The Energy Complexity of Diameter and Minimum Cut Computation in Bounded-Genus Networks

Abstract

This paper investigates the energy complexity of distributed graph problems in multi-hop radio networks, where the energy cost of an algorithm is measured by the maximum number of awake rounds of a vertex. Recent works revealed that some problems, such as broadcast, breadth-first search, and maximal matching, can be solved with energy-efficient algorithms that consume only ${{\text {poly}}}\log n$ energy. However, there exist some problems, such as computing the diameter of the graph, that require $\varOmega (n)$ energy to solve. To improve energy efficiency for these problems, we focus on a special graph class: bounded-genus graphs. We present algorithms for computing the exact diameter, the exact global minimum cut size, and a $(1\pm \epsilon )$-approximate s-t minimum cut size with $\tilde{O}(\sqrt{n})$ energy for bounded-genus graphs. Our approach is based on a generic framework that divides the vertex set into high-degree and low-degree parts and leverages the structural properties of bounded-genus graphs to control the number of certain connected components in the subgraph induced by the low-degree part.

1 Introduction

We consider the multi-hop radio network model [16] of distributed computing, where a communication network is modeled as a graph $G=(V,E)$: Each vertex $v\in V$ is a computing device and each edge $\{u, v\} \in E$ indicates that u and v are within the transmission range of each other. The graph topology of the underlying network G is initially unknown to all devices, except that two parameters $n = |V|$ and $\varDelta = \max _{v \in V} \deg (v)$ are global knowledge.

Communication proceeds in synchronized rounds. All devices agree on the same start time. In each round, each device can choose to do one of the following three operations: (i) listen to the channel, (ii) transmit a message, or (iii) stay idle. We do not allow a device to simultaneously transmit and listen, and we assume that there is no message size constraint.

Each transmitting or idle device does not receive any feedback from the communication channel, so a transmitting device u does not know whether its message is successfully received by any of its neighbors N(u). A listening device v successfully receives a message from a transmitting device $u \in N(v)$ if u is the only transmitting device in N(v). If the number of transmitting devices in N(v) is zero, then a listening device v hears $\textsf{silence}$. For the case the number of transmitting devices in N(v) is greater than one, then the feedback that a listening device v receives depends on the underlying model. In the $\textsf{No}{\text{- }}\textsf{CD}$ model (without collision detection), v still hears $\textsf{silence}$. In the $\textsf{CD}$ model (with collision detection), v hears $\textsf{collision}$. All our algorithms presented in this paper work in the $\textsf{No}{\text{- }}\textsf{CD}$ model.

We assume that each device has access to an unlimited local random source. We say that an event occurs with high probability (w.h.p.) if the event occurs with probability $1 - 1/{{\text {poly}}}(n)$. In this paper, we only consider Monte Carlo algorithms that succeed w.h.p. If we let each vertex $v \in V$ locally assign themselves $O(\log n)$-bit identifiers ${\text {ID}}(v)$, then they are distinct with high probability, so we assume that each device has a distinct identifier of length $O(\log n)$.

Complexity Measures. Time and energy are the two main complexity measures of a distributed algorithm in radio networks. The time complexity of an algorithm is the number of rounds of the algorithm. The energy complexity of an algorithm is the maximum energy cost of a device, where the energy cost of a device v is the number of rounds that v is non-idle. We only consider worst-case time and energy complexities. The motivation for studying energy complexity is that energy is a scarce resource in small battery-powered wireless devices, and such devices can save energy by entering a low-power sleep mode.

1.1 Prior Work

Most of the early work on the energy complexity focused on single-hop radio networks, which is the special case that $G=(V,E)$ is a complete graph. Over the last two decades, there is a long line of research to optimize the energy complexity of leader election and its related problems in single-hop radio networks [6, 8, 9, 11, 13, 14, 29,30,31,32, 34, 35, 39].

This line of research was recently extended to multi-hop radio networks [10, 12, 19, 20]. Chang et al. [10] considers the problem of broadcasting a message from one device to all other devices in a multi-hop radio network. They showed that broadcasting can be done in ${{\text {poly}}}\log n$ energy. Specifically, they presented randomized broadcasting algorithms for $\textsf{CD}$ and $\textsf{No}{\text{- }}\textsf{CD}$ using energy $O\left( \frac{\log n\log \log \varDelta }{\log \log \log \varDelta }\right)$ and $O(\log \varDelta \log ^2 n)$ w.h.p., respectively. They also proved that any algorithm transmitting a message from one endpoint to the other endpoint of an n-vertex path costs $\varOmega (\log n)$ energy in expectation. The lower bound applies even to the $\textsf{LOCAL}$ model of distributed computing.

Chang et al. [12] showed that breadth-first search can be done w.h.p. using $2^{O\left( \sqrt{\log n \log \log n}\right) }$ energy in $\textsf{No}{\text{- }}\textsf{CD}$. Their algorithm is based on a hierarchical clustering using the low-diameter decomposition algorithm of Miller, Peng, and Xu [38]. The energy complexity of breadth-first search was recently improved to ${{\text {poly}}}\log n$ by Dani and Thomas [20]. Combining the polylogarithmic-energy breadth-first search algorithm with the diameter approximation algorithm of Roditty and Williams [41], an approximation $\tilde{D}$ of the diameter D such that $\tilde{D} \in \left[ \lfloor 2D/3\rfloor , D\right]$ can be computed with $\tilde{O}(\sqrt{n})$ energy w.h.p. [12]. The notation $\tilde{O}(\cdot )$ suppresses any ${{\text {poly}}}\log n$ factor.

Dani et al. [19] showed that a maximal matching can be computed in $O(\varDelta \log n)$ time and $O(\log \varDelta \log n)$ energy w.h.p. in $\textsf{No}{\text{- }}\textsf{CD}$. There exists a family of graphs such that these time and energy bounds are simultaneously optimal up to polylogarithmic factors.

1.2 Our Contribution

Not all problems admit energy-efficient algorithms in multi-hop radio networks. It was shown in [12] that any algorithm that computes a $(1.5-\epsilon )$-approximation of the diameter requires $\tilde{\varOmega }(n)$ energy w.h.p. The lower bound holds even on graphs with arboricity $O(\log n)$ and treewidth $O(\log n)$.

To improve energy efficiency for diameter computation, we focus on the class of bounded-genus graphs. We show that the diameter of the graph can be computed using $\tilde{O}(\sqrt{n})$ energy w.h.p. in bounded-genus graphs.

Theorem 1

There is an algorithm that computes the diameter in $\tilde{O}(n^{1.5})$ time and $\tilde{O}(\sqrt{n})$ energy w.h.p. for bounded-genus graphs in $\textsf{No}{\text{- }}\textsf{CD}$.

Our approach is based on a generic framework that divides the vertex set into high-degree and low-degree parts. We then classify the connected components of the subgraph induced by the low-degree part into several types. We will leverage the structural properties of bounded-genus graphs to upper-bound the number of connected components of one type. For the remaining connected components, we will design energy-efficient algorithms that extract all the necessary information from these connected components for the purpose of diameter computation.

Our approach is sufficiently general so that it is applicable to other problems as well. Using the same approach, we show that the exact global minimum cut size and a $(1\pm \epsilon )$-approximate s-t minimum cut size can also be computed using $\tilde{O}(\sqrt{n})$ energy w.h.p. in bounded-genus graphs.

Theorem 2

There is an algorithm that computes the minimum cut size in $\tilde{O}(n^{1.5})$ time and $\tilde{O}(\sqrt{n})$ energy w.h.p. for bounded-genus graphs in $\textsf{No}{\text{- }}\textsf{CD}$.

Theorem 3

There is an algorithm that computes an $(1 \pm \epsilon )$-approximate s–t minimum cut size in $\tilde{O}(n^{1.5}) \cdot \epsilon ^{-O(1)}$ time and $\tilde{O}(\sqrt{n}) \cdot \epsilon ^{-O(1)}$ energy w.h.p. for bounded-genus graphs in $\textsf{No}{\text{- }}\textsf{CD}$.

To complement these algorithmic results, we show that any algorithm that computes the exact size of an s–t minimum cut or a global minimum cut requires $\varOmega (n)$ energy. The lower bound for the s–t minimum cut holds even for planar bipartite graphs, so it is necessary that we consider approximation algorithms for this problem. These lower bounds apply to both $\textsf{No}{\text{- }}\textsf{CD}$ and $\textsf{CD}$.

Theorem 4

For any randomized algorithm that computes the s–t minimum cut size of a planar bipartite graph w.h.p. in $\textsf{CD}$, the energy complexity of the algorithm is $\varOmega (n)$.

Theorem 5

For any randomized algorithm that computes the minimum cut size of a unit-disc graph w.h.p. in $\textsf{CD}$, the energy complexity of the algorithm is $\varOmega (n)$.

1.3 Additional Related Work

There are numerous works studying energy-aware distributed computing in multi-hop networks from different perspectives. In radio networks, the power of a signal received is proportional to $O(1/d^{\alpha })$, where d is the distance to the sender, and $\alpha$ is a constant related to environmental factors. Kirousis et al. [33] studied the optimization problem of assigning transmission ranges of devices subject to some connectivity and diameter constraints so as to minimize the total power consumption. See [2, 17, 43] for related work.

There are several works [7, 22, 42] on the subject of reducing the number of rounds or transmissions required to complete a specific communication task. In the setting of known network topology, Gasieniec et al. [23] designed a randomized protocol for broadcasting in $O(D + kn^{1/(k-2)}\log ^2 n)$ rounds such that each device transmits at most k times.

The energy complexity has recently been studied in the well-known $\textsf{LOCAL}$ and $\textsf{CONGEST}$ models of distributed computing [3, 5, 15, 21, 27].

There is a large body of research on distributed graph algorithms in planar networks, bounded-genus networks, or more broadly H-minor-free networks: distributed approximation [1, 18, 36, 44], low-congestion shortcuts and its applications [24,25,26, 28], and other planar graph algorithms [37, 40].

1.4 Organization

In Sect. 2, we present the basic tools. In Sect. 3, we present our lower bounds. In Sect. 4, we present our decomposition of bounded-genus networks. In Sect. 5 and Appendix A, we present our algorithm for diameter computation. In Appendix B, we present our algorithm for minimum cut computation. In Appendix C, we present the proof details for our basic tools.

2 Tools

2.1 Communication Between Two Sets of Vertices

Let $\mathcal {S}$ and $\mathcal {R}$ be two vertex sets that are not necessarily disjoint. The task $\textsf{SR}{\text{- }}\textsf{comm}$ [10] is defined as follows. Each vertex $u \in \mathcal {S}$ holds a message $m_u$ that it wishes to send, and each vertex $v \in \mathcal {R}$ wants to receive a message from vertices in $N^+(v) \cap \mathcal {S}$, where $N^+(v) = N(v) \cup \{v\}$ is the inclusive neighborhood of v.

Specifically, the task $\textsf{SR}{\text{- }}\textsf{comm}$ requires that for each vertex $v \in \mathcal {R}$ with $N^+(v) \cap \mathcal {S} \ne \emptyset$, vertex v receives a message $m_u$ from at least one vertex $u \in N^+(v) \cap \mathcal {S}$ w.h.p. Several variants of $\textsf{SR}{\text{- }}\textsf{comm}$ are defined as follows.

All messages: $\textsf{SR}{\text{- }}\textsf{comm}^{{\text {all}}}$.:

The task $\textsf{SR}{\text{- }}\textsf{comm}^{{\text {all}}}$ requires that each vertex $v \in \mathcal {R}$ receives the message $m_u$ for each $u \in N^+(v) \cap \mathcal {S}$ w.h.p.

Approximate sum: $\textsf{SR}{\text{- }}\textsf{comm}^{{\text {apx}}}$.:

Suppose the message $m_u$ sent from each vertex $u \in \mathcal {S}$ is an integer within the range [W]. The task $\textsf{SR}{\text{- }}\textsf{comm}^{{\text {apx}}}$ requires each vertex $v \in \mathcal {R}$ computes an $(1 \pm \epsilon )$-factor approximation of the summation $\sum _{u \in N^+(v) \cap \mathcal {S}} m_u$ w.h.p.

Minimum and maximum: $\textsf{SR}{\text{- }}\textsf{comm}^{{\text {min}}}$ and $\textsf{SR}{\text{- }}\textsf{comm}^{{\text {max}}}$.:

The message $m_u$ sent from each vertex $u \,{\in }\, \mathcal {S}$ contains a key $k_u$ from the key space $[K] = \{1, 2, \ldots , K\}$. For $\textsf{SR}{\text{- }}\textsf{comm}^{{\text {min}}}$, it is required that w.h.p., each vertex $v \in \mathcal {R}$ with $N^+(v) \cap \mathcal {S} \ne \emptyset$ receives a message $m_u$ from a vertex $u \in N^+(v) \cap \mathcal {S}$ such that $k_u = \min _{u' \in N^+(v) \cap \mathcal {S}} k_{u'}$. The task $\textsf{SR}{\text{- }}\textsf{comm}^{{\text {max}}}$ is defined analogously by replacing minimum with maximum.

Multiple messages: $\textsf{SR}{\text{- }}\textsf{comm}^{{\text {multi}}}$.:

Consider the setting where each vertex $u \in \mathcal {S}$ holds a set of messages $\mathcal {M}_u$. For each message m, all vertices holding the same message m have access to some shared random bits associated with m. We assume that for each $v \in \mathcal {R}$, the number of distinct messages in $\bigcup _{u \in N^+(v) \cap \mathcal {S}} \mathcal {M}_u$ is upper bounded by a number M that is known to all vertices. The task $\textsf{SR}{\text{- }}\textsf{comm}^{{\text {multi}}}$ requires that each vertex $v \in \mathcal {R}$ receives all distinct messages in $\bigcup _{u \in N^+(v) \cap \mathcal {S}} \mathcal {M}_u$ w.h.p.

Table 1 summarizes the time and energy complexities of our algorithms for these tasks. For $\textsf{SR}{\text{- }}\textsf{comm}^{{\text {all}}}$, the parameter $\varDelta '$ can be any known upper bound on $|\mathcal {S} \cap N(v)|$, for each $v \in \mathcal {R}$. For example, we may set $\varDelta ' = \varDelta$ if no better upper bound is known. The proofs for these results are left to Appendix C.

Table 1. The time and energy complexities of $\textsf{SR}{\text{- }}\textsf{comm}$ and its variants.

2.2 Communication via a Good Labeling

A good labeling is a vertex labeling $\mathcal {L}: V(G) \mapsto \{0, \ldots , n-1\}$ such that each vertex v with $\mathcal {L}(v) > 0$ has a neighbor u with $\mathcal {L}(u)=\mathcal {L}(v)-1$ [10]. A vertex v is called a layer-i vertex if $\mathcal {L}(v) = i$. Observe that if there is a unique layer-0 vertex r, then $\mathcal {L}$ represents a tree T rooted at r, so we call r the root of the good labeling $\mathcal {L}$. Since a vertex might have multiple choices of the parent, the tree T is not unique in general. The following lemma was proved in [10].

Lemma 1

([10]). A good labeling $\mathcal {L}$ with a unique layer-0 vertex can be constructed in $O(n \log \varDelta \log ^2 n)$ time and $O(\log \varDelta \log ^2 n)$ energy w.h.p.

The following lemma shows that a good labeling allows the vertices in the graph to broadcast messages in an energy-efficient manner.

Lemma 2

Suppose that we are given a good labeling $\mathcal {L}$ with a unique layer-0 vertex. Then we can achieve the following.

• 1. It takes $O(n \varDelta \log n)$ time and $O(\varDelta \log n)$ energy for each vertex to broadcast a message to the entire network w.h.p.

• 2. It takes $O(nx \log \varDelta \log ^2 n)$ time and $O(x \log \varDelta \log ^2 n)$ energy for x vertices to broadcast messages to the entire network w.h.p.

