Sensor network synchronization

Transcription

1 Sensor network synchronization Andrea Fusiello University of Udine joint work with Federica Arrigoni

2 Introduction In a network of nodes, each node has an unknown state and measures of differences (or ratios) of states are available. Example in Z:?? 7? - 6? - - 7? -? The goal is to guess the unknown states from the available measures.

3 Introduction In a network of nodes, each node has an unknown state and measures of differences (or ratios) of states are available. Example in Z: The goal is to guess the unknown states from the available measures.

4 adding a constant to the solution yields another valid solution not every edge labeling produces a solvable problem: circuits must have zero sum. Kirchhoff s voltage law (the directed sum of the electrical potential differences around any closed network is zero).

5 This is an istance of the synchronization problem. In general, states can be elements of any group, possibly with noisy or wrong measures. A related problem is that of localization, where the state is a position in D space and measures are relative displacements. Displacements can be known only partially, as directions or distances.

6 Consistent labelling Let Σ be a group with unit element Σ. States are elements of Σ. Let G = (V, E) be a finite simple digraph, with n = V vertices and m = E edges. Consider a labelling of the edges z : E Σ such that if (u, v) E then (v, u) E and z(v, u) = z(u, v). We say that Γ = (V, E, z) is a Σ-labelled graph.

7 A cycle in a undirected graph is a subgraph in which every vertex has even degree. A circuit is a connected cycle where every vertex has degree two. Definition. (Null cycle) We say that a cycle v v, v v,... v l v in Γ is null if and only if z(v, v ) z(v, v )... z(v l, v ) = Σ. x z, z,6 z, x z,6 x 6 x non-null cycle z, z, z, z, Σ z, x z, z,5 z,5 x 5 z5,6 null cycle z,5 z 5,6 z 6, = Σ

8 Definition. (Consistent labelling) Let Γ = (G, z) be a Σ-labelled graph for G = (V, E). Let x : V Σ be a vertex labeling. We say that x is a consistent labelling iff z(e) = x(u) x(v) e = (u, v) E x x z, = x x z, = xx z,6 = x 6 x z,6 = x 6 x x 6 x z, = xx z, = x x z,5 = x 5 x z5,6 = x6x 5 x z,5 = x 5 x x 5

9 Proposition. Let Γ = (G, z) be a Σ-labelled graph. Then there exists a polynomial algorithm which either finds a non-null cycle in Γ or finds a consistent labelling of Γ. Proof. (sketch) Use a spanning tree to label vertices: this is a consistent labelling by construction. Then add one by one the edges not belonging to the spanning tree, thereby creating a cycle. If the cycle is null the edge can be added and leave the labelling consistent, otherwise a non-null cycle has been found. Corollary. The graph Γ has a consistent labeling if and only if all its cycles are null.

10 . Finding non-null cycles: GFES Definition.5 (GFES) Let Γ = (G, z) be a Σ-labelled graph with labelling z. The Group Feedback Edge Set (GFES) problem is defined as follows: on input (Γ, k) for some k N, decide whether there exists a subset of the edges S E with S k such that the labelled graph of the remaining edges Γ = (V, E \ S, z) does not contain a non-null cycle (i.e. it has a consistent labelling). The interpretation is that S identifies edges with outlying labels that prohibit a ground truth consistent labelling to be found. Solutions: from graph theory community: (Cygan et al., 0; Wahlström, 0) from CV community: based on outliers rejection heuristics: (Enqvist et al., 0); (Arrigoni et al., 0b); based on RANSAC: (Govindu, 006); based on bayesian inference: (Zach et al., 00).

11 . Cycle basis Viewing cycles as vectors indexed by edges, addition of cycles corresponds to modulo- sum of vectors, and the cycles of a graph form a vector space in Z m. = = 0 (,) (,) (,) (,) (,5) (5,) Fig. : The sum of two cycles is a cycle where the common edges vanish.

