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Probabilistic Inference Lecture 4 – Part 2

Probabilistic Inference Lecture 4 – Part 2. M. Pawan Kumar pawan.kumar@ecp.fr. Slides available online http:// cvc.centrale-ponts.fr /personnel/ pawan /. Outline. Integer Programming Formulation LP Relaxation and its Dual Convergent Solution for Dual Properties and Computational Issues.

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Probabilistic Inference Lecture 4 – Part 2

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

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

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

  4. Integer Programming Formulation 2 0 4 Unary Potentials Label l1 1 2 1 b;0 = 2 a;0 = 5 Label l0 b;1 = 4 5 0 2 a;1 = 2 Vb Va Labelling f(a) = 1 ya;0 = 0 ya;1 = 1 yb;0 = 1 yb;1 = 0 f(b) = 0 Any f(.) has equivalent boolean variables ya;i

  5. Integer Programming Formulation 2 0 4 Unary Potentials Label l1 1 2 1 b;0 = 2 a;0 = 5 Label l0 b;1 = 4 5 0 2 a;1 = 2 Vb Va Labelling f(a) = 1 ya;0 = 0 ya;1 = 1 yb;0 = 1 yb;1 = 0 f(b) = 0 Find the optimal variables ya;i

  6. Integer Programming Formulation 2 0 4 Unary Potentials Label l1 1 2 1 b;0 = 2 a;0 = 5 Label l0 b;1 = 4 5 0 2 a;1 = 2 Vb Va Sum of Unary Potentials ∑a ∑i a;i ya;i ya;i  {0,1}, for all Va, li ∑i ya;i = 1, for all Va

  7. Integer Programming Formulation 2 0 4 Pairwise Potentials Label l1 1 2 1 ab;01 = 1 ab;00 = 0 Label l0 ab;11 = 0 5 0 2 ab;10 = 1 Vb Va Sum of Pairwise Potentials ∑(a,b) ∑ik ab;ik ya;iyb;k ya;i  {0,1} ∑i ya;i = 1

  8. Integer Programming Formulation 2 0 4 Pairwise Potentials Label l1 1 2 1 ab;01 = 1 ab;00 = 0 Label l0 ab;11 = 0 5 0 2 ab;10 = 1 Vb Va Sum of Pairwise Potentials ∑(a,b) ∑ik ab;ik yab;ik yab;ik =ya;i yb;k ya;i  {0,1} ∑i ya;i = 1

  9. Integer Programming Formulation min ∑a ∑i a;i ya;i + ∑(a,b) ∑ik ab;ik yab;ik ya;i  {0,1} ∑i ya;i = 1 yab;ik =ya;i yb;k

  10. Integer Programming Formulation min Ty ya;i  {0,1} ∑i ya;i = 1 yab;ik =ya;i yb;k  = [ … a;i …. ; … ab;ik ….] y = [ … ya;i …. ; … yab;ik ….]

  11. ya;1 ya;0 One variable, two labels y = [ ya;0 ya;1]  = [ a;0a;1] ya;0 + ya;1 = 1 ya;1  {0,1} ya;0  {0,1}

  12. Two variables, two labels ya;0 + ya;1 = 1 ya;1  {0,1} ya;0  {0,1} yb;0 + yb;1 = 1 yb;1  {0,1} yb;0  {0,1} yab;00 = ya;0 yb;0 yab;01 = ya;0 yb;1 yab;10 = ya;1 yb;0 yab;11 = ya;1 yb;1 • = [ a;0a;1 b;0b;1 ab;00ab;01 ab;10ab;11] y = [ ya;0ya;1 yb;0yb;1 yab;00yab;01 yab;10yab;11]

  13. In General Marginal Polytope

  14. 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 ya;i  {0,1} ∑i ya;i = 1 yab;ik =ya;i yb;k

  15. Integer Programming Formulation min Ty ya;i  {0,1} ∑i ya;i = 1 yab;ik =ya;i yb;k  = [ … a;i …. ; … ab;ik ….] y = [ … ya;i …. ; … yab;ik ….]

  16. Integer Programming Formulation min Ty ya;i  {0,1} ∑i ya;i = 1 yab;ik =ya;i yb;k Solve to obtain MAP labelling y*

  17. Integer Programming Formulation min Ty ya;i  {0,1} ∑i ya;i = 1 yab;ik =ya;i yb;k But we can’t solve it in general

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

  19. Linear Programming Relaxation min Ty ya;i  {0,1} ∑i ya;i = 1 yab;ik =ya;i yb;k Two reasons why we can’t solve this

  20. Linear Programming Relaxation min Ty ya;i  [0,1] ∑i ya;i = 1 yab;ik =ya;i yb;k One reason why we can’t solve this

  21. Linear Programming Relaxation min Ty ya;i  [0,1] ∑i ya;i = 1 ∑k yab;ik =∑kya;i yb;k One reason why we can’t solve this

