Group Announcement Logic

Transcription

1 Group Announcement Logic Thomas Ågotnes joint work with: Philippe Baliani Hans van Ditmarsch Palo Sean

2 Elevator pitch Group Announcement Logic extends pulic announcement logic with: G φ : Group G can make an announcement after which φ is true

3 Pulic Announcement Logic (Plaza, 1989) ϕ ::= p K i ϕ ϕ ϕ 1 ϕ 2 ϕ 1 ϕ 2 φ 1 is true, and φ 2 is true after φ 1 is announced Formally: M =(S, 1,..., n,v) i equivalence rel. over S M, s = K i φ t i s M, t = φ M, s = φ 1 φ 2 M, s = φ 1 and M φ 1,s = φ 2 The model resulting from removing states where φ 1 is false

4 Adding quantification: APAL M, s = φ 1 φ 2 M, s = φ 1 and M φ 1,s = φ 2 Idea (van Benthem, Analysis 64, 2004): interpret the modal diamond as there is an announcement such that.. Aritrary announcement logic (APAL) (Baliani et al., TARK 2007): ϕ ::= p K i ϕ ϕ ϕ 1 ϕ 2 ϕ 1 ϕ 2 φ M, s = φ ψ M, s = ψ φ

5 Quantification in APAL M, s = φ ψ M, s = ψ φ Note: the quantification includes announcements that cannot e truthfully made in the system

6 Quantification: announcements y an agent M, s = i φ ψ M, s = K i ψ φ

7 Quantification: announcements y an agent M, s = i φ ψ M, s = K i ψ φ

8 Quantification: announcements y a group M, s = G φ {ψ i : i G} M, s = i G K iψ φ Group Announcement Logic (GAL): ϕ ::= p K i ϕ ϕ ϕ 1 ϕ 2 ϕ 1 ϕ 2 G φ Similar to coalitional aility operators of Coalition Logic (Pauly, 2002) and ATL (Alur, Henzinger, Kupferman, 1997), with actions = pulic announcements But GAL is not a Coalition Logic

9 H. van Ditmarsch, The Russian Cards Prolem, Studia Logica 75, 2003 Example: The Russian Cards Prolem From a pack of seven known cards 0,1,2,3,4,5,6 Anne and Bill each draw three cards and Cath gets the remaining card. How can Anne and Bill openly inform each other aout their cards, without Cath learning who holds any of their cards?

10 H. van Ditmarsch, The Russian Cards Prolem, Studia Logica 75, 2003 Example: The Russian Cards Prolem From a pack of seven known cards 0,1,2,3,4,5,6 Anne and Bill each draw three cards and Cath gets the remaining card. How can Anne and Bill openly inform each other aout their cards, without Cath learning who holds any of their cards? Formalisation: 012 a : Ann has cards 0,1 and 2 (one) ijk (ijk K a ijk ) (two) ijk (ijk a K ijk a ) (three) 6 q=0 ((q a K c q a ) (q K c q ))

11 H. van Ditmarsch, The Russian Cards Prolem, Studia Logica 75, 2003 Example: The Russian Cards Prolem From a pack of seven known cards 0,1,2,3,4,5,6 Anne and Bill each draw three cards and Cath gets the remaining card. How can Anne and Bill openly inform each other aout their cards, without Cath learning who holds any of their cards? Formalisation: Known solution: 012 a : Ann has cards 0,1 and 2 (one) ijk (ijk K a ijk ) (two) ijk (ijk a K ijk a ) (three) 6 q=0 ((q a K c q a ) (q K c q )) anne 012 a 034 a 056 a 135 a 246 a ill

12 H. van Ditmarsch, The Russian Cards Prolem, Studia Logica 75, 2003 Example: The Russian Cards Prolem From a pack of seven known cards 0,1,2,3,4,5,6 Anne and Bill each draw three cards and Cath gets the remaining card. How can Anne and Bill openly inform each other aout their cards, without Cath learning who holds any of their cards? Formalisation: Known solution: PAL: 012 a : Ann has cards 0,1 and 2 (one) ijk (ijk K a ijk ) (two) ijk (ijk a K ijk a ) (three) 6 q=0 ((q a K c q a ) (q K c q )) anne 012 a 034 a 056 a 135 a 246 a ill K a anne K ill (one two three)

13 H. van Ditmarsch, The Russian Cards Prolem, Studia Logica 75, 2003 Example: The Russian Cards Prolem From a pack of seven known cards 0,1,2,3,4,5,6 Anne and Bill each draw three cards and Cath gets the remaining card. How can Anne and Bill openly inform each other aout their cards, without Cath learning who holds any of their cards? Formalisation: Known solution: PAL: GAL: 012 a : Ann has cards 0,1 and 2 (one) ijk (ijk K a ijk ) (two) ijk (ijk a K ijk a ) (three) 6 q=0 ((q a K c q a ) (q K c q )) anne 012 a 034 a 056 a 135 a 246 a ill K a anne K ill (one two three) a (one two three)

14 Quantification: sequences of announcements APAL: φ φ announcing in sequence ψ, χ announcing ψ [ψ]χ GAL: G G φ G φ?

15 Quantification: sequences of announcements APAL: φ φ announcing in sequence ψ, χ announcing ψ [ψ]χ GAL: G G φ G φ? Ki ψ, K i χ Ki ψ [ K i ψ] K i χ

16 Quantification: sequences of announcements APAL: φ φ announcing in sequence ψ, χ announcing ψ [ψ]χ GAL: G G φ G φ? Ki ψ, K i χ Ki ψ [ K i ψ] K i χ not a group announcement

17 Quantification: sequences of announcements APAL: φ φ announcing in sequence ψ, χ announcing ψ [ψ]χ GAL: G G φ G φ? Ki ψ, K i χ Ki ψ [ K i ψ] K i χ not a group announcement Theorem: Yes.

