Time Complexity of Broadcast and Consensus for Randomized Oblivious Message Adversaries

TL;DR

This paper studies broadcast and consensus in stochastic dynamic networks, showing that these tasks can be completed rapidly with high probability when communication occurs along random rooted trees, contrasting with worst-case models.

Contribution

It introduces a randomized oblivious message adversary model and proves fast completion times for broadcast and consensus in such stochastic networks.

Findings

Broadcast completes in O(k + log n) rounds.

Consensus completes in O(k + log n) rounds for k ≤ 0.1n.

Random rooted trees enable rapid information dissemination.

Abstract

Broadcast and consensus are most fundamental tasks in distributed computing. These tasks are particularly challenging in dynamic networks where communication across the network links may be unreliable, e.g., due to mobility or failures. Indeed, over the last years, researchers have derived several impossibility results and high time complexity lower bounds (i.e., linear in the number of nodes $n$) for these tasks, even for oblivious message adversaries where communication networks are rooted trees. However, such deterministic adversarial models may be overly conservative, as many processes in real-world settings are stochastic in nature rather than worst case. This paper initiates the study of broadcast and consensus on stochastic dynamic networks, introducing a randomized oblivious message adversary. Our model is reminiscent of the SI model in epidemics, however, revolving around trees (which renders the analysis harder due to the apparent lack of independence). In particular, we show that if information dissemination occurs along random rooted trees, broadcast and consensus complete fast with high probability, namely in logarithmic time. Our analysis proves the independence of a key variable, which enables a formal understanding of the dissemination process. More formally, for a network with $n$ nodes, we first consider the completely random case where in each round the communication network is chosen uniformly at random among rooted trees. We then introduce the notion of randomized oblivious message adversary, where in each round, an adversary can choose $k$ edges to appear in the communication network, and then a rooted tree is chosen uniformly at random among the set of all rooted trees that include these edges. We show that broadcast completes in $O(k+\log n)$ rounds, and that this it is also the case for consensus as long as $k \le 0.1n$.

Equations

$$\left(\prod _{i\in [m]}{f}_i\right)n^{n-2-|E|}$$

$$\pi :{\mathcal{T}}_n$$

$$T$$

$$\alpha :G\times X$$

$${\sigma }_i^k:\mathbb{Z}/k\mathbb{Z}$$

$$x$$

$$\left|G.x\right|=\frac{\left|G\right|}{\left|G_x\right|}$$

$$\mathbb{P}(N_{t+1}-N_t=x)=\sum _{J\in A(S_t,x)}\mathbb{P}\left(\bigcap _{y\in J}(P_{t+1}(y)\in I_t)\bigcap _{y\in S_t\setminus J}(P_{t+1}(y)\notin I_t)\right)$$

$$\mathbb{P}\left(\bigcap _{y\in J}(P_{t+1}(y)\in I_t)\right)$$

$$=\sum _{a\in [N_t{]}^{|J|}}\frac{\left|\{T\in {\mathcal{T}}_n:P_T(y)=i_{a_y},\forall y\in J\}\right|}{\left|{\mathcal{T}}_n\right|}$$

$$\mathbb{P}\left(\bigcap _{y\in J}(P_{t+1}(y)\in I_t)\right)=\sum _{a\in [N_t{]}^{|J|}}\frac{n^{n-1-|J|}}{n^{n-1}}={\left(\frac{N_t}{n}\right)}^{|J|}$$

$$\mathbb{P}(N_{t+1}-N_t=x)$$

$$=\left(\genfrac{}{}{0ex}{}{n-N_t}{x}\right){\left(\frac{N_t}{n}\right)}^x{\left(1-\frac{N_t}{n}\right)}^{n-N_t-x}$$

$$X_0$$

$$X_{t+1}$$

$$X_{u_{t+1}}$$

$$=n-n{\left(\frac{n-1}{n}\right)}^{2^t}+{\left(\frac{n-1}{n}\right)}^{2^t}\left(n-n{\left(\frac{n-1}{n}\right)}^{2^t}\right)$$

$$=n-n{\left(\frac{n-1}{n}\right)}^{2^{t+1}}$$

$$X_{u_{2\ln n}}$$

$$\ge n-n\exp \left(n^{2\ln 2}\left(\frac{n-1}{n}-1\right)\right)=n-n\exp \left(-n^{2\ln 2-1}\right)>n-1$$

$$\mathbb{P}(B\ge mp)>\frac{1}{4}$$

$$N_t+1+\frac{1}{N_t-1}\le n-2+1+\frac{1}{n-3}<n$$

$$\mathbb{P}(Y\le x)\le \exp \left(-2t{\left(p-\frac{x}{t}\right)}^2\right)$$

$$\mathbb{P}(Z_{\le t}^B\le 2\ln n)\le \exp \left(-2t{\left(\frac{1}{4}-\frac{2\ln n}{t}\right)}^2\right)=\exp \left(-2\cdot 16\ln n{\left(\frac{1}{4}-\frac{2}{16}\right)}^2\right)=n^{-2}$$

$$\mathbb{E}[N_{t+1}|N_t]=N_t+\frac{N_t}{n}\cdot (n-N_t)=2N_t-\frac{N_t^2}{n}$$

$$\mathbb{E}[N_{t+1}]=\mathbb{E}\left[\mathbb{E}[N_{t+1}|N_t]\right]=2\mathbb{E}[N_t]-\frac{\mathbb{E}[N_t^2]}{n}$$

$$\mathbb{E}[N_{t+1}]\le 2\mathbb{E}[N_t]-\frac{\mathbb{E}[N_t{]}^2}{n}$$

$$X_{u_{t+1}}=2X_{u_t}-\frac{X_{u_t}^2}{n}$$

$$\mathbb{P}(N_{\log n}\ge n)\le \frac{\mathbb{E}[N_{\log n}]}{n}=1-\frac{1}{4}$$

$$\mathbb{P}\left(\bigcup _{i\in [n]}{N}_{16\ln n}^{(i)}<n\right)\le n^{-1}$$

$$\mathbb{P}\left(\bigcap _{i\in [n]}{N}_{16\ln n}^{(i)}=n\right)=1-\mathbb{P}\left(\bigcup _{i\in [n]}{N}_{16\ln n}^{(i)}<n\right)\ge 1-n^{-1}$$

$$\mathbb{P}(\widehat{Y_1}\ge \widehat{Y_2})=1$$

$$\mathbb{P}\left(\left|\widehat{Y_1}\right|\ge \left|\widehat{Y_2}\right|\right)=1$$

$$\mathbb{P}\left({\cap }_{j\in [x]}(P_t(s_j)\in I_{t-1})\right)=\frac{\eta (\eta +{\sum }_{j\in [x]}{\eta }_j{)}^{x-1}}{n^x}=\frac{\eta (N_{t-1}{)}^{x-1}}{n^x}$$

Full text

This paper initiates the study of broadcast and consensus on stochastic dynamic networks, introducing a randomized oblivious message adversary. Our model is reminiscent of the SI model in epidemics, however, revolving around trees (which renders the analysis harder due to the apparent lack of independence). In particular, we show that if information dissemination occurs along random rooted trees, broadcast and consensus complete fast with high probability, namely in logarithmic time. Our analysis proves the independence of a key variable, which enables a formal understanding of the dissemination process.

More formally, for a network with $n$ nodes, we first consider the completely random case where in each round the communication network is chosen uniformly at random among rooted trees. We then introduce the notion of randomized oblivious message adversary, where in each round, an adversary can choose $k$ edges to appear in the communication network, and then a rooted tree is chosen uniformly at random among the set of all rooted trees that include these edges. We show that broadcast completes in $O(k+\log n)$ rounds, and that this it is also the case for consensus as long as $k\leq 0.1n$.

1 Introduction

Broadcast and consensus are most fundamental operations in distributed computing which, in large-scale systems, typically have to be performed over a network. These networks are likely to be dynamic and change over time due, e.g., to link failures, interference, or mobility. Understanding how information disseminates across such dynamic networks is hence important for developing and analyzing efficient distributed systems.

Over the last years, researchers have derived several important insights into information dissemination in dynamic networks. A natural and popular model assumes an oblivious message adversary which controls the information flow between a set of $n$ nodes, by dropping an arbitrary set of messages sent by some nodes in each round. Specifically, the adversary is defined by a set of directed communication graphs, whose edges determine which node can successfully send a message to which other node in a given round. Concretely, based on this set of graphs, the oblivious message adversary chooses a sequence of graphs over time, one per round, in such a way that the time complexity of the information dissemination task at hand is maximized. This model is appealing because it is conceptually simple and still provides a highly dynamic network model: The set of allowed graphs can be arbitrary, and the nodes that can communicate with one another can vary greatly from one round to the next. It is, thus, well-suited for settings where significant transient message loss occurs, such as in wireless networks. As information dissemination is faster on dense networks, we focus in this paper on sparse networks, in particular, on rooted trees, similar to prior work on the oblivious message adversary [13, 28].

Unfortunately, information dissemination can be slow in trees: broadcast can take time linear in the number of nodes under the oblivious message adversary[13, 28], even for constant-height trees (see Appendix A); and consensus can even take super-polynomial time until termination, if it completes at all [6, 18]. While this is bad news, one may argue that while the deterministic adversary model is useful in malicious environments, in real-word applications, the dynamics of communication networks is often more stochastic in nature. Accordingly, the worst-case model considered in existing literature may be overly conservative.

This motivates us, in this paper, to study information dissemination, and in particular broadcast and consensus tasks, initially in the case where the communication network is purely stochastic: in each round, the communication network is chosen uniformly at random among all rooted trees. We then initiate the study of an extension of this model to a setting where an adversary has some limited control over the communication network, which we call the randomized oblivious message adversary. More specifically, we study the setting where first a worst-case adversary chooses $k$ directed edges in the dynamic $n$-node network for $0\leq k<n$, and then a rooted tree is chosen uniformly at random among the set of all rooted trees that include these edges. With our parameterized approach, we can get a smooth transition between the purely stochastic model ($k=0$) and the completely deterministic adversary ($k=n-1$) typically studied in prior work.

We show that under our randomized oblivious message adversary broadcast completes in $O(k+\log n)$ time with high probability. Note that for $k=O(\log n)$ this is an exponential improvement over the deterministic setting. Furthermore, we also show that consensus completes and is fast, with high probability: namely in $O(k+\log n)$ time for $k\leq 0.1n$, and it only requires messages of constant size along each edge in each round (only 1 bit).

It is useful to put our model into perspective with the SI model in epidemics [11]: while in the SI model interactions occur on a network that equals a clique, our model revolves around trees which are chosen by an adversary. This tree structure renders the analytical understanding of the information dissemination process harder, due to the lack of independence between the edges in the network in a particular round. A key insight from our paper is that we can prove the independence of a key variable, which is crucial for our analysis. Our proof further relies on stochastic dominance, which makes it robust to the specific adversarial objective, and applies to any adversary definition (e.g., whether it aims to maximize the minimal or expected number of rounds until the process completes).

Model.

In a first model, that we call the Uniformly Random Trees model, let $n$ be the number of nodes, and let each node have a unique identifier from $[n]$. Let $\mathcal{T}_{n}$ be the set of all directed rooted trees on $n$ vertices (where all edges are pointed away from the root). Time proceeds in a sequence of rounds $t=1,2,\dots,$ such that in each round $t$ a network is chosen uniformly at random from $\mathcal{T}_{n}$ independently from other rounds, and that network will be the communication network for the corresponding round. In each round, every node sends a message to all of its out-neighbors before receiving one from its in-neighbor. There is no message size restriction. In this setting, we will study broadcast, all-to-all broadcast and consensus.

Broadcast.

In the Broadcast on Uniformly Random Trees problem, we start by giving a message to one node, and broadcast completes when that node has forwarded this message to all other nodes. Each node that received the message can replicate it as many times as needed, and start forwarding it as well. Communication networks are chosen according to the Uniformly Random Trees model.

We prove the following theorem:

Theorem 1.1**.**

Broadcast on Uniformly Random Trees completes within $16\ln n$ rounds with probability $p>1-\frac{1}{n^{2}}$.

We also show that this result is asymptotically tight. Indeed, we cannot hope for a similar probability for a number of rounds that is $o(\ln n)$:

Theorem 1.2**.**

If $n\geq 2$, then the probability that Broadcast on Uniformly Random Trees fails to complete within $\log n$ rounds is at least $\frac{1}{4}$.

All-to-All Broadcast.

In the All-to-All Broadcast on Uniformly Random Trees problem, we start by giving a distinct message to each node, and each node must forward this message to all other nodes. In each round, each node forwards all the messages it has received in previous rounds to all its out-neighbors. Communication networks are chosen according to the Uniformly Random Trees model. We prove the following theorem:

Theorem 1.3**.**

All-to-All Broadcast on Uniformly Random Trees completes within $16\ln n$ rounds with probability $p>1-\frac{1}{n}$.

Consensus.

In the Consensus on Uniformly Random Trees problem, we start by giving a value $v_{p}\in\{0,1\}$ to each node $p$, and each node must decide on a value in $\{0,1\}$. This should satisfy the following conditions:

• •

Agreement: No two nodes decide differently.

• •

Termination: Every node eventually decides.

• •

Validity: The value the nodes agree on should be one of the input values $v_{p}$.

Communication networks are chosen according to the Uniformly Random Trees model. We prove the following theorem:

Theorem 1.4**.**

There exists a protocol for Consensus on Uniformly Random Trees that satisfies Agreement and Validity, terminates within $16\ln n$ rounds with probability $p>1-\frac{2}{n^{2}}$, and only requires messages of 1 bit over each edge in each round.

In our second model an adversary can influence the network that is chosen in each round. The setting where the adversary completely determines the tree was studied in [28, 17] and Broadcast in that model was recently solved: The required number of rounds is $\Theta(n)$ [17, 13], while Consensus is unsolvable [6]. We generalize this model and consider the Randomized Oblivious Message Adversary model, where the power of the adversary is controlled by a parameter $k$. In that model, to construct the communication network for a round, the adversary chooses $k$ directed edges to appear in the tree, and a rooted tree is chosen uniformly at random among the trees from $\mathcal{T}_{n}$ that include those $k$ edges. Note that the case $k=n-1$ is exactly the case where the adversary chooses all edges in the tree for each round, while the case $k=0$ is where the adversary has no influence. We allow the adversary to access the random tree of all rounds $t^{\prime}<t$ before choosing its edges for round $t$. In this model we analyze Broadcast and Consensus.

Broadcast with a Randomized Oblivious Message Adversary

In the Broadcast with a Randomized Oblivious Message Adversary of parameter $k$ problem, we start by giving a different message to each node, and the message of one of those nodes (no matter which one) must be forwarded to all other nodes. Each node can replicate and start forwarding any message it has received, and it forwards as many messages as it wants in any given round. Communication networks are chosen according to the Randomized Oblivious Message Adversary of parameter $k$. We prove the following theorem:

Theorem 1.5**.**

Broadcast with a Randomized Oblivious Message Adversary of parameter $k$ completes within $O(k+\log n)$ rounds with probability $p\geq 1-\frac{2}{n^{2}}$.

