Spherical Euclidean Distance Embedding of a Graph
Hou-Duo Qi
University of Southampton
Presented at Isaac Newton Institute Polynomial Optimization
August 9, 2013 Spherical Embedding Problem The Problem: Given n points in
Pre-distance matrix (dissimilarity matrix):
I D is symmetric
I Dii = 0 (zero diagonals)
I Dij ≥ 0 (non-negativities)
n n n S , S+, and Sh (Hollow subspace): n I S := {all n × n symmetric matrices} , n n I Sh := {X ∈ S : Xii = 0 ∀ i} n n I S+ := the set of all PSD matrices in S . 1. Squared Euclidean Distance Matrix (EDM)
A n × n matrix D is a (squared) Euclidean Distance r Matrix (EDM) if there exist points p1,..., pn in < such that
2 Dij = kpi − pjk ∀ i, j.
? Squared pairwise distances are used. ? I The set of all n × n EDMs is a closed convex cone. 3. Characterization of EDM: Schoenberg (1935), Young and Householder (1938) I Schoenberg in (Ann. Math. 1935), and (independently) Young and Householder in (Psychometrika, 1938). n D is EDM ⇐⇒ D ∈ Sh and − (JDJ) 0, where 1 J = I − eeT /n or (J = I − ). n I Furthermore, let 1 B = − JDJ, 2 and B has the following decomposition: B = PP T , with P ∈ Let pi = P (i, :), we have 2 Dij = kpi − pjk . 3. Characterization of EDM: Schoenberg (1935), Young and Householder (1938) Remarks (R1) The Schoenberg-Young-Householder characterization has two steps: The first step is to versify whether a given matrix is EDM. The second step is the embedding step by computing a spectral decomposition. (R2) It has become a major method for data dimension reduction – the classical Multidimensional Scaling (cMDS). (R3) The matrix JDJ has zero as its eigenvalue. Therefore, the Slater condition is never satisfied for the constraint: −JDJ 0. 4. Partial Distances among 50 Sensors 5. Algorithm: Isomap I Many methods are available I Euclidean distance matrix completion (Laurent (1997), Wolkowicz, Anjos et. al from 1999 –) I Y. Ye and his co-authors on Semi-Definite Programming (SDP) Relaxations (from 2004 –) I Kim et. al on Sparse Full SDP (2009, 2012) I Mor´eand Wu (DGSOL package, Argonne National Laboratory, 1999). I Several more packages (e.g., PENNON). I Isomap by Tenenbaum, Silva, and Langford (Science 2000). I Regard the problem as a network (graph) problem. Length of the edge is the distance (not necessarily accurate) I Replace the missing distances by the shortest path distances in the graph. I Use the Schoenberg-Young-Householder method to recover the locations of the nodes. 4 Points Embedding 4 Points Embedding by Isomap Computing Nearest EDM I Given a pre-distance matrix D, find a true EDM matrix Y that is the nearest to D: min kY − Dk2 s.t. Y is EDM rank(JYJ) ≤ r (embedding dimension constraint) I By the Schoenberg-Young-Householder characterization, we have n Y is EDM ⇐⇒ Y ∈ Sh , −JYJ 0. I We have a convex quadratic SDP. 4 Points Embedding by EMBED (Q. and Yuan 2012) 6. Another Characterization of EDM I Hayden and Wells (SIMAX, 1990) and Gaffke and Mathar (Metrika,1989): n n D is EDM ⇐⇒ D ∈ Sh ∩ (−K+), where n n n T ⊥o K+ := A ∈ S : x Ax ≥ 0, x ∈ e . n Note: K+ is a closed convex cone. n I K+ as projected spectrahedra: n T K+ = A | (A, t ≥ 0) such that A − tee 0 n I K+ as set-copositive cone. I Conic Formulation of the nearest EDM (Q. and Yuan 2012): 2 n n min kY −Dk