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Knowledge Based / Artificial Intelligence Approaches

5.2 Rapid Simulation Model Development: Current Status

5.2.3 Knowledge Based / Artificial Intelligence Approaches

A token-forwarding algorithm for solving the gossip problem is an algorithm that does not manip-ulate the tokens in any way except storing and forwarding them. Specifically, the algorithm must satisfy the following conditions. Let𝑠𝐺𝑒(π‘Ÿ)denote the message broadcast by node 𝑒in round π‘Ÿ, when the algorithm is executed in dynamic graph𝐺= (𝑉, 𝐸).

1. 𝑠𝐺𝑒(π‘Ÿ)∈ 𝒯 βˆͺ {βŠ₯}for all roundπ‘Ÿand nodes𝑒.

2. Nodes can only learn new tokens by receiving them, either in their input or in a message from another node. Formally, let𝑅𝐺𝑒(π‘Ÿ) :={

𝑠𝐺𝑣(π‘Ÿ)∣ {𝑒, 𝑣} ∈𝐸(π‘Ÿ)}

denote the set of messages 𝑒receives in roundπ‘Ÿ, and let

𝐴𝐺𝑒(π‘Ÿ) :=𝐼(𝑒)βˆͺ (π‘Ÿβˆ’1

βˆͺ

π‘Ÿβ€²=0

𝑅𝐺𝑒(π‘Ÿβ€²) )

. We require the following.

βˆ™ 𝑠𝐺𝑒(π‘Ÿ)βˆˆπ΄πΊπ‘’(π‘Ÿ)βˆͺ {βŠ₯}for all nodes𝑒and roundsπ‘Ÿ, and

βˆ™ If node𝑒terminates in roundπ‘Ÿ, then𝐴𝐺𝑒(π‘Ÿ) =𝐼.

We omit the superscript𝐺when it is obvious from the context.

6.1 Ξ©(𝒏logπ’Œ) Lower Bound for Centralized π’Œ-Gossip in 1-Interval Connected Graphs

For this lower bound we assume that in each roundπ‘Ÿ, some central authority provides each node𝑒 with a value𝑑𝑒(π‘Ÿ)βˆˆπ΄π‘’(π‘Ÿ)to broadcast in that round. The centralized algorithm can see the state and history of the entire network, but it does not know which edges will be scheduled in the current round. Centralized algorithms are more powerful than distributed ones, since they have access to more information. To simplify, we begin with each of theπ‘˜tokens known to exactly one node.

This restriction is not essential. The lower bound holds as long as there is constant fraction of the nodes that still need to learnπ‘˜π›Ώtokens for some positive constant𝛿.

We observe that while the nodes only know a small number of tokens, it is easy for the algorithm to make progress; for example, in the first round of the algorithm at leastπ‘˜nodes learn a new token, because connectivity guarantees thatπ‘˜nodes receive a token that was not in their input. As nodes learn more tokens, it becomes harder for the algorithm to provide them with tokens they do not already know. Accordingly, our strategy is to charge a cost of1/(π‘˜βˆ’π‘–)for the𝑖-th token learned by each node: the first token each node learns comes at a cheap1/π‘˜, and the last token learned costs dearly (1). Formally, the potential of the system in roundπ‘Ÿ is given by

Ξ¦(π‘Ÿ) :=βˆ‘

π‘’βˆˆπ‘‰

βˆ£π΄π‘’(π‘Ÿ)βˆ£βˆ’1

βˆ‘

𝑖=0

1 π‘˜βˆ’π‘–.

In the first round we haveΞ¦(0) = 1, becauseπ‘˜nodes know one token each. If the algorithm terminates in roundπ‘Ÿ then we must haveΞ¦(π‘Ÿ) =π‘›β‹…π»π‘˜ = Θ(𝑛logπ‘˜), because all𝑛nodes must know all π‘˜ tokens. We construct an execution in which the potential increase is bounded by a constant in every round; this gives us anΞ©(𝑛logπ‘˜)bound on the number of rounds required.

