Probabilistic Analysis of Power Assignments

Transcription

1 Probabilistic Analysis of Power Assignments Maurits e Graaf 1,2 an Boo Manthey 1 1 University of Twente, Department of Applie Mathematics, Enschee, Netherlans m.egraaf/b.manthey@utwente.nl 2 Thales Neerlan B. V., Huizen, Netherlans arxiv: v1 [cs.ds] 24 Mar 2014 March 25, 2014 A funamental problem for wireless a hoc networks is the assignment of suitable transmission powers to the wireless evices such that the resulting communication graph is connecte. The goal is to minimize the total transmit power in orer to maximize the life-time of the network. Our aim is a probabilistic analysis of this power assignment problem. We prove complete convergence for arbitrary combinations of the imension an the istance-power graient p. Furthermore, we prove that the expecte approximation ratio of the simple spanning tree heuristic is strictly less than its worst-case ratio of 2. Our main technical novelties are two-fol: First, we fin a way to eal with the unboune egree that the communication network inuce by the optimal power assignment can have. Minimum spanning trees an traveling salesman tours, for which strong concentration results are known in Eucliean space, have boune egree, which is heavily exploite in their analysis. Secon, we apply a recent generalization of Azuma-Hoeffing s inequality to prove complete convergence for the case p for both power assignments an minimum spanning trees (MSTs). As far as we are aware, complete convergence for p > has not been prove yet for any Eucliean functional. 1 Introuction Wireless a hoc networks have receive significant attention ue to their many applications in, for instance, environmental monitoring or emergency isaster relief, where wiring is ifficult. Unlike wire networks, wireless a hoc networks lack a backbone infrastructure. Communication takes place either through single-hop transmission or by relaying through intermeiate noes. We consier the case that each noe can ajust its transmit power for the purpose of power conservation. In the assignment of transmit powers, two conflicting effects have to be taken into account: if the transmit powers are too low, the resulting network may be isconnecte. If the transmit powers are too high, the noes run out of energy quickly. The goal of the power assignment problem is to assign transmit powers to the transceivers such that the resulting network is connecte an the sum of transmit powers is minimize [13]. 1

2 1.1 Problem Statement an Previous Results We consier a set of vertices X [0,1], which represent the sensors, X = n, an assume that u v p, for some p R (calle the istance-power graient or path loss exponent), is the power require to successfully transmit a signal from u to v. This is calle the powerattenuation moel, where the strength of the signal ecreases with 1/r p for istance r, an is a simple yet very common moel for power assignments in wireless networks [19]. In practice, we typically have 1 p 6 [16]. A power assignment pa : X [0, ) is an assignment of transmit powers to the noes in X. Given pa, we have an ege between two noes u an v if both pa(x),pa(y) x y p. If the resulting graph is connecte, we call it a PA graph. Our goal is to fin a PA graph an a corresponing power assignment pa that minimizes v X pa(v). Note that any PA graph G = (X,E) inuces a power assignment by pa(v) = max u X:{u,v} E u v p. PA graphs can in many aspects be regare as a tree as we are only intereste in connecteness, but it can contain more eges in general. However, we can simply ignore eges an restrict ourselves to a spanning tree of the PA graph. The minimal connecte power assignment problem is NP-har for 2 an APX-har for 3 [3]. For = 1, i.e., when the sensors are locate on a line, the problem can be solve by ynamic programming [10]. A simple approximation algorithm for minimum power assignments is the minimum spanning tree heuristic (MST heuristic), which achieves a tight worst-case approximation ratio of 2 [10]. This has been improve by Althaus et al. [1], who evise an approximation algorithm that achieves an approximation ratio of 5/3. A first average-case analysis of the MST heuristic was presente by e Graaf et al. [4]: First, they analyze the expecte approximation ratio of the MST heuristic for the (non-geometric, nonmetric) case of inepenent ege lengths. Secon, they prove convergence of the total power consumption of the assignment compute by the MST heuristic for the special case of p =, but not of the optimal power assignment. They left as open problems, first, an average-case analysis of the MST heuristic for ranom geometric instances an, secon, the convergence of the value of the optimal power assignment. Other power assignment problems stuie inclue the k-station network coverage problem of Funke et al. [5], where transmit powers are assigne to at most k stations such that X can be reache from at least one sener, or power assignments in the SINR moel [7,9]. 1.2 Our Contribution In this paper, we conuct an average-case analysis of the optimal power assignment problem for Eucliean instances. The points are rawn inepenently an uniformly from the -imensional unit cube [0,1]. We believe that probabilistic analysis is a better-suite measure for performance evaluation in wireless a hoc networks, as the positions of the sensors in particular if eploye in areas that are ifficult to access are naturally ranom. Roughly speaking, our contributions are as follows: 1. We show that the power assignment functional has sufficiently nice properties in orer to apply Yukich s general framework for Eucliean functionals [25] to obtain concentration results (Section 3). 2. Combining these insights with a recent generalization of the Azuma-Hoeffing boun[24], we obtain concentration of measure an complete convergence for all combinations of an p 1, even for the case p (Section 4). In aition, we obtain complete 2