We show that this overhead of $k$ compared to the case where the adversary has no control is inevitable, as the adversary can always delay Broadcast for at least $\Omega(k)$ rounds:

Theorem 1.6**.**

If the adversary controls $k$ edges in each round, then there exists a strategy that, with probability $1$, guarantees that at least $\frac{k}{2}-1$ rounds are required.

Consensus with a Randomized Oblivious Message Adversary

In the Consensus with a Randomized Oblivious Message Adversary problem, we start by giving a value $v_{p}\in\{0,1\}$ to each node $p$, and each node must decide on a value in $\{0,1\}$. This should satisfy Validity, Agreement and Termination as defined above. Communication networks are chosen according to the Randomized Oblivious Message Adversary of parameter $k$. We prove the following theorem:

Theorem 1.7**.**

There exists a protocol for Consensus with a Randomized Oblivious Message Adversary that satisfies Agreement and Validity, and terminates in $O(k+\log n)$ rounds with probability $p\geq 1-\frac{2}{n^{2}}$, and only requires messages of 1 bit over each edge in each round, as long as $k\leq 0.1n$.

Organisation

The paper is organized as follows: we first introduce combinatorial results on rooted trees in Section 2. We then explore the fully random case in Section 3. In Section 4, we explore the case where the adversary controls $k$ edges in each round. We review related work in Section 5, then conclude in Section 6. Appendix A gives a lower bound for deterministic broadcast in constant-height trees. In Appendix B, we give some probability theory results that are useful throughout the paper. Finally, in Appendix C, we include omitted proofs from Section 4.

2 Counting Trees

In this section, we will present previously known and new results on the number of rooted trees that satisfy given properties. This will be helpful for computing probabilities in later sections. Namely, we are particularly interested in the two following results:

Theorem 2.1** (Lemma 1 of [26]).**

Let us be given a directed rooted forest $F$ on $n$ vertices, and let $\left|E\right|$ be the number of edges in $F$. Then, the number of directed rooted trees $T$ over $n$ vertices, such that $F$ is contained by $T$, is $n^{n-1-\left|E\right|}$.

Theorem 2.2**.**

Let us be given a directed rooted forest $F$ on $n$ vertices, let $v\in[n]$ be a vertex with no parent in $F$, and $f$ be the number of vertices of the component of $F$ containing $v$ (note that we can have $f=1$ if $v$ is an isolated vertex). Then the number of directed rooted trees $T$ on $n$ vertices, such that $F$ is contained in $T$, and such that $v$ is the root of $T$, is $fn^{n-2-\left|E\right|}$.

In this section, we will give a different proof to Theorem 2.1, as an analysis similar to that different proof will allow us to prove Theorem 2.2. To do so, we start by recalling Cayley’s formula [2]:

Theorem 2.3** (Cayley’s formula).**

The number of undirected trees on $n$ vertices is $n^{n-2}$.

As a corollary of this theorem, we can compute the number of rooted trees on $n$ vertices, as choosing a rooted tree is equivalent to choosing an undirected tree, and then choosing a root:

Corollary 2.4**.**

The number of rooted trees on $n$ vertices is $n^{n-1}$.

Throughout this section we use $F$ to denote an undirected or directed forest and $C_{1},C_{2},\ldots,C_{m}$ of $f_{1},\ldots,f_{m}$ vertices with integer $m\geq 1$ to denote the connected components of (the undirected version of) $F$. The next theorem on undirected trees gives the number of undirected trees which respect a set of fixed edges. It was shown by Lu, Mohr and Székely [24].

Theorem 2.5** (Lemma 6 of [24]).**

Let us be given an undirected forest $F$ on $n$ vertices, with connected components $C_{1},C_{2},\ldots,C_{m}$ of $f_{1},\ldots,f_{m}$ vertices with integer $m\geq 1$. Let $\left|E\right|$ be the number of edges in $F$. Then, the number of undirected trees $T$ on $n$ vertices, such that $F$ is contained in $T$, is:

[TABLE]

In the rooted case the formula is simpler, as one can drop the product of $f_{i}$. For this, let us first recall the definition of a directed rooted forest:

Definition 2.6** (Directed Rooted Forest).**

A directed rooted forest is a collection of disjoint directed rooted trees.

See 2.1

As stated above, we will give a new proof for this theorem. For simplicity, we will always require that $\sum_{i\in[m]}f_{i}=n$, which is always achievable by putting isolated vertices in trivial components. For any directed graph $G$, $u(G)$ will represent its undirected version. For any directed rooted tree $T$, its root is denoted by $r(T)$. We will also use the following bijection. Recall that $\mathcal{T}_{n}$ is the set of all directed rooted trees on $n$ vertices. We use $T_{n}$ to denote the set of all undirected trees on $n$ vertices.

Definition 2.7**.**

Let $T_{n}$ be the set of all undirected trees on $n$ vertices. We define $\pi$ to be the following bijection:

[TABLE]

To prove Theorem 2.1, we will first look at all the rooted trees that agree with $F$ if edge directions are ignored. Choosing such a tree is equivalent to choosing an undirected tree that contains $F$, then choosing a root. This results in $\left(\prod_{i\in[m]}f_{i}\right)n^{n-1-E}$ trees. However, while all of them agree with $F$ on the undirected edges, the direction of those edges will not correspond for a majority of them. We will then partition this set of trees such that only one element of each set of the partition agrees with $F$ on the directed edges, and counting the number of sets in the partition will yield the desired result. To do so, we will use group actions.

Definition 2.8** (Group action).**

If $G$ is a group with identity element $e$, and $X$ is a set, then a (left) group action $\alpha$ of $G$ on $X$ is a function

[TABLE]

that satisfies the following two axioms:

• •

Identity: $\alpha(e,x)=x,\forall x\in X$, where $e$ is the identity element of $G$.

• •

Compatibility: $\alpha(g,\alpha(h,x))=\alpha(gh,x),\forall g,h\in G,\forall x\in X$

Definition 2.9** (Rotations).**

Let $k>0$ be an integer and let $R_{k}$ be the group of all rotations of $[k]$, that is, the set of functions:

[TABLE]

Definition 2.10**.**

Let $F$ be a forest with vertices in $[n]$ (rooted and directed or undirected), and $T$ a tree with vertices in $[n]$ (rooted and directed or undirected). We say that they are undirected-compatible if $u(F)\subseteq u(T)$, where $u(G)$ represents the undirected version of graph $G$. If $F$ and $T$ are both rooted and directed or both undirected, we say that they are compatible if $F\subseteq T$.

Definition 2.11**.**

Let us be given a directed rooted forest $F$ with vertices in $[n]$. $A_{F}$ is the set of directed rooted trees on $n$ vertices that are undirected-compatible with $F$.

The following lemma follows almost immediately from Theorem 2.5.

Lemma 2.12**.**

Let $F$ be a directed rooted forest with $n$ vertices and $E$ edges. Then $\left|A_{F}\right|=\left(\prod_{i\in[m]}f_{i}\right)n^{n-1-\left|E\right|}$.

Proof.

Let $B_{F}$ be the set of all undirected rooted trees that are undirected-compatible with $F$. $\pi$ induces a bijection between $A_{F}$ and $B_{F}\times[n]$. Therefore, $\left|A_{F}\right|=\left|B_{F}\right|\cdot n$. By Theorem 2.5, $\left|B_{F}\right|=\left(\prod_{i\in[m]}f_{i}\right)n^{n-2-\left|E\right|}$. ∎

Definition 2.13**.**

For any $i\in[m]$, there exists a bijection between $\mathbb{Z}/f_{i}\mathbb{Z}$ and $C_{i}$. Let $b_{i}$ be that bijection.

Let $R=R_{f_{1}}\times\ldots\times R_{f_{k}}$. Note that $R$ is a group as a cartesian product of groups. We now define a group action of $R$ on $A_{F}$. This group action will allow us to partition $A_{F}$ as desired.

Definition 2.14** (Group Action of $R$ on $A_{F}$).**

Given a forest $F$ with connected components $C_{i}$ with $1\leq i\leq m$ and corresponding bijections $b_{i}$, let $\alpha$ be the group action of $R$ on $A_{F}$ defined as follows: Let $\sigma=(\sigma_{a_{1}}^{f_{1}},\dots,\sigma_{a_{m}}^{f_{m}})$ for some $(a_{1},\dots,a_{m})\in\mathbb{Z}/f_{1}\mathbb{Z}\times\dots\times\mathbb{Z}/f_{m}\mathbb{Z}$ be an element of $R$ and let $T\in A_{F}$. Then $\alpha(\sigma,T)$ is obtained from $T$ by making the following modifications to $\pi(T)=(u(T),r(T))$:

• •

For every $i$ such that $r(T)\notin C_{i}$, there is one (and only one) path from $r(T)$ to $C_{i}$ in $u(T)$. Let $(x,y)$ be the only edge on that path such that $x\notin C_{i},y\in C_{i}$. Replace edge $(x,y)$ with edge $(x,b_{i}\sigma_{a_{i}}^{f_{i}}b_{i}^{-1}(y))$.

• •

For $i$ such that $r(T)\in C_{i}$ for some $i\in[m]$, set $r(\alpha(\sigma,T))$ to $b_{i}\sigma_{a_{i}}^{f_{i}}b_{i}^{-1}(r(T))$.

The group action returns this modified tree rooted at $b_{i}\sigma_{a_{i}}^{f_{i}}b_{i}^{-1}(r(T))$.

To prove that this is indeed a group action, we need to verify (1) that $\alpha(\sigma,T)$ is indeed in $A_{F}$, (2) that the identity element $e=(\sigma_{0}^{f_{1}},\dots,\sigma_{0}^{f_{m}})$ of $R$ verifies $\alpha(e,T)=T$ for any $T\in S$, and (3) that for any two $\sigma,\tau\in R$, for any $T\in S$, we have $\alpha(\sigma,\alpha(\tau,T))=\alpha(\sigma\tau,T)$. The second condition being trivial as $\sigma^{f_{i}}_{0}$ is the identity function for any value of $f_{i}$, we only prove the other two.

Lemma 2.15**.**

$\alpha(\sigma,T)\in A_{F}$.

Proof.

Let us first show that $u(\alpha(\sigma,T))$ is an undirected tree. As it has $n-1$ edges, we only need to show that it is connected. Let $v$ be a vertex. We need to show that it can be reached from $r(T)$. Let $P$ be the (only) path from $r(T)$ to $v$ in $T$, written as a sequence of vertices. Then we can split up $P$ into $P=P_{1}P_{2}\dots P_{z}$, where each $P_{j}$ is a sequence of vertices that all belong to the same $C_{i}$ for some $i\in[m]$. We will now replace each of the $P_{j}$ by another path to make a path from $r(T)$ to $v$ in $u(\alpha(\sigma,T))$.

Consider every edge $(x,y)$ where $x$ is the last vertex of $P_{j}$ for some $j$, and $y$ is the first vertex of $P_{j+1}$. There exists some $k$ such that $y\in C_{k}$. Then $P_{1}P_{2}\dots P_{j}y$ is the path from $r(T)$ to $C_{k}$ in $u(T)$. Then $(x,b_{k}^{-1}\sigma_{a_{k}}^{f_{k}}b_{k}(y))\in u(\alpha(\sigma,T))$. Replace $y$ by $b_{k}^{-1}\sigma_{a_{k}}^{f_{k}}b_{k}(y)$ in $P$.

Let us now look at a particular $P_{j}$, and let $i$ be such that all of the vertices of $P_{j}$ belong to $C_{i}$, then its first vertex has been changed to another vertex of $C_{i}$, while all others are unchanged. Hence, the first and last vertex still belong to $C_{i}$. As $C_{i}$ is connected in $u(\alpha(\sigma,T))$ since no edge inside $C_{i}$ has been modified, there exists a path $P^{\prime}_{j}$ in $u(\alpha(\sigma,T))$ that connects that first and last vertex of $P_{j}$. We can thus replace $P_{j}$ by $P^{\prime}_{j}$.

The new path now correctly connects $r(T)$ and $v$ in $u(\alpha(\sigma,T))$, which shows that it is connected. Rooting $u(\alpha(\sigma,T))$ at $r(\alpha(\sigma,T))$ gives $\alpha(\sigma,T)$. Hence $\alpha(\sigma,T)$ is a tree. Since no edge in any particular $C_{i}$ has been modified, $\alpha(\sigma,T)$ is compatible with $F$. ∎

Lemma 2.16**.**

For any $T\in A_{F}$, and $\sigma,\tau\in R$ we have that $\alpha(\sigma,\alpha(\tau,T))=\alpha(\sigma\tau,T)$.

Proof.

Let $\sigma=(\sigma_{a_{1}}^{f_{1}},\dots,\sigma_{a_{m}}^{f_{m}})$ and $\tau=(\sigma_{c_{1}}^{f_{1}},\dots,\sigma_{c_{m}}^{f_{m}})$. Let $k$ be such that $r(T)\in C_{k}$, then $r(\alpha(\tau,T))=b_{k}\sigma_{c_{k}}^{f_{k}}b_{k}^{-1}(r(T)),r(\alpha(\sigma\tau,T))=b_{k}\sigma_{a_{k}}^{f_{k}}\sigma_{c_{k}}^{f_{k}}b_{k}^{-1}(r(T))$, and $r(\alpha(\sigma,\alpha(\tau,T)))=b_{k}\sigma_{a_{k}}^{f_{k}}b_{k}^{-1}b_{k}\sigma_{c_{k}}^{f_{k}}b_{k}^{-1}(r(T))=b_{k}\sigma_{a_{k}}^{f_{k}}\sigma_{c_{k}}^{f_{k}}b_{k}^{-1}(r(T))=r(\alpha(\sigma\tau,T))$. And then, for every $i\in[m]\setminus\{k\}$, the path from any of those roots to $C_{i}$ in $T$ will include the path from $C_{k}$ to $C_{i}$ which in turn will include the edge $(x,y)$ such that $x\notin C_{i},y\in C_{i}$, then the corresponding edge is $(x,b_{i}\sigma_{c_{i}}^{f_{i}}b_{i}^{-1}(y))$ in $\alpha(\tau,T)$, and $(x,b_{i}\sigma_{a_{i}}^{f_{i}}\sigma_{c_{i}}^{f_{i}}b_{i}^{-1}(y))$ in $\alpha(\sigma\tau,T)$. Hence, it is $(x,b_{i}\sigma_{a_{i}}^{f_{i}}b_{i}^{-1}b_{i}\sigma_{c_{i}}^{f_{i}}b_{i}^{-1}(y))$ in $\alpha(\sigma,\alpha(\tau,T))$. We thus have $\alpha(\sigma,\alpha(\tau,T))=\alpha(\sigma\tau,T)$. ∎

As we plan to use Lagrange’s theorem for group actions, we now compute the stabilizer of a tree $T$, which is the set of all rotations that do not modify the tree:

Lemma 2.17**.**

$R_{T}:=\{\sigma\in R:\alpha(\sigma,T)=T\}=\{e\}$, for every $T\in A_{F}$.