Theorem 6.1. Any centralized algorithm for π‘˜-gossip in 1-interval connected graphs requires Ξ©(𝑛logπ‘˜)rounds to complete in the worst case.

Proof. We construct the communication graph for each roundπ‘Ÿin three stages.

Stage I: Adding the free edges. An edge{𝑒, 𝑣}is said to be free if𝑑𝑒(π‘Ÿ) βˆˆπ΄π‘£(π‘Ÿ)and 𝑑𝑣(π‘Ÿ)∈ 𝐴𝑒(π‘Ÿ); that is, if we connect 𝑒and 𝑣, neither node learns anything new. Let𝐹(π‘Ÿ) denote the set of free edges in roundπ‘Ÿ; we add all of them to the graph. Let 𝐢1, . . . , 𝐢ℓ denote the connected components of the graph(𝑉, 𝐹(π‘Ÿ)). Observe that any two nodes𝑒and𝑣in different components must send different values, otherwise we would clearly have𝑑𝑒(π‘Ÿ)βˆˆπ΄π‘£(π‘Ÿ)and𝑑𝑣(π‘Ÿ)βˆˆπ΄π‘’(π‘Ÿ)and 𝑒and𝑣would be in the same component.

We choose representatives 𝑣1 ∈ 𝐢1, . . . , 𝑣ℓ ∈ 𝐢ℓ from each component arbitrarily. Our task 𝑣 , . . . , 𝑣

thatβ„“β‰₯6, otherwise we can connect the nodes arbitrarily for a constant cost. Letmissing(𝑒) :=

π‘˜βˆ’ βˆ£π΄π‘’(π‘Ÿ)∣denote the number of tokens node𝑒does not know at the beginning of roundπ‘Ÿ.

Stage II: We split the nodes into two sets Top, Bottom according to the number of tokens they know, with nodes that know many tokens β€œon top”: Top := {π‘£π‘–βˆ£missing(𝑣𝑖)≀ℓ/6} and consequentlyBottom :={π‘£π‘–βˆ£missing(𝑣𝑖)> β„“/6}.

Since top nodes know many tokens, connecting to them could be expensive. We will choose our edges in such a way that no top node will learn a new token, and each bottom node will learn at most three new tokens. We begin by bounding the size ofTop.

To that end, notice thatβˆ‘

π‘’βˆˆTopmissing(𝑒)β‰₯(∣Top∣

2

): for all𝑖, 𝑗such that𝑒, 𝑣 ∈Top, either 𝑑𝑒(π‘Ÿ) βˆ•βˆˆ 𝐴𝑣(π‘Ÿ) or 𝑑𝑣(π‘Ÿ) βˆ•βˆˆ 𝐴𝑒(π‘Ÿ), otherwise {𝑒, 𝑣} would be a free edge and 𝑒, 𝑣 would be in the same component; therefore each pair𝑒, 𝑣 ∈ Top contributes at least one missing token to the sum. On the other hand, since each node in Top is missing at mostβ„“/6 tokens, it follows that

βˆ‘

π‘’βˆˆTopmissing(𝑒) ≀ ∣Top∣ β‹…(β„“/6). Putting the two facts together we obtain∣Top∣ ≀ℓ/3 + 1, and consequently also∣Bottom∣=β„“βˆ’ ∣Top∣ β‰₯2β„“/3βˆ’1.

Stage III: Connecting the nodes. The bottom nodes are relatively cheap to connect to, so we connect them in an arbitrary line. In addition we want to connect each top node to a bottom node, such that no top node learns something new, and no bottom node is connected to more than one top node (see Fig. 1. That is, we are looking for a matching using only the edges 𝑃 :=

{{𝑒, 𝑣} βˆ£π‘’βˆˆTop, 𝑣 ∈Bottom and𝑑𝑣 βˆˆπ΄π‘’(π‘Ÿ)}.