3 convergence for p for minimum-weight spanning trees. As far as we are aware, complete convergence for p has not been prove yet for such functionals. The only exception we are aware of are minimum spanning trees for the case p = [25, Sect. 6.4]. 3. We provie a probabilistic analysis of the MST heuristic for the geometric case. We show that its expecte approximation ratio is strictly smaller than its worst-case approximation ratio of 2 [10] for any an p (Section 5). Our main technical contributions are two-fol: First, we introuce a transmit power reistribution argument to eal with the unboune egree that graphs inuce by the optimal transmit power assignment can have. The unbouneness of the egree makes the analysis of the power assignment functional PA challenging. The reason is that removing a vertex can cause the graph to fall into a large number of components an it might be costly to connect these components without the remove vertex. In contrast, the egree of any minimum spanning tree, for which strong concentration results are known in Eucliean space for p, is boune for every fixe, an this is heavily exploite in the analysis. (The concentration result by e Graaf et al. [4] for the power assignment obtaine from the MST heuristic also exploits that MSTs have boune egree.) Secon, we apply a recent generalization of Azuma-Hoeffing s inequality by Warnke [24] to prove complete convergence for the case p for both power assignments an minimum spanning trees. We introuce the notion of typically smooth Eucliean functionals, prove convergence of such functionals, an show that minimum spanning trees an power assignments are typically smooth. In this sense, our proof of complete convergence provies an alternative an generic way to prove complete convergence, whereas Yukich s proof for minimum spanning trees is tailore to the case p =. In orer to prove complete convergence with our approach, one only nees to prove convergence in mean, which is often much simpler than complete convergence, an typically smoothness. Thus, we provie a simple metho to prove complete convergence of Eucliean functionals along the lines of Yukich s result that, in the presence of concentration of measure, convergence in mean implies complete convergence [25, Cor. 6.4]. 2 Definitions an Notation Throughout the paper, (the imension) an p (the istance-power graient) are fixe constants. For three points x,y,v, we by xv the line through x an v, an we enote by (x,v,y) the angle between xv an yv. A Eucliean functional is a function F p for p > 0 that maps finite sets of points in [0,1] to some non-negative real number an is translation invariant an homogeneous of orer p [25, page 18]. From now on, we omit the superscript p of Eucliean functionals, as p is always fixe an clear from the context. PA B is the canonical bounary functional of PA (we refer to Yukich [25] for bounary functionals of other optimization problems): given a hyperrectangle R R with X R, this means that a solution is an assignment pa(x) of power to the noes x X such that x an y are connecte if pa(x),pa(y) x y p, x is connecte to the bounary of R if the istance of x to the bounary of R is at most pa(x) 1/p, an the resulting graph, calle a bounary PA graph, is either connecte or consists of connecte components that are all connecte to the bounary. 3

4 Then PA B (X,R) is the minimum value for x X pa(x) that can be achieve by a bounary PA graph. Note that in the bounary functional, no power is assigne to the bounary. It is straight-forwar to see that PA an PA B are Eucliean functionals for all p > 0 accoring to Yukich [25, page 18]. For a hyperrectangle R R, let iamr = max x,y R x y enote the iameter of R. For a Eucliean functional F, let F(n) = F({U 1,...,U n }), where U 1,...,U n are rawn uniformly an inepenently from [0,1]. Let E ( F(n) ) γ,p F = lim n (In principle, γ,p F nee not exist, but it oes exist for all functionals consiere in this paper.) Asequence(R n ) n N ofranomvariablesconverges in mean toaconstantγ iflim n E( R n γ ) = 0. The sequence (R n ) n N converges completely to a constant γ if we have n p P ( R n γ > ε ) < n=1 for all ε > 0. Besies PA, we consier two other Eucliean functions: MST(X) enotes the length of the minimum spanning tree with lengths raise to the power p. PT(X) enotes the total power consumption of the assignment obtaine from the MST heuristic, again with lengths raise to the power p. The MST heuristic procees as follows: First, we compute a minimum spanning tree of X. The let pa(x) = max{ x y p {x,y} is an ege of the MST}. By construction an a simple analysis, we have MST(X) PA(X) PT(X) 2 MST(X) [10]. For n N, let [n] = {1,...,n}. 3 Properties of the Power Assignment Functional After showing that optimal PA graphs can have unboune egree an proviing a lemma that helps solving this problem, we show that the power assignment functional fits into Yukich s framework for Eucliean functionals [25]. 3.1 Degrees an Cones As oppose to minimum spanning trees, whose maximum egree is boune from above by a constant that epens only on the imension, a technical challenge is that the maximum egree in an optimal PA graphs cannot be boune by a constant in the imension. This hols even for the simplest case of = 1 an p > 1. We conjecture that the same hols also for p = 1, but proving this seems to be more ifficult an not to a much. Lemma 3.1. For all p > 1, all integers 1, an for infinitely many n, there exists instances of n points in [0,1] such that the unique optimal PA graph is a tree with a maximum egree of n 1. Proof. Let n be o, an let 2m+1 = n. Consier the instance X m = {a m,a m+1,...,a 0,...,a m 1,a m }. 4

5 that consists of m positive integers a 1,...,a m, m negative integers a i = a i for 1 i m, an a 0 = 0. We assume that a i+1 a i for all i. By scaling an shifting, we can achieve that X fits into the unit interval. A possible solution pa : X m R + is assigning power a p i to a i an a i for 1 i m an power a p m to 0. In this way, all points are connecte to 0. We claim that this power assignment is the unique optimum. As a m = a m a i for i < m, the ominant term in the power consumption Ψ m is 3a p m (the power of a m, a m, an a 0 = 0). Note that no other term in the total power consumption involves a m. We show that a m an a m must be connecte to 0 in an optimal PA graph. First, assume that a m an a m are connecte to ifferent vertices. Then the total power consumption increases to about 4a p m because a ±m is very large compare to a i for all i < m (we say that a m is ominant). Secon, assume that a m an a m are connecte to a i with i 0. Without loss of generality, we assume that i > 0 an, thus, a i > 0. Then the total power consumption is at least 2 (a m +a i ) p +(a m a i ) p 3a p m+2a p 1 m a i. Because a m is ominant, this is strictly more than Ψ m because it contains the term 2a p 1 m a i, which contains the very large a m because p > 1. From now on, we can assume that 0 = a 0 is connecte to a ±m. Assume that there is some point a i that is connecte to some a j with i,j 0. Assumewithoutloss of generality that i > 0 an i j. Assume further that i is maximal in the sense that there is no k > i such that a k is connecte to some vertex other than 0. We set a i s power to a p i an a j s power to a j p. Then both are connecte to 0 as 0 has alreay sufficient power to sen to both. Furthermore, the PA graph is still connecte: All vertices a k with k > i are connecte to 0 by the choice of i. If some a k with k i an k i,j was connecte to a i before, then it has also sufficient power to sen to 0. The power balance remains to be consiere: If j = i, then the energy of both a i an a j has been strictly ecrease. Otherwise, j < i. The power of a i was at least (a i a j ) p before an is now a p i. The power of a j was at least (a i a j ) p before an is now a p j. Since a i ominates all a j for j < i, this ecreases the power. The unbouneness of the egree of PA graphs make the analysis of the functional PA challenging. The technical reason is that removing a vertex can cause the PA graph to fall into a non-constant number of components. The following lemma is the crucial ingreient to get over this egree hurle. Lemma 3.2. Let x,y X, let v [0,1], an assume that x an y have power pa(x) x v p an pa(y) y v p, respectively. Assume further that x v y v an that (x,v,y) α with α π/3. Then the following hols: (a) pa(y) x y p, i.e., y has sufficient power to reach x. (b) If x an y are not connecte (i.e., pa(x) < x y p ), then y v > sin(2α) sin(α) x v. Proof. Because α π/3, we have y v y x. This implies (a). The point x has sufficient power to reach any point within a raius of x v of itself. By (a), point y has sufficient power to sen to x. Thus, if y is within a istance of x v of x, then also x can sen to y an, thus, x an y are connecte. We project x, y, an v into the two-imensional subspace spanne by the vectors x v an y v. This yiels a situation as epicte in Figure 1. Since pa(x) x v p, point x can sen to all points in the light-gray region, thus in particular to all ark-gray points in the cone roote at v. In particular, x 5

