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Introduction Wireless Ad-Hoc Network  Set of transceivers communicating by radio.

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Presentation on theme: "Introduction Wireless Ad-Hoc Network  Set of transceivers communicating by radio."— Presentation transcript:

1 Introduction Wireless Ad-Hoc Network  Set of transceivers communicating by radio

2 Introduction Wireless Ad-Hoc Network  Each transceiver has a transmission power  which results in a transmission range

3 Introduction Wireless Ad-Hoc Network  Transceiver receives transmission from only if

4 Introduction Wireless Ad-Hoc Network  As a result a directed communication graph is  induced

5 Model & Problems Definition  A set of transceivers

6 Model & Problems Definition  A set of transceivers  is the power assignment

7 Model & Problems Definition  A set of transceivers  is the power assignment

8 Model & Problems Definitions  A set of transceivers  is the communication graph  is the power assignment

9 Model & Problems Definitions  A set of transceivers  is the communication graph  is the power assignment  is the cost of the assignment

10 Outline Connectivity problems Bounded hop broadcast Spanners Interference-free broadcast

11  paths connecting to  A graph is k-vertex-connected if for  any two nodes there exist k-vertex-disjoint Connectivity Definitions 2-vertex-connected

12  there exists so that  strongly connected and for each  For graph, a subset is a  connected backbone if restricted to is Connectivity Definitions Connected backbone

13 Connectivity Problem 1 (k-vertex-connectivity) Input:A set of transceivers, and a parameter Output:A power assignment with minimal possible cost, where is k-vertex connected

14 Connectivity Problem 1 (k-vertex-connectivity) Input:A set of transceivers, and a parameter Output:A power assignment with minimal possible cost, where is k-vertex connected -approximation algorithm

15 Connectivity Problem 2 (connected backbone) Input:A set of transceivers Output:A subset of and a power assignment with minimal possible cost, where (restricted to ) is strongly connected, and for each, there exists, such that

16 Connectivity Problem 2 (connected backbone) Input:A set of transceivers Output:A subset of and a power assignment with minimal possible cost, where (restricted to ) is strongly connected, and for each, there exists, such that Constant-factor approximation algorithm in time

17  nodes to Fault-Tolerant Power Assignment Definitions  For each, let be a set of closest

18  nodes to Fault-Tolerant Power Assignment Definitions  For each, let be a set of closest

19  nodes to Fault-Tolerant Power Assignment Definitions  For each, let be a set of closest  Let

20  Compute an of  Assign each the range (denote ) Fault-Tolerant Power Assignment The algorithm

21  Compute an of  Assign each the range (denote ) Fault-Tolerant Power Assignment The algorithm

22  For each edge of increase the range  of the nodes in such that each node Fault-Tolerant Power Assignment The algorithm  can reach all nodes in, and vice versa (denote )

23  For each edge of increase the range  of the nodes in such that each node Fault-Tolerant Power Assignment The algorithm  can reach all nodes in, and vice versa (denote )

24  Let  In each is assigned at most Fault-Tolerant Power Assignment Proof sketch  Case 1:

25  Let  In each is assigned at most Fault-Tolerant Power Assignment Proof sketch  Case 1:

26  Let  In each is assigned at most Fault-Tolerant Power Assignment Proof sketch  Case 2:

27  Let  In each is assigned at most Fault-Tolerant Power Assignment Proof sketch  Case 2:

28  Let  In each is assigned at most Fault-Tolerant Power Assignment Proof sketch  Easy to see

29  Let  In each is assigned at most Fault-Tolerant Power Assignment Proof sketch  Easy to see  Kirousis et al. proved

30  Let  In each is assigned at most Fault-Tolerant Power Assignment Proof sketch  Easy to see  Kirousis et al. proved  As a result and since degree of MST is constant

31  Given the of, for any node, let Connected Backbone Power Assignment Definitions  be the size of the longest edge adjacent to

32  Given the of, for any node, let Connected Backbone Power Assignment Definitions  be the size of the longest edge adjacent to

33  Compute an of Connected Backbone Power Assignment The algorithm

34  Compute an of Connected Backbone Power Assignment The algorithm

35  Compute an of Connected Backbone Power Assignment The algorithm  Let be the set of all internal nodes of

36  Compute an of Connected Backbone Power Assignment The algorithm  Let be the set of all internal nodes of  Assign each with (denote )

37  Compute an of Connected Backbone Power Assignment The algorithm  Let be the set of all internal nodes of  Assign each with (denote )

