Initializing Sensor Networks of Non-uniform Density in the Weak Sensor Model

Abstract

Assumptions about node density in the sensor networks literature are frequently too strong. Neither adversarially chosen nor uniform random deployment seem realistic in many intended applications of sensor nodes. We define smooth distributions of sensor nodes to be those where the minimum density is guaranteed to achieve connectivity in random deployments, but higher densities may appear in certain areas. We study basic problems for smooth distribution of nodes. Most notably, we present a Weak Sensor Model-compliant distributed protocol for hop-optimal network initialization (NI), a fundamental problem in sensor networks. In order to prove lower bounds, we observe that all nodes must communicate with some other node in order to join the network, and we call the problem of achieving such a communication the group therapy (GT) problem. We show a tight lower bound for the GT problem in radio networks for any class of protocols, and a stronger lower bound for the important class of randomized uniform-oblivious protocols. Given that any NI protocol also solves GT, these lower bounds apply to NI. We also show that the same lower bound holds for a related problem that we call independent set , when nodes are distributed uniformly, even in expectation.

Appendices

Appendix A: Implementation of the First Phase of the Disc Covering Scheme

In this section we describe how to distributedly implement the first phase of the disk covering scheme, i.e., to define an $ar/2$ radius disk-layout where each disk is centered in an uncovered node. As in [9], this phase can be achieved distributedly by means of a Maximal Independent Set (MIS) computation with nodes transmitting in a range of $ar/2$. However, given that in our setting the maximum number of nodes in any disk of diameter $r$ is $\varDelta$, we change appropriately the probabilities and counters used. We give the details in Algorithm 3.

Due to the interference that neighboring nodes running different phases produce, there is a potential circular argument in the analysis. As in [9], to break this circularity the following two lemmas are proved together by induction on the time slot in which a node joins the MIS with ties broken arbitrarily.

Lemma 4

Given any node that joins the MIS in a given time slot, the counter of all neighboring nodes is at most $\lceil \delta _3 \varDelta \log n\rceil - \lfloor \delta _2\log n\rfloor$ in the same time slot w.h.p.

Lemma 5

Every MIS node transmits its MIS status message successfully in the $\lfloor \delta _2\log n \rfloor$ time slots after it joins the MIS w.h.p.

Proof

Base case: Consider the first node within the whole network, call it $\mu _1$, that joins the MIS at time $t_1$. For the sake of contradiction, assume that there is a node $x$ contained in $\mu _1$’s neighborhood whose counter is greater than $L=\lceil \delta _3 \varDelta \log n\rceil - \lfloor \delta _2\log n\rfloor$ at $t_1$. By the definition of the algorithm, $\mu _1$ has transmitted at time $t_1 - \lceil \delta _3\varDelta \log n\rceil$ and $x$ has transmitted within the next $\lfloor \delta _2\log n\rfloor$ time slots. Afterwards, neither $\mu _1$ nor $x$ have received from each other. Otherwise, one of their counters would have been reset. Let $E(k)$ denote the event that neither $\mu _1$ nor $x$ have received from each other within $k$ time slots. As shown in the proof of Theorem 4, there are at most $\delta _6\varDelta$ nodes within the 2-hop neighborhood of $\mu _1$ for some constant $\delta _6>1$. Then,

$$\begin{aligned} Pr[E(L)] \le \left[ 1-2\frac{1}{\delta _1\varDelta }\left( 1-\frac{1}{\delta _1\varDelta }\right) ^{\delta _6\varDelta }\right] ^{\lceil \delta _3\varDelta \log n\rceil - \lfloor \delta _2\log n\rfloor }. \end{aligned}$$

Using Fact 1 for $n=\varDelta$ and $x=-1/\delta _1$ (which is valid because $|x|=1/\delta _1 <\varDelta$),

$$\begin{aligned} Pr[E(L)]\le \left[ 1-2\frac{1}{\delta _1\varDelta }\left( \frac{1}{e^{1/\delta _1}} \left( 1-\frac{1}{\delta _1^2\varDelta }\right) \right) ^{\delta _6}\right] ^{\lceil \delta _3 \varDelta \log n\rceil - \lfloor \delta _2\log n\rfloor }. \end{aligned}$$