Since each top node is missing at mostβ„“/6tokens, and each bottom node broadcasts a different value, for each top node there are at least∣Bottom∣ βˆ’β„“/6edges in𝑃 to choose from. But since we assumeβ„“β‰₯6,∣Top∣ ≀ℓ/3 + 1≀ ∣Bottom∣ βˆ’β„“/6; thus, each top node can be connected to a different bottom node using𝑃-edges.

What is the total cost of the graph? Top nodes learn no tokens, and bottom nodes learn at most two tokens from other bottom nodes and at most one token from a top node. Thus, the total cost is bounded by

βˆ‘

π‘’βˆˆBottom

min{3,missing(𝑒)}

βˆ‘

𝑖=1

1

missing(𝑒)βˆ’(π‘–βˆ’1) ≀ ∣Bottom∣ β‹… 6

β„“ 6

≀ℓ⋅36 β„“ = 36.

6.2 Ξ©(𝒏 + 𝒏2/𝑻) lower bound against knowledge-based token-forwarding algo-rithms

In this section we describe a lower bound against a restricted class of randomized token-forwarding algorithms. We represent randomness as a random binary string provided to each node at the beginning of the execution. In every round, the nodes may consume a finite number of random bits, and use them to determine their message for that round and their next state. In every execution nodes only use finitely many coin tosses; we use an infinite string when modelling the algorithm in order to avoid

𝑣𝑖1 𝑣𝑖2 𝑣𝑖3 𝑣𝑖4

𝑣𝑖5 𝑣𝑖6 𝑣𝑖7 𝑣𝑖8 𝑣𝑖1 𝑣𝑖2

missing≀ℓ/6tokens missing> β„“/6tokens Top

Bottom

Figure 1: Illustration for the proof of theΞ©(𝑛logπ‘˜)lower bound

A token-forwarding algorithm is said to be knowledge-based if it can be represented as a col-lection of functions{π‘“π‘’βˆ£π‘’βˆˆ 𝒰} βŠ† 𝒫(𝒯)βˆ—Γ— {0,1}βˆ— β†’ 𝐷(𝒯), such that in every round π‘Ÿ, if𝑅 is the sequence of coin-tosses for node𝑒 up to roundπ‘Ÿ (inclusive), the distribution according to which node𝑒decides which token to broadcast is given by𝑓𝑒(𝐴𝑒(0). . . , 𝐴𝑒(π‘Ÿ), 𝑅).

We say that two dynamic graphs 𝐺 = (𝑉, 𝐸) and 𝐺′ = (𝑉′, 𝐸′) are equal up to roundπ‘Ÿ if 𝑉 =𝑉′and for allπ‘Ÿβ€² < π‘Ÿwe have𝐸(π‘Ÿβ€²) =𝐸′(π‘Ÿβ€²). Let𝐷𝑒(π‘Ÿ)denote the probability distribution for node𝑒in roundπ‘Ÿ. Knowledge-based algorithms have the following property.

Lemma 6.2. Let𝐺, 𝐺′ be two dynamic graphs that are equal up to roundπ‘Ÿ, and let(𝑉, 𝐼)be an instance of gossip. If 𝑒 is a node such that𝐴𝐺𝑒(π‘Ÿ) = 𝐼, then for any round π‘Ÿβ€² β‰₯ 0 and string π‘…βˆˆ {0,1}πœ”we have𝐷𝑒𝐺(π‘Ÿβ€², 𝑅) =𝐷𝑒𝐺′(π‘Ÿβ€², 𝑅).

Proof. Since𝐺and𝐺′are equal up to roundπ‘Ÿ, the sequences𝐴𝐺𝑒(0). . . 𝐴𝐺𝑒(π‘Ÿ)and𝐴𝐺𝑒′(0). . . 𝐴𝐺𝑒′(π‘Ÿ) are equal, and in particular𝐴𝐺𝑒(π‘Ÿ) =𝐴𝐺𝑒′(π‘Ÿ) =𝐼.