38  Compute an of Connected Backbone Power Assignment The algorithm  Let be the set of all internal nodes of  Assign each with (denote )

39  Construct a power assignment for which Connected Backbone Power Assignment Proof sketch  it holds and,  as a result obtaining  is derived from

40  For each node let be the transmission  Let be the connected backbone in Connected Backbone Power Assignment Proof sketch  range of in

41  For each node let be all the nodes  within distance from Connected Backbone Power Assignment Proof sketch

42  For each node let be all the nodes  within distance from Connected Backbone Power Assignment Proof sketch

43  For each node compute of Connected Backbone Power Assignment Proof sketch  For each node let be all the nodes  within distance from

44  For each node compute of Connected Backbone Power Assignment Proof sketch  For each node let be all the nodes  within distance from

45 Connected Backbone Power Assignment Proof sketch  In : Each node is assigned

46 Connected Backbone Power Assignment Proof sketch  In : Each node is assigned

47 Connected Backbone Power Assignment Proof sketch  In : Each node is assigned

48 Connected Backbone Power Assignment Proof sketch  Carmi et al. showed that

49 Connected Backbone Power Assignment Proof sketch  Carmi et al. showed that

50 Connected Backbone Power Assignment Proof sketch  Carmi et al. showed that

51 Connected Backbone Power Assignment Proof sketch  Carmi et al. showed that + + +

52 Connected Backbone Power Assignment Proof sketch  Carmi et al. showed that Using this and is at least longest edge in we obtain

53 Connected Backbone Power Assignment Proof sketch  Kirousis et al. proved that given an assigning each node with yields a 2-factor approximation for strong-connectivity (denote )

54 Connected Backbone Power Assignment Proof sketch  Kirousis et al. proved that given an assigning each node with yields a 2-factor approximation for strong-connectivity Using this fact we obtain (denote )

55 Connected Backbone Power Assignment Proof sketch  Therefore,

56  at if there is a path from to any Broadcast  A graph is a broadcast graph rooted

57  at if there is a path from to any Broadcast  A graph is a broadcast graph rooted

58  graph rooted at if there is a path from to any Broadcast  A graph is a h-bounded-hop broadcast  and the number of hops is limited by 4-bounded-hop broadcast

59  it remains h-bounded-hop broadcast graph Broadcast  A graph is a k-h-broadcast graph if 2-4-bounded-hop broadcast  even with the removal of up to nodes

60 Broadcast 2-vertex disjoint paths under 4 hops  it remains h-bounded-hop broadcast graph  A graph is a k-h-broadcast graph if  even with the removal of up to nodes

61 Broadcast 2-vertex disjoint paths under 4 hops  it remains h-bounded-hop broadcast graph  A graph is a k-h-broadcast graph if  even with the removal of up to nodes

62 root node and parameters Problem 3 (k-h-bounded broadcast) Input:A set of transceivers in, Output:A power assignment so that is k-h-broadcast and is minimized

63  is 1-h-bounded hop graph The Algorithm Planar Case  Take a power assignment so that

64  is 1-h-bounded hop graph The Algorithm Planar Case  Take a power assignment so that  Let be a directed spanning tree of Max distance – h hops

65  is 1-h-bounded hop graph The Algorithm Planar Case  Take a power assignment so that  Let be a directed spanning tree of Max distance – h hops

66 The Algorithm Planar Case  Add edges from to its grandchildren

67  Remove edges from the children of The Algorithm Planar Case  Add edges from to its grandchildren

68  Remove edges from the children of The Algorithm Planar Case  Add edges from to its grandchildren Max distance – h-1 hops  Denote the resulting tree

69 The Algorithm Planar Case  No power is assigned yet!  We have a skeleton with a bounded cost

70 The Algorithm Planar Case  Assign

71 The Algorithm Planar Case  Assign

72 The Algorithm Planar Case  For each directed edge in  increase the range of all nodes in to  reach all nodes in

73 The Algorithm Planar Case  For each directed edge in  increase the range of all nodes in to  reach all nodes in

74 The Algorithm Planar Case  For each directed edge in  increase the range of all nodes in to  reach all nodes in

75 The Algorithm Planar Case  For each directed edge in  increase the range of all nodes in to  reach all nodes in

76 The Algorithm Planar Case  For each directed edge in  increase the range of all nodes in to  reach all nodes in

77 The Algorithm Planar Case  For each directed edge in  increase the range of all nodes in to  reach all nodes in

78 The Algorithm Planar Case  For each directed edge in  increase the range of all nodes in to  reach all nodes in

79 The Algorithm Planar Case  For each directed edge in  increase the range of all nodes in to  reach all nodes in

80 The Algorithm Planar Case  For each directed edge in  increase the range of all nodes in to  reach all nodes in

81 The Algorithm Planar Case  For each directed edge in  increase the range of all nodes in to  reach all nodes in

82 The Algorithm Planar Case  Denote the resulting power assignment

83 The Algorithm Planar Case  Denote the resulting power assignment  Along each path in there are  vertex-disjoint paths in of at most hops