Using Fact 1 for $n=\varDelta$ and $x=-2\left( \left( 1-1/(\delta _1^2\varDelta )\right) /e^{1/\delta _1} \right) ^{\delta _6}/\delta _1$, (which is valid because $1/(\delta _1^2\varDelta ) < 1$ and then $|x|=2\left( \left( 1-1/(\delta _1^2\varDelta )\right) /e^{1/\delta _1}\right) ^{\delta _6}/\delta _1\le 2/\delta _1 < \varDelta$)

$$\begin{aligned} Pr[E(L)]&\le \left[ \exp \left( -2\frac{1}{\delta _1}\left( \frac{1}{e^{1/\delta _1}} \left( 1-\frac{1}{\delta _1^2\varDelta }\right) \right) ^{\delta _6}\right) \right] ^{(\lceil \delta _3\varDelta \log n\rceil - \lfloor \delta _2\log n\rfloor )/\varDelta }\\&\le \left[ \exp \left( -2\frac{1}{\delta _1}\left( \frac{1}{e^{1/\delta _1}}\left( 1-\frac{1}{\delta _1^2\varDelta }\right) \right) ^{\delta _6}\right) \right] ^{\log n(\delta _3 - \delta _2/\varDelta )}\\&= n^{-\frac{2}{\delta _1}\left( \frac{1}{e^{1/\delta _1}} \left( 1-\frac{1}{\delta _1^2\varDelta }\right) \right) ^{\delta _6}\frac{\delta _3 - \delta _2/\varDelta }{\ln 2}}. \end{aligned}$$

Then, $Pr[E(L)] \in O(n^{-\gamma })$, for some constant $\gamma >0$, because $1/(\delta _1^2\varDelta )< 1$ and $\delta _3>\delta _2/\varDelta$.

Now we must additionally prove that within $\lfloor \delta _2\log n \rfloor$ time slots of $\mu _1$joining the MIS, all nodes within its range receive a message declaring its MIS status. For at least $\lfloor \delta _2\log n \rfloor$ time slots after the node $\mu _1$ joins the MIS, no other nodes in its neighborhood join the MIS w.h.p. as shown. If in this time its MIS status message is received by all its neighbors, then they will all stop counting and transition into the covered state. We will now show that this message is received by all its neighbors w.h.p. Let $E(k)$ denote the event that $\mu _1$ does not transmit without collision in $k$ consecutive time slots. Then,

$$\begin{aligned} Pr[E(\lfloor \delta _2\log n \rfloor )] = \left[ 1 - \frac{1}{\delta _4}\left( 1 - \frac{1}{\delta _1 \varDelta }\right) ^{\delta _6\varDelta }\right] ^{\lfloor \delta _2\log n \rfloor }. \end{aligned}$$

Using Fact 1 for $n=\delta _6\varDelta$ (which is valid because $\delta _6\varDelta >1$) and $x=-\delta _6/\delta _1$ (which is valid because $|x|=\delta _6/\delta _1 <\delta _6\varDelta$),

$$\begin{aligned} Pr[E(\lfloor \delta _2\log n \rfloor )] \le \left[ 1 - \frac{1}{\delta _4 e^{\delta _6/\delta _1}}\left( 1 - \frac{\delta _6}{\delta _1^2 \varDelta }\right) \right] ^{\lfloor \delta _2\log n \rfloor }. \end{aligned}$$

