Minimizing the Maximum Interference is Hard

Minimizing the Maxim um In terference is Hard Kevin Buc hin ∗ Abstract W e consider the following interference mo del for wireless sensor and ad ho c net works: the receiv er interference of a node is the num b er of transmission ranges it lies in. W e model transmission ranges as disks. F or this case we sho w that c ho osing transmission radii whic h minimize the maxim um in terference while maintaining a connected symmetric commu- nication graph is NP-complete. 1 In tro duction Limiting the interference b etw een nodes in a sensor netw ork is substantial for the energy-efficiency of the net work. A common approac h to reduce interference is top olo gy c ontr ol , i.e., restricting the communication graph (see [2, 4]). A theoretical problem in top ology control which has b een stated as essen tial to understanding sensor netw orks is the following. Problem 1 (Lo cher, v on Rick enbac h, W attenhofer [5]) . Given n no des in the plane. Conne ct the no des by a sp anning tr e e. F or e ach no de v we c onstruct a disk c entering at v with r adius e qual to the distanc e to v ’s furthest neighb or in the sp anning tr e e. The interfer enc e of a no de v is then define d as the numb er of disks that include no de v (not c ounting the disk of v itself ). Find a sp anning tr e e that minimizes the maximum interfer enc e. The c hoice of the radii as given in the problem statemen t guarantees that the symmetric comm unication graph con tains a spanning tree, i.e., that the symmetric communication graph is connected. The symmetric c ommunic ation gr aph is the undirected graph on the no des with edges b etw een nodes which b oth lie in each others tr ansmission r anges , i.e., in each others circles. W e refer to the radii of the circles as tr ansmission r adii . W e pro ve that Problem 1 is NP-hard. So far no lo w er bounds for the problem w ere known. Halld´ orsson and T okuyama [3] give an algorithm which yields a maxim um in terference in O ( √ n ). An op en problem that remains is to narrow the gap betw een this upper bound and our low er b ound. F or the case of p oin ts ∗ Department of Mathematics and Computer Science, Univ ersity of T ec hnology Eindhov en, kbuchin@tue.nl . I would like to thank Roger W attenhofer and Y uezhou Lv for helpful com- ments. 1 on a line there is a 4 √ n -appro ximation algorithm [7]. In a generalized version of the problem there is a p ositive real v alue asso ciated with each (ordered) pair of no des, and the first node can send a message to the second node (but will also in terfere with it) if its transmission p ow er is ab ov e this v alue. In this version an appro ximation within less than a logarithmic factor in polynomial time is not p ossible unless NP has slightly sup erp olynomial time algorithms [1]. 2 NP-Completeness In this section we pro ve that deciding whether the maximum interference of a net work is at most 3 is NP-complete. Strictly speaking, this implies that the in terference in Problem 1 cannot b e approximated within a factor less than 4 / 3 efficien tly , since it is not possible to distinguish betw een interference 3 and 4 in p olynomial time unless P = N P . W e prov e the NP-hardness b y a polynomial reduction from the problem of finding a Hamilton p ath in a grid gr aph of maxim um degree 3. A (vertex- induced) grid graph is a graph for whic h the v ertex set is a finite subset of the t wo-dimensional integer grid Z × Z and there is an edge b etw een t wo vertices x, y exactly if x, y are neighbors on the grid, i.e., k x − y k = 1. W e iden tify the corresp onding edge with the line segment from x to y . A Hamilton path in a graph is a path in the graph with every v ertex lying exactly once on the path. Deciding whether a Hamilton path in a grid graph with maximum degree 3 exists is NP-hard [6]. F or the reduction w e need for an y grid graph of maxim um degree 3 a p oly- nomial construction of a set of nodes suc h that there is a Hamilton path in the grid graph exactly if there is a spanning tree with maximum interference at most 3. W e ma y assume that the grid graph has no isolated v ertex b ecause in that case there is no Hamilton path and we can c heck this in linear time. A vertex x ∈ Z × Z of the grid graph is represented by a set of no des (which w e call a vertex gadget ) containing the follo wing nodes: • a c enter no de : a node at position x , • satel lite no des : three further no des at three disjoint positions from the set { x ± (0 , 1 / 4) , x ± (1 / 4 , 0) } . The satellites are c hosen such that the vertex gadget has a satellite no de on every edge at x of the grid graph. F or every v ertex of degree less than 3, the remaining satellite is