84 Analysis  increase of is bounded by:  For a single edge in the power

85 Analysis  increase of is bounded by:  For a single edge in the power

86 Analysis  increase of is bounded by:  For a single edge in the power Power assignment in

87 Analysis Planar Case  increase of is bounded by:  For a single edge in the power  Node can be in many -s

88 Analysis Planar Case  increase of is bounded by:  For a single edge in the power  Node can be in many -s, with many edges

89 Analysis Planar Case  increase of is bounded by:  For a single edge in the power  Node can be in many -s, with many edges  But eventually only one ‘dominates’ the bound

90 Analysis Planar Case  A node can be dominated only by the  outgoing edges of in

91  nodes (those in )  A single edge can dominate at most Analysis Planar Case  A node can be dominated only by the  outgoing edges of in

92  nodes (those in )  A single edge can dominate at most Analysis  A node can be dominated only by the  outgoing edges of in  Recall,

93  nodes (those in )  A single edge can dominate at most Analysis  A node can be dominated only by the  outgoing edges of in  As a result,

94 Analysis

95

96 Due to

97 Analysis PTAS due to Funke and Laue [24]

98 Analysis  for the k-h-broadcast problem  Let be the optimal power assignment  From,

99 Analysis  for the k-h-broadcast problem  Let be the optimal power assignment We need to bound  From,

100 Analysis  node has at least neighbors  Let be a power assignment so that each  Clearly,

101 Analysis -  Hamiltonian cycle based power  assignment for the k-(n-1)-broadcast  problem, so that

102  Hamiltonian cycle based power  assignment for the k-(n-1)-broadcast  problem, so that  In each node has at least neighbors Analysis -

103  Hamiltonian cycle based power  assignment for the k-(n-1)-broadcast  problem, so that  In each node has at least neighbors  From, Analysis – ( can be shown)

104 The Algorithm k-(n-1)-broadcast

105 The Algorithm  Compute an MST of k-(n-1)-broadcast

106 The Algorithm  Construct a Hamiltonian cycle with cost  Compute an MST of k-(n-1)-broadcast

107 The Algorithm  Construct a Hamiltonian cycle with cost  Compute an MST of  Assign each node to reach nodes  in both directions of the cycle  Example: k=4 k-(n-1)-broadcast

108 The Algorithm  Construct a Hamiltonian cycle with cost  Compute an MST of  Assign each node to reach nodes  in both directions of the cycle   As a result, k-(n-1)-broadcast

109 Hamiltonian Cycle Stage  Compute an MST of k-(n-1)-broadcast

110 Hamiltonian Cycle Stage  Compute an MST of  Apply MST-Augmentation (Calinescu and Wan) k-(n-1)-broadcast

111 Hamiltonian Cycle Stage  Compute an MST of  Apply MST-Augmentation (Calinescu and Wan) k-(n-1)-broadcast

112 Hamiltonian Cycle Stage  Compute an MST of  Apply MST-Augmentation (Calinescu and Wan) k-(n-1)-broadcast

113 Hamiltonian Cycle Stage  Compute an MST of  Apply MST-Augmentation (Calinescu and Wan) k-(n-1)-broadcast

114 Hamiltonian Cycle Stage  Compute an MST of  Apply MST-Augmentation (Calinescu and Wan) k-(n-1)-broadcast

115 Hamiltonian Cycle Stage  Compute an MST of  Apply MST-Augmentation (Calinescu and Wan) k-(n-1)-broadcast

116 Hamiltonian Cycle Stage  Compute an MST of  Apply MST-Augmentation (Calinescu and Wan) 2-strongly connected undirected graph k-(n-1)-broadcast

117 Hamiltonian Cycle Stage  Compute an MST of  Apply MST-Augmentation (Calinescu and Wan)  Apply TSP-Approx (Bender and Checkuri) Square of every biconnected graph is Hamiltonian (Fleischner) k-(n-1)-broadcast

