The epistemic gossip problem Sequential and parallel protocols To - PowerPoint PPT Presentation

The epistemic gossip problem Sequential and parallel protocols To appear in Discrete Mathematics (342)3, March 2019 https://authors.elsevier.com/c/1Y8W5,H-cE-W9 Martin C. Cooper, Andreas Herzig, Faustine Maffre, Frédéric Maris, Pierre Régnier December 5, 2018 IRIT – University of Toulouse 1 1/25

Gossip problem – Epistemic variants

The gossip problem [Baker & Shostak, 1972] • n friends, each with a secret s i • they can call each other to exchange every secret they know • how many calls to spread all secrets among all friends? (picture from [v. Ditmarsch&Kooi 2015]) 2 2/25

The gossip problem variants Centralized variants of Gossip problem: • Complete / incomplete graph • Positive / negative goals • Sequential / parallel protocols All these variants have been widely studied... • 1428 citations for the survey paper [Hedetniemi et al., 1988] • 1752 citations for [Bavelas, 1950] ...but only at the first epistemic level. 3 3/25

The gossip problem epistemic variants Epistemic level d = 1: • K i s j : agent i knows the secret of j Epistemic level d ≥ 2: • for k agents { i 1 . . . i k } , K i 1 . . . K i k s j : agent i 1 knows that . . . agent i k knows the secret of j � For r ≥ 1, T r = K i 1 . . . K i r − 1 s i r i 1 ,..., i r ∈{ 1 ,..., n } Proposition [Herzig & Maffre, 2015] For a n -vertex complete graph G , if n ≥ 4 and d ≥ 1 then any instance of Gossip-pos G ( d ) has a solution of length no greater than ( d + 1)( n − 2) calls. 4 4/25

Minimising the number of calls Proposition [Cooper et al., 2019] If the graph G is connected, then for n ≥ 2 and d ≥ 1, any instance of Gossip-pos G ( d ) has a solution of length no greater than d (2 n − 3) calls. • It is easy to find a spanning tree . . . 5 5/25

Minimising the number of calls Proposition [Cooper et al., 2019] If the graph G is connected, then for n ≥ 2 and d ≥ 1, any instance of Gossip-pos G ( d ) has a solution of length no greater than d (2 n − 3) calls. • It is easy to find a spanning tree . . . Proposition [Cooper et al., 2019] If the graph G has a Hamiltonian path, then any instance of Gossip-pos G ( d ) has a solution of length no greater than 1 + ( d + 1)( n − 2). • Determining the existence of a Hamiltonian path is known to be NP-complete [Garey & Johnson, 1979]. 5 5/25

Minimising the number of calls 1 ✥ ❵❵❵❵❵❵❵❵❵❵❵❵❵❵ ✘ ❳❳❳❳❳❳❳❳❳❳❳ ✥ ✘ ✟ • ❍❍❍❍❍ ✥ ✥ ✘ ✥ ✘ ✟ � ❅ ✥ ✘ ✥ ✘ ✟ ✥ ✥ ✘ ✥ ✘ ✟ � ❅ ✥ ✘ ✥ ✘ ✟ ✥ 3 ✥ ✘ ✘ ✟ � ❅ ❍ ❳ ❵ ✥✥✥✥✥✥✥✥✥✥✥✥✥✥ ❳ ❳ ✘✘✘✘✘✘✘✘✘✘✘ ✘ p ❵ q • • · · · • ❍ • • ✟ • · · · • • ❵ ❳ ❵ ❳ ❍ ✟✟✟✟✟ ❵ ❅ � ❳ ❵ ❳ ❍ 4 ❵ ❵ ❳ ❵ ❳ ❍ ❅ � ❵ ❳ ❵ ❳ ❍ ❵ ❵ ❳ ❅ ❍ ❵ ❳ � • 2 Proposition [Cooper et al., 2019] For n ≥ 4, if the n -vertex graph G has K 2 , n − 2 as a subgraph, then any instance of Gossip-pos G ( d ) has a solution of length no greater than ( d + 1)( n − 2). • For n ≥ 4, the complete graph has K 2 , n − 2 as a subgraph. • Detecting whether any graph G has K 2 , n − 2 as a subgraph can be achieved in polynomial time. 6 6/25

