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Probabilistic Inference Lecture 4 Part 2 M. Pawan Kumar pawan.kumar@ecp.fr Slides available online http://cvc.centrale-ponts.fr/personnel/pawan/

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Integer Programming Formulation LP Relaxation and its Dual Convergent Solution for Dual Properties and Computational Issues Outline

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Things to Remember Forward-pass computes min-marginals of root BP is exact for trees Every iteration provides a reparameterization

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Integer Programming Formulation VaVa VbVb Label l 0 Label l 1 2 5 4 2 0 1 1 0 2 Unary Potentials a;0 = 5 a;1 = 2 b;0 = 2 b;1 = 4 Labelling f(a) = 1 f(b) = 0 y a;0 = 0y a;1 = 1 y b;0 = 1y b;1 = 0 Any f(.) has equivalent boolean variables y a;i

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Integer Programming Formulation VaVa VbVb 2 5 4 2 0 1 1 0 2 Unary Potentials a;0 = 5 a;1 = 2 b;0 = 2 b;1 = 4 Labelling f(a) = 1 f(b) = 0 y a;0 = 0y a;1 = 1 y b;0 = 1y b;1 = 0 Find the optimal variables y a;i Label l 0 Label l 1

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Integer Programming Formulation VaVa VbVb 2 5 4 2 0 1 1 0 2 Unary Potentials a;0 = 5 a;1 = 2 b;0 = 2 b;1 = 4 Sum of Unary Potentials a i a;i y a;i y a;i {0,1}, for all V a, l i i y a;i = 1, for all V a Label l 0 Label l 1

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Integer Programming Formulation VaVa VbVb 2 5 4 2 0 1 1 0 2 Pairwise Potentials ab;00 = 0 ab;10 = 1 ab;01 = 1 ab;11 = 0 Sum of Pairwise Potentials (a,b) ik ab;ik y a;i y b;k y a;i {0,1} i y a;i = 1 Label l 0 Label l 1

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Integer Programming Formulation VaVa VbVb 2 5 4 2 0 1 1 0 2 Pairwise Potentials ab;00 = 0 ab;10 = 1 ab;01 = 1 ab;11 = 0 Sum of Pairwise Potentials (a,b) ik ab;ik y ab;ik y a;i {0,1} i y a;i = 1 y ab;ik = y a;i y b;k Label l 0 Label l 1

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Integer Programming Formulation min a i a;i y a;i + (a,b) ik ab;ik y ab;ik y a;i {0,1} i y a;i = 1 y ab;ik = y a;i y b;k

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Integer Programming Formulation min T y y a;i {0,1} i y a;i = 1 y ab;ik = y a;i y b;k = [ a;i . ; ab;ik .] y = [ y a;i . ; y ab;ik .]

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One variable, two labels y a;0 y a;1 y a;0 {0,1} y a;1 {0,1} y a;0 + y a;1 = 1 y = [ y a;0 y a;1 ] = [ a;0 a;1 ]

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Two variables, two labels = [ a;0 a;1 b;0 b;1 ab;00 ab;01 ab;10 ab;11 ] y = [ y a;0 y a;1 y b;0 y b;1 y ab;00 y ab;01 y ab;10 y ab;11 ] y a;0 {0,1} y a;1 {0,1} y a;0 + y a;1 = 1 y b;0 {0,1} y b;1 {0,1} y b;0 + y b;1 = 1 y ab;00 = y a;0 y b;0 y ab;01 = y a;0 y b;1 y ab;10 = y a;1 y b;0 y ab;11 = y a;1 y b;1

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In General Marginal Polytope

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In General R (|V||L| + |E||L| 2 ) y {0,1} (|V||L| + |E||L| 2 ) Number of constraints |V||L| + |V| + |E||L| 2 y a;i {0,1} i y a;i = 1y ab;ik = y a;i y b;k

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Integer Programming Formulation min T y y a;i {0,1} i y a;i = 1 y ab;ik = y a;i y b;k = [ a;i . ; ab;ik .] y = [ y a;i . ; y ab;ik .]

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Integer Programming Formulation min T y y a;i {0,1} i y a;i = 1 y ab;ik = y a;i y b;k Solve to obtain MAP labelling y*

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Integer Programming Formulation min T y y a;i {0,1} i y a;i = 1 y ab;ik = y a;i y b;k But we cant solve it in general

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Integer Programming Formulation LP Relaxation and its Dual Convergent Solution for Dual Properties and Computational Issues Outline

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Linear Programming Relaxation min T y y a;i {0,1} i y a;i = 1 y ab;ik = y a;i y b;k Two reasons why we cant solve this

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Linear Programming Relaxation min T y y a;i [0,1] i y a;i = 1 y ab;ik = y a;i y b;k One reason why we cant solve this

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Linear Programming Relaxation min T y y a;i [0,1] i y a;i = 1 k y ab;ik = k y a;i y b;k One reason why we cant solve this

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Linear Programming Relaxation min T y y a;i [0,1] i y a;i = 1 One reason why we cant solve this = 1 k y ab;ik = y a;i k y b;k

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Linear Programming Relaxation min T y y a;i [0,1] i y a;i = 1 k y ab;ik = y a;i One reason why we cant solve this

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Linear Programming Relaxation min T y y a;i [0,1] i y a;i = 1 k y ab;ik = y a;i No reason why we cant solve this * * memory requirements, time complexity

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One variable, two labels y a;0 y a;1 y a;0 {0,1} y a;1 {0,1} y a;0 + y a;1 = 1 y = [ y a;0 y a;1 ] = [ a;0 a;1 ]