Proof.

Let $\sigma\in R$ be a rotation such that $\alpha(\sigma,T)=T$. We obviously have that $r(T)=r(\alpha(\sigma,T))$. Let $i\in[m]$,

• •

Either $r(T)\in C_{i}$, in which case $r(\alpha(\sigma,T))=b_{i}\sigma_{a_{i}}^{f_{i}}b_{i}^{-1}(r(T))=r(T)$, which implies that $\sigma_{a_{i}}^{f_{i}}b_{i}^{-1}(r(T))=b_{i}^{-1}(r(T))$, therefore $b_{i}^{-1}(r(T))=b_{i}^{-1}(r(T))+a_{i}$, and hence $a_{i}=0$.

• •

Or $r(T)\notin C_{i}$. In that case, we look at the path from $r(T)$ to $C_{i}$ in both $T$ and $\alpha(\sigma,T)$. These two paths must be the same. However, if the first element of that path in $T$ that is in $C_{i}$ is some vertex $x$, then in $\alpha(\sigma,T)$, it is $b_{i}\sigma_{a_{i}}^{f_{i}}b_{i}^{-1}(x)$. We conclude that $b_{i}\sigma_{a_{i}}^{f_{i}}b_{i}^{-1}(x)=x$ and thus $a_{i}=0$.

We therefore have that $a_{i}=0$ for every $i\in[m]$, which proves that $\sigma=(\sigma^{f_{1}}_{0},\dots,\sigma^{f_{m}}_{0})=e$. ∎

We now take a look at the orbit $R\cdot T$ of a tree $T\in A_{F}$. The group action ensures that the orbits in $A_{F}$ form a partition of $A_{F}$.

Theorem 2.18** (Corollary 10.23 of [27]).**

Let $G$ be a group, $X$ a set and $\alpha$ a group action of $G$ on $X$. Let $x$ be an element of $X$, $G_{x}:=\{g\in G:\alpha(g,x)=x\}$ and $G.x:=\{y\in X:\exists g\in G,y=\alpha(g,x)\}$. Then we have that:

[TABLE]

Lemma 2.19**.**

Let, for every $T\in A_{F}$, $R\cdot T:=\{T^{\prime}\in A_{F}:\exists\sigma\in R,\alpha(\sigma,T)=T^{\prime}\}$. Then $\left|R\cdot T\right|=\prod_{i\in[m]}f_{i}$.

Proof.

By Theorem 2.18, we have that $\left|R\cdot T\right|=\frac{\left|R\right|}{R_{T}}=\frac{\prod_{i\in[m]}f_{i}}{1}$ ∎

We now show that exactly one tree in each orbit is compatible with $F$.

Lemma 2.20**.**

Let $T\in A_{F}$. Then there exists exactly one $T^{\prime}\in R\cdot T$ such that $T^{\prime}$ is compatible with $F$.

Proof.

Let $T^{\prime}\in R\cdot T$ be a tree such that $T^{\prime}$ is compatible with $F$, and let $\sigma$ be the rotation such that $T^{\prime}=\alpha(\sigma,T)$. Let, for every $i\in[m]$, $r_{i}$ be the root of $C_{i}$ in $F$ and let $k$ be such that $r(T)\in C_{k}$. Then we must have that $r_{k}=r(T^{\prime})=b_{k}\sigma_{a_{k}}^{f_{k}}b_{k}^{-1}r(T)$ and thus $a_{k}=b_{k}^{-1}r_{k}-b_{k}^{-1}r(T)$.

For every $i$ such that $r(T)\notin C_{i}$, look at the path from $r(T)$ to $C_{i}$ in $T$, and its corresponding path in $T^{\prime}$, computed similarly to the proof of Lemma 2.15. In $T^{\prime}$, the first vertex of that path in $C_{i}$ must be $r_{i}$, but it also is $b_{i}\sigma_{a_{i}}^{f_{i}}b_{i}^{-1}(y)$, where $y$ is the first vertex of the path in $T$. Hence $a_{i}=b_{i}^{-1}r_{i}-b_{i}^{-1}(y)$.

These conditions uniquely determine $\sigma$, and, thus, $T^{\prime}$. Conversely, setting $\sigma$ with each $a_{i}$ defined as above gives a tree $T^{\prime}$ that is compatible with $F$. ∎

We can now prove Theorem 2.1, which we recall below:

See 2.1

Proof.

Consider set $A_{F}$ as defined in Definition 2.11. We know that every directed rooted spanning tree $T$ in $K_{n}$ such that $F$ is contained by $T$ is in $A_{F}$. We can partition $A_{F}$ in orbits of the group action defined in Definition 2.14. By Lemma 2.19, each orbit has $\prod_{i\in[m]}f_{i}$ elements, and thus we have $\frac{\left|A_{F}\right|}{\prod_{i\in[m]}f_{i}}$ orbits, which is equal to $n^{n-1-E}$ by Lemma 2.12. Lemma 2.20 ensures that exactly one element in each orbit is a directed rooted spanning tree $T$ in $K_{n}$ such that $F$ is contained by $T$. ∎

Using a very similar technique, we prove next Theorem 2.2:

See 2.2

Proof.

Let $\hat{A}_{F}$ be the set of all undirected trees on $n$ vertices that are undirected-compatible with $F$. If $C_{1},\dots,C_{m}$ are the components of $F$, with respective cardinality $f_{1},\dots,f_{m}$, where $v\in C_{1}$. This implies that $f=f_{1}$. By Theorem 2.5, $\left|\hat{A}_{F}\right|=n^{n-2-E}\prod_{i\in[2,m]}f_{i}$. Rooting all of those trees at $v$ creates the set of all rooted trees on $n$ vertices that are undirected-compatible with $F$, rooted at $v$. Defining the group action as above (by using rotations on all $C_{i}$ for $i>1$), we can partition $\hat{A}_{F}$ into orbits. Each orbit has size $\frac{\left|R\right|}{R_{T}}=\prod_{i\in[2,m]}f_{i}$, so we have $f_{1}n^{n-2-\left|E\right|}=fn^{n-2-\left|E\right|}$ orbits, and in each orbit, exactly one tree is compatible with $F$, hence the result. ∎

3 The Uniformly Random Trees Model

We will now be able to give a precise description of how information flows in the random network over time. Indeed, the theorems of the previous section will allow us to find the probability that a set of edges exists in a uniformly chosen random tree. Since all nodes are symmetric, we will at each step, divide the nodes into two sets: the set $I$ of nodes that have received the message, called informed nodes, and the set $S$ of remaining nodes, called uninformed nodes. We study how $I$ grows over time.

For the rest of the section, $I_{t}$ and $S_{t}$ will, respectively, be the set of nodes that are informed and uninformed after round $t$. We set $I_{0}=\{f\}$ and $S_{0}=[n]-\{f\}$, where $f$ is the node that initially holds the message, $N_{t}=\left|I_{t}\right|$ to be the number of informed nodes after $t$ rounds, and $T_{t}$ to be the tree chosen at random in round $t$. For a tree $T$, for each node $p$, $P_{T}(p)$ is the (unique) parent of node $p$ in $T$, unless $p$ is the root of $T$, in which case $P_{T}(p)=p$. Simplifying the notation, we also use $P_{t}(p)$ to denote $P_{T_{t}}(p)$. We use ${A(S,x)}$ where $S$ is a set and $x$ an integer to represent the set of subsets of $S$ of size $x$.

The central lemma of the proof is the following lemma, which characterizes how many new nodes get informed in each round, depending on how many were informed after the previous round. This lemma shows that uninformed nodes get informed independently from each other.

Lemma 3.1**.**

For any $t>0$, $N_{t+1}-N_{t}$ follows a binomial distribution with parameters $\left(\frac{N_{t}}{n},n-N_{t}\right)$.

The proof of this lemma shows that every uninformed node has probability $\frac{N_{t}}{n}$ of having an informed parent in round $t+1$, independently of whether the other uninformed nodes have an uninformed parent.

Proof.

Let $I_{t}=\{i_{1},\dots,i_{N_{t}}\}$ and $S_{t}=\{s_{1},\dots,s_{n-N_{t}}\}$. We then have, for any integer $x$:

[TABLE]

Our goal is to show that the events $P_{t}(y)\in I_{t}$ for different $y\in S_{t}$ are mutually independent. Let us look at the event $\bigcap_{y\in J}(P_{t}(y)\in I_{t})$ for any $J\subseteq S_{t}$ (note that we do not require that $J$ has a specific size here). We can then write, indexing $a$ on $J$:

[TABLE]

Now consider the forest that is composed of stars whose centers are the $i_{a_{y}}$ and whose leaves are the nodes $y$. More specifically, consider the forest that contains the edges $(i_{a_{y}},y),\forall y\in J$. Note that $\left|\{T\in\mathcal{T}_{n}:P_{T}(y)=i_{a_{y}},\forall y\in J\}\right|$ equals the number of rooted trees that are compatible with this forest. By Theorem 2.1, we have that $\left|\{T\in\mathcal{T}_{n}:P_{T}(y)=i_{a_{y}},\forall y\in J\}\right|=n^{n-1-\left|J\right|}$. This allows us to compute the above probability as follows:

[TABLE]

This proves that the events $P_{t+1}(y)\in I_{t}$ for any two $y\in S_{t}$ are mutually independent (Definition B.7), each having probability $\frac{N_{t}}{n}$. Going back to the first equation of this proof, we can now compute, using Lemma B.8:

[TABLE]

∎

Our next goal is to show that $N_{t}=n$ with high probability for all $t\geq 16\ln n$. To do so we introduce a random variable $X_{t}$ that we use to lower bound $N_{t}$.

Definition 3.2**.**

Let $X_{t}$ be the random variable that is defined as follows:

[TABLE]

Lemma 3.3**.**

For every $t\in\mathbb{N}$, we have that $n\geq N_{t}\geq X_{t}$.

Proof.

Note that $N_{t}$ cannot go higher than $n$ because it is the number of nodes informed after round $t$, which is at most $n$.

We will prove the rest by induction on $t$. For the induction basis note that by definition $N_{0}=1=X_{0}$. For the induction step let us assume that $n\geq N_{t}\geq X_{t}$ for some $t\in\mathbb{N}$. Consider first the case that $N_{t+1}-N_{t}<(n-N_{t})\cdot\frac{N_{t}}{n}$. Since no informed node can become uninformed, we have that $N_{t+1}\geq N_{t}\geq X_{t}=X_{t+1}$, as desired. Next consider the case that $N_{t+1}-N_{t}\geq(n-N_{t})\cdot\frac{N_{t}}{n}$. Then $N_{t+1}\geq N_{t}+(n-N_{t})\cdot\frac{N_{t}}{n}$ and $X_{t+1}=X_{t}+(n-X_{t})\cdot\frac{X_{t}}{n}$. As the function $x\mapsto x+(n-x)\frac{x}{n}$ is strictly increasing for $x\leq n$, this proves that $N_{t+1}\geq X_{t+1}$, as desired. ∎

Lemma 3.4**.**

For every $t\in\mathbb{N}$, we have that $X_{t}\geq 1$.

Proof.

We will again show this by induction. For the induction basis note that by definition $1=X_{0}$. For the induction step let us assume that $X_{t}\geq 1$ for some $t\in\mathbb{N}$. We then have two cases, either $X_{t+1}=X_{t}$ and the result holds trivially, or $X_{t+1}=X_{t}+(n-X_{t})\cdot\frac{X_{t}}{n}$. Since $1\leq X_{t}\leq n$, we have that $X_{t+1}\geq X_{t}\geq 1$. ∎

Lemma 3.5**.**

For every $t\in\mathbb{N}$, we have that $n>X_{t}$, if $n>1$.

Proof.

We show this claim by induction on $t$. As $n>1$ and $X_{1}=1$, it is trivially true for $t=1$. Assume it is true for $t\in\mathbb{N}$. Then $X_{t+1}\leq X_{t}+(n-X_{t})\cdot\frac{X_{t}}{n}=n(\frac{X_{t}}{n}+\frac{n-X_{t}}{n}\frac{X_{t}}{n})<n$, where the last inequality holds by noting that $(\frac{X_{t}}{n}+\frac{n-X_{t}}{n}\frac{X_{t}}{n})$ is a convex combination of $1$ and $\frac{X_{t}}{n}$, the latter of which being strictly smaller than $1$. ∎

Essentially, this means that $X_{t}$ never reaches $n$, and thus that $X_{t+1}$ is always strictly larger than $X_{t}$ if $N_{t+1}-N_{t}\geq(n-N_{t})\cdot\frac{N_{t}}{n}$:

Corollary 3.6**.**

We have that $X_{t+1}>X_{t}$ if and only if $N_{t+1}-N_{t}\geq(n-N_{t})\cdot\frac{N_{t}}{n}$.

Lemma 3.7**.**

Let $u_{t}\in\mathbb{N}$ be the $t$-th round such that $X_{u_{t}+1}>X_{u_{t}}$ and let $u_{0}=0$. Then $X_{u_{t}}=n-n\left(\frac{n-1}{n}\right)^{2^{t}}$. Moreover, we have that $X_{u_{t+1}}=X_{u_{t}}+(n-X_{u_{t}})\cdot\frac{X_{u_{t}}}{n}$.

Proof.

We show the claim by induction on $t$. By definition of $u_{0}$ we have that $X_{u_{0}}=1$. Thus the induction basis $X_{u_{0}}=1=n-n\left(\frac{n-1}{n}\right)^{2^{0}}$ follows.

For the induction step assume next the result is true for some $t\in\mathbb{N}$. Note that for every $t\in\mathbb{N}$, it holds that $X_{u_{t+1}}=X_{u_{t}}+(n-X_{u_{t}})\cdot\frac{X_{u_{t}}}{n}$. Indeed, we have that $X_{u_{t+1}}=X_{u_{t+1}-1}=\dots=X_{u_{t}+1}=X_{u_{t}}+(n-X_{u_{t}})\cdot\frac{X_{u_{t}}}{n}$. Thus,

[TABLE]

∎

Lemma 3.8**.**

If $t\geq u_{2\ln n}$, then $N_{t}=n$.

Proof.

Since $N_{t}$ is non-decreasing and upper-bounded by $n$, it suffices to show that $N_{u_{2\ln n}}=n$. We will do so by using its lower bound $X_{u_{2\ln n}}$. We have that:

[TABLE]

Where we used that $\ln(x)\leq 1-x$ and for $x\geq 1,x\exp(-x^{2\ln 2-1})<1$. Since $n\geq N_{u_{t}}\geq X_{u_{t}}$ by Lemma 3.3, and since $N_{t}\in\mathbb{N}$, we have that $N_{u_{t}}=n$. ∎

We now state a result due to Greenberg and Mohri [19], that will give us an estimate of the probability of $X_{t}$ strictly increasing in a given round.