Minimising the number of calls odd passes: 1 ✛ ✥ ❵❵❵❵❵❵❵❵❵❵❵❵❵❵ ✘ ❳❳❳❳❳❳❳❳❳❳❳ ✥ ✘ ✟ • ❍❍❍❍❍ ✥ ✥ ✘ ✥ ✘ ✟ � ❅ ✥ ✘ ✥ ✘ ✟ ✥ ✥ ✘ ✥ ✘ ✟ � ❅ ✥ ✘ ✥ ✘ ✟ ✥ 3 ✥ ✘ ✘ ✟ � ❅ ❍ ❳ ❵ ✥✥✥✥✥✥✥✥✥✥✥✥✥✥ ❳ ❳ ✘✘✘✘✘✘✘✘✘✘✘ ✘ p ❵ q • • · · · • ❍ • • ✟ • · · · • • ❵ ❳ ❵ ❳ ❍ ✟✟✟✟✟ ❵ ❅ � ❳ ❵ ❳ ❍ 4 ❵ ❵ ❳ ❵ ❳ ❍ ❅ � ❵ ❳ ❵ ❳ ❍ ❵ ❳ ❵ ❳ ❵ ❅ ❍ � ✲ • 2 even passes: 1 ✛ ✘ ✥ ❵❵❵❵❵❵❵❵❵❵❵❵❵❵ ❳❳❳❳❳❳❳❳❳❳❳ ✥ • ✘ ✟ ❍❍❍❍❍ ✥ ✘ ✥ ✥ ✘ ✟ � ❅ ✥ ✘ ✥ ✘ ✟ ✥ ✥ ✘ ✥ ✘ ✟ � ❅ ✥ ✘ ✥ ✘ ✟ ✥ 3 ✥ ✘ ✘ ✟ � ❅ ❍ ❳ ❵ ✥✥✥✥✥✥✥✥✥✥✥✥✥✥ ❳ ❳ ✘✘✘✘✘✘✘✘✘✘✘ ✘ p ❵ q • • · · · • ❍ • • ✟ • · · · • • ❵ ❳ ❵ ❳ ❍ ✟✟✟✟✟ ❵ ❅ � ❳ ❵ ❳ ❍ 4 ❵ ❵ ❳ ❵ ❳ ❍ ❅ � ❵ ❳ ❵ ❳ ❍ ❵ ❵ ❳ ✲ ❍ ❅ ❵ ❳ � • 2 7 7/25

Minimising the number of calls Proposition [Cooper et al., 2019] The number of calls required to solve Gossip G ( d ) (for any graph G ) is at least ( d + 1)( n − 2). 8 8/25

Minimising the number of calls Proposition [Cooper et al., 2019] The number of calls required to solve Gossip G ( d ) (for any graph G ) is at least ( d + 1)( n − 2). Proof. H: at least ( r + 1)( n − 2) calls are required to establish T r +1 . At least 2( n − 2) calls to establish T 2 [Baker & Shostak, 1972]. Suppose that K j T r was false before the last call to establish T r +1 . This call establishes not only T r +1 , but also K j T r +1 and K i T r +1 . To establish T r +2 , it is necessary to distribute T r +1 from i and j to other agents and this takes at least n − 2 calls. At least ( r + 2)( n − 2) calls are required to establish T r +2 . 8 8/25

One-way communications If the directed graph G is strongly connected, the minimal number of calls for Directional-gossip-pos G (1) is 2 n − 2 [Harary & Schwenk, 1974]. 9 9/25

One-way communications If the directed graph G is strongly connected, the minimal number of calls for Directional-gossip-pos G (1) is 2 n − 2 [Harary & Schwenk, 1974]. Let G be the graph with the same n vertices as the directed graph G but with an edge between i and j if and only if G contains the two directed edges ( i , j ) and ( j , i ). Proposition [Cooper et al., 2019] For all d ≥ 1, if G contains a Hamiltonian path, then any instance of Directional-gossip-pos G ( d ) has a solution of length no greater than ( d + 1)( n − 1). 9 9/25