12 The dimension of such a space is m n+c, where c denotes the number of connected components in G = (V, E). A cycle basis is a minimal set of circuits such that any cycle can be written as linear combination of the circuits in the basis. If we stack the indicator vectors 5 of the circuits of a basis in a matrix C (by rows) we 5 obtain the cycle basis matrix. 5 (a) Graph G = (V, E). 5 (b) Cycle Basis. Fig. : Example of a cycle basis associated to a given graph G = (V, E). In general, a cycle basis is not unique. 5

13 Synchronization Find an approximately consistent labelling. Let us assume now that we are given a symmetric positive definite function f : Σ R with a unique minimum at Σ and f ( Σ ) = 0. Definition. (Consistency error) Let Γ = (G, z) be a Σ-labelled graph for G = (V, E). Let x : V Σ be a vertex labeling. The consistency error of x is defined as ɛ( x) = f ( z(u, v) z(u, v) ) () (u,v) E where z is the edge labeling induced by x: z(u, v) = x(u) x(v). A (vertex) labeling is consistent iff it has zero consistency error.

14 Definition. (Synchronization) The synchronization problem consists in finding a vertex labeling of G with minimum consistency error, given a (edge) labeling z. The interpretation is that one wants to recover unknown group elements (vertex labels) given a redundant set of noisy measurements of their ratios (edge labels). x =? x =? z, xx z, x x z, x x z, xx x =? z,6 x 6 x z,6 x 6 x z,5 x 5 x x 6 =? z5,6 x6x 5 x =? z,5 x 5 x x 5 =?

15 Synchronization can be solved as soon as the graph is connected, but errors compensation happens only within cycles. Some instances: Σ = R time synchronization: (Giridhar and Kumar, 006) Σ = Z sign synchronization: (Cucuringu, 05) Σ = R d state synchronization / translation synchronization: (Russel et al., 0; Tron et al., 05) Σ = SO() rotation synchronization (averaging): (Martinec and Pajdla, 007; Singer, 0; Hartley et al., 0; Chatterjee and Govindu, 0); (Arrigoni et al., 0a) Σ = SE() motion synchronization (averaging): (Govindu, 00; Torsello et al., 0); (Arrigoni et al., 06c) Σ = SL(d) homography synchronization: (Schroeder et al., 0) Σ = S d permutation synchronization: (Pachauri et al., 05).

16 . Synchronization in (R, +) In Σ = R a vertex labeling x : V R is consistent with a given edge labeling z : E R iff x(v) x(u) = z(u, v) (u, v) E () Let us denote the incidence vector of the edge (u, v) with b (u,v) = (0,...,,...,,..., 0) T () u Let x be the vector containing all the vertex labels and z the vector containing the edge labels (ordered as in B). Equation () can be written as v x T b (u,v) = z T () Let B be the n m incidence matrix of G, which has the b (u,v) as columns; it is easy to see that for all the edges the equation above writes x T B = z T, or B T x = z (5)

17 Considering f ( ) = : Σ R +, the consistency error of the synchronization problem writes ɛ(x) = x(v) x(u) z(u, v) (6) (u,v) E The least squares solution of B T x = z solves the synchronization problem. Oss. If c is the indicator vector of a cycle, the cycle is null iff c T z = 0. Proposition. (Russel et al. (0)) If ˆx is the least-squares solution of (5), then the induced edge labeling ẑ = B Tˆx solves the following constrained minimization problem min z z z s.t. C z = 0 (7) where C is the cycle basis matrix. The interpretation is that: the edge labels produced by the synchronization are the closest to the input edge labels among those that yield null-cycles.

18 .. Time synchronization. In a wireless network, nodes must often act in coordinated or synchronized fashion. This requires global clock synchronization, wherein all nodes in the network are synchronized to a common clock. The network is modeled as a directed graph of n + nodes {0,,,..., n}, where each edge represents the ability to transmit and receive packets between the corresponding pair of nodes. Each node i has a clock c i (t), where t represents the reference time variable of node 0, which is the designated reference node, and c i ( ) is some unknown function. A simple first order model of such a function would be c i (t) = t + o i. The clock synchronization problem would then consist of finding the offset o i of each node i from the reference node 0.

