Probabilistic Inference
Lecture 4 – Part 2
M. Pawan Kumar
[email protected]
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
Things to Remember
• BP is exact for trees
• Every iteration provides a reparameterization
• Forward-pass computes min-marginals of root
Integer Programming Formulation
Unary Potentials
a;0 = 5
b;0 = 2
a;1 = 2
b;1 = 4
Label l1
2
0
1
Label l0
5
Va
4
1
0
2
2
Vb
Labelling
f(a) = 1 ya;0 = 0 ya;1 = 1
f(b) = 0 yb;0 = 1 yb;1 = 0
Any f(.) has equivalent boolean variables ya;i
Integer Programming Formulation
Unary Potentials
a;0 = 5
b;0 = 2
a;1 = 2
b;1 = 4
Label l1
2
0
1
Label l0
5
Va
Labelling
f(a) = 1 ya;0 = 0 ya;1 = 1
f(b) = 0 yb;0 = 1 yb;1 = 0
Find the optimal variables ya;i
4
1
0
2
Vb
2
Integer Programming Formulation
Unary Potentials
a;0 = 5
b;0 = 2
a;1 = 2
b;1 = 4
Label l1
2
0
1
Label l0
5
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
4
1
0
2
Vb
2
Integer Programming Formulation
Pairwise Potentials
ab;00 = 0 ab;01 = 1
ab;10 = 1 ab;11 = 0
Label l1
2
0
1
Label l0
5
Va
Sum of Pairwise Potentials
∑(a,b) ∑ik ab;ik ya;iyb;k
ya;i  {0,1}
∑i ya;i = 1
4
1
0
2
Vb
2
Integer Programming Formulation
Pairwise Potentials
ab;00 = 0 ab;01 = 1
ab;10 = 1 ab;11 = 0
Label l1
2
0
1
Label l0
5
Va
4
1
0
2
2
Vb
Sum of Pairwise Potentials
∑(a,b) ∑ik ab;ik yab;ik
ya;i  {0,1} yab;ik = ya;i yb;k
∑i ya;i = 1
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
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 ….]
One variable, two labels
ya;1
ya;0
 = [ a;0 a;1]
ya;0  {0,1}
ya;1  {0,1}
y = [ ya;0 ya;1]
ya;0 + ya;1 = 1
Two variables, two labels
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
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]
In General
Marginal Polytope
In General

2
(|V||L|
+
|E||L|
)
R
y
2
(|V||L|
+
|E||L|
)
{0,1}
Number of constraints
|V||L| + |V| +
2
|E||L|
ya;i  {0,1}
yab;ik = ya;i yb;k
∑i ya;i = 1
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 ….]
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*
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
Outline
• Integer Programming Formulation
• LP Relaxation and its Dual
• Convergent Solution for Dual
• Properties and Computational Issues
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
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
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
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
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
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
One variable, two labels
ya;1
ya;0
 = [ a;0 a;1]
ya;0  {0,1}
ya;1  {0,1}
y = [ ya;0 ya;1]
ya;0 + ya;1 = 1
One variable, two labels
ya;1
ya;0
 = [ a;0 a;1]
ya;0  [0,1]
ya;1  [0,1]
y = [ ya;0 ya;1]
ya;0 + ya;1 = 1
Two variables, two labels
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
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]
Two variables, two labels
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
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]
Two variables, two labels
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
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]
Two variables, two labels
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
 = [ 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]
In General
Local
Polytope
Marginal Polytope
In General

2
(|V||L|
+
|E||L|
)
R
y
2
(|V||L|
+
|E||L|
)
[0,1]
Number of constraints
|V||L| + |V| + |E||L|
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
Linear Programming Relaxation
Extensively studied
Optimization
Schlesinger, 1976
Koster, van Hoesel and Kolen, 1998
Theory
Chekuri et al, 2001
Machine Learning
Wainwright et al., 2001
Archer et al, 2004
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
Chekuri et al., 2001
But we are interested in NP-hard cases
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
Linear Programming Relaxation
Many interesting properties - Dual
• A computationally useful dual
Optimal value of dual = Optimal value of primal
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
Dual of the LP Relaxation
Wainwright et al., 2001
Va
Vd
Vg
Vb
Ve
Vh
Vc
Vf
1
Va
Vb
Vc
1
2
Vd
Ve
Vf
2
3
Vg
Vh
Vi
3
Vi

i ≥ 0
 ii = 
4
5
6
Va
Vb
Vc
Vd
Ve
Vf
Vg
Vh
Vi
4
5
6
Dual of the LP Relaxation
Wainwright et al., 2001
Va
Vd
Vg
Vb
Ve
Vh
q*(1)
Va
Vb
Vc
1
q*(2)
Vd
Ve
Vf
2
q*(3)
Vg
Vh
Vi
3
Vc
Vf
Vi

Dual of LP
max  i q*(i)
 ii = 
q*(4) q*(5) q*(6)
i ≥ 0
Va
Vb
Vc
Vd
Ve
Vf
Vg
Vh
Vi
4
5
6
Dual of the LP Relaxation
Wainwright et al., 2001
Va
Vd
Vg
Vb
Ve
Vh
q*(1)
Va
Vb
Vc
1
q*(2)
Vd
Ve
Vf
2
q*(3)
Vg
Vh
Vi
3
Vc
Vf
Vi

Dual of LP
max  i q*(i)
 ii  
q*(4) q*(5) q*(6)
i ≥ 0
Va
Vb
Vc
Vd
Ve
Vf
Vg
Vh
Vi
4
5
6
Dual of the LP Relaxation
Wainwright et al., 2001
max  i q*(i)
 ii  
I can easily compute q*(i)
I can easily maintain reparam constraint
So can I easily solve the dual?
Continued in Lecture 5 …