s.t. Y ∈ Sh ∩(−K+), and rank(JXJ) ≤ r. 7. Dealing with Spherical Constraints I We now want to place n points on a sphere: kxik = R. I We assume the center to be the (n + 1)th point xn+1 so that 2 2 kxi − xn+1k = R , ∀ i = 1,..., 2. I The formulation of optimization problems with spherical constraints takes the following form: min / max f(Y ) n n+1 s.t. Y ∈ Sh ∩ (−K+ ) rank(JXJ) ≤ r Y1(n+1) = Yi(n+1), i = 2, . . . , n. I When there are no rank constraint, the problem is often convex (many such problems from geometric embedding of graphs). 8. Smallest Hypersphere Representation of a Grpah I Def. Let G = (V,E) be a graph with |V | = n.A unit-distance representation of g is a system of n vectors (p1,..., pn) in a Euclidean space such that kpi − pjk = 1 ∀ (i, j) ∈ E. I Def. If furthermore, kpik = kpjk ∀ i, j ∈ V the system is called a hypersphere representation of G. Unit-distance realization of Petersen graph on plane 8. Smallest Hypersphere Representation of a Graph I Finding the smallest radius of a hypersphere representation (Lov´asz(’09), Silva and Tuncel (’10)) th(G) := min t s.t. diag(X) = te Xii − 2Xij + Xjj = 1, ∀ (i, j) ∈ E n X ∈ S+, t ∈ <. I It is known 1 2th(G) + = 1. ϑ(G) I EDM formulation: min Y1(n+1) n n+1 s.t. Y ∈ Sh ∩ (−K+ ) Yij = 1 ∀ (i, j) ∈ E Y1(n+1) = Yi(n+1), i = 2, . . . , n. 9. Lov´asz-thetaFunction I Def. An orthonormal representation of G is a system {p1,..., pn} of unit vectors in a Euclidean distance space such that hpi, pji = 0 ∀ (i, j) 6∈ E. Theorem 5, Lov´asz(’79): Let (p1,..., pn) range over all orthonormal representations of G and d over all unit vectors. Then n X 2 ϑ(G) = max (hd, pii) . i=1 I SDP formulation: ϑ(G) = max hJ, Xi s.t. hI,Xi = 1 Xij = 0, ∀ (i, j) ∈ E X 0. From Projection to Euclidean Distance I We have 2 2 2 kd − pik = kdk + kpik − 2hd, pii = 2 − 2hd, pii. Hence 1 2 (hd, p i)2 = 1 − kd − p k2 . i 4 i 0 I Under the condition (part of Schrijver’s ϑ function): hd, pii ≥ 0, we have 1 2 max (hd, p i)2 ⇐⇒ min kd − p k2 i 4 i I This leads to the following EDM problem 1 Pn 22 p(G) := min 4 i=1 kd − pik s.t. kpik = 1; kdk = 1; hpi, pji = 0 ∀ (i, j) ∈ E. A Quantity that may be interesting I For a given graph, define the quantity q(G) such that √ pp(G) + pq(G) = n. I Let SOL(G) denote the solution set of the SDP of ϑ function. Define b(ϑ) τ := √ , ϑ n where Pn √ b(ϑ) := max i=1 Bii s.t. B ∈ SOL(G). I For vertex-transitive graphs τϑ = 1. Bound that measures Distortion I Define p rϑ := n/ϑ(G) and 1 p t := r − (r − 1)2 + 2r(1 − τ). ϑ τ I Define the distortion constant dϑ by dϑ := tϑτϑ I Claim (Bound of Distortion): 2 dϑϑ(G) ≤ q(G) ≤ ϑ(G). I Remark: dϑ hard to calculate. But for vertex-transitive graphs, we have dϑ = 1. Is Triangle Inequality `2 Metric? I One can add triangle inequalities to SDP to strengthen ϑ function: Xik + Xjk ≤ Xij + Xkk (i, j, k ∈ V ). I Let X 0 admit the Gram representation: X = P T P. Therefore, 2 kpi − pkk = Xii + Xkk − 2Xik. 2 I ` -metric: 2 2 2 kpi − pkk + kpj − pkk ≥ kpi − pkk which implies 1 X + X ≤ X + (X + X ). ik jk ik 2 kk jj A Wild Guess I The close τϑ to 1, the less room that adding cut (triangle) inequalities can strengthen ϑ(G). I Example is vertex-transitive graphs. I We can measure this by computing the ratio: ϑ(G) q(G) Both are convex problems.