Proof

Let r be the root of $\mathcal {L}$. For the first task, consider the following algorithm. We relay the message of each vertex to the root r using the following convergecast algorithm. For $i = n-1$ down to 1, do $\textsf{SR}{\text{- }}\textsf{comm}^{{\text {all}}}$ with $\mathcal {S}$ being the set of all layer-i vertices and $\mathcal {R}$ being the set of all layer-$(i-1)$ vertices. For each execution of $\textsf{SR}{\text{- }}\textsf{comm}^{{\text {all}}}$, each vertex in $\mathcal {S}$ transmits not only its message but also all other messages that it has received so far. Although we perform $\textsf{SR}{\text{- }}\textsf{comm}^{{\text {all}}}$ $n-1$ times, each vertex only participates at most twice. By Lemma 22, the cost of the convergecast algorithm is $O(n \varDelta \log n)$ time and $O(\varDelta \log n)$ energy.

At the end of the convergecast algorithm, the root r has gathered all messages sent during the algorithm. After that, the root r then broadcasts this information to all vertices via the following divergecast algorithm. For $i = 0$ to $n - 2$, do $\textsf{SR}{\text{- }}\textsf{comm}$ with $\mathcal {S}$ being the set of all layer-i vertices and $\mathcal {R}$ being the set of all layer-$(i+1)$ vertices. Similarly, although we perform $\textsf{SR}{\text{- }}\textsf{comm}$ for $n-1$ times, each vertex only participates at most twice. By Lemma 21, the cost of the divergecast algorithm is $O(n \log \varDelta \log n)$ time and $O(\log \varDelta \log n)$ energy. At the end of the divergecast algorithm, all vertices have received all messages.

For the rest of the proof, we consider the second task. Let X be the set of x vertices that attempt to broadcast a message. We solve this task similarly in two steps:

• We first do a convergecast, using $\textsf{SR}{\text{- }}\textsf{comm}^{{\text {multi}}}$ with $M=x$, to gather all x messages to the root. By Lemma 23, $\textsf{SR}{\text{- }}\textsf{comm}^{{\text {multi}}}$ costs $O(x\log \varDelta \log ^2 n)$ time, so the convergecast costs $O(nx \log \varDelta \log ^2 n)$ time and $O(x \log \varDelta \log ^2 n)$ energy.

• After that, we do a divergecast based on $\textsf{SR}{\text{- }}\textsf{comm}$ to broadcast these messages from root to everyone. The divergecast costs $O(n \log \varDelta \log n)$ time and $O(\log \varDelta \log n)$ energy.

In order to use $\textsf{SR}{\text{- }}\textsf{comm}^{{\text {multi}}}$, the initial holder of each message m needs to first generate a sufficient number of random bits and attach them to the message. These random bits serve as the shared randomness associated with the message m, which is needed in the definition of $\textsf{SR}{\text{- }}\textsf{comm}^{{\text {multi}}}$.   $\square$

Lemma 3

There is an algorithm that lets all vertices learn the entire graph topology in $O(n \varDelta \log n)$ time and $O(\varDelta \log n)$ energy w.h.p.

Proof

We first let each vertex v learn the list of identifiers in N(v) by doing $\textsf{SR}{\text{- }}\textsf{comm}^{{\text {all}}}$ with $\mathcal {S}=\mathcal {R}=V$, where the message of each vertex v is ${\text {ID}}(v)$. By Lemma 22, this step takes $O(\varDelta \log n)$ time and energy. After that, we apply Lemma 1 to construct a good labeling with a unique layer-0 vertex, and then we apply Lemma 2(1) to let all vertices learn the entire network topology by having each v broadcasting ${\text {ID}}(v)$ and the list of identifiers in N(v). This step takes $O(n \varDelta \log n)$ time and $O(\varDelta \log n)$ energy.    $\square$

3 Lower Bounds

In this section, we prove the two lower bounds: Theorems 4 and 5.

Theorem 4

For any randomized algorithm that computes the s–t minimum cut size of a planar bipartite graph w.h.p. in $\textsf{CD}$, the energy complexity of the algorithm is $\varOmega (n)$.

Proof

Suppose that there is a randomized algorithm $\mathcal {A}$ that computes the exact s–t minimum cut size of any planar bipartite graph with high probability and using o(n) energy. Let G be a complete bipartite graph $K_{2,\varDelta }$ with the bipartition $\{s, t\}$ and $\{v_1, \ldots , v_{\varDelta }\}$. Set $X = \varDelta /5$. We select $\varDelta$ to be sufficiently large so that it is guaranteed that both s and t use at most X unit of energy in an execution of $\mathcal {A}$ on G.

Let $G'$ be the result of removing $v_{\varDelta }$ from G. The size of a s–t minimum cut of G is $\varDelta$, and the size of a s–t minimum cut of $G'$ is $\varDelta - 1$. Therefore, $\mathcal {A}$ lets s correctly distinguish between G and $G'$ with high probability.

Consider an execution of $\mathcal {A}$ on G. Let S be the subset of $\{v_1, \ldots , v_{\varDelta }\}$ such that $v_i \in S$ if there is a time slot $\tau$ where (i) $v_i$ transmits, (ii) the number of vertices in $\{v_1, \ldots , v_{\varDelta }\}$ that transmit is at most 2, and (iii) at least one of s and t listens.

We claim that $|S| \le 4X = 4\varDelta /5$. Let T be the set of all time slots $\tau$ such that the above conditions (i), (ii), and (iii) hold for at least one $v_i \in \{v_1, \ldots , v_{\varDelta }\}$. In view of condition (ii), we must have $|T| \ge |S|/2$. In view of condition (iii), if $\tau \in T$, then at least one of s and t must listen at time $\tau$, so the energy cost of one of s and t must be at least $|T|/2 \ge |S|/ 4$, which implies $X \ge |S|/4$.

Let $\mathcal {E}$ be the event that $v_{\varDelta } \notin S$ in an execution of $\mathcal {A}$ on G. Whether or not $\mathcal {E}$ occurs depends only on the local randomness stored in the vertices $\{s, t\}$ and $\{v_1, \ldots , v_{\varDelta }\}$. Since $|S| \le 4\varDelta /5$, at least 1/5 fraction of the vertices in $\{v_1, \ldots , v_{\varDelta }\}$ are not in S. Since the probability that $v_{i} \notin S$ is identical for all $v_i \in \{v_1, \ldots , v_{\varDelta }\}$, we have ${\text {Pr}}[\mathcal {E}] \ge 1/5$.

Consider the following scenario. All vertices in $\{s, t\}$ and $\{v_1, \ldots , v_{\varDelta }\}$ have decided their random bits in advance. With probability 1/2, we run $\mathcal {A}$ on G. With probability 1/2, we run $\mathcal {A}$ on $G'$. If $\mathcal {E}$ occurs, then the execution of $\mathcal {A}$ on both G and $G'$ is completely identical from the point of view of each vertex, except for $v_{\varDelta }$. Therefore, conditioning on event $\mathcal {E}$, the probability that vertex s correctly decides whether the underlying graph is G or $G'$ is at most 1/2, as s can only guess randomly.

Since ${\text {Pr}}[\mathcal {E}] \ge 1/5$, the probability that vertex s fails to correctly decide whether the underlying graph is G or $G'$ is at least $(1/2) \cdot (1/5) = 1/10$, so s fails to correctly calculate the s–t minimum cut with probability at least 1/10 in the above scenario. This contradicts the assumption that $\mathcal {A}$ is able to compute the s–t minimum cut with high probability.   $\square$

Theorem 5

For any randomized algorithm that computes the minimum cut size of a unit-disc graph w.h.p. in $\textsf{CD}$, the energy complexity of the algorithm is $\varOmega (n)$.

Proof

Consider the case where the underlying graph is $K_n$ with probability 1/2, and is $K_n - e$ with probability 1/2, where the edge e is chosen uniformly at random from the set of all edges in $K_n$. Let $\mathcal {A}$ be any randomized algorithm that computes the size of a minimum cut exactly with high probability. Observe that the size of a minimum cut of $K_n$ is $n-1$ and the size of a minimum cut of $K_n - e$ is $n-2$, so $\mathcal {A}$ is able to distinguish between $K_n$ and $K_n - e$ with high probability. It was shown in [12] that any algorithm that distinguishes between $K_n$ and $K_n - e$ with success probability at least 3/4 necessarily has energy cost $\varOmega (n)$ in both $\textsf{CD}$ and $\textsf{No}{\text{- }}\textsf{CD}$, so the energy complexity of $\mathcal {A}$ is $\varOmega (n)$.   $\square$

4 Graph Partitioning

The genus of a graph G is the minimum number g such that G can be drawn on an oriented surface of g handles without crossing. For example, planar graphs are the graphs with genus zero, and the graphs that can be drawn on a torus without crossing are the graphs with genus at most one. A class of graphs is called bounded-genus if the genus of all graphs in the class can be upper bounded by some constant $g = O(1)$. In this section, we consider a classification of the connected components of the subgraph induced by the low-degree vertices in a bounded-genus graph. Our algorithms, which will be presented in subsequent sections, make use of the classification.

Let $G = (V,E)$ be any bounded-genus graph. Let $V_H$ be the set of vertices that have degree at least $\sqrt{n}$. Let $V_L = V \setminus V_H$. Since bounded-genus graphs have arboricity O(1), we have $|E| = O(n)$, which implies $|V_H| = O(\sqrt{n})$.

From now on, we assume $|V_H| \ge 1$, since otherwise G has maximum degree $\varDelta \le \sqrt{n}$, so we can already solve all problems using $O(n \varDelta \log \varDelta \log n)) = \tilde{O}(n^{1.5})$ time and $O(\varDelta \log \varDelta \log n)) = \tilde{O}(\sqrt{n})$ energy by learning the entire graph topology using the algorithm of Lemma 3.

Given a set of vertices S, we write G[S] to denote the subgraph of G induced by S and write $G^+[S]$ to denote the subgraph of G induced by all edges that have at least one endpoint in S. We divide the connected components of $G[V_L]$ into three types.

Type 1.:

A connected component S of $G[V_L]$ is of type-1 if $|S| < \sqrt{n}$ and $|\bigcup _{w \in S} N(w) \cap V_H| = 1$. For each vertex $u \in V_H$, we write C(u) to denote the set of type-1 components S such that $\bigcup _{w \in S} N(w) \cap V_H = \{u\}$.

Type 2.:

A connected component S of $G[V_L]$ is of type-2 if $|S| < \sqrt{n}$ and $|\bigcup _{w \in S} N(w) \cap V_H| = 2$. For each pair of two distinct vertices $\{u,v\} \subseteq V_H$, we write C(u, v) to denote the set of type-2 components S such that $\bigcup _{w \in S} N(w) \cap V_H = \{u,v\}$.

Type 3.:

A connected component S of $G[V_L]$ is of type-3 if it is neither of type-1 nor of type-2.

A connected component S of $G[V_L]$ is of type-3 if $|S| \ge \sqrt{n}$ or $|\bigcup _{w \in S} N(w) \cap V_H| \ge 3$. The number of type-3 components S with $|S| \ge \sqrt{n}$ is clearly at most $|V|/\sqrt{n} = \sqrt{n}$. We will show that the number of type-3 components with $|\bigcup _{w \in S} N(w) \cap V_H| \ge 3$ is also $O(\sqrt{n})$.

Lemma 4

Let $G=(V,E)$ be a bipartite graph of genus at most g. Let $V = X \cup Y$ be the bipartition of G. If $\deg (v) \ge 3$ for each $v \in X$, then $|X| \le 2|Y| + 4(g-1)$.

Proof

Consider any embedding of G into a surface of genus g, and let F be set of faces. In a bipartite graph, each face has at least four edges, and each edge appears in at most two faces, so $|E| \ge 2|F|$. Combining this inequality with Euler’s polyhedral formula $|V|-|E|+|F| \ge 2 - 2g$, we obtain that

$$\begin{aligned} 2\,V| - |E| \ge 4(1-g). \end{aligned}$$

Since $\deg (v) \ge 3$ for each $v \in X$, we have $|E| \ge 3|X|$, so

$$\begin{aligned} 2\,V| - |E| = 2(|X| +|Y|) - |E| \le 2(|X| +|Y|) - 3|X| = 2|Y| - |X|. \end{aligned}$$

Combining these upper and lower bounds of $2|V| - |E|$, we obtain that $2|Y| - |X| \ge 4(1-g)$, so $|X| \le 2|Y| + 4(g-1)$, as required.   $\square$

Note that Lemma 4 is precisely the reason that our algorithms only apply to bounded-genus graphs and do not work on an arbitrary H-minor-free graph. Consider a complete bipartite graph with the bipartition X and Y such that $|Y| = 3$. Such a graph does not contain $K_5$ as a minor, regardless of the size of X. Therefore, $K_5$-minor-freeness does not allow us to upper bound |X| by any function of |Y|. Therefore, the bounded-genus requirement in Lemma 4 cannot be relaxed to H-minor-freeness for an arbitrary H.

Lemma 5

If G is a bounded-genus graph, then the number of type-3 components is $O(\sqrt{n})$.

Proof

A connected component S of $G[V_L]$ is of type-3 if $|S| \ge \sqrt{n}$ or $|\bigcup _{w \in S} N(w) \cap V_H| \ge 3$. As discussed earlier, the number of type-3 components S with $|S| \ge \sqrt{n}$ is at most $\sqrt{n}$, so we just need to prove that the number of type-3 components S with $|\bigcup _{w \in S} N(w) \cap V_H| \ge 3$ is also $O(\sqrt{n})$. Consider a bipartite graph $G^*=(V^*, E^*)$ with the bipartition $V^*= X^*\cup Y^*$ defined as follows.

• $X^*$ is the set of all type-3 components S such that $|\bigcup _{w \in S} N(w) \cap V_H| \ge 3$.

• $Y^*= V_H$.

• For each component $S \in X^*$ and each vertex $v \in Y^*$, $\{S,v\} \in E^*$ if $v \in \bigcup _{w \in S} N(w)$.

Alternatively, $G^*$ can be constructed from G by the following steps.

• Remove all type-1, type-2, and type-3 components S with $|\bigcup _{w \in S} N(w) \cap V_H| \le 2$.

• For each type-3 component S with $|\bigcup _{w \in S} N(w) \cap V_H| \ge 3$, contract S into a vertex.

As $G^*$ can be obtained from G via a sequence of edge contractions and vertex removals, $G^*$ is a bounded-genus graph. Observe that $\deg (S) \ge 3$ for each $S \in X^*$ in $G^*$, so we may apply Lemma 4, which shows that the number $|X^*|$ of type-3 components S such that $|\bigcup _{w \in S} N(w) \cap V_H| \ge 3$ satisfies $|X^*| \le 2|Y^*| + O(1) = 2|V_H| + O(1) = O(\sqrt{n})$.   $\square$

We write $G_H$ to denote the graph defined by the vertex set $V_H$ and the edge set $\{\{u,v\} \, : \, |C(u,v)| > 0\}$. The following observation is useful.

Lemma 6

If G is a bounded-genus graph, then $G_H$ is also a bounded-genus graph, so the number of edges in $G_H$ is $O(\sqrt{n})$ and there exists an edge orientation of $G_H$ such that each vertex has outdegree O(1).

Proof

The graph $G_H$ can be obtained from G via a sequence of edge contractions and vertex removals, so $G_H$ is a bounded-genus graph. As bounded-genus graphs have arboricity O(1), and so the number of edges in $G_H$ is at most linear in the number of vertices in $G_H$, which is $O(\sqrt{n})$, and we can orient the edges of $G_H$ in such a way that each vertex has outdegree O(1).   $\square$

5 Diameter

In this section, we show that for bounded-genus graphs, the diameter can be computed using $\tilde{O}(\sqrt{n})$ energy. We begin with discussing the high-level proof idea. First of all, using Lemma 2, learning the entire graph topology of the subgraph induced by $V_H$ and all type-3 components is doable using $\tilde{O}(\sqrt{n})$ energy. Intuitively, this is due to the following facts: (i) $|V_H| = O(\sqrt{n})$, (ii) $\deg (v) = O(\sqrt{n})$ for each $v \in V_L$, and (iii) the number of type-3 components is $O(\sqrt{n})$.

