Probabilistic Inference Lecture 4 – Part 2 M. Pawan Kumar Slides available online

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

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

Integer Programming Formulation VaVa VbVb Label l 0 Label l 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

Integer Programming Formulation VaVa VbVb 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

Integer Programming Formulation VaVa VbVb 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

Integer Programming Formulation VaVa VbVb 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

Integer Programming Formulation VaVa VbVb 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

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

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 ….]

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 ]

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

In General Marginal Polytope

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

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 ….]

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*

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 can’t solve it in general

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

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 can’t solve this

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 can’t solve this

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 can’t solve this

Linear Programming Relaxation min  T y y a;i  [0,1] ∑ i y a;i = 1 One reason why we can’t solve this = 1 ∑ k y ab;ik = y a;i ∑ k y b;k

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 can’t solve this

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 can’t solve this * * memory requirements, time complexity

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 ]

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 ]

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

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

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

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

In General Marginal Polytope Local Polytope

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

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 can’t solve this

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

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

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

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

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

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 11 22 33 44 55 66 11 22 33 44 55 66   i  i =   i ≥ 0

Dual of the LP Relaxation Wainwright et al., 2001 11 22 33 44 55 66 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

Dual of the LP Relaxation Wainwright et al., 2001 11 22 33 44 55 66 q*(  1 )  ii   ii   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

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

Continued in Lecture 5 …