# S2-Chordal Graphs ```Bayesian Networks, Winter 2009-2010
Yoav Haimovitch &amp; Ariel Raviv
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Chordal Graph
Warm up
Theorem 7
Perfect Vertex Elimination Scheme
Maximal cliques Tree
Bibliography
• M.C.Golumbic Algorithmic Graph Theory and
Perfect Graphs. Second edition
• Pearl, J. Probabilistic Reasoning in Intelligent
Systems.
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Definition: Undirected graph G is chordal
if every cycle of length greater than 3 has a
chord.
 Remark: Chordality is a hereditary property,
All induced subgraphs of a chordal graph are
chordal!
 Chordal graphs are also called triangulated
and perfect elimination graphs.

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4


A vertex x of G is called simplicial if its
(not necessarily maximal)
Lemma 1[Dirac 1961]: Each chordal graph
has a simplicial vertex and if G is not a clique
it has two non adjacent simplicial vertices.
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

If G is a clique, done.
If not, assume that be has two non-adjacent
nodes a and b and that the lemma is true to
all graphs with fewer vertices than G. Let S be
a minimal vertex separator for a and b.
Let Ga and Gb be the
connected components
of a and b respectively.
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S is a clique – if not, then there exists x and y
such that there is no edge between them, and
because S is minimal vertex separator there is
a cycle &lt;a,…,x,…,b,…,y,…,a&gt;. It is easy to
show that there exist a minimal cycle with
length no
less then 4 with no
chords in it,

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S is a clique. GA+S is smaller than G therefore
by induction the lemma holds, i.e. GA+S is a
clique or has two non adjacent simplicial
vertices, one of each must be in GA. Any
simplicial vertex in GA is a simplicial vertex in G
because all elements of Adj(A) are inside GA+S.
Thus from GA and GB we get two simplicial
vertices in G.

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
Definition: The following are equivalent:
1. G is chordal
2. The edges of G can be directed acyclically
so that every pair of converging arrows
3. G has a perfect vertex elimination
scheme.
4. There is a tree T with maximal cliques of
G as vertices. Any two cliques containing
v are either adjacent in T or connected by
a path of cliques that contain v.
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
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
By lemma 1 every chordal graph has a
simplicial vertex
Direct all of its edges to it
Repeat the process with the rest of the graph
(can be done because being chordal is
hereditary)
From the definition of simplicial vertex
corollary that each two converging edges
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
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Let there be a cycle of size larger than 3.
There exist a valid direction which is acyclic
There are two converging edges, which come
The cycle has a chord.
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

Given a Markov network, we would like to
calculate a probability function over it’s
variables
What if we want to calculate the probability of
one variable?
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P(X1,X2,X3,X4,X5)=K∙f1(X1,X2)f2(X2,X3)f3(X3,X4)f4(X4,X5)f5(X5,X1)
P(X1,X2,X3,X4)=ΣX5K∙f1(X1,X2)f2(X2,X3)f3(X3,X4)f4(X4,X5)f5(X5,X1)=
=K∙f1(X1,X2)f2(X2,X3)f3(X3,X4) ΣX5f4(X4,X5)f5(X5,X1)
= K`∙f1(X1,X2)f2(X2,X3)f3(X3,X4) F(X4, X1)

Explanation: Each time we sum over a
(creating a clique)
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
Here is an example of how bad (i.e. big
cliques) things can happen with
incorrect order of summing
P(X1,X2,X3,X4,X5)=ΣX0K∙f1(X0,X1)f2(X0,X2)f3(X0,X3)f4(X0,X4)f5(X0,X5)
= K`∙F(X1,X2,X3,X4,X5)

Explanation: For wrong order of
summing we get big cliques and thus
huge tables of probabilities to store,
which are exponential in number of
variables.
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
Let G=(V,E) be an undirected graph and let
σ=[v1,v2,…,vn] be an ordering of the vertices.
We say that σ is a perfect vertex elimination
scheme if for each i: Xi = { vj in Adj(vi) | j&gt;I }
is complete.
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
Definition: The following are equivalent:
1. G is chordal
2. The edges of G can be directed acyclically
so that every pair of converging arrows
3. G has a perfect vertex elimination
scheme.
4. There is a tree T with maximal cliques of
G as vertices. Any two cliques containing
v are either adjacent in T or connected by
a path of cliques that contain v.
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
By lemma 1, G has a simplicial vertex.

The subgraph induced after removing this
vertex is also chordal.

By induction, there is an elimination order.
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
Let us assume a cycle of length 4 or greater.

Let x be the first vertex on the elimination
scheme of this cycle.

x is simplicial at this point.

Two vertices adjacent to x from the cycle are
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3. All vertices of G can be deleted by
arranging them in separate piles, one for
each clique, and then repeatedly applying
the following two operations:

Delete a vertex that occurs in only one pile.

Delete a pile if all its vertices appear in another pile.
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A
B
C
D
B
A
C
C
B
D
D
E
E
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A
B
A
C
B
A C
D
B
D
C
D
D
E
E
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
Definition: The following are equivalent:
1. G is chordal
2. The edges of G can be directed acyclically
so that every pair of converging arrows
3. G has a perfect vertex elimination
scheme.
4. There is a tree T with maximal cliques of
G as vertices. Any two cliques containing
v are either adjacent in T or connected by
a path of cliques that contain v.
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a
a
b
C1
b
C1
c
c
C2
C2
b
C3
e
d
c
d
C3
b
e
d
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C1
a
a
C1
b
C2
C2
C4
c
C3
d
b
C4
a
b
c
d
c
d
C3
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
Lemma 2: A vertex is simplicial iff it belongs
to exactly one clique.

Let G =(V,E) be a chordal graph and |V|=k

Proof by induction: Assume the claim holds
for all graphs of size &lt; k.

By Lemma 1 G has a simplicial vertex v.

By Lemma 2 v belongs to exactly one clique C.
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
For G’ =(V\{v},E’) there exists a tree T’ which
satisfies the claim. Split into two cases:
◦ C’ = C – {v} is maximal in G’. Add v to C’ to
build T from T’. T is a Clique-Tree as
needed.
a
C1
b
C1
a
c
b
e
C2
C2
b
c
e
d
d
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◦ C’ = C – {v} is not maximal in G’. There is a
maximal clique P in G’ such that C’⊂P. Add
C and (C, P) to T’ to build T. T is a CliqueTree as needed.
a
c
C2
C1
b
d
C1
a
c
C2
a
d
b
d
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

Let T be a Clique-Tree of graph G=(V,E) and
|V|=k.
Proof by induction: Assume the claim holds
for all graphs of size &lt; k.

Let L be a leaf in T, and P its parent in T.

Let v∈L\P (exists from maximality).

v cannot be in any other clique.

by Lemma 2, v is simplicial.
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
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Let T’ be T with v removed. T’ is a CliqueTree.
By assumption G’ =(V\{v},E’) is chordal.
By part 3 of Theorem 7, there exists σ’, a
Perfect Elimination Order for G’.
Let σ = [v, σ’] be a perfect elimination order
for G.
By the converse of part 3, we can conclude
that G is chordal
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