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

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Gossip problem – Epistemic variants

The gossip problem [Baker & Shostak, 1972]

• n friends, each with a secret si
• 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]) 2/25

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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.

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The gossip problem epistemic variants

Epistemic level d = 1:

• Kisj: agent i knows the secret of j

Epistemic level d ≥ 2:

• for k agents {i1 . . . ik}, Ki1 . . . Kiksj:

agent i1 knows that . . . agent ik knows the secret of j For r ≥ 1, Tr =

• i1,...,ir∈{1,...,n}

Ki1 . . . Kir−1sir Proposition [Herzig & Maffre, 2015] For a n-vertex complete graph G, if n ≥ 4 and d ≥ 1 then any instance of Gossip-posG(d) has a solution of length no greater than (d + 1)(n − 2) calls.

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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-posG(d) has a solution of length no greater than d(2n − 3) calls.

• It is easy to find a spanning tree . . .

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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-posG(d) has a solution of length no greater than d(2n − 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-posG(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].

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Minimising the number of calls

• · · ·

· · ·

❅ ❅ ❅

• ❍

❍ ❍ ❍ ❍ ❍ ✟✟✟✟✟ ✟ ❵ ❵ ❵ ❵ ❵ ❵ ❵ ❵ ❵ ❵ ❵ ❵ ❵ ❵ ✥✥✥✥✥✥✥✥✥✥✥✥✥✥ ❳ ❳ ❳ ❳ ❳ ❳ ❳ ❳ ❳ ❳ ❳ ❳ ✘✘✘✘✘✘✘✘✘✘✘ ✘

• ❅

❅ ❅ ✟ ✟ ✟ ✟ ✟ ✟ ❍❍❍❍❍ ❍ ✥ ✥ ✥ ✥ ✥ ✥ ✥ ✥ ✥ ✥ ✥ ✥ ✥ ✥ ❵❵❵❵❵❵❵❵❵❵❵❵❵❵ ✘ ✘ ✘ ✘ ✘ ✘ ✘ ✘ ✘ ✘ ✘ ✘ ❳❳❳❳❳❳❳❳❳❳❳ ❳

p q 1 3 4 2 Proposition [Cooper et al., 2019] For n ≥ 4, if the n-vertex graph G has K2,n−2 as a subgraph, then any instance of Gossip-posG(d) has a solution of length no greater than (d + 1)(n − 2).

• For n ≥ 4, the complete graph has K2,n−2 as a subgraph.
• Detecting whether any graph G has K2,n−2 as a subgraph can

be achieved in polynomial time.

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Minimising the number of calls

• · · ·

· · ·

❅ ❅ ❅

• ❍

❍ ❍ ❍ ❍ ❍ ✟✟✟✟✟ ✟ ❵ ❵ ❵ ❵ ❵ ❵ ❵ ❵ ❵ ❵ ❵ ❵ ❵ ❵ ✥✥✥✥✥✥✥✥✥✥✥✥✥✥ ❳ ❳ ❳ ❳ ❳ ❳ ❳ ❳ ❳ ❳ ❳ ❳ ✘✘✘✘✘✘✘✘✘✘✘ ✘

• ❅

❅ ❅ ✟ ✟ ✟ ✟ ✟ ✟ ❍❍❍❍❍ ❍ ✥ ✥ ✥ ✥ ✥ ✥ ✥ ✥ ✥ ✥ ✥ ✥ ✥ ✥ ❵❵❵❵❵❵❵❵❵❵❵❵❵❵ ✘ ✘ ✘ ✘ ✘ ✘ ✘ ✘ ✘ ✘ ✘ ✘ ❳❳❳❳❳❳❳❳❳❳❳ ❳

p q 1 3 4 2

• dd passes:

✲ ✛

• · · ·

· · ·

❅ ❅ ❅

• ❍

❍ ❍ ❍ ❍ ❍ ✟✟✟✟✟ ✟ ❵ ❵ ❵ ❵ ❵ ❵ ❵ ❵ ❵ ❵ ❵ ❵ ❵ ❵ ✥✥✥✥✥✥✥✥✥✥✥✥✥✥ ❳ ❳ ❳ ❳ ❳ ❳ ❳ ❳ ❳ ❳ ❳ ❳ ✘✘✘✘✘✘✘✘✘✘✘ ✘

