1 Introduction

The model and motivations Opportunistic Mobile Ad-hoc NETworks (in short MANETs), are a special class of mobile ad hoc networks where agent density might be so low that the network is disconnected at virtually any time. Communication is possible even in such a challenging environment by exploiting the so-called store-carry-and-forward mechanism, according to which a packet is stored in agent \(A\)’s buffer and carried around by the agent till a communication opportunity with another agent \(B\) arises (whence the name of this class of networks); at this point, the packet can be forwarded from \(A\) to \(B\), and the process is repeated until the packet is eventually delivered to the destination.

Opportunistic networks are receiving increasing attention in the research community [13, 15, 23, 26], since many emerging application scenarios can be considered as instances of opportunistic networks. This is the case, for instance, of vehicular networks (at least, when traffic density is medium to low), of certain types of mobile sensor networks, and of pocket switched networks. There are several interesting problems in MANETs, here we are interested in the speed of information spreading, a question that can model the spread of disease, the broadcast of files, of memes in social networks, etc.

Models of MANETs are typically given by an (agent) mobility model that determines the mobility region, the feasible paths, the path-selection criterium, and the speed of the agents. Recent important theoretical studies focused on Markovian mobility models: every agent independently selects his next destination and/or his next geometric trajectory according to some Markovian process such as Random Walks or Random Waypoints [12, 16, 19, 25]. If the Markovian process is ergodic, then it always converges toward a unique stationary probability distribution defined over the set of all possible configurations. A configuration at time \(t\) is the \(n\)-component vector describing the state of each of the \(n\) agents at time \(t\). The state of an agent at time \(t\) depends on the particular Markovian process, however it typically encodes position, direction\(/\)destination, and speed of the agent at that time. We recall that if the configuration of the process at time \(t\) is random with the stationary distribution then the next configuration will still be random with the stationary distribution. Since agents are independent, the probability distribution of the configuration is given by the cartesian product of \(n\) copies of the probability distribution of the agent state (over the set of all possible agent states). In this case, we can thus restrict our attention to the stationary distribution of the single agent.

According to almost all previous works [7, 9, 16, 24, 25], in this paper we will assume that the initial network configuration is random with the stationary distribution. Not surprisingly, the properties of the random network during the stationary phase (e.g. network connectivity and expected agent-degree) have a deep impact on the behavior of communication protocols running over the network. For this reason, stationary properties of such Markovian networks have been extensively studied [4, 5, 9, 19].

As for probabilistic mobility models, analytical results for the speed of information spreading in MANETs are only available for the Random-Walk model [7, 9, 16, 24, 25], but a very recent paper [11] on the Random-Waypoint model that we will discuss later. The stationary (agent) spatial probability distribution of the Random-Walk model is almost uniform: the probability that an agent lies in a given position according to the agent stationary distribution is almost the same for any choice of the position. The stationary (agent) destination probability distribution gives the probability that an agent, being in a position \((x_0,y_0)\), has a destination \((x,y)\); As for the Random Walk, the distribution is uniform over a disk centered on \((x_0,y_0)\) and it is zero elsewhere.

A very popular mobility model is the Random Waypoint (RWP) [4, 5, 19]. In the basic version of this model, each agent chooses independently and uniformly at random a destination over all the square. Then, he starts traveling at speed \(\text{ v}\) towards the destination along a simple path. When he reaches the destination (a way-point), he chooses another destination, and so on.

In this work, we consider the variant of the RWP, called Manhattan Random Waypoint (MRWP) model, where the path followed by an agent from a given point to a destination point is randomly chosen between the two Manhattan 2-segment shortest paths (in short Manhattan paths) connecting the two points (see Sect. 2 for a formal definition). This variant of the RWP is inspired by scenarios where agents follows rectilinear streets of an urban zone while keeping the chosen route as short as possible [10, 12, 19]. The MRWP provides a reasonable, basic model to approximate the trajectories followed by vehicles in a urban zone. Even though this approximation is rather rough, it is much better than those provided by the Random Walks and by the standard RWP where the feasible paths are the straight segments connecting each possible pairs \((source,destination)\). So the standard RWP paths fully disregard any constraint arising from the typical presence of street “grids” in urban zones.

A further interest in this mobility model lies in its stationary properties (such as the spatial and the destination distributions): on one hand, they are similar to those exhibited by several other Random-Trip Models [19] and, on the other hand, they are very different from those of the Random-Walk model.

Explicit formulas of the stationary spatial and destination MRWP probability distributions have been derived in [10, 12]. The stationary distributions of some other variants of the RWP have been obtained in [4, 6, 18, 20]. In general, the knowledge of such distributions is crucial to achieve perfect simulation [5, 19], to derive connectivity properties [12], and for the study of information spreading [7, 9, 12].

The analytical study of the MRWP model leads to address a set of new probabilistic issues that we believe to be relevant in the study of any reasonable model of vehicular mobility in urban zones. Indeed, differently from the case of Random-Walk model, it turns out that the stationary probability distributions yielded by the MRWP are very far from being uniform. As for the stationary spatial distribution, in the four regions close to the corners of the square, the probability density function is asymptotically much lower than that in the Central Zone (see Figs. 1,  2). These four corner regions form the Suburb (The Central Zone and the Suburb are formally defined later in Definition 4). The area of the Suburb is not negligible, being a constant fraction of the entire area. Agent’s density concentration towards the central zone is an important feature exhibited by several more realistic mobility models adopted to simulate agent mobility in urban zones [5, 21]. The stationary destination distribution is a further crucial difference with respect to the Random-Walk models and it is rather complex (see Eq. 3, Fig. 2). Finally, while in the Random-Walk model, agents perform a random choice at every time step, in the MRWP (like in almost all models for human/vehicular mobility), agents spend much time to travel along deterministic trajectories.

Fig. 1
figure 1

The probability density function of the MRWP stationary spatial distribution

Full size image
Fig. 2
figure 2

The probability density function of the MRWP stationary spatial distribution is shown by a gradation of gray (white corresponds to the maximum density and black corresponds to the minimum density). Moreover, the density function of the MRWP stationary destination distribution over the cross of agent position \((x_0, y_0) = (L/3, L/4)\) is shown

Full size image

We consider \(n\) agents that move independently at constant speed \(\text{ v}>0\) over a square of side length \(L\) according to the MRWP model. Agents can exchange data if they are at (Euclidean) distance at most \(R\) within each other, where \(R>0\) is the agent transmission radius. According to all previous analytical works in the literature [3, 9, 25], the network evolution works in discrete time steps. At every time step \(t\), the snapshot at time \(t\) is the symmetric disk graph \(G_t\) determined by the agent positions at time \(t\) and their relative transmission disks. The connectivity threshold is the smallest \(R\) such that \(G_t\) is connected. In [12], the connectivity threshold is derived for the graph \(G_t\) which is random with the MRWP stationary distribution: when \(L = \sqrt{n}\), it is \(\Omega (n^c)\) for some constant \(0<c<1/2\). Thus, it is exponentially higher than that yielded by mobility models having uniform stationary spatial distribution (such as the Random-Walk model and some RWP variants over toroidal spaces): this threshold being \(\Theta (\sqrt{\log n})\) for \(L= \sqrt{n}\) [14, 22].

The flooding mechanism is a simple broadcast protocol where every informed agent transmits the source message at discrete time steps to all its dynamic neighborhood. An agent is said to be informed if he knows the source message. Hence, a non-informed agent \(\textsc {a}\) gets informed at time step \(t\) if and only if, during \(t\), at least one informed agent \(b \) is within distance \(R\) from \(\textsc {a}\). We do not consider here the problem of transmission collisions due to interferences. The flooding time is the first time step in which all agents are informed. The flooding time is a general lower bound for the (worst-case) completion time of any broadcast protocol and, thus, it is often used to evaluate the relative efficiency of alternative protocols, especially in networks with unknown dynamic topology.

Our results We study the flooding time in the MRWP model with parameter ranges \(R \geqslant c_1L \sqrt{\log n/n} \) and \(\text{ v}\leqslant \ R / c_2 \), where \(c_1\) and \(c_2\) are positive constants. Observe that, from the above discussion on graph connectivity, the first assumption on \(R\) does not guarantee network connectivity [12]: the snapshot yielded by the stationary distribution could be highly disconnected in the Suburb. The second assumption means we are considering a slow-mobility scenario, i.e., when an agent movement, in a time unit, cannot be longer than the transmission radius. Observe that the second assumption implies the lower bound \(\Omega (L/R)\) for flooding time. We prove that flooding time is with high probability bounded by

$$\begin{aligned} O\left(\frac{L}{R} + \frac{S}{\text{ v}}\right) \end{aligned}$$
(1)

where

$$\begin{aligned} S = \Theta \left(\frac{L^3 \log n}{R^2 n}\right) \end{aligned}$$

is the (Euclidean) diameter of each of the four corner regions of the Suburb. As usual, we say event \(\mathcal E \) holds with high probability (w.h.p.) if \(\mathbf P \left( \mathcal E \right) \geqslant 1 - 1/ n^{c}\) for some \(c >0\). Informally speaking, our bound says that flooding time is asymptotically bounded by the sum of two consecutive time spans: the time to traverse the square at “speed” \(R\) and the time to traverse the Suburb at speed \(\text{ v}\).

