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Line-of-Sight Networks

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.

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Line-of-Sight Networks

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  1. Line-of-Sight Networks Frieze, Kleinberg, Ravi, DebanySODA 2007 Cheng-Chung Li2006/12/06

  2. Outline • Introduction • Connectivity • The Existence of a Giant Component • Finding Paths Between Nodes • Relay Placement: An Approximation Algorithm • Open Questions 52

  3. 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 52

  4. 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 52

  5. 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 52

  6. Suppose there are obstacles • Sensors A,B can not communicate. Need another model B A 52

  7. 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 52

  8. 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 52

  9. 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 kwhp (here, whp is equal to with high probability) 52

  10. Connectivity • Theorem 1.1 Suppose that /lnninf where =n, <6/(8k+7)Let k1 be a fixed positive integer and let p=[(1-0.5)lnn+k(lnlnn)/2+cn]/2. Thenlimninf Pr(G is k-connected)= 0,cn-inf e-k, cnc 1,cninfwhere k=[2k-2(1-0.5)ke-2c]/(k-1)! 52

  11. Some point of the torus that sees nodes in both J and K Come lose to see one another K J Connectivity: The First Part • Here we consider the case when cnc. 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 |S|<k S 52 H

  12. 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 firststage • We then independently place a node at each point with probability p2~1/(2lnn) 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 52

  13. Some Definitions Arm: the set contains  points that are visible from it in a single direction  • 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 10 segment length: /10 2 1 52

  14. Not mighty 1000 Not mighty 2 Not mighty 1 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 52 x

  15. 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 42n2(,1000)p11000n-1000/400=o(1) (x has four arms, and the nodes at  have two arms orthogonal to ) 52

  16. weak deg(v)<lnlnn v w 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 52

  17. Proof of Lemma 2.3 • The probability that H contains such a pair v,w is bounded by n2p1t=1 to lnlnn(4,t)p1t(1-p1)4-t(2n-1/400)=o(1) (There are two arms at  that are orthogonal to vw) v 52

  18. 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 n2p1t=0 to k-1(4,t)p1t(1-p1)4-t(4p2)=o(1) (The red part only considers the red vertices) 52

  19. 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 n2p2t=0 to k-1(4,t)p1t(1-p1)4-t=o(1) 52

  20. 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 xL, let vKx denote the node in K that is closet to x in L1 distance 52

  21. The relation between K and L L HS x3 12 K3 K1 vK1x3 x2 9 vK1x2 K2 x1 6 vK1x1 3 52 3 6 9 12

  22. Where is vKx ? • Lemma 2.6 vKx lies within the * box Bx centered at x x3 12 HS 9 K1 x1  6 vK1x1  3 3 6 52

  23. Proof of Lemma 2.6-1 • Let a red node be pink if it is not in S • We prove this lemma by contradiction v=vK0=(a,b) x=0 52

  24. Proof of Lemma 2.6-2 • By 3, v has at least one arm containing a pink node w w v=vK0=(a,b) x=0 52

  25. Proof of Lemma 2.6-3 • If the deg(v)<lnlnn, then we can use the non-occurrence of 2to argue that two arms of w orthogonal to vw are mighty mighty deg(v)<lnlnn v w  mighty 52

  26. Proof of Lemma 2.6-4 • If deg(v)≥lnlnn, then we can use the non-occurrence of 1to argue that there is a choice of lnlnn-4000 w’s such that the two arms of w orthogonal to vw are mighty w w mighty deg(v)≥lnlnn w w w v w  w mighty w 52

  27. Proof of Lemma 2.6-5 • Case 1: is the South arm of v • It follows that |a’|+|c|a+b+.1-.4, contradiction !  v=vK0=(a,b) |c-b|.1 z=(a’,c) x=0  y=(a,-b’’) w=(a’,-b’’)  ,: mighty a-a’[.4,.5] 52

  28. 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 !  y=(a,b’)  w=(a’,b’) z=(a’,b’’) |b’’-b|.1 v  x=0 a-a’[.4,.5] ,: mighty 52

  29. Proof of Lemma 2.6-6 • Case 2b-1:  is the North arm of v and a</2 w=(a’,b’) y=(a,b’)  |b-b’|.7  z=(a’,b’’) v=(a,b)  |b’’-b’|[.9,] x=0 ,: mighty 52 |a-a’|.1

  30. Proof of Lemma 2.6-7 • Case 2b-2:  is the North arm of v and a</2 w=(a’,b’) y=(a,b’)  |b-b’|>.7  q=(a’,b’’) v=(a,b) z=(a’’,b’’)  |b’-b’’|≥.9  p=(a’’,b’’’) |b’’-b’’’|[.5,.6]  x=0 ,,,: mighty 52 |a-a’|.1 |a’’-a’|.1

  31. 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 52

  32. 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 52

  33. Final Lemma • Lemma 2.7 The points z(J,K,x) and z(J,K,y) are distinct, for distinct points x and y L HS K J 12 y vJy vKy z(J,K,y) vJx 9 x vKx z(J,K,x) 6 3 3 6 52

  34. 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 n2/92 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)^(n2/92)  e^[-n2-3/(20lnn)] 52

  35. Finishing the Proof-2 • There are at most 2 components, since for any fixed point xL, 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 52

  36. 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 cninf the whp there are no vertices of degree less than k (the expected number tends to zero) • And, the argument cnc implies that G will be k-connected whp 52

  37. 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 contains a unique component with (1-o(1))(1-x2c)n2/ vertices, 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) 52

  38. Finding Paths Between Nodes • Theorem Let p=Clnn/ for a constant C3. 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 52

  39. 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 52

  40. Problem Model • As before, we are given an nn 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 xT, and for a set XT we define c(X)=xXcx • 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[XY] 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 52

  41. 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 ) 52

  42. 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 52

  43. 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 52

  44. 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; 52

  45. 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 vX • cv, otherwise • Let cx(Y)=yYcx,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 52

  46. 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 3 5 52

  47. Approximation Algorithm&Analysis • Lemma 5.4 w(*)4c(Y*) 52

  48. 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(’) 52

  49. Approximation Algorithm&Analysis • Finally, we use Y’ as our Steiner set for X • Because c(Y’)w(’)w(*)4c(Y*) • We obtain a bound on c(Y’) relative to the optimum c(Y*) 52

  50. 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) 52

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