ICCV2009: MAP Inference in Discrete Models: Part 3

MAP Inference in Discrete Models

M. Pawan Kumar, Stanford University

The Problem

E(x) = ∑ fi (xi) + ∑ gij (xi,xj) + ∑ hc(xc) i ij c

Unary Pairwise Higher Order

Minimize E(x) ….. Done !!!

Problems worthy of attackProve their worth by fighting back

Energy Minimization Problems

Space of Problems

CSP

MAXCUTNP-Hard

NP-hard

The Issues

• Which functions are exactly solvable?Boros Hammer [1965], Kolmogorov Zabih [ECCV 2002, PAMI 2004] , Ishikawa [PAMI

2003], Schlesinger [EMMCVPR 2007], Kohli Kumar Torr [CVPR2007, PAMI 2008] ,

Ramalingam Kohli Alahari Torr [CVPR 2008] , Kohli Ladicky Torr [CVPR 2008, IJCV

2009] , Zivny Jeavons [CP 2008]

• Approximate solutions of NP-hard problemsSchlesinger [76 ], Kleinberg and Tardos [FOCS 99], Chekuri et al. [01], Boykov et al.

[PAMI 01], Wainwright et al. [NIPS01], Werner [PAMI 2007], Komodakis et al. [PAMI, 05

07], Lempitsky et al. [ICCV 2007], Kumar et al. [NIPS 2007], Kumar et al. [ICML 2008],

Sontag and Jakkola [NIPS 2007], Kohli et al. [ICML 2008], Kohli et al. [CVPR 2008,

IJCV 2009], Rother et al. [2009]

• Scalability and Efficiency Kohli Torr [ICCV 2005, PAMI 2007], Juan and Boykov [CVPR 2006], Alahari Kohli Torr

[CVPR 2008] , Delong and Boykov [CVPR 2008]

Popular Inference Methods

Iterated Conditional Modes (ICM)

Simulated Annealing

Dynamic Programming (DP)

Belief Propagtion (BP)

Tree-Reweighted (TRW), Diffusion

Graph Cut (GC)

Branch & Bound

Relaxation methods:

…

Classical Move making algorithms

Combinatorial Algorithms

Message passing

Convex Optimization(Linear Programming,

...)

Road Map9.30-10.00 Introduction (Andrew Blake)

10.00-11.00 Discrete Models in Computer Vision (Carsten Rother)

15min Coffee break

11.15-12.30 Message Passing: DP, TRW, LP relaxation (Pawan

Kumar)

12.30-13.00 Quadratic pseudo-boolean optimization (Pushmeet

Kohli)

1hour Lunch break

14:00-15.00 Transformation and move-making methods (Pushmeet

Kohli)

15:00-15.30 Speed and Efficiency (Pushmeet Kohli)

15min Coffee break

15:45-16.15 Comparison of Methods (Carsten Rother)

16:30-17.30 Recent Advances: Dual-decomposition, higher-order,

etc. (Carsten Rother + Pawan Kumar)

Notation

Labels - li, lj, ….

Labeling - f : {a,b,..} {i,j, …}

Random variables - Va, Vb, ….

Unary Potential - a;f(a)

Pairwise Potential - ab;f(a)f(b)

Notation

MAP Estimation

f* = arg min Q(f; )

Q(f; ) = ∑a a;f(a) + ∑(a,b) ab;f(a)f(b)

Min-marginals

qa;i = min Q(f; ) s.t. f(a) = i

Energy Function

Outline

• Reparameterization

• Belief Propagation

• Tree-reweighted Message Passing

Reparameterization

Va Vb

2

5

4

2

0

1 1

0

f(a) f(b) Q(f; )

0 0 7

0 1 10

1 0 5

1 1 6

2 +

2 +

- 2

- 2

Add a constant to all a;i

Subtract that constant from all b;k

Reparameterization

f(a) f(b) Q(f; )

0 0 7 + 2 - 2

0 1 10 + 2 - 2

1 0 5 + 2 - 2

1 1 6 + 2 - 2

Add a constant to all a;i

Subtract that constant from all b;k

Q(f; ’) = Q(f; )

Va Vb

2

5

4

2

0

0

2 +

2 +

- 2

- 2

1 1

Reparameterization

Va Vb

2

5

4

2

0

1 1

0

f(a) f(b) Q(f; )

0 0 7

0 1 10

1 0 5

1 1 6

- 3 + 3

Add a constant to one b;k

Subtract that constant from ab;ik for all ‘i’

