line-of-sight networks

Line-of-Sight Networks

Frieze, Kleinberg, Ravi, Debany SODA 2007

Cheng-Chung LiCheng-Chung Li2006/12/062006/12/06

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Outline

IntroductionConnectivityThe Existence of a Giant ComponentFinding Paths Between NodesRelay Placement: An Approximation

AlgorithmOpen Questions

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Wireless Sensor Network

Most of today’s approaches to wireless computing and communications are built on architecture where base stations connect to wireless devices to a supporting infrastructure

Such networks can be viewed as consisting of a collection of nodes, representing wireless devices, positioned at various points in some physical region

The wireless links of the network, joining pairs of nodes that can directly communicate with one another, are predominantly short-range and constrained by light-of-sight; this is an inevitable result of the scarcity of radio frequency spectrum and physical constraints on the propagation of RF and optical signals

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The new model for analysis

Given this framework, random geometric graphs have emerged as a dominant model for theoretical analysis for distributed wireless networks One places n points uniformly at random in a geometric

region, and then, for a range parameter r, one connects each pair of nodes that are within distance r of one another

However, there can generally be a large number of obstructions limiting communication between nearby nodes due to lack of line-of-sight contact

Here, the authors give a new light-of-sight model to analyze the related properties

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Modeling Wireless Networks

Sensors modeled as discs of a fixed size placed randomly in [0,1]2. Two discs can communicate if their range overlap

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Suppose there are obstacles

Sensors A,B can not communicate. Need another model

A

B

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Line of Sight Model

Sensors are at centres of crosses can can communicate with sensors lying on their armsA,B can communicate, but A,C cannot

A

B C

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Basic Definitions

T={0,1,…,n-1}2 is a toroidal grid

Distance: d((x,y),(x’,y’))=min(|x-x’|,n-|x-x’|)+min(|y-y’|,n-|y-y’|)

Two points are mutually visible if they are in the same row or column and within distance of each other

n

n

placement probability p>0

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Preliminaries

We assume =O(n) for a value of <1 to be specified below

We now study the random graph G that results if, for some placement probability p>0, we locate a node at each point of T independently with probability p, and then connect those pairs of nodes that are mutually visible

Our main results states, roughly, that the smallest value of p at which G becomes k-connected whp is asymptotically the same as the smallest value of p at which the minimum degree in G is k whp (here, whp is equal to with high probability)

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Connectivity

Theorem 1.1 Suppose that /lnninf where =n, <6/(8k+7)Let k1 be a fixed positive integer and let p=[(1-0.5)lnn+k(lnlnn)/2+cn]/2. Thenlimninf Pr(G is k-connected)= 0,cn-inf e-k, cnc 1,cninfwhere k=[2k-2(1-0.5)ke-2c]/(k-1)!

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Connectivity: The First Part

Here we consider the case when cnc. Let p=[(1-0.5)lnn+k(lnlnn)/2+c]/2

The overall outline of the proof is as follows Add nodes in two stages – most of the nodes in the first

stage, and few final nodes in the second stage Consider the graph H formed by the nodes added in the

first stage

H

S|S|<k

J K

Come lose to see one another

Some point of the torus that sees nodes in both J and K

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The Two stages

We imagine placing nodes at random according to the following two-stage process We place node at each point with probability p1=[(1-0.5)lnn+k(lnlnn)/2+c-(lnn)-1]/2 in the first stage

We then independently place a node at each point with probability p2~1/(2lnn) in the second stage

For ease of terminology, we say that a node is red if it was placed in the first stage, and we say that it is blue if it is placed in the second stage at a point not hit by the first stage. Let H denotes the subgraph of G consisting only of red nodes

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Some Definitions

A segment is said to be weak, otherwise strong, if it contains fewer than lnn/50 red nodes

An arm is said to be mighty if all its segments are strong

Arm: the set contains points that are visible from it in a single direction

1

2

10

length: /10

segment

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First Event(1)

Lemma 2.2 WHP there does not exist a red node which has an arm on which we can find 1000 red vertices, each having an arm orthogonal to which is not mighty

1

2

1000 Not mighty

Not mighty

Not mighty

x

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Proof of Lemma 2.2

For a fixed x and arm , the probability that the arm contains a weak segment can be bounded by 10P(Bin(/10,p1)lnn/50)e-(lnn)/400=n-1/400 (at most 10 weak segments)

So the probability that there is a red node as described in the statement is bounded by 42n2(,1000)p1

1000n-1000/400=o(1) (x has four arms, and the nodes at have two arms orthogonal to )

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Second Event(2)

Lemma 2.3 WHP H does not contain a vertex v of degree less than lnlnn that has a neighbor w such that w contains an arm orthogonal to vw which is not mighty

v w

weak

deg(v)<lnlnn

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Proof of Lemma 2.3

The probability that H contains such a pair v,w is bounded by n2p1t=1 to lnlnn(4,t)p1

t(1-p1)4-t

(2n-1/400)=o(1) (There are two arms at that are orthogonal to vw)

v

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Third Event(3)

