Properties of Regular Protocols

We now consider some properties of regular protocols and regular execution sequences. The following observation is convenient.

Observation 6.8. Let the epistemic calling condition for agent ai to call aj, for Learn New Secrets, Known Information Growth de Dicto and Known Information Growth de Re be ϕΠ1(ai, aj), ϕΠ2(ai, aj) and ϕΠ3(ai, aj), respectively. Recall those conditions are

as follows: ϕΠ1(ai, aj) =Kai¬KwaiAj (6.1) ϕΠ2(ai, aj) =Kai _ ak∈Ag (KwaiAk ∇KwajAk) (6.2) ϕΠ3(ai, aj) =Kai _ ak∈Ag Kai(KwaiAk ∇KwajAk) (6.3)

Given any gossip treeT_g, from Observation 4.26 we see that:

T_g |= ϕΠ1(ai, aj) → ϕΠ3(ai, aj) and Tg |= ϕΠ3(ai, aj) → ϕΠ2(ai, aj) (6.4)

Let us define the conditionϕ(ai, aj) as: ϕ(ai, aj) =

_

ak∈Ag

Note that for each of the conditions ϕΠ1(ai, aj), ϕΠ2(ai, aj) and ϕΠ3(ai, aj), we have

that:

Tg |=ϕΠ1(ai, aj)→ϕ(ai, aj) and Tg |=ϕΠ2(ai, aj)→ϕ(ai, aj)

and Tg |=ϕΠ3(ai, aj)→ϕ(ai, aj) (by veridicality) (6.6)

In other words: if the calling condition for agent ai to call aj is satisfied for any of the three protocols, then there is some secret Ak that is known by only one of those two agents. Furthermore,

T_g |=¬KwaiAj →ϕΠ1(ai, aj) (by negative Introspection) (6.7)

Hence, from (6.4):

Tg |=¬KwaiAj →ϕΠ2(ai, aj) and Tg |=¬KwaiAj →ϕΠ3(ai, aj)

(by propositional logic) (6.8) And, from (6.8) and (6.6):

T_g |=¬KwaiAj →ϕ(ai, aj) (by propositional logic) (6.9)

Proposition 6.9. Given any network of agents, Known Information Growth de Re, Known Information Growth de Dicto and Learn New Secrets are regular protocols. Proof. Consider any network of agents, and take any execution sequenceσ of any of the three protocols. Let the gossip tree for the protocol be T_g, and let σ = . . .;aiaj;. . ., whereaiaj is an arbitrary call in σ. Let σ0 be the longest history of calls preceding the aiaj call in σ. Then the calling condition for agent ai to call aj in the aiaj call must have been true atσ0. From Definition 6.4, it follows thatai andaj are neighbours on the network graph, and the epistemic calling condition for agentai to call agentaj holds at

(T_g, σ0). Hence from Observation 6.8 (expression (6.6)) it also follows that there is some secretAk known by exactly one of agentsai andaj at(Tg, σ0), such that the other agent learnt this secret Ak in theaiaj call, and hence this call is not redundant. Since the call aiaj is chosen arbitrarily fromσ, and since σ is an arbitrary execution sequence of any of the three given protocols, then none of the execution sequences of the three protocols contain redundant calls, and therefore these three protocols are regular.

Lemma 6.10. Given an epistemic gossip scenario consisting ofnagents, and given any network topology, then in any successful execution sequence, there are at mostn−1fresh calls.

Proof. Consider any network of agents, and consider any successful execution sequence σ. Suppose, towards a contradiction, that there are more than n−1 fresh calls in σ, say, without loss of generality, that there aren fresh calls in σ. Since each of the fresh

calls must contain at least one fresh caller, and the first call inσ must contain exactly two fresh callers (since it is the first time that pair of agents make any call), it follows that there must be at least n+ 1 agents in the scenario, which then contradicts our assumption that there arenagents in the scenario.