The main difficulty in the diameter computation is dealing with type-1 and type-2 components. For example, a vertex $u \in V_H$ can be connected to $\Theta (n)$ type-1 components in that $|C(u)| = \Theta (n)$. Since we aim for an algorithm with energy complexity $\tilde{O}(\sqrt{n})$, throughout the entire algorithm, u can only receive messages from at most $\tilde{O}(\sqrt{n})$ components in C(u). The challenge here is to show that the diameter can still be calculated with a limited amount of information about type-1 and type-2 components and show that such information can be extracted in an energy-efficient manner in the radio network model.

We will define a set of parameters of type-1 and type-2 components and show that with these parameters, the exact value of the diameter can be calculated. Based on this result, we will define a subgraph $G^\star$ of G such that the diameter of G equals the diameter of $G^\star$, and then in Appendix A we will design an energy-efficient algorithm to learn the graph topology of $G^\star$.

In the subsequent discussion, we write ${\text {eccentricity}}(u,S)$ to denote $\max _{v \in S} {\text {dist}}(u,v)$. By default, all distances are measured in the underlying network G. We use subscripts to describe distances that are measured in a vertex set, an edge set, or a subgraph.

Parameters for Type-1 Components. We first consider the type-1 components in C(u), for any vertex $u \in V_H$.

$(A_i[u], a_i[u])$.:

Let $A_1[u]$ be a component $S \in C(u)$ that maximizes ${\text {eccentricity}}(u,S)$, and let $A_2[u]$ be a component $S \in C(u) \setminus \{A_1[u]\}$ that maximizes ${\text {eccentricity}}(u,S)$. For $i \in \{1,2\}$, we write $a_i[u] = {\text {eccentricity}}(u,A_i[u])$.

(B[u], b[u]).:

Let B[u] be a component $S \in C(u)$ that maximizes $\max _{s, t \in S \cup \{u\}} {\text {dist}}(s,t)$, and we write $b[u] = \max _{s, t \in B[u] \cup \{u\}} {\text {dist}}(s,t)$.

In the above definitions, ties can be broken arbitrarily if there are multiple choices. Some of the above definitions become undefined when |C(u)| is too small. For example, if $|C(u)| = 1$, then $A_2[u]$ and $a_2[u]$ are undefined. In such a case, we set these parameters to their default values: zero or an empty set. For example, if $|C(u)| = 1$, then we set $A_2[u] = \emptyset$ and $a_2[u] = 0$.

Any path connecting a vertex in $\bigcup _{S \in C(u)} S$ to the rest of the graph must pass the vertex $u \in V_H$, so the amount of information we can afford to extract from $\bigcup _{S \in C(u)} S$ is limited. Intuitively, for the purpose of calculating the diameter, we only need the following information from $\bigcup _{S \in C(u)} S$:

• The longest distance between two vertices in $\bigcup _{S \in C(u)} S \cup \{u\}$, which is $\max \{b[u], a_1[u]+a_2[u]\}$.

• The longest distance between u and a vertex in $\bigcup _{S \in C(u)} S$, which is $a_1[u]$.

Regardless of the size of C(u), we only need to learn $a_1[u]$, $a_2[u]$, and b[u] from the components of C(u). Later we will show that these parameters can be learned efficiently via $\textsf{SR}{\text{- }}\textsf{comm}^{{\text {max}}}$.

Parameters for Type-2 Components. Next, we consider the type-2 components in C(u, v), for any two distinct vertices $u, v \in V_H$.

(R[u, v], r[u, v]).:

Let R[u, v] be a component $S \in C(u,v)$ that minimizes ${\text {dist}}_{G^+[S]}(u,v)$, and we write $r[u,v] = {\text {dist}}_{G^+[R[u,v]]}(u,v)$. In other words, R[u, v] is a component that contains a shortest path between u and v, among all u–v paths via the vertices in $\bigcup _{S \in C(u,v)} S$.

$(A_i^k[u,v], a_i^k[u,v])$.:

For each component $S \in C(u,v)$, we write $S^{u,k}$ to denote the set of vertices $\{w \in S \, : \, {\text {dist}}_{G^+[S]}(w,v) - {\text {dist}}_{G^+[S]}(w,u) \ge k\}$. In other words, $S^{u,k}$ is the set of all vertices in S whose distance to u in $G^+[S]$ is shorter than that to v by at least k.

Let $A_1^k[u,v]$ be a component $S \in C(u,v)$ that maximizes ${\text {eccentricity}}_{G^+[S]}(u, S^{u,k})$, and let $A_2^k[u,v]$ be a component $S \!\in \! C(u,v){\setminus } \{A_1^k[u,v]\}$ that maximizes ${\text {eccentricity}}_{G^+[S]}(u, S^{u,k})$. We write $a_i^k[u,v] ={\text {eccentricity}}_{G^+[A_i^k[u,v]]} (u,A_i^k[u,v])$. We only consider $k \in \{-\sqrt{n}, \ldots , \sqrt{n}\}$.

$(B^l[u,v], b^l[u,v])$.:

For a component $S \in C(u,v)$, we write $G^{l}[S]$ to denote the graph resulting from adding to $G^+[S]$ a path of length l connecting u and v, and we write $\phi ^{l}(S)$ to denote the maximum value of ${\text {dist}}_{G^{l}[S]}(s,t)$ among all pairs of vertices $s, t \in S \cup \{u,v\}$. A useful observation here is that if ${\text {dist}}_{V \setminus S}(u,v) = l$, then $\phi ^{l}(S)$ equals the maximum value of ${\text {dist}}_G(s,t)$ among all pairs of vertices $s, t \in S \cup \{u,v\}$.

Let $B^l[u,v]$ be a component $S \in C(u,v) \setminus \{ R[u,v] \}$ that maximizes $\phi ^{l}(S)$, and we write $b^l(u,v) = \phi ^{l}(B^l[u,v])$. We only consider $l \in \{1, \ldots , \sqrt{n}\}$.

Similar to the parameters of type-1 components, all the above parameters are set to their default values if they are undefined. Note that the definitions of $a_i^k[u,v]$ and $A_i^k[u,v]$ are asymmetric in the sense that we might have $a_i^k[u,v] \ne a_i^k[v,u]$ and $A_i^k[u,v] \ne A_i^k[v,u]$. All remaining parameters for type-2 components are symmetric.

We briefly explain how the above parameters can be used in the diameter calculation. Let $P=(s, \ldots , t)$ be an s–t shortest path in G whose length equals the diameter. There are three possible ways that P intersects the vertex set $\bigcup _{S \in C(u,v)} S$.

• Suppose s and t are within $G^+[S]$, for a component $S \in C(u,v)$. In this case, if ${\text {dist}}_{V \setminus S}(u,v) = l$, then the length of P equals $\phi ^{l}(S) = b^l[u,v]$.

• Suppose there is a subpath $P' = (u, \ldots , v)$ of P whose intermediate vertices are all in $\bigcup _{S \in C(u,v)} S$. In this case, the length of $P'$ equals r[u, v].

• Suppose there is a component $S \in C(u,v)$ such that $s \in S$ and $t \notin S \cup \{u,v\}$. Suppose u is the first vertex of P that is not in S. Consider the subpath $P' = (s, \ldots , u)$ of P. If ${\text {dist}}(t,v) - {\text {dist}}(t,u) = k$, then we must have $s \in S^{u,k}$, since otherwise ${\text {dist}}(s,v) + {\text {dist}}(v,t)$ is smaller than the length of P, violating the assumption that P is an s–t shortest path. If $t \notin A_1^k[u,v]$, then the length of $P'$ equals $a_1^k[u,v]$. If $t \in A_1^k[u,v]$, then the length of $P'$ equals $a_2^k[u,v]$.

Intuitively, the above discussion shows that the parameters described above capture all the necessary information needed to be extracted from the type-2 components for the purpose of diameter computation. We have $O(\sqrt{n})$ parameters for each C(u, v). We will later show that all these parameters can be learned using $O(\sqrt{n})$ energy.

The graph $G^\star$. We define $G^\star$ as the subgraph induced by the union of (i) $V_H$, (ii) all type-3 components, (iii) $A_1[u]$, $A_2[u]$, and B[u], for each $u \in V_H$, and (iv) $A_i^k[u,v]$, $A_i^k[v,u]$, $B^l[u,v]$, and R[u, v], for each pair of distinct vertices $\{u,v\} \subseteq V_H$, $i \in \{1,2\}$, $k \in \{-\sqrt{n}, \ldots , \sqrt{n}\}$, and $l \in \{1, \ldots , \sqrt{n}\}$. In the subsequent discussion, we prove that the diameter of G equals the diameter of $G^\star$, so the task of computing the diameter of G is reduced to learning the topology of $G^\star$. We will show that the following two statements are correct.

(S1):

For each pair of vertices $\{s,t\}$ in graph $G^\star$, we have ${\text {dist}}_G(s,t) = {\text {dist}}_{G^\star }(s,t)$.

(S2):

For each pair of vertices $\{s,t\}$ in graph G, there exists a pair of vertices $\{s',t'\}$ in graph $G^\star$ satisfying ${\text {dist}}_G(s,t) \le {\text {dist}}_{G}(s',t')$.

These statements imply that G and $G^\star$ have the same diameter. We first prove that (S1) is true.

Lemma 7

For any two vertices s and t in $G^\star$, we have ${\text {dist}}_{G}(s,t) = {\text {dist}}_{G^\star }(s,t)$.

Proof

We choose P to be an s–t path in G whose length is ${\text {dist}}_{G}(s,t)$ that uses the minimum number of vertices not in $G^\star$. If P is entirely in $G^\star$, then we are done. For the rest of the proof, we assume that P is not entirely in $G^\star$. Then P contains a subpath $P' = (u, \ldots , v)$ whose intermediate vertices are all within a type-2 component $S \in C(u, v)$ that is not included to $G^\star$. By the definition of R[u, v], the length of $P'$ is at least r[u, v], which is the shortest path length between u and v via R[u, v]. Therefore, replacing $P'$ with a shortest u–v path in $R_j[u,v]$, which is entirely in $G^\star$, does not increase the path length. This contradicts our choice of P. Hence P is entirely in $G^\star$.   $\square$

Lemma 8

Let S be a type-1 or type-2 component that is not included in $G^\star$. Let $s \in S$. Let t be any vertex in G that does not belong to $G^+[S]$. Then there exists a vertex $s'$ in $G^\star$ such that ${\text {dist}}_{G}(s',t) \ge {\text {dist}}_{G}(s,t)$.

Proof

Let P be an s–t shortest path in G. Suppose that $S \in C(u)$ is of type-1. Because S is not included in $G^\star$, we must have $|C(u)| \ge 3$, so both $A_1[u] \ne S$ and $A_2[u] \ne S$ are not $\emptyset$. Let $i \in \{1,2\}$ be an index such that t is not in $A_i[u]$. Consider the subpath $\tilde{P} = (s, \ldots , u)$ of P. By the definition of $a_i[u]$ and $A_i[u]$, the length of $\tilde{P}$ is at most $a_i[u]$, and there exists a vertex $s' \in A_i[u]$ such that the length of the shortest path between $s'$ and u equals $a_i[u]$. Thus, we have

$${\text {dist}}_{G}(s',t) = {\text {dist}}_{G}(s',u) + {\text {dist}}_{G}(u,t) \ge {\text {dist}}_{G}(s,u) + {\text {dist}}_{G}(u,t) = {\text {dist}}_{G}(s,t).$$

Next, consider the case that $S \in C(u,v)$ is of type-2. The path P must contain at least one of u and v. Without loss of generality, assume that u is the first vertex of P that is not in S, so there is a subpath $\tilde{P} = (s, \ldots , u)$ of P such that all vertices in $\tilde{P}$ other than u are in S. The length of P equals ${\text {dist}}_{G^+[S]}(s,u) + {\text {dist}}_{G}(u,t)$.

Let $k = {\text {dist}}_{G^+[S]}(s,v) - {\text {dist}}_{G^+[S]}(s,u)$, so $S^{u,k} \supseteq \{s\} \ne \emptyset$. Since S is not of type-3, $|S| < \sqrt{n}$, so $k \in \{-\sqrt{n}, \ldots , \sqrt{n}\}$. Because S is not included in $G^\star$, both $A_1^k[u,v] \ne S$ and $A_1^k[u,v] \ne S$ are not $\emptyset$. At least one of $A_1^k[u,v]$ and $A_2^k[u,v]$ does not contain t. We choose $S' = A_i^k[u,v]$ as any one of them that does not contain t. We choose $s' \in S'$ as a vertex such that ${\text {dist}}_{G^+[S']}(s',u) = a_i^k[u,v]$ and ${\text {dist}}_{G^+[S']}(s',v) - {\text {dist}}_{G^+[S']}(s',u) \ge k$. The existence of such a vertex $s'$ is guaranteed by the definition of $A_i^k[u,v]$.

Our plan is to show that (i) $a_i^k[u,v] + {\text {dist}}_{G}(u,t) \ge {\text {dist}}_{G}(s,t)$ and (ii) ${\text {dist}}_{G}(s',t) = a_i^k[u,v] + {\text {dist}}_{G}(u,t)$. Combining these two inequalities give us the desired result: ${\text {dist}}_{G}(s',t) \ge {\text {dist}}_{G}(s,t)$.

Proof of (i). By the definition of $A_i^k[u,v]$, we must have

$$\begin{aligned} {\text {dist}}_{G^+[S']}(s',u) = a_i^k[u,v] \ge {\text {dist}}_{G^+[S]}(s,u), \end{aligned}$$

so we have

$$\begin{aligned} a_i^k[u,v] + {\text {dist}}_{G}(u,t) \ge {\text {dist}}_{G^+[S]}(s,u) + {\text {dist}}_{G}(u,t) = {\text {dist}}_{G}(s,t). \end{aligned}$$

Proof of (ii). Suppose that (ii) is not true. Then any shortest path between $s'$ and t must contain a subpath $P' = (s', \ldots , v)$ such that u is not in $P'$, and so we have:

$${\text {dist}}_{G}(s',t) = {\text {dist}}_{G^+[S']}(s',v) + {\text {dist}}_{G}(v,t) < {\text {dist}}_{G^+[S']}(s',u) + {\text {dist}}_{G}(u,t).$$

Combining this inequality with the known fact ${\text {dist}}_{G^+[S']}(s',v) - {\text {dist}}_{G^+[S']} (s',u) \ge k$, we have:

$${\text {dist}}_{G}(u,t) - {\text {dist}}_{G}(v,t) > {\text {dist}}_{G^+[S']}(s',v) - {\text {dist}}_{G^+[S']}(s',u) \ge k,$$

which implies that ${\text {dist}}_{G}(v,t) < {\text {dist}}_{G}(u,t) - k$ ($\star$). We calculate an upper bound of ${\text {dist}}_{G}(s,t)$:

$$\begin{aligned} {\text {dist}}_{G}(s,t)&\le {\text {dist}}_{G^+[S]}(s,v) + {\text {dist}}_{G}(v,t)\\&= (k + {\text {dist}}_{G^+[S]}(s,u)) + {\text {dist}}_{G}(v,t)&\text {by definition of }k.\\&< (k + {\text {dist}}_{G^+[S]}(s,u)) + ({\text {dist}}_{G}(u,t) - k)&\text {by }(\star ).\\&= {\text {dist}}_{G^+[S]}(s,u) + {\text {dist}}_{G}(u,t). \end{aligned}$$

This contradicts the assumption that P is a shortest path between s and t in G, as the length of P equals ${\text {dist}}_{G^+[S]}(s,u) + {\text {dist}}_{G}(u,t)$.   $\square$

The following lemma shows that (S2) is true.