Lemma 2.4 WHP H does not contain a red vertex with at most k-1 red neighbors and at least one blue neighbor

Proof of Lemma 2.4 The probability that H contains such a vertex v is bounded by n2p1t=0 to k-1(4,t)p1

t(1-p1)4-t(4p2)=o(1) (The red part only considers the red vertices)

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Fourth Event(4)

Lemma 2.5 WHP H does not contain a blue vertex with fewer than k red neighbors

Proof of Lemma 2.5 The probability that H contains such a vertex v is bounded by n2p2t=0 to k-1(4,t)p1

t(1-p1)4-t=o(1)

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Additional Definition

Recall that H is the subgraph of G consisting only of the red nodes

Let S be an arbitrary set of k-1 red vertices, and let Hs=H-S. Our goal is to show that if Hs has multiple connected components, then whp they will all be linked up by the addition of the blue nodes

Let L be the set of points in T with coordinates (i,j), where each of i and j is a multiple of 3

For each connected component K of Hs, and for each point xL, let vKx denote the node in K that is closet to x in L1 distance

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The relation between K and L

3 6 9 12

3

6

9

12

L

K1

K3

K2

HS

x1

vK1x1

vK1x2

x2

x3

vK1x3

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Where is vKx ?

Lemma 2.6 vKx lies within the * box Bx centered at x

3 6

3

6

9

12HS

x1

vK1x1

x3

K1

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Proof of Lemma 2.6-1

Let a red node be pink if it is not in SWe prove this lemma by contradiction

x=0

v=vK0=(a,b)

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By 3, v has at least one arm containing a pink node w

x=0

v=vK0=(a,b)

w

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If the deg(v)<lnlnn, then we can use the non-occurrence of 2 to argue that two arms of w orthogonal to vw are mighty

w

mighty

deg(v)<lnlnn

mighty

v

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If deg(v)≥lnlnn, then we can use the non-occurrence of 1 to argue that there is a choice of lnlnn-4000 w’s such that the two arms of w orthogonal to vw are mighty

v w

mighty

deg(v)≥lnlnn

mighty

w

w

ww w

w

w

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Case 1: is the South arm of v It follows that |a’|+|c|a+b+.1-.4,

contradiction !

x=0

v=vK0=(a,b)

y=(a,-b’’)w=(a’,-b’’)

z=(a’,c)

a-a’[.4,.5]

|c-b|.1

,: mighty

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Case 2a: is the North arm of v and a≥/2 It follows that |a’|+|b’’|a+b+.1-.4,

contradiction !

x=0

y=(a,b’)

v

w=(a’,b’)

a-a’[.4,.5]

z=(a’,b’’)|b’’-b|.1

,: mighty

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Case 2b-1: is the North arm of v and a</2

x=0

v=(a,b)

y=(a,b’)

w=(a’,b’)

|a-a’|.1,: mighty

|b-b’|.7

z=(a’,b’’)

|b’’-b’|[.9,]

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Case 2b-2: is the North arm of v and a</2

x=0

v=(a,b)

y=(a,b’)

w=(a’,b’)

|a-a’|.1,,,: mighty

|b-b’|>.7q=(a’,b’’)

|b’-b’’|≥.9

z=(a’’,b’’)

|a’’-a’|.1

p=(a’’,b’’’)|b’’-b’’’|[.5,.6]

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In case 2b-1, it follows that |a’|+|b’’|a+b+.1+.7-.9, contradiction

In case 2b-2, it follows that |a’’|+|b’’’|a+b++.1-.9+.1-.5, contradiction

The case is the west arm is dealt with as in case 1 and the case where is the east arm is dealt with as in case 2

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Derandomization

The proof of lemma 2.9 is the first example we have seen of derandomization – where we take a randomized algorithm or computation, and diminish or entirely remove the randomness in it

This is often a useful technique for the design of deterministic algorithms

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Final Lemma Lemma 2.7 The points z(J,K,x) and z(J,K,y) are distinct, for distinct poi

nts x and y

3 6

3

6

9

12

L

K

J

HS

vKx

x

vJx

z(J,K,x)

y vJyvKy z(J,K,y)

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Finishing the Proof-1

Note that of a node is placed at z(J,K,x), then it will be a neighbor both of a point in J and K, and hence J and K will belong to the same component in G

In the second stage of node placement, a blue node will be placed at each point z(J,K,x) will probability p2

By lemma 2.7, there are n2/92 such points for a fixed pair of components J,K, and so the probability that no blue point is placed at any of them is bounded by (1-p2)^(n2/92) e^[-n2-3/(20lnn)]

