𝐃 PA𝐔𝐌 𝐐𝐁𝐔𝐌∗ SAT long-standing

pen problem

Both SAT and Model Checking well studied Model Checking Stochastic Multi-agent Systems

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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)

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Alternating-Time µ-Calculus (AMC)

Syntax

φ ::= p | ¬p | Z | φ ∨ φ | φ ∧ φ | µZ.φ | νZ.φ | Aφ | [[A]]φ

Semantics

pξ

M = {q ∈ Q | p ∈ λ(q)};

¬pξ

M = Q \ pξ M;

Zξ

M = ξ(Z);

φ1 ∧ φ2ξ

M = φ1ξ M ∩ φ2ξ M;

φ1 ∨ φ2ξ

M = φ1ξ M ∪ φ2ξ M;

µZ.φξ

M = {Q′ ⊆ Q | φξ[Q′/Z] M

⊆ Q′}; νZ.φξ

M = {Q′ ⊆ Q | φξ[Q′/Z] M

⊇ Q′}; Aφξ

M = {q ∈ Q | ∃υA ∈ ΥA, ∀υA ∈ ΥA, ∀π ∈ O υA ,υA q

, π1 ∈ φξ

M};

[[A]]φξ

M = {q ∈ Q | ∀υA ∈ ΥA, ∃υA ∈ ΥA, ∀π ∈ O υA ,υA q

, π1 ∈ φξ

M}. 9 / 53

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Probabilistic Alternating-Time µ-Calculus (PAMC)

Syntax

φ ::= p | ¬φ | Z | φ ∧ φ | φ ∨ φ | µZ.φ | νZ.φ | A⊲⊳kφ | [[A]]⊲⊳kφ

Semantics

A⊲⊳kφξ

M =

q ∈ Q | ∃υA ∈ ΥA, ∀υA ∈ ΥA : Pr

υA ,υA q

({π ∈ O

υA ,υA q

| π1 ∈ φξ

M}) ⊲⊳ k

;

[[A]]⊲⊳kφξ

M =

q ∈ Q | ∀υA ∈ ΥA, ∃υA ∈ ΥA :

Pr

υA ,υA q

({π ∈ O

υA ,υA q

| π1 ∈ φξ

M}) ⊲⊳ k

.

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Recall The PBN Example

M = (Ag, Act, Q, Γ, δ, λ, q0) q3 q0 q1 q2 a1, b1, 0.9 a1, b2, 0.1 a1, b1, 0.7 a1, b2, 0.3 a2, b2, 0.5 a2, b1, 0.5 a2, b2, 0.6 a1, b2, 0.4

A Probabilistic Boolean Network

Ag := {1, 2} Q := {q0, q1, q2, q3} Act := {a1, a2, b1, b2} Γ1(q0) := {a1} Γ2(q0) := {b1, b2} Γ1(q1) := {a1} Γ2(q1) := {b1, b2} Γ1(q2) := {a2} Γ2(q2) := {b1, b2} Γ1(q3) := {a1, a2} Γ2(q3) := {b2} λ(q0) = λ(q2) = {pred} λ(q1) = λ(q3) = {pblue}

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PBN with PAMC

PBN 𝒘 = (𝒘𝟐, 𝒘𝟑 … 𝒘𝒏) 𝝑 𝟏, 𝟐 𝒏

Control inputs

Satisfies some properties?

C-PBN

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PBN with PAMC

PBN 𝒘 = (𝒘𝟐, 𝒘𝟑 … 𝒘𝒏) 𝝑 𝟏, 𝟐 𝒏

Control inputs

Satisfies some properties?

C-PBN

For A ⊆ {n + 1, ..., n + m} achievement property: µZ.(p ∨ A≥0.8Z) maintenance property: νZ.(p ∧ A>0.9Z)

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Main Contributions

Theorem

PAMC subsumes AMC and PµTL. PAMC and PATL/PATL∗ are incomparable.

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Main Contributions

Theorem

PAMC subsumes AMC and PµTL. 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.

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Main Contributions

Theorem

PAMC subsumes AMC and PµTL. 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

ccurring in φ,

the number of actions is bounded by |FL∃(φ)| + |FL∀(φ)| + 1, the outdegree is bounded by |FL∃(φ)| + 2.

