Rennes Position Estimation in Sensor Networks Brian D O Anderson Research School of Information Sciences and Engineering, Australian National University and National ICT Australia (Work with A S Morse, D Goldenberg, T Eren)

Rennes OUTLINE Aim of Presentation Control and Sensor Networks The sensor network localization problem Rigidity and Global rigidity Computational Complexity of Localization Conclusions and open problems Aim of presentation

Rennes AIM OF PRESENTATION To introduce problems involving control and sensor networks To explain the problem of position estimation of sensors (sensor network localization) To introduce tools of rigidity To use tools of rigidity theory to understand the essence of the sensor localization problem Motivations: printers in a building, underwater acoustic sensors, sensors in dense foliage, etc

Rennes OUTLINE Aim of Presentation Control and Sensor Networks The sensor network localization problem Rigidity and Global Rigidity Computational Complexity of Localization Conclusions and open problems

Rennes Sensor Networks A collection of sensors is given, in two or three dimensions. Warning: the earth is not flat! Typically, the absolute position of some of the sensors (beacons) is known, eg via GPS Sensors acquire some other position information, eg reciprocally measure distance to neighbours, ie those within a radius r. Sensors also measure something else-- biotoxins, water pressure, fire temperature, etc

Rennes Control problems and Sensor Networks Covering a region with sensors each may see 3 or 4 others sensors may fail exact positioning may not be possible region may have irregular boundaries and/or interior obstacles Scanning with moving sensors There may be an evader Evader may destroy sensors Sensors with different capabilities Dynamic network A priori or adaptive strategies? Management of energy usage Sensing radius depends on power level Control architecture for swarm What needs to be sensed to control a moving swarm (eg birds, fish, UAVs)? Allow for robustness In warfare, may constrain architecture to avoid disclosure of position when transmitting

Rennes OUTLINE Aim of Presentation Control and Sensor Networks The sensor network localization problem Rigidity and Global Rigidity Computational Complexity of Localization Conclusions and open problems

Rennes Sensor Networks Depicts sensors with sensing radius r r Sensor

Rennes Sensor Networks Sensor graph, with connection between two sensors if closer than r

Rennes Beacon sensor Normal sensor Sensor Networks Beacon sensor positions known absolutely Inter-neighbour distances known (edge distance for each edge of graph) plus inter-beacon distances

Rennes Sensor Networks Beacon sensor positions known absolutely Inter-neighbour distances known (edge distance for each edge of graph) plus inter-beacon distances Beacon sensor Normal sensor

Rennes Sensor Networks-Questions What are the conditions for network localizability, ie ability to determine the absolute position of all sensors--in first instance from NOISELESS data? What is the computational complexity of network localization? The first question is an old one (Cayley, Menger, chemists)

Rennes Sensor Networks-Questions Need to work with a notion of generic solvability--need solvability for all values of distance round nominal Could formulate other problems with different inter-sensor information (eg interval of distance values, or direction) Interest exists in two and three dimensions Not yet studying dynamic networks

Rennes Sensor Networks and Formations A point formation is a set of points together with a set of links and values for the lengths of the links. A formation determines a graph G = (V,L) of vertices and edges, and lengths of the edges. A formation is like a sensor network with the absolute beacon positions thrown away A formation that is exactly determined by its graph and distance function is globally rigid. Any other formation with the same data is congruent, ie is determinable by translation and/or rotation and/or reflection.

Rennes Congruent Formations Translation Reflection Original position Absolute beacon positions eliminate this residual uncertainty Rotation

Rennes Two dimensional rigidity examples Not rigid--distances do not determine precise shape. Globally Rigid-- distances determine shape to within reflection, rotation or translation Absolute beacon positions eliminate the reflection etc uncertainty

Rennes Sensor Networks and Formations Suppose: m beacons, n-m ordinary nodes, for 2 dimensions there are at least 3 beacons in general position, and in 3 dimensions at least 4 beacons in general position. Then: the sensor localization problem is solvable if and only if the associated formation is globally rigid Henceforth, we will focus on formations and their global rigidity

Rennes Aim of Presentation Control and Sensor Networks The sensor localization problem Rigidity and Global Rigidity Computational Complexity of Localization Conclusions and open problems OUTLINE

Rennes Let F be a formation with vertex and edge sets V and L. Imagine it is moving. Let q i denote the position at time t of the i-th vertex. For each edge (i,j) in L, let  (i,j) denote the fixed distance. Then: (q i - q j ) (Dq i - Dq j )= 0 Can write this equation for every edge: R(F)(Dq) = 0 Here R(F) is the rigidity matrix. For a rigid formation: 1. One rotation and two translations give nullspace of dimension 3 in two dimensions 2. Three rotations and three translations give nullspace of dimension 6 in two dimensions Rigidity