Lemma 9

For any two vertices s and t in graph G, there exist two vertices $s'$ and $t'$ in graph $G^\star$ such that ${\text {dist}}_G(s,t) \le {\text {dist}}_{G}(s',t')$.

Proof

If both s and t are already in $G^\star$, then we are done by setting $s' = s$ and $t' = t$. In the subsequent discussion, we focus on the case that at least one of s and t is not in $G^\star$. By symmetry, we will assume that s is not in $G^\star$, so there is a type-1 or a type-2 component S that is not included in $G^\star$ such that $s \in S$.

Case 1: t belongs to $G^+[S]$. If $S \in C(u)$ for some $u \in V_H$, then there exist two vertices $s'$ and $t'$ in the component $B[u] \in C(u)$ such that ${\text {dist}}_G(s',t') = b[u] \ge {\text {dist}}_G(s,t)$ by the definition of B[u].

The remaining case is that $S \in C(u,v)$ for some $u,v \in V_H$. Let $l = {\text {dist}}_{V \setminus S}(u,v)$. We observe that $l \le r[u,v]$. The reason is that the existence of a component $S \ne R[u,v]$ guarantees that $R[u,v] \ne \emptyset$, which implies that $l = {\text {dist}}_{V \setminus S}(u,v) \le {\text {dist}}_{G^+[R[u,v]]}(u,v) = r[u,v]$, as $G^+[R[u,v]]$ is a subgraph of $G[V \setminus S]$.

Since R[u, v] is of type-2, we have $r[u,v] \le |R[u,v]|+1 \le \sqrt{n}$, so $l \in \{1, \ldots , \sqrt{n}\}$ Consider the component $B^l[u,v] \in C(u,v)$. We observe that $l = {\text {dist}}_{V \setminus B^l[u,v]}(u,v)$, since the shortest u–v path length via R[u, v] is at most the length of any u–v path via S or $B^l[u,v]$, by our choice of R[u, v]. More precisely, we have $l = {\text {dist}}_{V \setminus S}(u,v) = {\text {dist}}_{V}(u,v) = {\text {dist}}_{V \setminus B^l[u,v]}(u,v)$, as the above discussion implies that including S and excluding $B^l[u,v]$ in the subscript does not change the shortest u–v path length. Here we use the fact that $B^l[u,v] \ne R[u,v]$, which is due to the definition of $B^l[u,v]$.

Since $l = {\text {dist}}_{V \setminus B^l[u,v]}(u,v)$, by the definition of $B^l[u,v]$, there exist two vertices $s'$ and $t'$ in $G^+[B^l[u,v]]$ such that ${\text {dist}}_{G}(s',t') \ge {\text {dist}}_{G}(s,t)$, since otherwise we would have selected $B^l[u,v] = S$.

Case 2: t does not belong to $G^+[S]$. We apply Lemma 8 to find a vertex $s'$ in $G^\star$ such that ${\text {dist}}_G(s,t) \le {\text {dist}}_{G}(s',t)$. If t is already in $G^\star$, then we are done. Otherwise, there is a type-1 or a type-2 component $S'$ that is not included in $G^\star$ such that $t \in S'$. There are two sub-cases.

• Suppose $s'$ belongs to $G^+[S']$. Then we may apply the same argument for Case 1 above to find two vertices $s''$ and $t''$ in $G^\star$ such that ${\text {dist}}_G(s,t) \le {\text {dist}}_{G}(s',t) \le {\text {dist}}_G(s'', t'')$.

• Suppose $s'$ does not belong to $G^+[S']$. Then we may apply Lemma 8 again to find a vertex $t'$ in $G^\star$ such that ${\text {dist}}_G(s,t) \le {\text {dist}}_{G}(s',t) \le {\text {dist}}_{G}(s',t')$.

In both sub-cases, we can find two vertices in $G^\star$ whose distance in G is at least ${\text {dist}}_G(s,t)$.   $\square$

Lemma 10

The diameter of G equals the diameter of $G^\star$.

Proof

Lemma 7 shows that (S1) is true. Lemma 9 shows that (S2) is true. These two results together imply that G and $G^\star$ have the same diameter. Statement (S1) implies that the diameter of $G^\star$ is at most the diameter of G. For the other direction, let s and t be two vertices in G such that ${\text {dist}}(s,t)$ equals the diameter of G. By (S2), there exist two vertices $s'$ and $t'$ in $G^\star$ such that ${\text {dist}}_G(s,t) \le {\text {dist}}_{G}(s',t')$. By (S1), ${\text {dist}}_G(s',t') = {\text {dist}}_{G^\star }(s',t')$, so the diameter of $G^\star$ is at least the diameter of G.   $\square$

In Appendix A, we will design an energy-efficient algorithm to learn the graph topology of $G^\star$. This algorithm, combined with Lemma 10, allows us to prove Theorem 1.

Appendices

A Learning the Topology of $G^\star$

By Lemma 10, the task of computing the diameter of a bounded-genus graph G is reduced to computing the diameter of $G^\star$. In this section, we show that all vertices can learn the graph topology of $G^\star$ using $\tilde{O}(\sqrt{n})$ energy.

Recall that $G_H$ is the graph defined by the vertex set $V_H$ and the edge set $\{\{u,v\} \, : \, |C(u,v)| > 0\}$. By Lemma 6, we know that $E(G_H) = O(\sqrt{n})$ and there exists an assignment $F: E(G_H) \mapsto V_H$ mapping each pair $\{u,v\} \in E(G_H)$ to one vertex in $\{u,v\}$ such that each $w \in V_H$ is mapped to at most O(1) times. Let $\mathcal {A}'$ be any deterministic centralized algorithm that finds such an assignment F, and we fix $F^\star$ to be the outcome of $\mathcal {A}'$ on the input $G_H$. If each vertex $v \in V$ already knows the graph $G_H$, then v can locally calculate $F^\star$ by simulating $\mathcal {A}'$.

To learn $G^\star$, we will let each vertex $u \in V$ learn the following information:

Basic information $\mathcal {I}_0(u)$.:

For each vertex $u \in V$, $\mathcal {I}_0(u)$ contains the following information: (i) whether $u \in V_H$ or $u \in V_L$, (ii) the list of vertices in $N(u) \cap V_H$, and (iii) the set of all pairs $\{u',v'\} \in E(G_H)$.

If u is in a connected component S of $G[V_L]$, then $\mathcal {I}_0(u)$ contains the following additional information: (i) the list of vertices in S, and (ii) the topology of the subgraph $G^+[S]$.

Information about type-1 components $\mathcal {I}_1(u)$.:

For each $u \in V_H$, $\mathcal {I}_1(u)$ contains the graph topology of $G^+[S']$, for each $S' = A_1[u]$, $A_2[u]$, and B[u].

Information about type-2 components $\mathcal {I}_2(u)$.:

For each $u \in V_H$, $\mathcal {I}_2(u)$ contains the following information. For each pair $\{u,v\} \in E(G_H)$ such that $F^\star (\{u,v\}) = u$, $\mathcal {I}_2(u)$ includes the graph topology of $G^+[S']$, for each $S' = A_i^k[u,v]$, $A_i^k[v,u]$, $B^l[u,v]$, and R[u, v], for each $i \in \{1,2\}$, $k \in \{-\sqrt{n}, \ldots , \sqrt{n}\}$, and $l \in \{1, \ldots , \sqrt{n}\}$.

The information $\mathcal {I}_0(u)$ contains the graph topology of $G_H$, allowing each vertex u to calculate $F^\star$ locally. Note that $\mathcal {I}_1(u)$ and $\mathcal {I}_2(u)$ contain nothing if $u \in V_L$. The following lemma shows that the graph topology of $G^\star$ can be learned efficiently given that each vertex $u \in V$ already knows $\mathcal {I}_0(u)$, $\mathcal {I}_1(u)$, and $\mathcal {I}_2(u)$.

Lemma 11

Given that each $u \in V$ already knows $\mathcal {I}_0(u)$, $\mathcal {I}_1(u)$, and $\mathcal {I}_2(u)$, using $\tilde{O}(n^{1.5})$ time and $\tilde{O}(\sqrt{n})$ energy, we can let all vertices in G learn the graph topology of $G^\star$ w.h.p.

Proof

To learn $G^\star$, it suffices to know the following information: (i) $\mathcal {I}_1(u)$ and $\mathcal {I}_2(u)$ for each $u \in V_H$, (ii) the graph topology of $G^+[S]$ for each type-3 component S, and (iii) the graph topology of the subgraph induced by $V_H$. For each type-3 component S, let $r_S$ be the smallest ID vertex in S. In view of the above, to let each vertex learn the topology of $G^\star$, it suffices to let the following $O(\sqrt{n})$ vertices broadcast the following information:

• For each $u \in V_H$, u broadcasts $\mathcal {I}_1(u)$, $\mathcal {I}_2(u)$, and the list of vertices $N(u) \cap V_H$, which is contained in $\mathcal {I}_0(u)$.

• For each $u \in V_L$ such that $u = r_S$ for a type-3 component S, u broadcasts the graph topology of $G^+[S]$. Note that each vertex $u \in V_L$ can decide locally using the information in $\mathcal {I}_0(u)$ whether or not u itself is $r_S$ for a type-3 component S.

Since $|V_H| = O(\sqrt{n})$ and the number of type-3 components is also $O(\sqrt{n})$ by Lemma 5, the number of vertices that need to broadcast is $O(\sqrt{n})$. We run the algorithm of Lemma 1 to find a good labeling $\mathcal {L}$ of G, and then we use Lemma 2(2) with $x = O(\sqrt{n})$ to let the above $O(\sqrt{n})$ vertices broadcast their information. This can be done in time $\tilde{O}(n^{1.5})$, and energy $\tilde{O}(\sqrt{n})$. After that, all vertices know the graph topology of $G^\star$.   $\square$

Next, we consider the task of learning the basic information $\mathcal {I}_0(u)$.

Lemma 12

Using $\tilde{O}(\sqrt{n})$ time and energy, we can let all vertices $v\in V$ learn the following information w.h.p.

• Each $v \in V$ learns whether $v \in V_H$ or $v \in V_L$.

• If $v \in V_H$, then v also learns the list of vertices in $N(v) \cap V_H$.

• If $v \in V_L$, then v also learns the two lists of vertices $N(v) \cap V_L$ and $N(v) \cap V_H$.

Proof

First, we run $\textsf{SR}{\text{- }}\textsf{comm}^{{\text {apx}}}$ with $W=1$, $\epsilon = 1$, $\mathcal {S}=\mathcal {R}=V$, and $m_u = 1$, for each $u \in \mathcal {S}$. This step lets each $v \in V$ estimate $\deg (v)$ up to a factor of 2. This step costs ${{\text {poly}}}\log n$ time, by Lemma 27.

After that, we run $\textsf{SR}{\text{- }}\textsf{comm}^{{\text {all}}}$ with $\mathcal {S} = V$ and $\mathcal {R}$ being the set of all vertices v whose estimate of $\deg (v)$ is at most $2 \sqrt{n}$. The message $m_v$ for each vertex v is ${\text {ID}}(v)$, and we use the bound $\varDelta ' = 4\sqrt{n}$ for $\textsf{SR}{\text{- }}\textsf{comm}^{{\text {all}}}$. Recall that $V_L$ is the set of vertices of degree at most $\sqrt{n}$, so we must have $V_L \subseteq \mathcal {R}$. The algorithm of $\textsf{SR}{\text{- }}\textsf{comm}^{{\text {all}}}$ allows each vertex $v \in \mathcal {R}$ to calculate $\deg (v)$ precisely. Therefore, after this step, each vertex $v \in V$ has enough information to decide whether $v \in V_H$ or $v \in V_L$. Furthermore, if $v \in V_L$, then v knows the list of all vertices N(v). This step takes $\tilde{O}(\sqrt{n})$ time, by Lemma 22.

In order for each vertex to learn all the required vertex lists, we run $\textsf{SR}{\text{- }}\textsf{comm}^{{\text {all}}}$ again with the following parameters: $\mathcal {S} = V_H$, $\mathcal {R} = V$, and the message $m_v$ for each vertex $v \in \mathcal {S}$ is its ${\text {ID}}(v)$. This time we may use the bound $\varDelta ' = \sqrt{n} \ge |V_H|$. After the algorithm of $\textsf{SR}{\text{- }}\textsf{comm}^{{\text {all}}}$, each vertex $v \in V$ knows the list of vertices in $N(v) \cap V_H$. For each $v \in V_L$, since v already knows the list of all vertices N(v), it can locally calculate the list $N(v) \cap V_L$. This step also takes $\tilde{O}(\sqrt{n})$ time.   $\square$

Lemma 13

Using $\tilde{O}(n^{1.5})$ time and $\tilde{O}(\sqrt{n})$ energy, we can let all vertices v in all connected components S of $G[V_L]$ learn (i) the vertex set S and (ii) the graph topology of $G^+[S]$ w.h.p.

Proof

First, we apply Lemma 12 to let all vertices $v \in V_L$ learn the two lists $N(v) \cap V_L$ and $N(v) \cap V_H$. To let all vertices learn the required information in the lemma statement, it suffices to let each vertex $v \in S$ broadcast the two lists $N(v) \cap V_L$ and $N(v) \cap V_H$ to all other vertices in S, for all connected components S of $G[V_L]$.

We do the above broadcasting task in parallel, for all connected components S of $G[V_L]$. We use Lemma 1 to let each component S compute a good labeling, and then we use Lemma 2(1) to let each vertex $v \in S$ broadcast the two lists $N(v) \cap V_L$ and $N(v) \cap V_H$ to all other vertices in S. Recall that the degree of any vertex in $V_L$ is less than $\sqrt{n}$, so the algorithm of Lemma 2(1) costs $\tilde{O}(n^{1.5})$ time and $\tilde{O}(\sqrt{n})$ energy.   $\square$

For each connected component S of $G[V_L]$, at the end of the algorithm of Lemma 13, each vertex $w \in S$ is able to determine the type of S. If S is of type-1, w knows the vertex $u \in V_H$ such that $S \in C(u)$. If S is of type-2, w knows the two vertices $u, v \in V_H$ such that $S \in C(u,v)$. Given such information, in the following lemma, we design an algorithm for learning the topology of $G_H$.

Lemma 14

Suppose that each vertex in each type-2 component S already knows (i) the vertex set S and (ii) the graph topology of $G^+[S]$. Using $\tilde{O}(n^{1.5})$ time and $\tilde{O}(\sqrt{n})$ energy, all vertices in the graph can learn the set of all pairs $\{u,v\} \in E(G_H)$ w.h.p.

Proof

First of all, we let all vertices in $V_H$ agree on a fixed ordering $V_H = \{v_1, \ldots , v_{|H|}\}$ as follows. We use Lemma 1 to compute a good labeling of G, and then we use Lemma 2(2) with $x = \sqrt{n}$ to let each vertex $v \in V_H$ broadcast ${\text {ID}}(v)$. After that, we may order $V_H = \{v_1, \ldots , v_{|H|}\}$ by increasing ordering of ${\text {ID}}$. This step takes $\tilde{O}(n^{1.5})$ time and $\tilde{O}(\sqrt{n})$ energy.

Next, we consider the task of letting each $u \in V_H$ learn the list of all $v \in V_H$ such that $C(u,v) \ne \emptyset$. We solve this task by $|V_H|$ invocations of $\textsf{SR}{\text{- }}\textsf{comm}$. Given a type-2 component $S \in C(u,v)$, we define $z_{u,S}$ as the smallest-ID vertex in $N(v) \cap S$. The vertex $z_{u,S}$ will be responsible for letting v know that $C(u,v) \ne \emptyset$. For $i = 1$ to $|V_H|$, we do an $\textsf{SR}{\text{- }}\textsf{comm}$ with $\mathcal {R} = V_H$ and $\mathcal {S}$ being the set of all vertices $z_{v_i,S}$ such that S is a type-2 component S with $v_i \in G^+[S]$. Observe that a vertex $u \in V_H$ receives a message during the ith iteration if and only if $C(u, v_i)\ne \emptyset$, i.e., $\{u, v_i\} \in E(G_H)$. By Lemma 21, this step takes $|V_H| \cdot {{\text {poly}}}\log n = \tilde{O}(\sqrt{n})$ time.

