3eme Cours : Chordal Graphs MPRI 2011 2012habib/Documents/cours3-2011.pdf · 3eme Cours : Chordal...

3eme Cours : Chordal Graphs MPRI 2011–2012

3eme Cours : Chordal GraphsMPRI 2011–2012

Michel [email protected]

http://www.liafa.jussieu.fr/~habib

Chevaleret, septembre 2011

http://www.liafa.jussieu.fr/~habib


Schedule

What we already know

More structural insights of chordal graphs

Properties of reduced clique graphs

Interval graphs

Exercises



Fundamental objects to play with

Definition

A graph is chordal iff it has no chordless cycle of length ≥ 4.

Maximal Cliquesunder inclusion

Minimal SeparatorsA subset of vertices S is a minimal separator if Gif there exist a, b ∈ G with ab /∈ G , such that a andb are not connected in G − S .and S is minimal for inclusion with this property .



An example

a

b c ef

d3 minimal separators {b} for f and a, {c} for a and e and {b, c}

for a and d .



If G = (V ,E ) is connected then for every a, b ∈ V such thatab /∈ Ethen there exists at least one minimal separator.But there could be an exponential number of minimal separators.



VIN : Maximal Clique trees

A maximal clique tree (clique tree for short) is a tree T thatsatisfies the following three conditions :

I Vertices of T are associated with the maximal cliques of G

I Edges of T correspond to minimal separators.

I For any vertex x ∈ G , the cliques containing x yield a subtreeof T .



Helly Property

Definition

A subset family {Ti}i∈I satisfies Helly property if∀J ⊆ I et ∀i , j ∈ J Ti ∩ Tj 6= ∅ implies ∩i∈ITi 6= ∅

Exercise

Subtrees in a tree satisfy Helly property.



Demonstration.

Suppose not. Consider a family of subtrees that pairwise intersect.For each vertex x of the tree T , it exists at least one subtree of thefamily totally contained in one connected component of T − x .Else x would belong to the intersection of the family, contradictingthe hypothesis.Direct exactly one edge of T from x to this part.We obtain a directed graph G , which has exactly n vertices and ndirected edges. Since T is a tree, it contains no cycle, therefore itmust exist a pair of symmetric edges in G , which contradicts thepairwise intersection.



Back to chordal graphs

Chordal graph recognition

1. Apply a LexBFS on G O(n + m)

2. Check if the reverse ordering is a simplicial elimination schemeO(n + m)

3. In case of failure, exhibit a certificate : i.e. a cycle of length≥ 4, without a chord. O(n)



Subtrees in a tree

Using results of Dirac 1961, Fulkerson, Gross 1965, Buneman1974, Gavril 1974 and Rose, Tarjan and Lueker 1976 :

For a connected graph, the following statements are equivalentand characterize chordal graphs :

(i) G has a simplicial elimination scheme

(ii) Every minimal separator is a clique

(iii) G admits a maximal clique tree.

(iv) G is the intersection graph of subtrees in a tree.

(v) Any MNS (LexBFS, LexDFS, MCS) provides asimplicial elimination scheme.



Two subtrees intersect iff they have at least one vertex in common.By no way, these representations can be uniquely defined !



An example



Proof of the chordal characterization theorem

I Clearly (iii) implies (iv)

I For the converse, each vertex of the tree corresponds to aclique in G .But the tree has to be pruned of all its unnecessary nodes,until in each node some subtree starts or ends. Then nodescorrespond to maximal cliques.

I We need now to relate these new conditions to chordal graphs.(iii) implies (i) since a maximal clique tree yields a simplicialelemination scheme.(iv) implies chordal since a cycle without a chord generates acycle in the tree.(iv) implies (ii) since each edge of the tree corresponds to aminimal separator which is a clique



from (i) to (iv)

Demonstration.