- 3

Reparameterization

Va Vb

2

5

4

2

0

1 1

0

f(a) f(b) Q(f; )

0 0 7

0 1 10 - 3 + 3

1 0 5

1 1 6 - 3 + 3

- 3 + 3

- 3

Q(f; ’) = Q(f; )



Reparameterization

Va Vb

2

5

4

2

3 1

0

1

2

Va Vb

2

5

4

2

3 1

1

0

1

- 2

- 2

- 2 + 2+ 1

+ 1

+ 1

- 1

Va Vb

2

5

4

2

3 1

2

1

0 - 4 + 4

- 4

- 4

’a;i = a;i ’b;k = b;k

’ab;ik = ab;ik

+ Mab;k

- Mab;k

+ Mba;i

- Mba;i

Q(f; ’)

= Q(f; )

Reparameterization

Q(f; ’) = Q(f; ), for all f

’ is a reparameterization of , iff

’

’b;k = b;k

’a;i = a;i

’ab;ik = ab;ik

+ Mab;k

- Mab;k

+ Mba;i

- Mba;i

Equivalently Kolmogorov, PAMI, 2006

Va Vb

2

5

4

2

0

0

2 +

2 +

- 2

- 2

1 1

Recap

MAP Estimation

f* = arg min Q(f; )

Q(f; ) = ∑a a;f(a) + ∑(a,b) ab;f(a)f(b)

Min-marginals

qa;i = min Q(f; ) s.t. f(a) = i

Q(f; ’) = Q(f; ), for all f ’