At the end of the above algorithm, each $u \in V_H$ knows the list of all $v \in V_H$ such that $C(u,v) \ne \emptyset$. In order to let all vertices in G learn the topology of $G_H$, it suffices to let all $u \in V_H$ broadcast this information. This can be done using Lemma 2(2) with $x = \sqrt{n}$, which costs $\tilde{O}(n^{1.5})$ time and $\tilde{O}(\sqrt{n})$ energy.   $\square$

Lemma 15

In $\tilde{O}(n^{1.5})$ time and $\tilde{O}(\sqrt{n})$ energy, we can let all $u \in V$ learn $\mathcal {I}_0(u)$ w.h.p.

Proof

This follows from Lemma 13 and Lemma 14.   $\square$

Next, we consider the task of learning $\mathcal {I}_1(u)$ and $\mathcal {I}_2(u)$.

Lemma 16

Suppose that each $v \in V$ knows $\mathcal {I}_0(v)$. Using $\tilde{O}(n^{1.5})$ time and $\tilde{O}(\sqrt{n})$ energy, we can let all vertices $u \in V_H$ learn $\mathcal {I}_1(u)$ and $\mathcal {I}_2(u)$ w.h.p.

Proof

Consider any vertex $u \in V_H$. For each component $S \in C(u)$, we let $r_{S,u}$ be the smallest-ID vertex in the set $S \cap N(u)$. For each $v \in V_H$ such that $F^\star (\{u,v\}) = u$, and for each component $S \in C(u,v)$, we similarly let $r_{S,u}$ be the smallest-ID vertex in the set $S \cap N(u)$. As we will later see, $r_{S,u}$ will be the vertex in S responsible for sending the graph topology $G^+[S]$ to u in case $G^+[S]$ belongs to $\mathcal {I}_1(u)$ or $\mathcal {I}_2(u)$.

Recall that $\mathcal {I}_1(u)$ and $\mathcal {I}_2(u)$ consist of the graph topology $G^+[S']$ of some selected type-1 and type-2 component $S'$ such that u belongs to $G^+[S']$. We will present a generic approach that lets $u \in V_H$ learn one graph topology in $\mathcal {I}_1(u)$ and $\mathcal {I}_2(u)$. As we will later see, the cost of learning one graph topology is ${{\text {poly}}}\log n$ time and energy. If the graph topology to be learned is in C(u), then only u and the vertices $r_{S,u}$ for all $S \in C(u)$ need to participate in the algorithm for learning the graph topology. If the graph topology to be learned is in C(u, v), then only u and the vertices $r_{S,u}$ for all $S \in C(u,v)$ need to participate in the algorithm for learning the graph topology. We only describe the algorithms that let $u \in V_H$ learn $A_1[u]$ and $A_2[u]$. The algorithms for learning the remaining graph topologies are analogous.

• Learning $A_1[u]$. Recall that $A_1[u]$ is a component $S' \in C(u)$ that maximizes ${\text {eccentricity}}(u,S')$. To learn $A_1[u]$, we use $\textsf{SR}{\text{- }}\textsf{comm}^{{\text {max}}}$ with $\mathcal {S} = \{r_{S, u} \, : \, S \in C(u)\}$ and $\mathcal {R} = \{u\}$. The message $m_v$ of $v = r_{S, u}$ is the graph topology of $G^+[S]$, and the key of $v = r_{S, u}$ is $k_v = {\text {eccentricity}}(u,S)$. Since each type-1 and type-2 component satisfies $|S| \le \sqrt{n}$, the maximum possible value of ${\text {eccentricity}}(u,S)$ is $\sqrt{n}$, so the size of the key space for $\textsf{SR}{\text{- }}\textsf{comm}^{{\text {max}}}$ is $K = \sqrt{n}$.

  If $|C(u)| > 0$, then the message that u receives from $\textsf{SR}{\text{- }}\textsf{comm}^{{\text {max}}}$ is the topology of $G^+[S']$, for a component $S' \in C(u)$ that attains the maximum value of ${\text {eccentricity}}(u,S')$ among all components in C(u), so u may set $A_1[u] = S'$. If $|C(u)| = 0$, the vertex u receives nothing from $\textsf{SR}{\text{- }}\textsf{comm}^{{\text {max}}}$, so u may set $A_1[u] = \emptyset$. By Lemma 24, the cost of $\textsf{SR}{\text{- }}\textsf{comm}^{{\text {max}}}$ is $O(\log K \log \varDelta \log n) = {{\text {poly}}}\log n$.

• Learning $A_2[u]$. The procedure for learning $A_2[u]$ is almost exactly the same as that for $A_1[u]$, with only one difference. Recall that $A_2[u]$ is a component $S' \in C(u) \setminus \{A_1[u]\}$ that maximizes ${\text {eccentricity}}(u,S')$, so we need to exclude the component $A_1[u]$ from participating. To do so, before we apply $\textsf{SR}{\text{- }}\textsf{comm}^{{\text {max}}}$, we use one round to let u send ${\text {ID}}(r_{A_1[u],u})$ to all vertices $\{r_{S, u} \, : \, S \in C(u)\}$. This allows each $r_{S,u}$ to learn whether or not $S = A_1[u]$.

For each $u \in V_H$, the number of pairs $\{u,v\}$ such that $F^\star (\{u,v\}) = u$ is O(1), so the number of graph topologies needed to be learned in $\mathcal {I}_1(u)$ and $\mathcal {I}_2(u)$ by u is $O(\sqrt{n})$. The total number of graph topologies needed to be learned, for all $u \in V_H$, is at most $|V_H| \cdot O(\sqrt{n}) = O(n)$. We fix any ordering of these learning tasks and solve them sequentially. For each of these tasks, we use the above generic approach to solve the task, so the time and energy cost for learning one graph topology is ${{\text {poly}}}\log n$. Since there are O(n) tasks, the overall time complexity is $O(n) \cdot {{\text {poly}}}\log n = \tilde{O}(n)$. Each vertex participates in $O(\sqrt{n})$ tasks, so the overall energy complexity is $O(\sqrt{n}) \cdot {{\text {poly}}}\log n = \tilde{O}(\sqrt{n})$.   $\square$

Lemma 17

Using $\tilde{O}(n^{1.5})$ time and $\tilde{O}(\sqrt{n})$ energy, we can let all vertices in G learn the graph topology of $G^\star$ w.h.p.

Proof

The lemma follows from combining Lemmas 11, 15 and 16.   $\square$

Now we are ready to prove Theorem 1.

Theorem 1

There is an algorithm that computes the diameter in $\tilde{O}(n^{1.5})$ time and $\tilde{O}(\sqrt{n})$ energy w.h.p. for bounded-genus graphs in $\textsf{No}{\text{- }}\textsf{CD}$.

Proof

The theorem follows from combining Lemmas 10 and 17.   $\square$

B Minimum Cut

In this section, we apply the approach introduced in Sect. 5 to show that (i) the exact global minimum cut size and (ii) an approximate s–t minimum cut size of any bounded-genus graph can be computed in $\tilde{O}(\sqrt{n})$ energy. We follow the same approach introduced in Sect. 5. That is, we still decompose the vertex set into $V_H$ and $V_L$, and we categorize the connected components of $G[V_L]$ into three types. The only difference here is the information that we extract from type-1 and type-2 components.

Given a cut $\mathcal {C} = (X, V \setminus X)$ of $G=(V,E)$, the two vertex sets $X \ne \emptyset$ and $V \setminus X \ne \emptyset$ are called the two parts of $\mathcal {C}$, and the cut edges of $\mathcal {C}$ are defined as $\{ \{u,v\} \, : \, u \in X, v \in V \setminus X\}$. The size of a cut $\mathcal {C}$, which we denote as $|\mathcal {C}|$, is defined as the number of cut edges of $\mathcal {C}$. A minimum cut of a graph is a cut $\mathcal {C}$ that minimizes $|\mathcal {C}|$ among all possible cuts. An s–t minimum cut of a graph is a cut $\mathcal {C}$ the minimizes $|\mathcal {C}|$ among all possible cuts subject to the constraint that s and t belong to different parts. We consider the following definitions:

c(S).:

For any type-1 component S, let c(S) be the minimum cut size of $G^+[S]$.

$c'(S)$.:

For any type-2 component $S \in C(u,v)$, let $c'(S)$ be the u–v minimum cut size of $G^+[S]$.

$c''(S)$.:

For any type-2 component $S \in C(u,v)$, let $c''(S)$ be the minimum cut size of $G^+[S]$ among all cuts such that both u and v are within the same part of the cut.

We make the following observations.

Lemma 18

Let $\mathcal {C} = (X, V \setminus X)$ be any minimum cut of G. For any vertex $u \in V_H$, one of the following statements is true:

• One part of the cut contains all vertices in $\bigcup _{S \in C(u)} S \cup \{u\}$.

• the size of the cut is $\min _{S \in C(u)} c(S)$.

Proof

Suppose that the first statement is false. Then there exists a component $S' \in C(u)$ such that $S' \cup \{u\}$ intersects both parts of the cut, so $\mathcal {C}' = (X \cap (S' \cup \{u\}), (V \setminus X) \cap (S' \cup \{u\}))$ is a cut of $G^+[S']$. Therefore, $\min _{S \in C(u)} c(S) \le c(S') \le |\mathcal {C}'| \le |\mathcal {C}|$. To prove that the second statement is true, we just need to show that $|\mathcal {C}| \le \min _{S \in C(u)} c(S)$. This inequality follows from the observation that for any component $S \in C(u)$, any cut of $G^+[S]$ can be extended to a cut of G of the same size by adding all vertices in $V \setminus (S \cup \{u\})$ to the part of the cut that contains u.    $\square$

Lemma 19

Let $\mathcal {C} = (X, V \setminus X)$ be any minimum cut of G. For two distinct vertices $u,v \in V_H$, one of the following statements is true:

• One part of the cut contains all vertices in $\bigcup _{S \in C(u,v)} S \cup \{u,v\}$.

• The size of the cut is $\min _{S \in C(u,v)} c''(S)$.

• u and v belong to different parts of the cut, and the number of cut edges that have at least one endpoint in $\bigcup _{S \in C(u,v)} S'$ is $\sum _{S \in C(u,v)} c'(S)$.

Proof

Suppose that the first statement is false. We first focus on the case where u and v belong to the same part of the cut $\mathcal {C}$. In this case, there exists a component $S' \in C(u,v)$ such that $S' \cup \{u,v\}$ intersects both parts of the cut, so $\mathcal {C}' = (X \cap (S' \cup \{u,v\}), (V \setminus X) \cap (S' \cup \{u,v\}))$ is a cut of $G^+[S]$ such that u and v belong to the same part of the cut. Therefore, $\min _{S \in C(u,v)} c''(S) \le c''(S') \le |\mathcal {C}'| \le |\mathcal {C}|$. Similar to the proof of Lemma 18, we also have $|\mathcal {C}| \le \min _{S \in C(u,v)} c''(S)$, as any cut of $G^+[S]$ such that u and v belong to the same part of the cut can be extended to a cut of G of the same size. Therefore, we must have $|\mathcal {C}| = \min _{S \in C(u,v)} c''(S)$, that is, the second statement is true.

For the rest of the proof, we consider the case where u and v belong to different parts of the cut $\mathcal {C}$. For each component $S \in C(u,v)$, we write $Z_{S}$ to denote the number of cut edges of $\mathcal {C}$ that have at least one endpoint in S. Then we must have $Z_{S} = c'(S)$, since otherwise $\mathcal {C}$ is not a minimum cut. Therefore, the number of cut edges that have at least one endpoint in $\bigcup _{S \in C(u,v)} S'$ is $\sum _{S \in C(u,v)} c'(S)$, that is, the third statement is true.   $\square$

Using the above two observations, we prove Theorem 2.

Theorem 2

There is an algorithm that computes the minimum cut size in $\tilde{O}(n^{1.5})$ time and $\tilde{O}(\sqrt{n})$ energy w.h.p. for bounded-genus graphs in $\textsf{No}{\text{- }}\textsf{CD}$.

Proof

Bounded-genus graphs have bounded arboricity. The minimum degree of any graph of arboricity $\alpha$ is at most $2 \alpha - 1$. The minimum cut size of any graph is at most the minimum degree of the graph. Therefore, there is a constant $\lambda _0$ such that the minimum cut size of G is at most $\lambda _0$.

Define the graph $G^\diamond$ as the result of applying the following operations to G:

• Remove all type-1 components.

• For each pair $\{u,v\}$ of distinct vertices in $V_H$ with $|C(u,v)| > 0$, replace C(u, v) with $\min \{\lambda _0, \sum _{S \in C(u,v)} c'(S)\}$ multi-edges between u and v.

By Lemmas 18 and 19, the minimum cut size of G is the minimum of the following numbers:

• The minimum value of $\min _{S \in C(u)} c(S)$ among all $u \in V_H$ such that $|C(u)|>0$.

• The minimum value of $\min _{S \in C(u,v)} c''(S)$ among all $u,v \in V_H$ such that $|C(u,v)|>0$.

• The minimum cut size of $G^\diamond$.

For each vertex $u \in V$, we define $\mathcal {I}^\diamond _0(u)$, $\mathcal {I}^\diamond _1(u)$, and $\mathcal {I}^\diamond _2(u)$ as follows.

• $\mathcal {I}^\diamond _0(u)$ is the same as the basic information $\mathcal {I}_0(u)$ defined in Sect. 5.

• $\mathcal {I}^\diamond _1(u)$ contains the number $\min _{S \in C(u)} c(S)$.

• $\mathcal {I}^\diamond _2(u)$ contains $\min _{S \in C(u,v)} c''(S)$ and $\min \{\lambda _0, \sum _{S \in C(u,v)} c'(S)\}$, for all pairs $\{u,v\} \in E(G_H)$ such that $F^\star (\{u,v\}) = u$.

Similar to the proof of Theorem 1, $\mathcal {I}_1(u)$ and $\mathcal {I}_2(u)$ contain nothing if $u \in V_L$.

As $\mathcal {I}^\diamond _0(u) = \mathcal {I}_0(u)$, we may use the algorithm of Lemma 15 to let all vertices $u \in V$ learn the information $\mathcal {I}^\diamond _0(u)$ using $\tilde{O}(n^{1.5})$ time and $\tilde{O}(\sqrt{n})$ energy.

The algorithm of Lemma 16 can be modified to allow all vertices $u \in V_H$ learn the information $\mathcal {I}^\diamond _1(u)$ and $\mathcal {I}^\diamond _2(u)$. Specifically, the number $\min _{S \in C(u)} c(S)$ can be learned by the same algorithm for learning $A_1[u]$ described in the proof of Lemma 15 by replacing $\textsf{SR}{\text{- }}\textsf{comm}^{{\text {max}}}$ with $\textsf{SR}{\text{- }}\textsf{comm}^{{\text {min}}}$ letting $v = r_{S_u}$ use the key $k_v = c(S)$. The algorithm for learning $\min _{S \in C(u,v)} c''(S)$ is similar.

For each pair $\{u,v\} \in E(G_H)$ such that $F^\star (\{u,v\}) = u$, to let u learn $\min \{\lambda _0, \sum _{S \in C(u,v)} c'(S)\}$, we use $\textsf{SR}{\text{- }}\textsf{comm}^{{\text {apx}}}$ with the following parameters:

• $\mathcal {S} = \{r_{S, u} \, : \, S \in C(u,v)\}$.

• $\mathcal {R} = \{u\}$.

• $\epsilon = 1/(2\lambda _0 + 1)$.

