Chordal Graphs: Theory and Algorithms
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Chordal graphs
Chordal graph : Every cycle of four or more vertices has a chord in it,
i.e. there is an edge between two non consecutive vertices of the cycle.
Also called rigid circuit graphs, Perfect Elimination Graphs,
Triangulated Graphs, monotone transitive graphs.
2
Subclasses of Chordal Graphs
Trees
K-trees ( 1-tree is tree)
Kn: Complete Graph
Block Graphs
Bipartite Graph with bipartition X and Y ( not
necessarily Chordal )
Split Graphs: Make one part of Bipartite graph
complete
Interval Graphs
3
Subclasses of Chordal Graphs …
Rooted Directed Path Graphs
Directed Path Graphs
Path Graphs
Strongly Chordal Graphs
Doubly Chordal Graphs
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Research Issues in Trees
Computing Achromatic number in tree is NP-Hard
Conjecture: Every tree is Graceful
L(2,1)-labeling number of a tree with maximum
degree is either +1 or +2. Characterize trees
having L(2,1)-labeling number +1
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Why Study a Special Graph Class?
Reasons
The Graph Class arises from applications
The graph class posses some interesting structures that
helps solving certain hard but important problems
restricted to this class
some interesting meaningful theory can be developed
in this class.
All of these are true for chordal Graphs.
Hence study Chordal Graphs.
See the book: Graph Classes (next slide)
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7
Characterizing Property: Minimal separators
are cliques ( Dirac 1961)
Characterizing Property: Every minimal a-b vertex
separator is a clique ( Complete Subgraph)
Minimal a-b Separator: S V is a minimal vertex
separator if a and b lie in different components of
G-S and no proper subset of S has this property.
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Lemma 1
A vertex x of G is called simplicial if its adjacency
set Adj(x) is a clique
(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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Proof of Lemma 1
If G is a clique, done.
If not, assume that G 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.
1
0
Proof of Lemma 1 (Cont.)
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 <a,…,x,…,b,…,y,…,a>. It is easy to show
that there exist a minimal cycle with length no
less then 4 with no
chords in it,
contradiction.
1
1
Proof of Lemma 1 (Cont.)
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.
1
2
Characterizing Property: PEO ( Fulkerson and
Gross 1965)
Lemma: (Dirac 1961) Every chordal graph has a simplicial
vertex. If G is not complete, then it has two non-adjacent
simplicial vertices. Simplicial: G[NG(v)] is complete
Theorem: G is chordal iff it has a Perfect Elimination
Ordering
2
4
1
3
(1,2,3,4,5) is a PEO of G
5
G ,…,v ) is a PEO if v is a simplicial vertex of
PEO: (v1,v
2
n
i
G[{vi,vi+1,…,vn}], 1 ≤ i ≤n.
13
PEO Illustration
Perfect Elimination order : necessary and sufficient
condition for Chordal graphs
Simplicial vertex : along with their neighbors form
a clique
14
Theorem 7
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 comes
from two adjacent vertices.
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.
1
5
Proof of Theorem 7 (1→2)
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 coming from
adjacent vertices.
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6
Proof of Theorem 7 (2→1)
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 from
two adjacent vertices
The cycle has a chord.
1
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Perfect Vertex Elimination
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>I } is complete.
1
8
Theorem 7 (Cont.)
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 comes from two
adjacent vertices.
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.
1
9
Proof of Theorem 7 (1→3)
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.
2
0
Proof of Theorem 7 (3→1)
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 also
adjacent between themselves.
2
1
Theorem 7 (Cont.)
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 comes from two
adjacent vertices.
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.
2
2
Clique Tree Example
a
a
b
C1
b
C1
c
c
C2
C2
b
C3
e
d
c
d
C3
b
e
d
2
3
Proof of Theorem 7 (1→4)
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 < k.
By Lemma 1 G has a simplicial vertex v.
By Lemma 2 v belongs to exactly one clique C.
2
4
Proof of Theorem 7 (1→4)
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
2
5
Proof of Theorem 7 (1→4)
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 Clique-Tree as needed.
a
b
C2
c
C1
d
C1
a
c
C2
a
d
b
d
2
6
Proof of Theorem 7 (4→1)
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 < 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.
2
7
Proof of Theorem 7 (4→1)
Let T’ be T with v removed. T’ is a Clique- Tree.
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
2
8
PEO is a Key in designing algorithms and
obtaining structural properties
The following classical Graph Optimization problems are
NP-Hard for general Graphs
Finding
(G) :clique number (the size of maximum size clique )
(G): Chromatic number ( minimum number of colors needed in a
proper coloring of G)
(G): independence number( maximum size of an independent set)
(G): clique covering number ( minimum number of cliques
needed to cover V(G) )
All these four problems can be solved in linear time in
Chordal Graphs given a PEO as part of input
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Polynomial Algorithms for chordal Graphs:
Clique Number
Given a PEO of a Chordal Graph G, maximum
size clique can be found as follows:
Let (v1,v2,…,vn) be a PEO of G.