Theorem 3.9** (Theorem 1 of [19]).**

For any positive integer $m$ and any probability $p$ such that $p>\frac{1}{m}$, let $B$ be a binomial random variable of parameters $(p,m)$. Then, the following inequality holds:

[TABLE]

Lemma 3.10**.**

If $n>4$, for every $t\in\mathbb{N}$, we have that $\mathbb{P}\left(X_{t+1}>X_{t}\right)\geq\frac{1}{4}$

Proof.

By Corollary 3.6, we have that $X_{t+1}>X_{t}$ if and only if $N_{t+1}-N_{t}\geq(n-N_{t})\cdot\frac{N_{t}}{n}$. This implies that $\mathbb{P}\left(X_{t+1}>X_{t}\right)=\mathbb{P}\left(N_{t+1}-N_{t}\geq(n-N_{t})\cdot\frac{N_{t}}{n}\right)$. By Lemma 3.1, $N_{t+1}-N_{t}$ follows a binomial distribution of parameters $\left(\frac{N_{t}}{n},n-N_{t}\right)$ for any $t>0$. Thus the expected value of $N_{t+1}-N_{t}$ is $\left(n-N_{t}\right)\frac{N_{t}}{n}$. We have multiple cases to consider:

Case 1: If $2\leq N_{t}\leq n-2$, then $N_{t+1}-N_{t}$ has expected value $\left(n-N_{t}\right)\frac{N_{t}}{n}$. We will show below that $\frac{N_{t}}{n}\geq\frac{1}{n-N_{t}}$, implying that, by Theorem 3.9, the result holds.

The function $x\mapsto x+1+\frac{1}{x-1}$ is strictly increasing between $2$ and $n-2$, using that $\frac{1}{n-3}<1$ when $n>4$, we have that:

[TABLE]

Therefore $N_{t}+1+\frac{1}{N_{t}-1}<n$, which implies that $n>\frac{N_{t}^{2}}{N_{t}-1}$. This further implies that $-n>N_{t}(N_{t}-n)$ and therefore $\frac{N_{t}}{n}>\frac{1}{n-N_{t}}$.

Case 2: If $N_{t}=1$, then $(n-N_{t})\cdot\frac{N_{t}}{n}=\frac{n-1}{n}$. Therefore $\mathbb{P}(N_{t+1}-N_{t}\geq(n-N_{t})\cdot\frac{N_{t}}{n})=\mathbb{P}(N_{t+1}-N_{t}\geq 1)=1-\mathbb{P}(N_{t+1}-N_{t}=0)=1-(\frac{n-1}{n})^{n-1}>\frac{1}{4}$ since $n>4$.

Case 3: If $N_{t}=n-1$, then $\mathbb{P}(N_{t+1}-N_{t}\geq(n-N_{t})\cdot\frac{N_{t}}{n})=\mathbb{P}(N_{t+1}-N_{t})\geq\frac{n-1}{n})=\frac{n-1}{n}>\frac{1}{4}$ since $n>4$.

Case 4: If $N_{t}=n$, then $\mathbb{P}(N_{t+1}-N_{t}\geq(n-N_{t})\cdot\frac{N_{t}}{n})=\mathbb{P}(N_{t+1}-N_{t}\geq 0)=1>\frac{1}{4}$.

∎

Let $(B_{t})_{t\in\mathbb{N}}$ be Bernoulli independent random variables of parameter $\frac{1}{4}$. Let $Z^{B}_{\leq t}=\sum_{z\in[t]}B_{z}$ and $Z_{\leq t}=\sum_{z\in[t]}\mathbbl{1}\left(X_{z+1}>X_{z}\right)$.

Corollary 3.11**.**

For any $\ell\in\mathbb{N}$, we have that $\mathbb{P}(Z_{\leq t}\leq\ell)\leq\mathbb{P}(Z_{\leq t}^{B}\leq\ell)$.

Lemma 3.12** (Hoeffding’s inequality for binomial distributions  [21]).**

Let $Y$ be a binomial random variable with parameters $(t,p)$. We then have, for any $x\leq tp$:

[TABLE]

Lemma 3.13**.**

Let $t=16\ln n$. Then $\mathbb{P}(Z_{\leq t}\leq 2\ln n)\leq\frac{1}{n^{2}}$.

Proof.

Note that $Z_{\leq t}^{B}$ is a binomial distribution of parameters $(t,\frac{1}{4})$. Using Hoeffding’s inequality, we have that:

[TABLE]

Corollary 3.11 then gives the desired result. ∎

We now have all the tools to prove Theorem 1.1, which we recall here:

See 1.1

Proof.

By Lemma 3.13, we have that, with probability $p\leq 1-\frac{1}{n^{2}}$, $X_{t+1}>X_{t}$ for at least $2\ln n$ many rounds within the $16\ln n$ first rounds. Recall that $u_{2\ln n}$ is the $2\ln n$-th round where $X_{t+1}>X_{t}$. We thus have that $\mathbb{P}\left(u_{2\ln n}\leq 16\ln n\right)\geq 1-n^{-2}$. But, by Lemma 3.8 the event $u_{2\ln n}\leq 16\ln n$ implies the event $N_{16\ln n}=n$, therefore $\mathbb{P}\left(N_{16\ln n}=n\right)\geq 1-n^{-2}$. ∎

We now show that this result is asymptotically tight. Indeed, we can show that if at most $\log n$ rounds are allowed, then with probability $q\geq\frac{1}{4}$, Broadcast does not complete:

See 1.2

Proof.

We will first show by induction that $\mathbb{E}(N_{t})\leq X_{u_{t}}$ for every $t\in\mathbb{N}$. We will then conclude using Markov’s inequality.

The induction basis is clear as $N_{0}=X_{0}=1$. For the induction step, assume that for some $t\in\mathbb{N}$, we have that $\mathbb{E}(N_{t})\leq X_{u_{t}}$. Let us show that this implies that $\mathbb{E}(N_{t+1})\leq X_{u_{t+1}}$. Indeed, by Lemma 3.1, $N_{t+1}-N_{t}$ has a binomial distribution of parameters $\frac{N_{t}}{n}$ and $n-N_{t}$. This implies that:

[TABLE]

Therefore:

[TABLE]

As $Var(N_{t})=\mathbb{E}[N_{t}^{2}]-\mathbb{E}[N_{t}]^{2}\geq 0$, we have that $-\mathbb{E}[N_{t}^{2}]\leq-\mathbb{E}[N_{t}]^{2}$. This implies:

[TABLE]

Note that we have that, by Lemma 3.7:

[TABLE]

Since $x\mapsto 2x-\frac{x^{2}}{n}$ is strictly increasing between [math] and $n$, with both $X_{t}$ and $\mathbb{E}[N_{t}]$ falling in that range (Lemmata 3.3 and 3.4), the induction hypothesis implies that $2\mathbb{E}[N_{t}]-\frac{\mathbb{E}[N_{t}]^{2}}{n}\leq 2X_{u_{t}}-\frac{X_{u_{t}}^{2}}{n}$. This implies $\mathbb{E}[N_{t+1}]\leq X_{u_{t+1}}$.

We know the value of $X_{u_{t}}$ from Lemma 3.7. We can thus give the upper bound $\mathbb{E}[N_{\log n}]\leq X_{u_{\log n}}=n(1-((n-1)/n)^{n})\leq n(1-\frac{1}{4})$, since $n\geq 2$. Using Markov’s inequality, we thus have:

[TABLE]

∎

We now use this result to get a similar result for All-to-all Broadcast. Using a union-bound, we obtain:

See 1.3

Proof.

Let $N_{t}^{(i)}$ be the random variable that represents the number of nodes that are informed after round $t$ of the message given to node $i$. By Theorem 1.1, we know that $\mathbb{P}\left(N_{16\ln n}^{(i)}<n\right)\leq n^{-2}$ for every $i\in[n]$. Using a union-bound, we get that:

[TABLE]

And thus:

[TABLE]

∎

We now finally recall Theorem 1.4, that states a result on Consensus:

See 1.4

Proof.

Algorithm 1 is an algorithm where everyone agrees on $v_{1}$, the input to node $1$, and where only $v_{1}$ is passed along. Thus every node outputs either $v_{1}$ or $\bot$. However, if $v_{1}$ has broadcast within the first $16\ln n$ rounds, then everyone outputs $v_{1}$. This happens with probability $p\geq 1-n^{-2}$, by Theorem 1.1. ∎

Note that Algorithm 1 can be adapted to different variants for Consensus. To keep our presentation concise, we do not explore them further in detail. For example, the version given here satisfies the condition that no node continues to communicate after it has decided on a value, but Consensus does not complete with probability $1$ after everyone has decided as some nodes might output $\bot$. A different definition of Consensus could allow each node to send messages after it decides on a value, in which case a different version of the algorithm could be given, where each node can decide as soon as it receives the value $v_{1}$.

4 The Randomized Oblivious Message Adversary

In this section, we consider a more general model where a parametrized adversary controls a certain number of edges in every round, and the others are chosen randomly. More specifically, in each round, the adversary $A$ chooses $k$ edges such that the resulting graph is a directed rooted forest $F$, and then a tree is chosen uniformly at random among the rooted trees that are compatible with $F$. We consider the model where the adversary has access to the randomly chosen trees of all previous rounds, but has no information on the random coin flips of the current and future rounds.

Let us start by understanding how Broadcast works in this model. Here, we start by giving each node a message, and in each round each node can make copies of all messages it has previously received and send them to all its out-neighbors. There is no restriction on the number of copies nor the size/number of messages that can be sent per round. The goal of the adversary is to delay the number of rounds until one message is broadcast to all nodes.

Note that in the case $k=n-1$, this is the deterministic case where in each round the adversary gets to exactly choose which tree is the communication network of the round. This is exactly the model studied in [13], where it was shown that the adversary cannot delay broadcast for more than $\left\lceil(1+\sqrt{2})n\right\rceil\approx 2.4n$.

Theorem 4.1** (Theorem 3.6 of [13]).**

The adversary cannot delay broadcast for more than $\left\lceil(1+\sqrt{2})n\right\rceil$ rounds.

We will prove the following theorem:

Theorem 4.2**.**

If the adversary controls $k$ edges in each round, then with probability $p\geq 1-2n^{-2}$, broadcast completes within $O(k+\log n)$ rounds.

In order to understand how tight this bound is, we first give a lower bound on how many rounds the adversary can delay broadcast:

See 1.6

Proof.

Let the adversary choose the set of edges $(1,2),\ldots(k,k+1)$ in all of the rounds. Then for any node $p\in[2,k+1]$, every message it has received must have been received by $p-1$ in a strictly smaller round, unless that message is the one given initially to $p$. Let $m$ be a message that has been broadcast. In particular, $m$ has been received by all nodes in $[k+1]$. If $m$ was given initially to some node $p$ such that $p\leq\left\lceil\frac{k}{2}\right\rceil$ or $p>k+1$, then $m$ must have needed $\left\lfloor\frac{k}{2}\right\rfloor$ rounds to travel from node $\left\lceil\frac{k}{2}\right\rceil+1$ to node $k+1$. If on the other hand, it was a message initially given to a node $\left\lceil\frac{k}{2}\right\rceil+1\leq p\leq k+1$, then $m$ must have needed $\left\lceil\frac{k}{2}\right\rceil-1$ rounds to travel from node $1$ to node $\left\lceil\frac{k}{2}\right\rceil$. ∎

Let us now concentrate on the upper bound. We will consider two cases, one case where $k$ is large, and where we will use Theorem 4.1, and one where $k$ is small, where we will use a similar analysis to Section 3.

Lemma 4.3**.**

If $k\geq\frac{n}{10}$, then the adversary cannot delay broadcast for more than $\left\lceil 10(1+\sqrt{2})k\right\rceil=O(k)$ rounds.

Proof.

By Theorem 4.1, the adversary cannot delay broadcast for more than $\left\lceil(1+\sqrt{2})n\right\rceil$ rounds even if the adversary controls all edges. Since $n\leq 10k$, we have the result. ∎

We will now prove the result for $k\leq n/10$. To do so, we will introduce an alternative adversary $A^{\prime}$ whose goal is to maximize the number of rounds until node $1$ has broadcast, independently of whether other nodes have broadcast or not. Clearly, this is in favour of the adversary and will not result in a smaller number of rounds than against $A$. Thus any upper bound on the number of rounds needed by $A^{\prime}$ is also an upper bound for $A$.

For the rest of the section, $I_{t}$ and $S_{t}$ will, respectively, be the set of nodes that are informed and uninformed after round $t$. We set $I_{0}=\{1\}$ and $S_{0}=[n]-\{1\}$, $N_{t}=\left|I_{t}\right|$ to be the number of informed nodes after $t$ rounds, and $T_{t}$ to be the tree chosen at random in round $t$. For a tree $T$, for each vertex $p$, $P_{T}(p)$ is the (unique) parent of node $p$ in $T$, unless $p$ is the root of $T$, in which case $P_{T}(p)=p$. Simplifying the notation, we also use $P_{t}(p)$ to denote $P_{T_{t}}(p)$.

We start by finding the best strategy $A^{\prime}$ could use and then analyze that strategy.

4.1 Best Strategy for the Alternative Adversary A’

To find the best strategy the adversary $A^{\prime}$ can use, we will use the notion of stochastic dominance. Intuitively, if a strategy yields more informed nodes than another one, then the adversary will choose the latter one. Stochastic dominance is the tool we use to formalize this.

Definition 4.4** (Stochastic Dominance).**

We say that a real random variable $Y_{1}$ stochastically dominates another real random variable $Y_{2}$, if, for every $x\in\mathbb{R}$, we have that $\mathbb{P}(Y_{1}\geq x)\geq\mathbb{P}(Y_{2}\geq x)$.

For any set $S$, let $\mathcal{P}(S)$ be the set of all subsets of $S$.

Definition 4.5** (Stochastic dominance).**

We say that a random variable $Y_{1}$ with values in $\mathcal{P}([n])$ stochastically dominates another random variable $Y_{2}$ with values in $\mathcal{P}([n])$, if, for every $x\in\mathbb{N}$, we have that $\mathbb{P}(\left|Y_{1}\right|\geq x)\geq\mathbb{P}(\left|Y_{2}\right|\geq x)$.

With stochastic dominance, we will use a related notion, that is coupling. Coupling is a useful tool to compare two random variables, and in particular, it helps translate probabilistic events into deterministic ones, which are easier to analyze.

Definition 4.6** (Coupling).**

A coupling of two random variables $Y_{1},Y_{2}$ is a third random variable $(\hat{Y_{1}},\hat{Y_{2}})$ such that $Y_{1}$ has the same distribution as $\hat{Y_{1}}$, and $Y_{2}$ has the same distribution as $\hat{Y_{2}}$.