• $W = \lambda _0$.

• For each $S \in C(u,v)$, the message $m_v$ of the representative $v = r_{S, u}$ of S is $\min \{\lambda _0, c'(S)\}$.

After the algorithm of $\textsf{SR}{\text{- }}\textsf{comm}^{{\text {apx}}}$, u learns an $(1\pm \epsilon )$-approximation of

$$\begin{aligned} \sum _{v \in N^+(u) \cap \mathcal {S}} m_v = \sum _{S \in C(u,v)} \min \{\lambda _0, c'(S)\}. \end{aligned}$$

We claim that this allows u to calculate $\min \{\lambda _0, \sum _{S \in C(u,v)} c'(S)\}$ precisely. To prove this claim, we break the analysis into two cases. Let x be the approximation of $\sum _{S \in C(u,v)} \min \{\lambda _0, c'(S)\}$ computed by $\textsf{SR}{\text{- }}\textsf{comm}^{{\text {apx}}}$.

If $\min \{\lambda _0, \sum _{S \in C(u,v)} c'(S)\} = \lambda _0$, then

$$\begin{aligned} \sum _{v \in N^+(u) \cap \mathcal {S}} m_v = \sum _{S \in C(u,v)} \min \{\lambda _0, c'(S)\} \ge \lambda _0, \end{aligned}$$

which implies

$$\begin{aligned} x \ge (1-\epsilon )\lambda _0 > \lambda _0 - 1/2. \end{aligned}$$

If $\min \{\lambda _0, \sum _{S \in C(u,v)} c'(S)\} = \sum _{S \in C(u,v)} c'(S)$, then

$$\begin{aligned} \sum _{v \in N^+(u) \cap \mathcal {S}} m_v = \sum _{S \in C(u,v)} \min \{\lambda _0, c'(S)\} = \sum _{S \in C(u,v)} c'(S), \end{aligned}$$

which implies

$$\begin{aligned} x&\in \left[ (1-\epsilon )\sum _{S \in C(u,v)} c'(S), (1+\epsilon )\sum _{S \in C(u,v)} c'(S)\right] \\ {}&\subseteq \left[ \left( \sum _{S \in C(u,v)} c'(S)\right) - \frac{1}{2}, \left( \sum _{S \in C(u,v)} c'(S)\right) +\frac{1}{2}\right] . \end{aligned}$$

Therefore, u can calculate $\min \{\lambda _0, \sum _{S \in C(u,v)} c'(S)\}$ precisely from x. By Lemma 27, the cost for u to calculate $\min \{\lambda _0, \sum _{S \in C(u,v)} c'(S)\}$ via $\textsf{SR}{\text{- }}\textsf{comm}^{{\text {apx}}}$ is ${{\text {poly}}}\log n$ time.

For each $u \in V_H$, the number of pairs $\{u,v\}$ such that $F^\star (\{u,v\}) = u$ is O(1), so the number of parameters needed to be learned in $\mathcal {I}^\diamond _1(u)$ and $\mathcal {I}^\diamond _2(u)$ by u is O(1). The total number of parameters needed to be learned, for all $u \in V_H$, is at most $|V_H| \cdot O(1) = O(\sqrt{n})$. We fix any ordering of these learning tasks and solve them sequentially. The time and energy cost for learning one parameter is ${{\text {poly}}}\log n$. Since there are $O(\sqrt{n})$ tasks, the overall time complexity for learning $\mathcal {I}^\diamond _1(u)$ and $\mathcal {I}^\diamond _2(u)$ for all $u \in V_H$ is $O(\sqrt{n}) \cdot {{\text {poly}}}\log n = \tilde{O}(\sqrt{n})$.

In view of the above discussion, the minimum cut size of G can be calculated from the following information: (i) $\mathcal {I}^\diamond _1(u)$ and $\mathcal {I}^\diamond _2(u)$ for all $u \in V_H$, (ii) the topology of $G^+[S]$ for each type-3 component S, and (iii) the topology of the subgraph induced by $V_H$. By replacing $\mathcal {I}_1(u)$ and $\mathcal {I}_2(u)$ with $\mathcal {I}^\diamond _1(u)$ and $\mathcal {I}^\diamond _2(u)$ in the description of the algorithm of Lemma 11, we obtain an algorithm that lets all vertices learn this information using $\tilde{O}(n^{1.5})$ time and $\tilde{O}(\sqrt{n})$ energy.   $\square$

The proof of Theorem 3 is similar to that of Theorem 2. The main difference for the setting of s–t minimum cut is that if s or t happens to be within a type-1 or a type-2 component S, then we additionally need to learn the topology of $G^+[S]$. Any type-1 component that does not contain s or t is irrelevant to the s–t minimum cut size.

In the subsequent discussion, we fix s and t to be any two distinct vertices of G. for each $x \in \{s,t\}$, let $S_x$ be the type-1 or type-2 component containing x. In case x is not contained in any type-1 or type-2 component, we let $S_x = \emptyset$. We define $G^\bullet$ as the result of applying the following operations to G.

• Remove all type-1 components, except for $S_s$ and $S_t$.

• For each pair $\{u,v\}$ of distinct vertices in $V_H$ with $|C(u,v) \setminus \{S_s, S_t\}| > 0$, replace all components in $C(u,v) \setminus \{S_s, S_t\}$ with $\sum _{S \in C(u,v) \setminus \{S_s, S_t\}} c'(S)$ multi-edges between u and v.

Similar to Lemmas 18 and 19, we have the following observation.

Lemma 20

Both G and $G^\bullet$ have the same minimum s–t cut size.

Proof

Fix $\mathcal {C} = (X, V \setminus X)$ to be any minimum s–t cut of G, where $s \in X$ and $t \in V\setminus X$. To show that both G and $G^\bullet$ have the same minimum s–t cut size, it suffices to show the following two statements:

• For each type-1 component S that is not $S_s$ and $S_t$, we must have either $S \subseteq X$ or $S \subseteq V\setminus X$.

• For each pair $\{u,v\}$ of distinct vertices in $V_H$ with $|C(u,v) \setminus \{S_s, S_t\}| > 0$, if u and v belong to different parts of cut $\mathcal {C}$, then the number of cut edges of $\mathcal {C}$ with at least one endpoint in $\bigcup _{S \in C(u,v) \setminus \{S_s, S_t\}} S$ equals $\sum _{S \in C(u,v) \setminus \{S_s, S_t\}} c'(S)$.

The first statement follows from the observation that for each $u \in V_H$, all vertices in $\bigcup _{S \in C(u) \setminus \{S_s, S_t\}} S$ must belong to the part of cut $\mathcal {C}$ that u belongs to, since otherwise $\mathcal {C}$ is not a minimum s–t cut, as moving all vertices in $\bigcup _{S \in C(u) \setminus \{S_s, S_t\}} S$ to the part of cut that u belongs to reduces the number of cut edges.

To show the second statement, consider a pair $\{u,v\}$ of distinct vertices in $V_H$ with $|C(u,v) \setminus \{S_s, S_t\}| > 0$ such that u and v belong to different parts of cut $\mathcal {C}$. Similar to the proof of Lemma 19, for each component $S \in C(u,v) \setminus \{S_s, S_t\}$, we write $Z_{S}$ to denote the number of cut edges of $\mathcal {C}$ that have at least one endpoint in S. Then we must have $Z_{S} = c'(S)$, since otherwise $\mathcal {C}$ is not a minimum cut. Therefore, the number of cut edges of $\mathcal {C}$ that have at least one endpoint in $\bigcup _{S \in C(u,v) \setminus \{S_s, S_t\}} S'$ is $\sum _{S \in C(u,v) \setminus \{S_s, S_t\}} c'(S)$.   $\square$

We are ready to prove Theorem 3.

Theorem 3

There is an algorithm that computes an $(1 \pm \epsilon )$-approximate s–t minimum cut size in $\tilde{O}(n^{1.5}) \cdot \epsilon ^{-O(1)}$ time and $\tilde{O}(\sqrt{n}) \cdot \epsilon ^{-O(1)}$ energy w.h.p. for bounded-genus graphs in $\textsf{No}{\text{- }}\textsf{CD}$.

Proof

Let $\tilde{G}^\bullet$ be any graph such that for each pair of vertices $\{u,v\}$, the number of mult-edges in $\tilde{G}^\bullet$ is within a $(1 \pm \epsilon )$ factor of the number of mult-edges in $G^\bullet$. By Lemma 20, the minimum s–t cut size in $\tilde{G}^\bullet$ is a $(1 \pm \epsilon )$-approximation of the minimum s–t cut size of G. Therefore, the task of computing the minimum s–t cut size of G is reduced to computing such a graph $\tilde{G}^\bullet$.

For each $u \in V_H$, we let $\mathcal {I}_2^{\bullet }(u)$ contain the number $\sum _{S \in C(u,v) \setminus \{S_s, S_t\}} c'(S)$ for all pairs $\{u,v\} \in E(G_H)$ with $F^\star (\{u,v\}) = u$. The same algorithm for learning $\mathcal {I}_2^{\diamond }(u)$ presented in the proof of Theorem 2 can be applied here to let all $u \in V_H$ approximately learn each number in $\mathcal {I}_2^{\bullet }(u)$ within a $(1 \pm \epsilon )$ factor, by using $\textsf{SR}{\text{- }}\textsf{comm}^{{\text {apx}}}$ with parameter $\epsilon$. We can tolerate this approximation factor because here our goal is to learn $\tilde{G}^\bullet$.

In view of the above, a $(1\pm \epsilon )$-approximation of the minimum s–t cut size of G can be calculated from the following information: (i) $\mathcal {I}_2^\bullet (u)$ for all $u \in V_H$, (ii) the topology of $G^+[S]$ for $S = S_s$, $S = S_t$, and each type-3 component S, and (iii) the topology of the subgraph induced by $V_H$. Same as the proof of Theorem 2, we can obtain an algorithm that lets all vertices learn this information using $\tilde{O}(n^{1.5}) \cdot \epsilon ^{-O(1)}$ time and $\tilde{O}(\sqrt{n}) \cdot \epsilon ^{-O(1)}$. Here the extra $\epsilon ^{-O(1)}$ is due to the use of $\textsf{SR}{\text{- }}\textsf{comm}^{{\text {apx}}}$, which requires $\epsilon ^{-O(1)} \cdot {{\text {poly}}}\log n$ time and energy.   $\square$

C Algorithms for Communication Between Two Sets of Vertices

In this section, we present our algorithms for $\textsf{SR}{\text{- }}\textsf{comm}$ and its variants. Recall that $\textsf{SR}{\text{- }}\textsf{comm}$ requires that each vertex $v \in \mathcal {R}$ with $N^+(v) \cap \mathcal {S} \ne \emptyset$ receives a message $m_u$ from at least one vertex $u \in N^+(v) \cap \mathcal {S}$ w.h.p.

Lemma 21

( [4]) $\textsf{SR}{\text{- }}\textsf{comm}$ can be solved in time $O(\log \varDelta \log n)$ and energy $O(\log \varDelta \log n)$.

Proof

By the definition of $\textsf{SR}{\text{- }}\textsf{comm}$, each vertex $v \in \mathcal {S} \cap \mathcal {R}$ is not required to receive any message from other vertices, as we already have $v \in N^+(v) \cap \mathcal {S}$. Therefore, in the subsequent discussion, we assume that $\mathcal {S} \cap \mathcal {R} = \emptyset$.

The task $\textsf{SR}{\text{- }}\textsf{comm}$ with $\mathcal {S} \cap \mathcal {R} = \emptyset$ can be solved using the well-known decay algorithm of [4], which repeats the following routine for $C \log n$ times: For $i = 1$ to $\log \varDelta$, let each vertex $u \in \mathcal {S}$ transmit with probability $2^{-i}$. Each $v \in \mathcal {R}$ is always listening throughout the procedure. Here $C > 0$ is some large enough constant to be determined.

Consider a vertex $v \in \mathcal {R}$ such that $N(v) \cap \mathcal {S} \ne \emptyset$. Let $i^\star$ be the largest integer i such that $2^{i} \le 2|N(v) \cap \mathcal {S}|$. Consider a time slot t where each vertex $u \in \mathcal {S}$ transmits with probability $2^{-i^\star }$. For notational simplicity, we write $n' = |N(v) \cap \mathcal {S}|$ and $p' = 2^{-i^\star }$. Our choice of $i^\star$ implies that $1/n' \ge p' \ge 1/(2n')$. The probability of the event that exactly one vertex in the set $N(v) \cap \mathcal {S}$ transmits equals $n' p' (1 - p')^{n'-1} \ge 1/(2e)$. The calculation follows from the inequalities $n' p' \ge 1/2$ and $(1 - p')^{n'-1} \ge (1 - 1/n')^{n'-1} \ge 1/e$.

If the above event occurs, then v successfully receives a message $m_u$ from a vertex $u \in N(v) \cap \mathcal {S}$. The probability that v does not receive any message from vertices in $N(v) \cap \mathcal {S}$ throughout the entire algorithm is at most $(1 - 1/(2e))^{C \log n} = n^{-\varOmega (C)}$. By setting C to be a large enough constant, the algorithm successfully solves $\textsf{SR}{\text{- }}\textsf{comm}$ w.h.p., and the time and energy complexities of the algorithm are $O(\log \varDelta \log n)$.   $\square$

Recall that the goal of $\textsf{SR}{\text{- }}\textsf{comm}^{{\text {all}}}$ is to let each vertex $u \in \mathcal {S} \cap N^+(v)$ deliver a message $m_u$ to $v \in \mathcal {R}$, for each $v \in \mathcal {R}$.

Lemma 22

$\textsf{SR}{\text{- }}\textsf{comm}^{{\text {all}}}$ can be solved in time $O(\varDelta ' \log n)$ and energy $O(\varDelta ' \log n)$, where $\varDelta '$ is an upper bound on $|\mathcal {S} \cap N(v)|$, for each $v \in \mathcal {R}$.

Proof

Consider the algorithm which repeats the following routine for $C \cdot \varDelta ' \log n$ rounds, for some sufficiently large constant $C > 0$. In each round, each vertex $u \in \mathcal {S}$ sends $m_u$ with probability $1 / \varDelta '$. For each $u \in \mathcal {R}$, if u does not send in this round, then u listens.

Let $e = \{u,v\}$ be any edge with $u \in \mathcal {S}$ and $v \in \mathcal {R}$. In one round of the above algorithm, u successfully sends a message to v if (i) all vertices in $\{v\} \cup (\mathcal {S} \cap N(v)) \setminus \{u\}$ do not send, and (ii) u sends. Therefore, the probability that u successfully sends a message to v is

$$ (1 - 1/\varDelta ')^{|\mathcal {S} \cap N(v)|-1} \cdot (1/\varDelta ') \ge (1 - 1/\varDelta ')^{\varDelta '-1} \cdot (1/\varDelta ') \ge 1/(e\varDelta ') $$

The probability that u does not successfully send a message to v throughout all $C \cdot \varDelta ' \log n$ rounds is at most $(1 - 1/(e\varDelta '))^{C \cdot \varDelta ' \log n} = n^{-\varOmega (C)}$. Selecting a large enough constant C, by a union bound for all $u \in \mathcal {S} \cap N(v)$ and all $v \in \mathcal {R}$, we conclude that the algorithm solves $\textsf{SR}{\text{- }}\textsf{comm}^{{\text {all}}}$ w.h.p. The time and energy complexities are $O(\varDelta ' \log n)$.   $\square$

Recall that the task $\textsf{SR}{\text{- }}\textsf{comm}^{{\text {multi}}}$ requires that each vertex $v \in \mathcal {R}$ receive all distinct messages in $\bigcup _{u \in N^+(v) \cap \mathcal {S}} \mathcal {M}_u$, where is the $\mathcal {M}_u$ is the set of messages hold by u.