18 Quantification: sequences of announcements G G φ G φ M, s = G φ there is an announcement y G, after which φ

19 Quantification: sequences of announcements G G φ G φ M, s = G φ there is an announcement y G, after which φ there is a sequence of announcements y G, after which φ

20 Example: Russian Cards (ctnd.) K a anne K ill (one two three) a (one two three) a (one two three)

21 Being ale to without knowing it p s a p t s = a p K a a p

22 Being ale to, knowing that, ut not knowing how p,q u p, q v p,q s a p,q t φ = K q ( K p ˆK a (K p K q)) s = K a q φ t = K a p φ t = a φ s = a φ s = Ka a φ

23 Being ale to, knowing that, ut not knowing how p,q u p, q v p,q s a p,q t φ = K q ( K p ˆK a (K p K q)) s = K a q φ t = K a p φ t = a φ s = a φ s = Ka a φ s = φ

24 Being ale to, knowing that, ut not knowing how p,q u p, q v p,q s a p,q t φ = K q ( K p ˆK a (K p K q)) s = K a q φ t = K a p φ t = a φ s = a φ s = Ka a φ

25 Being ale to, knowing that, ut not knowing how p,q u p, q v p,q s a p,q t φ = K q ( K p ˆK a (K p K q)) s = K a q φ t = K a p φ t = a φ s = a φ s = Ka a φ t = φ

26 Being ale to, knowing that, ut not knowing how p,q u p, q v p,q s a p,q t φ = K q ( K p ˆK a (K p K q)) s = K a q φ t = K a p φ t = a φ s = a φ s = Ka a φ

27 Being ale to, knowing that, ut not knowing how p,q u p, q v p,q s a p,q t φ = K q ( K p ˆK a (K p K q)) s = K a q φ t = K a p φ t = a φ s = a φ s = Ka a φ s = K a p φ t = K a q φ

28 Being ale to, knowing that, ut not knowing how u p,q p, q v φ = K q ( K p ˆK a (K p K q)) s = K a q φ t = K a p φ s = K a p φ t = K a q φ p,q s s = a φ t = a φ The same does not a know how p,q to t s = Ka a φ announcement will not achieve the goal in oth s and t - a achieve it

29 Expressing knowledge de dicto/de re Aility Knowledge of aility, de dicto Knowledge of aility, de re ψ s = K a ψ φ s a t ψ t = K a ψ φ ψ s a tt = K a ψ φ s = a φ s = K a a φ??

30 Expressing knowledge de dicto/de re Aility Knowledge of aility, de dicto Knowledge of aility, de re ψ s = K a ψ φ s a t ψ t = K a ψ φ ψ s a tt = K a ψ φ s = a φ s = K a a φ?? s = a K a φ ψ s = K a ψ K a φ

31 Expressing knowledge de dicto/de re Aility Knowledge of aility, de dicto Knowledge of aility, de re ψ s = K a ψ φ s a t ψ t = K a ψ φ ψ s a tt = K a ψ φ s = a φ s = K a a φ?? s = a K a φ ψ s = K a ψ K a φ

32 Expressing knowledge de dicto/de re Aility Knowledge of aility, de dicto Knowledge of aility, de re ψ s = K a ψ φ s a t ψ t = K a ψ φ ψ s a tt = K a ψ φ s = a φ s = K a a φ?? Depends on (1) the fact that actions are announcements (2) the S5 properties s = a K a φ ψ s = K a ψ K a φ

33 Example: Russian Cards (ctnd.) Ann and Bill knows how to exectute a successful protocol: a K a (two three K (one two three))

34 Some logical properties [G H]φ [G][H]φ G [G]φ [G] G φ (Church-Rosser) G [H]φ [H] G φ (..generalised)

35 Axiomatisation S5 n axioms and rules P AL axioms and rules [G]φ [ i G K iψ i ]φ where ψ i L el From φ, infer [G]φ From φ [θ][ i G K ip i ]ψ, infer φ [θ][g]ψ where p i Θ φ Θ θ Θ ψ Theorem: Sound & complete.

36 Model Checking The model checking prolem: Given M, s and φ, does M, s = φ hold? Theorem: The model checking prolem is PSPACE-complete (also extends to APAL)

37 Directions More general actions Coalition Announcement Logic a coalition logic there are announcements y G such that for all announcements y the other agents,... Pulic Announcement Games -> Hans talk

38 For more details: T. Ågotnes and H. van Ditmarsch, Coalitions and Announcements, Proc. AAMAS 2008 T. Ågotnes, P. Baliani, H. van Ditmarsch and P. Sean, Group Announcement Logic, Journal of Applied Logic 8(1), 2010