19 Estimates of clock differences between pair of nodes connected by an edge can be obtained by exchanging of time-stamped packets The differences between the time of transmission and the time of reception is the sum of the propagation delay d, the clock offsets of the respective clocks, and random effects (eliminated by averaging). The effect of propagation delay can be eliminated by subtracting the one-way difference from the reverse-way difference. o = (t t + t t)/

20 At the end, we are able to obtain for each edge ij in the network, an estimate o ij of the true clock offset. These estimates must then be processed by the network to obtain estimates of the clock offsets of all nodes from the common reference clock od node 0. This is a simple synchronization in (R d, +) where x i = o i and z ij = o ij. Alternately, one could employ a two parameter model for the node clock: c i (t) = d i t + o i where the skew parameter d i represent the clock drift relative to node 0. This is an istance of syncronization in the group of -d affinities, a subgroup of GL(). In reality, these parameters will be time-varying. Thus, it is necessary to recompute their values periodically in order to maintain a bounded estimation error.

21 . Synchronization in (R d, +) In Σ = R d a vertex labeling x : V R d is consistent with a given edge labeling z : E R d iff x(v) x(u) = z(u, v) (u, v) E (8) Reasoning as in the scalar case, this becomes XB = Z (9) where the columns of X and Z are vectors of R d and B is the n m incidence matrix of G. Equivalently, using the Kronecker product: (B T I d ) vec X = vec Z. (0) This is indeed a synchronization problem where the incidence matrix B gets inflated by the Kronecker product with I d in oder to cope with the vector representation of the group elements.

22 . Synchronization in (R\{0}, ) In Σ = (R\{0}, ) a vertex labeling x : V R is consistent with a given edge labeling z : E R iff z(u, v) = x(u) x(v) (u, v) E () The consistency constraint can be expressed in an equivalent compact matrix form. Let x be the vector containing the vertex labels and let Z be the matrix containing the edge labels x = x x..., Z = xn z... z n z... z n () z n z n...

23 For a complete graph, the consistency constraint rewrites Z = xx T () where xx T contains the edge labels induced by x (with a little abuse of notation, we define x as the vector containing the inverse of each element of x Σ). If the graph G is not complete (missing measures) we shall write the constraint as (Z A) }{{} = (xx T ) A () Z A where A is the (0-) adjacency matrix of G and is the Hadamard product. The synchronization problem rewrites: Z A = A (xx T ) Z A x = Dx D Z A x = x (D Z A )x = 0 (5) where D = diag(a) is the degree matrix of the graph. The solution x is the eigenvector of D Z A associated to eigenvalue ; The solution x belongs to the null-space of (D Z A ).

24 Considering f ( ) = : Σ R +, the consistency error of the synchronization problem writes ɛ(x) = z(u, v) x(u) x(v) (6) (u,v) E In matrix form, the cost function of the synchronization rewrites ɛ(x) = (Z xx T ) A F (7) To the best of our knowledge, no general results are known linking the algebraic solutions (eigen and SVD) to the synchronization cost function, in the noisy case. Note also that the two algebraic solutions do not coincide in this case.

25 . Synchronization in GL(d) In this case Σ is the group of d d invertible matrices over R. Hence the (multiplicative) synchronization problem can be instantiated with: X I X... X n X = X..., Y = [ X, X,... X n, ], Z = X I... X n (8) Xn X n X n... I The consistency constraint rewrites (compete graph): The solution X is recovered as (general graph): the d top eigenvectors of (D I d ) Z A ; the d-dimensional null-space of (D I d ) Z A. where Z A = Z (A d d ). Z = XY (9)

26 Brief recap. If the group is additive, solve the linear system B T x = z. The measures (state differences) are stacked in z according to the edge ordering; B is the incidence matrix of the graph. If the group is multiplicative, solve the homogeneous system (Z A D)x = 0. The measures (state ratios) are in a matrix Z A and the matrix (D Z A ) = (D A) Z is related to the graph Laplacian (D A). This holds for scalar states; if the state is a matrix, everything must be suitably inflated with a wise use of Kronecker products.

27 Motion synchronization. Motion synchronization is an instance of the GL(d) synchronization with Σ = SE(). Each group element is described by a homogeneous rigid transformation M i = ( ) Ri x i 0 SE() M ij = ( ) Rij x ij 0 SE() (0) where R i, R ij SO() and x i, x ij R represent the rotation and translation components of the isometry. The vertex labeling is consistent iff M ij = Mi M j, which is equivalent to R ij = R T i R j () x ij = R T i x j R T i x i () by considering separately the rotation and translation terms.