Parallel protocols

Minimising the number of steps with parallel communications Proposition [Cooper et al., 2019] For n ≥ 2, if the n -vertex graph G has the complete bipartite graph K ⌈ n / 2 ⌉ , ⌊ n / 2 ⌋ as a subgraph, then any instance of Parallel-gossip-pos G ( d ) has a solution with d ( ⌈ log 2 n ⌉ − 1) + 1 time steps if n is even, or d ⌈ log 2 n ⌉ + 1 time steps if n is odd. 10 10/25

Minimising the number of steps with parallel communications Proposition [Cooper et al., 2019] For n ≥ 2, if the n -vertex graph G has the complete bipartite graph K ⌈ n / 2 ⌉ , ⌊ n / 2 ⌋ as a subgraph, then any instance of Parallel-gossip-pos G ( d ) has a solution with d ( ⌈ log 2 n ⌉ − 1) + 1 time steps if n is even, or d ⌈ log 2 n ⌉ + 1 time steps if n is odd. Partition of the vertex set of G : • V 1 of size ⌈ n / 2 ⌉ • V 2 of size ⌊ n / 2 ⌋ such that G has an edge { i , j } for each i ∈ V 1 and j ∈ V 2 . Number agents by elements of the ring Z n = { 1 , . . . , n } so that: • ∀ i ∈ Z , 2 i + 1 ∈ V 1 • ∀ i ∈ Z , 2 i + 2 ∈ V 2 10 where arithmetic here and throughout the sequel is modulo n . 10/25

Minimising the number of steps for positive goals For even n , consider the protocol: First pass : For each step s from 1 to ⌈ log 2 n ⌉ : ∀ i ∈ { 0 , . . . , ( n 2 − 1) } , C(2 i + 1 , 2 i + 2 s ) Subsequent passes : Reorder even agents according to the permutation π given by π (2 i + 2 ⌈ log 2 n ⌉ ) = 2 i + 2; Proceed as in the first pass but only for steps s from 2 to ⌈ log 2 n ⌉ 11 11/25

Minimising the number of steps with parallel communications V 1 V 2 V 1 V 2 V 1 V 2 V 1 V 2 ❍❍ • • • • • • • • 1 2 1 2 1 2 1 2 ✂ ✂ ❏ ✁ ❍❍ • • • • • • • • 3 4 3 4 3 4 3 4 ✂ ❏ ❏ ✁ ✁ ❍❍ • • • • • • • • 5 6 5 6 5 6 5 6 ✂ ❏ ✁ ❏ ✁ ❏ ❏ ✁ ❍❍ • • • • • • • • 7 8 7 8 7 8 7 8 ✂ ❏ ✁ ✁ ❏ ✁ ❏ ❏ ❍❍ • • • • • • • • 9 10 9 10 9 10 9 10 ✂ ✁ ✁ ❏ ❏ ❏ ❍❍ • • • • • • • • 11 12 11 12 11 12 11 12 ✂ ✁ ❏ ❏ • • • • • • • • 13 14 13 14 13 14 13 14 12 12/25

Minimising the number of steps with parallel communications For odd n : • V first is the set of the first 2 ⌊ log 2 n ⌋ agents ; • V last is the set of the last n − 2 ⌊ log 2 n ⌋ agents. Consider the protocol: Preliminary step : Each agent in V 1 ∩ V last calls one agent in V 2 ∩ V first , and each agent in V 2 ∩ V last calls one agent in V 1 ∩ V first . Subsequent passes : Proceed in V first as for the first pass of even case in Z 2 ⌊ log2 n ⌋ ; Each agent in V 1 ∩ V last calls one agent in V 2 ∩ V first , and each agent in V 2 ∩ V last calls one agent in V 1 ∩ V first . 13 13/25