By definition, for all π‘Ÿβ€² β‰₯ π‘Ÿ we have 𝐴𝐺𝑒(π‘Ÿ) βŠ† 𝐴𝐺𝑒(π‘Ÿβ€²) and 𝐴𝐺𝑒′(π‘Ÿ) βŠ† 𝐴𝐺𝑒′(π‘Ÿβ€²); there-fore, 𝐴𝐺𝑒(π‘Ÿβ€²) = 𝐴𝐺𝑒′(π‘Ÿβ€²) = 𝐼 for all π‘Ÿβ€² β‰₯ π‘Ÿ. Consequently, for all π‘Ÿβ€² β‰₯ 0, the sequences 𝐴𝐺𝑒(0). . . 𝐴𝐺𝑒(π‘Ÿβ€²)and𝐴𝐺𝑒′(0). . . 𝐴𝐺𝑒′(π‘Ÿβ€²)are equal, and the claim follows.

Theorem 6.3. Any knowledge-based token-forwarding algorithm forπ‘˜-input gossip in𝑇-interval connected graphs over𝑛nodes requiresΞ©(𝑛+π‘›π‘˜/𝑇)rounds to succeed with probability at least 1/2. Further, ifβˆ£π’°βˆ£ = Ξ©(𝑛2π‘˜/𝑇), then for sufficiently large𝑛, deterministic algorithms require Ξ©(𝑛+π‘›π‘˜/𝑇)rounds even when each node begins with at most one token.

Proof. A lower bound ofΞ©(𝑛)is demonstrated trivially in a static line network where at least one π‘˜ >1.

Let{𝑓𝑒} be an knowledge-based token-forwarding algorithm for π‘˜-gossip. We use the UID space as the token domain, and choose nodes𝑒1, . . . , 𝑒𝑛: for randomized algorithms we choose the UIDs arbitrarily, but for deterministic algorithms we must choose them carefully (see the last part of the proof). If the algorithm is randomized, we choose an input assignment where some node𝑒1starts with allπ‘˜tokens, and all other nodes𝑒𝑖 βˆ•=𝑒1 start with a set𝐼(𝑒𝑖) βŠ† {𝑒1, 𝑒𝑖}. For deterministic algorithms, we later show that we can reach this state from some input assignment where each node starts with at most one token. For now let us suppose that we have reached some roundπ‘Ÿ0 in which𝐴𝑒1(π‘Ÿ0) =𝐼 and for all𝑒𝑖 βˆ•=𝑒1we have𝐴𝑒𝑖 βŠ† {𝑒1, 𝑒𝑖}. In this starting state there areπ‘›βˆ’2nodes that do not know each tokenπ‘‘βˆ•=𝑒1. We abuse notation by using𝐼to denote the set of all tokens𝑒1, . . . , π‘’π‘˜as well as the input assignment𝐼(𝑒𝑖)to each node𝑒𝑖.

Letπ‘Ÿ1 :=π‘Ÿ0+ (π‘›βˆ’2)(π‘˜βˆ’2)/(4𝑇). For a tokenπ‘‘βˆˆπΌ, letE [#𝑑]denote the expected number of times token𝑑is broadcast by𝑒between roundsπ‘Ÿ0andπ‘Ÿ1(exclusive). We have

βˆ‘

π‘‘βˆˆπΌ

E [#𝑑] =βˆ‘

π‘‘βˆˆπΌ π‘Ÿ1βˆ’1

βˆ‘

π‘Ÿ=π‘Ÿ0+1

Pr [𝑑is broadcast in roundπ‘Ÿ] =π‘Ÿ1βˆ’π‘Ÿ0βˆ’2<(π‘›βˆ’2)(π‘˜βˆ’2)/(4𝑇).