28 Reference M Reference M Reference M n M n World reference system M Reference Reference M M Reference Reference n M M n Reference n Fig. : Motion synchronization.

29 When the synchronization problem is solved in GL(), the blocks in X are not guaranteed to belong to SE(). In order to project the solution onto SE(): compute a linear combination of the columns of X such that every fourth row becomes [0 0 0 ] project each block onto SO() through SVD.

30 . Rotation synchronization It is also known as multiple rotation averaging (Hartley et al., 0). Rotation synchronization is a particular case of the synchronization problem in the group of rotations SO() = {R R s.t. R T R = I, det(r) = }. In matrix form: X = R T R..., Y = [ R, R,... R n, ], Z = Rn T I R... R n R I... R n () R n R n... I With respect to the general case, SO() is a special one, for Y = X T symmetric and positive semidefinite. and Z is Therefore, the consistency constraint R ij = R T i R j becomes: Z = XY = XX T. ()

31 The consistency error for rotation synchronization is ɛ = R ij Ri T R j F (5) (i,j) E Singer (0); Arie-Nachimson et al. (0) prove that taking the top eigenvectors of (D I ) Z A corresponds to solving a relaxed rotation synchronization with orthonormality of the columns of X (instead of each block.) At the end, each block of X is projected onto SO() through SVD. Usual scheme: relax the constraints and subsequently project.

32 . Translation synchronization. The consistency constraint for translations (): can be written equivalently as x ij = R T i x j R T i x i (6) R i x ij = x j x i := u ij (7) where x i is the centre of the i-th camera and u ij is the baseline. By juxtaposing all the m baselines u ij in one m matrix U, we obtain an instance of syncronization in (R d, +), where the node labels are the position of the cameras: (B T I ) vec X = vec U. (8)

33 Localization The goal of localization is to compute the position of n nodes in d-space given measures on the edges. As in a translation synchronization problem the states of the nodes x i are positions, but the available measures are not differences of the states, in general. Localization Synchronization Direction: x i x j x i x j Distance: x i x j Translation: x i x j SO() SE() distance-based: only magnitute of translation is known; direction-based: only direction of translation is measured (AOA, bearings);

34 . Bearing-based localization The goal is to recover the position of n nodes in d-space, where pairs of nodes can measure the direction of the line joining their locations. x? x? u x x x x u x x x x x? u6 x x 6 x x 6 u 6 x x 6 x x 6 x 6? u x x x x u x x x x u5 x x 5 x x 5 u56 x 5 x6 x5 x6 x? u 5 x x 5 x x 5 x 5? The solution is defined up to translation and scale.

35 .. Incidence-based formulation Let us multiply the translation synchronization equation: (B T I ) vec X = vec U (9) by the block diagonal matrix Ŝ = blkdiag ({[û ij ] } (i,j) E ) yielding Ŝ(B T I ) vec X = Ŝ vec U = 0 (0) This step has the effect of substituting U, which is unknown, with Û) which is known instead. Ŝ (derived from The solution is the null space of Ŝ(BT I ).

36 Requiring that the underlying graph G = (V, E) is connected is not sufficient for unique localizability. In general, the graph must be parallel rigid (Whiteley, 997). Parallel Rigid Flexible

37 .. Parallel rigidity The classical characterization of parallel rigidity is given in terms of the rank of the parallel rigidity matrix, which is defined with reference to a particular embedding x (coordinates) of the graph G = (V, E): Q p,g = Ŝx(B T I ) where x R nd (n = V ) is an embedding of the graph, and Ŝx contains the induced bearings. Whiteley (997) demonstrates that a point formation (i.e., graph + embedding) in d-dimensions is parallel rigid iff rank(q p,g ) = d V (d + ) (in our case, n ) The notion of generic parallel rigidity, instead, is a property of the graph G without reference to a specific embedding; A graph G = (V, E) is generically parallel rigid in d dimensions iff max x is generic rank(q x,g) = d V (d + ) where x is a generic (i.e., no algebraic dependency among coordinates) embedding of the graph.