Theorem 4.7** (Stochastic Dominance and Coupling, Theorem 7.1 of [7]).**

If a real random variable $Y_{1}$ stochastically dominates another real random variable $Y_{2}$, then there exists a coupling $(\hat{Y_{1}},\hat{Y_{2}})$ of $Y_{1}$ and $Y_{2}$ such that

[TABLE]

Theorem 4.8** (stochastic dominance and coupling, Theorem 7.8 of [7]).**

If a random variable $Y_{1}$ with values in $[n]$ stochastically dominates another random variable $Y_{2}$ with values in $[n]$, then there exists a coupling $(\hat{Y_{1}},\hat{Y_{2}})$ of $Y_{1}$ and $Y_{2}$ such that

[TABLE]

Lemma 4.9**.**

[Distribution Domination] Let $t$ be a round. Let $E_{1},E_{2}$ be two sets of edges the adversaries could choose for round $t$. Let $N^{(1)}_{t}$ (resp. $I^{(1)}_{t}$) be the number (resp. set) of informed nodes after round $t$ if $E_{1}$ is chosen, and $N^{(2)}_{t}$ (resp. $I^{(2)}_{t}$) if $E_{2}$ is chosen. Then if $\mathbb{P}(N^{(1)}_{t}\geq m)\geq\mathbb{P}(N^{(2)}_{t}\geq m)$ for every $m\in\mathbb{N}$ (that is, if $N_{t}^{(1)}$ stochastically dominates $N^{(2)}_{t}$), then choosing $E_{2}$ is a better strategy for the adversary than choosing $E_{1}$.

Intuitively, the way to prove this is to build, for any strategy the adversary might use after choosing $E_{1}$, another strategy that would work better if used after choosing $E_{2}$. To prove that it is indeed the case, we couple these two strategies to prove that after any round, the number of informed nodes in one strategy stochastically dominates the number of informed nodes in the other one. The full details of the proof can be found in Appendix C.

The next step is to show that the adversary will never force an edge from an informed node to an uninformed one. Indeed, intuitively, this means the adversary forces a node to be informed, which is against its interests. To do so, we introduce the notions of non-increasing and increasing trees, and show that $A^{\prime}$ will never choose an increasing tree.

Definition 4.10**.**

A rooted tree $U$ in a round $t$ is said to be non-increasing in round $t$ if all edges in $U$ whose source is in $I_{t-1}$ have their target in $I_{t-1}$ as well. Otherwise a tree is (information)-increasing in round $t$.

To show that the adversary never uses an increasing tree, we introduce the notion of a correction of an increasing tree, which will be non-increasing, and show that choosing the correction is a better strategy for the adversary than choosing the increasing tree.

Definition 4.11** (Isomorphism).**

We say that a rooted tree $U$ on $n$ nodes is isomorphic to a rooted tree $U^{\prime}$ on $n$ nodes if there exists a bijection $b$ from $[n]$ to $[n]$ such that for every (directed) edge $(u,v)\in U$, we have that $(b(u),b(v))\in U^{\prime}$, and for every (directed) edge $(u,v)\in U^{\prime}$, we have that $(b^{-1}(u),b^{-1}(v))\in U$.

In particular, if $r$ is the root of $U$, then $b(r)$ is the root of $U^{\prime}$.

Definition 4.12**.**

A correction of a tree $U$ that is increasing in a round $t$ is a tree $U^{\prime}$ over the same nodes as $U$ that is non-increasing in round $t$, is isomorphic to $U$, and whose root is a node $u\in S_{t-1}$ such that $P_{U}(u)\in I_{t-1}$.

Lemma 4.13**.**

For any increasing tree $U$, there exists a correction $U^{\prime}$.

Proof.

Let $V(U)$ be the set of nodes of $U$ and let $|V(U)\cap S_{t-1}|=\ell$. To show the lemma we will give a bijection $b$ that maps the $\ell$ uninformed nodes of $V(U)$ to the $\ell$ first nodes of $U$ in bfs-order and the informed nodes to the remaining nodes of $U$. The resulting tree will be the correction $U^{\prime}$. As a result of this bijection every uninformed node of $U^{\prime}$ has only uninformed ancestors and, thus, $U^{\prime}$ is non-increasing.

More formally, if $U$ is increasing, then there exists an edge $(i,s)$ such that $i\in I_{t-1},s\in S_{t-1}$. Let $\pi$ be a bijection from $[\left|V(U)\right|]$ to $V(U)$ such that $\pi(1)=s,\{\pi(2),\dots,\pi(\ell)\}\subset S_{t-1}$, and $\{\pi(\ell+1),\dots,\pi(\left|V(U)\right|)\}\subseteq I_{t-1}$. On another hand, let $\rho$ be a bijection from $[\left|V(U)\right|]$ to $V(U)$ such that $\rho(j)$ is the $j$-th node encountered in a breadth-first traversal starting at the root of $U$. Then let $b=\pi\circ\rho^{-1}$. Note that the tree $U^{\prime}$, whose set of edges is $\{(b(u),b(v)):(u,v)\in U\}$ is a correction of $U$. Indeed, it is clearly a tree as a relabeling of $U$, over the same nodes as $U$, and for every $(b(u),b(v))\in U^{\prime}$, $u$ is encountered in a BFS before $v$ in $U$, therefore $\rho^{-1}(u)<\rho^{-1}(v)$, and therefore if $\rho^{-1}(u)\geq\ell$, we also have $\rho^{-1}(v)\geq\ell$. This means that if $b(u)\in I_{t-1}$, then $b(v)\in I_{t-1}$. ∎

Lemma 4.14**.**

Let $t$ be a round and $N_{t-1}$ be the number of informed nodes after round $t-1$. Let $E_{1},E_{2}$ be two sets of edges that the adversary could choose for round $t$ such that

$E_{1}$* is a collection of rooted trees such that at least one tree $U$ is information-increasing, and* 2. 2.

$E_{2}$* is obtained from $E_{1}$ by replacing $U$ with a correction $U^{\prime}$ of $U$.*

Let $N^{(1)}_{t}$ be the number of informed nodes after round $t$ if $E_{1}$ is chosen, and let $N^{(2)}_{t}$ be that number if $E_{2}$ is chosen. Then choosing $E_{2}$ is a better strategy for the adversary than choosing $E_{1}$.

The proof can be found in Appendix C. This lemma proves that the adversary will never choose a set of edges such that one (or more) component is increasing. Indeed, if such components existed, then the adversary would have replaced all of them with non-increasing ones, as this will lead to no fewer and potentially more rounds. Therefore, we can assume in the following that all components are non-increasing.

The next step is to show that if the adversary chooses a forest, all edges will be used in one component. For that, we introduce the notion of merging trees, and show that if the adversary chooses a forest with 2 or more non-trivial components, then merging two of those non-trivial components will yield a better strategy for the adversary.

Lemma 4.15**.**

Let $t$ be a round, let $E$ be the set of $k$ edges forming a directed rooted forest over $[n]$ which the adversary chooses in round $t$ such that each component of $E$ is non-increasing, and let $s_{1},\dots,s_{x}$ be uninformed nodes that are roots of their component (which might have size only 1). Note that $\{s_{1},\dots,s_{x}\}$ needs not be the set of the roots of all components, simply a collection of some of them. Let $\eta_{1},\dots,\eta_{x}$ be the number of informed nodes in the component of $s_{1},\dots,s_{x}$ respectively, and $\eta$ the number of informed nodes outside the components of $s_{1},\dots,s_{x}$. Then we have that:

[TABLE]

Proof.

We have that:

[TABLE]

However, many terms of that sum are equal to [math]. Indeed, for example, if $a_{1}$ is one of the $\eta_{1}$ informed nodes in the component of $s_{1}$, then $\mathbb{P}(P_{t}(s_{1})=a_{1})=0$. More generally, if the choice of $a$ is so that $E\cup\bigcup_{j\in[i]}(a_{j},s_{j})$ contains an (undirected cycle), in other words, is incompatible with a rooted tree, then $\mathbb{P}(P_{t}(s_{1})=a_{1})=0$. If, on the other hand, the choice of $a$ is compatible with a rooted tree, then, applying Theorem 2.1, we have:

[TABLE]

We now have to count how many choices of $a$ are compatible with a rooted tree. Let us first assume that none of the $\eta_{j}$ nor $\eta$ is equal to [math]. Let $\alpha$ denote the set of all such values of $a$, and define $\beta$ as follows: create a forest $F$ with $x+1$ (directed) line graphs, each line having respectively $\eta_{1},\dots,\eta_{x},\eta$ nodes. Then $\beta$ is the set of all rooted trees that are compatible with $F$, and whose root is the root of the last tree of $F$.

To determine $|\alpha|$, we show that there is a bijection between $\alpha$ and $\beta$ and determine $|\beta|$. To create the bijection first take an arbitrary but fixed bijection $b$ that maps every informed node from $I_{t-1}$ to a node from $F$, such that an informed node from the component of $s_{j}$ is mapped to a node of the $j-th$ line of $F$. Then we can map a choice of $a\in\alpha$ to a tree $T\in\beta$ by setting the parent in $T$ of the root of the $j-th$ line to be $b(a_{j})$ for every $j$. Note that this uniquely identifies a tree of $\beta$. Conversely, to find a choice $a\in\alpha$ from a tree $T\in B$, set $a_{j}=b^{-1}(p_{j})$ where $p_{j}$ is the parent of the root of the $j$-th line of $F$ in $T$. Now note that $\beta$ is the set of all rooted trees that are compatible with $F$, and whose root is the root of the last tree of $F$. By Theorem 2.2, $\left|\beta\right|=\eta(\eta+\sum_{j\in[x]}\eta_{j})^{x-1}$, which concludes the proof.

If $\eta=0$, it is easy to see that no choice of $a$ is compatible with a rooted tree.

If there exists some values of $j$ such that $\eta_{j}=0$, then assume wlog that $\eta_{1}=\dots=\eta_{\ell}=0$, and $\eta_{j}>0$ for every $j>\ell$. As seen above, there will be $\eta(\eta+\sum_{j\in[x]}\eta_{j})^{x-\ell-1}$ choices for $(a_{\ell+1},\dots,a_{x})$. Once this choice is made, for every $1\leq j\leq\ell$, $a_{j}$ can take any value in $I_{t-1}$, where $\left|I_{t-1}\right|=\eta+\sum_{j\in[x]}\eta_{j})$. The total number of choices for $a$ is $\eta(\eta+\sum_{j\in[x]}\eta_{j})^{x-1}$. ∎

The following merge operation combines two trees such as to make a non-informed root the root of the merged tree, if at least one of the roots is non-informed.

Definition 4.16**.**

We say that we merge two non-trivial trees $U$ and $U^{\prime}$ with respective roots $r$ and $r^{\prime}$ in round $t$ when we apply the following operation:

• •

If $r\in I_{t-1}$, then for every $p\in U$ with $(r,p)\in U$, replace edge $(r,p)$ with the edge $(r^{\prime},p)$.

• •

If $r\notin I_{t-1}$, then for every $p\in U^{\prime}$ with $(r^{\prime},p)\in U^{\prime}$, replace edge $(r^{\prime},p)$ with the edge $(r,p)$.

Lemma 4.17**.**

Let $t$ be a round and $N_{t-1}$ be the number of informed nodes after round $t-1$. Let $E_{1},E_{2}$ be two sets of edges that the adversary could choose for round $t$, as follows: let $E_{1}$ be a collection of rooted trees such that every tree is non-increasing, with at least two non-trivial components $U$ with root $r$ and $U^{\prime}$ with root $r^{\prime}$, and let $E_{2}$ be obtained from $E_{1}$ by merging $U$ and $U^{\prime}$. Let $N^{(1)}_{t}$ be the number of informed nodes after round $t$ if $E_{1}$ is chosen, and $N^{(2)}_{t}$ if $E_{2}$ is chosen. Then choosing $E_{2}$ is a better strategy for the adversary than choosing $E_{1}$.

The proof of this lemma being fairly technical, we delay it to Appendix C

This lemma implies that the adversary will never choose a set of edges with more than one non-trivial component, i.e., the adversary will choose one tree with $k+1$ nodes. We already showed that the adversary will only choose non-increasing components. Therefore, we are left with analyzing the case where the adversary chooses one non-trivial non-increasing tree with $k+1$ nodes.

Lemma 4.18**.**

Let $t$ be a round and $N_{t}$ be the number of informed nodes after round $t$. Let $U$ be a non-increasing tree over $k+1$ nodes in round $t+1$. Let $\sigma$ be the number of uninformed nodes in $U$ and $\eta$ the number of informed nodes in $U$. Then the distribution of $N_{t+1}-N_{t}$ equals the sum of of $n-N_{t}-\sigma$ independent Bernoulli random variables of parameter $\frac{N_{t}}{n}$ plus one Bernoulli random variable of parameter $\frac{N_{t}-\eta}{n}$

As this proof is similar to the proof of Lemma 3.1, we delay it to Appendix C.

Corollary 4.19**.**

Let $t$ be a round and $N_{t}$ be the number of informed nodes after round $t$. Let $U$ be a non-increasing tree over $k+1$ nodes in round $t+1$ and let $\eta$ be its number of informed nodes in $U$. The optimal strategy for the adversary is to minimize $\eta$ in every round.

Proof.

Note that we always have $\sigma+\eta=k+1$. Let us consider two non-increasing trees $U$ and $U^{\prime}$ over $k+1$ nodes. Let $\eta_{1}$ (resp. $\sigma_{1}$) be the number of informed (resp. uninformed) nodes in $U$, and $\eta_{2}$ (resp. $\sigma_{2}$) be the number of informed (resp. uninformed) nodes in $U^{\prime}$. Assume wlog that $\eta_{1}>\eta_{2}\geq 0$. Then $\sigma_{1}<\sigma_{2}$. Let $N^{(1)}_{t+1}-N_{t}$ and $N^{(2)}_{t+1}-N_{t}$ be the number of newly informed nodes after round $t+1$ if the adversary chooses respectively tree $U$ or $U^{\prime}$. The distribution of $N^{(1)}_{t+1}-N_{t}$ is the sum of at least $n-N_{t}-\sigma_{1}$ independent Bernoulli variables of parameter $\frac{N_{t}}{n}$, while $N^{(2)}_{t+1}-N_{t}$ is the sum of at most $n-N_{t}-\sigma_{2}+1$ independent Bernoulli variables of parameter at most $\frac{N_{t}}{n}$. The first distribution clearly dominates the second, and by the Distribution Domination Lemma (Lemma 4.9), the result holds. ∎

This shows that the optimal strategy for the adversary is always to choose $\sigma=k+1$ for the tree $U$ it chooses, unless $N_{t-1}$ is so large that the number of available uninformed nodes is smaller than $k+1$, in which case $\sigma=n-N_{t-1}$. As the number $N_{t}$ of informed nodes never decreases, this leads to the following partitioning of the rounds into two phases: one phase which contains all rounds $t$ with $n-N_{t-1}\geq k+1$, in which case $\sigma=k+1$, and another phase which contains all rounds $t$ with $n-N_{t-1}<k+1$, in which case $\sigma=n-N_{t-1}$. We will show that the first phase takes $O(\log n)$ rounds, while the second one takes $O(k+\log n)$ rounds.