  22. Linear Programming Relaxation min Ty ya;i  [0,1] ∑i ya;i = 1 ∑k yab;ik =ya;i∑k yb;k = 1 One reason why we can’t solve this

  23. Linear Programming Relaxation min Ty ya;i  [0,1] ∑i ya;i = 1 ∑k yab;ik =ya;i One reason why we can’t solve this

  24. Linear Programming Relaxation min Ty ya;i  [0,1] ∑i ya;i = 1 ∑k yab;ik =ya;i No reason why we can’t solve this * *memory requirements, time complexity

  25. ya;1 ya;0 One variable, two labels y = [ ya;0 ya;1]  = [ a;0a;1] ya;0 + ya;1 = 1 ya;1  {0,1} ya;0  {0,1}

  26. ya;1 ya;0 One variable, two labels y = [ ya;0 ya;1]  = [ a;0a;1] ya;0 + ya;1 = 1 ya;1  [0,1] ya;0  [0,1]

  27. Two variables, two labels ya;0 + ya;1 = 1 ya;1  {0,1} ya;0  {0,1} yb;0 + yb;1 = 1 yb;1  {0,1} yb;0  {0,1} yab;00 = ya;0 yb;0 yab;01 = ya;0 yb;1 yab;10 = ya;1 yb;0 yab;11 = ya;1 yb;1 • = [ a;0a;1 b;0b;1 ab;00ab;01 ab;10ab;11] y = [ ya;0ya;1 yb;0yb;1 yab;00yab;01 yab;10yab;11]

  28. Two variables, two labels ya;0 + ya;1 = 1 ya;1  [0,1] ya;0  [0,1] yb;0 + yb;1 = 1 yb;1  [0,1] yb;0  [0,1] yab;00 = ya;0 yb;0 yab;01 = ya;0 yb;1 yab;10 = ya;1 yb;0 yab;11 = ya;1 yb;1 • = [ a;0a;1 b;0b;1 ab;00ab;01 ab;10ab;11] y = [ ya;0ya;1 yb;0yb;1 yab;00yab;01 yab;10yab;11]

  29. Two variables, two labels ya;0 + ya;1 = 1 ya;1  [0,1] ya;0  [0,1] yb;0 + yb;1 = 1 yb;1  [0,1] yb;0  [0,1] yab;00 + yab;01 = ya;0 yab;10 = ya;1 yb;0 yab;11 = ya;1 yb;1 • = [ a;0a;1 b;0b;1 ab;00ab;01 ab;10ab;11] y = [ ya;0ya;1 yb;0yb;1 yab;00yab;01 yab;10yab;11]

  30. Two variables, two labels ya;0 + ya;1 = 1 ya;1  [0,1] ya;0  [0,1] yb;0 + yb;1 = 1 yb;1  [0,1] yb;0  [0,1] yab;00 + yab;01 = ya;0 yab;10 + yab;11 = ya;1 • = [ a;0a;1 b;0b;1 ab;00ab;01 ab;10ab;11] y = [ ya;0ya;1 yb;0yb;1 yab;00yab;01 yab;10yab;11]

  31. In General Local Polytope Marginal Polytope

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

  33. Linear Programming Relaxation min Ty ya;i  [0,1] ∑i ya;i = 1 ∑k yab;ik =ya;i No reason why we can’t solve this

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

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

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

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

  38. Dual of the LP Relaxation Wainwright et al., 2001 min Ty Va Vb Vc Vd Ve Vf ya;i  [0,1] Vg Vh Vi ∑i ya;i = 1  ∑k yab;ik =ya;i

  39. Dual of the LP Relaxation Wainwright et al., 2001 1 1 Va Vb Vc Va Vb Vc 2 Vd Ve Vf 2 Vd Ve Vf 3 Vg Vh Vi 3 Vg Vh Vi 4 5 6  Va Vb Vc i ≥ 0 Vd Ve Vf Vg Vh Vi  ii =  4 5 6

  40. Dual of the LP Relaxation Wainwright et al., 2001 1 q*(1) Va Vb Vc Va Vb Vc Vd Ve Vf 2 q*(2) Vd Ve Vf Vg Vh Vi 3 q*(3) Vg Vh Vi q*(4) q*(5) q*(6)  Va Vb Vc i ≥ 0 Dual of LP Vd Ve Vf max  iq*(i) Vg Vh Vi  ii =  4 5 6

  41. Dual of the LP Relaxation Wainwright et al., 2001 1 q*(1) Va Vb Vc Va Vb Vc Vd Ve Vf 2 q*(2) Vd Ve Vf Vg Vh Vi 3 q*(3) Vg Vh Vi q*(4) q*(5) q*(6)  Va Vb Vc i ≥ 0 Dual of LP Vd Ve Vf max  iq*(i) Vg Vh Vi  ii   4 5 6

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

  43. Continued in Lecture 5 …

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