Distance constraints in Euclidean geometry Leo Liberti IBM - PowerPoint PPT Presentation

Distance constraints in Euclidean geometry Leo Liberti IBM Research, Yorktown Heights LIX, Ecole Polytechnique, France Joint work with : C. Lavor (IMECC-UNICAMP), N. Maculan (COPPE-UFRJ), A. Mucherino (Univ. Rennes) J. Lee (Univ. Michigan), B. Masson (INRIA), M. Nilges (Inst. Pasteur), T. Malliavin (Inst. Pasteur) OSE Workshop 2013 – p. 1/63

At a glance 0 1 8 4 3 6 3 1 6 8 2 9 9 2 4 5 0 7 3 7 1 2 4 5 0 6 6 8 9 7 2 1 6 9 5 1 5 0 3 8 4 4 7 0 8 3 5 2 9 7 Which graph has most symmetries? OSE Workshop 2013 – p. 2/63

How does a weighted graph look? 1 3 4 1 2 2 1 Like this? 3 OSE Workshop 2013 – p. 3/63

How does a weighted graph look? 1 3 4 1 2 2 1 Like this? 3 1 1 2 1 2 3 4 2 3 1 4 Perhaps like this? 3 OSE Workshop 2013 – p. 3/63

Motivation I: Don’t confuse a graph with its drawing OSE Workshop 2013 – p. 4/63

Completing partial matrices Schoenberg’s theorem: Euclidean Distance Matrix Completion Problem ⇔ Positive Semidefinite Matrix Completion Problem Low-rank matrix completion relaxations Covariance/correlation matrix completions OSE Workshop 2013 – p. 5/63

Motivation II: Drawing conclusions from partial data OSE Workshop 2013 – p. 6/63

Many applications Applications : Phase retrieval — 1D Wireless sensor network localization — 2D Molecular conformation — 3D Multidimensional scaling — (whatever)D OSE Workshop 2013 – p. 7/63

Motivation III: Variety: a new dimension, a new application! OSE Workshop 2013 – p. 8/63

A nonlinear system Given a simple weighted undirected graph G = ( V, E ) with a distance function d : E → R + , solve the constraint system: (1) ∀{ u, v } ∈ E � x u − x v � = d uv Obtain an embedding x : V → R 2 Computationally OK up to 5-10 vertices OSE Workshop 2013 – p. 9/63

Global optimization Reformulate (1) to � ( � x u − x v � 2 − d 2 uv ) 2 (2) min x { u,v }∈ E G has an embedding ⇔ optimum x ∗ of (2) has value 0. Eq (2) is nonconvex in x Computationally OK up to 100 vertices Surveys: [ITOR(2010), EJOR(2012), SIREV(to appear)] OSE Workshop 2013 – p. 10/63

Large-scale methods: Exploiting the combinatorial structure OSE Workshop 2013 – p. 11/63

The number of embeddings Uncountably many (incongruent) embeddings 3 4 1 2 OSE Workshop 2013 – p. 12/63

The number of embeddings Uncountably many (incongruent) embeddings Finitely many 3 3 4 4 1 2 1 2 4’ OSE Workshop 2013 – p. 12/63

The number of embeddings Uncountably many (incongruent) embeddings Finitely many At most one 3 3 4 4 3 4 1 2 1 2 4’ 1 2 Cannot have countably infinitely many solutions OSE Workshop 2013 – p. 12/63

Trilateration u 2 u 1 v �� �� �� �� �� �� �� �� u 3 v has ≥ K + 1 adjacencies with known general positions ⇒ If system has a solution, find x v in polytime OSE Workshop 2013 – p. 13/63

A linear system Let v ∈ V be adjacent to 1 , 2 , 3 x 1 , x 2 , x 3 known, find x v ∈ R 2 � x v � 2 − 2 x v · x 1 + � x 1 � 2 d 2 1 v (3) = � x v − x 1 � 2 d 2 = 1 v � x v � 2 − 2 x v · x 2 + � x 2 � 2 d 2 ⇒ 1 v (4) � x v − x 2 � 2 d 2 = = 2 v � x v � 2 − 2 x v · x 3 + � x 3 � 2 d 2 � x v − x 3 � 2 d 2 = 1 v (5) = 3 v (5)-(3)      ( � x 1 � 2 − � x 3 � 2 ) − ( d 2 1 v − d 2  2( x 1 − x 3 ) 3 v ) (5)-(4) ⇒  x v =  ( � x 2 � 2 − � x 3 � 2 ) − ( d 2 2 v − d 2 2( x 2 − x 3 ) 3 v ) Solve K × K system in polytime ⇒ but �⇐ : Cannot detect infeasibility OSE Workshop 2013 – p. 14/63

Combinatorial iterative approach V = { 1 , 2 , 3 , 4 , 5 , 6 , 7 } 5 3 6 7 4 1 2 OSE Workshop 2013 – p. 15/63

Combinatorial iterative approach V = { 1 , 2 , 3 , 4 , 5 , 6 , 7 } 5 3 6 7 4 1 2 OSE Workshop 2013 – p. 15/63

Combinatorial iterative approach V = { 1 , 2 , 3 , 4 , 5 , 6 , 7 } 5 3 6 7 4 1 2 OSE Workshop 2013 – p. 15/63

Combinatorial iterative approach V = { 1 , 2 , 3 , 4 , 5 , 6 , 7 } 5 3 6 7 4 1 2 OSE Workshop 2013 – p. 15/63

Combinatorial iterative approach V = { 1 , 2 , 3 , 4 , 5 , 6 , 7 } 5 3 6 7 4 1 2 OSE Workshop 2013 – p. 15/63