• ❅

❅ ❅ ✟ ✟ ✟ ✟ ✟ ✟ ❍❍❍❍❍ ❍ ✥ ✥ ✥ ✥ ✥ ✥ ✥ ✥ ✥ ✥ ✥ ✥ ✥ ✥ ❵❵❵❵❵❵❵❵❵❵❵❵❵❵ ✘ ✘ ✘ ✘ ✘ ✘ ✘ ✘ ✘ ✘ ✘ ✘ ❳❳❳❳❳❳❳❳❳❳❳ ❳

p q 1 3 4 2 even passes:

✛ ✲

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Minimising the number of calls

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

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Minimising the number of calls

Proposition [Cooper et al., 2019] The number of calls required to solve GossipG(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 Tr+1. At least 2(n − 2) calls to establish T2 [Baker & Shostak, 1972]. Suppose that KjTr was false before the last call to establish Tr+1. This call establishes not only Tr+1, but also KjTr+1 and KiTr+1. To establish Tr+2, it is necessary to distribute Tr+1 from i and j to

• ther agents and this takes at least n − 2 calls.

At least (r + 2)(n − 2) calls are required to establish Tr+2.

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One-way communications

If the directed graph G is strongly connected, the minimal number

• f calls for Directional-gossip-posG(1) is 2n − 2

[Harary & Schwenk, 1974].

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One-way communications

If the directed graph G is strongly connected, the minimal number

• f calls for Directional-gossip-posG(1) is 2n − 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

• f Directional-gossip-posG(d) has a solution of length no greater

than (d + 1)(n − 1).

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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-posG(d) has a solution with d(⌈log2 n⌉ − 1) + 1 time steps if n is even, or d⌈log2 n⌉ + 1 time steps if n is odd.

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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-posG(d) has a solution with d(⌈log2 n⌉ − 1) + 1 time steps if n is even, or d⌈log2 n⌉ + 1 time steps if n is odd. Partition of the vertex set of G:

• V1 of size ⌈n/2⌉
• V2 of size ⌊n/2⌋

such that G has an edge {i, j} for each i ∈ V1 and j ∈ V2. Number agents by elements of the ring Zn = {1, . . . , n} so that:

• ∀i ∈ Z, 2i + 1 ∈ V1
• ∀i ∈ Z, 2i + 2 ∈ V2

where arithmetic here and throughout the sequel is modulo n.

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Minimising the number of steps for positive goals

For even n, consider the protocol: First pass: For each step s from 1 to ⌈log2 n⌉: ∀i ∈ {0, . . . , (n

2 − 1)}, C(2i + 1, 2i + 2s)

Subsequent passes: Reorder even agents according to the permutation π given by π(2i + 2⌈log2 n⌉) = 2i + 2; Proceed as in the first pass but only for steps s from 2 to ⌈log2 n⌉

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Minimising the number of steps with parallel communications

V1 V2

• 13

11 9 7 5 3 1 14 12 10 8 6 4 2 V1 V2

• ✂

✂ ✂ ✂ ✂ ✂ ✂ ❍❍ ❍❍ ❍❍ ❍❍ ❍❍ ❍❍

13 11 9 7 5 3 1 14 12 10 8 6 4 2 V1 V2

• ✁

✁ ✁ ✁ ✁ ✁ ✁ ✁ ✁ ✁ ✁ ✁ ❏ ❏ ❏ ❏ ❏ ❏ ❏ ❏ ❏ ❏ ❏ ❏ ❏ ❏ ❏ ❏

13 11 9 7 5 3 1 14 12 10 8 6 4 2 V1 V2

• 13

11 9 7 5 3 1 14 12 10 8 6 4 2

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Minimising the number of steps with parallel communications

For odd n:

• Vfirst is the set of the first 2⌊log2 n⌋ agents ;
• Vlast is the set of the last n − 2⌊log2 n⌋ agents.

Consider the protocol: Preliminary step: Each agent in V1 ∩ Vlast calls one agent in V2 ∩ Vfirst, and each agent in V2 ∩ Vlast calls one agent in V1 ∩ Vfirst. Subsequent passes: Proceed in Vfirst as for the first pass of even case in Z2⌊log2 n⌋; Each agent in V1 ∩ Vlast calls one agent in V2 ∩ Vfirst, and each agent in V2 ∩ Vlast calls one agent in V1 ∩ Vfirst.

