Finding k-best MAP Solutions Using LP Relaxations Amir Globerson - PowerPoint PPT Presentation

The Marginal Polytope M ( G ) � � max µ ij ( x i , x j ) θ ij ( x i , x j ) Marginal µ µ ∈ M ( G ) x i ,x j ij ∈ E Polytope

The Marginal Polytope M ( G ) � � max µ ij ( x i , x j ) θ ij ( x i , x j ) Marginal µ µ ∈ M ( G ) x i ,x j ij ∈ E Polytope There exists a p(x) s.t. p ( x i , x j ) = µ ij ( x i , x j )

The Marginal Polytope M ( G ) � � max µ ij ( x i , x j ) θ ij ( x i , x j ) Marginal µ µ ∈ M ( G ) x i ,x j ij ∈ E Polytope There exists a p(x) s.t. p ( x i , x j ) = µ ij ( x i , x j ) Difficult set to characterize. Easy to outer bound

The Marginal Polytope M ( G ) � � max µ ij ( x i , x j ) θ ij ( x i , x j ) Marginal µ µ ∈ M ( G ) x i ,x j ij ∈ E Polytope There exists a p(x) s.t. p ( x i , x j ) = µ ij ( x i , x j ) Difficult set to characterize. Easy to outer bound The vertices have integral values and correspond to assignments on x

Relaxing the MAP LP � � � max θ ij ( x i , x j ) = max µ ij ( x i , x j ) θ ij ( x i , x j ) µ ∈ M ( G ) x x i ,x j ij ij ∈ E M ( G )

Relaxing the MAP LP � � � max θ ij ( x i , x j ) = max µ ij ( x i , x j ) θ ij ( x i , x j ) µ ∈ M ( G ) x x i ,x j ij ij ∈ E Exact but Hard! M ( G )

Relaxing the MAP LP � � � max θ ij ( x i , x j ) ≤ max µ ij ( x i , x j ) θ ij ( x i , x j ) µ ∈ S x ij x i ,x j ij ∈ E M ( G ) S

Relaxing the MAP LP � � � max θ ij ( x i , x j ) ≤ max µ ij ( x i , x j ) θ ij ( x i , x j ) µ ∈ S x ij x i ,x j ij ∈ E If optimum is an integral vertex, MAP is solved M ( G ) S

Relaxing the MAP LP � � � max θ ij ( x i , x j ) ≤ max µ ij ( x i , x j ) θ ij ( x i , x j ) µ ∈ S x ij x i ,x j ij ∈ E If optimum is an integral vertex, MAP is solved M ( G ) Possible outer bound: Pairwise consistency S

Relaxing the MAP LP � � � max θ ij ( x i , x j ) ≤ max µ ij ( x i , x j ) θ ij ( x i , x j ) µ ∈ S x ij x i ,x j ij ∈ E If optimum is an integral vertex, MAP is solved M ( G ) Possible outer bound: Pairwise consistency S k � j � � � µ ij ( x i , x j ) = µ jk ( x j , x k ) i � x i x k

Relaxing the MAP LP � � � max θ ij ( x i , x j ) ≤ max µ ij ( x i , x j ) θ ij ( x i , x j ) µ ∈ S x ij x i ,x j ij ∈ E If optimum is an integral vertex, MAP is solved M ( G ) Possible outer bound: Pairwise consistency S k � j � � � µ ij ( x i , x j ) = µ jk ( x j , x k ) Exact for trees i � x i x k

Relaxing the MAP LP � � � max θ ij ( x i , x j ) ≤ max µ ij ( x i , x j ) θ ij ( x i , x j ) µ ∈ S x ij x i ,x j ij ∈ E If optimum is an integral vertex, MAP is solved M ( G ) Possible outer bound: Pairwise consistency S k � j � � � µ ij ( x i , x j ) = µ jk ( x j , x k ) Exact for trees i � x i x k Efficient message passing schemes for solving the resulting (dual) LP

Outline LP formulation of the MAP problem LP for 2 nd best General (intractable) exact formulation Tractable formulation for tree graphs Approximations for non-tree graphs Experiments

The 2 nd best problem and LP 2 nd best MAP

The 2 nd best problem and LP 2 nd best MAP max f ( x ) x

The 2 nd best problem and LP 2 nd best MAP x � = x (1) f ( x ) max max f ( x ) x

The 2 nd best problem and LP 2 nd best MAP x � = x (1) f ( x ) max max f ( x ) x µ ∈ M ( G ) µ · θ max

The 2 nd best problem and LP 2 nd best MAP x � = x (1) f ( x ) max max f ( x ) x (1) x µ ∈ M ( G, x (1) ) µ · θ µ ∈ M ( G ) µ · θ max max

The 2 nd best problem and LP 2 nd best MAP x � = x (1) f ( x ) max max f ( x ) x (1) x µ ∈ M ( G, x (1) ) µ · θ µ ∈ M ( G ) µ · θ max max Approximations:

The 2 nd best problem and LP 2 nd best MAP x � = x (1) f ( x ) max max f ( x ) x (1) x µ ∈ M ( G, x (1) ) µ · θ µ ∈ M ( G ) µ · θ max max Approximations:

The 2 nd best problem and LP 2 nd best MAP x � = x (1) f ( x ) max max f ( x ) x (1) x µ ∈ M ( G, x (1) ) µ · θ µ ∈ M ( G ) µ · θ max max Approximations:

A new marginal polytope Given an assignment z , define the Assignment Excluding Marginal Polytope: M ( G, z )

A new marginal polytope Given an assignment z , define the Assignment Excluding Marginal Polytope: M ( G, z ) M ( G, z )