4.2 Phase 1

As the analysis of this phase is very similar to Section 3, we delay the proofs to Appendix C.1. We however state the main result here:

Lemma 4.20**.**

If $n-k>4$ then Phase 1 ends within $8(3+\sqrt{5})\log n$ rounds with probability $p\geq 1-n^{-2}$.

4.3 Phase 2

Phase two starts when there are only $k$ more nodes to infect. This essentially means that the adversary can protect all uninformed nodes but one, as the trees they will choose will have an uninformed root and might get informed in this round, but all uninformed nodes below it will not become informed in the current round.

Lemma 4.21**.**

Let $\gamma=\frac{65}{32}+\frac{5\sqrt{105}}{32}\approx 3.63$. Phase 2 ends within $\gamma(\log n+k)$ rounds with probability $p\geq 1-n^{-2}$, if $n\geq 10$.

Proof.

In each round, by Lemma 4.18, the root of the tree the adversary chooses gets informed with probability $\frac{n-k-1}{n}\geq\frac{8}{10}$, where the inequality holds as $k\leq n/10$ and $n\geq 10$. Assimilating this to a flip of a coin where the coin has probability $\frac{8}{10}$ of landing on heads, and flipping the coins $\gamma(k+\log n)$ times, we are asking what is the probability $p$ of the coin landing on heads at least $k$ times. Again, using Hoeffding’s inequality (Lemma 3.12), we have that:

[TABLE]

∎

4.4 Combining Phase 1 and 2

We first combine the results for Phases 1 and 2 to show that broadcast completes in $O(\log n+k)$ rounds if $k\leq\frac{n}{10}$:

Theorem 4.22**.**

If the adversary can control $k\leq n/10$ edges in each round, broadcast completes within $(24+\gamma+8\sqrt{5})\log n+\gamma k$ rounds with probability $p\geq 1-2n^{-2}$

Proof.

This is a direct result of Lemmata 4.20 and 4.21 ∎

And then combine this result with Lemma 4.3, that dealt with the case $k\geq\frac{n}{10}$, to give the general result:

Theorem 4.23**.**

If the adversary can control $k$ edges in each round, broadcast completes within $O(\log n+k)$ rounds, with probability $p\geq 1-2n^{-2}$.

Proof.

This is a consequence of Theorem 4.22 and Lemma 4.3 ∎

4.5 Consensus

Finally, we see that a direct application of Theorem 4.22 gives us a reliable algorithm for Consensus with a Randomized Oblivious Message Adversary of parameter $k$, as long as $k\leq\frac{n}{10}$:

See 1.7

Proof.

By Theorem 4.22, node 1 broadcasts within $(24+\gamma+8\sqrt{5})\log n+\gamma k$ rounds with probability $p\geq 1-2n^{-2}$. Therefore, Algorithm 1 achieves consensus within $(24+\gamma+8\sqrt{5})\log n+\gamma k$ rounds with probability $p\geq 1-2n^{-2}$. ∎

5 Related Work

Information dissemination in general and broadcasting in particular are fundamental topics in distributed computing, also because of the crucial role they play for consensus [20]. In contrast to this paper, most classic literature on network broadcast as well as on related tasks such as gossiping, considers a static setting, e.g., where in each round each node can send information to one neighbor [22, 16].

Kuhn, Lynch and Oshman [23] explore the all-to-all data dissemination problem (gossiping) in an undirected dynamic network, where nodes do not know beforehand the total number of nodes and must decide on that number. Ahmadi, Kuhn, Kutten, Molla and Pandurangan [1] study the message complexity of broadcast in an undirected dynamic setting, where the adversary pays up a cost for changing the network.

In dynamic networks, the oblivious message adversary is a commonly considered model, especially for broadcast and consensus problems, first introduced by Charron-Bost and Schiper [4]. The broadcast problem under oblivious message adversaries has been studied for many years. A first key result for this problem was the $n\log n$ upper bound by Zeiner, Schwarz, and Schmid [28] who also gave a $\left\lceil\frac{3n-1}{2}\right\rceil-2$ lower bound. Another important result is by Függer, Nowak, and Winkler [17] who presented an $O(\log\log n)$ upper bound if the adversary can only choose nonsplit graphs; combined with the result of Charron-Bost, Függer, and Nowak [3] that states that one can simulate $n-1$ rounds of rooted trees with a round of a nonsplit graph, this gives the previous $O(n\log\log n)$ upper bound for broadcasting on trees. Dobrev and Vrto [9, 8] give specific results when the adversary is restricted to hypercubic and tori graphs with some missing edges. El-Hayek, Henzinger, and Schmid [12, 13] recently settled the question about the asymptotic time complexity of broadcast by giving a tight $O(n)$ upper bound, also showing the upper bound still holds in more general models. Regarding consensus, Coulouma, Godard and Peters in [6] presented a general characterization on which dynamic graphs consensus is solvable, based on broadcastability. Winkler, Rincon Galeana, Paz, Schmid, and Schmid [18] recently presented an explicit decision procedure to determine if consensus is possible under a given adversary, enabling a time complexity analysis of consensus under oblivious message adversaries, both for a centralized decision procedure as well as for solving distributed consensus. They also showed that reaching consensus under an oblivious message adversary can take exponentially longer than broadcasting.

In contrast to the above works, in this paper we study a more randomized message adversary, considering a stochastic model where adversarial graphs are partially chosen uniformly at random. While a randomized perspective on dynamic networks is natural and has been considered in many different settings already, existing works on random dynamic communication networks, e.g., on the radio network model [14], on rumor spreading [5], as well as on epidemics [11], do not consider oblivious message adversaries. Note, however, that the information dissemination considered in this paper is similar to the SI model for virus propagation, with results having implications in both directions [15]. For example, Doerr and Fouz [10] introduced an information dissemination protocol inspired by epidemics. More generally, randomized information dissemination protocols can be well-understood from an epidemiological point-of-view, and are very similar to the SI model which has been very extensively studied. In contrast to the typical SI models considered in the literature [25], however, our model in this paper revolves around tree communication structures which introduce additional technical challenges. Furthermore, existing literature often provides results in expectation, while we in this paper provide tail bounds.

6 Conclusion

We studied the fundamental problems of broadcast and consensus on dynamic networks from a randomized perspective, studying randomized oblivious message adversaries with parameter $k$. We showed that for small values of $k$ information dissemination is significantly faster compared to the deterministic setting.

We believe that our work opens several interesting avenues for future research. In particular, it would be interesting to extend our study of randomized oblivious message adversaries to other information dissemination problems and network topologies. We also believe that our techniques can be useful to analyze other dynamic models, including the SI model in epidemics.

Appendix A Lower Bound for Deterministic Broadcast in Constant Height Trees

In this section, we consider a very similar model to [13], the only difference being that the adversary is restricted to choosing trees of height at most $2$.

Model

. We are given $n$ nodes, and these nodes can communicate in synchronous rounds. Each node has a distinct I.D., and aims to share this I.D. with as many nodes as possible. In the beginning, each node only knows its own I.D.. An adversary chooses for each round a directed network along which nodes can communicate, among a set $A$ of allowed networks. In each round, each node sends all I.D.s it has received in previous rounds to each one of its out-neighbors. The adversary’s goal it to maximize the number of rounds until broadcast, that is, until one I.D. has been received by everyone. The question is: how many rounds can the adversary delay broadcast, depending on $A$?

Authors in [13] have shown that if $A$ is the set of rooted trees, then the adversary can delay broadcast for a linear number of rounds. Since a linear number of rounds is easily achievable by the adversary simply by taking a line graph $L$, and using $L$ as the communication network in each round, one would think that the height of the trees allowed play an important role to determine broadcast time. We give in Figure 1 a counter example, where $A$ is the set of rooted trees of height at most $2$, and where broadcast needs at least a linear number of rounds.

In this example, in round $t$, for $t<n-2$, the adversary chooses the tree rooted at node $1$, with edges $(1,t+1)$ and $(t+1,i)$ for every $i\in[n]\setminus\{1,t+1\}$. Since node $1$ never has an in-neighbor, broadcast completes when the I.D. of node $1$ is shared to every node. It is easy to see that this only happens after round $n-2$.

Appendix B Probabilities tools

Lemma B.1**.**

Let $X_{1},\dots,X_{m}$ and $Y_{1},\dots Y_{m}$ be binary random variables such that for every $I\subseteq[m]$ we have that $\mathbb{P}(\cap_{i\in I}(X_{i}=1)\cap_{i\notin I}(X_{i}=0))=\mathbb{P}\left(\cap_{i\in I}(Y_{i}=1)\cap_{i\notin I}(Y_{i}=0)\right)$, then the probability distribution of $\sum_{i\in[m]}X_{i}$ is equal to the probability distribution of $\sum_{i\in[m]}Y_{i}$.

Proof.

We have, for every $k\in[m]$:

[TABLE]

∎

Lemma B.2**.**

Let $X_{1},\dots,X_{m}$ and $Y_{1},\dots Y_{m}$ be binary random variables such that for every $\ell\in\mathbb{N}$, $\sum_{\left|I\right|=\ell}\mathbb{P}(\cap_{i\in I}(X_{i}=1)\cap_{i\notin I}(X_{i}=0))=\sum_{\left|I\right|=\ell}\mathbb{P}(\cap_{i\in I}(Y_{i}=1)\cap_{i\notin I}(Y_{i}=0))$, then the probability distribution of $\sum_{i\in[m]}X_{i}$ is equal to the probability distribution of $\sum_{i\in[m]}Y_{i}$.

Proof.

We have, for every $k\in[m]$:

[TABLE]

∎

Lemma B.3**.**

Let $X_{1},\dots,X_{m}$ and $Y_{1},\dots Y_{m}$ be binary random variables such that for every $I\subset[m],\mathbb{P}(\cap_{i\in I}(X_{i}=1))=\mathbb{P}(\cap_{i\in I}(Y_{i}=1))$, then the probability distribution of $\sum_{i\in[m]}X_{i}$ is equal to the probability distribution of $\sum_{i\in[m]}Y_{i}$.

Proof.

We start by proving by induction on the size of $J$, $\mathbb{P}(\cap_{i\in I}(X_{i}=1)\cap_{j\in J}(X_{j}=0))=\mathbb{P}(\cap_{i\in I}(Y_{i}=1)\cap_{j\in J}(Y_{j}=0))$ for any $I,J\subseteq[n]$ such that $I\cap J=\varnothing$. This is clear for $\left|J\right|=0$.

Let $I,J\subseteq[n]$ such that $I\cap J=\varnothing$ and $\left|J\right|>0$. Let $a$ be an element of $J$. Then we have:

[TABLE]

Similarly:

[TABLE]

By induction hypothesis, we have:

[TABLE]

Hence:

[TABLE]

The result follows from Lemma B.1, when we take $J=[m]\setminus I$ ∎

Lemma B.4**.**

Let $X_{1},\dots,X_{m}$ and $Y_{1},\dots Y_{m}$ be binary random variables such that $\sum_{\left|I\right|=\ell}\mathbb{P}(\cap_{i\in I}(X_{i}=1))=\sum_{\left|I\right|=\ell}\mathbb{P}(\cap_{i\in I}(Y_{i}=1))$ for every $\ell\in\mathbb{N}$, then the probability distribution of $\sum_{i\in[m]}X_{i}$ is equal to the probability distribution of $\sum_{i\in[m]}Y_{i}$.

Proof.

We start by proving by induction on $k$, that for every $\ell,k\in\mathbb{N},\sum_{\left|I\right|=\ell}\mathbb{P}(\cap_{i\in I}(X_{i}=1)\cap_{j\in J_{I}}(X_{j}=0))=\sum_{\left|I\right|=\ell}\mathbb{P}(\cap_{i\in I}(Y_{i}=1)\cap_{j\in J_{I}}(Y_{j}=0))$ for any choice of $J_{I}\subseteq[n]$ such that $I\cap J_{I}=\varnothing$ and $\left|J_{I}\right|=k$. This is clear for $k=0$.

For the induction case, let us assume, that for $k>1$, we have that for every $\ell\in\mathbb{N},\sum_{\left|I\right|=\ell}\mathbb{P}(\cap_{i\in I}(X_{i}=1)\cap_{j\in J_{I}}(X_{j}=0))=\sum_{\left|I\right|=\ell}\mathbb{P}(\cap_{i\in I}(Y_{i}=1)\cap_{j\in J_{I}}(Y_{j}=0))$ for any choice of $J_{I}\subseteq[n]$ such that $I\cap J_{I}=\varnothing$ and $\left|J_{I}\right|=k-1$.

Let us fix $\ell$, and for every $I\subseteq[m]$ such that $\left|I\right|=\ell$, let $J_{I}\subseteq[m]$ be such that $I\cap J_{I}=\varnothing$ and $\left|J_{I}\right|=k>0$. Let $a_{I}$ be an element of $J_{I}$. Then we have:

[TABLE]

Similarly:

[TABLE]

By induction hypothesis, we have:

[TABLE]

Hence:

[TABLE]

This concludes the induction step.

The result follows from Lemma B.2, when we take for every $I$, $J_{I}=[m]\setminus I$, for $k=m-\ell$. ∎

Lemma B.5**.**

Let $X_{1},\dots,X_{m}$ and $Y_{1},\dots Y_{m}$ be binary random variables, $\alpha\in\mathbb{R},\alpha\geq 1$ and $r\in\mathbb{N}$ such that for any $I\subseteq[m]\setminus\{r\},\mathbb{P}(\cap_{i\in I}(X_{i}=1))=\mathbb{P}(\cap_{i\in I}(Y_{i}=1))$, and $\mathbb{P}(\cap_{i\in I\cup\{r\}}(X_{i}=1))=\alpha\mathbb{P}(\cap_{i\in I\cup\{r\}}(Y_{i}=1))$ then $\sum_{i\in[m]}X_{i}$ stochastically dominates $\sum_{i\in[m]}Y_{i}$.

Proof.

We start by proving by induction on the size of $J$, for every $I,J\subset[m]\setminus\{r\}$ such that $I\cap J=\varnothing,\mathbb{P}(\cap_{i\in I}(X_{i}=1)\cap_{j\in J}(X_{j}=0))=\mathbb{P}(\cap_{i\in I}(Y_{i}=1)\cap_{j\in J}(Y_{j}=0))$. This is clear for $\left|J\right|=0$.

Let $I,J\subseteq[n]$ such that $I\cap J=\varnothing$ and $\left|J\right|>0$. Let $a$ be an element of $J$. Then we have:

[TABLE]

Similarly:

[TABLE]

By induction hypothesis, we have:

[TABLE]

Hence:

[TABLE]

We then show by induction on the size of $J$, that for every $I,J\subset[m]$ such that $r\in I$, $I\cap J=\varnothing,\mathbb{P}(\cap_{i\in I}(X_{i}=1)\cap_{j\in J}(X_{j}=0))=\alpha\mathbb{P}(\cap_{i\in I}(Y_{i}=1)\cap_{j\in J}(Y_{j}=0))$ for any $I,J\subseteq[n]$ . This is clear for $\left|J\right|=0$.