Lemma 23

$\textsf{SR}{\text{- }}\textsf{comm}^{{\text {multi}}}$ can be solved in time $O(M \log \varDelta \log ^2 n)$ and energy $O(M \log \varDelta \log ^2 n)$, where M is an upper bound on the number of distinct messages in $\bigcup _{u \in N^+(v) \cap \mathcal {S}}$, for each $v \in \mathcal {R}$.

Proof

Consider the algorithm which repeatedly runs $\textsf{SR}{\text{- }}\textsf{comm}$ for $C \cdot M \log n$ times, where in each iteration, the sets $(\mathcal {S}',\mathcal {R'})$ for $\textsf{SR}{\text{- }}\textsf{comm}$ are chosen randomly as follows. We select $\mathcal {R'}$ as a random subset of $\mathcal {R}$ such that each $v \in \mathcal {R}$ joins $\mathcal {R'}$ with probability 1/2. We select $\mathcal {S}'$ as a random subset of $\mathcal {S} \setminus \mathcal {R'}$ such that for each message m, all vertices in $\mathcal {S} \setminus \mathcal {R'}$ that hold m join $\mathcal {S}'$ with probability 1/M, using the shared randomness associated with the message m.

Due to the shared randomness, if $u \in \mathcal {S} \setminus \mathcal {R'}$ joins $\mathcal {S}'$ due to message m, then all vertices in $\mathcal {S} \setminus \mathcal {R'}$ holding the same message m also joins $\mathcal {S}'$. Note that a vertex $u \in \mathcal {S} \setminus \mathcal {R'}$ might hold more than one message in that $|\mathcal {M}_u| > 1$. The probability that $u \in \mathcal {S} \setminus \mathcal {R'}$ joins $\mathcal {S}'$ equals ${\text {Pr}}[{\text {Binomial}}(|\mathcal {M}_u|, 1/M) \ge 1]$, because each message $m \in \mathcal {M}_u$ lets u join $\mathcal {S}'$ with probability 1/M independently.

To analyze the algorithm, we focus on one vertex $v \in \mathcal {R}$ in one iteration of the above algorithm. Consider any message $m \in \bigcup _{u \in N(v) \cap \mathcal {S}} \mathcal {M}_u \setminus \mathcal {M}_v$. Observe that v receives m if the following three events $\mathcal {E}_1$, $\mathcal {E}_2$, and $\mathcal {E}_3$ occur:

• $\mathcal {E}_1$ is the event that v joins $\mathcal {R}'$.

• $\mathcal {E}_2$ is the event that at least one vertex $u \in N(v) \cap \mathcal {S}$ with $m \in \mathcal {M}_u$ does not join $\mathcal {R}'$.

• $\mathcal {E}_3$ is the event that the subset of vertices of $N(v) \cap \mathcal {S} \setminus \mathcal {R'}$ joining $\mathcal {S}'$ is exactly the set of all vertices $u \in N(v) \cap \mathcal {S} \setminus \mathcal {R'}$ with $m \in \mathcal {M}_u$.

If $\mathcal {E}_1$, $\mathcal {E}_2$, and $\mathcal {E}_3$ occur, then $v \in \mathcal {R}'$, $N(v) \cap \mathcal {S}' \ne \emptyset$, and all vertices $u \in N(v) \cap \mathcal {S}'$ satisfy $m \in \mathcal {M}_u$. Therefore, conditioning on $\mathcal {E}_1$, $\mathcal {E}_2$, and $\mathcal {E}_3$, $\textsf{SR}{\text{- }}\textsf{comm}$ in this iteration allows v to receive message m.

The way $\mathcal {R}'$ is selected implies that ${\text {Pr}}[\mathcal {E}_1] = 1/2$ and ${\text {Pr}}[\mathcal {E}_2] \ge 1/2$. Observe that $\mathcal {E}_1$ and $\mathcal {E}_2$ are independent events. The way $\mathcal {S}'$ is selected implies that ${\text {Pr}}[\mathcal {E}_3 | \mathcal {E}_1 \cap \mathcal {E}_2] \ge {\text {Pr}}[{\text {Binomial}}(M, 1/M)=1] = (1/M) \cdot (1 - 1/M)^{M-1} \ge 1/(eM)$. Therefore, the probability that v receives m in this iteration is at least 1/(4eM).

The probability that v does not receive m in all iterations is at most $(1 - 1/(4eM))^{C \cdot M \log n} = n^{-\varOmega (C)}$. Selecting a large enough constant C, by a union bound for all $v \in \mathcal {R}$ and all $m \in \bigcup _{u \in N(v) \cap \mathcal {S}} \mathcal {M}_u \setminus \mathcal {M}_v$, we conclude that the algorithm solves $\textsf{SR}{\text{- }}\textsf{comm}^{{\text {all}}}$ w.h.p. The time and energy complexities are $O(M \log \varDelta \log ^2 n)$, as the number of iterations is $O(M \log n)$ and the time complexity of each iteration is $O(\log \varDelta \log n)$ by Lemma 21.   $\square$

Consider the setting where the message $m_u$ sent from each vertex $u \in \mathcal {S}$ contains a key $k_u$ from the key space $[K] = \{1, 2, \ldots , K\}$. Recall that $\textsf{SR}{\text{- }}\textsf{comm}^{{\text {min}}}$ requires that each vertex $v \in \mathcal {R}$ with $N^+(v) \cap \mathcal {S} \ne \emptyset$ receives a message $m_u$ from a vertex $u \in N^+(v) \cap \mathcal {S}$ such that $k_u = \min _{u' \in N^+(v) \cap \mathcal {S}} k_{u'}$.

Lemma 24

Both $\textsf{SR}{\text{- }}\textsf{comm}^{{\text {min}}}$ and $\textsf{SR}{\text{- }}\textsf{comm}^{{\text {max}}}$ can be solved in time $O(K \log \varDelta \log n)$ and energy $O(\log K \log \varDelta \log n)$. For the special case of $\mathcal {S} \cap \mathcal {R} = \emptyset$ and $|\mathcal {R} \cap N(u)| \le 1$ for each $u \in \mathcal {S}$, the time complexity can be improved to $O(\log K \log \varDelta \log n)$.

Proof

We only prove the lemma for $\textsf{SR}{\text{- }}\textsf{comm}^{{\text {min}}}$, as the proof for $\textsf{SR}{\text{- }}\textsf{comm}^{{\text {max}}}$ is the same. The proof presented here is analogous to the analysis of a deterministic version of $\textsf{SR}{\text{- }}\textsf{comm}$ in [10]. Observe that we can do $\textsf{SR}{\text{- }}\textsf{comm}$ once to let each $v \in \mathcal {R}$ test whether or not $N^+(v) \cap \mathcal {S} \ne \emptyset$. If a vertex $v \in \mathcal {R}$ knows that $N^+(v) \cap \mathcal {S} = \emptyset$, then v may remove itself from $\mathcal {R}$. Thus, in the subsequent discussion, we assume $N^+(v) \cap \mathcal {S} \ne \emptyset$ for each $v \in \mathcal {R}$.

Let $v \in \mathcal {R}$, and we define $f_v = \min _{u \in N^+(v) \cap \mathcal {S}} k_u$. The high-level idea of the algorithm is to conduct a binary search to determine all $\log K$ bits of the binary representation of $f_v$.

General Case. Suppose at some moment each vertex $v \in \mathcal {R}$ already knows the first x bits of $f_v$. The following procedure allows each $v \in \mathcal {R}$ to learn the $(x+1)$th bit of $f_v$. For each $(x+1)$-bit binary string s, we do $\textsf{SR}{\text{- }}\textsf{comm}$ with the following choices of $(\mathcal {S}',\mathcal {R}')$:

• $\mathcal {S}'$ is the set of vertices $u \in \mathcal {S}$ such that the first $x+1$ bits of $k_u$ equal s.

• $\mathcal {R}'$ is the set of vertices $v \in \mathcal {R}$ such that the first x bits of $f_v$ equal the first x bits of s.

In this procedure, we perform $2^{x+1}$ times of $\textsf{SR}{\text{- }}\textsf{comm}$ in total, but each vertex only participates in at most three of them, as each vertex joins $\mathcal {S}'$ at most once and joins $\mathcal {R}'$ at most twice. Thus, the procedure costs $O(2^x \log \varDelta \log n)$ time and $O(\log \varDelta \log n)$ energy, by Lemma 21. For each $v \in \mathcal {R}$, the messages that v receive during the procedure allows v to determine the $(x+1)$th bit of $f_v$.

We will run the above procedure for $\log K$ iterations from $x=0$ to $x = \log K - 1$. Observe that in the last iteration, each vertex $v \in \mathcal {R}$ is guaranteed to receive a message $m_u$ from a vertex $u \in N^+(v) \cap \mathcal {S}$ such that $k_u = f_v = \min _{w \in N^+(v) \cap \mathcal {S}} k_{w}$, so this algorithm allows us to solve $\textsf{SR}{\text{- }}\textsf{comm}^{{\text {min}}}$. The overall time complexity of the algorithm is

$$\begin{aligned} \sum _{x=0}^{\log K - 1} O(2^x \log \varDelta \log n) = O(K \log \varDelta \log n), \end{aligned}$$

and the overall energy complexity of the algorithm is

$$\begin{aligned} \sum _{x=0}^{\log K - 1} O(\log \varDelta \log n) = O(\log K \log \varDelta \log n). \end{aligned}$$

Special Case. For the rest of the proof, we focus on the special case of $\mathcal {S} \cap \mathcal {R} = \emptyset$ and $|\mathcal {R} \cap N(u)| \le 1$ for each $u\in \mathcal {S}$. These assumptions imply that the family of sets $(\mathcal {S} \cap N(v)) \cup \{v\}$ for all $v \in \mathcal {R}$ are disjoint. The high-level idea is that for each $v \in \mathcal {R}$, we may let the set of vertices $(\mathcal {S} \cap N(v)) \cup \{v\}$ jointly conduct a binary search to determine all bits of $f_v = \min _{u \in N(v) \cap \mathcal {S}} k_u$, in parallel for all $v \in \mathcal {R}$.

Suppose that for each vertex $v \in \mathcal {R}$, all vertices in the set $(\mathcal {S} \cap N(v)) \cup \{v\}$ already know the first x bits of $f_v$. We present a more efficient algorithm that let all vertices in the set $(\mathcal {S} \cap N(v)) \cup \{v\}$ learn the $(x+1)$th bit of $f_v$.

• Step 1. Perform $\textsf{SR}{\text{- }}\textsf{comm}$ with the following choices of $(\mathcal {S}',\mathcal {R}')$:

  • $\mathcal {R}' = \mathcal {R}$.

  • $\mathcal {S}'$ is the subset of $\mathcal {S}$ that contains all vertices $u \in \mathcal {S}$ satisfying the following conditions:

    • The first x bits of $k_u$ equal the first x bits of $f_v$, where v is the unique vertex in $\mathcal {R} \cap N(u)$.

    • The $(x+1)$th bit of $k_u$ is 0.

  This step allows each $v \in \mathcal {R}$ to learn the $(x+1)$th bit of $f_v$. If $v \in \mathcal {R}$ receives a message in $\textsf{SR}{\text{- }}\textsf{comm}$, then v knows that the $(x+1)$th bit of $f_v$ is 0. Otherwise, v knows that the $(x+1)$th bit of $f_v$ is 1.

• Step 2. Perform $\textsf{SR}{\text{- }}\textsf{comm}$ with the following choices of $(\mathcal {S}',\mathcal {R}')$:

  • $\mathcal {R}' = \mathcal {S}$.

  • $\mathcal {S}' = \mathcal {R}$.

  This step lets each $v \in \mathcal {R}$ send the $(x+1)$th bit of $f_v$ to all vertices in $\mathcal {S} \cap N(v)$.

The time and energy complexities of this algorithm are asymptotically the same as that of $\textsf{SR}{\text{- }}\textsf{comm}$, which are $O(\log \varDelta \log n)$. As discussed earlier, to solve $\textsf{SR}{\text{- }}\textsf{comm}^{{\text {min}}}$, all we need to do is to run the above algorithm from $x=0$ to $x = \log K - 1$. The overall time and energy complexities of the algorithm for $\textsf{SR}{\text{- }}\textsf{comm}^{{\text {min}}}$ are $O(\log K \log \varDelta \log n)$, as there are $\log K$ iterations.   $\square$

For the rest of the section, we consider the task $\textsf{SR}{\text{- }}\textsf{comm}^{{\text {apx}}}$, which requires each vertex $v \in \mathcal {R}$ to compute an $(1 \pm \epsilon )$-factor approximation of the summation $\sum _{u \in N^+(v) \cap \mathcal {S}} m_u$. We need the following fact, whose correctness can be verified by means of a simple calculation.

Lemma 25

There exist three universal constants $0< \epsilon _0 < 1$, $N_0 \ge 1$, and $c_0 \ge 1$ such that the following statement holds: For any pair of numbers $(N,\epsilon )$ such that $N \ge N_0$ and $\epsilon _0 \ge |\epsilon | \ge c_0 / \sqrt{N}$,

$$\begin{aligned} e^{-1}(1 - 0.51 \epsilon ^2) \le (1+\epsilon ) (1 - (1+\epsilon )/N)^{N-1} \le e^{-1}(1 - 0.49 \epsilon ^2). \end{aligned}$$

Note that the parameter $\epsilon$ in Lemma 25 can be either positive or negative. For the rest of the section, we assume that the message $m_u$ sent from each vertex $u \in \mathcal {S}$ is an integer within the range [W]. We first consider the special case of $\textsf{SR}{\text{- }}\textsf{comm}^{{\text {apx}}}$ with $W = 1$. In this case, $\textsf{SR}{\text{- }}\textsf{comm}^{{\text {apx}}}$ is the same as the approximate counting problem whose goal is to let each $v \in \mathcal {R}$ compute $|N^+(v) \cap \mathcal {S}|$, up to a $(1 \pm \epsilon )$-factor error.

Lemma 26

For $W = 1$, $\textsf{SR}{\text{- }}\textsf{comm}^{{\text {apx}}}$ can be solved in $O((1/\epsilon ^5) \log \varDelta \log n)$ time and energy.

Proof

In this proof, we will focus on a slightly different task of estimating $|N(v) \cap \mathcal {S}|$ within a $(1 \pm \epsilon )$-factor approximation, for each $v \in \mathcal {R}$. If each $v \in \mathcal {R}$ knows such an estimate of $|N(v) \cap \mathcal {S}|$, then v can locally calculate an estimate of $|N^+(v) \cap \mathcal {S}|$ within a $(1 \pm \epsilon )$-factor approximation, thereby solving $\textsf{SR}{\text{- }}\textsf{comm}^{{\text {apx}}}$ for the case of $W=1$.

Basic Setup. Let $C > 0$ be a sufficiently large constant. Let $\epsilon _0, N_0$, and $c_0$ be the constants in Lemma 25. We assume that $\epsilon \le \epsilon _0$. If this is not the case, then we may reset $\epsilon = \epsilon _0$.

The algorithm consists of two phases. The first phase of the algorithm aims to achieve the following goals: For each $v \in \mathcal {R}$, either (i) v learns the number $|N(v) \cap \mathcal {S}|$ exactly or (ii) v detects that $\epsilon \ge 10 c_0 / \sqrt{|N(v) \cap \mathcal {S}|}$. For each vertex $v \in \mathcal {R}$ that calculates the number $|N(v) \cap \mathcal {S}|$ exactly in the first phase, we remove v from $\mathcal {R}$. The second phase of the algorithm then solves $\textsf{SR}{\text{- }}\textsf{comm}^{{\text {apx}}}$ for the remaining vertices in $\mathcal {R}$. These vertices $v \in \mathcal {R}$ satisfy $\epsilon \ge 10 c_0 / \sqrt{|N(v) \cap \mathcal {S}|}$.

The First Phase. We define $Z = (10 c_0 / \epsilon )^2$. The algorithm consists of $C \cdot Z \log n$ rounds, where we do the following in each round:

• Each vertex $u \in \mathcal {S} \cup \mathcal {R}$ flips a biased coin that produces head with probability 1/Z.

