Angular Synchronization and its application in Phase Retrieval - PowerPoint PPT Presentation

Angular Synchronization and its application in Phase Retrieval Afonso S. Bandeira PACM, Princeton University joint work with Amit Singer (Princeton), Daniel A. Spielman (Yale), Boris Alexeev (Princeton), Matthew Fickus (AFIT), and Dustin G. Mixon (AFIT) OSL 2013, Les Houches. January 11, 2013 http//www.math.princeton.edu/~ajsb

Spectral Clustering – Cheeger Inequality � � G = V, E, ( W ) ij = w ij Cheeger Constant: h G = min S ⊂ V h G ( S ) cut( S, S c ) h G ( S ) = min { vol( S ) , vol( S c ) } 2/23

Spectral Clustering – Cheeger Inequality � � G = V, E, ( W ) ij = w ij Cheeger Constant: h G = min S ⊂ V h G ( S ) cut( S, S c ) h G ( S ) = min { vol( S ) , vol( S c ) } Graph Laplacian D = diag ( d i ) L 0 = I − D − 1 / 2 WD − 1 / 2 L 0 = D − W and x T L 0 x � ij w ij | x i − x j | 2 x T Dx = 1 2 � i d i x 2 i 2/23

Spectral Clustering – Cheeger Inequality � � G = V, E, ( W ) ij = w ij Cheeger Constant: h G = min S ⊂ V h G ( S ) cut( S, S c ) h G ( S ) = min { vol( S ) , vol( S c ) } Graph Laplacian D = diag ( d i ) L 0 = I − D − 1 / 2 WD − 1 / 2 L 0 = D − W and x T L 0 x � ij w ij | x i − x j | 2 x T Dx = 1 2 � i d i x 2 i Theorem (Cheeger Inequality (Alon 86)) 1 � 2 λ 2 ( L 0 ) ≤ h G ≤ 2 λ 2 ( L 0 ) 2/23

Problem Relaxation f a function that takes values in [0 , 1] . Want to minimize it over a (discrete) set “comb”. 3/23

Problem Relaxation f a function that takes values in [0 , 1] . Want to minimize it over a (discrete) set “comb”. Relax the problem to a continuous set “relax” that contains “comb” and on which minimizing f is easier. 3/23

Problem Relaxation f a function that takes values in [0 , 1] . Want to minimize it over a (discrete) set “comb”. Relax the problem to a continuous set “relax” that contains “comb” and on which minimizing f is easier. “rounding” procedure ( that takes elements in “relax” and sends them to “comb”) on which, say, the value of f never more than doubles. 3/23

Problem Relaxation f a function that takes values in [0 , 1] . Want to minimize it over a (discrete) set “comb”. Relax the problem to a continuous set “relax” that contains “comb” and on which minimizing f is easier. “rounding” procedure ( that takes elements in “relax” and sends them to “comb”) on which, say, the value of f never more than doubles. opt relax ≤ opt comb ≤ 2 ( opt relax ) 3/23

The Synchronization Problem Problem Determine a potential on the set V of vertices of a graph, with values on a group G g : V → G i → g i given a few, possibly noisy, of the pairwise offset measurements (corresponding to the edges E of the graph) ρ : E → G ρ ij ≈ g i g − 1 ( i, j ) → j . 4/23

Examples... G = O (1) = Z 2 5/23

Examples... G = O (1) = Z 2 5/23

Examples... G = O (1) = Z 2 When all edges are red this is essentially Max-Cut 5/23

Examples... G = O (1) = Z 2 Orientation of a Manifold. ρ ij = det( O ij ) 6/23

Examples... G = SO (2) 7/23

Examples... G = SO (2) 7/23

Examples... G = SO (2) 7/23

Examples... G = SO (2) 7/23

Solution to the “frustration free” case 8/23

Solution to the “frustration free” case 8/23

Solution to the “frustration free” case 8/23

Solution to the “frustration free” case 8/23

The Angular Synchronization Problem Problem Determine an angular potential on the set V of vertices of a graph, e iθ · = v : V θ · : V → [0 , 2 π ) → T ⊂ C i → θ i i → v i given a few, possibly noisy, of the relative angle measurements (corresponding to the edges E of the graph) e iθ ·· = ρ : E θ ·· : E → [0 , 2 π ) → T ⊂ C ρ ij ≈ v i v − 1 ( i, j ) → θ ij ≈ θ i − θ j . ( i, j ) → j . 9/23

The Angular Synchronization Problem Problem Determine an angular potential on the set V of vertices of a graph, e iθ · = v : V θ · : V → [0 , 2 π ) → T ⊂ C i → θ i i → v i given a few, possibly noisy, of the relative angle measurements (corresponding to the edges E of the graph) e iθ ·· = ρ : E θ ·· : E → [0 , 2 π ) → T ⊂ C ρ ij ≈ v i v − 1 ( i, j ) → θ ij ≈ θ i − θ j . ( i, j ) → j . 9/23

The Angular Synchronization Problem Problem Determine an angular potential on the set V of vertices of a graph, e iθ · = v : V θ · : V → [0 , 2 π ) → T ⊂ C i → θ i i → v i given a few, possibly noisy, of the relative angle measurements (corresponding to the edges E of the graph) e iθ ·· = ρ : E θ ·· : E → [0 , 2 π ) → T ⊂ C ρ ij ≈ v i v − 1 ( i, j ) → θ ij ≈ θ i − θ j . ( i, j ) → j . Minimize: � ij w ij | v i − ρ ij v j | 2 1 � w ij | v i − ρ ij v j | 2 . η G = min v : V → T η ( v ) = = � i d i | v i | 2 vol( G ) ij 9/23

