Probabilistic Inference - M. Pawan...
Transcript of Probabilistic Inference - M. Pawan...
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Probabilistic Inference Lecture 5
M. Pawan Kumar [email protected]
Slides available online http://cvc.centrale-ponts.fr/personnel/pawan/
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• Open Book – Textbooks – Research Papers – Course Slides – No Electronic Devices
• Easy Questions – 10 points
• Hard Questions – 10 points
What to Expect in the Final Exam
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Easy Question – BP Compute the reparameterization constants for (a,b) and (c,b) such that the unary potentials of b are equal to its min-marginals.
Va Vb
2
5 5 -3 Vc
6 12 -6
-5
-2
9
-2 -1 -4 -3
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Hard Question – BP Provide an O(h) algorithm to compute the reparameterization constants of BP for an edge whose pairwise potentials are specified by a truncated linear model.
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Easy Question – Minimum Cut Provide the graph corresponding to the MAP estimation problem in the following MRF.
Va Vb
2
5 5 -3 Vc
6 12 -6
-5
-2
9
-2 -1 -4 -3
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Hard Question – Minimum Cut Show that the expansion algorithm provides a bound of 2M for the truncated linear metric, where M is the value of the truncation.
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Easy Question – Relaxations Using an example, show that the LP-S relaxation is not tight for a frustrated cycle (cycle with an odd number of supermodular pairwise potentials).
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Hard Question – Relaxations Prove or disprove that the LP-S and SOCP-MS relaxations are invariant to reparameterization.
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Recap
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Integer Programming Formulation
min ∑a ∑i θa;i ya;i + ∑(a,b) ∑ik θab;ik yab;ik
ya;i ∈ {0,1}
∑i ya;i = 1
yab;ik = ya;i yb;k
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Integer Programming Formulation
min θTy
ya;i ∈ {0,1}
∑i ya;i = 1
yab;ik = ya;i yb;k
θ = [ … θa;i …. ; … θab;ik ….] y = [ … ya;i …. ; … yab;ik ….]
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Linear Programming Relaxation
min θTy
ya;i ∈ {0,1}
∑i ya;i = 1
yab;ik = ya;i yb;k
Two reasons why we can’t solve this
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Linear Programming Relaxation
min θTy
ya;i ∈ [0,1]
∑i ya;i = 1
yab;ik = ya;i yb;k
One reason why we can’t solve this
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Linear Programming Relaxation
min θTy
ya;i ∈ [0,1]
∑i ya;i = 1
∑k yab;ik = ∑kya;i yb;k
One reason why we can’t solve this
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Linear Programming Relaxation
min θTy
ya;i ∈ [0,1]
∑i ya;i = 1
One reason why we can’t solve this
= 1 ∑k yab;ik = ya;i∑k yb;k
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Linear Programming Relaxation
min θTy
ya;i ∈ [0,1]
∑i ya;i = 1
∑k yab;ik = ya;i
One reason why we can’t solve this
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Linear Programming Relaxation
min θTy
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
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Dual of the LP Relaxation Wainwright et al., 2001
Va Vb Vc
Vd Ve Vf
Vg Vh Vi
θ
Va Vb Vc
Vd Ve Vf
Vg Vh Vi
Va Vb Vc
Vd Ve Vf
Vg Vh Vi
θ1
θ2
θ3 θ4 θ5 θ6
∑ θi = θ
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Dual of the LP Relaxation Wainwright et al., 2001
q*(θ1)
∑ θi = θ
q*(θ2)
q*(θ3) q*(θ4) q*(θ5) q*(θ6)
∑ q*(θi) Dual of LP
θ
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 max
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Dual of the LP Relaxation Wainwright et al., 2001
q*(θ1)
∑ θi ≡ θ
q*(θ2)
q*(θ3) q*(θ4) q*(θ5) q*(θ6)
Dual of LP
θ
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 ∑ q*(θi) max
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Dual of the LP Relaxation Wainwright et al., 2001
∑ θi ≡ θ
max ∑ q*(θi)
I can easily compute q*(θi)
I can easily maintain reparam constraint
So can I easily solve the dual?