Reparameterization

Outline



– Exact MAP for Chains and Trees

– Approximate MAP for general graphs

– Computational Issues and Theoretical Properties


Belief Propagation

• Belief Propagation gives exact MAP for chains

• Some MAP problems are easy

• Exact MAP for trees

• Clever Reparameterization

Two Variables

Va Vb

2

5 2

1

0

Va Vb

2

5

40

1

Choose the right constant ’b;k = qb;k



Va Vb

2

5 2

1

0

Va Vb

2

5

40

1


a;0 + ab;00 = 5 + 0

a;1 + ab;10 = 2 + 1minMab;0 =

Two Variables

Va Vb

2

5 5

-2

-3

Va Vb

2

5

40

1


Two Variables

Va Vb

2

5 5

-2

-3

Va Vb

2

5

40

1


f(a) = 1

’b;0 = qb;0

Two Variables

Potentials along the red path add up to 0

Va Vb

2

5 5

-2

-3

Va Vb

2

5

40

1


a;0 + ab;01 = 5 + 1

a;1 + ab;11 = 2 + 0minMab;1 =

Two Variables

Va Vb

2

5 5

-2

-3

Va Vb

2

5

6-2

-1


f(a) = 1

’b;0 = qb;0

f(a) = 1

’b;1 = qb;1

Minimum of min-marginals = MAP estimate

Two Variables

Va Vb

2

5 5

-2

-3

Va Vb

2

5

6-2

-1


f(a) = 1

’b;0 = qb;0

f(a) = 1

’b;1 = qb;1

f*(b) = 0 f*(a) = 1

Two Variables

Va Vb

2

5 5

-2

-3

Va Vb

2

5

6-2

-1


f(a) = 1

’b;0 = qb;0

f(a) = 1

’b;1 = qb;1

We get all the min-marginals of Vb

Two Variables

Recap

We only need to know two sets of equations

General form of Reparameterization

’a;i = a;i

’ab;ik = ab;ik

+ Mab;k

- Mab;k

+ Mba;i

- Mba;i

’b;k = b;k

Reparameterization of (a,b) in Belief Propagation

Mab;k = mini { a;i + ab;ik }

Mba;i = 0

Three Variables

Va Vb

2

5 2

1

0

Vc

4 60

1

0

1

3

2 3

Reparameterize the edge (a,b) as before

l0

l1

Va Vb

2

5 5-3

Vc

6 60

1

-2

3


f(a) = 1

f(a) = 1

-2 -1 2 3

Three Variables

l0

l1

Va Vb

2

5 5-3

Vc

6 60

1

-2

3


f(a) = 1

f(a) = 1


-2 -1 2 3

Three Variables

l0

l1

Va Vb

2

5 5-3

Vc

6 60

1

-2

3

Reparameterize the edge (b,c) as before

f(a) = 1

f(a) = 1


-2 -1 2 3

Three Variables

l0

l1

Va Vb

2

5 5-3

Vc

6 12-6

-5

-2

9


f(a) = 1

f(a) = 1


f(b) = 1

f(b) = 0

-2 -1 -4 -3

Three Variables

l0

l1

Va Vb

2

5 5-3

Vc

6 12-6

-5

-2

9


f(a) = 1

f(a) = 1


f(b) = 1

f(b) = 0

qc;0

qc;1-2 -1 -4 -3

Three Variables

l0

l1

Va Vb

2

5 5-3

Vc

6 12-6

-5

-2

9

f(a) = 1

f(a) = 1

f(b) = 1

f(b) = 0

qc;0

qc;1

f*(c) = 0 f*(b) = 0 f*(a) = 1

Generalizes to any length chain

-2 -1 -4 -3

Three Variables

l0

l1

Va Vb

2

5 5-3

Vc

6 12-6

-5

-2

9

f(a) = 1

f(a) = 1

f(b) = 1

f(b) = 0

qc;0

qc;1

f*(c) = 0 f*(b) = 0 f*(a) = 1

Only Dynamic Programming

-2 -1 -4 -3

Three Variables

l0

l1

Why Dynamic Programming?

3 variables 2 variables + book-keeping

n variables (n-1) variables + book-keeping

Start from left, go to right

Reparameterize current edge (a,b)


’ab;ik = ab;ik+ Mab;k - Mab;k’b;k = b;k

Repeat

Why Dynamic Programming?





Repeat

Messages Message Passing

Why stop at dynamic programming?

Va Vb

2

5 5-3

Vc

6 12-6

-5

-2

9

Reparameterize the edge (c,b) as before

-2 -1 -4 -3

Three Variables

l0

l1

Va Vb

2

5 9-3

Vc

11 12-11

-9

-2

9

Reparameterize the edge (c,b) as before

-2 -1 -9 -7

’b;i = qb;i

Three Variables

l0

l1

Va Vb

2

5 9-3

Vc

11 12-11

-9

-2

9

Reparameterize the edge (b,a) as before

-2 -1 -9 -7

Three Variables

l0

l1

Va Vb

9

11 9-9

Vc

11 12-11

-9

-9

9

Reparameterize the edge (b,a) as before

-9 -7 -9 -7

’a;i = qa;i

Three Variables

l0

l1

Va Vb

9

11 9-9

Vc

11 12-11

-9

-9

9

Forward Pass Backward Pass

-9 -7 -9 -7

All min-marginals are computed

Three Variables

l0

l1

Belief Propagation on Chains





Repeat till the end of the chain

Start from right, go to left

Repeat till the end of the chain

Belief Propagation on Chains

• A way of computing reparam constants

• Generalizes to chains of any length

• Forward Pass - Start to End

• MAP estimate

• Min-marginals of final variable

• Backward Pass - End to start

• All other min-marginals

Won’t need this .. But good to know

Computational Complexity

• Each constant takes O(|L|)

• Number of constants - O(|E||L|)

O(|E||L|2)

• Memory required ?

O(|E||L|)

Belief Propagation on Trees

Vb

Va

Forward Pass: Leaf Root

All min-marginals are computed

Backward Pass: Root Leaf

Vc

Vd Ve Vg Vh

Outline







Belief Propagation on Cycles

Va Vb

Vd Vc

Where do we start? Arbitrarily

a;0

a;1

b;0

b;1

d;0

d;1

c;0

c;1

Reparameterize (a,b)


Va Vb

Vd Vc

a;0

a;1

’b;0

’b;1

d;0

d;1

c;0

c;1



Va Vb

Vd Vc

a;0

a;1

’b;0

’b;1

d;0

d;1

’c;0

’c;1



Va Vb

Vd Vc

a;0

a;1

’b;0

’b;1

’d;0

’d;1

’c;0

’c;1



Va Vb

Vd Vc

’a;0

’a;1

’b;0

’b;1

’d;0

’d;1

’c;0

’c;1


- a;0

- a;1

Did not obtain min-marginals


Va Vb

Vd Vc

’a;0

’a;1

’b;0

’b;1

’d;0

’d;1

’c;0

’c;1

- a;0

- a;1

Reparameterize (a,b) again


Va Vb

Vd Vc

’a;0

’a;1

’’b;0

’’b;1

’d;0

’d;1

’c;0

’c;1


But doesn’t this overcount some potentials?