By induction on the number of vertices. Let x be a simplicialvertex of G .By induction G − x can be represented with a family of subtreeson a tree T .N(x) is a clique and using Helly property, the subtreescorresponding to N(x) have a vertex in common α.To represent G we just add a pending vertex β adjacent to α.x being represented by a path restricted to the vertex β, and weadd to all the subtrees corresponding to vertices in N(x) the edgeαβ.



Playing with the representation

Easy Exercises :

1. Find a minimum Coloring (resp. a clique of maximum size) ofa chordal graph in O(n + m).Consequences : chordal graphs are perfect.At most n maximal cliques.

2. Find a minimum Coloring (resp. a clique of maximum size) ofan interval graph in O(n)using the interval representation.



Exercises

1. Can we use efficiently this representation of chordal graphs asintersection of subtrees ?

2. Same question for path graphs ? (intersection graph of pathsin a tree)



Which kind of representation to look for for special classesof graphs ?

I Easy to manipulate (optimal encoding, easy algorithms foroptimisation problems)

I Geometric in a wide meaning (ex : permutation graphs =intersection of segments between two lines)

I Examples : disks in the plane, circular genomes . . .



First remark

Proposition

Every undirected graph can be obtained as the intersection of asubset family

Proof

G = (V ,E )Let us denote by Ex the set of edges adjacent to x .xy ∈ E iff Ex ∩ Ey 6= ∅We could also have taken the set Cx of all maximal cliques whichcontains x .Cx ∩ Cy 6= ∅ iff ∃ one maximal clique containing both x and y



Starting from a graph in some application, find its characteristic :

1. 2-intervals on a line (biology), intersection of disks (orhexagons) in the plane (radio frequency), filament graphs,trapezoid graphs . . .

2. A whole book on this subject :J. Spinrad, Efficient Graph Representations, Fields InstituteMonographs, 2003.

3. A website on graph classes :http ://www.graphclasses.org/



Clique treeclique tree of G = a minimum size tree model of G

for a clique tree T of G :I vertices of T correspond precisely to the maximal cliques of GI for every maximal cliques C ,C ′, each clique on the path in T

from C to C ′ contains C ∩ C ′

I for each edge CC ′ of T , the set C ∩ C ′ is a minimal separator(an inclusion-wise minimal set separating two vertices)

Note : we label each edge CC ′ of T with the set C ∩ C ′.



Consequences of maximal clique tree

Theorem

Every minimal separator belongs to every maximal clique tree.

Lemma

Every minimal separator is the intersection of at least 2 maximalcliques of G .

Corollary

There are at most n minimal separators.



Theorem

Every minimal separator belongs to every maximal clique tree.

Lemma

Every minimal separator is the intersection of at least 2 maximalcliques of G .



Proof of the lemma

Demonstration.

We know that S is a clique. Let us consider G1 a connectedcomponent of G − S . Let x1, . . . , xk be the vertices of G1 having amaximal neighbourhood in S .If k = 1 then x1 must be universal to S , since S is a minimalseparator.Else, consider a shortest path µ in G1 from x1 to xk . Necessarily x1(resp. xk) has a private neighbour z (resp. t) in S .Then the cycle [x1, µ, xk , t, z ] has no chord, a contradiction.Therefore x1 ∪ S is a clique, and is contained in some maximalclique C in G1. We finish the proof by taking another connectedcomponent of G − S .



Proof of the theorem

Demonstration.

Therefore S = C ′ ∩ C ′′. These two maximal cliques belong to anymaximal clique tree T of G . Let us consider the unique path µ inT joigning C ′ to C ′′.All the internal maximal cliques in µ must contain S . Suppose thatall the edges of µ are labelled with minimal separators strictlycontaining S , then we can construct a path in G from C ′ − S toC ′′ − S avoiding S , a contradiction. So at least one edge of µ islabelled with S .



