Minimizing Interference in Wireless Networks. Yves Brise, ETH Zürich, Joint work with Marek, Michael, and Tobias.

Transcription

1 Minimizing Interference in Wireless Networks Yves Brise, ETH Zürich, Joint work with Marek, Michael, and Tobias.

2 The Problem

3 The Problem

4 Formal Description Input: Output: Set V of n vertices in R d Assignment of radii r v, v V Communication Network: Graph G =(V, E) symmetric E := {{v, w} r v v w and r w v w } E := {(v, w) r v v w } asymmetric Interference: Conditions: I(v) := {w V \{v} r w v w } 1. G is (strongly) connected 2. Minimize max v V I(v)

5 Summary 1D, symmetric Exponential Node Chain: For d = 1 and some n N, setv := {2 i } n 1 i=0. (log scale) v 1... v 2 v n von Rickenbach, Schmid, Wattenhofer, Zollinger, 2005: For d = 1, I n is a worst-case lower bound, and I O( n) is an upper bound. Haldórsson, Tokuyama, 2008: For d = 2, there is an algorithm, achieving I O( n)

6 Upper Bound C V ; while ( C > 1) { c C connect to nearest component from sink(c); update C; } The sink of a component is the set of vertices that can be reached from anywhere in the component. The number of rounds is at most log n, andineach round the interference is increased by 2 at most.

7 Sink Algorithm Remarks 1. In the end, we just increase the radius of one vertex in the final sink (adding one more interference). 2. Works in any dimension. The bound depends on the Kissing number, K d. It is something like log n + K d The Kissing number is the number of unit spheres that can simultaneously touch another unit sphere without intersecting.

8 Sink Alg is not Optimal Alternating Exponential Node Chain: For d = 1 and some n N, setv := {( 2) i } n 1 i=1 {0} i 2 i 2 i Delay the activation of group i until we are ready to draw the arc from i 1. Sink Alg produces Ω(log n) interferencevs.o(1).

9 Upper Bound is tight Cantor Construction: Repeatedly delete middle third of all intervals After k iterations there are n =2 k+1 vertices. Lemma: For d 1 there exist point sets of size n that require Ω(log(n)) interference to be connected.

10 For d>1 the Problem is hard Theorem (Buchin, 2008): Deciding whether a set of vertices in the plane allows for a symmetric network with interference at most 3 is NP-complete. Theorem: Deciding whether a set of vertices in the plane allows for an asymmetric network with interference at most 4 is NP-complete.

11 NP-completeness Grid graph G =(V, E) with V Z Z and (v, w) E iff v w = 1. Maximum degree (G) 3. Find Hamilton path! Lemma (Papadimitriou, Vazirani, 1984): In a grid graph G with (G) 3, it is NP-hard to decide whether there exists a Hamilton path.

12 NP-completeness Vertex gadget: Nearest neighbors gives interference 3. Theorem (Buchin, 2008): Deciding whether a set of vertices in the plane allows for a network with interference at most 3 is NP-complete.

13 NP-completeness If there is a Hamilton path interference at most 3 Connect along the path using satellites. If the interference is at most 3 Hamilton path 1. Gadgets may not connect other than through partners. 2. A gadget may connect to at most two neighbors.

14 NP-completeness Problem: Blowing up the radius of main vertices to 3/4 gives a connected network with interference 3. Solution: More complicated gadget! Inhibitor Essentially the same result, but basic interference is 4.

15 Summary & Open Problems d =1 Worst-case: n Algorithm: O( n) Approximation: O( 4 n) Worst-case: log n 1 Algorithm: log n + 4 Approximation: O(log n) d 2 NP-complete Algorithm: O( n) (only d = 2) Hard to approximate in 4/3 NP-complete Algorithm: O(log n + K d + 2) Hard to approximate in 5/4 Complexity of the 1d problem Proper approximation algorithm

16 This is the end!

17 Naive Approach Exponential Node Chain: For d = 1 and some n N, setv := {2 i } n 1 i=0. Try connecting linearly! v 1... v 2 v n (log scale) We call this approach, the linearly connected network, or LIN for short. I(v 1 )=I(LIN) =n 2

18 Hub Based Approach G =(V, E) :=(V, ); h v 1 ; for (i = 2; i n; ++i) { E E {{h, v i }}; if (I(G) increased){ h v i ; } } return G; A hub is any vertex that has an incident edge to the right in the resulting graph. All hubs interfere with v 1. Each new hub connects to one more vertices that are to the right of it. h v i v 1 v 2 v 3 v 4 v 5 v 6 v 7 v 8 v 9 v 10 v 11 I v 12

19 Hub Based Approach Lemma (von Rickenbach, Schmid, Wattenhofer, Zollinger, 2005): For the exponential node chain, the hub based approach results in an interference I(HUB) O( n). Proof Let the current graph have interference I, then,immediately after the determination of a new hub, this hub can be connected to I 1 vertices to its right, before increasing the interference again. n = I 1 i=1 i +2= 1 2 I2 1 2 I +2 I O( n)

20 Hub is Optimal Lemma (von Rickenbach, Schmid, Wattenhofer, Zollinger, 2005): For the exponential node chain, with V = n, n is a lower bound for the interference. Proof Partition V = H S into hubs and non-hubs. Property 1: I(G) H 1 Property 2: I(G) (G) Assume G with I(G) n 1 H n. Also, S n( n 3) + 2 H + S n 2 n + 2. Contradiction to H + S = n.

21 Restraining the Radii Theorem (von Rickenbach, Schmid, Wattenhofer, Zollinger, 2005): For d = 1, there is an algorithm, achieving I O( ) Theorem (von Rickenbach, Schmid, Wattenhofer, Zollinger, 2005): For d = 1, there is an algorithm, that approximates the optimal interference within a factor of O( 4 ) Theorem (von Haldórsson, Tokuyama, 2008): For d = 2, there is an algorithm, achieving I O( ) The parameter is the maximum degree of the initial graph.