Thus, there are at least two tokens 𝑑 βˆ•= 𝑑′ such that E [#𝑑],E [#𝑑′] < (π‘›βˆ’2)/(4𝑇). Assume w.l.o.g. thatπ‘‘βˆ•=𝑒1. From Markov’s inequality, node𝑒1broadcasts𝑑less than(π‘›βˆ’2)/(2𝑇)times with probability at least1/2in any execution fragment starting from roundπ‘Ÿ0 and ending before round π‘Ÿ1, regardless of the dynamic graph we choose. The idea in the proof is to use 𝑒1 as a buffer between the nodes that have already learned𝑑and those that have not; since𝑒1broadcasts 𝑑 infrequently with high probability, in this manner we can limit the number of nodes that learn𝑑.

We divide the rounds betweenπ‘Ÿ0 andπ‘Ÿ1 into segments 𝛼1, . . . , π›Όπ‘š. The graph remains static during each segment, but changes between segments. For each segment𝛼𝑖we define two sets of nodes,𝐢𝑖 and𝐷𝑖, whereπΆπ‘–βˆ©π·π‘– ={𝑒1}. The nodes in𝐷𝑖 are β€œcontaminated nodes” that might know token𝑑at the beginning of the segment; we connect them in a clique. The nodes in𝐢𝑖are

β€œclean”: initially, except for𝑒1, these nodes do not know𝑑 (some of them might learn𝑑 during the segment). The only way the nodes in𝐢𝑖 can learn𝑑is if𝑒1 broadcasts it. In the first segment 𝐢𝑖 is arranged in a line with𝑒1 at one end; in subsequent segments we β€œclose”𝐢𝑖to form a ring.

Initially𝐷1 ={𝑒1, 𝑑}and𝐢1 =𝑉 βˆ– {𝑑}(recall that𝑑, in addition to being a token, is also the UID of a node).

There are two types of segments in our construction.

βˆ™ Quiet segments are ones in which𝑒1does not broadcast𝑑until the last round in the segment.

In the last round of a quiet segment, 𝑒1 broadcasts 𝑑, and some nodes in the ring become contaminated. The first segment𝛼1is a quiet segment.

βˆ™ After every quiet segment there follows one or more active segments, in which we clean up the ring and move contaminated nodes from 𝐢𝑖 to 𝐷𝑖. We have to do this in a way that preserves 𝑇-interval connectivity. Each active segment is triggered by𝑒1 broadcasting𝑑in the previous segment; if in some active segment𝑒1 does not broadcast𝑑, the next segment will be quiet.

An active segment lasts exactly𝑇rounds, and a quiet segment lasts until the first time𝑒1broadcasts 𝑑(including that round).

Next we define in detail the construction of the communication graph in each segment. We maintain the following property:

(β˜…) At the beginning of each active segment𝛼𝑖, of all the nodes in𝐢𝑖, only𝑒1 and at most 𝑇 nodes in the𝑇-neighborhood of𝑒1in the ring know token𝑑. Further, all the nodes that know 𝑑are on the same side of𝑒1. We refer to the side of𝑒1where these nodes are located as the contaminated side of𝑒1.

(β˜…β˜…) At the beginning of each quiet segment𝛼𝑖, node𝑒1is the only node in the ring that knows token𝑑.

Let𝑣1, . . . , π‘£π‘›βˆ’2be some ordering of the nodes in𝐢1βˆ– {𝑒1}(nodes that initially do not know 𝑑). In each segment𝑖the nodes in𝐢𝑖 will be some contiguous subset𝑣𝐿𝑖, . . . , 𝑣𝑅𝑖, where𝐿𝑖+1 β‰₯ 𝐿𝑖 β‰₯1and𝑅𝑖+1 ≀𝑅𝑖 β‰€π‘›βˆ’2for all𝑖. We place𝑒1between𝑣𝐿𝑖 and𝑣𝑅𝑖 in the ring. Formally, the edges in any roundπ‘Ÿβˆˆπ›Όπ‘–where𝑖 >1are given by

𝐸(π‘Ÿ) :=𝐷(2)𝑖 βˆͺ {{𝑣𝑗, 𝑣𝑗+1} βˆ£πΏπ‘– ≀𝑗 < 𝑅𝑖} βˆͺ {{𝑒1, 𝑣𝐿𝑖},{𝑒1, 𝑣𝑅𝑖}}.