Rennes Two dimensional rigidity examples Not rigid. One degree of freedom “floppiness”. R(F) has 4 dimensional nullspace Rigid. R(F) has 3 dimensional nullspace

Rennes Three dimensional rigidity examples Not rigid. R(F) has 7 dimensional nullspace Rigid. R(F) has 6 dimensional nullspace

Rennes Rigid Formations v1v1 v2v2 v3v3 v4v4 (1,2)x 1 - x 2 y 1 - y 2 00 (1,3)x 1 - x 2 y 1 - y (1,4)000x 1 - x 2 y 1 - y 2 (2,3)0x 1 - x 2 y 1 - y 2 0 (2,4)0x 1 - x 2 y 1 - y 2 0 (3,4)00x 1 - x 2 y 1 - y 2 Sample two dimensional Rigidity Matrix--a Matrix Net ∑ x i M i +y i N i in coordinates x i and y i of points.

Rennes More on rigidity Rank R(F) for a fixed graph will have the same value for almost all lengths One has to focus on genericity issues and work with generic rigidity In two dimensions, there is a combinatorial characterization of generically rigid graphs- Laman’s theorem, with fast algorithm for testing No such result is available in three dimensions. (Partial results exist)

Rennes Rigidity versus global rigidity a d c a c d b b Both formations are rigid. Neither can be changed into the other by translation, rotation or reflection.They have the same edge lengths. So they are not globally rigid! It is possible to have a strictly finite number greater than one of solutions to the formation realization problem --this connotes rigidity but not global rigidity.

Rennes Rigidity versus global rigidity We can fix the previous problem if we fix the distance between b and a. This makes the graph redundantly rigid (and 3-connected, see next slide) a c d b

Rennes Rigidity versus global rigidity Formally, a graph is redundantly rigid if the removal of any single edge gives a graph that is also generically rigid. A graph is k-connected if the removal of any set of less than k vertices means that it is still connected. Equivalently, it is k-connected if for any pair of vertices, one can find k paths joining them, with no common vertices except the end vertices. Theorem: In two dimensions a graph with at least 4 vertices is generically globally rigid if and only if it is 3- connected and redundantly rigid. This connects a global property needed to solve estimation problem to a local property holding almost everywhere, for 2D graphs.

Rennes D Global rigidity--examples Theorem: In two dimensions a graph with at least 4 vertices is generically globally rigid if and only if it is 3- connected and redundantly rigid. Nontrivial consequence: 6-connectivity is sufficient for global rigidity in two dimensions. “Wheel” graphs with at least four vertices are globally rigid

Rennes D Global rigidity --examples Theorem: Let G=(V,E) be a 2-connected graph. Let G 2 = (V,E  E 2 ) be the graph formed from G by adding an edge between any two vertices with a common neighbor vertex in G. Then G 2 is globally rigid. One gets G 2 by doubling sensor radius! Example where G is a cycle G G2G2

Rennes D global rigidity In three dimensions: If a graph is generically globally rigid, then it is redundantly rigid and at least 4-connected. There is a counterexample to the converse: bipartite graph K 5,5 Necessary and sufficient conditions for 3D global rigidity are not known! In three dimensions, if a particular formation (graph plus distances) is globally rigid, it is not known whether almost all formations with the same graph are globally rigid. 12-connected 3D graphs might be always globally rigid

Rennes Two dimensional trilateration

Rennes Trilateration One way to construct globally rigid formations: add a new node to a globally rigid formation, connecting it to d + 1 nodes of the existing formation in general position (d = spatial dimension). Then the new formation is generically globally rigid. A trilateration graph G in dimension d is one with an ordering of the vertices 1,…d+1,d+2,….n such that the complete graph on the initial d+1 vertices is in G and from every vertex j > d+1, there are at least d+1 edges to vertices earlier in the sequence. Trilateration graphs are generically globally rigid.

