Network Identification via Node Knock-out
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This document presents a method for network identification via node knock-out. It discusses using input-output data from a network under different node knockout conditions to identify the network structure. Specifically, it shows how the characteristic polynomials of the original and grounded networks relate to the network's generating function and Laplacian matrix. This allows exact recovery of the network from system identification. Applications to fault detection are also discussed, where changes in characteristic polynomials after node removal can localize broken links.
![Uncontrollable grounded consensus =⇒ Disconnected Gv
w.l.o.g |I| = 1 and the grounded node v has the last index.
ˆ
We can also rewrite B as B = [ B T , 0 ].
PBH test
The original graph is controllable − − − −
− − − → (z = 0 and λ)
such that L(G)z = λz and z T B = 0.
Therefore for any choice of q ∈ Rn , ∃ p such that
p T = [p1 , p2 , p3 ] and
T T T
L(Gv ) + ∆v − λI δv ˆ
B
T p = q. (1)
δv deg v − λ 0
PBH test
The grounded consensus is uncontrollable − − − − ∃ (s = 0
− − −→
ˆ = 0) such that (L(Gv ) + ∆v )s = λs and s T B = 0.
and λ ˆ ˆ
The claim: ∃ a vector r = 1 ∈ R(n−1)×1 such that
L(Gv )r = 0. Prove that r = p1 .
Network Identification via Node Knock-out, slide 14/1](https://image.slidesharecdn.com/netid-120826203533-phpapp01/85/Network-Identification-via-Node-Knock-out-14-320.jpg)
![Exact network ID is facilitated by the following two results
Lemma
Let v ∈ G. Then
φv (s)
G
PG (s)vv =
φG (s)
Lemma
Let u, v ∈ G. Then
[ΨG (s)]uv
PG (s)uv =
φG (s)
where
[ΨG (s)]2 = φu (s)φv (s) − φG (s)φuv (s)
uv G G G
Since (sI + L(G))−1 = PG (s), L(G) can be determined
Network Identification via Node Knock-out, slide 16/1](https://image.slidesharecdn.com/netid-120826203533-phpapp01/85/Network-Identification-via-Node-Knock-out-16-320.jpg)
Network Identification via Node Knock-out
- 1. Network Identification via Node Knock-out Marzieh Nabi A., Mehran Mesbahi Aeronautics & Astronautics University of Washington February 11th, 2011 Network Identification via Node Knock-out, slide 1/1
- 2. Can one hear the shape of a drum? Network Identification via Node Knock-out, slide 2/1
- 3. 1966 - Mark Kac stated the problem. Almost immediately - John Milnor proved the existence of a pair of 16D tori. 1992 - Gordon, Webb, and Wolpert constructed a pair of regions in the plane. ... Steve Zelditch proved that the answer to Kac’s question is positive by restrictions to certain convex planar regions with analytic boundary. 2011 - open problem for non-convex analytic domains. Network Identification via Node Knock-out, slide 3/1
- 4. Can one hear the shape of a graph? Play eigenvalue scale Play eigenvalue scale 1 1 http://www.math.ucsd.edu/ fan/hear/index.html Network Identification via Node Knock-out, slide 4/1
- 5. Inverse Problems Forward Problem Forward Problem True Model Model Data Appraisal Problem Data Inverse Problem Estimated Model Estimation Problem geophysics medical imaging remote sensing ocean acoustic tomography ... Network Identification via Node Knock-out, slide 5/1
- 6. 1 Input signal Input signal Basic idea behind the procedure: 3 2 4 A network with unknown 5 6 communication structure Sensor 7 9 Steer the network through a set of 8 10 input nodes 11 13 Let the network run a simple 12 consensus protocol Sensor Sensor Observe a set of nodes Input signal The set of input nodes I = {2, 3, 12} The set of output nodes O = {5, 11, 13} Network Identification via Node Knock-out, slide 6/1
- 7. Modelling x(t) = A(G)x(t) + Bu(t) ˙ y (t) = Cx(t) where A(G) = −Lw (G) ∈ Rn×n is the weighted Laplacian. −1 0 1 0 0 −1 3 −1 −1 0 A(G) = − 0 −1 3 −1 −1 , 2 4 0 −1 −1 3 −1 0 0 −1 −1 2 1 3 1 0 0 1 5 1 0 0 0 0 B = 0 0 , C = . 0 0 0 0 0 1 0 0 0 Network Identification via Node Knock-out, slide 7/1
- 8. Network identification ... (partial) network identification via its characteristic polynomial (partial) network identification via a graph sieve method (exact) network identification via node knock-out and generating functions Network Identification via Node Knock-out, slide 8/1
- 9. System ID Set of Input- Output Linear realization System ID Data of the system (A, B, C) The estimated triplet (A, B, C ) is a realization of (A(G), B(G), C (G)). There exists a similarity transformation T , such that A = TA(G)T −1 B = TB(G) C = C (G)T Question: are we able to find the transformation matrix T from (A, B, C )? Network Identification via Node Knock-out, slide 9/1
- 10. Controllability/ observability of the network p=2 ln(n)/n+0.2 1 Percent of graphs controllable from at least one node The controllability/ observabil- 0.95 Planar Random Graphs ity conjecture in algebraic graph 0.9 theory (Godsil): for large values Bipartite Random Graphs 0.85 of n, the ratio of graphs with n nodes that are not controllable 0.8 to the total number of graphs 0.75 on n nodes approaches zero as 0.7 Random Graphs (n → ∞) 0.65 0 50 100 150 200 250 300 350 # of Vertices (n) Network Identification via Node Knock-out, slide 10/1