Let us discuss the consequences of our bound in the “standard” case \(L = \sqrt{n}\) (clearly, similar consequences hold for different values of \(L\)). If \(R= \Theta (\sqrt{\log n})\), our bound becomes \(O(L/\text{ v})\): this is optimal whenever \(\text{ v}= \Theta (R)\). In general, our bound is optimal whenever the speed \(\text{ v}\) falls into the range \( {\log n}/ R \leqslant \text{ v}\leqslant R\). For instance, if \(R = \Theta (\log n)\), our bound becomes optimal provided that \(\text{ v}\) is larger than an absolute constant. We observe that, under such conditions, the network in the Suburb is sparse and highly-disconnected since, as mentioned above, the connectivity threshold in the Suburb is exponentially larger than \(\log n\) [12]. Nevertheless, our bound says that flooding succeeds over the Suburb as well and, even more, its completion time is similar to that in the Central Zone where the network is fully-connected. This phenomenon holds even when the agent speed \(\text{ v}\) is very low.

We do not know whether our bound is optimal for all the range of the network parameters. In particular, when \(\text{ v}=\omega (R)\), we suspect that the optimal bound could be \(\Theta (L/\text{ v})\). Moreover, our upper bound cannot be improved to \(O(L/R)\). Indeed, we prove that, for some ranges of \(R, n\), and \(L\), flooding time is asymptotically larger than \(L/R\) and, moreover, it must depend on \(\text{ v}\).

Related works Flooding time of some classes of Markovian evolving graphs [1] has been recently studied in [3, 7–9, 16, 24, 25]. We observe that in all those classes, the agent behavior is characterized by the presence of independent random choices at every time step. On the contrary, in the RWP model (and, in particular, in the MRWP), the adopted relatively-long deterministic trajectories make the agent behavior strongly dependent on the past. This crucial difference makes our analysis of flooding time rather far from the previous ones.

No analytical results are known on any variant of the RWP model but a very recent work [11] (more recent than the conference version of this paper) shows an upper bound on the flooding time. However, our bound is still better for the “slow” case, i.e. whenever \(\text{ v}= o(R)\). The technical arguments in [11] exploit the mixing time of the RWP and are very different from ours.

Organization of the paper Section 2 provides the formal definition of the MRWP model ant the relative stationary probability distributions. Section 3 defines the flooding process and states the main theorem while, in Sect.  4, we present the proof of the main theorem. In Sect. 5, a lower bound on the flooding time is given and, finally, in Sect. 6 we discuss the obtained results and some open questions.

2 The Manhattan Random-Waypoint model

In this section, we formally present the MRWP model. Consider a square of edge length \(L>0\). A set of \(n\) independent agents move over this square according to the following stochastic rule. Starting from an initial position \((x_0,y_0)\), every agent selects a destination \((x,y)\) uniformly at random in the square. Then, the agent chooses uniformly at random between the two feasible Manhattan paths

$$\begin{aligned} \text{ P}_1 = ((x_0,y_0) \rightarrow (x_0,y) \rightarrow (x,y)) \ \text{ and} \ \text{ P}_2 = ( (x_0,y_0) \rightarrow (x,y_0) \rightarrow (x,y) ) \end{aligned}$$

Once the feasible path is selected, the agent starts following the chosen route with constant speed determined by the parameter \(\text{ v}\). An agent, once arrived at the selected destination, re-applies the process described above again and again. This Markovian process yields the MRWP model. We assume that all agents have the same speed \(\text{ v}\) that represents the travelled distance by an agent in the time unit. The time unit is the time interval between two consecutive discrete time instants in which we observe the agents.

The stationary probability distributions of the MRWP have been recently analytically derived. The stationary (agent) spatial distribution gives the probability that an agent lies in a position \((x,y)\) and it has been derived in [12]. The stationary (agent) destination distribution gives the probability that an agent, conditioned to be in position \((x_0,y_0)\), is traveling toward destination \((x,y)\) and it has been determined in [10]. An informal representation of the two distributions is given in Fig. 2.

Theorem 1

[12] The probability density function of the stationary spatial distribution is

$$\begin{aligned} f(x,y) \ = \ \frac{3}{L^3}(x+y)-\frac{3}{L^4}\left({x}^2+{y}^2\right) \end{aligned}$$
(2)

Theorem 2

[10] The probability density function of the stationary destination distribution is

$$\begin{aligned} f_{(x_0,y_0)}(x,y) \!=\!\left\{ \!\! \begin{array}{ll} \frac{2L \!-\! x_0 \!-\! y_0}{4L(L(x_0\!+\!y_0) \!-\!(x_0^2 \!+\!y_0^2 ))}&\text{ if}\; x\!<\!x_0 \quad \text{ and}\quad y<y_0\\ \frac{ x_0\!+\!y_0}{4L(L(x_0\!+\!y_0) \!-\!(x_0^2 \!+\!y_0^2 ))}&\text{ if}\; x\!>\!x_0\quad \text{ and}\quad y>y_0\\ \frac{ L-x_0 + y_0 }{4L(L(x_0\!+\!y_0)\! -\!(x_0^2 +y_0^2 ))}&\text{ if}\; x\!<\!x_0\quad \text{ and}\quad y>y_0\\ \frac{ L + x_0 -y_0}{4L(L(x_0+y_0) \!-\!(x_0^2 \!+\!y_0^2 ))}&\text{ if}\; x\!>\!x_0\quad \text{ and}\quad y<y_0\\ + \infty&\quad \text{ otherwise.} \end{array}\right.\nonumber \\ \end{aligned}$$
(3)

In order to get the destination distribution where the probability density function is infinite, one has to consider the four segments outgoing from \((x_0,y_0)\) and parallel to the axes (see Fig. 2). For every segment \(s \in \{ \text{ S}, \text{ W}, \text{ N}, \text{ E} \}\), the probability that an agent in node \((x_0,y_0)\), has destination lying on the segment \(s\), is \(\phi ^{s}_{(x_0,y_0)}\). It has been proved in [10] that

$$\begin{aligned} \phi ^\mathrm{{S}}_{(x_0,y_0)}&= \phi ^\mathrm{{N}}_{(x_0,y_0)} \ = \ \frac{y_0(L-y_0)}{4L(x_0+y_0) - 4(x_0^2 + y_0^2)}\end{aligned}$$
(4)
$$\begin{aligned} \phi ^\mathrm{{W}}_{(x_0,y_0)}&= \phi ^\mathrm{{E}}_{(x_0,y_0)} \ = \ \frac{x_0(L-x_0)}{4L(x_0+y_0) - 4(x_0^2 + y_0^2)} \end{aligned}$$
(5)

We observe that the sum of the above four probabilities (i.e. the probability that the agent has destination over the cross centered on \((x_0,y_0)\)) is not zero (it is equal to \(1/2\)) despite the fact that this region (i.e. the cross) has area 0. This is due to the shape of the Manhattan paths and it will be used in our analysis of flooding over the Suburb.

3 Flooding over Manhattan

In this section we give a formal definition of the flooding and then we state the main theorem. The flooding process starts from a source agent that wants to broadcast his message to all agents. An agent is said to be informed if he knows the source message. At the starting time, only the source agent is informed. Then, an agent \(\textsc {a}\) gets informed at time step \(t = 1,2, \ldots \) if and only if, during that time, there is at least one informed agent \(b \) that travels at distance not larger than \(R >0\), where \(R\) is the transmission radius valid for all agents.

The \(n\) agents move over the square \(L \times L\) (with \(L >0\)) according to the MRWP model. We want to provide an upper bound on the flooding time, i.e., the maximal number of time steps required by the flooding process to inform all the agents, for any possible choice of the source agent. We can now state our main result.

Theorem 3

Consider a MANET of \(n\) agents moving over a square of size \(L\) according to the MRWP model. Let \(R\) be the transmission radius and \(\text{ v}\) be the agent speed. Assume that \(R \geqslant c_1 L \sqrt{\log n / n}\) and \(\text{ v}\leqslant R/c_2 \) for sufficiently large constants \(c_1\) and \(c_2\). Then, for sufficiently large \(n\), the flooding time in the stationary phase is w.h.p. bounded by

$$\begin{aligned} O\left( \frac{L}{R} + \frac{L^3\log n }{\text{ v}R^2 n} \right) \end{aligned}$$

4 Proof of theorem 3

First, we outline the proof technique of Theorem 3. The stationary spatial probability distribution shows a central region of high density (the Central Zone) and four corner regions (the Suburb) of low density (see Figs. 1, 2). High density means that the expected number of agents in any disk of radius \(R\) is \(\Omega (R^2)\) and that the resulting local network is w.h.p. connected.