• Each $u \in \mathcal {S}$ sends ${\text {ID}}(u)$ if the outcome of its coin flip is head.

• Each vertex $v \in \mathcal {R}$ listens if the outcome of its coin flip is tail.

For each vertex $v \in \mathcal {R}$, there are two cases:

• Suppose that there is a vertex $u \in N(v) \cap \mathcal {S}$ such that the number of messages that v receives from is smaller than $0.5 \cdot (C \log n)/e$. Then v decides that $\epsilon \ge 10 c_0 / \sqrt{|N(v) \cap \mathcal {S}|}$ and proceeds to the second phase.

• Suppose that for all vertices $u \in N(v) \cap \mathcal {S}$, the number of messages that v receives from is at least $0.5 \cdot (C \log n)/e$. Then v calculate $|N(v) \cap \mathcal {S}|$ by the number of distinct ${\text {ID}}$s that v receives.

The time complexity of the first phase of the algorithm is $C \cdot Z \log n = O((1/\epsilon ^{2}) \log n)$.

Analysis. To analyze the algorithm, let $e = \{u,v\}$ be any edge such that $u \in \mathcal {S}$ and $v \in \mathcal {R}$. In one round of the above algorithm, u successfully sends a message to v if and only if (i) the outcome of u’s coin flip is head, and (ii) the outcome of the coin flips of all vertices in $(N(v)\cap \mathcal {S}) \cup \{v\} \setminus \{u\}$ are all tails. This event occurs with probability $p^\star = (1 - 1/Z)^{|N(v) \cap \mathcal {S}|} \cdot (1/Z)$. Let X be the number of times v receives a message from u. To prove the correctness of the algorithm, we show the following three concentration bounds:

• If $v \in \mathcal {R}$ satisfies $\epsilon \le 10 c_0 / \sqrt{|N(v) \cap \mathcal {S}|}$, then ${\text {Pr}}[X \ge 0.8 \cdot (C \log n)/e] = 1 - n^{-\varOmega (C)}$.

• If $v \in \mathcal {R}$ satisfies $\epsilon \ge 20 c_0 / \sqrt{|N(v) \cap \mathcal {S}|}$, then ${\text {Pr}}[X \le 0.2 \cdot (C \log n)/e] = 1 - n^{-\varOmega (C)}$.

• If $v \in \mathcal {R}$ satisfies $\epsilon \le 20 c_0 / \sqrt{|N(v) \cap \mathcal {S}|}$, then ${\text {Pr}}[X \ge 1] = 1 - n^{-\varOmega (C)}$.

We show the correctness of the algorithm given these concentration bounds. For the case $\epsilon \ge 20 c_0 / \sqrt{|N(v) \cap \mathcal {S}|}$, the second bound implies that the number of messages that v receives from u is greater than $0.5 \cdot (C \log n)/e$ w.h.p., so v correctly decides that $\epsilon \ge 10 c_0 / \sqrt{|N(v) \cap \mathcal {S}|}$ and proceeds to the second phase. For the case $\epsilon \le 20 c_0 / \sqrt{|N(v) \cap \mathcal {S}|}$, the third bound implies that v receives at least one message from each vertex in $N(v) \cap \mathcal {S}$ w.h.p., so v can calculate $|N(v) \cap \mathcal {S}|$ precisely. The only remaining thing to show is that when $\epsilon$ is at most $10 c_0 / \sqrt{|N(v) \cap \mathcal {S}|}$, w.h.p. v does not decide that $\epsilon \ge 10 c_0 / \sqrt{|N(v) \cap \mathcal {S}|}$. This follows from the first bound, which implies that the number of messages that v receives from u is greater than $0.5 \cdot (C \log n)/e$ w.h.p.

We prove the three concentration bounds as follows:

• Suppose that vertex $v \in \mathcal {R}$ satisfies $\epsilon \le 10 c_0 / \sqrt{|N(v) \cap \mathcal {S}|}$. We show that in this case the number of messages that v receives from $u \in N(v) \cap \mathcal {S}$ is at least $0.8 \cdot (C \log n)/e$, with probability $1 - n^{-\varOmega (C)}$. In this case, we have $Z = (10 c_0 / \epsilon )^2 \ge |N(v) \cap \mathcal {S}|$, so $p^\star = (1 - 1/Z)^{|N(v) \cap \mathcal {S}|} \cdot (1/Z) \ge (1 - 1/Z)^{Z} \cdot (1/Z) \ge 0.9 /(e Z)$. The expected value $\mu$ of X satisfies $\mu = C \cdot Z \log n \cdot p^\star \ge 0.9 (C \log n)/e$. By a Chernoff bound, ${\text {Pr}}[X \le 0.8 \cdot (C \log n)/e] \le \exp (-\varOmega (C \log n)) = n^{-\varOmega (C)}$.

• Suppose that vertex $v \in \mathcal {R}$ satisfies $\epsilon \ge 20 c_0 / \sqrt{|N(v) \cap \mathcal {S}|}$. We show that in this case the number of messages that v receives from $u \in N(v) \cap \mathcal {S}$ is at most $0.2 \cdot (C \log n)/e$, with probability $1 - n^{-\varOmega (C)}$. In this case, we have $Z = (10 c_0 / \epsilon )^2 \le |N(v) \cap \mathcal {S}| / 4$, so $p^\star = (1 - 1/Z)^{|N(v) \cap \mathcal {S}|} \cdot (1/Z) \le (1 - 1/Z)^{4Z} \cdot (1/Z) \le 1/(e^4 Z)$. The expected value $\mu$ of X satisfies $\mu = C \cdot Z \log n \cdot p^\star \le (C \log n)/ e^4 < 0.1 (C \log n)/ e$. By a Chernoff bound, ${\text {Pr}}[X \ge 0.2 \cdot (C \log n)/e] \le \exp (-\varOmega (C \log n)) = n^{-\varOmega (C)}$.

• Suppose that vertex $v \in \mathcal {R}$ satisfies $\epsilon \le 20c_0 / \sqrt{|N(v) \cap \mathcal {S}|}$. We show that in this case the number of messages that v receives from $u \in N(v) \cap \mathcal {S}$ is at least 1, with probability $1 - n^{-\varOmega (C)}$. In this case, we have $Z = (10 c_0 / \epsilon )^2 \ge |N(v) \cap \mathcal {S}| / 4$, so $p^\star = (1 - 1/Z)^{|N(v) \cap \mathcal {S}|} \cdot (1/Z) \ge (1 - 1/Z)^{4Z} \cdot (1/Z) \ge 0.9 /(e^4 Z)$. We have ${\text {Pr}}[X < 1] = (1-p^\star )^{CZ \log n} \le (1 - 0.9 /(e^4 Z))^{CZ \log n} = n^{-\varOmega (C)}$.

The Second Phase. For each vertex $v \in \mathcal {R}$ that have already calculated the number $|N(v) \cap \mathcal {S}|$ exactly in the first phase, v removes itself from $\mathcal {R}$. We know that all the remaining vertices in $\mathcal {R}$ satisfy $\epsilon \ge 10 c_0 / \sqrt{|N(v) \cap \mathcal {S}|}$.

We consider the sequence of sending probabilities: $p_1 = 2/\varDelta$, and $p_i = \min \{1, p_{i-1} \cdot (1+ \epsilon )\}$ for $i > 1$. We let $i^\star = O((1/\epsilon ) \log \varDelta )$ be the smallest index i such that $p_i = 1$.

The second phase of the algorithm consists of $i^\star$ iterations, where the ith iteration repeats the following procedure for $C \cdot (1/\epsilon ^4) \log n$ times for all vertices $v \in \mathcal {S} \cup \mathcal {R}$:

• v flips a fair coin.

• If the outcome of the coin flip is head and $v \in \mathcal {S}$, then v sends with probability $p_i$.

• If the outcome of the coin flip is tail and $v \in \mathcal {R}$, then v listens to the channel.

After finishing the algorithm, each vertex $v \in \mathcal {R}$ finds an index $i'$ such that the number of messages that v successfully receives during the $i'$th iteration is the highest. Then v decides that $2 / p_{i'}$ is an estimate of $|N(v) \cap \mathcal {S}|$ within a factor of $(1\pm \epsilon )$. The time complexity of the second phase of the algorithm is $i^\star \cdot C \cdot (1/\epsilon ^4) \log n = O((1/\epsilon ^5) \log \varDelta \log n)$.

Analysis. To show the correctness of the above algorithm, in the subsequent discussion, we focus on a vertex $v \in \mathcal {R}$ in the ith iteration. We say that i is good for v if $p_i/2$ is within a $(1\pm 0.6\epsilon )$-factor of $1/|N(v) \cap \mathcal {S}|$, and we say that i is bad for v if $p_i/2$ is not within a $(1\pm \epsilon )$-factor of $1/|N(v) \cap \mathcal {S}|$. Our choice of the sequence $(p_1, p_2, \ldots )$ implies that there must be at least one good index i for v.

We write $p_i^{\text {suc}}$ to denote the probability that v successfully receives a message in one round of the ith iteration. From the description of the algorithm, we have

$$\begin{aligned} p_i^{\text {suc}} = (1/2) \cdot |N(v) \cap \mathcal {S}| \cdot (p_i / 2) \cdot (1-(p_i / 2))^{|N(v) \cap \mathcal {S}|-1}. \end{aligned}$$

We define

$$\begin{aligned} p_{\text {good}} = (1/2) \cdot e^{-1}(1 - 0.51 (0.6 \epsilon )^2) \ \ \ \text {and} \ \ \ p_{\text {bad}} = (1/2) \cdot e^{-1}(1 - 0.49 \epsilon ^2). \end{aligned}$$

We claim that (i) $p_i^{\text {suc}} \ge p_{\text {good}}$ if i is good for v and (ii) $p_i^{\text {suc}} \le p_{\text {bad}}$ if i is bad for v.

We first prove this claim for the case that i is good for v. For simplicity, we write $N = |N(v) \cap \mathcal {S}|$. Since i is good, $p_i / 2 = (1 + \epsilon ')/|N(v) \cap \mathcal {S}|$ for some $\epsilon ' \in [-0.6 \epsilon , 0.6 \epsilon ]$. Using the new notations, we may rewrite $p_i^{\text {suc}}$ as

$$\begin{aligned} p_i^{\text {suc}}&= (1/2) \cdot |N(v) \cap \mathcal {S}| \cdot (p_i / 2) \cdot (1-(p_i / 2))^{|N(v) \cap \mathcal {S}|-1}\\&= (1/2) \cdot (1 + \epsilon ') \cdot (1-(1 + \epsilon '))^{N-1}. \end{aligned}$$

By Lemma 25, we infer that $p_i^{\text {suc}} \ge (1/2) \cdot e^{-1}(1 - 0.51 (\epsilon ')^2) \ge e^{-1}(1 - 0.51 (0.6 \epsilon )^2) = p_{\text {good}}$.

Now consider the case i is bad for v. Again, we write $N = |N(v) \cap \mathcal {S}|$. Since i is bad, $p_i / 2 = (1 + \epsilon ')/|N(v) \cap \mathcal {S}|$ for some $\epsilon ' \notin (-\epsilon , \epsilon )$. The above formula for $p_i^{\text {suc}}$ still applies to this case, and Lemma 25 implies that $p_i^{\text {suc}} \le (1/2) \cdot e^{-1}(1 - 0.49 (\epsilon ')^2) \le e^{-1}(1 - 0.49 \epsilon ^2) = p_{\text {bad}}$.

Let X be the number of messages that v receives in the ith iteration of the algorithm. The expected value of X is $\mu = p_i^{\text {suc}} \cdot C \cdot (1/\epsilon ^4) \log n$. For the case i is good for v, we have $\mu \ge p_{\text {good}} \cdot C \cdot (1/\epsilon ^4) \log n$, so by a Chernoff bound, we have:

$$\begin{aligned} {\text {Pr}}[X \le (1-0.01 \epsilon ^2) p_{\text {good}} \cdot C \cdot (1/\epsilon ^4) \log n] = e^{-\varOmega (\epsilon ^4 \cdot C \cdot (1/\epsilon ^4) \log n)} = n^{-\varOmega (C)}. \end{aligned}$$

For the case i is bad for v, we have $\mu \le p_{\text {bad}} \cdot C \cdot (1/\epsilon ^4) \log n$, so by a Chernoff bound, we have:

$$\begin{aligned} {\text {Pr}}[X \ge (1+0.01 \epsilon ^2) p_{\text {bad}} \cdot C \cdot (1/\epsilon ^4) \log n] = e^{-\varOmega (\epsilon ^4 \cdot C \cdot (1/\epsilon ^4) \log n)} = n^{-\varOmega (C)}. \end{aligned}$$

Since $(1-0.01 \epsilon ^2) p_{\text {good}} > (1+0.01 \epsilon ^2) p_{\text {bad}}$, we conclude that w.h.p. the index $i'$ selected by v must be good, which implies that the estimate $2/p_{i'}$ calculated by v is within a $(1\pm \epsilon )$-factor of $|N(v) \cap \mathcal {S}|$, as we know that $p_{i'}/2$ is within a $(1\pm 0.6\epsilon )$-factor of $1/|N(v) \cap \mathcal {S}|$, as $i'$ is good.   $\square$

In the following lemma, we extend Lemma 26 to any value of W.

Lemma 27

$\textsf{SR}{\text{- }}\textsf{comm}^{{\text {apx}}}$ can be solved in $O((1/\epsilon ^6) \log W \log \varDelta \log n)$ time and energy.

Proof

We let $\epsilon ' = \Theta (\epsilon )$ be chosen such that $(1+\epsilon ')^2 < 1+\epsilon$ and $(1-\epsilon ')^2 > 1-\epsilon$. We consider the following sequence: $w_1 = 1$ and $w_i = \min \{ W, (1+\epsilon ')w_{i-1}\}$ for $i > 1$. Let $i^\star$ be the smallest index i such that $w_i = W$.

From $i = 1$ to $i^\star$, we run the algorithm of Lemma 26 with the following setting:

• $\mathcal {S}'$ is the vertices $u \in \mathcal {S}$ with $m_u \in (w_{i-1}, w_{i}]$.

• $\mathcal {R}' = \mathcal {R}$.

• The error parameter is $\epsilon '$.

The algorithm of Lemma 26 lets each $v \in \mathcal {R}'$ compute an $(1 \pm \epsilon ')$-factor approximation of $|N^+(v) \cap \mathcal {S}'|$ using $O((1/\epsilon ^5) \log \varDelta \log n)$ time and energy.

For each $v \in \mathcal {R}$, we write $N_i$ to denote the number of vertices $u \in N^+(v) \cap \mathcal {S}$ such that $m_u \in (w_{i-1}, w_{i}]$, and we write $\tilde{N}_i$ to denote the estimate of $|N^+(v) \cap \mathcal {S}'|$ computed by v in the ith iteration. We have the following observations:

• $\tilde{N}_i$ is an $(1\pm \epsilon ')$-factor approximation of $N_i$.

• $\sum _{i=1}^{i^\star } w_i N_i$ is an $(1\pm \epsilon ')$-factor approximation of $\sum _{u \in N^+(v) \cap \mathcal {S}} m_u$.

Thus, $\sum _{i=1}^{i^\star } w_i \tilde{N}_i$, which can be calculated locally at v at the end of the algorithm, is an $(1\pm \epsilon )$-factor approximation of $\sum _{u \in N^+(v) \cap \mathcal {S}} m_u$, by our choice of $\epsilon '$.

By Lemma 26, the time and energy complexities for each iteration are $O((1/\epsilon ^5)\log \varDelta \log n)$. The total number of iterations is $i^\star = O((1/\epsilon ) \log W)$. Thus, the overall time and energy complexities are $O((1/\epsilon ^6) \log W \log \varDelta \log n)$.

   $\square$