6 v α α x β α z Figure 1: Point x can sen to all points in the gray area as it can sen to v. In particular, x can sen to all points that are no further away from v than z. This inclues all points to the left of the otte line. The otte line consists of points at a istance of sin(2α) sin(α) x v of v. can sen to all points that are no further away from v than the point z. The triangle vxz is isosceles. Thus, also the angle at z is α an the angle at x is β = π 2α. Using the law of sines together with sin(β) = sin(2α) yiels that z v = sin(2α) sin(α) x v, which completes the proof of (b). For instance, α = π/6 results in a factor of 3 = sin(π/3)/sin(π/6). In the following, we invoke this lemma always with α = π/6, but this choice is arbitrary as long as α < π/3, which causes sin(2α)/ sin(α) to be strictly larger than Deterministic Properties In this section, we state properties of the power assignment functional. Subaitivity (Lemma 3.3), superaitivity (Lemma 3.4), an growth boun (Lemma 3.5) are straightforwar. Lemma 3.3 (subaitivity). PA is subaitive [25, (2.2)] for all p > 0 an all 1, i.e., for any point sets X an Y an any hyperrectangle R R with X,Y R, we have PA(X Y) PA(X)+PA(Y)+O ( (iamr) p). Proof. Let T X an T Y be optimal PA graphs for X an Y, respectively. We connect these graphs by an ege of length at most iamr. This yiels a solution for X Y, i.e., a PA graph, an the aitional costs are boune from above by the length of this ege to the power p, which is boune by (iamr) p. Lemma 3.4 (superaitivity). PA B is superaitive for all p 1 an 1 [25, (3.3)], i.e., for any X, hyperrectangle R R with X R an partition of R into hyperrectangles R 1 an R 2, we have PA p B (X,R) PAp B (X R 1,R 1 )+PA p B (X R 2,R 2 ). Proof. Let T be an optimal bounary PA graph for (X,R). This graph restricte to R 1 an R 2 yiels bounary graphs T 1 an T 2 for (X R 1,R 1 ) an (X R 2,R 2 ), respectively. The sum of the costs of T 1 an T 2 is upper boune by the costs of T because p 1 (splitting an ege at the borer between R 1 an R 2 results in two eges whose sum of lengths to the power p is at most the length of the original ege to the power p). 6

7 Lemma 3.5 (growth boun). For any X [0,1] an 0 < p an 1, we have ( }) PA B (X) PA(X) O max {n p,1. Proof. This follows from the growth boun for the MST [25, (3.7)], because MST(X) PA(X) 2MST(X) for all X [10]. The inequality PA B (X) PA(X) hols obviously. The following lemma shows that PA is smooth, which roughly means that aing or removing a few points oes not have a huge impact on the function value. Its proof requires Lemma 3.2 to eal with the fact that optimal PA graphs can have unboune egree. Lemma 3.6. The power assignment functional PA is smooth for all 0 < p [25, (3.8)], i.e., PA p (X Y) PA p (X) ( ) = O for all point sets X,Y [0,1]. Y p Proof. One irection is straightforwar: PA(X Y) PA(X) is boune by Ψ = O ( Y p ), because the optimal PA graph for Y has a value of at most Ψ by Lemma 3.5. Then we can take the PA graph for Y an connect it to the tree for X with a single ege, which costs at most O(1) Ψ because p. For the other irection, consier the optimal PA graph T for X Y. The problem is that the egrees eg T (v) of vertices v Y can be unboune (Lemma 3.1). (If the maximum egree were boune, then we coul argue in the same way as for the MST functional.) The iea is to exploit the fact that removing v Y also frees some power. Roughly speaking, we procee as follows: Let v Y be a vertex of possibly large egree. We a the power of v to some vertices close to v. The graph obtaine from removing v an istributing its energy has only a constant number of components. To prove this, Lemma 3.2 is crucial. We consier cones roote at v with the following properties: The cones have a small angle α, meaning that for every cone C an every x,y C, we have (x,v,y) α. We choose α = π/6. Every point in [0,1] is covere by some cone. There is a finite number of cones. (This can be achieve because is a constant.) Let C 1,...,C m be these cones. By abusing notation, let C i also enote all points x C i (X Y \{v}) that are ajacent to v in T. For C i, let x i be the point in C i that is closest to v an ajacent to v (breaking ties arbitrarily), an let y i be the point in C i that is farthest from v an ajacent to v (again breaking ties arbitrarily). (For completeness, we remark that then C i can be ignore if C i X =.) Let l i = y i v be the maximum istance of any point in C i to v, an let l = max i l i. We increase the power of x i by l p /m. Since the power of v is at least l p an we have m cones, we can account for this with v s power because we remove v. Because α = π/6 an x i is closest to v, any point in C i is closer to x i than to v. Accoring to Lemma 3.2(a), every point in C i has sufficient power to reach x i. Thus, if x i can reach a point z C i, then there is an establishe connection between them. 7