A key-issue here is that there is strong stochastic dependence between consecutive snapshots: if we observe and bound some function of the snapshot at time \(t\), e.g. the number of informed nodes at that time, then the next snapshot is conditioned by the observed bound and, thus, it cannot be anymore considered random with the stationary distribution. So, even though we can say that each of such snapshots (individually) enjoys w.h.p. some good properties (such as high density and connectivity in the Central Zone), we cannot directly exploit them during the flooding process.

Informally speaking, this technical issue is solved by proving that any sufficiently-long sequence of snapshots (observed during the flooding process) is w.h.p. formed by conditioned snapshots, all having high-density in the Central Zone similar to that enjoyed by the (individual) snapshot which is random with the stationary distribution. Then, the Central Zone is partitioned in square cells of suitable size \(\Theta (R)\) and the flooding process on the agents is viewed as a propagation of information from cells to their adjacent ones (see Lemmas 7, 8, and 9). Thanks to the above high-density property, we prove this propagation takes \(O(L/R)\) time to “infect” all the Central Zone. We introduced a similar technique in [9] for the Random-Walk model. Here, we carefully adapt it for the particular shape of the Central Zone and for the different mobility (i.e. the MRWP model).

The analysis of the flooding over the Suburb is much harder. Indeed, in addition to the above key-issue, in the sparse and highly disconnected Suburb, we cannot exploit any high-density property of the snapshots. Moreover, it is possible to prove that, with non negligible probability, there are agents that will not visit the Central Zone for a (too) long time. We resort to the properties of the stationary destination probability distribution (combined with those of the spatial distribution) to prove that a sufficiently large number of informed agents, coming from the Central Zone, well spread over the Suburb. So, an agent traveling for a “long” time in the Suburb will meet one of such informed agents within the time claimed by our bound. More precisely, the proof is organized in the following key-steps. First, we show that an agent, traveling in the Suburb for a long time, w.h.p. makes “few” waypoints (i.e. he makes “few” random choices of his destinations). This implies that he will make “few” turns and, so, he will follow at least one “sufficiently-long” segment towards the Central Zone (this behavior is proved in Lemmas 12 and 14). Then, we prove that, along this sufficiently-long segment, the Suburb agent will meet one of the agents coming from the Central Zone. A special care is required in order to show that this agent meeting arises even when conditioned to the following two events: the Central-Zone agent has been informed sometime before the meeting and this meeting happens within time bounded by Theorem 3 (this fact is proved in Lemma 17).

4.1 Preliminaries

In the sequel, we assume that \(R \leqslant \sqrt{2} \, L\). Observe that if \(R >\sqrt{2} \, L\), then the bound on the flooding time is trivial. We partition the square into \(m \times m\) square cells of side length \(\ell \) with

$$\begin{aligned} \frac{R}{1 + \sqrt{5}} \leqslant \ell \leqslant \frac{R}{\sqrt{5}} \end{aligned}$$
(6)

Notice that \(\ell \) is chosen in order to guarantee that an agent inside a cell \(C\) can transmit to any agent lying in any of the four adjacent cells of \(C\). The core of a cell \(C\) is the central subsquare of \(C\) with side length \(\ell /3\). We assume the agent transmission radius and the agent speed satisfy the following bounds

$$\begin{aligned} R&\ge 200 L \sqrt{\frac{\log n}{n}}\end{aligned}$$
(7)
$$\begin{aligned} \text{ v}&\le \frac{R}{ 3(1 + \sqrt{5}) } \end{aligned}$$
(8)

Observe that the above condition on \(\text{ v}\) guarantees that an agent, lying in the core of a cell \(C\) at time \(t\), will remain in \(C\) at time \(t+1\) as well.

From Eq. 2, the probability that an agent lies in a cell \(C\) is given by

$$\begin{aligned} \int \limits _C f(x,y) dx dy \end{aligned}$$

We can now define the Central Zone and the Suburb.

Definition 4

(Central Zone and Suburb) The Central Zone is the subset \({\textsc {CZ}}\) of those cells \(C\) such that

$$\begin{aligned} \int \limits _C f(x,y) dx dy \ \geqslant \ \frac{3}{8}\, \frac{\log n}{n} \end{aligned}$$

The complement set of the Central Zone is called Suburb.

From the above definition, it turns out that a cell of the Central Zone has an \(\Omega (\log n)\) expected number of agents according to the stationary spatial distribution.

Note In order to make the technical arguments more readable, our asymptotic analysis definitely does not optimize the constants in the upper bound and in the assumption on \(R\). A more refined analysis in order to get better involved constants is possible but it is out of the goal of this work.

4.2 Flooding in the central zone

From the definition of the density function \(f(x,y)\) (Eq. 2), for any cell \(C\) having its South-West (SW) corner in position \((x_0,y_0)\), it holds that

$$\begin{aligned} \int \limits _C f(x,y) dx dy = \frac{3\ell ^2}{L^4} \Bigg ( \frac{\ell }{3} (3L - 2\ell ) + x_0(L-\ell - x_0) +\, y_0(L -\ell -y_0) \Bigg ) \end{aligned}$$
(9)

Moreover, from Ineq. 6, we get

$$\begin{aligned} \int \limits _C f(x,y) dx dy \geqslant \ \frac{\ell ^3 (3L -2 \ell )}{L^4}\geqslant \left(\frac{R}{(1+\sqrt{5}) L}\right)^3 \end{aligned}$$
(10)

From Eq. 9, it is easy to verify that the constant 3/8 in Definition 4 guarantees the following

Lemma 5

The number of rows (and columns as well) of cells that belong to the Central Zone is at least \(m / \sqrt{2}\).

We say that the density condition holds at time \(t\) if, for every cell \(C\) of the Central Zone, the number of agents in the core of \(C\) at time \(t\) is at least \(\gamma \log n\), for a suitable positive constant \(\gamma \). Let \(\mathcal D \) be the following event: the density condition holds for every time step \(t = 0, 1,\ldots , n\).

The proof of the following lemma easily derives from the definition of the Central Zone by a simple union-bound argument.

Lemma 6

(Density) The probability of event \(\mathcal D \) is at least \(1-1/n^4\).

In the analysis of the flooding over the Central Zone, we will tacitly assume that event \(\mathcal D \) holds. Thanks to the previous lemma, since we are conditioning w.r.t. an event that holds w.h.p., the corresponding unconditional probabilities are affected by a negligible factor only.

We say that a cell \(C\) is informed at time \(t\) if all agents visiting \(C\) at time \(t\) are informed. Consider an informed cell \(C\) of the Central Zone at time \(t\). By the density condition, its core contains at least one informed agent \(\textsc {a}\). Thanks to the Ineq.  8, agent \(\textsc {a}\) will remain inside \(C\) till the end of time step \(t+1\). Then, after the agent transmission of time \(t+1\), all agents lying in \(C\) or in its adjacent cells at time \(t+1\), will get informed. We have thus shown the following.

Lemma 7

(Stability) For any \(0 \le t \le n\), If a cell \(C\) of the Central Zone is informed at time \(t>0\), then \(C\) and all its adjacent cells (in the Central Zone) will be informed at time \(t +1\) with probability at least \(1 - 1/n^4\).

For any subset \(B\) of cells of the Central Zone, define the boundary \(\partial B\) of \(B\) as follows

$$\begin{aligned} \partial B = \{ C \;|\; C \in {\textsc {CZ}}\setminus B \;\wedge \; \exists C^{\prime }\in B: C^{\prime } {\text{ is} \text{ adjacent} \text{ to}}\;\; C\}. \end{aligned}$$

We now provide a lower bound on the expansion of any cell subset of the Central Zone. The result is similar to that in [7], however, in our case, the proof must be adapted for the particular shape of the Central Zone.

Lemma 8

(Boundary) Let \(B\) be any cell subset of the Central Zone. It holds that

$$\begin{aligned} |\partial B| \;\geqslant \; \sqrt{\min \{|B|, |{\textsc {CZ}}| - |B|\}}. \end{aligned}$$

Proof

For the sake of convenience, we say that a cell in \(B\) is a black cell and all the cells not in \(B\) are white cells. In the sequel, with the term row (column) we mean the subrow (subcolumn) of cells that belong to the Central Zone. Observe that, according to this definition, rows and columns have variable length. We say that a row is black if all the cells of the row are black. Similarly, we define a black column. Moreover, a row or a column which contains both at least one black cell and at least one white cell is said to be gray.   

\(\square \)

We observe that the following inequalities hold

$$\begin{aligned} \sqrt{\min \{|B|, |{\textsc {CZ}}| - |B|\}} \ \le \ \sqrt{\frac{ |{\textsc {CZ}}| }{2} } \ \le \ \frac{ m}{\sqrt{2}} \end{aligned}$$
(11)

Let \(b_r\) and \(b_c\) be, respectively, the number of black rows and the number of black columns. In order to prove the lemma, we distinguish four cases.

  • \(b_r = 0 \wedge b_c \geqslant 1\): In this case, from Lemma 5, the number of gray rows is at least \(m / \sqrt{2}\). This implies every gray row contains at least one cell in \(\partial B\). Then, \( |\partial B | \ge m / \sqrt{2}\) and the lemma follows from Ineq. 11.

