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The use of an inertial aiding system, in order to obtain two pieces of complementary data. For example, this could be achieved with a two-dimensional accelerometer or gyroscope (or both) to de ne the horizontality of the mobile terminal. The combined use of a one-dimensional accelerometer and a satellite measurement that allows us to obtain the distance separating a satellite from the mobile terminal. The implementation of some orientation devices to obtain the absolute orientation of the terminal (namely magnetometers). We know that electronic compasses can be disturbed in certain environments, depending on the amount of metallic mass near the compass. The fact that the terminal is built in a known environment (a car for instance), where the horizontality is de ned (the two DOAs allow us to calculate the position of the car). Certainly some other possibilities, depending on speci c application requirements.

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We proceed now to develop an on-line local routing algorithm for unit wireless communication networks Observe first that UW(Pn) is not necessarily planar For instance if Pn consists of 12 points contained within a circle of radius 1, UW(Pn) is not planar In order to use the results presented in the previous section, we should be able to extract a planar subnetwork from any UW(Pn) Two requirements must be satisfied by the method we use to extract the planar subgraph to fully ensure its functionality for real-life applications: 1 If a cellular communication network is connected, the resulting planar subgraph must be connected 2.

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We must have a local protocol so that each node of the network can decide in a consistent manner which neighbor connections to keep, and ensure that, collectively, and without the need to communicate, the set of edges chosen individually by the nodes of the network form a planar graph The necessity for the second condition follows from our desire to have fully distributed protocols that avoid the use of any kind of centralized protocols The problem of extracting or even deciding if a graph contains a planar connected subgraph is a well-known NP-complete problem [16] Fortunately, UW(Pn) networks always have such a subgraph and, in fact, finding it is relatively straightforward The key to our result arises from the use of Gabriel graphs [1].

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Given two points p and q on the plane, let C(p, q) be the circle passing through them such that the line segment joining p to q is a diameter of C(p, q) Given a set of n points Pn = {p1, , pn} on the plane, the Gabriel graph of Pn is the graph whose set of vertices is Pn, in which two points u and v of Pn are adjacent iff the C(p, q) contains no other points of Pn Let G (Pn) be the graph with vertex set Pn such that two vertices p and q are adjacent in G (Pn) iff C(p, q) contains no other points of Pn and p and q are adjacent in UW(Pn), that is G (Pn) is the intersection of the Gabriel graph of Pn with UW(Pn) The following result was proved in [3]: Theorem 12.

IMPLICATIONS OF THE ASSUMPTIONS FOR THE BEHAVIOR OF THE RESPONSE VARIABLE y 1. Based on the zero-mean assumption, we have E(y) = E ( 0 + 1 x + ) = E( 0 ) + E( 1 x) + E( ) = 0 + 1 x That is, for each value of x, the mean of the y s lies on the regression line. 2. Based on the constant-variance assumption, we have the variance of y, Var(y), given as Var(y) = Var ( 0 + 1 x + ) = Var( ) = 2 That is, regardless of which value is taken by the predictor x, the variance of the y s is always constant. 3. Based on the independence assumption, it follows that for any particular value of x, the values of y are independent as well. 4. Based on the normality assumption, it follows that y is also a normally distributed random variable. In other words, the values of the response variable yi are independent normal random variables, with mean 0 + 1 x and variance 2 .

1 If UW(Pn) is connected then G (Pn) is also connected The easiest proof of this result proceeds as follows Let p and q be such that they are adjacent in UW(Pn) and there is no path connecting them in G (Pn) Suppose further that their distance is the smallest possible among all such pairs of points in Pn Since p and q are not connected in G (Pn), C(p, q) contains at least a third point r Pn Observe that the distances from r to p and q are smaller than the distance from p to q, and thus there is a path P in G (Pn) connecting r to p and a path P connecting r to q The concatenation of these paths produces a path from p to q in G (Pn) Our result follows.

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