8 From this an increasing x i s power to at least l p /m, there is an ege between x i an every point z C i that has aistance of at most l/ p mfrom v. We recall thatmanpareconstants. Now let z 1,...,z k C i be the vertices in C i that are not connecte to x i because x i has too little power. We assume that they are sorte by increasing istance from v. Thus, z k = y i. We can assume that no two z j an z j are in the same component after removal of v. Otherwise, we can simply ignore one of the eges {v,z j } an {v,z j } without changing the components. Sincez j an z j+1 wereconnecte tov an they arenot connecte toeach other, we can apply Lemma 3.2(b), which implies that z j+1 v 3 z j v. Furthermore, z 1 v l/ p m by assumption. Iteratingthisargument yielsl = z k v 3 k 1 z 1 v 3 k 1 l/ p m. This implies k log 3 ( p m)+1. Thus, removing v an reistributing its energy as escribe causes thepagraphtofallintoatmostaconstantnumberofcomponents. Removing Y pointscauses the PA graph to fall into at most O( Y ) components. These components can be connecte with costs O( Y p ) by choosing one point per component an applying Lemma 3.5. Lemma 3.7. PA B is smooth for all 1 p [25, (3.8)]. Proof. Theieaisessentially ientical totheproofoflemma3.6, anweusethesamenotation. Again, one irection is easy. For the other irection, note that every vertex of G = (X,E), with E inuce by pa is connecte to at most one point at the bounary. We use the same kin of cones as for Lemma 3.6. Let v G be a vertex that we want to remove. We ignore v s possible connection to the bounary an procee with the remaining connections. In this way, we obtain a forest with O( G ) components. We compute a bounary PA graph for one vertex of each component an are one because of Lemma 3.5 an in the same way as in the proof of Lemma 3.6. Crucial for convergence of PA is that PA, which is subaitive, an PA B, which is superaitive, are close to each other. Then both are close to being both subaitive an superaitive. The following lemma states that inee PA an PA B o not iffer too much for 1 p <. Lemma 3.8. PA is point-wise close to PA B for 1 p < [25, (3.10)], i.e., for every set X [0,1] of n points. PA p (X) PA p B (X,[0,1] ) = o ( n p ) Proof. Let T be an optimal bounary PA graph for X. Let Q X be the set of points that have a connection to the bounary of T an let Q be the corresponing points on the bounary. If we remove the connections to the bounary, we obtain a graph T. We can assume that Q contains exactly one point per connecte component of the graph T. We use the same yaic ecomposition as Yukich [25, proof of Lemma 3.8]. This yiels that the sum of transmit powers use to connect to the bounary is boune by the maximum of O(n p 1 1 ) an O(logn) for p 1 an by a constant for p ( 1,). We omit the proof as it is basically ientical to Yukich s proof. Let Q X be the points connecte to the bounary, an let Q be the points where Q connects to the bounary. We compute a minimum-weight spanning tree Z of Q. (Note that we inee compute an MST an not a PA. This is because the MST has boune egree an PA an MST iffer by at most a factor of 2.) This MST Z has a weight of ( }) ) O max {n 1 p 1,1 = o (n p 8

9 accoring to the growth boun for MST [25, (3.7)]. an because > p. If two points q, q Q are connecte in this tree, then we connect the corresponing points q,q Q. The question that remains is by how much the power of the vertices in Q has to be increase in orer to allow the connections as escribe above. If q,q Q are connecte, then an upper boun for their power is given by the p-th power of their istances to the bounary points q an q plus the length of the ege connecting q an q. Applying the triangle inequality for powers of metrics twice, the energy neee for connecting q an q is at most 4 p = O(1) times the sum of these istances. Since the egree of Z is boune, every vertex in Q contributes to only a constant number of eges an, thus, only to the power consumption of a constant number of other vertices. Thus, the total aitional power neee is boune by a constant times the power of connecting Q to the bounary plus the power to use Z as a PA graph. Because of the triangle inequality for powers of metrics, the boune egree of every vertex of Q in Z, an because of the yaic ecomposition mentione above, the increase of power is in compliance with the statement of the lemma. Remark 3.9. Lemma 3.8 is an analogue of its counterpart for MST, TSP, an matching [25, Lemma 3.7] in terms of the bouns. Namely, we obtain PA(X) PA B (X) 3.3 Probabilistic Properties O( X p 1 1 ) if 1 p < 1, O(log X ) if p = 1 1, O(1) if 1 < p < or p = 1 = 1. For p >, smoothe is not guarantee to hol, an for p, point-wise closeness is not guarantee to hol. But similar properties typically hol for ranom point sets, namely smoothness in mean (Definition 3.14) an closeness in mean (Definition 3.16). In the following, let X = {U 1,...,U n }. Recall that U 1,...,U n are rawn uniformly an inepenently from [0,1]. Before proving smoothness in mean, we nee a statement about the longest ege in an optimal PA graph an bounary PA graph. The boun is asymptotically equal to the boun for the longest ege in an MST [6,11,17]. To prove our boun for the longest ege in optimal PA graphs (Lemma 3.12), we nee the following two lemmas. Lemma 3.10 is essentially equivalent to a result by Kozma et al. [11], but they o not state the probability explicitly. Lemma 3.11 is a straight-forwar consequence of Lemma Variants of both lemmas are known [6,17,18,23], but, for completeness, we state an prove both lemmas in the forms that we nee. Lemma For every β > 0, there exists a c ball = c ball (β,) such that, with a probability of at least 1 n β, every hyperball of raius r ball = c ball (logn/n) 1/ an with center in [0,1] contains at least one point of X in its interior. Proof. We sketch the simple proof. We cover [0,1] with hypercubes of sie length Ω(r ball ) such that every ball of raiusr ball even if its center is in a corner (for a point on the bounary, still at least a 2 = Θ(1) fraction is within [0,1] ) contains at least one box. The probability that such a box oes not contain a point, which is necessary for a ball to be empty, is at most ( 1 Ω(rball ) ) n n Ω(1) by inepenence of the points in X an the efinition of r ball. The rest of the proof follows by a union boun over all O(n/logn) boxes. 9