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• TRW Message Passing
• Dual Decomposition
Outline
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Things to Remember
• Forward-pass computes min-marginals of root
• BP is exact for trees
• Every iteration provides a reparameterization
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TRW Message Passing Kolmogorov, 2006
Va Vb Vc
Vd Ve Vf
Vg Vh Vi
Va Vb Vc
Vd Ve Vf
Vg Vh Vi
θ1
θ2
θ3
θ4 θ5 θ6
∑ θi ≡ θ ∑ q*(θi)
Pick a variable Va
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TRW Message Passing Kolmogorov, 2006
∑ θi ≡ θ ∑ q*(θi)
Vc Vb Va
θ1c;0
θ1c;1
θ1b;0
θ1b;1
θ1a;0
θ1a;1
Va Vd Vg
θ4a;0
θ4a;1
θ4d;0
θ4d;1
θ4g;0
θ4g;1
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TRW Message Passing Kolmogorov, 2006
θ1 + θ4 + θrest ≡ θ q*(θ1) + q*(θ4) + K
Vc Vb Va Va Vd Vg
Reparameterize to obtain min-marginals of Va
θ1c;0
θ1c;1
θ1b;0
θ1b;1
θ1a;0
θ1a;1
θ4a;0
θ4a;1
θ4d;0
θ4d;1
θ4g;0
θ4g;1
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TRW Message Passing Kolmogorov, 2006
θ’1 + θ’4 + θrest
Vc Vb Va
θ’1c;0
θ’1c;1
θ’1b;0
θ’1b;1
θ’1a;0
θ’1a;1
Va Vd Vg
θ’4a;0
θ’4a;1
θ’4d;0
θ’4d;1
θ’4g;0
θ’4g;1
One pass of Belief Propagation
q*(θ’1) + q*(θ’4) + K
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TRW Message Passing Kolmogorov, 2006
θ’1 + θ’4 + θrest ≡ θ
Vc Vb Va Va Vd Vg
Remain the same
q*(θ’1) + q*(θ’4) + K
θ’1c;0
θ’1c;1
θ’1b;0
θ’1b;1
θ’1a;0
θ’1a;1
θ’4a;0
θ’4a;1
θ’4d;0
θ’4d;1
θ’4g;0
θ’4g;1
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TRW Message Passing Kolmogorov, 2006
θ’1 + θ’4 + θrest ≡ θ
min{θ’1a;0,θ’1
a;1} + min{θ’4a;0,θ’4
a;1} + K
Vc Vb Va Va Vd Vg
θ’1c;0
θ’1c;1
θ’1b;0
θ’1b;1
θ’1a;0
θ’1a;1
θ’4a;0
θ’4a;1
θ’4d;0
θ’4d;1
θ’4g;0
θ’4g;1
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TRW Message Passing Kolmogorov, 2006
θ’1 + θ’4 + θrest ≡ θ
Vc Vb Va Va Vd Vg
Compute average of min-marginals of Va
θ’1c;0
θ’1c;1
θ’1b;0
θ’1b;1
θ’1a;0
θ’1a;1
θ’4a;0
θ’4a;1
θ’4d;0
θ’4d;1
θ’4g;0
θ’4g;1
min{θ’1a;0,θ’1
a;1} + min{θ’4a;0,θ’4
a;1} + K
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TRW Message Passing Kolmogorov, 2006
θ’1 + θ’4 + θrest ≡ θ
Vc Vb Va Va Vd Vg
θ’’a;0 = θ’1a;0+ θ’4
a;0
2
θ’’a;1 = θ’1a;1+ θ’4
a;1
2
θ’1c;0
θ’1c;1
θ’1b;0
θ’1b;1
θ’1a;0
θ’1a;1
θ’4a;0
θ’4a;1
θ’4d;0
θ’4d;1
θ’4g;0
θ’4g;1
min{θ’1a;0,θ’1
a;1} + min{θ’4a;0,θ’4
a;1} + K
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TRW Message Passing Kolmogorov, 2006
θ’’1 + θ’’4 + θrest
Vc Vb Va Va Vd Vg
θ’1c;0
θ’1c;1
θ’1b;0
θ’1b;1
θ’’a;0
θ’’a;1
θ’’a;0
θ’’a;1
θ’4d;0
θ’4d;1
θ’4g;0
θ’4g;1
θ’’a;0 = θ’1a;0+ θ’4
a;0
2
θ’’a;1 = θ’1a;1+ θ’4
a;1
2
min{θ’1a;0,θ’1
a;1} + min{θ’4a;0,θ’4
a;1} + K
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TRW Message Passing Kolmogorov, 2006
θ’’1 + θ’’4 + θrest ≡ θ
Vc Vb Va Va Vd Vg
θ’1c;0
θ’1c;1
θ’1b;0
θ’1b;1
θ’’a;0
θ’’a;1
θ’’a;0
θ’’a;1
θ’4d;0
θ’4d;1
θ’4g;0
θ’4g;1
θ’’a;0 = θ’1a;0+ θ’4
a;0
2
θ’’a;1 = θ’1a;1+ θ’4
a;1
2
min{θ’1a;0,θ’1
a;1} + min{θ’4a;0,θ’4
a;1} + K
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TRW Message Passing Kolmogorov, 2006
Vc Vb Va Va Vd Vg
2 min{θ’’a;0, θ’’a;1} + K
θ’1c;0
θ’1c;1
θ’1b;0
θ’1b;1
θ’’a;0
θ’’a;1
θ’’a;0
θ’’a;1
θ’4d;0
θ’4d;1
θ’4g;0
θ’4g;1
θ’’1 + θ’’4 + θrest ≡ θ
θ’’a;0 = θ’1a;0+ θ’4
a;0
2
θ’’a;1 = θ’1a;1+ θ’4
a;1
2
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TRW Message Passing Kolmogorov, 2006
Vc Vb Va Va Vd Vg
θ’1c;0
θ’1c;1
θ’1b;0
θ’1b;1
θ’’a;0
θ’’a;1
θ’’a;0
θ’’a;1
θ’4d;0
θ’4d;1
θ’4g;0
θ’4g;1
min {p1+p2, q1+q2} min {p1, q1} + min {p2, q2} ≥ 2 min{θ’’a;0, θ’’a;1} + K
θ’’1 + θ’’4 + θrest ≡ θ
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TRW Message Passing Kolmogorov, 2006
Vc Vb Va Va Vd Vg
Objective function increases or remains constant
θ’1c;0
θ’1c;1
θ’1b;0
θ’1b;1
θ’’a;0
θ’’a;1
θ’’a;0
θ’’a;1
θ’4d;0
θ’4d;1
θ’4g;0
θ’4g;1
2 min{θ’’a;0, θ’’a;1} + K
θ’’1 + θ’’4 + θrest ≡ θ
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TRW Message Passing
Initialize θi. 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