118 Hamiltonian Cycle Stage  Compute an MST of  Apply MST-Augmentation (Calinescu and Wan)  Apply TSP-Approx (Bender and Checkuri) The cost of the Hamiltonian cycle  As a result, k-(n-1)-broadcast

119  A simple approximation due to: Analysis -  For any it holds: Back to k-h-broadcast

120 Analysis -  Take as before Back to k-h-broadcast

121 Analysis -  The most distant node at most hops away  Take as before Back to k-h-broadcast

122 Analysis -  The most distant node at most hops away  Take as before  Assign the root to reach all! Back to k-h-broadcast

123 Spanners What is a spanner?  A spanning subgraph that approximates  some measure of the original graph

124 Spanners What is a spanner?  A spanning subgraph that approximates  some measure of the original graph E.g., Euclidean distance

125 Spanners What is a spanner?  A spanning subgraph that approximates  some measure of the original graph E.g., Euclidean distance

126 Spanners What is a spanner?  A spanning subgraph that approximates  some measure of the original graph E.g., Euclidean distance times longer than in Shortest path is at most

127 Spanners What is a spanner?  A spanning subgraph that approximates  some measure of the original graph E.g., Euclidean distance times longer than in Shortest path is at most stretch factor

128 Spanners  We propose two spanner optimization measures  Distance – reducing transmission latency  Energy – increasing network lifetime

129 Spanner optimization measures The original graph  Let be the wireless nodes in the plane  Let be a weighted complete  graph  Weight function: The Euclidean distance

130 Spanner optimization measures The original graph  Let be the wireless nodes in the plane  Let be a weighted complete  graph  Weight function: Proportional to the energy required to transmit from to The Euclidean distance

131 Spanner optimization measures The original graph  Let be the wireless nodes in the plane  Let be a weighted complete  graph  Weight function:

132 Spanner optimization measures The spanner  Let p be a power assignment

133 Spanner optimization measures The spanner  Let p be a power assignment  is an induced directed graph,  where

134 Spanner optimization measures The spanner  Let p be a power assignment  is an induced directed graph,  where  The cost:

135 Spanner optimization measures Energy measure (stretch factor)  The energy of some path is its weight

136 Spanner optimization measures Energy measure (stretch factor)  The energy of some path is its weight  The minimum energy from to in 

137 Spanner optimization measures Energy measure (stretch factor)  The energy of some path is its weight  The minimum energy from to in 

138 Spanner optimization measures Energy measure (stretch factor)  The energy stretch factor of

139 Spanner optimization measures Energy measure (stretch factor)  The energy stretch factor of   We aim to minimize both and

140 Spanner optimization measures Energy measure (stretch factor)  The energy stretch factor of   Clear benefits  Prolonged network lifetime  Low cost  Low interference…  We aim to minimize both and

141 Spanner optimization measures Distance measure (stretch factor)  The distance of some path

142  The minimum distance from to in Spanner optimization measures Distance measure (stretch factor)  The distance of some path 

143 Spanner optimization measures Distance measure (stretch factor)  The distance stretch factor of 

144 Spanner optimization measures Distance measure (stretch factor)  The distance stretch factor of   We aim to minimize both and

145 Spanner optimization measures Distance measure (stretch factor)  The distance stretch factor of   Clear benefits  Low delay in message delivery  Low cost  We aim to minimize both and

146 Main results Preliminaries  We consider a random, independent, and  uniform node distribution in a unit square The probability of our results converges to 1 as the number of nodes, n, increases

147 Main results Preliminaries  Spanners make sense only if the induced  graph is strongly connected  uniform node distribution in a unit square  We consider a random, independent, and

148 Main results Preliminaries  Spanners make sense only if the induced  graph is strongly connected  uniform node distribution in a unit square Otherwise, the stretch factor is infinity Path does not exist  We consider a random, independent, and

149 Main results Preliminaries  Spanners make sense only if the induced  graph is strongly connected  uniform node distribution in a unit square Þ The cost of any spanner is at least Þ the minimum cost of strong connectivity  We consider a random, independent, and

150 Main results Preliminaries  Spanners make sense only if the induced  graph is strongly connected  uniform node distribution in a unit square Þ The cost of any spanner is at least Þ the minimum cost of strong connectivity Þ (denote this cost )  We consider a random, independent, and