Let $I,J\subseteq[n]$ such that $r\in I$, $I\cap J=\varnothing$ and $\left|J\right|>0$. Let $a$ be an element of $J$. Then we have:

[TABLE]

Similarly:

[TABLE]

By induction hypothesis, we have:

[TABLE]

Hence:

[TABLE]

We now show that for any $x\in\mathbb{N}$, we have that $\mathbb{P}\left(\sum_{i\in[m]}X_{i}\geq x\right)\geq\mathbb{P}\left(\sum_{i\in[m]}Y_{i}\geq x\right)$. Indeed, we have:

[TABLE]

∎

Lemma B.6**.**

Let $X$ be a random variable that has a binomial distribution of parameters $(m,p)$. Then if $0<p\leq\frac{1}{m}$, we have that $\mathbb{P}(X\geq mp)\geq\frac{1}{4}$ as soon as $p\geq 1-\left(\frac{3}{4}\right)^{\frac{1}{m}}$.

In particular, it suffices for $p$ to be larger than $\frac{1}{3m}$.

Proof.

If $p\leq\frac{1}{m}$ then $0<mp\leq 1$ which means that the events $X\geq mp$ and $X\geq 1$ are the same since the binomial distribution takes only integer values. Hence:

[TABLE]

As the function $\frac{1}{3m}-1+\left(\frac{3}{4}\right)^{\frac{1}{m}}$ is positive for $m=1$ and strictly decreasing towards [math] with increasing $m$, it is always positive and thus the second claim holds. ∎

Definition B.7** (Mutually independent events).**

Let $A_{1},\dots,A_{n}$ be events. They are said to be mutually independent if and only if, for every subset $I\subseteq[n]$, we have that:

[TABLE]

Lemma B.8**.**

If $A_{1},\dots,A_{n}$ are mutually independent events, then we have that, for every subsets $I,J\subseteq[n]$, $I\cap J=\varnothing$:

[TABLE]

Proof.

We will show by induction on the size of $J$ that this holds for every $I\subseteq[n]$ such that $I\cap J=\varnothing$. It is clear for the case $\left|J\right|=0$.

Let $J$ be a nonempty subset of $[n]$ and $I$ a subset of $[n]$ such that $I\cap J=\varnothing$. Let $a\in J$. Then we have that, by the induction hypothesis:

[TABLE]

However, we also have that:

[TABLE]

Again, by the induction hypothesis, we have that

[TABLE]

Piecing everything together, we get that:

[TABLE]

∎

Appendix C Ommitted proofs of Section 4

See 4.9

Proof.

Saying that $N_{t}^{(1)}$ stochastically dominates $N^{(2)}_{t}$ is equivalent to saying that $I_{t}^{(1)}$ stochastically dominates $I^{(2)}_{t}$.By Theorem 4.8, we introduce a coupling $(\hat{I}^{(1)}_{t},\hat{I}_{t}^{(2)})$ of $I_{t}^{(1)}$ and $I_{t}^{(2)}$ such that $\mathbb{P}\left(\left|\hat{I}^{(1)}_{t}\right|\geq\left|\hat{I}_{t}^{(2)}\right|\right)=1$.

Let $\beta$ be a bijection from $[n]$ to $[n]$ such that $\beta(\hat{I}_{t}^{(2)})\subseteq\hat{I}^{(1)}_{t}$. For any $t^{\prime}\geq t$, let $E^{t^{\prime}}_{1}$ and $E^{t^{\prime}}_{2}$ be respectively the edges chosen by the adversary in round $t^{\prime}$ after choosing $E_{1}$, respectively $E_{2}$, in round $t$. Let $T^{t^{\prime}}_{1}$ and $T^{t^{\prime}}_{2}$ be respectively the trees in round $t^{\prime}$ containing $E^{t^{\prime}}_{1}$ and $E^{t^{\prime}}_{2}$, respectively. We introduce a coupling $(\hat{T}^{t^{\prime}}_{1},\hat{T}^{t^{\prime}}_{2})$ such that if $E^{t^{\prime}}_{1}=\beta(E^{t^{\prime}}_{2})$ then $\mathbb{P}(\hat{T}^{t^{\prime}}_{1}=\beta(\hat{T}^{t^{\prime}}_{2}))=1$ as follows: If $E^{t^{\prime}}_{1}=\beta(E^{t^{\prime}}_{2})$, then note that $\beta$ induces a bijection between $\mathcal{T}_{n}^{(2)}=\{T\in\mathcal{T}_{n}:E^{t^{\prime}}_{2}\subseteq T\}$ and $\mathcal{T}_{n}^{(1)}=\{T\in\mathcal{T}_{n}:E^{t^{\prime}}_{1}\subseteq T\}$. In this case to define the coupling choose $\hat{T}^{t^{\prime}}_{2}$ uniformly at random from $\mathcal{T}_{n}^{(2)}$ and set $\hat{T}^{t^{\prime}}_{1}=\beta(\hat{T}^{t^{\prime}}_{2})$. Otherwise to define the coupling choose $\hat{T}^{t^{\prime}}_{2}$ uniformly at random from $\mathcal{T}_{n}^{(2)}$ and, independently, $\hat{T}^{t^{\prime}}_{1}$ uniformly at random from $\mathcal{T}_{n}^{(1)}$. Let $\hat{I}_{1}^{t^{\prime}}$ and $\hat{I}_{2}^{t^{\prime}}$ be the set of informed nodes we get after round $t^{\prime}$ if the trees in round $t^{\prime}$ were $\hat{T}^{t^{\prime}}_{1},$ respectively $\hat{T}^{t^{\prime}}_{2}$.

Let us now assume that the sequence $(E^{t^{\prime}}_{1})_{t^{\prime}>t}$ is an optimal strategy for the adversary after choosing $E_{1}$ in round $t$. Then consider the sequence $\sigma_{2}$ which chooses $E_{2}$ in round $t$ then $\beta^{-1}(E^{t^{\prime}}_{1})$ in rounds $t^{\prime}\geq t$ and compare it with the sequence $\sigma_{1}$ that chooses $E_{1}$ in round $t$ then $E^{t^{\prime}}_{1}$ in rounds $t^{\prime}>t$. Recall that $\sigma_{1}$ is optimal after choosing $E_{1}$ in round $t$. We show by induction that $\sigma_{2}$ is a better overall strategy for the adversary than $\sigma_{1}$. Indeed, we can show by induction on the number of rounds $t^{\prime}\geq t$ that $\mathbb{P}\left(\beta(\hat{I}_{2}^{t^{\prime}})\subseteq\hat{I}_{1}^{t^{\prime}}\right)=1$. The induction basis is trivial as $\beta(\hat{I}_{2}^{t})\subseteq\hat{I}_{1}^{t}$. For the induction step, note that for any $v\in\hat{I}_{2}^{t^{\prime}}$, either (1) $v\in\hat{I}_{2}^{t^{\prime}-1}$ which, using the induction assumption implies with probability 1 that $\beta(v)\in\beta(\hat{I}_{2}^{t^{\prime}-1})\subseteq\hat{I}_{1}^{t^{\prime}-1}\subseteq\hat{I}_{1}^{t^{\prime}}$, or (2) $P_{\hat{T}_{2}^{t^{\prime}}}(v)\in\hat{I}_{2}^{t^{\prime}-1}$. Case (2) implies (by the definition of the coupling) that with probability $1$, $\beta(P_{\hat{T}_{2}^{t^{\prime}}}(v))=P_{\hat{T}_{1}^{t^{\prime}}}(\beta(v))$. As $P_{\hat{T}_{2}^{t^{\prime}}}(v)\in\hat{I}_{2}^{t^{\prime}-1}$ and by induction, with probability 1 $\beta(\hat{I}_{2}^{t^{\prime}-1})\subseteq\hat{I}_{1}^{t^{\prime}-1}$, it follows that with probability 1, it holds that $P_{\hat{T}_{1}^{t^{\prime}}}(\beta(v))=\beta(P_{\hat{T}_{2}^{t^{\prime}}}(v))\in\hat{I}_{1}^{t^{\prime}-1}$ and, thus, $\beta(v)\in\hat{I}_{1}^{t^{\prime}}$. Hence, in both cases, with probability 1, $\beta(I_{2}^{t^{\prime}})\in\hat{I}_{1}^{t^{\prime}}$.

Now note that $\mathbb{P}\left(\beta(\hat{I}_{2}^{t^{\prime}})\subseteq\hat{I}_{1}^{t^{\prime}}\right)=1$ implies that if $t^{\prime}$ is the smallest round at which broadcast completes after the adversary chooses $E_{2}$, that is, if $\left|\hat{I}_{2}^{t^{\prime}-1}\right|=n$, then broadcast completes in a no later round if the adversary chooses $E_{1}$. ∎

See 4.14

Proof.

We will build a bijection $\pi$ from $\mathcal{T}_{n}^{(2)}=\{T\in\mathcal{T}_{n}:E_{2}\subseteq T\}$ to $\mathcal{T}_{n}^{(1)}=\{T\in\mathcal{T}_{n}:E_{1}\subseteq T\}$ such that for every $s\in S_{t-1}$ and any $T\in\mathcal{T}_{n}^{(2)}$ with $P_{T}(s)\in I_{t-1}$, we have that $P_{\pi(T)}(s)\in I_{t-1}$. Hence, $\pi(T)$ has more uninformed nodes that become informed than $T$. We will use this property to show that $N_{t}^{(1)}$ stochastically dominates $N_{t}^{(2)}$.

To do so, let $b$ be the bijection that achieves the isomorphism of the proof of Lemma 4.13 from $U$ to $U^{\prime}$. $\pi(T)$ is constructed in a way such that all nodes have the same parents as in $T$, unless they are in $U^{\prime}$. More specifically, we let $\pi(T):=\pi_{b}(T)$ where $\pi_{b}(T)$ is the tree obtained from $T$ by replacing every edge $(u,v)\in T$ as follows:

• •

if $u,v\in U^{\prime}$, then replace it with the edge $(b^{-1}(u),b^{-1}(v))$.

• •

if $u\notin U^{\prime}$, $v\notin U^{\prime}$, then keep it the same.

• •

if $u\in U^{\prime}$, $v\notin U^{\prime}$, then keep it the same.

• •

if $u\notin U^{\prime},v\in U^{\prime}$, then replace it with $(u,b^{-1}(v))$.

We clearly have that $U\subseteq E_{1}\subset\pi(T)$ and $U^{\prime}\subseteq E_{2}\subset T$. Also, for any node $v$, the path from the root to $v$ in $T$ can be transformed into a path from the root in $\pi(T)$ by replacing the subpath $P=u_{0},\dots,u_{\ell}$ that is in $U^{\prime}$ with the path from $b^{-1}(u_{0})$ to $u_{\ell}$ in $U$. Hence $\pi(T)\in\mathcal{T}_{n}^{(1)}$. Since $\pi_{b^{-1}}$ is clearly an inverse of $\pi_{b}$, we have that $\pi$ is a bijection.

Let $s\in S_{t-1}$ be such that $P_{T}(s)\in I_{t-1}$. If $s\notin U^{\prime}$, then it has the same parent in $T$ and in $\pi(T)$. If $s\in U^{\prime}$, which is a non-increasing tree, then by the fact that the parent of $s$ in $T$ belongs to $I_{t-1}$ it follows that $s$ is the root of $U^{\prime}$, and, thus, that its parent does not belong to $U^{\prime}$. By the definition of a correction it follows that the parent of $s$ in $U$ is informed. As $U\subseteq\pi(T)$ the parent of $s$ in $\pi(T)$ is a node of $I_{t-1}$.

We, therefore, have that, for every $x\in\mathbb{N}$:

[TABLE]

The lemma now follows from the Distribution Domination Lemma (Lemma 4.9). ∎

See 4.17

Proof.

We will show that for any $x\in\mathbb{N}$, we have that $\mathbb{P}(N^{(1)}_{t}-N_{t-1}=x)\geq\mathbb{P}(N^{(2)}_{t}-N_{t-1}=x)$. Then the result will follow from the Distribution Domination Lemma.

In the following let $S$ be a set of uninformed nodes $s_{1},\dots,s_{|S|}$, let $\eta_{j}$ for $1\leq j\leq|S|$ be the number of informed nodes in the connected component of $s_{j}$ in $E_{1}$ and let $\eta(S)$ be the number of informed nodes that do not belong to the connected component of any $s_{j}$.

We will analyze the value of $\mathbb{P}(\cap_{s\in S}P_{t}(s)\in I_{t-1})$ when the adversary chooses $E_{1}$, and when it chooses $E_{2}$. Then two cases can arise: Either the value of $\sum_{\left|S\right|=\ell}\mathbb{P}(\cap_{s\in S}P_{t}(s)\in I_{t-1})$ is equal whether the adversary chooses $E_{1}$ or $E_{2}$, and this for every $\ell$, and the result will follow Lemma B.4, or $\mathbb{P}(\cap_{s\in S}P_{t}(s)\in I_{t-1})$ will be the same whether the adversary chooses $E_{1}$ or $E_{2}$ for every set $S$ except if $S$ includes a particular node, where there will be a constant factor difference between the two values, and the result will then follow from Lemma B.5.

As all trees in $E_{1}$ (respectively $E_{2}$) are non-increasing, the parent in $E_{1}$ (respectively $E_{2}$) of every non-root node $s\in S$ is uniformed. Thus, if there exists a node in $S$ that is not a root of $E_{1}$ (respectively $E_{2}$) then $\mathbb{P}(\cap_{s\in S}P_{t}(s)\in I_{t-1})=0.$ Hence, we only need to analyze the setting where all nodes of $S$ are roots in $E_{1}$.

Case A: Let us first consider the case $r\in I_{t-1}$ in which case the merge of $U$ and $U^{\prime}$ makes all children of $r$ to children of $r^{\prime}$. In that case, $r\notin S$ and let $\gamma=\left|U\right|-1$. We have two subcases: (A1) If $r^{\prime}\notin S$, then the number of informed nodes in none of the components with roots in $S$, $\eta(S)$, remains unchanged. It follows from Lemma 4.15 that $\mathbb{P}(\cap_{s\in S}P_{t}(s)\in I_{t-1})$ is the same whether the adversary chooses $E_{1}$ or $E_{2}$. (A2) If $r^{\prime}\in S$, wlog assume that $r^{\prime}=s_{1}$. Then we have that $\mathbb{P}(\cap_{s\in S}P_{t}(s)\in I_{t-1})=\frac{\eta(S)(N_{t-1})^{\left|S\right|-1}}{n^{\left|S\right|}}$ if the adversary chooses $E_{1}$, while $\mathbb{P}(\cap_{s\in S}P_{t}(s)\in I_{t-1})=\frac{(\eta(S)-\gamma)(N_{t-1})^{\left|S\right|-1}}{n^{\left|S\right|}}$ if the adversary chooses $E_{2}$. Applying Lemma B.5 where we set $X_{s}=P_{t}(s)\in I_{t-1}$ if the adversary chooses $E_{1}$, and $Y_{s}=P_{t}(s)\in I_{t-1}$ if the adversary chooses $E_{2}$, and $\alpha=\frac{\eta(S)}{\eta(S)-\gamma}$, we have that $N_{t}^{(1)}-N_{t-1}=\sum_{s\in S_{t-1}}X_{s}$ stochastically dominates $N_{t}^{(2)}-N_{t-1}=\sum_{s\in S_{t-1}}Y_{s}$. The result follows from the Distribution Domination Lemma.