10 We also nee the following lemma, which essentially states that if z an z are sufficiently far away, then there is with high probability always a point y between z an z in the following sense: the istance of y to z is within a preefine upper boun 2r ball, an y is closer to z than z. Lemma For every β > 0, with a probability of at least 1 n β, the following hols: For every choice of z,z [0,1] with z z 2r ball, there exists a point y X with the following properties: z y 2r ball. z y < z z. Proof. The set of caniates for y contains a ball of raius r ball, namely a ball of this raius whose center is at a istance of r ball from z on the line between z an z. This allows us to use Lemma Lemma 3.12 (longest ege). For every constant β > 0, there exists a constant c ege = c ege (β) such that, with a probability of at least 1 n β, every ege of an optimal PA graph an an optimal bounary PA graph PA B is of length at most r ege = c ege (logn/n) 1/. Proof. We restrict ourselves to consiering PA graphs. The proof for bounary PA graphs is almost ientical. Let T be any PA graph. Let c ege = 4k 1/p c ball /(1 3 p ) 1/p, where k is an upper boun for the number of vertices without a pairwise connection at a istance between r an r/ 3 for arbitrary r. It follows from Lemma 3.2 an its proof, that k is a constant that epens only on p an. Note that c ege > 2c ball. We are going to show that if T contains an ege that is longer than r ege, then we can fin a better PA graph with a probability of at least 1 n β, which shows that T is not optimal. Let v be a vertex incient to the longest ege of T, an let r big > r ege be the length of this longest ege. (The longest ege is unique with a probability of 1. The noe v is not unique as the longest ege connects two points.) We ecrease the power of v to r big / 3. This implies that v loses contact to some points otherwise, the power assignment was clearly not optimal. The number c ball epens on the exponent β of the lemma. Let x 1,...,x k with k k be the points that were connecte to v but are in ifferent connecte components than v after ecreasing v s power. This is because the only noes that might lose their connection to v are within a istance between r big / 3 an r big, an there are at most k such noes without a pairwise connection. Consier x 1. Let z 0 = v. Accoring to Lemma 3.11, there is a point z 1 that is closer to x 1 an at most 2r ball away from v. Iteratively for i = 1,2,..., we istinguish three cases until this process stops: (i) z i belongs to the same component as x j for some j (z i is closer to x 1 than z i 1, but this oes not imply j = 1). We increase z i s power such that z i is able to sen to z i 1. If i > 1, then we also increase z i 1 s power accoringly. (ii) z i belongs to the same component as v. Then we can apply Lemma 3.11 to z i an x 1 an fin a point z i+1 that is closer to x 1 than z i an at most at a istance of 2r ball of z i. 10

11 (iii) z i is within a istance of at most 2r ball of some x j. In this case, we increase the energy of z i such that z i an x j are connecte. (The energy of x j is sufficiently large anyhow.) Running this process once ecreases the number of connecte components by one an costs at most 2(2r ball ) p = 2 p+1 r p ball aitional power. We run this process k k times, thus spening at most k2 p+1 r p ball of aitional power. In this way, we obtain a vali PA graph. We have to show that the new PA graph inee saves power. To o this, we consier the power save by ecreasing v s energy. By ecreasing v s power, we save an amount of r p big (r big/ 3) p > (1 3 p ) r p ege. By the choice of c ege, the save amount of energy excees the aitional amount of k2 p+1 r p ball. This contraicts the optimality of the PA graph with the ege of length r big > r ege. Remark Since the longest ege has a length of at most r ege with high probability, i.e., with a probability of 1 n Ω(1), an any ball of raius r ege contains roughly O(logn) points ue to Chernoff s boun [15, Chapter 4], the maximum egree of an optimum PA graph of a ranom point set is O(log n) with high probability contrasting Lemma 3.1. Yukich gave two ifferent notions of smoothness in mean [25, (4.13) an (4.20) & (4.21)]. We use the stronger notion, which implies the other. Definition 3.14 (smooth in mean [25, (4.20), (4.21)]). A Eucliean functional F is calle smooth in mean if, for every constant β > 0, there exists a constant c = c(β) such that the following hols with a probability of at least 1 n β : an for all 0 k n/2. F(n) F(n±k) ck (logn n FB (n) F B (n±k) = ck (logn n ) p/ ) p/. Lemma PA B an PA are smooth in mean for all p > 0 an all. Proof. The boun PA(n + k) PA(n) + O ( k (logn )p ) n follows from the fact that for all k aitional vertices, with a probability of at least 1 n β for any β > 0, there is a vertex among the first n within a istance of at most O ( (logn/n) 1/) accoring to Lemma 3.10 (β influences the constant hien in the O). Thus, we can connect any of the k new vertices with costs of O ( (logn/n) p/) to the optimal PA graph for the n noes. Let us now show the reverse inequality PA(n) PA(n+k)+O ( )p k (logn ) n. To o this, we show that with a probability of at least 1 n β, we have ( (logn ) )p PA(n) PA(n+1)+O. (1) n Then we iterate k times to obtain the boun we aim for. The proof of (1) is similar to the analogous inequality in Yukich s proof [25, Lemma 4.8]. The only ifference is that we first have to reistribute the power of the point U n+1 to its closest neighbors as in the proof of Lemma 3.6. In this way, removing U n+1 results in a constant number of connecte components. The longest ege incient to U n+1 has a length 11