Va Vb

Vd Vc

’a;0

’a;1

’’b;0

’’b;1

’d;0

’d;1

’c;0

’c;1


Yes. But we will do it anyway


Va Vb

Vd Vc

’a;0

’a;1

’’b;0

’’b;1

’d;0

’d;1

’c;0

’c;1

Keep reparameterizing edges in some order

No convergence guarantees

Belief Propagation

• Generalizes to any arbitrary random field

• Complexity per iteration ?

O(|E||L|2)

• Memory required ?

O(|E||L|)

Outline







Computational Issues of BP

Complexity per iteration O(|E||L|2)

Special Pairwise Potentials ab;ik = wabd(|i-k|)

i - k

d

Potts

i - k

d

Truncated Linear

i - k

d

Truncated Quadratic

O(|E||L|) Felzenszwalb & Huttenlocher, 2004


Memory requirements O(|E||L|)

Half of original BP Kolmogorov, 2006

Some approximations exist

But memory still remains an issue

Yu, Lin, Super and Tan, 2007

Lasserre, Kannan and Winn, 2007


Order of reparameterization

Randomly

Residual Belief Propagation

In some fixed order

The one that results in maximum change

Elidan et al. , 2006

Summary of BP

Exact for chains

Exact for trees

Approximate MAP for general cases

Convergence not guaranteed

So can we do something better?

Outline




– Integer Programming Formulation

– Linear Programming Relaxation and its Dual

– Convergent Solution for Dual


Integer Programming Formulation

Va Vb

Label l0

Label l12

5

4

2

0

1 1

0

2

Unary Potentials

a;0 = 5

a;1 = 2

b;0 = 2

b;1 = 4

Labeling

f(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

2

Unary Potentials

a;0 = 5

a;1 = 2

b;0 = 2

b;1 = 4

Labeling

f(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

2

Unary 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

2

Pairwise 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

2

Pairwise 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 ….]


min Ty

ya;i {0,1}

∑i ya;i = 1

yab;ik = ya;i yb;k

Solve to obtain MAP labeling 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

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

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

i i =

i ≥ 0


1

2

3

4 5 6

q*( 1)

i i =

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)

i i

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


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?

Outline








TRW Message PassingKolmogorov, 2006

Va Vb Vc

Vd Ve Vf

Vg Vh Vi

VaVb Vc

VdVe Vf

VgVh Vi

1

2

3

1

2

3

4 5 6

4 5 6

i i

i q*( i)

Pick a variable Va


i i

i q*( i)