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Minimising the number of steps with parallel communications

V1 V2 Vfirst Vlast

• 13

11 9 7 5 3 1 12 10 8 6 4 2 V1 V2

• ✡

✡ ✡ ✡ ❍❍ ❍❍ ❍❍

13 11 9 7 5 3 1 12 10 8 6 4 2 V1 V2

• ✟✟

✟✟ ✟✟ ❏ ❏ ❏ ❏

13 11 9 7 5 3 1 12 10 8 6 4 2 V1 V2

• ✁

✁ ✁ ✁ ✁ ✁ ✁ ✁ ✁ ✁ ✁ ✁ ❆ ❆ ❆ ❆ ❆ ❆ ❆ ❆

13 11 9 7 5 3 1 12 10 8 6 4 2

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Minimising the number of steps with parallel communications

Proposition [Cooper et al., 2019] The number of steps required to solve Parallel-gossipG(d) for any graph G with n ≥ 2 vertices is at least d(⌈log2 n⌉ − 1) + 1 if n is even, or d⌈log2 n⌉ + 1 if n is odd.

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Minimising the number of steps with parallel communications

Proposition [Cooper et al., 2019] The number of steps required to solve Parallel-gossipG(d) for any graph G with n ≥ 2 vertices is at least d(⌈log2 n⌉ − 1) + 1 if n is even, or d⌈log2 n⌉ + 1 if n is odd. Proof. For even n: H: at least r(⌈log2 n⌉ − 1) + 1 steps are required to establish Tr+1. At least ⌈log2 n⌉ steps to establish T2 [Bavelas, 1950] Suppose that KjTr was false before the last step to establish Tr+1. This step establishes not only Tr+1, but also KjTr+1 and KiTr+1. To establish Tr+2, it is necessary to distribute Tr+1 from i and j to

• ther agents and this takes at least ⌈log2 n⌉ − 1 steps.

At least (r + 1)(⌈log2 n⌉ − 1) + 1 steps are required to establish Tr+2.

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Minimising the number of steps with parallel communications

Proposition [Cooper et al., 2019] The number of steps required to solve Parallel-gossipG(d) for any graph G with n ≥ 2 vertices is at least d(⌈log2 n⌉ − 1) + 1 if n is even, or d⌈log2 n⌉ + 1 if n is odd. Proof. For odd n: At least one more step is required for each epistemic level r because at least one agent doesn’t communicate his knowledge in the first step to establish Tr+1. Hence, it takes at least a sequence of ⌈log2 n⌉ + 1 steps for knowledge from all n agents to reach each others, and the lower bound is d⌈log2 n⌉ + 1.

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One-way parralel communications

If the directed graph G is strongly connected, the minimal number

• f calls for Directional-gossip-posG(1) is 2n − 2

[Harary & Schwenk, 1974].

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One-way parralel communications

If the directed graph G is strongly connected, the minimal number

• f calls for Directional-gossip-posG(1) is 2n − 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 n ≥ 4, if the n-vertex graph G has the complete bipartite graph K⌈n/2⌉,⌊n/2⌋ as a subgraph, then Parallel-directional-gossip(d) has a solution with d(⌈logϕn⌉ + 2) + 1 time steps.

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One-way parralel communications

For even n, consider the protocol: Preliminary step: For each i ∈ {0, . . . , (n

2 − 1)}, C(2i + 1, 2i + 2).

First pass: For each step s from 1 to ⌈logϕ n⌉: ∀i ∈ {0, . . . , (n

2 − 1)},

if s is odd, C(2i + 2, 2(i + Fs−1) + 1); if s is even, C(2i + 1, 2(i + Fs−1) + 2). Subsequent passes: If ⌈logϕ n⌉ is odd, reorder all agents according to the permutation π1 given by π1(2(i + F⌈logϕ n⌉+1) + 1) = 2i + 2 and π1(2i + 2) = 2i + 1; If ⌈logϕ n⌉ is even, reorder even agents according to the permutation π2 given by π2(2(i + F⌈logϕ n⌉+1) + 2) = 2i + 2; Proceed as in the first pass for steps s from 1 to ⌈logϕ n⌉.