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PAMC: Expressiveness

Proof sketch:

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PAMC: Expressiveness

Proof sketch:

1

For every CGS M, AMC φ, φM = t(φ)M. t(p) = p, t(¬p) = ¬p, t(Z) = Z t(µZ.φ) = µZ.t(φ), t(νZ.φ) = νZ.t(φ) t(Aφ) = A≥1t(φ), t([[A]]φ) = [[A]]≥1t(φ) t(φ1 ◦ φ2) = t(φ1) ◦ t(φ2)

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PAMC: Expressiveness

Proof sketch:

1

For every CGS M, AMC φ, φM = t(φ)M. t(p) = p, t(¬p) = ¬p, t(Z) = Z t(µZ.φ) = µZ.t(φ), t(νZ.φ) = νZ.t(φ) t(Aφ) = A≥1t(φ), t([[A]]φ) = [[A]]≥1t(φ) t(φ1 ◦ φ2) = t(φ1) ◦ t(φ2)

2

For every MC M, PµTL φ, φM = tµ(φ)M. tµ(p) = p, tµ(¬p) = ¬p, tµ(Z) = Z tµ(µZ.φ) = µZ.tµ(φ),tµ(νZ.φ) = νZ.tµ(φ) tµ([Xφ]⊲⊳k) = ∅⊲⊳ktµ(φ) tµ(φ1 ◦ φ2) = tµ(φ1) ◦ tµ(φ2)

3

PµTL and PCTL are incomparable (Liu et al. 2015).

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PAMC: Expressiveness

Proof sketch:

1

For every CGS M, AMC φ, φM = t(φ)M. t(p) = p, t(¬p) = ¬p, t(Z) = Z t(µZ.φ) = µZ.t(φ), t(νZ.φ) = νZ.t(φ) t(Aφ) = A≥1t(φ), t([[A]]φ) = [[A]]≥1t(φ) t(φ1 ◦ φ2) = t(φ1) ◦ t(φ2)

2

For every MC M, PµTL φ, φM = tµ(φ)M. tµ(p) = p, tµ(¬p) = ¬p, tµ(Z) = Z tµ(µZ.φ) = µZ.tµ(φ),tµ(νZ.φ) = νZ.tµ(φ) tµ([Xφ]⊲⊳k) = ∅⊲⊳ktµ(φ) tµ(φ1 ◦ φ2) = tµ(φ1) ◦ tµ(φ2)

3

PµTL and PCTL are incomparable (Liu et al. 2015).

4

AMC is strictly more expressive than ATL∗.

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PAMC: Model Checking

Idea: a recursive procedure (adapted from AMC Model Checking)

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PAMC: Model Checking

Idea: a recursive procedure (adapted from AMC Model Checking) Challenge: A⊲⊳kφ, [[A]]⊲⊳kφ

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PAMC: Model Checking

Idea: a recursive procedure (adapted from AMC Model Checking) Challenge: A⊲⊳kφ, [[A]]⊲⊳kφ Solution: take ⊲⊳∈ {>, ≥} and A⊲⊳kφ for example: q ∈ A⊲⊳kφξ iff         sup

f∈Fq

A

inf

g∈Fq

A

d∈D

Prf,g(d) ·

q′∈φξ

δ(q, d, q′)          ⊲⊳ k

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PAMC: Model Checking

Idea: a recursive procedure (adapted from AMC Model Checking) Challenge: A⊲⊳kφ, [[A]]⊲⊳kφ Solution: take ⊲⊳∈ {>, ≥} and A⊲⊳kφ for example: q ∈ A⊲⊳kφξ iff         sup

f∈Fq

A

inf

g∈Fq

A

d∈D

Prf,g(d) ·

q′∈φξ

δ(q, d, q′)          ⊲⊳ k

Lemma

For any PAMC formula φ, PCGS M and valuation ξ, φξ

M = eval(φ, ξ). Moreover,

the procedure eval runs in O((|φ| · |M|)c·dep(φ)) time for some constant c.

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PAMC: Satisfiability Checking

SAT Checking

f

𝝔

Solving Two-player Parity Game

𝓗𝝔

φ is satisfiable iff Player-0 has a winning strategy in Gφ.