Using Fact 1 for $n=1$ and $x=-\left( 1 - \delta _6/(\delta _1^2 \varDelta )\right) /(\delta _4 e^{\delta _6/\delta _1})$ (which is valid because $\delta _6/( \delta _1^2\varDelta )<1$ and then $|x|=\left( 1 - \delta _6/(\delta _1^2 \varDelta )\right) /(\delta _4 e^{\delta _6/\delta _1}) < 1$ for $\delta _4> 1$),

$$\begin{aligned} Pr[E(\lfloor \delta _2\log n \rfloor )]&\le \left[ \exp \left( -\frac{1}{\delta _4 e^{\delta _6/\delta _1}}\left( 1 - \frac{\delta _6}{\delta _1^2 \varDelta }\right) \right) \right] ^{\lfloor \delta _2\log n \rfloor }\\&\le \left[ \exp \left( -\frac{1}{\delta _4 e^{\delta _6/\delta _1}}\left( 1 - \frac{\delta _6}{\delta _1^2 \varDelta }\right) \right) \right] ^{\delta _2\log n -1}\\&=\exp \left( \frac{1}{\delta _4 e^{\delta _6/\delta _1}}\left( 1 -\frac{\delta _6}{\delta _1^2 \varDelta }\right) \right) n^{-\frac{\delta _2}{\delta _4 e^{\delta _6/\delta _1}\ln 2}\left( 1 - \frac{\delta _6}{\delta _1^2 \varDelta }\right) }. \end{aligned}$$

Then, $Pr[E(\lfloor \delta _2\log n \rfloor )] \in O(n^{-\gamma })$ for some constant $\gamma >0$, because $\delta _2>0$ and $\delta _6/( \delta _1^2\varDelta )<1$.

This shows that $\mu _1$ sends its MIS status message without collision successfully in $\lfloor \delta _2\log n\rfloor$ time slots w.h.p.

Inductive Step:Consider the $i$th node $\mu _i$, $i>1$, that joins the MIS at time $t_i$.

Inductive hypothesis: For all nodes $\mu _j$ such that $j<i$, joining the MIS at time $t_j$, the counters of all nodes in the neighborhood of $\mu _j$ are at most $\lceil \delta _3\varDelta \log n\rceil - \lfloor \delta _2\log n\rfloor$ at time $t_j$ w.h.p. Additionally all nodes $\mu _j$ transmit their MIS status message successfully within the interval $t_j, \ldots , t_j+\lfloor \delta _2\log n\rfloor$ w.h.p.

Therefore, by time $t_j+\lfloor \delta _2\log n\rfloor$ all nodes in the range of all MIS nodes $\mu _1,\ldots , \mu _{i-1}$ will be in the covered state. From the previous statements of the inductive hypothesis we can conclude that none of the MIS nodes $\mu _j$ (where $j<i$) are neighbors of each other w.h.p.

We want to show that the counters of all nodes in the neighborhood of $\mu _i$ are at most $\lceil \delta _3\varDelta \log n\rceil - \lfloor \delta _2\log n\rfloor$ at time $t_i$ w.h.p. and that all neighbors of $\mu _i$ are in the covered state by time $t_i +\lfloor \delta _2\log n\rfloor$ w.h.p.

If $\mu _i$ is out of the two-hop neighborhood of all the previous MIS members, the claim can be easily proved using the same argument as in the base case. Otherwise, $\mu _i$ is within a two-hop neighborhood of some MIS members. Since all nodes that previously joined the MIS are not in range of each other, $\mu _i$ is within the two-hop neighborhood of at most $12$ other MIS members. This is true because a regular polygon with side of length at least $r$ and distance from the center to the vertices at most $2r$ has at most $12$ sides.

For the sake of contradiction, assume that there is a node $y$ contained in $\mu _i$’s neighborhood whose counter is greater than $L=\lceil \delta _3 \varDelta \log n\rceil - \lfloor \delta _2\log n\rfloor$ at $t_i$. By the definition of the algorithm, $\mu _i$ has first transmitted at time $t_i - \lceil \delta _3\varDelta \log n\rceil$ and $y$ has first transmitted within the next $\lfloor \delta _2\log n\rfloor$ time slots. Afterwards, neither $\mu _i$ nor $y$ have sent without collision otherwise one of their counters would have been reset. Let $E(k)$ be the event that neither $\mu _i$ or $y$ send without collision for $k$ consecutive time slots. Then,

$$\begin{aligned} Pr[E(L)] \le \left[ 1-2\frac{1}{\delta _1\varDelta } \left( 1-\frac{1}{\delta _1\varDelta }\right) ^ {\delta _6\varDelta }\left( 1-\frac{1}{\delta _4}\right) ^{12}\right] ^ {\lceil \delta _3\varDelta \log n\rceil - \lfloor \delta _2\log n\rfloor }. \end{aligned}$$

