Probabilistic Inference Lecture 4 – Part 2

Probabilistic InferenceLecture 4 – Part 2

M. Pawan [email protected]

Slides available online http://cvc.centrale-ponts.fr/personnel/pawan/

• 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

Va Vb

Label l0

Label l12

5

4

2

0

1 1

0

2Unary Potentials

a;0 = 5

a;1 = 2

b;0 = 2

b;1 = 4

Labellingf(a) = 1

f(b) = 0

ya;0 = 0 ya;1 = 1

yb;0 = 1 yb;1 = 0

Any f(.) has equivalent boolean variables ya;i


Va Vb

2

5

4

2

0

1 1

0

2Unary Potentials

a;0 = 5

a;1 = 2

b;0 = 2

b;1 = 4

Labellingf(a) = 1

f(b) = 0

ya;0 = 0 ya;1 = 1

yb;0 = 1 yb;1 = 0

Find the optimal variables ya;i

Label l0

Label l1


Va Vb

2

5

4

2

0

1 1

0

2Unary Potentials

a;0 = 5

a;1 = 2

b;0 = 2

b;1 = 4

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

Label l0

Label l1


Va Vb

2

5

4

2

0

1 1

0

2Pairwise Potentials

ab;00 = 0

ab;10 = 1

ab;01 = 1

ab;11 = 0

Sum of Pairwise Potentials

∑(a,b) ∑ik ab;ik ya;iyb;k

ya;i {0,1}

∑i ya;i = 1

Label l0

Label l1


Va Vb

2

5

4

2

0

1 1

0

2Pairwise Potentials

ab;00 = 0

ab;10 = 1

ab;01 = 1

ab;11 = 0

Sum of Pairwise Potentials

∑(a,b) ∑ik ab;ik yab;ik

ya;i {0,1}

∑i ya;i = 1

yab;ik = ya;i yb;k

Label l0

Label l1


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


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 ….]

One variable, two labels

ya;0

ya;1

ya;0 {0,1} ya;1 {0,1} ya;0 + ya;1 = 1

y = [ ya;0 ya;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 = [ ya;0 ya;1 yb;0 yb;1

yab;00 yab;01 yab;10 yab;11]

ya;0 {0,1} ya;1 {0,1} ya;0 + ya;1 = 1

yb;0 {0,1} yb;1 {0,1} yb;0 + yb;1 = 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

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

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


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 ….]


min Ty

ya;i {0,1}

∑i ya;i = 1

yab;ik = ya;i yb;k

Solve to obtain MAP labelling y*


min Ty

ya;i {0,1}

∑i ya;i = 1

yab;ik = ya;i yb;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 Ty

ya;i {0,1}

∑i ya;i = 1

yab;ik = ya;i yb;k

Two reasons why we can’t solve this


min Ty

ya;i [0,1]

∑i ya;i = 1

yab;ik = ya;i yb;k

One reason why we can’t solve this


min Ty

ya;i [0,1]

∑i ya;i = 1

∑k yab;ik = ∑kya;i yb;k



min Ty

ya;i [0,1]

∑i ya;i = 1


= 1∑k yab;ik = ya;i∑k yb;k


min Ty

ya;i [0,1]

∑i ya;i = 1

∑k yab;ik = ya;i



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


ya;0

ya;1

ya;0 {0,1} ya;1 {0,1} ya;0 + ya;1 = 1

y = [ ya;0 ya;1] = [ a;0 a;1]


ya;0

ya;1

ya;0 [0,1] ya;1 [0,1] ya;0 + ya;1 = 1

y = [ ya;0 ya;1] = [ a;0 a;1]


= [ a;0 a;1 b;0 b;1



ya;0 {0,1} ya;1 {0,1} ya;0 + ya;1 = 1

yb;0 {0,1} yb;1 {0,1} yb;0 + yb;1 = 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



ya;0 [0,1] ya;1 [0,1] ya;0 + ya;1 = 1

yb;0 [0,1] yb;1 [0,1] yb;0 + yb;1 = 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



ya;0 [0,1] ya;1 [0,1] ya;0 + ya;1 = 1

yb;0 [0,1] yb;1 [0,1] yb;0 + yb;1 = 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



ya;0 [0,1] ya;1 [0,1] ya;0 + ya;1 = 1

yb;0 [0,1] yb;1 [0,1] yb;0 + yb;1 = 1

yab;00 + yab;01 = ya;0

yab;10 + yab;11 = ya;1

In General

Marginal Polytope

LocalPolytope

In General

R(|V||L| + |E||L|2)

y [0,1](|V||L| + |E||L|2)

Number of constraints

|V||L| + |V| + |E||L|


min Ty

ya;i [0,1]

∑i ya;i = 1

∑k yab;ik = ya;i

No reason why we can’t solve this


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


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


• 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


• A computationally useful dual

Many interesting properties - Dual

Optimal value of dual = Optimal value of primal

Dual of the LP RelaxationWainwright et al., 2001

Va Vb Vc

Vd Ve Vf

Vg Vh Vi

min Ty

ya;i [0,1]

∑i ya;i = 1

∑k yab;ik = ya;i


Va Vb Vc

Vd Ve Vf

Vg Vh Vi

Va Vb Vc

Vd Ve Vf

Vg Vh Vi

Va Vb Vc

Vd Ve Vf

Vg Vh Vi

1

2

3

4 5 6

1

2

3

4 5 6

ii =

i ≥ 0


1

2

3

4 5 6

q*(1)

ii =

q*(2)

q*(3)

q*(4) q*(5) q*(6)

i q*(i)

Dual of LP

Va Vb Vc

Vd Ve Vf

Vg Vh Vi

Va Vb Vc

Vd Ve Vf

Vg Vh Vi

Va Vb Vc

Vd Ve Vf

Vg Vh Vi

i ≥ 0

max


1

2

3

4 5 6

q*(1)

ii

q*(2)

q*(3)

q*(4) q*(5) q*(6)

Dual of LP

Va Vb Vc

Vd Ve Vf

Vg Vh Vi

Va Vb Vc

Vd Ve Vf

Vg Vh Vi

Va Vb Vc

Vd Ve Vf

Vg Vh Vi

i ≥ 0

i q*(i)max


ii

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 …

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Probabilistic Inference Lecture 4 – Part 2