Finally to show that n−1 fresh calls are possible inσ, consider a case, given below, where a designated agent a1 calls each of all the other agents in turn at the beginning

of σ:

σ=a1a2;. . .;a1an;. . .

It is clear in the given exampleσ, that the first n−1calls are the fresh calls.

Lemma 6.11. For a gossip scenario, given any regular and successful execution sequence

σ, and any agentai ∈Ag, then exactly n−1 calls are required in σ to spread the secret of agent ai to all the other agents.

Proof. Prior to the first call thatai makes inσ, there aren−1agents that do not know the secret of agentai. Any call in σ in which the secret of ai is learnt for the first time must be between an agent who knows the secret of ai and another agent who does not know the secret of agent ai. Thus for the n−1 agents who did not initially know the secret of ai, we need n−1 calls to inform them all of the secret ofai.

Proposition 6.12. In a gossip scenario ofnagents, the maximum length of any regular and successful execution sequence isn(n−1)/2.

Proof. Let f : N → N be a function that returns the maximum length of any regular

and successful execution sequence of calls given the number n of agents in the gossip scenario. Then:

f(1) = 0; (no calls are made) f(2) = 1; (only one call is made)

Forn= 3, any two agents have to call each other in the first call; in the second call, for some agent to learn some new secret, one of the agents who called in the first call has to call with the other agent who did not take part in the first call; in the third call, for some agent to learn some new secret, one of the callers in the second call have to call with the agent who was not in the second call. After this third call, all three agents know all the secrets in the scenario, and so no further calls are required. So exactly three calls are made in any regular and successful execution sequence. Therefore forn= 3:

f(3) = 3;

Now suppose thatf(n−1)is the maximum length of a regular and successful execution sequence in a gossip scenario with n−1 agents, and there is an execution sequence σ consisting of f(n−1) calls after which all the n−1 agents know each other’s secrets. Suppose that we introduce a new agent an into such a scenario after executing σ, then

by Lemma 6.11, exactly n−1 additional calls are required to spread the secret ofanto all the other agents in the scenario, given that some agent must learn some new secret in each call. Note that the first call that followsσ must be a call involvingan, otherwise the call will be redundant. Note also that in that first call after σ, agent an learns the secret of every other agent. Furthermore, from Lemma 6.11, for each of the initialn−1

agents,n−1 calls are required for all the other agents to know their secret. But afterσ, those initialn−1agents know all of each other’s secrets, so again, from Lemma 6.11, for each agent a0 in the initialn−1 agents, σ must contain exactly n−2 calls in which an agent learnt the secret ofa0 for the first time. So, at the first call followingσ in whichan learnt the secret of all the initialn−1 agents, we obtain the total ofn−1 calls required by Lemma 6.11 for the secret of each of the initialn−1agents to be learnt by the other agents in the scenario. From the foregoing we see therefore that f(n) is given by the recursion:

f(n) =f(n−1) + (n−1) (6.10)

To solve Equation 6.10 we proceed as follows. f(n) = f(n−1) + (n−1) = f(n−2) + (n−2) + (n−1) .. . ... = f(n−k) + (n−k) + (n−(k−1)) + (n−(k−2)) +· · ·+ (n−1) .. . ... = f(n−(n−1)) + (n−(n−1)) + (n−(n−2)) +· · ·+ (n−1) = f(1) + 1 + 2 +· · ·+ (n−1) = f(1) + n−1 P i=1 i = 0 +n(n₂−1) = n(n₂−1)

Proposition 6.13. The maximum length of any execution sequence of a regular and successful epistemic gossip protocol is n(n−1)/2.

Proof. From Definitions 4.19, 4.20 and 6.3, any execution sequence of a regular and successful epistemic gossip protocol is also regular and successful. Therefore the proof of the proposition follows from Proposition 6.12.

Let us now continue with properties and proofs regarding the complete topology network, and then go on to those for the line, tree and circle topology networks, in turn.