Rennes Two dimensional trilateration

Rennes Nongeneric behaviour Globally rigid a b c da b c d’ But not globally rigid when a,b,c are collinear! Globally rigid

Rennes Trilateration Theorem: Let G=(V,E) be a connected graph. Let G 3 = (V,E  E 2  E 3 ) be the graph formed from G by adding an edge between any two vertices at the ends of a path of 1,2 or 3 edges. Then G 3 is a trilateration graph in 2 dimensions. Also G 4 is a trilateration graph in 3 dimensions. Hence if G(r) is connected, G(3r) is a trilateration in two dimensions, and G(4r) is a trilateration in three dimensions

Rennes Aim of Presentation Control and Sensor Networks The sensor localization problem Rigidity and Global Rigidity Computational Complexity of Localization Conclusions and open problems OUTLINE

Rennes Brute force: Minimize  {  (i,j) - || q i - q j ||} 2 (i,j)  E Computational Complexity of Localization Theorem: Trilateration graph is realizable in polynomial time. (Proof relies on finding a seed in polynomial time--choose 3 out of n--and then realizing starting with seed, which is linear time) Theorem: Realization for globally rigid weighted graphs (formations) that are realizable is NP-hard. (Proof relies on wheel graph and NP-hardness of set-partition -search problem. Heuristic argument on next slide)

Rennes Computational Complexity Reflection possibilities are linked with computational complexity Suppose all edge distances known for small triangles. Localization goes working out from any beacon. Triangle reflection possibilities grow exponentially…. …and reflection possibilities are only sorted out when one gets to another beacon

Rennes Trilateration localization protocol Sensors have 2 modes, localized and unlocalized Sensors determine distance from heard transmitter All sensors are pre-placed and listening Localized mode Broadcast position Unlocalized mode listen for broadcast IF broadcast from (x,y) heard, determine distance to (x,y) IF 3 broadcasts heard, determine position and switch to unlocalized mode Decentralized algorithm! But how fast?

Rennes Aim of Presentation Control and Sensor Networks The sensor localization problem Rigidity and Global Rigidity Computational Complexity of Localization Conclusions and open problems OUTLINE

Rennes Conclusions Rigidity is not enough; you need global rigidity to localize (+ beacons) Even then, computational complexity may be terrifying Polynomial or linear time localization is possible, given trilateration Change of sensing radius converts connectedness to global rigidity/trilateration For a class of random sensor graphs, there is not much difference between rigid, globally rigid and trilateration. Results for 3D are less developed.

Rennes Some Open Problems Three dimensional graphs Partial localizability Islands of localizability in random graphs Asymmetric sensing radii Angular sensing Measures of ‘health’: graphical, and geometric Motion of sensors Random graphs.

Rennes Random sensor networks Sensors may be deployed randomly. We are interested in localization. The tool is random graph theory (which has been heavily studied) The random geometric graphs G n (r) are the graphs associated with two dimensional formations with n vertices with all links of length less than r, where the vertices are points in [0,1] 2 generated by a two dimensional Poisson point process of intensity n

Rennes Random geometric graphs There is a phase transition at which the graph becomes connected with high probability: r = O(sqrt[(log n)/n]) Connected means: if G n (r) has a minimum vertex degree of k then with high probability it is k-connected. Since 6-connectivity guarantees global rigidity, r = O(sqrt[(log n)/n]) implies global rigidity with high connectivity.

Rennes D Random geometric graphs If G n (r) is 2-connected, then G n (2r) is globally rigid If G n (r) is connected, then G n (3r) is a trilateration Let r 1,r 2, r 3, r g and r t denote the radius at which G n (r) is connected, 2-connected, 3- connected, globally rigid and a trilateration with probability 1- . Then for large n, r 6  r g  r 3  r 2 and3r 1  r t  2r 2  r g

Rennes Illustration of phase transition Probability that G n (r) is k-connected or globally rigid

Rennes Random geometric graphs All the above have an underlying condition of type r = O(sqrt[(log n)/n]) If nr 2 /(log n) > 8, then with high probability G is a trilateration graph, and is localizable in linear time given the positions of 3 connected nodes. Key observation for proof: the density of nodes guarantees one can pick an initial triangle of 3, and then one at a time a new node connected to 3 of those already chosen It is also localizable in a sort of decentralized fashion.

Rennes Beacons and localization time Suppose sensors placed with Poisson intensity n and sensing radius r = O(sqrt[(log n)/n]). If 3 beacons are placed closer than r, can localize in O(sqrt[n/(log n)] steps If beacons are placed on the unit square by Poisson process with intensity O(n/log n), can localize in O(sqrt(log n)) steps (Key idea: probability that square of side O(r) has 3 beacons is constant p; so some such square has 3 with very high probability) If beacons are placed by Poisson process of intensity O(n), localization can be effected in O(1) time with very high probability

Rennes Illustration of phase transition Phase transition is sharper for bigger n (Beacons all sense one another) This and next graphs for 3D!

Rennes Theory vs simulation Sensing radius required to get 95% localization via trilateration (Beacons all sense one another)

Rennes Speed of localization Steps required to complete localization vs sensing radius (Beacons all sense one another)