Case B: Let us now look at the case where $r\notin I_{t-1}$. In this case the merge of $U$ and $U^{\prime}$ makes all children of $U^{\prime}$ children of $U$.

We consider again two cases: (B1) If $r^{\prime}\in I_{t-1}$, then this case is symmetric to Case (A1) and the same proof as above applies.

(B2) If $r^{\prime}\notin I_{t-1}$, for any $\ell\in\mathbb{N}$, we have that:

[TABLE]

We need to analyze the three sums.

For the first two sums, where both $r$ and $r^{\prime}$ or neither belong to $S$, the number of informed nodes in none of the components with root in $S$, $\eta(S)$, is not different in $E_{1}$ and in $E_{2}$ and, thus, Lemma 4.15 implies that $\mathbb{P}(\cap_{s\in S}P_{t}(s)\in I_{t-1})$ does not change whether the adversary chooses $E_{1}$ or $E_{2}$.

For the third sum, let $\gamma,\gamma^{\prime}$ be respectively the number of informed nodes in the component of $r,r^{\prime}$ in $E_{1}$. Let us first consider the case where the adversary chooses $E_{1}$. We have, by Lemma 4.15:

[TABLE]

and

[TABLE]

therefore:

[TABLE]

If the adversary chooses $E_{2}$, then $r$ has $\gamma+\gamma^{\prime}$ informed nodes in its component in $E_{2}$, while $r^{\prime}$ has [math] of them. by Lemma 4.15:

[TABLE]

and

[TABLE]

therefore:

[TABLE]

Therefore, $\sum_{\left|S\right|=\ell}\mathbb{P}(\cap_{s\in S}P_{t}(s)\in I_{t-1})$ has the same value whether the adversary chooses $E_{1}$ or $E_{2}$. Applying Lemma B.4 where we set $X_{s}=P_{t}(s)\in I_{t-1}$ if the adversary chooses $E_{1}$, and $Y_{s}=P_{t}(s)\in I_{t-1}$ if the adversary chooses $E_{2}$, we have that $N_{t}^{(1)}-N_{t-1}=\sum_{s\in S_{t-1}}X_{s}$ and $N_{t}^{(2)}-N_{t-1}=\sum_{s\in S_{t-1}}Y_{s}$ have the same distribution. The result follows from the Distribution Domination Lemma.

∎

See 4.18

Proof.

Let $I_{t}=\{i_{1},\dots,i_{N_{t}}\}$ and $S_{t}=\{s_{1},\dots,s_{n-N_{t}}\}$ such that $i_{1},\dots,i_{\eta}$ are nodes of $U$, $s_{1},\dots,s_{\sigma-1}$ are uninformed nodes of $U$ that are not the root, and $s_{\sigma}$ is the root of $U$. As $U$ is non-increasing $s_{1},\dots s_{\sigma-1}$ cannot get informed in round $t+1$. As the parent of $s_{\sigma}$ does not belong to $U$, it cannot belong to $i_{1},\dots,i_{\eta}$. We will show that the events uninformed node $s$ gets informed in round $t+1$ for different uninformed nodes $s\in[\sigma,n-N_{t}]$ are mutually independent. To do so we take some $J\subseteq[\sigma,n-N_{t}]$ and analyze the event $\bigcap_{y\in J}(P_{t}(s_{y})\in I_{t})$, We distinguish two cases.

Case 1: If $\sigma\notin J$, then it holds that

[TABLE]

By Theorem 2.1, we have that $\left|\{T^{\prime}\in\mathcal{T}_{n}:U\subset T^{\prime}\}\right|=n^{n-1-k}$, and $\left|\{T^{\prime}\in\mathcal{T}_{n}:P_{U}(s_{y})=i_{a_{y}},\forall y\in J\land U\subset T^{\prime}\}\right|=$ $n^{n-1-k-\left|J\right|}$. Therefore it follows that:

[TABLE]

Case 2: If $\sigma\in J$, we have to take extra care of node $s_{\sigma}$:

[TABLE]

By Theorem 2.1, we have that $\left|\{T^{\prime}\in\mathcal{T}_{n}:U\subset T^{\prime}\}\right|=n^{n-1-k}$, and $\left|\{T^{\prime}\in\mathcal{T}_{n}:P_{U}(s_{y})=i_{a_{y}},\forall y\in J\land U\subset T^{\prime}\}\right|=$ $n^{n-1-k-\left|J\right|}$. Therefore we have that:

[TABLE]

This proves that the events $P_{t+1}(s_{y})\in I_{t}$ are mutually independent for every $y\geq\sigma$, each having probability $\frac{N_{t}}{n}$, except if $y=\sigma$, which has probability $\frac{N_{t}-\eta}{n}$.

∎

C.1 Phase 1

In this phase, $N_{t}\leq n-k-1$ To understand how many rounds it takes until $N_{t}>n-k-1$, we will follow similar ideas to the ones presented in Section 3. Note that Corollary 4.19 implies that the adversary will choose $\sigma=k+1$ and $\eta=0$ in each tree and, thus, Lemma 3.1 implies that $N_{t+1}-N_{t}$ follows a binomial distribution with parameters $(n-N_{t}-k,\frac{N_{t}}{n})$.

Definition C.1**.**

Let $X_{t}$ be the following random variable:

[TABLE]

Lemma C.2**.**

For every $t\in\mathbb{N}_{0}$, we have that $n-k>N_{t}\geq X_{t}$.

Proof.

The value of $N_{t}$ cannot go higher than $n-k$ at the beginning of any round $t$ in Phase 1, because if it did we would switch to Phase 2. We will prove the rest by induction. By induction, we have that $N_{0}=1=X_{0}$. Let us assume that $n-k>N_{t}\geq X_{t}$ for some $t\in\mathbb{N}$. Then if $N_{t+1}-N_{t}<(n-N_{t}-k)\cdot\frac{N_{t}}{n}$, since no node goes from uninformed to informed, we have that $N_{t+1}\geq N_{t}\geq X_{t}=X_{t+1}$. If, however, $N_{t+1}-N_{t}\geq(n-N_{t}-k)\cdot\frac{N_{t}}{n}$, then $N_{t+1}\geq N_{t}+(n-N_{t}-k)\cdot\frac{N_{t}}{n}$ and $X_{t+1}=X_{t}+(n-X_{t}-k)\cdot\frac{X_{t}}{n}$. As the function $x\mapsto x+(n-x-k)\frac{x}{n}$ is strictly increasing for $x\leq n-k$, this proves that $n-k\geq N_{t+1}\geq X_{t+1}$. ∎

Lemma C.3**.**

For every $t\in\mathbb{N}_{0}$, we have that $X_{t}\geq 1$.

Proof.

We show this claim by induction on $t$. For the induction case note that by definition $X_{0}=1$. For the induction step let us assume that $X_{t}\geq 1$ for some $t\in\mathbb{N}$. We then have two cases, either $X_{t+1}=X_{t}$ and the result holds trivially, or $X_{t+1}=X_{t}+(n-k-X_{t})\cdot\frac{X_{t}}{n}$. Since $1\leq X_{t}\leq n-k$ by Lemma C.2, we have that $X_{t+1}\geq X_{t}\geq 1$ ∎

Lemma C.4**.**

For every $t\in\mathbb{N}_{0}$, we have that $n-k>X_{t}$, if $n-k>1$.

Proof.

We show this claim by induction on $t$. For the induction base note that as $n-k>1$, it holds that $n-k\geq 2>X_{0}=1.$ Thus, the claim is true for $t=0$. Next assume it is true for $t\in\mathbb{N}$. Then $X_{t+1}\leq X_{t}+(n-X_{t}-k)\cdot\frac{X_{t}}{n}\leq(n-k)(\frac{X_{t}}{n-k}+\frac{n-X_{t}-k}{n-k}\cdot\frac{X_{t}}{n})<n-k$, where the last inequality holds by noting that $(\frac{X_{t}}{n-k}+\frac{n-X_{t}-k}{n-k}\cdot\frac{X_{t}}{n})$ is a convex combination of $1$ and $\frac{X_{t}}{n}$, the latter of which being strictly smaller than $1$. ∎

Corollary C.5**.**

We have that $X_{t+1}>X_{t}$ if and only if $N_{t+1}-N_{t}\geq(n-N_{t}-k)\cdot\frac{N_{t}}{n}$.

Let $u_{t}\in\mathbb{N}_{0}$ be the $t$-th round such that $X_{u_{t}+1}>X_{u_{t}}$ and let $u_{0}=0$.

Lemma C.6**.**

If $n\geq 4$ then $X_{u_{4\log n}}>n-k-1$.

Proof.

Let $t^{\prime}=\min\{t\in\mathbb{N}:X_{u_{t}}\geq\frac{n}{2}\}$ as an intermediate point. We first show that $t^{\prime}\leq\log_{1.4}n$. Indeed, if $X_{u_{t}}\leq\frac{n}{2}$, then

[TABLE]

As by Lemma C.3 all $X_{t}\geq 1$, $X_{u_{t}}\geq 1.4^{t}$, which implies that $X_{u_{t}}\geq\frac{n}{2}$ for every $t\geq\log_{1.4}{\frac{n}{2}}$. This implies that $t^{\prime}\leq\log_{1.4}n$.

Next let $t^{\prime\prime}=\min\{t\in\mathbb{N}:X_{u_{t}}>n-k-1\}$. We show that $t^{\prime\prime}-t^{\prime}\leq\log n$. Indeed, we have, if $X_{u_{t}}\geq\frac{n}{2}$ then

[TABLE]

Thus the distance of $X_{u_{k}}$ to $n-k$ is halved between rounds $u_{t}$ and $u_{t+1}$ if $u_{t}\geq t^{\prime}$. Since $n-k-X_{u_{t^{\prime}}}\leq n-k-n/2\leq n/2$, we have that $t^{\prime\prime}-t^{\prime}\leq 1+\log\frac{n}{2}\leq\log n$. Hence it follows that $t^{\prime\prime}\leq 4\log n$.

∎

Recall the following theorem:

See 3.9

Lemma C.7**.**

If $n-k>4$, for every $t\in\mathbb{N}$, we have that $\mathbb{P}\left(X_{t+1}>X_{t}\right)\geq\frac{1}{4}$

Proof.

By Corollary C.5, we have that $X_{t+1}>X_{t}$ if and only if $N_{t+1}-N_{t}\geq(n-N_{t}-k)\cdot\frac{N_{t}}{n}$. Thus, $\mathbb{P}\left(X_{t+1}>X_{t}\right)=\mathbb{P}\left(N_{t+1}-N_{t}\geq(n-N_{t}-k)\cdot\frac{N_{t}}{n}\right)$. Recall that $N_{t+1}-N_{t}$ follows a binomial distribution with parameters $m=n-N_{t}-k$ and $p=\frac{N_{t}}{n}$. Thus $\mathbb{P}\left(N_{t+1}-N_{t}\geq(n-N_{t}-k)\cdot\frac{N_{t}}{n}\right)=\mathbb{P}\left(N_{t+1}-N_{t}\geq mp\right)$. As we are analyzing Phase 1, we are guaranteed that $N_{t}<n-k$, i.e., $m\geq 1$. We have multiple cases:

(1) If $N_{t}=n-k-1$, then $\mathbb{P}(N_{t+1}-N_{t}\geq(n-N_{t}-k)\cdot\frac{N_{t}}{n})=\mathbb{P}(N_{t+1}-N_{t}\geq\frac{n-k-1}{n})=\frac{n-k-1}{n}>\frac{1}{4}$ since $n>4$.

(2) If $n-k-2\geq N_{t}>\frac{n}{2}$, then the probability parameter $N_{t}/n$ of the binomial distribution fulfills $N_{t}/n>1/2\geq\frac{1}{n-N_{t}-k}.$ Thus applying Theorem 3.9 with $p=N_{t}/n$ and $m=n-N_{t}-k$ gives the result.

(3) If $N_{t}\leq\frac{n}{2}$, then $\frac{1}{n-N_{t}-k}\leq\frac{1}{0.4n}\leq\frac{N_{t}}{0.4n}=\frac{p}{0.4}$, where the last inequality holds as $N_{t}\geq 1$ for all $t$. Thus, $p\geq\frac{0.4}{m}\geq\frac{1}{3m}.$ If $\frac{N_{t}}{n}\leq\frac{1}{n-N_{t}-k}$, the result now follows from Lemma B.6. If $\frac{N_{t}}{n}>\frac{1}{n-N_{t}-k}$, applying Theorem 3.9 with $p=N_{t}/n$ and $m=n-N_{t}-k$ gives the result.

∎

Corollary C.8**.**

Let $(B_{t})_{t\in\mathbb{N}}$ be Bernoulli independent random variables of parameter $\frac{1}{4}$. Let $S^{B}_{t}=\sum_{s\in[t]}B_{s}$ and $S_{t}=\sum_{s\in[t]}\mathbbl{1}\left(X_{s+1}>X_{s}\right)$. Then for any $\ell\in\mathbb{N}$, we have that $\mathbb{P}(S_{t}\leq\ell)\leq\mathbb{P}(S_{t}^{B}\leq\ell)$.

Lemma C.9**.**

Let $t=8(3+\sqrt{5})\log n$. Then $\mathbb{P}(S_{t}\leq 4\log n)\leq\frac{1}{n^{2}}$.

Proof.

Using Hoeffding’s inequality (Lemma 3.12), we have that:

[TABLE]

∎

We now have all the tools to prove that phase 1 ends in fewer than $8(3+\sqrt{5})\log n$ rounds with high probability:

See 4.20

Proof.

By Lemma C.9, we have that, with probability $1-\frac{1}{n^{2}}$, $X_{t+1}>X_{t}$ for at least $4\log n$ many rounds within the $8(3+\sqrt{5})\log n$ first rounds. As $u_{4\log n}$ is the $4\log n$-th round where $X_{t+1}>X_{t}$, we thus have that $\mathbb{P}\left(u_{4\log n}\leq 8(3+\sqrt{5})\log n\right)\geq 1-n^{-2}$. But, by Lemmata C.6 and C.2 $N_{u_{4\log n}}>n-k-1$, i.e., Phase 1 ends. Thus, with probability $1-\frac{1}{n^{2}}$, Phase 1 ends within the $8(3+\sqrt{5})\log n$ first rounds. ∎