  • \(b_r \geqslant 1 \wedge b_c = 0\): This case is symmetric to the previous one.

  • \(b_r \geqslant 1 \wedge b_c \geqslant 1\): If there exist a black row and a black column of maximal length \(m\), then all non-black rows and columns are gray. Observe that each of the \(|{\textsc {CZ}}|-|B|\) white cells uniquely determines a pair formed by one gray row and one gray column. We thus get

$$\begin{aligned} |{\textsc {CZ}}|-|B| \le (m -b_c)(m-b_r) \end{aligned}$$

Without loss of generality, we assume that \(b_r \le b_c\), then the above inequality implies that

$$\begin{aligned} |{\textsc {CZ}}|-|B| \le (m-b_r)^2 \end{aligned}$$

The thesis follows from the fact that \(|\partial B| \ge m - b_r\). Now, we consider the case where no black row or black column of maximal length do exist. Without loss of generality, we assume that no black row of maximal length exists. Since there is at least a black column, then all rows of maximal length are gray. From Lemma 5, the number of gray rows is at least \(m / \sqrt{2}\). This implies every gray row contains at least one cell in \(\partial B\). Then, \( |\partial B | \ge m / \sqrt{2}\) and the lemma follows from Ineq. 11.

  • \(b_r = 0 \wedge b_c = 0\): Let \(y_r\) and \(y_c\) be, respectively, the number of gray rows and the number of gray columns. Since there are neither black rows nor black columns, it must be the case that every black cell belongs to both a gray row and a gray column. This implies that

$$\begin{aligned} y_r\cdot y_c \;\geqslant |B|. \end{aligned}$$

Without loss of generality, assume that \(y_r \geqslant y_c\). It follows that \(y_r^2 \geqslant |B|\) and thus \(y_r \geqslant \sqrt{|B|}\). Since every gray row contains at least a cell in \(\partial B\), it holds that \(|\partial B| \geqslant \sqrt{|B|} \geqslant \sqrt{\min \{|B|, m^2 - |B|\}}\).

By exploiting Lemmas 7, 8 we get the following bound.

Lemma 9

Assume that, at time \(t=0\), at least one informed agent lies in the Central Zone. Then, with probability at least \(1 - 1/n^2\), at every time \(t\) with \(18{L}/{R} \leqslant t \leqslant n\), all the cells in the Central Zone are informed.

Proof

For any \(t \geqslant 0\), let \(\mathcal Q _t\) be the set of informed cells at time \(t\) in the Central Zone. By hypothesis \(|\mathcal Q _0| \geqslant 1\). In virtue of Lemma 8, if all cells in \(\mathcal Q _t\) and all their adjacent cells get informed at time \(t+1\), then \( |\mathcal Q _{t+1}| \geqslant |\mathcal Q _t| + \sqrt{\min \{ |\mathcal Q _t|, |{\textsc {CZ}}| - |\mathcal Q _t| \} }\).

This implies that if the above inequality does not hold then a cell \(C\in \mathcal Q _t\) exists such that \(C\) or one adjacent cell of \(C\) is not informed at time \(t+1\). It follows that

$$\begin{aligned}&\mathbf P \left( |\mathcal Q _{t+1}| < |\mathcal Q _t| + \sqrt{\min \{ |\mathcal Q _t|, |{\textsc {CZ}}| - |\mathcal Q _t| \} } \right)\nonumber \\&\quad \leqslant \mathbf P \left( \exists C\in \mathcal Q _t : \mathcal E _{C, t+1} \right) \end{aligned}$$

where \(\mathcal E _{C, t+1}\) is the event that occurs if \(C\) or one of its adjacent cells in \({\textsc {CZ}}\) is not informed at time \(t+1\). By the union bound, it holds that

$$\begin{aligned} \mathbf P \left( \exists C\in \mathcal Q _t : \mathcal E _{C, t+1} \right)&\leqslant \sum _{C} \mathbf P \left( C\in \mathcal Q _t \wedge \mathcal E _{C, t+1} \right)\\&\!=\!&\sum _{C} \!\mathbf P \left( \! \;\left|\; \mathcal E _{C, t\!+\!1} \right.\right){ C\in \mathcal Q _t } \!\mathbf P \left( \! \right){C\!\in \! \mathcal Q _t}. \end{aligned}$$

From Lemma 7, for every cell \(C\), we have \(\mathbf P \left( \mathcal E _{C, t\!+\!1}\! \;\left|\; C\in \! \mathcal Q _t\! \right.\right) \leqslant 1/{n^4}\). It follows that

$$\begin{aligned}&\mathbf P \left( |\mathcal Q _{t+1}| < |\mathcal Q _t| + \sqrt{\min \{ |\mathcal Q _t|, |{\textsc {CZ}}| - |\mathcal Q _t| \} } \right)\\&\leqslant \sum _{C} \mathbf P \left( \mathcal E _{C, t+1} \;\left|\; C\in \mathcal Q _t \right.\right) \mathbf P \left( C\in \mathcal Q _t \right) \\&\leqslant \sum _{C} \mathbf P \left( \mathcal E _{C, t+1} \;\left|\; C\in \mathcal Q _t \right.\right) \leqslant \frac{|{\textsc {CZ}}|}{n^4} \;\leqslant \; \frac{1}{n^3}. \end{aligned}$$

Thus, by the union bound, with probability at least \(1 - \frac{1}{n^2}\), it holds that

$$\begin{aligned}&\forall t = 0,1,\ldots , n \qquad \\&\quad |\mathcal Q _{t+1}| \;\geqslant \; |\mathcal Q _t|+ \sqrt{\min \{ |\mathcal Q _t|, |{\textsc {CZ}}| - |\mathcal Q _t| \} }. \end{aligned}$$

Now we use the following claim proven in Lemma 9 of [7].

\(\square \)

Claim 10

Let \(\bar{q}\) be any integer s.t. \(\bar{q} \geqslant 1\) and let \(\{q_t \;|\; t\in \mathbb N \}\) be a sequence of integers such that \(q_0 \geqslant 1\), for every \(t\geqslant 0, q_t \leqslant \bar{q}\) and \(q_{t+1} \geqslant q_t + \sqrt{\min \{q_t, \bar{q} - q_t\}}\). Then, it holds that, for every \(t \geqslant 5\sqrt{\bar{q}}, q_t = \bar{q}\).

By applying the claim with \(q_t = |\mathcal Q _t |\) and \(\bar{q} = |{\textsc {CZ}}|\), we get \(|\mathcal Q _t| = |{\textsc {CZ}}|\) for every \(t\) with \(5\sqrt{|{\textsc {CZ}}|} \leqslant t \leqslant n \). The thesis follows since \(5 \sqrt{|{\textsc {CZ}}|} \leqslant 5 L/\ell \leqslant 18 L/R\).

Notice that, thanks to Ineq. 10, if

$$\begin{aligned} R \geqslant \frac{(1 + \sqrt{5})}{2} L \left(\frac{3\log n}{ n}\right)^{1/3} \end{aligned}$$

then all cells of the square belong to the Central Zone. Hence, from Lemma 9, we get

Corollary 11

(Large \(R\)) If \(R \geqslant \frac{(1 + \sqrt{5})}{2} L \left(\frac{3\log n}{ n}\right)^{1/3}\), with probability at least \(1 - 1/n^2\), the (overall) flooding time is \(18 L / R\).

4.3 Flooding over the Suburb

We now analyze the flooding process over the Suburb. Thanks to Corollary 11, we can assume that

$$\begin{aligned} R \leqslant \frac{1 + \sqrt{5}}{2} L \left(\frac{3\log n }{ n}\right)^{1/3} \end{aligned}$$
(12)

otherwise the Suburb would be empty. In this region, the agent density is not sufficiently high to adopt the same cell-partition technique. The new approach exploits the structure of the paths performed by an agent that walks for a long time in the Suburb and on the probability that he meets agents coming from the Central Zone.

We say an agent performs a choice when he has arrived to his destination and he randomly chooses the next one. Let \(\textsc {a}\) be an agent, for any time \(t\), the random variable \(H_{t,\tau }\) counts the number of choices performed by \(\textsc {a}\) during the time interval \([ t, t + \tau ]\). The next lemma shows that this number cannot be “too” large.

Lemma 12

Let \(t \ge 0\) and let \(\tau \) be such that

$$\begin{aligned} \frac{L}{n \text{ v}} \ \leqslant \ \tau \leqslant \frac{L}{4 \text{ v}} \end{aligned}$$

With probability at least \(1- 1/n^4\), it holds that

$$\begin{aligned} H_{t,\tau } \ \leqslant \ \frac{4 \log n}{ \log \left( \frac{L}{\text{ v}\tau }\right)} \end{aligned}$$

Proof

For the sake of simplicity, we will use the following probability notations. For an event \(\mathcal E \) and a r.v. \(X\), the notation \( \mathbf P \left( \mathcal E \;\left|\; X \right.\right) \ \leqslant p\) means that, for every possible value \(x\) of \(X\), it holds \( \mathbf P \left( \mathcal E \;\left|\; X=x \right.\right) \leqslant p \).