Using Fact 1 for $n=\delta _6\varDelta$ (which is valid because $\delta _6\varDelta >1$) and $x=-\delta _6/\delta _1$ (which is valid because $|x| = \delta _6/\delta _1 < \delta _6\varDelta$),

$$\begin{aligned} Pr[E(L)] \le \left[ 1-\frac{2}{\delta _1\varDelta e^{\delta _6/\delta _1}}\left( 1-\frac{\delta _6}{\delta _1^2\varDelta }\right) \left( 1-\frac{1}{\delta _4}\right) ^{12}\right] ^ {\lceil \delta _3\varDelta \log n\rceil - \lfloor \delta _2\log n\rfloor }. \end{aligned}$$

Using Fact 1 for $n=1$ and $x=-2\left( 1-\delta _6/\delta _1^2\right) \left( 1-1/\delta _4\right) ^{12}/(\delta _1\varDelta e^{\delta _6/\delta _1})$ (which is valid because $\delta _6/\delta _1^2< 1$ and then

$|x| = 2\left( 1-\delta _6/\delta _1^2\right) \left( 1-1/\delta _4\right) ^{12}/(\delta _1\varDelta e^{\delta _6/\delta _1}) \le 2/(\delta _1\varDelta e^{\delta _6/\delta _1}) < 1$),

$$\begin{aligned} Pr[E(L)]&\le \left[ \exp \left( -\frac{2}{\delta _1 e^{\delta _6/\delta _1}}\left( 1-\frac{\delta _6}{\delta _1^2\varDelta }\right) \left( 1-\frac{1}{\delta _4}\right) ^{12}\right) \right] ^{(\lceil \delta _3\varDelta \log n\rceil - \lfloor \delta _2\log n\rfloor )/\varDelta }\\&\le \left[ \exp \left( -\frac{2}{\delta _1 e^{\delta _6/\delta _1}} \left( 1-\frac{\delta _6}{\delta _1^2\varDelta }\right) \left( 1-\frac{1}{\delta _4}\right) ^{12}\right) \right] ^ {\log n(\delta _3 - \delta _2/\varDelta )}\\&= n^{-\frac{2(\delta _3 - \delta _2/\varDelta )}{\delta _1 e^{\delta _6/\delta _1}\ln 2}\left( 1-\frac{\delta _6}{\delta _1^2\varDelta }\right) \left( 1-\frac{1}{\delta _4}\right) ^{12}}. \end{aligned}$$

Then, $Pr[E(L)] \in O(n^{-\gamma })$ for some constant $\gamma >0$, because $\delta _3 > \delta _2/\varDelta$, $\delta _6/(\delta _1^2\varDelta )<1$, and $\delta _4>1$.

Now we will show that all neighbors of MIS node $\mu _i$ will be in the covered state by time slot $t_i+\lfloor \delta _2\log n\rfloor$. Any neighbor of an MIS node has a counter that lags the MIS node’s counter by at least $\lfloor \delta _2\log n\rfloor$. Additionally no MIS node can be within range of any other. Hence every MIS node can be subjected to interference by at most $18$ other MIS nodes (by a simple geometric packing argument). Let $E(k)$ denote the event that a neighbor of an MIS node does not receive its MIS status message for $k$ consecutive time slots. Then,

$$\begin{aligned} Pr[E(\lfloor \delta _2\log n\rfloor )] \le \left[ 1- \frac{1}{\delta _4}\left( 1-\frac{1}{\delta _4}\right) ^{18} \left( 1-\frac{1}{\delta _1\varDelta }\right) ^ {\delta _6\varDelta }\right] ^{\lfloor \delta _2\log n\rfloor }. \end{aligned}$$