The Angular Synchronization Problem Problem Determine an angular potential on the set V of vertices of a graph, e iθ · = v : V θ · : V → [0 , 2 π ) → T ⊂ C i → θ i i → v i given a few, possibly noisy, of the relative angle measurements (corresponding to the edges E of the graph) e iθ ·· = ρ : E θ ·· : E → [0 , 2 π ) → T ⊂ C ρ ij ≈ v i v − 1 ( i, j ) → θ ij ≈ θ i − θ j . ( i, j ) → j . The Frustration Constant: � ij w ij | v i − ρ ij v j | 2 1 � w ij | v i − ρ ij v j | 2 . η G = min v : V → T η ( v ) = = � i d i | v i | 2 vol( G ) ij 9/23

The Graph Connection Laplacian W 1 ∈ C n × n ( W 1 ) ij = w ij ρ ij ∈ C . The Graph Connection Laplacian is L 1 ∈ C n × n L 1 = D − W 1 10/23

The Graph Connection Laplacian W 1 ∈ C n × n ( W 1 ) ij = w ij ρ ij ∈ C . The Graph Connection Laplacian is L 1 ∈ C n × n L 1 = D − W 1 The Normalized Graph Connection Laplacian is L 1 ∈ C n × n L 1 = D − 1 / 2 L 1 D − 1 / 2 = I n − D − 1 / 2 W 1 D − 1 / 2 . Under certain conditions L 1 converges to the Connection Laplacian in Riemannian Geometry. 10/23

The Graph Connection Laplacian W 1 ∈ C n × n ( W 1 ) ij = w ij ρ ij ∈ C . The Graph Connection Laplacian is L 1 ∈ C n × n L 1 = D − W 1 The Normalized Graph Connection Laplacian is L 1 ∈ C n × n L 1 = D − 1 / 2 L 1 D − 1 / 2 = I n − D − 1 / 2 W 1 D − 1 / 2 . Under certain conditions L 1 converges to the Connection Laplacian in Riemannian Geometry. x T L 1 x ij w ij | x i − ρ ij x j | 2 � x T Dx = 1 = η ( x ) 2 � i d i | x i | 2 10/23

Partial Frustration Constants x T L 1 x ij w ij | x i − ρ ij x j | 2 � x T Dx = 1 = η ( x ) 2 � i d i | x i | 2 11/23

Partial Frustration Constants x T L 1 x ij w ij | x i − ρ ij x j | 2 � x T Dx = 1 = η ( x ) 2 � i d i | x i | 2 λ 1 ( L 1 ) = min x ∈ C n η ( x ) = min x : V → C η ( x ) η G = min v : V → T η ( v ) Question Can we relate η G to λ 1 ( L 1 ) ? 11/23

Partial Frustration Constants x T L 1 x ij w ij | x i − ρ ij x j | 2 � x T Dx = 1 = η ( x ) 2 � i d i | x i | 2 λ 1 ( L 1 ) = min x ∈ C n η ( x ) = min x : V → C η ( x ) η G = min v : V → T η ( v ) Question Can we relate η G to λ 1 ( L 1 ) ? NO! 11/23

Partial Frustration Constants x T L 1 x ij w ij | x i − ρ ij x j | 2 � x T Dx = 1 = η ( x ) 2 � i d i | x i | 2 λ 1 ( L 1 ) = min x ∈ C n η ( x ) = min x : V → C η ( x ) η G = min v : V → T η ( v ) Question Can we relate η G to λ 1 ( L 1 ) ? NO! Fix – Consider instead: η ∗ G = v : V → T ∪{ 0 } η ( v ) . min 11/23

Partial Frustration Constants x T L 1 x ij w ij | x i − ρ ij x j | 2 � x T Dx = 1 = η ( x ) 2 � i d i | x i | 2 λ 1 ( L 1 ) = min x ∈ C n η ( x ) = min x : V → C η ( x ) η G = min v : V → T η ( v ) Question Can we relate η G to λ 1 ( L 1 ) ? NO! Fix – Consider instead: η ∗ G = v : V → T ∪{ 0 } η ( v ) . min Theorem λ 1 ( L 1 ) ≤ η ∗ � G ≤ 10 λ 1 ( L 1 ) 11/23

Global Synchronization – What about η G ? Problematic case: 12/23

Global Synchronization – What about η G ? If G has a large spectral gap λ 2 ( L 0 ) Problematic case: (or, equivalently a large Cheeger Constant), this should not be a problem. 12/23

Global Synchronization – What about η G ? If G has a large spectral gap λ 2 ( L 0 ) Problematic case: (or, equivalently a large Cheeger Constant), this should not be a problem. ij w ij | v i − ρ ij v j | 2 � 1 η ( v ) = � i d i | v i | 2 2 ij w ij ( | v i | − | v j | ) 2 � 1 ≥ � i d i | v i | 2 2 12/23

Global Synchronization – What about η G ? If G has a large spectral gap λ 2 ( L 0 ) Problematic case: (or, equivalently a large Cheeger Constant), this should not be a problem. ij w ij | v i − ρ ij v j | 2 � 1 η ( v ) = � i d i | v i | 2 2 ij w ij ( | v i | − | v j | ) 2 � 1 ≥ � i d i | v i | 2 2 Theorem 1 � � λ 1 ( L 1 ) ≤ η G ≤ λ 2 ( L 0 ) O λ 1 ( L 1 ) . 12/23

Examples... G = SO (3) What about beyond Z / 2 Z = O (1) and SO (2) Synchronization? 13/23

Examples... G = SO (3) What about beyond Z / 2 Z = O (1) and SO (2) Synchronization? 13/23

Examples... G = SO (3) What about beyond Z / 2 Z = O (1) and SO (2) Synchronization? 13/23