Consider the choices performed by \(\textsc {a}\) after time \(t\). For any \(i = 1, 2, \ldots \), define \(X_i\) the distance travelled by agent \(\textsc {a}\) between the \(i\)-th choice and the \(i+1\)-th choice. We then consider the binary r.v. defined as follows

$$\begin{aligned} Y_i = 1 \ \text{ if} \ X_i \ \leqslant \ \text{ v}\tau \ \text{ and}\quad&0 \ \text{ otherwise} \end{aligned}$$

Observe that if \(X_i \leqslant \text{ v}\tau \) then the \((i+1)\)-choice point lies in the square centered at the \(i\)-th choice point, with diagonals parallel to the axes, and that has side length \(\sqrt{2} \, \text{ v}\tau \). So, it holds that

$$\begin{aligned} \mathbf P \left( Y_i =1 \;\left|\; X_1, \ldots , X_{i-1} \right.\right) \ \leqslant \ p \ \text{ where} \ p \ = \ \frac{ (\sqrt{2} \, \text{ v}\tau )^2 }{L^2} \end{aligned}$$

Notice that, since \(\tau \leqslant L / (4 \text{ v})\), then \(p < 1\). We now need the following standard probability bound (See [2]). \(\square \)

Claim 13

Let \(X_1, \dots , X_n\) be a sequence of random variables with values in an arbitrary domain, and let \(Y_1, \dots , Y_n\) be a sequence of binary random variables, with the property that \(Y_i = Y_i(X_1, \dots , X_i)\). If

$$\begin{aligned} \mathbf P \left( Y_i = 1 \;|\; X_1, \dots , X_{i-1} \right) \leqslant p \end{aligned}$$

then

$$\begin{aligned} \mathbf P \left( \sum Y_i \geqslant k \right) \leqslant \mathbf P \left( B(n,p) \geqslant k \right)\end{aligned}$$

where \(B(n,p)\) denotes the binomially distributed random variable with parameters \(n\) and \(p\).

For any \( h = 1,2, \ldots \), from the above Claim it holds that

$$\begin{aligned} \mathbf P \left( \sum _{i =1}^h Y_i \ = \ h \right) \ \leqslant \ \mathbf P \left( B\left(h, p \right) = h \right) \end{aligned}$$

So, it clearly holds that

$$\begin{aligned} \mathbf P \left( \sum _{i =1}^h Y_i \ = \ h \right) \ \leqslant \ p^h \end{aligned}$$
(13)

We observe that

$$\begin{aligned} \sum _{i = 1}^{H_{t,\tau } -1} X_i \ \leqslant \text{ v}\tau \end{aligned}$$

Hence, event “ \(H_{t,\tau } \geqslant h+1\) ” implies event “\( \sum _{i = 1}^{h} X_i \leqslant \text{ v}\tau \)” and so

$$\begin{aligned} \mathbf P \left( H_{t,\tau } \geqslant h+1 \right) \ \leqslant \ \mathbf P \left( \sum _{i = 1}^{h} X_i \leqslant \text{ v}\tau \right) \end{aligned}$$

Moreover, the following implication holds

$$\begin{aligned} \sum _{i = 1}^{h} X_i \ \leqslant \text{ v}\tau \ \Longrightarrow \ \sum _{i = 1}^{h} Y_i \ = \ h \end{aligned}$$

Hence, from Ineq. 13, we get

$$\begin{aligned} \mathbf P \left( H_{t,\tau } \geqslant h+1 \right)&\leqslant \mathbf P \left( \sum _{i = 1}^{h} X_i \leqslant \text{ v}\tau \right) \nonumber \\&\leqslant \mathbf P \left( \sum _{i =1}^h Y_i = h \right) \leqslant \ p^h \end{aligned}$$
(14)

Observe that for

$$\begin{aligned} h = \left\lceil \frac{4 \log n}{\log (1/p) }\right\rceil \end{aligned}$$

it holds that

$$\begin{aligned} \mathbf P \left( H_{t,\tau } \geqslant h+1 \right) \ \leqslant \ \frac{1}{n^4} \end{aligned}$$

We also have that

$$\begin{aligned} \frac{4 \log n}{\log (1/p)} \ = \ \frac{4 \log n}{ \log \left(\frac{L^2}{(\sqrt{2} \text{ v}\tau )^2} \right)} \ = \ \frac{2 \log n}{ \log \left(\frac{L }{\sqrt{2} \text{ v}\tau } \right)} \end{aligned}$$

Hence, with probability at least \(1 - 1/n^4\), it holds that

$$\begin{aligned}&H_{t,\tau } \leqslant \left\lceil \frac{2 \log n}{ \log \left(\frac{L }{\sqrt{2} \text{ v}\tau } \right)}\right\rceil \ \leqslant \ 1 + \frac{2 \log n}{ \log \left(\frac{L }{\sqrt{2} \text{ v}\tau } \right)} \\&\quad = \left( 1 + \frac{2 \log n}{ \log \left(\frac{L }{\sqrt{2} \text{ v}\tau } \right)} \right) \frac{4 \log n }{\log \left(\frac{L }{ \text{ v}\tau } \right) } \frac{ \log \left(\frac{L }{ \text{ v}\tau } \right) }{4 \log n } \\&\quad = \frac{4 \log n }{\log \left(\frac{L }{ \text{ v}\tau } \right) } \left( \frac{ \log \left(\frac{L }{ \text{ v}\tau } \right) }{4 \log n } + \frac{ 2 \log \left(\frac{L }{ \text{ v}\tau } \right)}{ 4 \log \left( \frac{L }{ \sqrt{2} \text{ v}\tau } \right) } \right) \\&(\text{ since} \ \frac{L}{\text{ v}\tau } \leqslant n) \leqslant \frac{4 \log n }{\log \left(\frac{L }{ \text{ v}\tau } \right) } \left( \frac{1 }{4 } + \frac{2}{4} \frac{ \log \left(\frac{L }{ \text{ v}\tau } \right)}{ \log \left( \frac{L }{ \text{ v}\tau } \right) - \frac{1}{2}} \right) \end{aligned}$$

Notice that

$$\begin{aligned} \frac{ \log \left(\frac{L }{ \text{ v}\tau } \right)}{ \log \left( \frac{L }{ \text{ v}\tau } \right) - \frac{1}{2}} \ = \ \frac{1}{1 - \frac{1}{2 \log \left(\frac{L }{ \text{ v}\tau } \right)}} \ \leqslant \ \frac{4}{3} \ \ (\text{ since} \ \frac{L}{\text{ v}\tau } \geqslant 4) \end{aligned}$$

Finally, we get

$$\begin{aligned} H_{t,\tau } \ \leqslant \ \frac{4 \log n }{\log \left(\frac{L }{ \text{ v}\tau } \right) } \left( \frac{1 }{4 } + \frac{2}{4} \cdot \frac{4}{3} \right) \ \leqslant \ \frac{4 \log n }{\log \left(\frac{L }{ \text{ v}\tau } \right) } \end{aligned}$$

The next lemma allows us to get high probability for the existence of a “good” segment, i.e., a sufficiently-long segment traveled by any agent in the Suburb toward the Central Zone.

In the sequel, we assume that agent \(\textsc {a}\) lies in the South-West subsquare of size \(L/2\), i.e., the subsquare \([0,L/2] \times [0,L/2]\). The position of \(\textsc {a}\) at time \(t\) is denoted as \((x_0,y_0)\). The analysis of the other three subsquares is symmetric.

Lemma 14

Let \(t \ge 0\) and let \(\tau \) be such that

$$\begin{aligned} \frac{ \max \{L/n, 4x_0,4 y_0 \}}{\text{ v}} \ \leqslant \ \tau \ \leqslant \ \frac{L}{4 \text{ v}} \end{aligned}$$

Then, during the time interval \([t,t+ \tau ]\), with probability at least \(1 -1/n^4\), agent \(\textsc {a}\) travels over a (horizontal or vertical) segment directed to the Central Zone and that has length at least

$$\begin{aligned} \frac{\text{ v}\, \tau \log \left( \frac{L}{\text{ v}\tau }\right)}{40 \log n } \end{aligned}$$

Proof

Let \(k= H_{t,\tau }\) be the number of \(\textsc {a}\)’s choices in the interval \([t,t+\tau ]\). For any \(i = 1, \ldots , k\), we define \((x_i,y_i)\) as the \(i\)-th choice position of agent \(\textsc {a}\). We denote the \(\textsc {a}\)’s position at time \(t+ \tau \) as \((x_{k+1},y_{k+1})\). For any \(j = 1, \ldots , k+1\), define \(h_j = x_j - x_{j-1}\) and \(v_j = y_j - y_{j-1}\).