Define L(i)= maximum j, j > i such that vivj
E(G). F[i]=minimum j j >I such vivj isan edge.
Let Max {L[i]-F[i]+1}=k=L[j]-F[j]+1.
Then k= (G) and C={vj,vj+1,…,v_{j+k-1}} is a
maximum size clique.
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Maximum Clique: Illustration
L[1]=3,L[2]=4,L[3]=5,L[4]=5,L[5]=5.
Max{L[i]-i+1}=3=L[1]-1+1=L[2]-2+1=L[3]-3+1.
Maximum size cliques are {1,2,3},{2,3,4}, {3,4,5}
2
4
1
G
3
(1,2,3,4,5) is a PEO of G
5
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Polynomial Algorithms for chordal
graphs: chromatic number [Gavril 1972]
Minimum-Coloring Algorithm
Greedy Algorithm
Scan the vertices in the reverse order of PEO and color each
vertex with the smallest color not used among its successors
Gives Optimal coloring
Complexity = Polynomial (given that PEO is already known)
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Minimum Coloring: Illustration
C(1)=1,C(2)=2,C(3)=3,C(4)=1,C(5)=2
(G)=3.
2
4
1
G
3
(5,4,3,2,1) is a PEO of G
5
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Why optimal?
Clearly, (G) (G)
Let (v1,v2,…,vn) be PEO. So (vn,…,v2,v1)
Is the reverse PEO. Now coloredNbr(vi)=|{vj|
vivj is an edge and j >i}| ≤ (G)
So (G) = (G) and hence Greedy algorithm is
optimal.
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Polynomial Algorithms for chordal graphs:
Independence Number
Maximum Independent Set
Greedy algorithm
Scan the vertices in the order of PEO, and for each vi,
add vi to I if none of its predecessor has been added to I
Complexity = O(n+m)
Predecessor of vi is vj if j < i and vjvi E
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Maximum Independent Set: Illustration
I={1,4}
2
4
1
G
3
(1,2,3,4,5) is a PEO of G
5
36
Clique cover
Xi={vj | vivj and j >i}
y1=v1; yi is the first vertex in which follows yi-1
and which is not in X1 X2 … Xi-1; all
vertices following yt are in X1 X2 … Xt.
Hence V={y1,y2,…yt} X1 X2 … Xt.
Theorem: {y1,y2,…yt} is a maximum independent
set and the collection of sets Yi={yi} Xi, 1≤ i ≤t
comprises a minimum clique cover of G.
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Proof
{y1,y2,…yt} is an independent set as yiyj E(G)
implies yj Xi which is a contradiction. So (G)
t. As Yi, 1 ≤ i≤ t, is a clique cover of G. So
(G) t. Hence (G) =t and the minimum clique
cover contains at least t cliques. Hence the
theorem.
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Perfect Elimination Order (PEO)
•
An ordered sequence of all vertices {V1, V2,…, Vn}
•
Successor (Vi) = {Vj: j>i and (i,j) ε E}
•
•
Predecessor (Vi) = {Vj : i>j and (i,j) ε E}
Sequence of vertices such that for each vertex Vi, successors
of Vi form a clique
•
A graph is chordal if and only if it has a Perfect Elimination
Order (PEO) [Fulkerson 1965]
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Simplicial Vertex
A vertex whose all neighbors form a clique
Every chordal graph has at least one simplical vertex
The first vertex in PEO is simplicial
If you remove the simplicial vertex, then the graph induced by
remaining nodes is also a chordal graph (hence must have a
simplicial vertex of its own)
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Simple method to find PEO
Given G = (V, E)
For (i=1 to N) {
Vi = Simplicial vertex (G);
G = Graph induced by vertex set (V-Vi);
}
What if no simplicial vertex exists at some point?
Complexity = O(N4)
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Finding a PEO (Maximum Cardinality
Method)
Rose and Tarjan [Rose 1975]
Cardinality Number = Number of neighbors picked up
Step1: Pick the node whose maximum number of neighbors have
already been chosen
Reverse the order
Step 2: In the end, verify the ordering obtained is indeed PEO
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Example
Seq = 2
E (1)
(0)
Seq = 6
(1)
FF(0)
F (2)
Seq = 8
A(0)
Seq =1
D(2)
D(0)
D(1)
Seq =3
G(2)
G(0)
G(1)
H
(1)
HH
H(3)
(2)
(0)
Seq = 7
The PEO is:
G, H, F, B, C, D, E , A
BB(2)
B(3)
B(0)
(1)
Seq = 5
C(0)
C(1)
C(2)
Seq = 4
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Observations
1.