Vc Vb Va

1c;0

1c;1

1b;0

1b;1

1a;0

1a;1

Va Vd Vg

4a;0

4a;1

4d;0

4d;1

4g;0

4g;1


1 1 + 4 4 + rest

1 q*( 1) + 4 q*( 4) + K

Vc Vb Va Va Vd Vg

Reparameterize to obtain min-marginals of Va

1c;0

1c;1

1b;0

1b;1

1a;0

1a;1

4a;0

4a;1

4d;0

4d;1

4g;0

4g;1


1 ’1 + 4 ’4 + rest

Vc Vb Va

’1c;0

’1c;1

’1b;0

’1b;1

’1a;0

’1a;1

Va Vd Vg

’4a;0

’4a;1

’4d;0

’4d;1

’4g;0

’4g;1

One pass of Belief Propagation

1 q*( ’1) + 4 q*( ’4) + K


1 ’1 + 4 ’4 + rest

Vc Vb Va Va Vd Vg

Remain the same

1 q*( ’1) + 4 q*( ’4) + K

’1c;0

’1c;1

’1b;0

’1b;1

’1a;0

’1a;1

’4a;0

’4a;1

’4d;0

’4d;1

’4g;0

’4g;1


1 ’1 + 4 ’4 + rest

1 min{ ’1a;0, ’1a;1} + 4 min{ ’4a;0, ’4a;1} + K

Vc Vb Va Va Vd Vg

’1c;0

’1c;1

’1b;0

’1b;1

’1a;0

’1a;1

’4a;0

’4a;1

’4d;0

’4d;1

’4g;0

’4g;1


1 ’1 + 4 ’4 + rest

Vc Vb Va Va Vd Vg

Compute weighted average of min-marginals of Va

’1c;0

’1c;1

’1b;0

’1b;1

’1a;0

’1a;1

’4a;0

’4a;1

’4d;0

’4d;1

’4g;0

’4g;1

1 min{ ’1a;0, ’1a;1} + 4 min{ ’4a;0, ’4a;1} + K


1 ’1 + 4 ’4 + rest

Vc Vb Va Va Vd Vg

’’a;0 = 1 ’1a;0+ 4 ’4a;0

1 + 4

’’a;1 = 1 ’1a;1+ 4 ’4a;1

1 + 4

’1c;0

’1c;1

’1b;0

’1b;1

’1a;0

’1a;1

’4a;0

’4a;1

’4d;0

’4d;1

’4g;0

’4g;1

1 min{ ’1a;0, ’1a;1} + 4 min{ ’4a;0, ’4a;1} + K


1 ’’1 + 4 ’’4 + rest

Vc Vb Va Va Vd Vg

’1c;0

’1c;1

’1b;0

’1b;1

’’a;0

’’a;1

’’a;0

’’a;1

’4d;0

’4d;1

’4g;0

’4g;1

1 min{ ’1a;0, ’1a;1} + 4 min{ ’4a;0, ’4a;1} + K

’’a;0 = 1 ’1a;0+ 4 ’4a;0

1 + 4

’’a;1 = 1 ’1a;1+ 4 ’4a;1

1 + 4


1 ’’1 + 4 ’’4 + rest

Vc Vb Va Va Vd Vg

1 min{ ’’a;0, ’’a;1} + 4 min{ ’’a;0, ’’a;1} + K

’1c;0

’1c;1

’1b;0

’1b;1

’’a;0

’’a;1

’’a;0

’’a;1

’4d;0

’4d;1

’4g;0

’4g;1

’’a;0 = 1 ’1a;0+ 4 ’4a;0

1 + 4

’’a;1 = 1 ’1a;1+ 4 ’4a;1

1 + 4


1 ’’1 + 4 ’’4 + rest

Vc Vb Va Va Vd Vg

( 1 + 4) min{ ’’a;0, ’’a;1} + K

’1c;0

’1c;1

’1b;0

’1b;1

’’a;0

’’a;1

’’a;0

’’a;1

’4d;0

’4d;1

’4g;0

’4g;1

’’a;0 = 1 ’1a;0+ 4 ’4a;0

1 + 4

’’a;1 = 1 ’1a;1+ 4 ’4a;1

1 + 4


1 ’’1 + 4 ’’4 + rest

Vc Vb Va Va Vd Vg

( 1 + 4) min{ ’’a;0, ’’a;1} + K

’1c;0

’1c;1

’1b;0

’1b;1

’’a;0

’’a;1

’’a;0

’’a;1

’4d;0

’4d;1

’4g;0

’4g;1

min {p1+p2, q1+q2} min {p1, q1} + min {p2, q2}≥


1 ’’1 + 4 ’’4 + rest

Vc Vb Va Va Vd Vg

Objective function increases or remains constant

’1c;0

’1c;1

’1b;0

’1b;1

’’a;0

’’a;1

’’a;0

’’a;1

’4d;0

’4d;1

’4g;0

’4g;1

( 1 + 4) min{ ’’a;0, ’’a;1} + K

TRW Message Passing

Initialize i. Take care of reparam constraint

Choose random variable Va

Compute min-marginals of Va for all trees

Node-average the min-marginals

REPEAT

Kolmogorov, 2006

Can also do edge-averaging

Example 1

Va Vb

0

1 1

0

2

5

4

2l0

l1

Vb Vc

0

2 3

1

4

2

6

3

Vc Va

1

4 1

0

6

3

6

4

2 =1 3 =11 =1

5 6 7

Pick variable Va. Reparameterize.

Example 1

Va Vb

-3

-2 -1

-2

5

7

4

2

Vb Vc

0

2 3

1

4

2

6

3

Vc Va

-3

1 -3

-3

6

3

10

7

2 =1 3 =11 =1

5 6 7

Average the min-marginals of Va

l0

l1

Example 1

Va Vb

-3

-2 -1

-2

7.5

7

4

2

Vb Vc

0

2 3

1

4

2

6

3

Vc Va

-3

1 -3

-3

6

3

7.5

7

2 =1 3 =11 =1

7 6 7

Pick variable Vb. Reparameterize.