151 Main results Energy spanner  Develop power assignment so that  where,,

152 Main results Distance spanner  Develop a power assignment so that = O(1)

153 Technical details Some bounds…  Using [Zhang and Hou ‘05] Lower bound on the cost of any spanner

154 Technical details Some bounds…  Using [Zhang and Hou ‘05]  From [Kirousis et al. ‘00]  Minimum spanning tree of G The weight of the tree

155 Technical details Some bounds…  Using [Zhang and Hou ‘05]  From [Kirousis et al. ‘00]   Using [Berend et al. ‘08] & [Penrose ‘97] Maximum length edge of MST

156 Technical details Energy spanner [power assignment]

157 Technical details Energy spanner [power assignment]  Find the minimum spanning tree (MST)

158 Technical details Energy spanner [power assignment]  Lemma: We can find nodes  so that any node is within  Find the minimum spanning tree (MST)  hops from some node in U

159 Technical details Energy spanner [power assignment]  Lemma: We can find nodes  so that any node is within  Find the minimum spanning tree (MST)  hops from some node in U Take diameter

160 Technical details Energy spanner [power assignment]  Lemma: We can find nodes  so that any node is within  Find the minimum spanning tree (MST)  hops from some node in U Take diameter Add the -th node to U

161 Technical details Energy spanner [power assignment]  Lemma: We can find nodes  so that any node is within  Find the minimum spanning tree (MST)  hops from some node in U Take diameter Add the -th node to U Remove first nodes from the diameter

162 Technical details Energy spanner [power assignment]  Lemma: We can find nodes  so that any node is within  Find the minimum spanning tree (MST)  hops from some node in U Take diameter Add the -th node to U Remove first nodes from the diameter

163 Technical details Energy spanner [power assignment]  Lemma: We can find nodes  so that any node is within  Find the minimum spanning tree (MST)  hops from some node in U Take diameter Add the -th node to U Remove first nodes from the diameter

164 Technical details Energy spanner [power assignment]  Lemma: We can find nodes  so that any node is within  Find the minimum spanning tree (MST)  hops from some node in U Take diameter Add the -th node to U Remove first nodes from the diameter

165 Technical details Energy spanner [power assignment]  Lemma: We can find nodes  so that any node is within  Find the minimum spanning tree (MST)  hops from some node in U Take diameter Add the -th node to U Remove first nodes from the diameter

166 Technical details Energy spanner [power assignment]  Lemma: We can find nodes  so that any node is within  Find the minimum spanning tree (MST)  hops from some node in U

167 Technical details Energy spanner [power assignment]  Lemma: We can find nodes  so that any node is within  Find the minimum spanning tree (MST)  hops from some node in U  Let be a LAST rooted at  LAST [Khuller et al. ’93] is a spanning tree T of G, rooted at some so that and

168 Technical details Energy spanner [power assignment]  Define the power assignment p so that

169 Technical details Energy spanner [power assignment]  Define the power assignment p so that  Let

170 Technical details Energy spanner [power assignment]  Define the power assignment p so that  Let   Finally, For technical reasons

171 Technical details Energy spanner [cost analysis]

172 Technical details Energy spanner [stretch analysis]  If, there is a path P in G, so that  and

173 Technical details Energy spanner [stretch analysis]  If, there is a path P in G, so that  and  Therefore, since for every u,  path P also exists in

174 Technical details Energy spanner [stretch analysis]  Otherwise,

175 Technical details Energy spanner [stretch analysis]  For any two nodes, s and t, the path in  first arrives at some LAST origin  by using the MST edges (denote P’)

176 Technical details Energy spanner [stretch analysis]  For any two nodes, s and t, the path in  first arrives at some LAST origin  by using the MST edges (denote P’)

177 Technical details Energy spanner [stretch analysis]  For any two nodes, s and t, the path in  first arrives at some LAST origin  by using the MST edges (denote P’)  second travels through the edges of  from to t (denote P’’)

178 Technical details Energy spanner [stretch analysis]  For any two nodes, s and t, the path in  first arrives at some LAST origin  by using the MST edges (denote P’)  second travels through the edges of  from to t (denote P’’)

179 Technical details Energy spanner [stretch analysis]  Otherwise,  We bound the weight of P’ and P’’  Maximum edge of MST Lemma

180 Technical details Energy spanner [stretch analysis]  Otherwise,  We bound the weight of P’ and P’’ A possible path goes through s

181 Technical details Energy spanner [stretch analysis]  Otherwise,  We bound the weight of P’ and P’’  Eventually,

182 Technical details Distance spanner [power assignment]  The general idea is that for uniformly  distributed nodes, we can always find  “good” relays between any pair of nodes