Combinatorial iterative approach V = { 1 , 2 , 3 , 4 , 5 , 6 , 7 } 5 3 6 7 4 1 2 OSE Workshop 2013 – p. 15/63

Does it work on my favourite application? OSE Workshop 2013 – p. 16/63

Proteins Proteins: backbone and side chains Backbone: total order < on a set V of atoms Assume known embedding for backbone; embedding side chains is known as S IDE C HAIN P LACEMENT P ROBLEM OSE Workshop 2013 – p. 17/63

Protein distances Covalent bond distances d v − 1 ,v are known H H O Angles between covalent bonds are known H H O ⇒ d v − 2 ,v is known for all v > 3 H H Distances d v − 3 ,v are always < 6 Å, so they can be measured using NMR techniques We assume these distances are exact: this is false in practice, but we can find orders for which this assumption holds (see later if I have time) NMR might give other distances too Atoms may be distant order-wise but closer than 6 ˚ A in space OSE Workshop 2013 – p. 18/63

Discretizable MDGP Protein backbones: 3 consecutive predecessors in 3D Weaken the condition ≥ K + 1 adjacent predecessors in R K to: ≥ K consecutive adjacent predecessors in R K DMDGP: complete an initial partial embedding in this setting NP -hard [Lavor et al. COAP 2012] OSE Workshop 2013 – p. 19/63

Adapt the iterative method? OSE Workshop 2013 – p. 20/63

Sphere intersections For given v > 3 , x v − 3 , x v − 2 , x v − 1 are known d v,v − 1 , d v,v − 2 , d v,v − 3 are known find x v Non-empty intersection of K spheres in R K contains 2 points in general OSE Workshop 2013 – p. 21/63

When does it fail? v v − 3 v − 1 v − 2 v OSE Workshop 2013 – p. 22/63

Branch-and-Prune v : rank of current atom x <v : partial embedding to rank v − 1 G : instance X : current pool of embeddings S ( y, r ) : R K sphere centered at y with radius r B RANCH A ND P RUNE ( v , x <v , G , X ): � Let S ← S ( x v − i , d v − i,v ) = ( { s 1 , s 2 } or ∅ ) i ∈{ 1 ,...,K } for s ∈ S do Extend current embedding to x = ( x <v , s ) if ∀ u ∈ AdjPred ( v ) � x u − x v � = d uv then if ( v = n ) then Let X ← X ∪ { x } else B RANCH A ND P RUNE ( v + 1 , x , G , X ) end if end if end for OSE Workshop 2013 – p. 23/63

BP properties BP: worst-case exponential time With probability 1, find all incongruent embeddings of G extending initial partial embedding Performs very efficiently (speed and accuracy) Embed 10,000 vertices in a 13 seconds of CPU time Two empirical observations: 1. the number of solutions it finds is always a power of two 2. | V | versus CPU time plots are always linear-like for PDB OSE Workshop 2013 – p. 24/63

Symmetry OSE Workshop 2013 – p. 25/63

BP root node symmetry [Lavor et al. COAP , to appear] e 2 4 is a reflection of x 4 x ′ x 1 w.r.t. the plane defined e 3 by x 1 , x 2 , x 3 ⇒ BP tree symmetric below level 3 x 2 x ′ Start branching from 4 x 3 level 4, not 3 x 4 e 1 OSE Workshop 2013 – p. 26/63

Number of solutions | X | Instance 2 1brv 4 1aqr 2 | X | 2erl Instance 2 1crn 2 16 mmorewu-2 1ahl 2 2 mmorewu-3 1ptq 4 4 mmorewu-4 1brz 2 4 1hoe mmorewu-5 4 2 mmorewu-6 1lfb For all tested DMDGP in- 2 1pht 4 2 lavor10 0 1jk2 stances, ∃ ℓ ∈ N such that 16 2 lavor15 0 1f39a 8 8 lavor20 0 1acz | X | = 2 ℓ 8 2 lavor25 0 1poa 2 2 lavor30 0 1fs3 64 2 lavor35 0 1mbn 2 2 lavor40 0 1rgs 2 2 lavor45 0 1m40 4096 2 lavor50 0 1bpm 64 2 lavor55 0 1n4w 64 2 lavor60 0 1mqq 2 1rwh 2 3b34 2 2e7z 2 1epw OSE Workshop 2013 – p. 27/63

A BP search tree example Typical BP search tree (embeddings = paths root → leaves) 1 2 3 4 29 5 17 30 42 6 13 18 22 31 38 43 47 7 12 14 15 19 21 23 24 32 37 39 40 44 46 48 49 8 16 20 25 33 41 45 50 9 26 34 51 10 27 35 52 11 28 36 53 Root node symmetry: | X | is even No evident reason why | X | should be a power of two OSE Workshop 2013 – p. 28/63

A BP search tree example Typical BP search tree (embeddings = paths root → leaves) Root node symmetry: | X | is even No evident reason why | X | should be a power of two (why not symmetric paths to level | V | from nodes 16 and 45?) OSE Workshop 2013 – p. 28/63

Discretization/pruning distances Let E D = {{ u, v } | | u − v | ≤ K } and E P = E � E D E D : discretization distances they guarantee that the instance is a DMDGP they allow the construction of the complete BP tree this tree has 2 | V |− 3 leaves, 2 | V |− 4 if we consider root node symmetry E P : pruning distances they allow pruning of the BP tree not clear why they should prune branches symmetrically OSE Workshop 2013 – p. 29/63

Symmetry by pruning distances [Liberti et al., LNCS (COCOA), 2011] Given embedding x , x = reflection w.r.t. hyperplane x v − K , . . . , x v − 1 R v x v − 3 x v − 1 x v − 2 OSE Workshop 2013 – p. 30/63