l0

l1

Example 1

Va Vb

-7.5

-7 -5.5

-7

7.5

7

8.5

7

Vb Vc

-5

-3 -1

-3

9

6

6

3

Vc Va

-3

1 -3

-3

6

3

7.5

7

2 =1 3 =11 =1

7 6 7

Average the min-marginals of Vb

l0

l1

Example 1

Va Vb

-7.5

-7 -5.5

-7

7.5

7

8.75

6.5

Vb Vc

-5

-3 -1

-3

8.75

6.5

6

3

Vc Va

-3

1 -3

-3

6

3

7.5

7

2 =1 3 =11 =1

6.5 6.5 7

Value of dual does not increase

l0

l1

Example 1

Va Vb

-7.5

-7 -5.5

-7

7.5

7

8.75

6.5

Vb Vc

-5

-3 -1

-3

8.75

6.5

6

3

Vc Va

-3

1 -3

-3

6

3

7.5

7

2 =1 3 =11 =1

6.5 6.5 7

Maybe it will increase for Vc

NO

l0

l1

Example 1

Va Vb

-7.5

-7 -5.5

-7

7.5

7

8.75

6.5

Vb Vc

-5

-3 -1

-3

8.75

6.5

6

3

Vc Va

-3

1 -3

-3

6

3

7.5

7

2 =1 3 =11 =1

Strong Tree Agreement

Exact MAP Estimate

f1(a) = 0 f1(b) = 0 f2(b) = 0 f2(c) = 0 f3(c) = 0 f3(a) = 0

l0

l1

Example 2

Va Vb

0

1 1

0

2

5

2

2

Vb Vc

1

0 0

1

0

0

0

0

Vc Va

0

1 1

0

0

3

4

8

2 =1 3 =11 =1

4 0 4

Pick variable Va. Reparameterize.

l0

l1

Example 2

Va Vb

-2

-1 -1

-2

4

7

2

2

Vb Vc

1

0 0

1

0

0

0

0

Vc Va

0

0 1

-1

0

3

4

9

2 =1 3 =11 =1

4 0 4

Average the min-marginals of Va

l0

l1

Example 2

Va Vb

-2

-1 -1

-2

4

8

2

2

Vb Vc

1

0 0

1

0

0

0

0

Vc Va

0

0 1

-1

0

3

4

8

2 =1 3 =11 =1

4 0 4

Value of dual does not increase

l0

l1

Example 2

Va Vb

-2

-1 -1

-2

4

8

2

2

Vb Vc

1

0 0

1

0

0

0

0

Vc Va

0

0 1

-1

0

3

4

8

2 =1 3 =11 =1

4 0 4

Maybe it will decrease for Vb or Vc

NO

l0

l1

Example 2

Va Vb

-2

-1 -1

-2

4

8

2

2

Vb Vc

1

0 0

1

0

0

0

0

Vc Va

0

0 1

-1

0

3

4

8

2 =1 3 =11 =1

f1(a) = 1 f1(b) = 1 f2(b) = 1 f2(c) = 0 f3(c) = 1 f3(a) = 1

f2(b) = 0 f2(c) = 1

Weak Tree Agreement

Not Exact MAP Estimate

l0

l1

Example 2

Va Vb

-2

-1 -1

-2

4

8

2

2

Vb Vc

1

0 0

1

0

0

0

0

Vc Va

0

0 1

-1

0

3

4

8

2 =1 3 =11 =1

Weak Tree Agreement

Convergence point of TRW

l0

l1

f1(a) = 1 f1(b) = 1 f2(b) = 1 f2(c) = 0 f3(c) = 1 f3(a) = 1

f2(b) = 0 f2(c) = 1

Obtaining the Labeling

Only solves the dual. Primal solutions?

Va Vb Vc

Vd Ve Vf

Vg Vh Vi

’ = i i

Fix the label

Of Va

Obtaining the Labeling

Only solves the dual. Primal solutions?

Va Vb Vc

Vd Ve Vf

Vg Vh Vi

’ = i i

Fix the label

Of Vb

Continue in some fixed order

Meltzer et al., 2006

Outline

• Problem Formulation








Computational Issues of TRW

• Speed-ups for some pairwise potentials

Basic Component is Belief Propagation

Felzenszwalb & Huttenlocher, 2004

• Memory requirements cut down by half

Kolmogorov, 2006

• Further speed-ups using monotonic chains

Kolmogorov, 2006

Theoretical Properties of TRW

• Always converges, unlike BP

Kolmogorov, 2006

• Strong tree agreement implies exact MAP

Wainwright et al., 2001

• Optimal MAP for two-label submodular problems

Kolmogorov and Wainwright, 2005

ab;00 + ab;11 ≤ ab;01 + ab;10

Summary

• Trees can be solved exactly - BP

• No guarantee of convergence otherwise - BP

• Strong Tree Agreement - TRW-S

• Submodular energies solved exactly - TRW-S

• TRW-S solves an LP relaxation of MAP estimation

ICCV2009: MAP Inference in Discrete Models: Part 3

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Transcript of ICCV2009: MAP Inference in Discrete Models: Part 3