Observe that when \(h_j\) or \(v_j\) are positive then the travelled segment is directed towards the Central Zone. It holds that

$$\begin{aligned} \sum _{j=1}^{k+1} | h_j| + \sum _{j=1}^{k+1} | v_j| = \text{ v}\, \tau \end{aligned}$$

Without loss of generality, we assume that the first sum is not smaller than the second one. So,

$$\begin{aligned} \sum _{j=1}^{k+1} | h_j| \geqslant \frac{\text{ v}\tau }{2} \end{aligned}$$

Now, define the following index subsets:

$$\begin{aligned} J^+ = \{ j \ | \ h_j > 0 \} \ \text{ and} J^- = \{ j \ | \ h_j \leqslant 0 \} \ \end{aligned}$$

Hence, we get

$$\begin{aligned} \sum _{j \in J^+} h_j \geqslant \frac{\text{ v}\tau }{2} + \sum _{j \in J^-} h_j \end{aligned}$$
(15)

Observe that

$$\begin{aligned} \sum _{j \in J^-} h_j + \sum _{j \in J^+} h_j = \sum _{j=1}^{k+1} h_j = x_{k+1} - x_0 \geqslant - x_0 \, \geqslant \, - \frac{\text{ v}\, \tau }{4} \end{aligned}$$

This implies

$$\begin{aligned} \sum _{j \in J^-} h_j \geqslant - \frac{\text{ v}\, \tau }{4} - \sum _{j \in J^+} h_j \end{aligned}$$

By combining the above equation with Eq. 15, we obtain

$$\begin{aligned} \sum _{j \in J^+} h_j \geqslant \frac{\text{ v}\tau }{2} - \frac{\text{ v}\tau }{4} - \sum _{j \in J^+} h_j \end{aligned}$$

So,

$$\begin{aligned} \sum _{j \in J^+} h_j \geqslant \frac{\text{ v}\tau }{8} \end{aligned}$$

Hence, we can say that an index \(\hat{j}\) exists such that

$$\begin{aligned} h_{\hat{j} } \geqslant \frac{ \text{ v}\tau }{ 8( k + 1) } \end{aligned}$$

From Lemma 12 and the fact that \(\tau \geqslant L /(n \text{ v})\), with probability at least \(1 - 1/n^4\) it holds that

$$\begin{aligned} h_{\hat{j} } \ \geqslant \frac{\text{ v}\tau }{ 8\left( \frac{4 \log n }{ \log \left( \frac{L}{\text{ v}\tau }\right) } +1 \right) } \geqslant \ \frac{\text{ v}\, \tau \log \left( \frac{L}{\text{ v}\tau }\right)}{40 \log n } \end{aligned}$$

\(\square \)

Observation 15

The previous two lemmas show the existence of a “good” segment by exploiting the fact that the Suburb agent makes few choices (and thus few “turns”). This is due to the adopted Manhattan paths but we believe a similar argument may also hold for a generalized variant of such paths where the agent can make a constant number of turns (rather than only one) between two consecutive waypoints. Indeed, adding a constant number of turns along one path would decrease the length of the good segments and/or its probability by a constant fraction.

For the sake of convenience, let \(S = \frac{3 L^3 \log n}{ 2\ell ^2 n}\).

Lemma 16

The diameter of south-west corner of the Suburb is \(\Theta (S)\). In particular, for every point \((x_0,y_0)\) in the south-west corner of the Suburb, it holds that both \(x_0\) and \(y_0\) are not larger than \(S\).

Proof

By replacing the coordinates \(x_0=S/2\) and \(y_0 = S/2\) in Eq. 9, we easily get that the corresponding cell probability is smaller than \((3/8) (\log n)/(n)\) and, thus, the cell is inside the Suburb. Now, let us prove that the Suburb diameter is at most \(S\). Let \((\overline{x},\overline{y})\) the SW corner of the cell containing \((x_0,y_0)\). We will prove the lemma’s bound for \(x_0\); the bound for \(y_0\) can be obtained by a symmetric argument. Since this cell does not belong to the Central Zone, from Eq. 9, we get

$$\begin{aligned}&\frac{3\ell ^2}{L^4} \left( \frac{\ell }{3} (3L - 2\ell ) + \overline{x} (L-\ell - \overline{x}) + \overline{y} (L -\ell -\overline{y}) \right) \ \leqslant \ \frac{3}{8} \frac{\log n}{n} \end{aligned}$$

From the above inequality, we get

$$\begin{aligned} \overline{x} (L - \ell - \overline{x}) \ \leqslant \ \frac{ L^4 \log n}{8 \ell ^2 n} \end{aligned}$$
(16)

Notice that, by definition of \(\ell \) and Ineq. 12, we get

$$\begin{aligned} \ell \ \leqslant \ \frac{R}{\sqrt{5}} \ \leqslant \ \frac{3}{40} \, L \end{aligned}$$
(17)

From Lemma 5, we obtain

$$\begin{aligned} \overline{x} \ \leqslant \ \frac{m-\frac{m}{\sqrt{2}}}{2 } \, \ell = \frac{2 - \sqrt{2}}{4} \, L \end{aligned}$$

From this inequality and Ineq. 17, we obtain

$$\begin{aligned} \ L - \ell - \overline{x} \ \geqslant \ L - \frac{3}{40} \, L - \frac{2 - \sqrt{2}}{4} \, L \ \geqslant \ \frac{3}{4} L \end{aligned}$$

From the above inequality and Ineq. 16, it holds that

$$\begin{aligned} \frac{3}{4} \, \overline{x} L \leqslant \overline{x} (L - \ell - \overline{x}) \ \leqslant \ \frac{ L^4 \log n}{8 \ell ^2 n} \end{aligned}$$

and

$$\begin{aligned} \overline{x} \ \leqslant \ \frac{ L^3 \log n}{6 \ell ^2 n} \end{aligned}$$
(18)

Since \(x_0 \leqslant \overline{x} + \ell \), we now bound \(\ell \). From Ineq. 12, we obtain

$$\begin{aligned} \ell \leqslant \frac{R}{ \sqrt{5}} \ \leqslant \frac{1 + \sqrt{5}}{2\sqrt{5}} L \left(\frac{3\log n}{n}\right)^{1/3} \end{aligned}$$

Then

$$\begin{aligned} \ell ^3 \ \leqslant \ \left(\frac{1 + \sqrt{5}}{2\sqrt{5}} \right)^3 3 L^3 \frac{\log n}{n} \end{aligned}$$

We thus get

$$\begin{aligned} \ell \ \leqslant \ \left( \frac{1 + \sqrt{5}}{2\sqrt{5}} \right)^3 \frac{ 3 L^3 \log n}{ \ell ^2 n} \ \leqslant \ 1.2 \, \frac{ L^3 \log n}{ \ell ^2 n} \end{aligned}$$

From Ineq. 18 and the last one, we finally obtain

$$\begin{aligned} x_0 \ \leqslant \ \overline{x} +&\ell \ \leqslant \ \frac{1}{6} \frac{ L^3 \log n}{ \ell ^2 n} + 1.2 \, \frac{ L^3 \log n}{ \ell ^2 n}\\&\quad \ \leqslant \ \frac{ 3 L^3 \log n}{2 \ell ^2 n}=S \end{aligned}$$

\(\square \)

Meeting agents coming from the Central Zone Two agents are said to meet each other at time \(t \geqslant 0\) if, at that time, their relative distance is not larger than \((3/4) R\). Observe that, due to Ineq. 8, if one informed agent meets another agent at some time, then within the next time unit, the latter will get informed.

There may be some non-informed agents that travel over the Suburb for a long period. For those agents, we evaluate the probability to meet agents coming from the Central Zone within relatively-small time. A symmetric argument will be applied to manage the case where the source will be in the Suburb for a long time.

We say that a point belongs to the Extended Suburb if the Manhattan distance between the point and the Suburb is not larger than \(2S\). Clearly, all points in the Suburb belong to the Extended Suburb.

The first property of the next lemma is used when the source lies in Central Zone. The second property is instead used when the source lies in the Suburb for a long time.

Lemma 17

Let \(\textsc {a}\) be an agent lying in the Extended Suburb at any time \(t \geqslant S / \text{ v}\). For sufficiently large \(n\), with probability at least \(1-1/n^2\), an agent \(b \) exists that has the following properties:

  1. 1.

    \(b \) was in the Central Zone at time \(t- S / \text{ v}\) and \(b \) will meet agent \(\textsc {a}\) within time \(T + \tau \), where \(\tau = 590 (S/\text{ v})\).

  2. 2.

    Agent \(b \), after meeting \(\textsc {a}\), will be in the Central ZoneFootnote 1 within time \(t + \tau + 3 (S/\text{ v})\).