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Finishing the Proof-2

There are at most 2 components, since for any fixed point xL, each component has a node in the * box around x

Thus, the probability that there exists a set S of size at most k-1 and components J,K of Hs, which are not connected in G by a blue vertex is at most 4e^[-n2-3/(20lnn)]n2k-2=o(1)

Thus, conditional on there being no vertices of degree k-1 or less, if we remove any set S of k-1 vertices, then whp the graph Hs has a component containing all of the red vertices

It follows from 4 that G-S is connected and so G itself is k connected whp

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Connectivity: The Second&Third Part

If cn-inf then one uses the Chebyshev inequality to show that whp there are vertices of degree less than k

If cninf the whp there are no vertices of degree less than k (the expected number tends to zero)

And, the argument cnc implies that G will be k-connected whp

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Giant Component

G will whp contain ~np2 vertices. A giant component is therefore one with (n2p) vertices

Theorem 3.1 If p=c/ where c>1 and inf the whp G contai

ns a unique component with (1-o(1))(1-x2c)n2/ v

ertices, where xc is the unique solution in (0,1) of xe-x =ce-c

If p=c/ where c<1(4e) and inf then whp the largest component in G has size O(lnn)

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Finding Paths Between Nodes

Theorem Let p=Clnn/ for a constant C3. There is a decentralized algorithm that whp, given nodes s and t, constructs an s-t path with O(d(s,t)/ +lnn) edges while involving O(d(s,t)/ + lnn) nodes in the computation

This bound is nearly optimal, since (d(s,t)/) is a simple lower bound on the number of edges and number of nodes involved in any s-t path

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Relay Placement

Relay Placement problem: given a set of nodes on a grid, we would like to add a small number of additional nodes so that the full set becomes connected

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Problem Model

As before, we are given an nn torus of points T. Let K=(T,E) be the graph defined on the points of T, in which we join two points by an edge if they can see each other

Also, we are given a cost cx for each point xT, and for a set XT we define c(X)=xXcx

Let X={x1,x2,…,xk} be a given set of points inT. We consider the problem of choosing a set of additional points Y={y1,y2,…,ys} such that K[XY] is connected

We call Y a Steiner set for X. Our goal is to find a Steiner set whose total cost as small as possible

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Relative to other problems

This is an instance of Node-Weighted Steiner tree problem in the graph K, with X as the set of given terminals and Y as the set of additional Steiner nodes whose total cost we want to minimize

In general, there is an (logn) hardness of approximation for this problem(Klein & Ravi, J. Algorithms, 1994 )

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Results of This Paper

However, the special structure of the graph K allows us to efficiently find a Steiner set whose cost is within a constant factor of minimum

Theorem 5.1 There is a polynomial-time algorithm that produces a Steiner set whose total cost is within a factor of 6.2 of optimal

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The crucial combinatorial property

The crucial combinatorial property of K that we use is captured by the following definition

We say that a graph H is d-cohesive if every connected subset of H has a spanning tree of maximum degree d That is, given any connected subset S of V(H),

we can choose a set F of edges, each with both ends in S, such that (S,F) is a tree of maximum degree d

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Graph K is 4-cohesive

Lemma 5.3 The graph K is 4-cohesive For each edge of K, define its length to be the number

of rows or columns of T that separate its ends Now, consider an arbitrary connected subset S of K,

and let (S,F) be a spanning tree of S whose total edge length is minimum

We claim that the maximum degree of (S,F) is four. For suppose not;

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Transform node weight to edge weight

We first define weights on the edges of K as follows First, let X-reduced cost cx,v of a node v

0, if vXcv, otherwise

Let cx(Y)=yYcx,y

For each edge e=(v,w) of K, we define its weight we to be max(cx,v, cx,w)

For a subgraph of K, let w() denote its total edge weight

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Approximation Algorithm&Analysis

Now, let Y* be a Steiner set for X of minimum cost, and let * be a Steiner tree for X of minimum total edge weight (Note that the Steiner nodes of * may be different from Y*)

10

3 5 3

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Lemma 5.4 w(*)4c(Y*)

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A Steiner tree whose edge weight is within a constant 1.55 of optimal can be computed in polynomial time via an algorithm

Let ’ be a Steiner tree for X computed using this algorithm

Let Y’ be the Steiner nodes of ’ By charging the costs of nodes in Y’ to the weights

of distinct incident edges in ’, we have Lemma 5.5 c(Y’)w(’)

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Finally, we use Y’ as our Steiner set for XBecause c(Y’)w(’)w(*)4c(Y*)We obtain a bound on c(Y’) relative to the o

ptimum c(Y*)

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Open Questions

Find the exact threshold for the existence of a giant component

Remove the restrictions on Study problems associated with points of G

moving (randomly)

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T HAN K YOU

line-of-sight networks

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