In the first segment, the edges are𝐸(π‘Ÿ) :=𝐷1(2)βˆͺ {{𝑣𝑗, 𝑣𝑗1} ∣1≀𝑗 < π‘›βˆ’2} βˆͺ {{𝑒1, 𝑣1}}(we do not close the ring; this is to ensure that (β˜…) holds for the first active segment).

If𝛼𝑖 is a quiet segment, then we define𝐢𝑖+1 := 𝐢𝑖 (and consequently 𝐷𝑖+1 := 𝐷𝑖); that is, the network does not change between𝛼𝑖and𝛼𝑖+1(except possibly for the closing of the ring after the first segment). However, if𝛼𝑖 is an active session, then𝑒1 has some neighbors in the ring that knows𝑑, and they might spread𝑑to other nodes even when𝑒1does not broadcast𝑑. We divide the nodes inπΆπ‘–βˆ– {𝑒1}into three subsets.

βˆ™ The red nodesred𝑖comprise the2𝑇 nodes adjacent to𝑒1on the contaminated side. The first 𝑇 of these (the ones closer to𝑒1) may know𝑑at the beginning of the segment; the other𝑇 may become contaminated if some of the first𝑇 broadcast token𝑑. To be safe, we treat all red nodes as though they know𝑑by the end of the session.

βˆ™ The yellow nodes yellow𝑖 comprise the𝑇 nodes adjacent to𝑒1on the uncontaminated side.

These nodes may learn𝑑during the segment, but only if𝑒1broadcasts it.

βˆ™ The green nodesgreen𝑖are all the other nodes in the ring. These nodes cannot become con-taminated during the segment, because their distance from any node that knows𝑑is greater than𝑇.

Our cleanup between segments𝛼𝑖and 𝛼𝑖+1 consists of moving all the red nodes into𝐷𝑖+1. For-mally, if𝑣𝐿𝑖 ∈red𝑖, then we define𝑣𝐿𝑖+1 :=𝑣𝐿𝑖+ 2𝑇 and𝑣𝑅𝑖+1 :=𝑣𝑅𝑖; otherwise, if𝑣𝑅𝑖 ∈red𝑖, then we define𝑣𝑅𝑖+1 := 𝑣𝑅𝑖 + 2𝑇 and 𝑣𝐿𝑖+1 := 𝑣𝐿𝑖. This satisfies (β˜…) and (β˜…β˜…): if𝑒1 does not broadcast 𝑑during segment 𝛼𝑖, then only the red nodes can know 𝑑at the end, and since we re-moved them from the ring, at the beginning of𝛼𝑖+1no node knows𝑑except𝑒1. The next segment will be quiet. Otherwise, if𝑒1does broadcast𝑑during𝛼𝑖, then at the beginning of the next session (which is active) only the yellow nodesyellow𝑖can know𝑑. These nodes then become red nodes in