Using Fact 1 for $n=\delta _6\varDelta$ (which is valid because $\delta _6\varDelta >1$) and $x=-\delta _6/\delta _1$ (which is valid because $|x| = \delta _6/\delta _1 < \delta _6\varDelta$),

$$\begin{aligned} Pr[E(\lfloor \delta _2\log n\rfloor )] \le \left[ 1- \frac{1}{\delta _4}\left( 1-\frac{1}{\delta _4}\right) ^{18} \frac{1}{e^{\delta _6/\delta _1}}\left( 1-\frac{\delta _6}{\delta _1^2\varDelta }\right) \right] ^{\lfloor \delta _2\log n\rfloor }. \end{aligned}$$

Using Fact 1 for $n=1$ and $x=-\left( 1-1/\delta _4\right) ^{18} \left( 1-\delta _6/(\delta _1^2\varDelta )\right) /(\delta _4 e^{\delta _6/\delta _1})$ (which is valid because $\delta _6/(\delta _1^2\varDelta )<1$ and then

$|x| = \left( 1-1/\delta _4\right) ^{18} \left( 1-\delta _6/(\delta _1^2\varDelta )\right) /(\delta _4 e^{\delta _6/\delta _1}) <1$),

$$\begin{aligned} Pr[E(\lfloor \delta _2\log n\rfloor )]&\le \left[ \exp \left( - \frac{1}{\delta _4}\left( 1-\frac{1}{\delta _4}\right) ^{18} \frac{1}{e^{\delta _6/\delta _1}}\left( 1-\frac{\delta _6}{\delta _1^2\varDelta }\right) \right) \right] ^{\lfloor \delta _2\log n\rfloor }\\&\le \left[ \exp \left( - \frac{1}{\delta _4}\left( 1-\frac{1}{\delta _4}\right) ^{18} \frac{1}{e^{\delta _6/\delta _1}}\left( 1-\frac{\delta _6}{\delta _1^2\varDelta }\right) \right) \right] ^{\delta _2\log n-1}\\&=\exp \left( \frac{1}{\delta _4}\left( 1-\frac{1}{\delta _4}\right) ^{18} \frac{1}{e^{\delta _6/\delta _1}}\left( 1-\frac{\delta _6}{\delta _1^2\varDelta } \right) \right) \\&\quad \cdot n^{- \frac{\delta _2}{\delta _4\ln 2} \left( 1-\frac{1}{\delta _4}\right) ^{18} \frac{1}{e^{\delta _6/\delta _1}}\left( 1-\frac{\delta _6}{\delta _1^2\varDelta }\right) }. \end{aligned}$$

Then, $Pr[E(\lfloor \delta _2\log n\rfloor )] \in O(n^{-\gamma })$ for some constant $\gamma >0$, because $\delta _4>1$, $\delta _6/(\delta _1^2\varDelta )<1$, and $\delta _2>0$. $\square$

Lemma 6

No two nodes belonging to the MIS are within the transmission ranges of each other w.h.p.

Proof

This is a direct conclusion of Lemmas 4 and  5. $\square$

Lemma 7

For any node running the MIS algorithm with radius $r$, there is at least one node, in its immediate $r/2$ neighborhood, that transmits without collision within $\lceil \delta _5 \varDelta \log n\rceil$ steps w.h.p., for some constant $\delta _5>0$.

Proof

Consider a node $A$ running the MIS algorithm (refer to Fig. 3). Since $A$ is active, there is at least one node active in $C$ at time $t$. From Lemma 6 it can be seen that no MIS nodes can be within range of each other, therefore there can be at most 9 MIS nodes within $D$ (If there were more then one of them would be in range of $A$). Let $E(k)$ denote the event that no node in $A$’s $r/2$ neighborhood (including $A$) transmits without collision in $k$ consecutive time slots. As shown in the proof of Theorem 4, there are at most $\delta _6\varDelta$ nodes within the 2-hop neighborhood of $\mu _1$ for some constant $\delta _6>1$. Then,

$$\begin{aligned} Pr[E(\lceil \delta _5 \varDelta \log n\rceil )] \le \left[ 1- \frac{1}{\delta _1 \varDelta } \left( 1 - \frac{1}{\delta _1 \varDelta }\right) ^{\delta _6\varDelta } \left( 1 - \frac{1}{\delta _4}\right) ^{9}\right] ^{\lceil \delta _5\varDelta \log n\rceil }. \end{aligned}$$