- 11. Network ID via input-output data When the network is controllable and observable, its realization provides access to the (Laplacian) characteristic equation of the underlying networked system φG (s) = det(sI + L(G)) = (s − λi (G)) n (1/n) i=2 λi (G) is the number of spanning trees in G |E| = 1/2 n λi , where |E| is the number of edges in i=1 the graph if an−1 = n, the underlying interconnection is a tree If T is a tree with n ≥ 2 vertices and only one positive Laplacian eigenvalue with multiplicity one, then T is the star K1,n−1 Network Identification via Node Knock-out, slide 11/1
- 12. Node knock-out: a mechanism for exact network characterization Exact network characterization is facilitated by “node knockout”: the input-output behavior can be observed after zeroing out the signal generated by a single vertex. G u Define: φG (s) = Characteristic equation −L(G) φu (s) = Characteristic equation of −(L(Gu) + ∆u ) G Gu: the graph after removing node u ∆u : the diagonal matrix capturing the links between u and the nodes in Gu Network Identification via Node Knock-out, slide 12/1
- 13. Controllability/ observability of the grounded network Gu Lemma The steered-and-observed grounded consensus remains controllable and observable as long as none of the input-output vertices are identical to the grounded node(s) if and only if the graph Gu stays connected. Main idea of the proof: The necessary condition follows from the definition of the controllability. The sufficient condition: if the grounded consensus is uncontrollable, the graph Gu has to be disconnected. Network Identification via Node Knock-out, slide 13/1
- 14. Uncontrollable grounded consensus =⇒ Disconnected Gv w.l.o.g |I| = 1 and the grounded node v has the last index. ˆ We can also rewrite B as B = [ B T , 0 ]. PBH test The original graph is controllable − − − − − − − → (z = 0 and λ) such that L(G)z = λz and z T B = 0. Therefore for any choice of q ∈ Rn , ∃ p such that p T = [p1 , p2 , p3 ] and T T T L(Gv ) + ∆v − λI δv ˆ B T p = q. (1) δv deg v − λ 0 PBH test The grounded consensus is uncontrollable − − − − ∃ (s = 0 − − −→ ˆ = 0) such that (L(Gv ) + ∆v )s = λs and s T B = 0. and λ ˆ ˆ The claim: ∃ a vector r = 1 ∈ R(n−1)×1 such that L(Gv )r = 0. Prove that r = p1 . Network Identification via Node Knock-out, slide 14/1
- 15. In fact, we will not identify the network directly, but find its generating function: ∞ χG (s) := s k (−L(G))k = (I + sL(G))−1 k=0 Useful observation: network input/output map: s −1 χG (s −1 ) = PG (s) Network Identification via Node Knock-out, slide 15/1
- 16. Exact network ID is facilitated by the following two results Lemma Let v ∈ G. Then φv (s) G PG (s)vv = φG (s) Lemma Let u, v ∈ G. Then [ΨG (s)]uv PG (s)uv = φG (s) where [ΨG (s)]2 = φu (s)φv (s) − φG (s)φuv (s) uv G G G Since (sI + L(G))−1 = PG (s), L(G) can be determined Network Identification via Node Knock-out, slide 16/1
- 17. Main Result Hence, L(G) can now be uniquely identified by running the system identification on the ungrounded and grounded system. (sI + L(G))−1 equivalent to s −1 χG (s −1 ), which according to the mentioned Lemmas, can be determined by the characteristic polynomials of the ungrounded and grounded consensus. Network Identification via Node Knock-out, slide 17/1
- 18. Network ID via a graph sieve method ˜˜˜ C AB = CAB Consider the structure of B and C Relabel the input-output nodes as the first r nodes The first equality leads to the first r × r block partition of the matrix L(G) Finding the degree of the input-output nodes, degvi for i = 1, . . . , r n ˜ degvi = trace(A) i=1 ˜ r Integer partitioning on s = trace(A) − i=1 degvi to find possible degree sequences Check whether these sequences are a graphical degree sequence Network Identification via Node Knock-out, slide 18/1
- 19. Sieve method continue Build all possible graphs based on the degree sequence and match with the known r × r known block Example nodes 1, 2, and 3 as the input-output nodes φG (s) = s 6 + 220s 5 + 190s 4 + 804s 3 + 1664s 2 + 1344s (a) a simple graph on 6 nodes, (b) a candidate graph with degree sequence {3, 4, 3, 4, 4, 4}, (c) {3, 4, 3, 4, 4, 4}, (d) {3, 4, 3, 5, 5, 2}. Network Identification via Node Knock-out, slide 19/1
- 20. Application: Fault detection Thus, as soon as a failure occurs in an edge in the network, the characteristic equation of modified network reflects this failure. The result of the identification process: the characteristic equation for the system and the number of edges in the network Explore the possibility of identifying the broken link from running the identification procedure on the network and its grounded version Network Identification via Node Knock-out, slide 20/1
- 21. Fault detection continue ... G Gu G = (V, E) and E = 2E Gv v Gv = (V, Ev ) and Ev = 2Ev u E − Eu − Ev + Euv = 0 if there is no edge between nodes u and v , E − Eu − Ev + Euv = 1 if there is an edge between u and v . Network Identification via Node Knock-out, slide 21/1
- 22. Example Node 3 Node 1 Consider G = (V, E) when |V| = 100, and |E| = 284. Node 2 E1 = 562, E2 = 560, and E1,2 = 554. E − E1 − E2 + E1,2 = 0 so there is no edge between nodes 1 and 2 Network Identification via Node Knock-out, slide 22/1
- 23. Concluding remarks Apply system ID in networked systems Find isomorphic graphs from the set of input-output data Application in fault detection More questions Applications in Random Walks on graphs, and Markov chains Biological networks Network Identification via Node Knock-out, slide 23/1