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Example 1
Va Vb
0
1 1
0
2
5
4
2 l0
l1
Vb Vc
0
2 3
1
4
2
6
3 Vc Va
1
4 1
0
6
3
6
4
5 6 7
Pick variable Va. Reparameterize.
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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
5 6 7
Average the min-marginals of Va
l0
l1
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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
7 6 7
Pick variable Vb. Reparameterize.
l0
l1
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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
7 6 7
Average the min-marginals of Vb
l0
l1
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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
6.5 6.5 7 Value of dual does not increase
l0
l1
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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
6.5 6.5 7 Maybe it will increase for Vc
NO
l0
l1
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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
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
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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
4 0 4
Pick variable Va. Reparameterize.
l0
l1
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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
4 0 4
Average the min-marginals of Va
l0
l1
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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
4 0 4 Value of dual does not increase
l0
l1
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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
4 0 4 Maybe it will decrease for Vb or Vc
NO
l0
l1
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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
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
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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
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
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Obtaining the Labelling
Only solves the dual. Primal solutions?
Va Vb Vc
Vd Ve Vf
Vg Vh Vi
θ’ = ∑ θi ≡ θ
Fix the label Of Va
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Obtaining the Labelling
Only solves the dual. Primal solutions?
Va Vb Vc
Vd Ve Vf
Vg Vh Vi
θ’ = ∑ θi ≡ θ
Fix the label Of Vb
Continue in some fixed order Meltzer et al., 2006
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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
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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
θab;00 + θab;11 ≤ θab;01 + θab;10
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Results Binary Segmentation Szeliski et al. , 2008
Labels - {foreground, background}
Unary Potentials: -log(likelihood) using learnt fg/bg models
Pairwise Potentials: 0, if same labels 1 - λexp(|da - db|), if different labels
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Results Binary Segmentation
Labels - {foreground, background}
Unary Potentials: -log(likelihood) using learnt fg/bg models
Szeliski et al. , 2008
Pairwise Potentials: 0, if same labels 1 - λexp(|da - db|), if different labels
TRW
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Results Binary Segmentation
Labels - {foreground, background}
Unary Potentials: -log(likelihood) using learnt fg/bg models
Szeliski et al. , 2008
Belief Propagation
Pairwise Potentials: 0, if same labels 1 - λexp(|da - db|), if different labels
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Results Stereo Correspondence Szeliski et al. , 2008
Labels - {disparities}
Unary Potentials: Similarity of pixel colours
Pairwise Potentials: 0, if same labels 1 - λexp(|da - db|), if different labels
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Results Szeliski et al. , 2008
Labels - {disparities}
Unary Potentials: Similarity of pixel colours
Pairwise Potentials: 0, if same labels 1 - λexp(|da - db|), if different labels
TRW
