line-of-sight networks
DESCRIPTION
Line-of-Sight Networks. Frieze, Kleinberg, Ravi, Debany SODA 2007. Cheng-Chung Li 2006/12/06. Outline. Introduction Connectivity The Existence of a Giant Component Finding Paths Between Nodes Relay Placement: An Approximation Algorithm Open Questions. Wireless Sensor Network. - PowerPoint PPT PresentationTRANSCRIPT
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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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Proof of Lemma 2.6-2
By 3, v has at least one arm containing a pink node w
x=0
v=vK0=(a,b)
w
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Proof of Lemma 2.6-3
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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Proof of Lemma 2.6-4
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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Proof of Lemma 2.6-5
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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Proof of Lemma 2.6-5
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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Proof of Lemma 2.6-6
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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Proof of Lemma 2.6-7
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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Proof of Lemma 2.6-8
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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Approximation Algorithm&Analysis
Lemma 5.4 w(*)4c(Y*)
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Approximation Algorithm&Analysis
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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Approximation Algorithm&Analysis
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