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One-way parralel communications

Fs = ϕs − (−ϕ)−s √ 5 ∼ ϕs √ 5 = ϕs−logϕ

√ 5

(Fs ≥ n) ⇔

• s ≥ logϕn + logϕ

√ 5

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One-way parralel communications

For odd n: m =

n+3

2

if n ≡ 1 (mod 4)

n+1 2

if n ≡ 3 (mod 4) Let define :

• V odd

first is the set of the first m agents ;

• V odd

last is the set of the last n − m agents ;

• V even

first is the set of the last m agents ;

• V even

last

is the set of the first n − m agents. The agents in the two latter sets are reordered following the permutation π given by π(j) = n + 1 − j for all j ∈ Zn

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One-way parralel communications

Fs = ϕs − (−ϕ)−s √ 5 ∼ ϕs √ 5 = ϕs−logϕ

√ 5

(Fs ≥ m) ⇔

• s ≥ logϕm + logϕ

√ 5

• Fs is greater than m = n+3

2

when s ≥ logϕ n+3

2

+ logϕ √ 5. For n ≥ 7, logϕ(1 + 3

n) + logϕ √ 5 2 < 1

then logϕ(n + 3) + logϕ

√ 5 2 < logϕn + 1 ≤ ⌈logϕ n⌉ + 1 = s.

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One-way parralel communications

For odd n, consider the protocol: Preliminary step: Each agent in V1 ∩ V odd

last calls one agent in V2 ∩ V odd first ,

and each agent in V2 ∩ V odd

last calls one agent in V1 ∩ V odd first

Subsequent passes: Proceed in V p

first as for the preliminary step and

the first pass of even case in Zm, where p ∈ {odd, even} is the parity of current pass; Each agent in V1 ∩ V p

last calls one agent in V2 ∩ V p first,

and each agent in V2 ∩ V p

last calls one agent in V1 ∩ V p first,

where p is the parity of next pass.

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One-way parralel communications

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Conclusion

Conclusion

We have generalised many results from the classical gossip problem to the epistemic version.

• 1. Sequential protocols :
• no protocol exists which solves GossipG(d) in less than

(d + 1)(n − 2) calls.

• a protocol which uses only this number of calls (for any graph

G containing K2,n−2 as a subgraph).

• optimal protocol generalised for one-way communications

((d+1)(n-1) calls)

• 2. Parallel protocols :
• optimal protocol generalised for parallel communications

(O(d log2 n) steps)

• protocol generalised for one-way parallel communications

(O(d logϕ n) steps)

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Open problems

Determine the minimum number of calls given

• any graph G;
• any set of goals.

One-way communications

• optimal protocol which does not require the existence of a

Hamiltonian path ;

• optimal protocol for parallel one-way communications.

The complexity of the problem of minimising the number of calls (whether two-way or one-way) in an arbitrary graph G is still open. Consider the generalised gossip problem, assuming that all agents are autonomous.

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Brenda Baker, Robert Shostak, Gossips and telephones, Discrete Mathematics, pp 191–193, 1972. Alex Bavelas, Communication patterns in task-oriented groups, The Journal of the Acoustical Society of America, pp 725–730, 1988. Tom Bylander, The Computational Complexity of Propositional STRIPS Planning, Artif. Intell. (69), pp 165–204, 1994. Michael R. Garey, David S. Johnson, Computers and Intractability: A Guide to the Theory of NP-Completeness, W.

• H. Freeman Ed., 1979.

Frank Harary, Allen J. Schwenk, The communication problem

• n graphs and digraphs, Journal of the Franklin Institute

(297), pp 491–495, 1974.

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Sandra M. Hedetniemi, Stephen T. Hedetniemi, Arthur L. Liestman, A survey of gossiping and broadcasting in communication networks, Networks, pp 319–349, 1988. Andreas Herzig, Faustine Maffre, How to share knowledge by gossiping, 3rd International Conference on Agreement Technologies, 2015. Martin C. Cooper, Andreas Herzig, Faustine Maffre, Frédéric Maris, Pierre Régnier, The epistemic gossip problem, Discrete Mathematics (342), volume 3, pp 654–663, 2019.

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