Stereo Correspondence
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Results Szeliski et al. , 2008
Labels - {disparities}
Unary Potentials: Similarity of pixel colours
Belief Propagation
Pairwise Potentials: 0, if same labels 1 - λexp(|da - db|), if different labels
Stereo Correspondence
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Results Non-submodular problems Kolmogorov, 2006
BP TRW-S
30x30 grid K50
BP TRW-S
BP outperforms TRW-S
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Code + Standard Data
http://vision.middlebury.edu/MRF
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• TRW Message Passing
• Dual Decomposition
Outline
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Dual Decomposition
minx ∑i gi(x) s.t. x ∈ C
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Dual Decomposition
minx,xi ∑i gi(xi)
s.t. xi ∈ C xi = x
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Dual Decomposition
minx,xi ∑i gi(xi)
s.t. xi ∈ C
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Dual Decomposition
minx,xi ∑i gi(xi) + ∑i λi
T(xi-x) s.t. xi ∈ C
maxλi
KKT Condition: ∑i λi = 0
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Dual Decomposition
minx,xi ∑i gi(xi) + ∑i λi
Txi s.t. xi ∈ C
maxλi
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Dual Decomposition
minxi ∑i (gi(xi) + λi
Txi) s.t. xi ∈ C
Projected Supergradient Ascent
maxλi
Supergradient s of h(z) at z0
h(z) - h(z0) ≤ sT(z-z0), for all z in the feasible region
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Dual Decomposition
minxi ∑i (gi(xi) + λi
Txi) s.t. xi ∈ C
Initialize λi0
= 0
maxλi
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Dual Decomposition
minxi ∑i (gi(xi) + λi
Txi) s.t. xi ∈ C
Compute supergradients
maxλi
si = argminxi ∑i (gi(xi) + (λi
t)Txi)
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Dual Decomposition
minxi ∑i (gi(xi) + λi
Txi) s.t. xi ∈ C
Project supergradients
maxλi
pi = si - ∑j sj/m
where ‘m’ = number of subproblems (slaves)
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Dual Decomposition
minxi ∑i (gi(xi) + λi
Txi) s.t. xi ∈ C
Update dual variables
maxλi
λit+1
= λit + ηt pi
where ηt = learning rate = 1/(t+1) for example
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Dual Decomposition Initialize λi
0 = 0
Compute projected supergradients
si = argminxi ∑i (gi(xi) + (λi
t)Txi)
pi = si - ∑j sj/m
Update dual variables
λit+1
= λit + ηt pi
REPEAT
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Dual Decomposition Komodakis et al., 2007
Va Vb Vc
Vd Ve Vf
Vg Vh Vi
Va Vb Vc
Vd Ve Vf
Vg Vh Vi
θ1
θ2
θ3
θ4 θ5 θ6
1 0
s1a =
1 0
s4a =
Slaves agree on label for Va
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Dual Decomposition Komodakis et al., 2007
Va Vb Vc
Vd Ve Vf
Vg Vh Vi
Va Vb Vc
Vd Ve Vf
Vg Vh Vi
θ1
θ2
θ3
θ4 θ5 θ6
1 0
s1a =
1 0
s4a =
0 0
p1a =
0 0
p4a =
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Dual Decomposition Komodakis et al., 2007
Va Vb Vc
Vd Ve Vf
Vg Vh Vi
Va Vb Vc
Vd Ve Vf
Vg Vh Vi
θ1
θ2
θ3
θ4 θ5 θ6
1 0
s1a =
0 1
s4a =
Slaves disagree on label for Va
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Dual Decomposition Komodakis et al., 2007
Va Vb Vc
Vd Ve Vf
Vg Vh Vi
Va Vb Vc
Vd Ve Vf
Vg Vh Vi
θ1
θ2
θ3
θ4 θ5 θ6
1 0
s1a =
0 1
s4a =
0.5
-0.5 p1
a =
-0.5
0.5 p4
a =
Unary cost increases
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Dual Decomposition Komodakis et al., 2007
Va Vb Vc
Vd Ve Vf
Vg Vh Vi
Va Vb Vc
Vd Ve Vf
Vg Vh Vi
θ1
θ2
θ3
θ4 θ5 θ6
1 0
s1a =
0 1
s4a =
0.5
-0.5 p1
a =
-0.5
0.5 p4
a =
Unary cost decreases
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Dual Decomposition Komodakis et al., 2007
Va Vb Vc
Vd Ve Vf
Vg Vh Vi
Va Vb Vc
Vd Ve Vf
Vg Vh Vi
θ1
θ2
θ3
θ4 θ5 θ6
1 0
s1a =
0 1
s4a =
0.5
-0.5 p1
a =
-0.5
0.5 p4
a =
Push the slaves towards agreement
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Comparison TRW DD
Fast Slow
Local Maximum Global Maximum
Requires Min-Marginals
Requires MAP Estimate
Other forms of slaves Tighter relaxations
Sparse high-order potentials