12 of O ( (logn/n) 1/) with a probability of at least 1 n β for any constant β > 0. Thus, we can connect these constant number number of components with extra power of at most O ( (logn/n) p/). The proof of ( ( logn PA(n) PA(n k) = O k n ) )p an the statement ( ( logn PA B (n) PA B (n±k) = O k n ) )p for the bounary functional are almost ientical. Definition 3.16 (close in mean [25, (4.11)]). A Eucliean functional F is close in mean to its bounary functional F B if ) E( F(n) F B (n) ) = o (n p. Lemma PA is close in mean to PA B for all an p 1. Proof. It is clear that PA B (X) PA(X) for all X. Thus, in what follows, we prove that PA(X) PA B (X) +o ( n p ) hols with a probability of at least 1 n β, where β influences the constant hien in the o. This implies closeness in mean. With a probability of at least 1 n β, the longest ege in the graph that realizes PA B (X) has a length of c ege (logn/n) 1/ (Lemma 3.12). Thus, with a probability of at least 1 n β for any constant β > 0, only vertices within a istance of at most c ege (logn/n) 1/ of the bounary are connecte to the bounary. As the -imensional unit cube is boune by 2 hyperplanes, the expecte number of vertices that are so close to the bounary is boune from above by c ege n2 (logn/n) 1/ = O ( (logn) 1/ ) n 1. With a probability of at least 1 n β for any β > 0, this number is exceee by no more than a constant factor. Removingthesevertices causesthebounarypa graphtofall intoat mosto ( (logn) 1/ n 1 components. We choose one vertex of every component an start the process escribe in the proof of Lemma 3.12 to connect all of them. The costs per connection is boune from above by O ( (logn/n) p/) with a probability of 1 n β for any constant β > 0. Thus, the total costs are boune from above by O ( (logn/n) p/) O ( (logn) 1/ ) ) n 1 = O ((logn) p 1 n 1 p = o ( n p ) with a probability of at least 1 n β for any constant β > 0. 4 Convergence 4.1 Stanar Convergence Our finings of Sections 3.2 yiel complete convergence of PA for p < (Theorem 4.1). Together with the probabilistic properties of Section 3.3, we obtain convergence in mean in a straightforwar way for all combinations of an p (Theorem 4.2). In Sections 4.2 an 4.3, we prove complete convergence for p. ) 12

13 Theorem 4.1. For all an p with 1 p <, there exists a constant γ,p PA converges completely to γ,p PA. PA p (n) n p such that Proof. This follows from the results in Section 3.2 together with results by Yukich [25, Theorem 4.1, Corollary 6.4]. Theorem 4.2. For all p 1 an 1, there exists a constant γ,p PA (equal to the constant of Theorem 4.1 for p < ) such that E ( PA p (n) ) E ( PA p B lim = lim (n)) = γ,p n n p n n p PA. Proof. This follows from the results in Sections 3.2 an 3.3 together with results by Yukich [25, Theorem 4.5]. 4.2 Concentration with Warnke s Inequality McDiarmi s or Azuma-Hoeffing s inequality are powerful tools to prove concentration of measure for a function that epens on many inepenent ranom variables, all of which have only a boune influence on the function value. If we consier smoothness in mean (see Lemma 3.15), then we have the situation that the influence of a single variable is typically very small (namely O((logn/n) p/ )), but can be quite large in the worst case (namely O(1)). Unfortunately, this situation is not covere by McDiarmi s or Azuma-Hoeffing s inequality. Fortunately, Warnke [24] prove a generalization specifically for the case that the influence of single variables is typically boune an fulfills a weaker boun in the worst case. The following theorem is a simplifie version (personal communication with Lutz Warnke) of Warnke s concentration inequality [24, Theorem 2], tailore to our nees. Theorem 4.3 (Warnke). Let U 1,...,U n be a family of inepenent ranom variables with U i [0,1] for each i. Suppose that there are numbers c goo c ba an an event Γ such that the function F : ([0,1] ) n R satisfies max i [n] max F(U 1,...,U n ) F(U 1,...,U i 1,x,U i+1,...,u k ) x [0,1] { c goo if Γ hols an c ba otherwise. (2) Then, for any t 0 an γ (0,1] an η = γ(c ba c goo ), we have P ( F(n) E(F(n)) t ) 2exp ( ) t 2 2n(c goo +η) + n 2 γ P( Γ). (3) Proof sketch. There are two ifferences of this simplifie variant to Warnke s result [24, Theorem 2]: First, the numbers c goo an c ba o not epen on the inex i but are chosen uniformly for all inices. Secon, an more importantly, the event B [24, Theorem 2] is not use in Theorem 4.3. In Warnke s theorem [24, Theorem 2], the event B plays only a briging 13

14 role: it is require that P(B) n i=1 1 γ i P( Γ) for some γ 1,...,γ n that show up in the tail boun as well. Choosing γ i = γ for all i yiels P(B) n γ P( Γ). Then yiels P ( F(n) E(F(n)+t an B ) ( t 2 ) exp 2n(c goo +η) 2 P ( F(n) E(F(n)) t ) ( 2exp t 2 2n(c goo +η) 2 ) + n γ P( Γ) by observing that a two-sie tail boun can be obtaine by symmetry an aing an upper boun for the probability of B to the right-han sie. Next, we efine typical smoothness, which means that, with high probability, a single point oes not have a significant influence on the value of F, an we apply Theorem 4.3 for typically smooth functionals F. The boun of c (logn/n) p/ in Definition 4.4 below for the typical influence of a single point is somewhat arbitrary, but works for PA an MST. This boun is also essentially the smallest possible, as for there can be regions of iameter c (logn/n) 1/ for some small constant c > 0 that contain no or only a single point. It might be possible to obtain convergence results for other functionals for weaker notions of typical smoothness. Definition 4.4 (typically smooth). A Eucliean functional F is typically smooth if, for every β > 0, there exists a constant c = c(β) such that max x [0,1],i [n] with a probability of at least 1 n β. F(U1,...,U n ) F(U 1,...,U i 1,x,U i+1,...,u n ) logn c ( n Theorem 4.5 (concentration of typically smooth functionals). Assume that F is typically smooth. Then P ( F(n) E(F(n)) t ) ) O(n β )+exp ( t2 n 2p 1 C(logn) 2p/ ) p/ for an arbitrarily large constant β > 0 an another constant C > 0 that epens on β. Proof. We use Theorem 4.3. The event Γ is that any point can change the value only by at most O ( (logn/n) p/ ). Thus, c goo = O ( (logn/n) p/ ) an c ba = O(1). The probability that we o not have the event Γ is boune by O(n β ) for an arbitrarily large constant β by typical smoothness. This only influences the constant hien in the O of the efinition of c goo. We choose γ = O ( (logn/n) p/ ). In the notation of Theorem 4.3, we choose η = O(γ), which is possible as c ba c goo c ba = Θ(1). Using the conclusion of Theorem 4.3 yiels P ( F(n) E(F(n)) t ) ( ) n γ P( Γ)+exp t2 n 2p/ nc(logn) ( 2p/ ) O(n β )+exp t2 n 2p/ nc(logn) 2p/ for some constant C > 0. Here, β can be chosen arbitrarily large. 14