Clique graphthe clique graph C(G ) of G = intersection graph of maximalcliques of G

G C(G )



Reduced clique graphthe reduced clique graph Cr (G ) of G = graph on maximal cliquesof G where CC ′ is an edge of Cr (G ) ⇐⇒ C ∩ C ′ is a minimalseparator.





Combinatorial structure of Cr (G )

Lemma 1 : M.H and C. Paul 95

If C1,C2,C3 is a cycle in Cr (G ), with S12,S23 and S23 be theassociated minimal separators then two of these three separatorsare equal and included in the third.

Lemma 2 : M.H. and C. Paul 95

Let C1,C2,C3 be 3 maximal cliques, ifC1 ∩ C2 = S12⊂S23 = C2 ∩ C3 then it yields a triangle in Cr (G )



Lemma 3 : Equality case

Let C1,C2,C3 be 3 maximal cliques, if S12 = S23 then :

I either the C1 ∩ C3 = S13 is a minimal separator

I or the edges C1C2 and C2C3 cannot belong together to amaximal clique tree of G .



a

bc

d fe

g

h

b,c,e

a,b,c

b,d,e e,g c,e,f

c,hb,e

e c,e

c

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b,c c

b,c,e

a,b,c

b,d,e e,g c,e,f

c,hb,e c,e

ce

b,c

b,c,e

a,b,c

b,d,e e,g c,e,f

c,hb,e

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Theorem (Gavril 87, Shibata 1988, Blayr and Payton 93)

The clique trees of G are precisely the maximum weight spanningtrees of C(G ) where the weight of an edge CC ′ is defined as|C ∩ C ′|.

Theorem (Galinier, Habib, Paul 1995)

The clique trees of G are precisely the maximum weight spanningtrees of Cr (G ) where the weight of an edge CC ′ is defined as|C ∩ C ′|.

Moreover, Cr (G ) is the union of all clique trees of G.



Applications

I All clique trees have exactly the same labels, includingrepetitions.



Maximal Cardinality Search : MCS

Data: a graph G = (V ,E ) and a start vertex s

Result: an ordering σ of V

Assign the label 0 to all verticeslabel(s)← 1for i ← n a 1 do

Pick an unumbered vertex v with largest labelσ(i)← vforeach unnumbered vertex w adjacent to v do

label(w)← label(w) + 1end

end



Maximum spanning trees

Maximal Cardinality Search can be seen as Prim algorithm forcomputing a maximal spanning tree of Cr (G ).



I How to compute a clique tree ?

I How to generate all simplicial elimination schemes ?



LexBFS, MCS or MNS visit maximal cliques “consecutively” (i.e.when the search explores a vertex x of a maximal clique C thatdoes not belong to any of the previously visited maximal cliquesthen all the unvisited vertices of C will appear consecutively justafter x).

Therefore when applying a search (LexBFS, MCS or MNS) one cancompute a clique tree, by considering the strictly increasingsequences of labels.



a

bc

d fe

g

h

b,c,e

a,b,c

b,d,e e,g c,e,f

c,hb,e

e c,e

c

c

ee

b,c c

b,c,e

a,b,c

b,d,e e,g c,e,f

c,hb,e c,e

ce

b,c

b,c,e

a,b,c

b,d,e e,g c,e,f

c,hb,e

e c,e

c

b,c

b, a, c, e, d , h, f , g is a LexBFS ordering.we can find the maximal cliques b, a, c then b, c , e then b, e, dthen c, h then c , e, f and e, g .



Simplicial elimination schemes

1. Choose a maximal clique tree T

2. While T is not empty doSelect a vertex x ∈ F − S in a leaf F of T ;F ← F − x ;If F = S delete F ;



Canonical simplicial elimination scheme

1. Choose a maximal clique tree T

2. While T is not empty doChoose a leaf F of T ;Select successively all vertices in F − Sdelete F ;



Remark

Does there exist other simplicial elimination scheme ?



Size of a maximal clique tree in a chordal graph

I Let G = (V ,E ) be a chordal graph.