Using Fact 1 for $n=\delta _6\varDelta$ (which is valid because $\delta _6\varDelta >1$) and $x=-\delta _6/\delta _1$ (which is valid because $|x| = \delta _6/\delta _1 < \delta _6\varDelta$),

$$\begin{aligned} Pr[E(\lceil \delta _5 \varDelta \log n\rceil )] \le \left[ 1- \frac{1}{\delta _1 \varDelta e^{\delta _6/\delta _1}}\left( 1 - \frac{\delta _6}{\delta _1^2 \varDelta }\right) \left( 1 - \frac{1}{\delta _4}\right) ^{9}\right] ^{\lceil \delta _5\varDelta \log n\rceil }. \end{aligned}$$

Using Fact 1 for $n=\varDelta$ (which is valid because $\varDelta >1$) and $x=-\left( 1 - \delta _6/(\delta _1^2 \varDelta )\right)$ $\left( 1 - 1/\delta _4\right) ^{9}/(\delta _1 e^{\delta _6/\delta _1})$ (which is valid because $\delta _6/(\delta _1^2\varDelta )<1$, $\delta _4>1$, and then $|x| = \left( 1 - \delta _6/(\delta _1^2 \varDelta )\right) \left( 1 - 1/\delta _4\right) ^{9}/(\delta _1 e^{\delta _6/\delta _1}) < \varDelta$),

$$\begin{aligned} Pr[E(\lceil \delta _5 \varDelta \log n\rceil )]&\le \left[ \exp \left( - \frac{1}{\delta _1 e^{\delta _6/\delta _1}}\left( 1 - \frac{\delta _6}{\delta _1^2 \varDelta }\right) \left( 1 - \frac{1}{\delta _4}\right) ^{9}\right) \right] ^{\lceil \delta _5\varDelta \log n\rceil /\varDelta }\\&\le \left[ \exp \left( - \frac{1}{\delta _1 e^{\delta _6/\delta _1}}\left( 1 - \frac{\delta _6}{\delta _1^2 \varDelta }\right) \left( 1 - \frac{1}{\delta _4}\right) ^{9}\right) \right] ^{\delta _5\log n}\\&= n^{- \frac{\delta _5}{\delta _1 e^{\delta _6/\delta _1}\ln 2}\left( 1 - \frac{\delta _6}{\delta _1^2 \varDelta }\right) \left( 1 - \frac{1}{\delta _4}\right) ^{9}}. \end{aligned}$$

Then, $Pr[E(\lceil \delta _5 \varDelta \log n\rceil )] \in O(n^{-\gamma })$ for some constant $\gamma >0$, because $\delta _4>1$, $\delta _6/(\delta _1^2\varDelta )<1$, and $\delta _5>0$. $\square$

Fig. 3

Illustration for Lemma 7

Theorem 6

For a given node running the MIS algorithm, at least one node within its transmission range joins the MIS in $O(\varDelta \log n)$ time slots and no two MIS nodes are within range of each other w.h.p.

Proof

The proof is illustrated in Fig. 4. In Lemma 7, it was shown that within a circle of radius $r/2$ centered on any node $x_1$, there will be a node $x_2$, transmitting without collision, in less than $\lceil \delta _5 \varDelta \log n\rceil$ steps w.h.p. After this single transmission, there is at least one node, namely $x_2$, within the neigborhood of $x_1$ increasing its counter. If $x_2$ joins the MIS after its counter reaches the value $\lceil \delta _3\varDelta \log n\rceil$, then the statement of the theorem is proved. Otherwise, some other node, call it $x_3$, within range of $x_2$, reaches this value and joins the MIS before. If $x_3$ is within range of $x_1$, then the statement of the theorem is proved. Otherwise, $x_3$ covers at least one node within the $r/2$ neighborhood of $x_1$, namely $x_2$, within the next $\lceil \delta _2\log n \rceil$ time slots w.h.p. (as shown in Lemma 5). Note that the distance between $x_1$ and $x_3$ satisfies the following relation

$$\begin{aligned} r<D(x_1,x_3)\le 3r/2\ . \end{aligned}$$

(4)