𝛼 𝑇

The cleanup step preserves𝑇-interval connectivity: assume thatred𝑖 ={𝑣𝐿𝑖, . . . , 𝑣𝐿𝑖+2𝑇}(the other case is similar). Then the line𝑣𝐿𝑖+2𝑇, 𝑣𝐿𝑖+2π‘‡βˆ’1, . . . , 𝑒1, 𝑣𝑅𝑖, 𝑣𝑅𝑖+1, . . . , 𝑣𝐿𝑖+2π‘‡βˆ’1 exists throughout both segment𝛼𝑖 and segment𝛼𝑖+1: in segment 𝛼𝑖 it exists as part of the ring, and in segment 𝛼𝑖+1, after we moved the red nodes into the clique 𝐷𝑖+1, the first part of the line 𝑣𝐿𝑖+2𝑇, 𝑣𝐿𝑖+2π‘‡βˆ’1, . . . , 𝑒1 exists in the clique and the second part 𝑒1, 𝑣𝑅𝑖, 𝑣𝑅𝑖+1, . . . , 𝑣𝐿𝑖+2π‘‡βˆ’1 exists in the ring. The nodes in𝐷𝑖are all connected to each other in both segments; thus, there is a static connected graph that persists throughout both segments𝛼𝑖, 𝛼𝑖+1, and in particular it exists in any𝑇rounds that start in𝛼𝑖. (Note that𝛼𝑖+1may be quiet, and in this case it can be shorter than𝑇 rounds. But in this case it will be followed by an active segment which has exactly the same edges and lasts𝑇 rounds.)

Notice that the number of uncontaminated nodes at the beginning of every active segment is at most2𝑇 less than in the previous active session. Therefore the total number of nodes that know𝑑 by roundπ‘Ÿ1 is at most2𝑇 times the number of active sessions, and this in turn is bounded by2𝑇 times the number of rounds in which𝑒1broadcasts𝑑. Since𝑒1broadcasts𝑑less than(π‘›βˆ’2)/(2𝑇) times with probability at least1/2, the algorithm is not finished by round π‘Ÿ1 with probability at least1/2.

Deterministic algorithms. If the algorithm is deterministic, we first show that there exists an input assignment in which each node begins with at most one token, from which either

1. the algorithm runs forΞ©(π‘›π‘˜/𝑇)rounds, or

2. we reach a round π‘Ÿ0 in which some node 𝑒1 has𝐴𝑒1(π‘Ÿ0) = 𝐼 and for all 𝑖 βˆ•= 1 we have 𝐴𝑒𝑖(π‘Ÿ0)βŠ† {𝑒1, 𝑒𝑖}.

In the case of (2), we then continue with the same proof as for the input assignment where some node starts with all tokens and the rest of the nodes have no tokens (see above). Since we are free to choose the input assignment, we restrict attention to instances in which the inputs toπ‘˜nodes are their own UIDs, and the inputs to the other tokens areβˆ….

For deterministic algorithms the function𝑓𝑒representing node𝑒’s behavior must return a distri-bution in which one token has probability 1. We abuse notation slightly by using𝑓𝑒(𝐴𝑒(0). . . , 𝐴𝑒(π‘Ÿβˆ’

1))to denote this token.

We say that a process 𝑒 ∈ 𝒰 fires in roundπ‘Ÿ if when process𝑒 receives{𝑒}as its input and hears nothing in the firstπ‘Ÿβˆ’1rounds, it will stay silent in those rounds and then spontaneously broadcast its token in roundπ‘Ÿ. Formally, process𝑒fires in roundπ‘Ÿif

1. For allπ‘Ÿβ€²< π‘Ÿwe have𝑓𝑒({βŠ₯}π‘Ÿβ€²) =βŠ₯, and 2. 𝑓𝑒({𝑒}π‘Ÿ) =𝑒.

If process𝑒 does not fire in any roundπ‘Ÿβ€² ≀ π‘Ÿ, we say that 𝑒is passive until roundπ‘Ÿ. (Note that nodes that receive no tokens in their input have no choice but to broadcast nothing until they receive a token from someone.)

Since βˆ£π’°βˆ£ = Ξ©(𝑛2π‘˜/𝑇), there exist constants 𝑐, 𝑛0 such that for all𝑛 β‰₯ 𝑛0 we have βˆ£π’°βˆ£ β‰₯ 𝑐𝑛2π‘˜+π‘›βˆ’1. Let𝑛β‰₯𝑛0. We divide into two cases.