I G admits at most |V | maximal cliques and therefore the treeis also bounded by |V | (vertices and edges).

I But some vertices can be repeated in the cliques. If weconsider a simplicial elimination ordering the size of a givenmaximal clique is bounded by the neighbourhood of the firstvertex of the maximal clique.

I Therefore any maximal clique tree is bounded by |V |+ |E |.



Size of CS(G )

Considering a star on n vertices,shows |CS(G )| ∈ O(n2)Not linear in the size of G



CS(G ) is not chordal !



CS(G ) is not chordal !



In fact CS(G ) is dually chordal (almost chordal)and CS(CS(G )) is chordal.



Cannonical representation

I For an interval graph, its PQ-tree represents all its possiblemodels and can be taken as a cannonical representation of thegraph (for example for graph isomorphism)

I But even path graphs are isomorphism complete. Therefore acanonical tree representation is not obvious for chordal graphs.

I Cr (G ) is a Pretty Structure to study chordalgraphs.To prove structural properties of all maximal clique trees of agiven chordal graph.


Interval graphs

Characterization Theorem for interval graphs

(i) G = (V ,E ) is interval graph.

(ii) It exists a total ordering τ of the vertices of V s.t.∀x , y , z ∈ G with x ≤τ y ≤τ z and xz ∈ E thenxy ∈ E .

(iii) G has a maximal clique path. (A maximal clique pathis just a maximal clique tree T, reduced to a path).

(iv) G is the intersection graph of a family of intervals ofthe real line


Interval graphs

To recognize an interval graph, we just have to compute amaximal clique tree and check if it is a path ?Difficulty : an interval graph has many clique trees among themsome are paths


Interval graphs


Interval graphs


Interval graphs

Many linear time algorithms already proposed for interval graphrecognition ....using nice algorithmic tools :graph searches, modular decomposition, partition refinement,PQ-trees . . .


Interval graphs

Linear time recognition algorithms for interval graphs

I Booth and Lueker 1976, using PQ-trees.

I Korte and Mohring 1981 using LexBFS and ModifiedPQ-trees.

I Hsu and Ma 1995, using modular decomposition and avariation on Maximal Cardinality Search.

I Corneil, Olariu and Stewart SODA 1998, using a series of 6consecutive LexBFS, published in 2010.

I M.H, McConnell, Paul and Viennot 2000, using LexBFS andpartition refinement on maximal cliques.


Interval graphs

A partition refinement algorithm working on maximalcliques

1. Compute a tree T using LexBFIf T is not a maximal clique tree ; then G is not chordal,neither interval.

2. Start form the last maximal clique Refine the cliques with theminimal separator.

3. Refine until each part is a singleton

4. If a part is not a singleton start recursively from the last cliqueof this part according to LexBFS.


Interval graphs


Interval graphs

Variations on these representations

I Path graphs = Paths in a treeIn between chordal and interval.

I Directed path graphs = directed paths in a rooted directedtreeappear in some polynomial CSP class.


Interval graphs

State of the Art

I We (Juraj Stacho, M.H.) study chordal representations foralgorithms

I A linear time algorithm for path graph recognition is stillmissing


Exercises

Exercise 1

Ends of a LexBFS

Many properties can be expressed on the last vertex of a LexBFS.Example : if G is a chordal graph, the last vertex is simplicial .

1. Show that the last maximal clique visited can be taken as theend of some chain of cliques if G is an interval graph.

2. Complexity of the following decision problem :

Data: a graph G = (V ,E ) and a given vertex x ∈ V

Result: Does there exist a LexBFS of G ending at x ?


Exercises

Exercise 2

If we consider the edges of the clique tree labelled with the size ofthe minimal separators, show that :for every maximal clique tree Tweight(T ) = Σ1≤i≤k |Ci | − n, where C1, . . .Ck are the maximalcliques of G .

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