All uncovered active nodes within the $r/2$ neighborhood of $x_1$ are still counting. Hence, the same argument can be repeatedly applied with the restriction that the next MIS node is at least at a distance of $r$ from $x_3$ (by Lemma 6). There can be at most $9$ MIS nodes around $x_1$ before $x_1$ or one of its neighbors joins the MIS, as explained in Lemma 7. Thus, this process terminates in at most $10(\lceil \delta _3 \varDelta \log n\rceil +\lceil \delta _5 \varDelta \log n\rceil +\lfloor \delta _2\log n \rfloor )$ time slots. $\square$

Fig. 4

Illustration for Theorem 6

Appendix B: Implementation of the Second and Third Phases of the Disc Covering Scheme

In this section we describe how to distributedly implement the second and third phases of the disk covering scheme, i.e., to add all edges of length at most $r$ among bridges, and to expand the disks to a bigger radius of $br/2$ defining in this manner which bridges cover each non-bridge node. For this purpose, it is enough to achieve non-colliding transmissions of each bridge in a radius of $r$ and $br/2$ respectively.

Let $\delta$ denote an upper bound on the number of interfering MIS neighbors. The algorithm is simple to describe: with probability $1/\beta _1$, each MIS node transmits its ID, within range $\beta _2 r$. Where $\beta _1>\sqrt{\delta }$, $\beta _2\le 1$, and $\delta >1$ are constants whose values depend on which of the aforementioned steps is implemented. For informing the non-MIS nodes about assignment, the transmission is made with $\beta _2=b/2$. For setting up connections with neighboring MIS nodes, the transmission is made with $\beta _2=1$. Then, the following theorem holds.

Lemma 8

Any MIS node running the broadcast algorithm achieves a transmission without collision within $O(\log n)$ steps w.h.p.

Proof

Let $Pr[\hbox {fail}]$ denote the probability that any node fails to transmit without collision after $\beta _3\log n$ steps for some constant $\beta _3$ to be defined later. Using the union bound, we have

$$\begin{aligned} Pr[\hbox {fail}] \le n\left( 1 - \frac{1}{\beta _1} \left( 1- \frac{1}{\beta _1}\right) ^{\delta }\right) ^{\beta _3\log n}. \end{aligned}$$

Using Fact 1 for $n=\delta$ (which is valid because $\delta > 1$) and $x=-\delta /\beta _1$ (which is valid because $|x|=\delta /\beta _1< \delta$),

$$\begin{aligned} Pr[\hbox {fail}] \le n\left( 1 - \frac{1}{\beta _1 e^{\delta /\beta _1}}\left( 1 - \frac{\delta }{\beta _1^2}\right) \right) ^{\beta _3\log n}. \end{aligned}$$

Using Fact 1 for $n=1$ and $x=-\left( 1 - \delta /\beta _1^2\right) /(\beta _1 e^{\delta /\beta _1})$ (which is valid because $\delta /\beta _1^2<1$ and then $|x|=\left( 1 - \delta /\beta _1^2\right) /(\beta _1 e^{\delta /\beta _1})<1$),

$$\begin{aligned} Pr[\hbox {fail}]&\le n\left( \exp \left( - \frac{1}{\beta _1 e^{\delta /\beta _1}} \left( 1 - \frac{\delta }{\beta _1^2}\right) \right) \right) ^{\beta _3\log n}\\&= n^{ 1 - \frac{\beta _3}{\beta _1 e^{\delta /\beta _1}\ln 2} \left( 1 - \frac{\delta }{\beta _1^2}\right) }. \end{aligned}$$

Then, making $\beta _3 > (\beta _1 e^{\delta /\beta _1}\ln 2) / (\left( 1 - \delta /\beta _1^2\right) )$, it is $Pr[\hbox {fail}] \in O(n^{-\gamma })$ for some constant $\gamma >0$. $\square$