Proof

Let \((x_0,y_0)\) be the position of \(\textsc {a}\) at time \(t\). Observe that \(L/n\leqslant S\) and, from Lemma 16 and the definition of Extended Suburb, both \(x_0\) and \(y_0\) are not larger than \(3S\). Thus it holds that

$$\begin{aligned} \frac{ \max \{L/n, 4x_0,4 y_0 \}}{\text{ v}} \ \leqslant \frac{12S}{\text{ v}} \ < \ \tau \end{aligned}$$
(19)

From Ineq. 6 and 7 we obtain

$$\begin{aligned} \frac{L}{\ell }\leqslant \frac{1+\sqrt{5}}{200}\sqrt{\frac{n}{\log n}} \end{aligned}$$
(20)

We thus get

$$\begin{aligned} \tau = 590 \frac{ S}{\text{ v}}&= \frac{L}{4\text{ v}}\left(3540\frac{L^2\log n}{\ell ^2 n}\right)\leqslant \frac{L}{4\text{ v}}\left(\frac{3540\cdot (1\!+\!\sqrt{5})^2}{200^2}\right) < \frac{L}{4\text{ v}} \end{aligned}$$

Due to the above inequality and Ineq. 19, we can apply Lemma 14 with \(\tau \) and \(t\) specified in the thesis. Hence, with probability at least \(1 - 1/n^4\), agent \(\textsc {a}\) will travel a good segment (i.e. toward the Central Zone) of length

$$\begin{aligned} d \ = \ \frac{ \text{ v}\, \tau \log \left( \frac{L}{\text{ v}\tau }\right)}{40 \log n } \ \text{ during} \text{ time} \text{ interval} \ [ T \ , \ T + \tau ] \end{aligned}$$

Observe that

$$\begin{aligned} d= \frac{590S\log \left(\frac{L}{590S}\right)}{40\log n}\ge 22\frac{L^3}{\ell ^2n}\log \left(\frac{\ell ^2n}{885L^2\log n}\right) \end{aligned}$$
(21)

Without loss of generality, we assume that the good segment is horizontal. Let \(t_\textsc {a}\) be the time in which \(\textsc {a}\) starts running the good segment and let \((x_\textsc {a},y_\textsc {a})\) be his position at that time. Consider the rectangle \(I\) such that: its SW vertex is point \((x_\textsc {a}+ d + D, y_\textsc {a})\) where \(D = d/4 + \text{ v}(t_\textsc {a}- t + S/ \text{ v})\), its horizontal size is \(d/2\), and its vertical size is \(\ell \). The next claim is the key-ingredient of the proof. \(\square \)

Claim 18

For sufficiently large \(n\), with probability at least \(1-1/n^2\), an agent \(b \) exists satisfying the following properties:

  1. 1.

    at time \(t - S/\text{ v}\) he is in some position \((x_b ,y_b ) \in I\) and has destination \((x, y_b )\), for some \( 0 \leqslant x \leqslant x_\textsc {a}+ d/2 \), and

  2. 2.

    his destination after next is in the Central Zone.

Proof of the Claim

Let \(\bar{P}(x_b ,y_b )\) be the probability that an agent \(b \), being in \((x_b ,y_b )\), has destination \((x, y_b )\) for some \( 0 \leqslant x \leqslant x_\textsc {a}+ d/2 \). From Eq. 5, it holds that

$$\begin{aligned} \bar{P}(x_b ,y_b )&= \frac{x_\textsc {a}+ d/2}{x_b } \phi ^\mathrm{{W}}_{(x_b ,y_b )}&= \ \frac{(x_\textsc {a}+ d/2) (L-x_b ) }{ 4L(x_b +y_b ) - 4(x_b ^2 + y_b ^2)} \end{aligned}$$

Let \(P_b \) be the probability that agent \(b \) satisfies Property 1. Then,

$$\begin{aligned} P_b = \int \limits _I \bar{P}(x,y) f(x,y) \text{ d}x\text{ d}y\end{aligned}$$

where \(f(x,y)\) is the probability density function of the spatial distribution in Eq. 2. It thus follows that

$$\begin{aligned} P_b&\!=\!&\int \limits _I \frac{(x_\textsc {a}\!+ \!d/2) (L\!-\!x) }{ 4L(x\!+\!y) \!-\! 4(x^2 \!+\! y^2)}\nonumber \\&\times \left( \frac{3}{L^3}(x\!+\!y)\!-\!\frac{3}{L^4}({x}^2\!+\!{y}^2) \right) \text{ d}x\text{ d}y\nonumber \\&= \frac{3(x_\textsc {a}+ d/2)}{4 L^4 } \int \limits _I (L-x) \text{ d}x\text{ d}y\nonumber \\&= \frac{3d \ell (x_\textsc {a}+ d/2) (L - x_\textsc {a}- (5/4)d - D)}{8 L^4 }\nonumber \\&\ge \frac{3d^2 \ell }{16 L^4 } (L - x_\textsc {a}- (5/4)d - D) \end{aligned}$$
(22)

Observe that

$$\begin{aligned}\begin{array}{lll}&x_\textsc {a}+ (5/4)d + D = x_\textsc {a}+ (3/2) d + \text{ v}(t_\textsc {a}- t + S/\text{ v})\\&\leqslant x_\textsc {a}+ (3/2) d +( \text{ v}\tau - d + S) \quad \text{ Since} x_\textsc {a}\leqslant x_0 + \text{ v}\tau - d \\&\leqslant x_0+ 2\text{ v}\tau +S \quad \text{ By} \text{ Lemma} \text{16} \\&\leqslant 3S+ 2\text{ v}\tau \\&\leqslant \frac{1775 L^3\log n}{\ell ^2 n} \quad \mathrm{By Ineq.} (20) \\&\leqslant \frac{L}{2} \end{array} \end{aligned}$$

From the above inequality, Ineq. 22 and 21, we obtain

$$\begin{aligned} P_b \ge \frac{3d^2 \ell }{32 L^3 }\ge 45\frac{L^3}{\ell ^3n^2}\log ^2\left(\frac{\ell ^2n}{885L^2\log n}\right) \end{aligned}$$
(23)

It is easy to verify that the right-hand side of the above inequality is a decreasing function of \(\ell \). So, in order to get a lower bound for that value, we evaluate it in an upper bound of \(\ell \).

Now from Ineqs. 6 and 12 we have that

$$\begin{aligned} \ell \leqslant \frac{R}{\sqrt{5} }&\leqslant \frac{(1 + \sqrt{5})}{2\sqrt{5}} L \left(\frac{3\log n}{ n}\right)^{1/3}\\&\leqslant \frac{16L}{15}\left(\frac{\log n}{n}\right)^{1/3} = \ell _{ub} \end{aligned}$$

Thus we have

$$\begin{aligned} P_b&\ge 45\frac{L^3}{\ell _{ub}^3n^2}\log ^2\left(\frac{\ell _{ub}^2n}{885L^2\log n}\right)\\&\ge \frac{37}{n\log n}\log ^2\left(\frac{13}{10^4}\left(\frac{n}{\log n}\right)^{1/3}\right)\\&\ge \frac{37}{n\log n}\log ^2\!\left(n^{1/4}\right)\!\ge 2.3\frac{\log n}{n} \text{ For} \text{ sufficiently} \text{ large}\, n. \end{aligned}$$

Since every destination is selected uniformly at random over the square and the Central Zone’s area is (by Ineq.  7) at least \((11/12) L^2\), the probability that agent \(b \) satisfies both properties of the claim is

$$\begin{aligned} P \geqslant \frac{11}{12} P_b \geqslant 2.1 \frac{\log n}{n} \end{aligned}$$

Since there are \(n-1\) independent agents, the probability that no agent satisfies both properties is

$$\begin{aligned} (1- P)^{n-1}\leqslant \left(1-2.1\frac{\log n}{n}\right)^{n-1} \leqslant e^{-2.1\log n \frac{n-1}{n}}\leqslant \frac{1}{n^{2}} \end{aligned}$$

where the last inequality holds for sufficiently large \(n\). \(\square \)

We now show that the two properties of Claim 18 imply the two properties of the lemma. Observe that Ineq.  7 implies that \(I\) fully belongs to the Central Zone. Let \(b \) be an agent satisfying the two properties of Claim 18 and let

$$\begin{aligned} \bar{t}=\frac{x_b -x_\textsc {a}+\text{ v}t_\textsc {a}-S -\text{ v}t}{2\text{ v}} \end{aligned}$$

It holds that

$$\begin{aligned} \bar{t}&= \frac{x_b -x_\textsc {a}+\text{ v}t_\textsc {a}-S -\text{ v}t}{2\text{ v}} \leqslant \frac{(5/4) d + 2 \text{ v}t_\textsc {a}- 2 \text{ v}}{2\text{ v}}\\&= \frac{5 d}{8 \text{ v}} + t_\textsc {a}-t = \tau - \frac{3d}{8 \text{ v}} < \tau \end{aligned}$$

Consider the horizontal coordinates \(\bar{x}_\textsc {a}\) and \(\bar{x}_b \), at time \(t + \bar{t}\), of agents \(\textsc {a}\) and \(b \), respectively. It holds that

$$\begin{aligned} \bar{x}_\textsc {a}&\!=\!&x_\textsc {a}\!+\!\text{ v}(\bar{t} \!+\! t\! -\! t_\textsc {a}) \!=\! \frac{x_\textsc {a}\!+\! x_b \!-\! \text{ v}(t_\textsc {a}\!- \!t \!+\! S / \text{ v})}{2}\\ \bar{x}_b&\!=\!&x_b \!-\!\text{ v}\left(t\!+\! \bar{t} \!-\! \left(t\!-\!\frac{S}{\text{ v}}\right)\right) \!=\! \frac{x_\textsc {a}\!+\! x_b \!-\! \text{ v}(t_\textsc {a}\!-\! t \!+\! S /\text{ v})}{2} \end{aligned}$$