A new marginal polytope Given an assignment z , define the Assignment Excluding Marginal Polytope: M ( G, z ) M ( G, z ) µ

A new marginal polytope Given an assignment z , define the Assignment Excluding Marginal Polytope: M ( G, z ) M ( G, z ) µ There exists a p(x) s.t. p ( x i , x j ) = µ ij ( x i , x j )

A new marginal polytope Given an assignment z , define the Assignment Excluding Marginal Polytope: M ( G, z ) M ( G, z ) µ There exists a p(x) s.t. p ( x i , x j ) = µ ij ( x i , x j ) and:

A new marginal polytope Given an assignment z , define the Assignment Excluding Marginal Polytope: M ( G, z ) M ( G, z ) µ There exists a p(x) s.t. p ( x i , x j ) = µ ij ( x i , x j ) and: p ( z ) = 0

A new marginal polytope Given an assignment z , define the Assignment Excluding Marginal Polytope: M ( G, z ) M ( G, z ) µ There exists a p(x) s.t. p ( x i , x j ) = µ ij ( x i , x j ) and: p ( z ) = 0

A new marginal polytope Given an assignment z , define the Assignment Excluding Marginal Polytope: M ( G, z ) M ( G, z ) µ M ( G ) There exists a p(x) s.t. p ( x i , x j ) = µ ij ( x i , x j ) and: p ( z ) = 0

A new marginal polytope Given an assignment z , define the Assignment Excluding Marginal Polytope: M ( G, z ) M ( G, z ) z µ M ( G ) There exists a p(x) s.t. p ( x i , x j ) = µ ij ( x i , x j ) and: p ( z ) = 0

LP for the 2 nd best problem The 2 nd best problem corresponds to x (1) the following LP: x � = x (1) f ( x ; θ ) = max max µ ∈ M ( G, x (1) ) µ · θ

LP for the 2 nd best problem The 2 nd best problem corresponds to x (1) the following LP: x � = x (1) f ( x ; θ ) = max max µ ∈ M ( G, x (1) ) µ · θ M ( G, x (1) ) Is there a simple characterization of ?

LP for the 2 nd best problem The 2 nd best problem corresponds to x (1) the following LP: x � = x (1) f ( x ; θ ) = max max µ ∈ M ( G, x (1) ) µ · θ M ( G, x (1) ) Is there a simple characterization of ? Is it plus one inequality? M ( G )

LP for the 2 nd best problem The 2 nd best problem corresponds to x (1) the following LP: x � = x (1) f ( x ; θ ) = max max µ ∈ M ( G, x (1) ) µ · θ M ( G, x (1) ) Is there a simple characterization of ? Is it plus one inequality? M ( G ) If so, what inequality?

Outline LP formulation of the MAP problem LP for 2 nd best General (intractable) exact formulation Tractable formulation for tree graphs Approximations for non-tree graphs Experiments

Adding inequalities to M ( G ) z z

Adding inequalities to M ( G ) Any valid inequality must separate z z from the other vertices

Adding inequalities to M ( G ) Any valid inequality must separate z z from the other vertices How about: (Santos 91) � µ i ( z i ) ≤ n − 1 i

Adding inequalities to M ( G ) Any valid inequality must separate z z from the other vertices How about: (Santos 91) � µ i ( z i ) ≤ n − 1 i RHS is n for z and or less for n − 1 other vertices

Adding inequalities to M ( G ) Any valid inequality must separate z z from the other vertices How about: (Santos 91) � µ i ( z i ) ≤ n − 1 i RHS is n for z and or less for n − 1 other vertices But: Results in fractional vertices, even for trees

Adding inequalities to M ( G ) Any valid inequality must separate z z from the other vertices How about: (Santos 91) � µ i ( z i ) ≤ n − 1 i RHS is n for z and or less for n − 1 other vertices But: Results in fractional vertices, even for trees

Adding inequalities to M ( G ) Any valid inequality must separate z z from the other vertices How about: (Santos 91) � µ i ( z i ) ≤ n − 1 i RHS is n for z and or less for n − 1 other vertices But: Results in fractional vertices, even for trees Only an outer bound on M ( G, z )

The tree case

The tree case Focus on the case where G is a tree

The tree case Focus on the case where G is a tree is given by pairwise consistency M ( G )

The tree case Focus on the case where G is a tree is given by pairwise consistency M ( G ) Define: � � I ( µ , z ) = (1 − d i ) µ i ( z i ) + µ ij ( z i , z j ) i ij ∈ G

The tree case Focus on the case where G is a tree is given by pairwise consistency M ( G ) Define: � � I ( µ , z ) = (1 − d i ) µ i ( z i ) + µ ij ( z i , z j ) i ij ∈ G Bethe: � � H ( µ ) = (1 − d i ) H i ( X i ) + H ( X i , X j ) i ij ∈ G

The tree case Focus on the case where G is a tree is given by pairwise consistency M ( G ) Define: � � I ( µ , z ) = (1 − d i ) µ i ( z i ) + µ ij ( z i , z j ) i ij ∈ G

The tree case Focus on the case where G is a tree is given by pairwise consistency M ( G ) Define: � � I ( µ , z ) = (1 − d i ) µ i ( z i ) + µ ij ( z i , z j ) i ij ∈ G Theorem: � I ( µ , z ) ≤ 0 � M ( G, z ) = µ | µ ∈ M ( G ) ,

The tree case Focus on the case where G is a tree is given by pairwise consistency M ( G ) Define: � � I ( µ , z ) = (1 − d i ) µ i ( z i ) + µ ij ( z i , z j ) i ij ∈ G Theorem: M ( G ) z � I ( µ , z ) ≤ 0 � M ( G, z ) = µ | µ ∈ M ( G ) ,