38 parallel rigid parallel rigid flexible flexible parallel rigid parallel rigid parallel rigid flexible 6...

39 .. Cycle-based formulation Let us rewrite the consistency constraint of synchronization over R in terms of directions û and magnitudes α. x i x j = u ij (i, j) E (B T I ) vec(x) = vec(u) x i x j = α ij û ij (i, j) E (B T I ) vec(x) = (I m Û)α where Û = [û... û ij...] and is the Khatri-Rao product. }{{} m Let us consider a cycle matrix C and multiply both sides by C I. (CB T I ) vec(x) = (C Û)α = CB T =0 0 = (C Û)α

40 Example.. Compute a cycle basis for the epipolar graph. 5 C = (,) (,) (,) (,) (,) (,5) (5,) Solve a homogeneous linear system. (C Û)α = 0 α û 0 0 û û û 5 û 5 α û 0 û û û 0 0 α = 0 0 û 0 û û 0 0 α α 5 α 5 α

41 Proposition. (Arrigoni et al. (05)) Node locations can be uniquely (up to translation and scale) determined from pairwise directions if and only if edge magnitues can be uniquely (up to scale) recovered from pairwise directions. parallel rigidity rank(c Û) = m

42 .. Laplacian formulation The least squares solution of bearing-based localization (both methods) is equivalent to minimize the following quadratic form where H = j [û,j] j [û,j]... min x = vec(x)t H vec(x). () j [û n,j] 0... [û,n ]..... () [û n, ]... 0 The matrix H resembles the definition of a Laplacian where the edge labels are the matrices on the diagonal blocks of Ŝ. This formulation has the same structure of a Laplacian eigenmap for distance-based localization, where the measures are the matrices [û i,j ] instead of distances.

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48 5 Khatri-Rao product The Khatri-Rao product (Khatri and Rao, 968), denoted by, is in some sense a partitioned Kronecker product, where by default the column-wise partitioning is considered. Let us consider two matrices A of order p r and B of order q r and denote the columns of A by a a r and the those of B by b b r. The Khatri-Rao product is defined to be the partitioned matrix of order pq r: where denotes the Kronecker product. If X is diagonal, then A B = [a b, a r b r ] () vec(axb) = (B T A) diag (X) () where diag returns a vector containing the diagonal elements of its argument.

49 With B = I one obtains It it is easy to see that vec(ax) = (I A) diag (X). (5) (I A) = blkdiag(a... a n ) (6) where a... a n are the columns of A and blkdiag is the operator that construct a block diagonal matrix with its arguments as blocks.

50 6 Graphs basics Let G = (V, E) a finite simple undirected graph with n nodes and m vertices. The adjacency matrix of G is defined as the n n matrix A(G) in which: {, if i and j are adjacent A(G) ij = 0, otherwise The incidence matrix of a finite simple directed graph G = (V, E) with n nodes and m edges is defined as:, if i is the head of e j B( G) ij =, if i is the tail of e j 0, otherwise The rows of the incidence matrix correspond to vertices of G and its columns to edges of G.

51 The degree matrix of the graph is the diagonal matrix defined as: { deg(v i ) = j D(G) ij = A(G) i,j, if i = j 0, otherwise or, equivalently: D = diag(z). A cycle in a undirected graph is a subgraph in which every vertex has even degree. A circuit is a connected cycle where every vertex has degree two.

52 Viewing cycles as vectors indexed by edges, addition of cycles corresponds to modulo- sum of vectors, and the cycles of a graph form a vector space in Z m. = = 0 (,) (,) (,) (,) (,5) (5,) Fig. : The sum of two cycles is a cycle where the common edges vanish.

53 A cycle basis is a minimal set of circuits such that any cycle can be written as linear combination of the circuits in the basis. If we stack the indicator vectors of the circuits of a basis in a matrix C (by rows) we obtain the cycle basis matrix. The dimension of the cycle space is m n + c, where c denotes the number of 5 5 connected components in G = (V, E). 5 (a) Graph G = (V, E). 5 (b) Cycle Basis. Fig. 5: Example of a cycle basis associated to a given graph G = (V, E). In general, a cycle basis is not unique. It can be proven that CB T = 0. 5