15 Choosing t = n p / log n yiels a nontrivial concentration result that suffices to prove complete convergence of typically smooth Eucliean functionals. Corollary 4.6. Assume that F is typically smooth. Then P ( F(n) E(F(n)) > n p /logn ) ( ( O n β n +exp C(logn) 2+2p for any constant β an C epening on β as in Theorem 4.5. )) (4) con- Proof. The proof is straightforwar by exploiting that the assumption that F(n)/n p verges in mean to γ,p F implies E(F(n)) = Θ(n p ). 4.3 Complete Convergence for p In this section, we prove that typical smoothness (Definition 4.4) suffices for complete convergence. This implies complete convergence of MST an PA by Lemma 4.8 below. Theorem 4.7. Assume that F is typically smooth an F(n)/n p Then F(n)/n p Proof. Fix any ε > 0. Since there exists an n 0 such that converges completely to γ,p F. lim E n ( ) F(n) E n p ( ) F(n) n p = γ,p F, [ γ,p F ε 2,γ,p F + ε ] 2 converges in mean to γ,p F. for all n n 0. Furthermore, there exists an n 1 such that, for all n n 1, the probability that F(n)/n p eviates by more than ε/2 from its expecte value is smaller than n 2 for all n n 1. To see this, we use Corollary 4.6 an observe that the right-han sie of (4) is O(n 2 ) for sufficiently large β an that the event on the left-han sie is equivalent to F(n) E(F(n)) ( ) 1 n p n p > O, logn where O(1/logn) < ε/2 for sufficiently large n 1 an n n 1. Let n 2 = max{n 0,n 1 }. Then ( ) PA(X) P > ε n 2 + n=1 n p n=n 2 +1 n 2 = n 2 +O(1) <. Although similar in flavor, smoothness in mean oes not immeiately imply typical smoothness or vice versa: the latter makes only a statement about single points at worst-case positions. The former only makes a statement about aing an removing several points at ranom positions. However, the proofs of smoothness in mean for MST an PA o not exploit this, an we can aapt them to yiel typical smoothness. 15

16 Lemma 4.8. PA an MST are typically smooth. Proof. We first consier PA. Replacing a point U k by some other (worst-case) point z can be moele by removing U k an aing z. We observe that, in the proof of smoothness in mean (Lemma 3.15, we i not exploit that the point ae is at a ranom position, but the proof goes through for any single point at an arbitrary position. Also the other way aroun, i.e., removing z an replacing it by a ranom point U k, works in the same way. Thus, PA is typically smooth. Closely examining Yukich s proof of smoothness in mean for MST [25, Lemma 4.8] yiels the same result for MST. Corollary 4.9. For all an p with p 1, MST(n)/n p to constants γ,p MST an γ,p PA, respectively. an PA(n)/n p converge completely Proof. Both MST an PA are typically smooth an converge in mean. Thus, the corollary follows from Theorem 4.7. Remark Instea of Warnke s metho of typical boune ifferences, we coul also have use Kutin s extension of McDiarmi s inequality [12, Chapter 3]. However, this inequality yiels only convergence for p 2, which is still an improvement over the previous complete convergence of p <, but weaker than what we get with Warnke s inequality. Furthermore, Warnke s inequality is easier to apply an a more natural extension in the following way: intuitively, one might think that we coul just take McDiarmi s inequality an a the probability that we are not in a nice situation using a simple union boun, but, in general, this is not true [24, Section 2.2]. 5 Average-Case Approximation Ratio of the MST Heuristic In this section, we show that the average-case approximation ratio of the MST heuristic for power assignments is strictly better than its worst-case ratio of 2. First, we prove that the average-case boun is strictly (albeit marginally) better than 2 for any combination of an p. Secon, we show a simple improve boun for the 1-imensional case. 5.1 The General Case The iea behin showing that the MST heuristic performs better on average than in the worst case is as follows: the weight of the PA graph obtaine from the MST heuristic can not only be upper-boune by twice the weight of an MST, but it is in fact easy to prove that it can be upper-boune by twice the weight of the heavier half of the eges of the MST [4]. Thus, we only have to show that the lighter half of the eges of the MST contributes Ω(n p ) to the value of the MST in expectation. For simplicity, we assume that the number n = 2m + 1 of points is o. The case of even n is similar but slightly more technical. We raw points X = {U 1,...,U n } as escribe above. Let PT(X) enote the power require in the power assignment obtaine from the MST. Furthermore, let H enote the m heaviest eges of the MST, an let L enote the m lightest eges of the MST. We omit the parameter X since it is clear from the context. Then we have H+L = MST PA PT 2H = 2MST 2L 2MST (5) 16

17 since the weight of the PA graph obtaine from an MST can not only be upper boune by twice the weight of a minimum-weight spanning tree, but it is easy to show that the PA graph obtaine from the MST is in fact by twice the weight of the heavier half of the eges of a minimum-weight spanning tree [4]. For istances raise to the power p, the expecte value of MST is (γ,p MST we can prove that the lightest m eges of the MST are of weight Ω(n p ±o(1)) n p. If ), then it follows that the MST power assignment is strictly less than twice the optimal power assignment. L is lower-boune by the weight of the lightest m eges of the whole graph without any further constraints. Let A = A(X) enote the weight of these m lightest eges of the whole graph. Note that both L an A take ege lengths to the p-power, an we have A L. Let c be a small constant to be specifie later on. Let v,r = π/2 r Γ( n +1) be the volume of a 2 -imensional ball of raius r. For compactness, we abbreviate c = π/2 Γ( n +1), thus v,r = c r. 2 Note that all c s are constants since is constant. The probability P k that a fixe vertex v has at least k other vertices within a istance of at most r = l 1/n for some constant l > 0 is boune from above by ( ) n 1 P k k v k,r nk (c r ) k k! = nk (c l n 1 ) k k! = ck k! for another constant c = l c. This follows from inepenence an a union boun. The expecte number of eges of a specific vertex that have a length of at most r is thus boune from above by n 1 P k k=1 n 1 k=1 c k k! k=1 c k k! = e c 1. By choosing l appropriately small, we can achieve that c 1/3. This yiels e c 1 < 1/2. By linearity of expectation, the total number of eges of length at most r in the whole graph is boune from above by m/2. Thus, at least m/2 of the lightest m eges of the whole graph have a length of at least r. Hence, the expecte value of A is boune from below by for some constant C,p A m 2 rp = m l p 2 lp n p n p 4 = C,p A n p. > 0. Then the expecte value of PT is boune from above by ( ) 2γ,p MST 2C,p A +o(1) n p by (5). From this an the convergence of PA, we can conclue the following theorem. Theorem 5.1. For any 1 an any p 1, we have for some constant C,p A γ,p MST γ,p PA 2γ,p MST 2C,p A < 2γ,p MST > 0 that epens only on an p. By exploiting that in particular PA converges completely, we can obtain a boun on the expecte approximation ratio from the above result. 17