Case I. There exist 𝑒1, . . . , 𝑒𝑛 ∈ 𝒰 that are all passive until round π‘π‘›π‘˜/𝑇. In this case we construct the static clique over 𝑒1, . . . , 𝑒𝑛 and let the algorithm run. During the first π‘π‘›π‘˜/𝑇 rounds, all nodes send onlyβŠ₯, and no node learns new tokens. Consequently all nodes𝑒𝑖 have 𝐴𝑒𝑖(π‘›π‘˜/𝑇) =in(𝑒𝑖)βˆ•=𝐼, and the algorithm cannot terminate by roundπ‘π‘›π‘˜/𝑇.

Case II. All butπ‘›βˆ’1processes fire no later than roundπ‘π‘›π‘˜/𝑇.

Sinceβˆ£π’°βˆ£ β‰₯𝑐(𝑛2π‘˜/𝑇+π‘›βˆ’1), by the pigeonhole principle there must exist a roundπ‘Ÿ0β‰€π‘π‘›π‘˜/𝑇 such that at least𝑛processes fire in roundπ‘Ÿ0. Let𝑒1, . . . , 𝑒𝑛be𝑛such processes. We choose the instance where each node𝑒𝑖receives as input{𝑒𝑖}ifπ‘–β‰€π‘˜, orβˆ…if𝑖 > π‘˜.

Let𝑆be the static star with𝑒1at the center: 𝑆 = (𝑉, 𝐸𝑆), where𝐸𝑆(π‘Ÿ) ={{𝑒1, 𝑒𝑖} βˆ£π‘– >1}

for allπ‘Ÿ. Because all nodes fire in roundπ‘Ÿ0, when the algorithm is executed in𝑆, the network is silent until roundπ‘Ÿ0. In roundπ‘Ÿ0all nodes that have a token broadcast it. Following roundπ‘Ÿ0we have𝐴𝑒1(π‘Ÿ0+ 1) =𝐼, and for all𝑖 >1,𝐴𝑒𝑖(π‘Ÿ0+ 1) =𝐼(𝑒𝑖)βˆͺ {𝑒1} βŠ† {𝑒1, 𝑒𝑖}. This is the state from which we start the main body of the proof above.

𝑑 𝑒1 𝑣1 𝑣2 π‘£π‘›βˆ’3 π‘£π‘›βˆ’2

(a) The network at the beginning of the execution. Nodes that may know token𝑑are indicated in solid blue.

𝑑

𝑣2𝑇+1 𝑣2𝑇 𝑣2π‘‡βˆ’1

𝑣2 𝑣𝑅=𝑣1

𝑒1 𝑣𝐿=π‘£π‘›βˆ’1

π‘£π‘›βˆ’π‘‡

(b) The network at the beginning of the first phase: the line is closed to form a ring. The dotted line indicates the edge we will add at the end of the phase to re-close the ring after we remove the red nodes; double lines indicate stable edges, along which𝑇-interval connec-tivity is preserved between phases.

𝑒1 𝑣3𝑇 𝑣2𝑇+2

𝑣2𝑇+1

π‘£π‘›βˆ’1

π‘£π‘›βˆ’π‘‡

(c) The network after the end of the first phase: the red nodes are removed from the ring and placed in the clique, and the ring is repaired by connecting𝑒1 to𝑣2𝑇+1. Double lines indicate stable edges along which𝑇-interval connectivity was preserved in the transition between the phases.

𝑒1 𝑣3𝑇 𝑣2𝑇+2

𝑣2𝑇+1

π‘£π‘›βˆ’1

π‘£π‘›βˆ’π‘‡

(d) If𝑒1 broadcast𝑑at any point during the first phase, we begin a new phase. The nodes that were yellow in the first phase become red, and the β€œclean” nodes on𝑒1’s other side become yel-low. Double lines indicate edges that will be stable through the next two phases.

Figure 2: Illustrations for the proof of theΞ©(𝑛+π‘›π‘˜/𝑇)lower bound,𝑇 = 3