placed such that the distance b etw een the remaining satellites of neighboring degree-2 v ertices is larger than 1. This can b e achiev ed by greedily placing the remaining satellites for each chain of degree-2 vertices. Tw o satellites (from different v ertex gadgets) on the same edge of the grid graph are called p artners . Figure 1 sho ws a grid graph of maximum degree 3 and a corresp onding no de set. F or the NP-hardness we need to prov e that the grid graph has a Hamilton path exactly if the corresp onding no de set has a spanning tree with interference 2 Figure 1: Example of a grid graph and a corresponding set of no des. at most 3. W e get one of the implications by constructing such a tree from a Hamilton path. Lemma 1. If a grid gr aph has a Hamilton p ath then the c orr esp onding set of no des has a sp anning tr e e with interfer enc e 3 . Pr o of. Giv en a grid graph with Hamilton path w e can construct a spanning tree with in terference 3 in the following wa y: F or an arbitrary Hamilton path • connect each cen ter no de to its satellites, • connect satellite partners if they lie on an edge of the Hamilton cycle. Cen ter nodes and satellite no des without a partner ha ve transmission radius 1 / 4, while satellites with partners ha ve transmission radius 1 / 2. This yields the following interferences: A center no de is in the transmission range of its satellites. It is not in the transmission range of any other no de since it has distance at least 3 / 4 to any other no de. Thus the interference at a center no de is 3. A satellite is in the transmission range of the cen ter node. It is in the transmission range of any (other) satellite in its v ertex gadget that connects to a partner. If it connects to its partner, it is in the transmission range of the partner. There can b e no further interference at a satellite since all other no des ha ve distance at least 3 / 4 to the satellite. In a (Hamilton) path every vertex of the grid graph is connected to at most t wo other vertices. Therefore in a v ertex gadget at most tw o satellites connect to their partners. This yields an interference of at most 3 at satellites. Next we show that if the interference induced by a spanning is at most 3 then in the spanning tree v ertex gadgets may only connect through partners. Lemma 2. Assume a grid gr aph has no isolate d vertic es. If a sp anning tr e e on the c orr esp onding set of no des has an e dge b etwe en two differ ent vertex gadgets other than an e dge b etwe en p artners then ther e is a no de with interfer enc e at le ast 4 . Pr o of. Supp ose a satellite connects to a node which is further aw a y than its partner. In this case it con tains at least one cen ter no de outside of its vertex gadget in its transmission range. Since this cen ter no de will also lie in the transmission ranges of its satellites, it will ha ve interference at least 4. 3 Lemma 3. If the no de set c orr esp onding to a grid gr aph without isolate d ver- tic es has a sp anning tr e e with interfer enc e at most 3 then the grid gr aph has a Hamilton p ath. Pr o of. Supp ose we ha ve a spanning tree in whic h from eac h vertex gadget at most tw o satellites connect to partners. Then this directly giv es us a Hamilton path in the corresp onding grid graph b y simply connecting the vertices in the same w a y as the v ertex gadgets. No w assume there is a spanning tree with interference at most 3 whic h is not of this t yp e. The spanning tree has a vertex gadget that connects to at least three other v ertex gadgets and by Lemma 2 these must be connections from satellites to their partners. Th us all three satellites in the vertex gadget connect to their partners. Now all three satellites lie in the transmission range of their partner, of the other t wo satellites, and of the cen ter node of the gadget. Therefore, the satellites ha v e interference at least 4 con tradicting the assumption of in terference 3. Theorem 1. De ciding whether a set of no des in the plane has a sp anning tr e e with interfer enc e at most 3 is NP-c omplete. Pr o of. The polynomial construction of the no de set from the grid graph together with Lemmas 1 and 3 directly yield the NP-hardness. T o v erify whether a spanning tree has a certain interference it suffices to p erform  n 2  in-circle tests. Th us, the problem is in NP . References [1] D. Bil` o and G. Proietti. On the complexity of minimizing in terference in ad-hoc and sensor netw orks. In Pr o c. 2nd Internat. Workshop A lgorithmic Asp e cts of Wir eless Sensor Networks (ALGOSENSOR) , vol. 4240 of LNCS , pp. 13–24, 2006. [2] K. Buchin and M. Buc hin. T opology con trol. In D. W agner and R. 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