Moreover observe that, by definition of rectangle \(I\), it holds that

$$\begin{aligned} x_\textsc {a}+ \frac{5}{8} d \leqslant \ \bar{x}_\textsc {a}\ = \ \bar{x}_b \ \leqslant x_\textsc {a}+ \frac{7}{8} d \end{aligned}$$

Hence agents \(\textsc {a}\) and \(b \) at time \(t+\bar{t}\) are at points \((\bar{x}_\textsc {a},y_\textsc {a})\) and \((\bar{x}_\textsc {a}, y_b )\), respectively. Their distance at that time is

$$\begin{aligned} |y_\textsc {a}-y_b | \ \leqslant \ \ell \ \leqslant \ \frac{R}{\sqrt{5}} \leqslant \frac{3}{4}R \end{aligned}$$

Hence, agents \(\textsc {a}\) and \(b \) will meet at time \(t+\bar{t}\) and this shows the first property of the Lemma.

As for the second property, we observe that agent \(b \) will reach position \((x_\textsc {a}+ \frac{d}{2}, y_b )\) within time \(t+\tau \). Two cases may arise.

  1. a)

    Position \((x_\textsc {a}+ \frac{d}{2}, y_b )\) lies in the Central Zone.

    • From the above observation, we immediately have that \(b \) will be in the Central Zone after meeting \(\textsc {a}\) within time \(t+\tau \).

  2. b)

    Position \((x_\textsc {a}+ \frac{d}{2}, y_b )\) lies in the Suburb.

    • Let \(\hat{x}_b \) be the horizontal coordinate of the \(b \)’s destination determined in the first property of Claim 18; it holds that \( x_\textsc {a}+ \frac{d}{2} - \hat{x}_b \leqslant S - \hat{x}_b \leqslant S\) Hence, \(b \) reaches this destination within time \(t + \tau + S/\text{ v}\). The second property of Claim 18 ensures that the next destination is in the Central Zone. It is easy to see that the maximal traveled distance to enter into the Central Zone is \(2S\). It follows that, within time \(t + \tau + 3 (S/\text{ v})\), agent \(b \) will be in the Central Zone.\(\square \)

We conclude this section by observing that similar arguments to those in Observation 15 hold for the above lemma as well: we thus believe that it is possible to extend the lemma to more general Manhattan paths where the number of turns is bounded by an arbitrary constant.

4.4 Combining the bounds for the central zone and the Suburb

Let time step \(0\) be the starting time of flooding: we assume the MANET is already in its stationary phase.

  • We first consider the case where the source lies in the Central Zone when flooding starts. From Lemma 9, with probability at least \(1- 1/n^2\), at time \(T_c = O(L/R)\), all the cells of the Central Zone are informed. Observe that if

    $$\begin{aligned} R \ge \frac{1 + \sqrt{5}}{2} L \left(\frac{3\log n }{ n}\right)^{1/3} \end{aligned}$$

    Corollary 11 implies that the Suburb is empty and, hence, the flooding is completed. In the rest of the proof, we can thus assume that Ineq. 12 holds and focus only on those agents that at time \(T_c\) are not in Central Zone. Consider any agent \(\textsc {a}\) among the latter agents. By definition of Extended Suburb, if an agent is not in the Central Zone at time \(T_c\), then he will necessarily be in the Extended Suburb at time \(T_c + S/\text{ v}\). So, by applying Lemma 17 to agent \(\textsc {a}\) with \(t = T_c + S/\text{ v}\), we obtain that, with probability at least \(1 - 1/n^2\), an agent \(b \) exists that was in the Central Zone at time \(T_c\) and he will meet \(\textsc {a}\) within time \(T_c + O(S/\text{ v})\). Within the latter time, agent \(\textsc {a}\) will be thus informed with probability \(1 - 1/n^2\). By using the union bound, we get that all such agents will be informed with high probability within time

    $$\begin{aligned} T_c + O\left( \frac{L^3\log n}{\text{ v}R^2 n} \right) \ \ \ \text{ since} \ \ S = \Theta \left(\frac{L^3 \log n}{ R^2 n}\right) \end{aligned}$$

  • We now consider the case where the source agent lies in Suburb when flooding starts. By applying Property 2 of Lemma 17 to the source agent with \(t = S /\text{ v}\), we get that, with probability at least \(1-1/n^2\), there is at least one agent \(b \) that meets the source agent and, after that, will be in the Central Zone within time \(O(S/\text{ v})\). The rest of the proof works as in the first case.

This concludes the proof of Theorem 3.

5 A lower bound for flooding time

We observe that our upper bound holds for arbitrarily small speed \(\text{ v}\) while if \(\text{ v}=0\), flooding never terminates whenever the Suburb is not empty. More generally, we can prove the following lower bound.

Theorem 19

If \(R = O(L/n^{1/3})\) then, with constant positive probability, the flooding time is \(\Omega (L/(\text{ v}n^{1/3}))\).

Proof

Let \(d\) be such that \(d = \Theta (L/n^{1/3})\) and \(d \geqslant R\) (since \(R = O(L/n^{1/3})\), such a \(d\) does exist). Let \(F\) and \(E\) be the subsquares having their SW corner in \((0,0)\) and side length \(d\) and \(3d\), respectively. From Eq. 9, it holds that the probability that a fixed agent lies in \(F\) is \(P_F = \Theta ((d/L)^3)\) and the probability that a fixed agent lies in \(E\) is \(P_E = \Theta ((d/L)^3)\). Consider the event \(B =\) “at time 0, at least one agent is in \(F\) and no agent is in \(E - F\)”. Let \(P\) be the probability that event \(B\) holds. Then,

$$\begin{aligned} P&\geqslant \sum _{i = 1}^n \mathbf P \left( \text{ agent} \ i \ \text{ is} \text{ in} F \right) \mathbf P \left( \text{ all} \text{ agents} \text{ are} \text{ not} \text{ in} E - F \right)\\&= n P_F (1 - P_E)^{n - 1} = \Theta (1) \end{aligned}$$

Hence, \(P\) is a constant positive probability.

If event \(B\) holds (and the source is not in \(F\)), an agent \(\textsc {a}\) in \(F\), at time 0, gets informed at time \(t\) only if there is an informed agent that, at time \(t\), is at distance at most \(R\) from \(\textsc {a}\). Since at time 0 the distance from \(\textsc {a}\) and any agent not in \(F\) is at least \(2d\), it takes at least a time span of \((2d - R)/(2\text{ v})\) so that \(\textsc {a}\) and an agent that was outside \(E\) could be at distance not larger than \(R\). Thus, the flooding time is at least

$$\begin{aligned} (2d - R)/(2\text{ v}) = \Omega (L/(\text{ v}n^{1/3})) \end{aligned}$$

\(\square \)

Let us see when the above lower bound is asymptotically larger than \(L/R\). A necessary and sufficient condition is \(R/(\text{ v}n^{1/3}) \rightarrow \infty \). From the theorem’s hypothesis, this is true only if \(L/(\text{ v}n^{2/3}) \rightarrow \infty \). For instance, if \(L = n^{1/2}\) then we need \(\text{ v}\) asymptotically smaller than \(1/n^{1/6}\).

If \(L = n\) and \(R = L/n^{1/3} = n^{2/3}\) (so a large transmission radius) then the lower bound above becomes \(\Omega (n^{2/3}/\text{ v})\): for \(\text{ v}= \Theta (1)\), this is larger than \(L/R\) for an \(n^{1/3}\) factor.

6 Conclusions

Our work presents the first analytical results on flooding time in a mobility model that strongly departs from the Random-Walk model and that exhibits some crucial typical features of network scenarios yielded by agent mobility in urban zones. We believe that our ideas and techniques can be adapted to analyze flooding over other variants of the RWP model and even over some versions of the more general Random Trip model [19]. Observation 15 provides some technical arguments to extend our analysis to the MRWP variant where the agent can make a constant number of turns (rather than only one) between two consecutive waypoints. The proof of this generalization would require some more complex technicalities without adding no new interesting proof ideas. However, such generalized paths may provide a better approximation for real trajectories performed by vehicles in urban streets.

By comparing the upper bound in Theorem 3 with the lower bound in Theorem 19 and the general lower bound \(\Omega (L/ (\text{ v}+ R))\), we get that our bound for the flooding time is not always tight: for instance, when \(L \sim n, \text{ v}= \Theta ( 1)\) and \(R= \Theta (\sqrt{n \log n})\), our upper bound becomes \(O(n)\) while the lower bound is \(\Omega (n^{2/3})\). Deriving the tight bound for such ranges is an interesting open question.