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SCE
SDE
SAE
SDF
SBD
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SCE
SDE
SAE
SDF
SBD
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SCE
SDE
SAE
SDF
SBD
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SCE
SDE
SAE
SDF
SBD
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SCE
SDE
SAE
SDF
SBD
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SCE
SDE
SAE
SDF
SBD
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SCE
SDE
SAE
SDF
SBD
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SCE
SDE
SAE
SDF
SBD
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SCE
SDE
SAE
SDF
SBD
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SCE
SDE
SAE
SDF
SBD
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SCE
SDE
SAE
SDF
SBD
Recap and Overview
Redraw any undirected graph into a clique graph
Initialize one separator potential as Z and all others as 1
For directed graphs, use moralization to transform into anundirected graph
If the clique graph is a tree, use breadth-firstadsorption from leaves to root and back down.
Got marginals for all clique and separator nodes
What if it is not a tree?
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Cycle breaking does not always work
We need: 

𝑝 𝑎,𝑏 =𝜙 𝑎,𝑏  𝑐   𝜙 𝑏,𝑐  𝑑   𝜙 𝑐,𝑑 𝜙 𝑎,𝑑
a,b
a,d
c,d
b,c
a
b
d
c
Message received at (a,b):

𝜙 𝑎,𝑏  𝑑  𝜙 𝑎,𝑑   𝑐   𝜙 𝑏,𝑐  𝑑  𝜙 𝑐,𝑑
Message received at (a,b):

𝜙 𝑎,𝑏  𝑐   𝜙 𝑏,𝑐    𝑑   𝜙 𝑐,𝑑  𝑎  𝜙 𝑎,𝑑
Cycle breaking does not always work
We need: 

𝑝 𝑎,𝑏 =𝜙 𝑎,𝑏  𝑐   𝜙 𝑏,𝑐  𝑑   𝜙 𝑐,𝑑 𝜙 𝑎,𝑑
a,b
a,d
c,d
b,c
a
b
d
c
Message received at (a,b):

𝜙 𝑎,𝑏  𝑑  𝜙 𝑎,𝑑   𝑐   𝜙 𝑏,𝑐  𝑑  𝜙 𝑐,𝑑
Message received at (a,b):

𝜙 𝑎,𝑏  𝑐   𝜙 𝑏,𝑐    𝑑   𝜙 𝑐,𝑑  𝑎  𝜙 𝑎,𝑑
Cycle breaking does not always work
We need: 

𝑝 𝑎,𝑏 =𝜙 𝑎,𝑏  𝑐   𝜙 𝑏,𝑐  𝑑   𝜙 𝑐,𝑑 𝜙 𝑎,𝑑
a,b
a,d
c,d
b,c
a
b
d
c
Message received at (a,b):

𝜙 𝑎,𝑏  𝑑  𝜙 𝑎,𝑑   𝑐   𝜙 𝑏,𝑐  𝑑  𝜙 𝑐,𝑑
The problem lies in the fact that the messages arevery different depending which branch you cut
Cycle breaking is only possible if the resultingsubtree has a special property – runningintersection property.
Message received at (a,b):

𝜙 𝑎,𝑏  𝑐   𝜙 𝑏,𝑐    𝑑   𝜙 𝑐,𝑑  𝑎  𝜙 𝑎,𝑑
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Given a clique graph, how can we find a junction sub-tree, ifexisted?
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Constructing Junction Tree
B
B
C
C
D
D
A
A
E
E
CDE
CDE
BCD
BCD
ABD
ABD
BD
BD
CD
CD
D
D
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Constructing Junction Tree
B
B
C
C
D
D
A
A
E
E
CDE
CDE
BCD
BCD
ABD
ABD
BD
BD
CD
CD
D
D
CDE
CDE
BCD
BCD
CD
CD
BD
BD
ABD
ABD
But what if the maximal weight spanning tree in NOT a junction tree?
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a,b
a,d
c,d
b,c
a
b
d
c
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G is decomposable  The clique graph corresponding to G has a junction tree
Proof: omitted
Loopy ladder