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PAMC: Satisfiability Checking

SAT Checking

f

𝝔

Solving Two-player Parity Game

𝓗𝝔

φ is satisfiable iff Player-0 has a winning strategy in Gφ. φ = 1, 2>0.5p1 ∧ 3>0.5(¬p1 ∨ p2) ∧ [[1, 3]]>0.4(¬p1 ∧ ¬p2)

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PAMC: Satisfiability Checking

{φ = 1, 2>0.5p1 ∧ 3>0.5(¬p1 ∨ p2) ∧ [[1, 3]]>0.4(¬p1 ∧ ¬p2)}

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PAMC: Satisfiability Checking

{φ = 1, 2>0.5p1 ∧ 3>0.5(¬p1 ∨ p2) ∧ [[1, 3]]>0.4(¬p1 ∧ ¬p2)} {φ = 1, 2>0.5p1 ∧ 3>0.5(¬p1 ∨ p2), [[1, 3]]>0.4(¬p1 ∧ ¬p2)}

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PAMC: Satisfiability Checking

{φ = 1, 2>0.5p1 ∧ 3>0.5(¬p1 ∨ p2) ∧ [[1, 3]]>0.4(¬p1 ∧ ¬p2)} {φ = 1, 2>0.5p1 ∧ 3>0.5(¬p1 ∨ p2), [[1, 3]]>0.4(¬p1 ∧ ¬p2)} {φ = 1, 2>0.5p1, 3>0.5(¬p1 ∨ p2), [[1, 3]]>0.4(¬p1 ∧ ¬p2)}

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PAMC: Satisfiability Checking

1

MISv = S1 ∪ S2: S1 = misv \

[[B]]⊲⊳k ϕ∈[[v]] misB v

S2 = {M ∪ {[[B]]⊲⊳kϕ} | M ∈ misB

v , [[B]]⊲⊳kϕ ∈ [[v]]}

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PAMC: Satisfiability Checking

1

MISv = S1 ∪ S2: S1 = misv \

[[B]]⊲⊳k ϕ∈[[v]] misB v

S2 = {M ∪ {[[B]]⊲⊳kϕ} | M ∈ misB

v , [[B]]⊲⊳kϕ ∈ [[v]]}

MISv =

{1, 2>0.5p1, 3>0.5(¬p1 ∨ p2)},

{3>0.5(¬p1 ∨ p2), [[1, 3]]>0.4(¬p1 ∧ ¬p2)}

.

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PAMC: Satisfiability Checking

1

MISv = S1 ∪ S2: S1 = misv \

[[B]]⊲⊳k ϕ∈[[v]] misB v

S2 = {M ∪ {[[B]]⊲⊳kϕ} | M ∈ misB

v , [[B]]⊲⊳kϕ ∈ [[v]]}

MISv =

{1, 2>0.5p1, 3>0.5(¬p1 ∨ p2)},

{3>0.5(¬p1 ∨ p2), [[1, 3]]>0.4(¬p1 ∧ ¬p2)}

.

2

Check satisfiability of probability constraints

S1: c1 = {{p1, ¬p1 ∨ p2}}, w1({p1, ¬p1 ∨ p2}) = 1

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PAMC: Satisfiability Checking

1

MISv = S1 ∪ S2: S1 = misv \

[[B]]⊲⊳k ϕ∈[[v]] misB v

S2 = {M ∪ {[[B]]⊲⊳kϕ} | M ∈ misB

v , [[B]]⊲⊳kϕ ∈ [[v]]}

MISv =

{1, 2>0.5p1, 3>0.5(¬p1 ∨ p2)},

{3>0.5(¬p1 ∨ p2), [[1, 3]]>0.4(¬p1 ∧ ¬p2)}

.

2

Check satisfiability of probability constraints

S1: c1 = {{p1, ¬p1 ∨ p2}}, w1({p1, ¬p1 ∨ p2}) = 1 S2: c1 = {{¬p1 ∨ p2, ¬p1 ∧ ¬p2}}, w1({¬p1 ∨ p2, ¬p1 ∧ ¬p2} = 1)

r c2 = {{¬p1 ∨ p2}, {¬p1 ∧ ¬p2}},

w2({¬p1 ∨ p2}) = 0.55, w2({¬p1 ∧ ¬p2}) = 0.45)

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