18 Corollary 5.2. For any 1 an p 1 an sufficiently large n, the expecte approximation ratio of the MST heuristic for power assignments is boune from above by a constant strictly smaller than 2. Proof. Theexpecte approximation ratio is E ( PT(n)/PA(n) ) = E ( PT(n)/n p ). We know that PA(n)/n p PA(n)/n p converges completely to γ,p PA. This implies that the probability that PA(n)/n p is o(1) for any ε > 0. eviates by more than ε > 0 from γ,p PA If PA(n)/n p [γ,p PA ε,γ,p PA +ε], then the expecte approximation ratio can be boune from above by 2γ,p MST 2C,p A. This is strictly smaller than 2 for a sufficiently small ε > 0. γ,p PA ε Otherwise, we boun the expecte approximation ratio by the worst-case ratio of 2, which contributes only o(1) to its expecte value. Remark 5.3. Complete convergence of the functional PT as well as smoothness an closeness in mean has been shown for the specific case p = [4]. We believe that PT converges completely for all p an. Since then γ,p PT 2γ,p MST 2C,p A < 2γ,p MST, we woul obtain a simpler proof of Corollary An Improve Boun for the One-Dimensional Case The case = 1 is much simpler than the general case, because the MST is just a Hamiltonian path starting at the left-most an ening at the right-most point. Furthermore, we also know precisely what the MST heuristic oes: assume that a point x i lies between x i 1 an x i+1. The MST heuristic assigns power PA(x i ) = max{ x i x i 1, x i x i+1 } p to x i. The example that proves that the MST heuristic is no better than a worst-case 2-approximation shows that it is ba if x i is very close to either sie an goo if x i is approximately in the mile between x i 1 an x i+1. In orer to show an improve boun for the approximation ratio of the MST heuristic for = 1, we introuce some notation. First we remark that for X = {U 1,...,U n } with high probability, there is no subinterval of length clogn/n of [0,1] that oes not contain any of the n points U 1,...,U n (see Lemma 3.10 for the precise statement). We assume that no interval of length clogn/n is empty for some sufficiently large constant c for the rest of this section. We procee as follows: Let x 0 = 0, x n+1 = 1, an let x 1... x n be the n points (sorte in increasing orer) that are rawn uniformly an inepenently from the interval [0, 1]. Now we istribute the weight of the power assignment PT(X) in the power assignment obtaine from the MST, an the weight of the MST as follows: For the power assignment, every point x i (for 1 i n) gets a charge of P i = max{x i x i 1,x i+1 x i } p. This is precisely the power that this point nees in the power assignment obtaine from the spanning tree. For the minimum spanningtree, we ivie the power of an ege (x i 1,x i ) (for 1 i n+1) evenly between x i 1 an x i. This means that the charge of x i is M i = 1 2 ((x i x i 1 ) p +(x i+1 x i ) p). The length of the minimum spanning tree is thus MST = n M i + 1 ((x 1 x 0 ) p +(x n+1 x n ) p). 2 i=1 }{{}}{{} =M M 18

19 The total power for the power assignment obtaine from this tree is PT = n i=1 P i }{{} P +(x 1 x 0 ) p +(x n+1 x n ) p. }{{} =P Note the following: If the largest empty interval has a length of at most clogn/n, then the terms P an M are negligible accoring to the following lemma. Thus, we ignore P an M afterwars to simplify the analysis. Lemma 5.4. Assume that the largest empty interval has a length of at most clogn/n. Then M = O ( M ) (logn)p n an P = O ( P ) (logn)p n. Proof. We have M (clogn/n) p an P 2(clogn/n) p because x 1 clogn/n an x n 1 clogn/n by assumption. Thus, M,P = O (( logn) p ) n. Furthermore, M = n i=1 1 2 (( x i x i 1 ) p + ( xi+1 x i ) p ). Since p 1, this function becomes minimal if we place x 1,...,x n equiistantly. Thus, M n i=1 ( ) 1 p ( ) 1 p = n = Ω ( n 1 p). n+1 n+1 With a similar calculation, we obtain P = Ω ( n 1 p) an the result follows. For simplicity, we assume from now on that n is even. If n is o, the proof procees in exactly the same way except for some changes in the inices. In orer to analyze M an P, we procee in two steps: First, we raw all points x 1,x 3,...,x n 1 (calle the o points). Given the locations of these points, x i for even i (x i is then calle an even point) is istribute uniformly in the interval [x i 1,x i+1 ]. Note that we o not really raw the o points. Instea, we let an aversary fix these points. But the aversary is not allowe to keep an interval of length c log n/n free (because ranomness woul not o so either with high probability). Then the sums n/2 M even = an i=1 P even = M 2i n/2 P 2i are sums of inepenent ranom variables. (Of course M 2i an P 2i are epenent.) Now let l 2i = x 2i+1 x 2i 1 be the length of the interval for x 2i. The expecte value of M 2i is i=1 E(M 2i ) = 1 l2i 1 (x p +(l 2i x) p) x = lp 2i l 2i 0 2 p+1. 19