183 Technical details Distance spanner [power assignment]  To find these relays, for any pair of nodes,  s and t, we start a recursive process

184 Technical details Distance spanner [power assignment]  To find these relays, for any pair of nodes,  s and t, we start a recursive process  At step i, we place adjacent disks along  the edge The diameter of a disk at step i is

185 Technical details Distance spanner [power assignment]  To find these relays, for any pair of nodes,  s and t, we start a recursive process  At step i, we place adjacent disks along  the edge The diameter of a disk at step i is

186 Technical details Distance spanner [power assignment]  To find these relays, for any pair of nodes,  s and t, we start a recursive process  At step i, we place adjacent disks along  the edge The diameter of a disk at step i is

187 Technical details Distance spanner [power assignment]  To find these relays, for any pair of nodes,  s and t, we start a recursive process  At step i, we place adjacent disks along  the edge The diameter of a disk at step i is The process ends when one of the disks has no relay nodes

188 Technical details Distance spanner [power assignment]  To find these relays, for any pair of nodes,  s and t, we start a recursive process  At step i, we place adjacent disks along  the edge  Finally, we use relay nodes to obtain a path We use an arbitrary node in each disk at the last non-empty step

189  The power assignment p is obtained by  ensuring that all paths are in Technical details Distance spanner [power assignment]

190  The power assignment p is obtained by  ensuring that all paths are in Technical details Distance spanner [power assignment]  Let be the constructed path from s to t

191  The power assignment p is obtained by  ensuring that all paths are in Technical details Distance spanner [power assignment]  Let be the constructed path from s to t  And be all  the edges from u in all the paths

192  The power assignment p is obtained by  ensuring that all paths are in Technical details Distance spanner [power assignment]  Let be the constructed path from s to t  And be all  the edges from u in all the paths  Finally,

193 Technical details Distance spanner [analysis]  Lemma: Let D be the maximum radius  disk which can be placed inside the unit  square, so there are no nodes in D Let r be the radius of D

194 Technical details Distance spanner [analysis]  Lemma: Let D be the maximum radius  disk which can be placed inside the unit  square, so there are no nodes in D Let r be the radius of D  Then,

195 Technical details Distance spanner [analysis]  From Lemma,

196 Technical details Distance spanner [analysis]  From Lemma,  Clearly,

197 Extended wireless network model Power assignment  The lifetime of node v is  Each node has an initial battery charge b(v)  Nodes have no fixed power supply  The network lifetime is

198 Wireless network model Power assignment  a power assignment p  Interference is a direct consequence of ?

199 Wireless network model Power assignment  Several interference models exist  a power assignment p  Number of nodes affected by transmission  Number of edges affected by transmission  Interference is a direct consequence of

200 Wireless network model Power assignment  Several interference models exist  a power assignment p  Number of nodes affected by transmission  Number of edges affected by transmission  We combine several common models by  defining the interference to be  Interference is a direct consequence of

201 Main results Contribution  We develop two power assignments: and

202 Main results Contribution  We develop two power assignments: and  can be computed in time and  where n is the number of nodes

203 Main results Contribution  We develop two power assignments: and  can be computed in time  where n is the number of nodes and

204 Technical details The construction  The first power assignment is local

205 Technical details The construction  The first power assignment is local  To compute simply assign  to all u  We use a Lemma from [Shpungin and Segal ’09]  to prove the correctness of this power assignment

206 Technical details The construction  The first power assignment is local  To compute simply assign  to all u  We use a Lemma from [Shpungin and Segal ’09]  to prove the correctness of this power assignment  (due to uniform distribution a path always exists)

207 Technical details The construction  The second power assignment is computed  by dividing the unit square into k grid cells

208 Technical details The construction  The second power assignment is computed  Then we compute a k shortest path trees rooted  at an arbitrary node in each cell  by dividing the unit square into k grid cells

209 Technical details The construction  The second power assignment is computed  Then we compute a k shortest path trees rooted  at an arbitrary node in each cell  The power assignment of nodes is increased to  assure all these k trees are included  by dividing the unit square into k grid cells

210 Technical details The construction  The second power assignment is computed  Then we compute a k shortest path trees rooted  at an arbitrary node in each cell  The power assignment of nodes is increased to  assure all these k trees are included  by dividing the unit square into k grid cells  The power assignment of nodes is increased again  to be at least


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