2.
Except for the first node, any node being picked
up has at least one neighbor which has already
been picked
Set of picked nodes always form a connected
graph
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Proof of Correctness
If a PEO does not exist, then algorithm does not
generate a PEO
If the PEO exists, then algorithm outputs some
PEO
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Proof of Correctness (Part 2)
Let {V1, V2, V3,…,Vi, …, Vj,…, Vk,…,Vn} be the ordering generated.
Let Vi is not in sequence. Then Vj and Vk be the two successors of Vi that
are not connected.
Case 1:
Then Vj and Vk are parts of two graphs which are connected only through
Vi
The order in which the nodes were picked is:
{Vn, …, Vk, …., Vj,…, Vi,…,V3, V2, V1}
Then we claim that label sequence must have been different
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Proof (Contd.)
Case 2:
Consider the path from k to j on which the nodes
were picked up.
All such nodes should be connected to i except
those nodes, which are directly connected to j.
Let m be such a vertex on that path. m must have
some vertices which are not connected to i.
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Proof (contd.)
Let w be such a vertex. Therefore w must be
connected to k. (through a path not involving i).
If i is not connected to w, then we have a non
chordal cycle i-k-w-m-j-i.
i must be connected to w as well. Therefore there is
no way that m can be picked before I since i is
connected to one extra node than m (Vk)
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Complexity Analysis (Step 2)
Check if the generated ordering is perfect
Trivial algorithm:
Number of vertices whose successors have to be checked: N
Time complexity to check if all the successors of a particular node
are connected: NC2 = N2
Total complexity = N3
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Step 2: Efficient Implementation
Efficient Way:
Given ordering: {V1, V2, V3, …, Vi, …, Vj,…, Vk,…,Vn}
1. for i = 1 to n do
2.
if Vi has successors
3.
Let u be the first successor of Vi
4.
For all w ∈ Successor (Vi), w ≠u
5.
Add (u, w) to TEST
6. Test whether all vertex-pairs in TEST are adjacent.
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Why?
If the algorithm detects ordering as non-perfect, then ordering must be non-
perfect
If the algorithm detects ordering as PEO, then ordering cannot be imperfect (By
contradiction)
complexity if implemented carefully: O ( N + E)
Maximum size of set TEST can be O(E)
Instead of O (N3)
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Implementation of step 1 (generating
perfect ordering)
Trivial Method:
Maintain a heap of cardinality numbers
Complexity of each for loop:
Find the maximum cardinality node and removing it: O(lg N)
Number of updates for each node picked: N
modifying the cardinality number and rearranging the heap: N lg N
Total Complexity = N* (lg N + N lg N) = N2 lg N
Can also be shown as E lg N
52
Efficient Implementation (Step 1)
With complexity O( N + E)
For each node, maintain list of neighbors. (Already given in this
format)
Maintain linked list of vertices (in order of cardinality number)
For each node, maintain its location in the linked list
Picking the vertex with max. cardinality number = O(1)
Finding all the neighbors and updating their cardinality number and
their location in linked list = O(E)
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Another definition for chordal graphs
Intersection Graph: A graph that represents the intersection of sets
Subtree Graph:
Break a tree into multiple subtrees (overlapping)
Now make a graph whose each vertex represents one subtree. An edge between two
vertices if they have a common vertex (between the subtrees they represent)
The graph thus obtained is known as Subtree Graph
As [Gavril 1974] showed, the subtree graphs are exactly the chordal graphs. So a
chordal graph can be represented as an intersection graph of subtrees.
54
Bibliography
[Gavril 1974]: Gavril, Fănică, "The intersection graphs of subtrees in trees
are exactly the chordal graphs”, Journal of Combinatorial Theory, Series B
16, 1974
[Gavril 1972]: Gavril, “Algorithms for Minimum Coloring, Maximum
Clique, Minimum Independent Set of a Chordal Graph”, SIAM J. Comp.,
Vol 1, 1972
[Rose 1975]: Rose, Tarjan, “Algorithmic Aspects of Vertex Elimination”,
Proc. 7th annual ACM Symposium on Theory of Computing (STOC), 1975
55
Bibliography
[Arnborg 1989]: Arnborg et. al., “Linear time algorithms for NP-hard
problems restricted to partial k-trees”, Discrete Applied Mathematics,
Vol. 23, Issue 1, 1989
[Fulkerson 1965]: Fulkerson, D. R.; Gross, O. A. (1965). "Incidence
matrices and interval graphs”, Pacific J. Math 15: 835–855.
M.C.Golumbic Algorithmic Graph Theory and Perfect Graphs. Second
edition
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