Split Decomposition and Circle Graph Recognition in Quasi-Linear Time

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Split Decomposition and Circle Graph Recognition in Quasi-Linear Time

Christophe Paul

CNRS - LIRMM - Universit´e Montpellier II, France

March 25, 2009

Joint work with D. Corneil(U. of Toronto),E. Gioan(CNRS LIRMM) andM. Tedder(U. of Toronto)

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Graph-labeled tree (GLT)

I a pair (T,F) withT a tree andF a set of graphs

I each nodev of degreek ofT is labelled by a graphG_v ∈ F

I a bijectionρv from the tree-edges incident tov toV(Gv)

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Accessiblity graph of a GLT

Given a GLT (T,F), define the graphG(T,F) with

I vertex set of G(T,F) = set of leaves ofT

I xy is an edge iffρv(uv)ρv(vw)∈E(Gv) for all tree-edgesuv, vw on thex,y-path inT

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A split in a graph

G

= (V

,E

) is a bipartition (A,

B) ofV

st.

I |A|>2,|B|>2 and

I forx∈A, y∈B,xy∈E iffx ∈N(B) andy ∈N(A)

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Every edgeof a GLT defines asplit

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N(q) N(r)

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split

join 1 q

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Theorem [Cunnigham’82 reformulated]

For any connected graphG, there exists a unique graph-labelled tree ST(G) = (T,F) with a minumun number of nodes such that

I G =G(T,F),

I any graph ofF is primeordegenerate.

Prime Degenerate

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Vertex incremental characterization

A vertexx is added to a graph G = (V,E) with neighborsS ⊆V. (S is represented inred)

I How to compute the split treeST(G+ (x,S)) fromST(G) ?

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States of marker vertices

Letqbe marker vertex (or a leaf)

LetL(q) be the set of leavesl such that the path betweenl andq contains the edge of the tree associated withq

I q isemptyifL(q)∩S =∅

I q isperfectif every leaf inL(q)∩S is accessible fromq

I q ismixedotherwise

q

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States of marker vertices

E

E E E

E

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States of marker vertices

P E

E

E E

P P

P

P P

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States of marker vertices

M E E

P P

P

E E

P M

M M

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Vertex Incremental Characterization

The split tree ofG+ (x,S) is obtained from ST(G) by 3 cases:

I 1 - T has an edgeewhose extremities are perfect

P P

I 2 -T has an edgee, one extremity of which is perfect and the other empty

I 3 - T has afully-mixednodeu

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Vertex Incremental Characterization

P P

x

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Vertex Incremental Characterization

P E

x

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Vertex Incremental Characterization

every marker vertex adjacent to the nodeuis mixed

M M

M

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Vertex Incremental Characterization

I 3 - T has afully-mixednodeu(the set of fully-mixed nodes is a subtree ofT)

Make a join

E E

M

P P M M

M

M M

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Vertex Incremental Characterization

E E

P P

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Vertex Incremental Characterization

Prime node unless x has a twin x

E E

P P

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Theorem: the incremental split decomposition algorithm runs in

O

((n +

m)α(n,m))

Ingredients for the complexity analysis

I

the vertex insertion ordering has to be a

LexBFS

ordering

I

a carefully amortizing analysing based on

discharging techniques

Previous results

I

[Cunningham 1982]

O

(nm)

I

[Ma, Spinrad 1994]

O(n²

)

I

[Dahlhaus 2000]

O(n

+

m) (tough !)

I

[Charbit, De Montgolfier, Raffinot 2009]

O(n

+

m) (today)

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Application to circle graphs recognition

I Acircle graph is the intersection graph of chords in a circle.

I A graph is circleiff all the prime nodes of its split tree are circle graphs

I A circle graph is prime for the split decomposition iff it has a unique (up to mirror) realizer

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Application to circle graphs recognition

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Application to circle graphs recognition

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Application to circle graphs recognition

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Extreme chord - vertex

A vertexx is extremeif its chordc(x) cut the realizer in a way that the chords of all the non-neighors ofx are either all on the right or all on the left ofc(x).

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First LexBFS Lemma

LetG be a circle graph andσbe a LexBFS ordering ofG ending atx.

Then there exists a realizer ofG in which the chordc(x) is extreme.

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First LexBFS Lemma

LetG be a circle graph andσbe a LexBFS ordering ofG ending atx.

Then there exists a realizer ofG in which the chordc(x) is extreme.

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I Consequence: we know in constant time where to insertx in a given realizer of G.

I Butmany realizers unless prime

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A second LexBFS Lemma

Letσbe a LexBFS ordering ofG =GS(T,F). The”induced”ordering σu of the marker vertices ofGu is a LexBFS ordering.

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A second LexBFS Lemma

Letσbe a LexBFS ordering ofG =GS(T,F). The”induced”ordering σ_u of the marker vertices ofG_u is a LexBFS ordering.

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I consequence: we can apply the first LexBFS lemma on each node of the split tree.

I what remains to do ?

handle the merging of the fully-mixed nodes before insertion.

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A second LexBFS Lemma

Letσbe a LexBFS ordering ofG =GS(T,F). The”induced”ordering σ_u of the marker vertices ofG_u is a LexBFS ordering.

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3 4

5

2 7

6

8

1

3 4

2 5

7 8

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6

5 7

I consequence: we can apply the first LexBFS lemma on each node of the split tree.

I what remains to do ?

handle the merging of the fully-mixed nodes before insertion.

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Circle graph recognition - ingredients

1. Insert vertices with respect to a LexBFS ordering.

2. Maintain the split tree in which each prime node are represented by its realizer.

3. Proceed the merging step on the realizers of the fully-mixed nodes according to the LexBFS Lemmas.

⇒need some small tricks for realizer representation

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Circle graph recognition - ingredients

1. Insert vertices with respect to a LexBFS ordering.

2. Maintain the split tree in which each prime node are represented by its realizer.

3. Proceed the merging step on the realizers of the fully-mixed nodes according to the LexBFS Lemmas.

⇒need some small tricks for realizer representation

Theorem: The circle graph recognition problem can be solved in time O((n+m)α(n,m))

→Previous complexity: O(n²) [Spinrad, J. of Alg. (16), 1994]

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Thank you...

I Int. Workshop on Graph Theoretical Concepts in Computer Science WG 2009

Montpellier, June 24-26th http://www.lirmm.fr/wg2009/

I Spring School on Fixed Parameter and Exact Algorithms AGAPE 2009

May 25-29, Lozari, Corsica

http://www-sop.inria.fr/mascotte/seminaires/AGAPE/