chordal graphs: theory and algorithmsweb.iitd.ac.in/~bspanda/chordal.pdf · 2 chordal graphs...

1

Chordal Graphs: Theory and Algorithms

2

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.

3

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

4

Subclasses of Chordal Graphs …

Rooted Directed Path Graphs

Directed Path Graphs

Path Graphs

Strongly Chordal Graphs

Doubly Chordal Graphs

5

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

6

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)

7

8

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.

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.

Lemma 1

9

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

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.

Proof of Lemma 1 (Cont.)

1

1

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.

Proof of Lemma 1 (Cont.)

1

2

13

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

PEO: (v1,v2,…,vn) is a PEO if vi is a simplicial vertex of

G[{vi,vi+1,…,vn}], 1 ≤ i ≤n.

1

2

3

4

5

(1,2,3,4,5) is a PEO of G

G

14

PEO Illustration

Perfect Elimination order : necessary and sufficient condition for Chordal graphs

Simplicial vertex : along with their neighbors form a clique

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.

Theorem 7

1

5

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.

Proof of Theorem 7 (1→2)

1

6

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.

Proof of Theorem 7 (2→1)

1

7

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.

Perfect Vertex Elimination

1

8

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.

Theorem 7 (Cont.)

1

9

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.

Proof of Theorem 7 (1→3)

2

0

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.

Proof of Theorem 7 (3→1)

2

1

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.

Theorem 7 (Cont.)

2

2

Clique Tree Example

2

3

a

c b

d e

C2

C3

C1

a

c b

C1

c b

d

b

d e

C2

C3

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.

Proof of Theorem 7 (1→4)

2

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.

Proof of Theorem 7 (1→4)

2

5

a

c b

d e

C2

C1 a

b

C1

b

d

C2

e c

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.

Proof of Theorem 7 (1→4)

2

6

b a

d c

C2

C1

C1

a

d

C2

c

b a

d

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.

Proof of Theorem 7 (4→1)

2

7

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

Proof of Theorem 7 (4→1)

2

8

29

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

30

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.

31

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}

1

2

3

4

5

(1,2,3,4,5) is a PEO of G

G

32

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)

33

Minimum Coloring: Illustration

C(1)=1,C(2)=2,C(3)=3,C(4)=1,C(5)=2

(G)=3.

1

2

3

4

5

(5,4,3,2,1) is a PEO of G

G

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.

34

35

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

36

Maximum Independent Set: Illustration

I={1,4}

1

2

3

4

5

(1,2,3,4,5) is a PEO of G

G

37

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.

38

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.

39

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]

40

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)

41

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)

42

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

43

Example

A(0)

B(0) C(0)

D(0)

E (0) F (0)

G(0)

Seq =1

B (1)

D(1)

C(1)

E (1)

Seq = 2

D(2)

F (1)

Seq =3

C(2) B (2)

F (2)

Seq = 4

B (3)

Seq = 5

Seq = 6

G(1)

H (0) H (1) H (2) H (3)

Seq = 7

G(2)

Seq = 8

The PEO is:

G, H, F, B, C, D, E , A

44

Observations

1. Except for the first node, any node being picked

up has at least one neighbor which has already

been picked

2. Set of picked nodes always form a connected

graph

45

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

46

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

47

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.

48

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)

49

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

50

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.

51

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)

52

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

53

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)

54

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.

55

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

56

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