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One variable, two labels y a;0 y a;1 y a;0 [0,1] y a;1 [0,1] y a;0 + y a;1 = 1 y = [ y a;0 y a;1 ] = [ a;0 a;1 ]

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Two variables, two labels = [ a;0 a;1 b;0 b;1 ab;00 ab;01 ab;10 ab;11 ] y = [ y a;0 y a;1 y b;0 y b;1 y ab;00 y ab;01 y ab;10 y ab;11 ] y a;0 {0,1} y a;1 {0,1} y a;0 + y a;1 = 1 y b;0 {0,1} y b;1 {0,1} y b;0 + y b;1 = 1 y ab;00 = y a;0 y b;0 y ab;01 = y a;0 y b;1 y ab;10 = y a;1 y b;0 y ab;11 = y a;1 y b;1

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Two variables, two labels = [ a;0 a;1 b;0 b;1 ab;00 ab;01 ab;10 ab;11 ] y = [ y a;0 y a;1 y b;0 y b;1 y ab;00 y ab;01 y ab;10 y ab;11 ] y a;0 [0,1] y a;1 [0,1] y a;0 + y a;1 = 1 y b;0 [0,1] y b;1 [0,1] y b;0 + y b;1 = 1 y ab;00 = y a;0 y b;0 y ab;01 = y a;0 y b;1 y ab;10 = y a;1 y b;0 y ab;11 = y a;1 y b;1

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Two variables, two labels = [ a;0 a;1 b;0 b;1 ab;00 ab;01 ab;10 ab;11 ] y = [ y a;0 y a;1 y b;0 y b;1 y ab;00 y ab;01 y ab;10 y ab;11 ] y a;0 [0,1] y a;1 [0,1] y a;0 + y a;1 = 1 y b;0 [0,1] y b;1 [0,1] y b;0 + y b;1 = 1 y ab;00 + y ab;01 = y a;0 y ab;10 = y a;1 y b;0 y ab;11 = y a;1 y b;1

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Two variables, two labels = [ a;0 a;1 b;0 b;1 ab;00 ab;01 ab;10 ab;11 ] y = [ y a;0 y a;1 y b;0 y b;1 y ab;00 y ab;01 y ab;10 y ab;11 ] y a;0 [0,1] y a;1 [0,1] y a;0 + y a;1 = 1 y b;0 [0,1] y b;1 [0,1] y b;0 + y b;1 = 1 y ab;00 + y ab;01 = y a;0 y ab;10 + y ab;11 = y a;1

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In General Marginal Polytope Local Polytope

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In General R (|V||L| + |E||L| 2 ) y [0,1] (|V||L| + |E||L| 2 ) Number of constraints |V||L| + |V| + |E||L|

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Linear Programming Relaxation min T y y a;i [0,1] i y a;i = 1 k y ab;ik = y a;i No reason why we cant solve this

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Linear Programming Relaxation Extensively studied Optimization Schlesinger, 1976 Koster, van Hoesel and Kolen, 1998 Theory Chekuri et al, 2001Archer et al, 2004 Machine Learning Wainwright et al., 2001

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Linear Programming Relaxation Many interesting properties Global optimal MAP for trees Wainwright et al., 2001 But we are interested in NP-hard cases Preserves solution for reparameterization Global optimal MAP for submodular energy Chekuri et al., 2001

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Linear Programming Relaxation Large class of problems Metric Labelling Semi-metric Labelling Many interesting properties - Integrality Gap Manokaran et al., 2008 Most likely, provides best possible integrality gap

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Linear Programming Relaxation A computationally useful dual Many interesting properties - Dual Optimal value of dual = Optimal value of primal

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Dual of the LP Relaxation Wainwright et al., 2001 VaVa VbVb VcVc VdVd VeVe VfVf VgVg VhVh ViVi min T y y a;i [0,1] i y a;i = 1 k y ab;ik = y a;i

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Dual of the LP Relaxation Wainwright et al., 2001 VaVa VbVb VcVc VdVd VeVe VfVf VgVg VhVh ViVi VaVa VbVb VcVc VdVd VeVe VfVf VgVg VhVh ViVi VaVa VbVb VcVc VdVd VeVe VfVf VgVg VhVh ViVi 11 22 33 44 55 66 11 22 33 44 55 66 i i = i 0

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Dual of the LP Relaxation Wainwright et al., 2001 11 22 33 44 55 66 q*( 1 ) i i = q*( 2 ) q*( 3 ) q*( 4 )q*( 5 )q*( 6 ) i q*( i ) Dual of LP VaVa VbVb VcVc VdVd VeVe VfVf VgVg VhVh ViVi VaVa VbVb VcVc VdVd VeVe VfVf VgVg VhVh ViVi VaVa VbVb VcVc VdVd VeVe VfVf VgVg VhVh ViVi i 0 max

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Dual of the LP Relaxation Wainwright et al., 2001 11 22 33 44 55 66 q*( 1 ) ii ii q*( 2 ) q*( 3 ) q*( 4 )q*( 5 )q*( 6 ) Dual of LP VaVa VbVb VcVc VdVd VeVe VfVf VgVg VhVh ViVi VaVa VbVb VcVc VdVd VeVe VfVf VgVg VhVh ViVi VaVa VbVb VcVc VdVd VeVe VfVf VgVg VhVh ViVi i 0 i q*( i ) max

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Dual of the LP Relaxation Wainwright et al., 2001 ii ii max i q*( i ) I can easily compute q*( i ) I can easily maintain reparam constraint So can I easily solve the dual?

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Continued in Lecture 5