AlgoPerm2012 - 05 Ioan Todinca

1/14

Introduction Optimization problems Recognition Encoding all realizers Conclusion

Permutation graphs — an introduction

Ioan Todinca

LIFO - Universite d’Orleans

Algorithms and permutations, february 2012

2/14


Permutation graphs

3

5 6

1 2 4

��

��

��

��

��

��

143

65

2

• Optimisation algorithmsuse, as input, the intersection model (realizer)

• Recognition algorithmsoutput the intersection(s) model(s)

3/14


Plan of the talk

1. Relationship with other graph classes

2. Optimisation problems :MaxIndependentSet/MaxClique/Coloring ;Treewidth

3. Recognition algorithm

4. Encoding all realizers via modular decomposition

5. Conclusion

4/14


Definition and basic properties

3

5 6

1 2 4

��

��

��

��

��

��

1

2

1

34561

42365

43

65

2

• Realizer : (π1, π2)

• One can ”reverse” the realizer upside-down or right-left :(π1, π2) ∼ (π2, π1) ∼ (π1, π2) ∼ (π2, π1)

• Complements of permutation graphs are permutation graphs.→ Reverse the ordering of the bottoms of the segments :(π1, π2)→ (π1, π2)

4/14



3

5 6

1 2 4

��

��

��

��

��

��

5

2

1

34561

4236




4/14



3

5 6

1 2 4

��

��

��

��

��

��

3

2 34561

1564 2




4/14



5 6

1 2 43

��

��

��

��

��

��

3

2 34561

1564 2




5/14


More intersection graph classes

Circle graphs Trapezoid graphs

Books on graph classes : [Golumbic ’80 ; Brandstadt, Le, Spinrad’99 ; Spinrad 2003]

6/14


MaxIndependentSet via Dynamic Programming

6

21 3 64 5

12 34 5

• Dynamic programming from left to right :

MIS [i ] = 1 + maxj left to i

MIS [j ]

• MaxIndependentSet corresponds to the longest increasingsequence in a permutation — O(n log n)

• MaxClique : longest decreasing sequence

• Coloring : chromatic number = max clique (perfect graphs)

7/14


Treewidth via dynamic programming on scanlines

6

1 2 43

5

23

��

��

��

��

��

25612 34

��

5

2

1

34561

4236

• Minimal separators correspond to scanlines

• Bags correspond to areas between two scanlines

• Treewidth can be solved in polynomial time [Bodlaender,Kloks, Kratsch 95 ; Meister 2010]

7/14



5 6

1 2 43

23

��

��

��

��

��

25612 34

��

5

2

1

34561

4236




7/14



5 6

1 432

23

��

��

��

��

��

25612 34

��

5 36

2

1

34561

42




8/14


Recognition algorithm

Theorem ([Pnueli, Lempel, Even ’71], see also [Golumbic ’80])

G is a permutation graph if and only if G and G are comparabilitygraphs.

Algorithm

1. Find a transitive orientation of G and one of G

2. Construct an intersection model for G

In O(n + m) time by [McConnell, Spinrad ’99]

9/14


permutation ⊆ comparability ∩ co-comparability

Transitive orientation of a permutation graph G : orient edgesaccording to the top endpoints of the segments.

3

5 6

1 2 4

��

��

��

��

��

��

5

2

1

34561

4236

If xy , yz ∈ E and π1(x) < π1(y) < π1(z) then xz ∈ E .

9/14


permutation ⊆ comparability ∩ co-comparability

Transitive orientation of a permutation graph G : orient edgesaccording to the top endpoints of the segments.

5 6

1 2 43

��

��

��

��

��

��

5

2

1

34561

4236

If xy , yz ∈ E and π1(x) < π1(y) < π1(z) then xz ∈ E .

10/14


comparability ∩ co-comparability ⊆ permutation

Let Etr be a transitive orientation of G and Ftr a transitiveorientation of its complement.

6

1 2 43

5

��

��

��

��

��

��

Lemma

Etr ∪ Ftr induces a total ordering π1(Etr ∪ Ftr ) on the vertex set.

rev(Etr ) ∪ Ftr induces another total ordering π2(rev(Etr ) ∪ Ftr ).

10/14


comparability ∩ co-comparability ⊆ permutation

Let Etr be a transitive orientation of G and Ftr a transitiveorientation of its complement.

6

1 2 43

5

��

��

��

��

��

��

Lemma

Etr ∪ Ftr induces a total ordering π1(Etr ∪ Ftr ) on the vertex set.rev(Etr ) ∪ Ftr induces another total ordering π2(rev(Etr ) ∪ Ftr ).

11/14


A realizer of G

Permutations π1(Etr ∪ Ftr ) and π2(rev(Etr ) ∪ Ftr ) form a realizerof G .

5

2

1

34561

4236

3

5 6

1 2 4

��

��

��

��

��

��

Segments x and y intersect iff (xy) ∈ Etr and (yx) ∈ rev(Etr ) orvice-versa ; equivalently, iff xy ∈ E .

12/14


Modules and common intervals

• Substituting a segment (vertex) by the realizer of apermutation graph (module) produces a new permutationgraph.

• A common interval of π1 and π2 forms a module in G• Strong modules correspond exactly to strong common

intervals [de Mongolfier 2003]

• A graph is a permutation graphs iff all prime nodes in themodular decomposition are permutation graphs.

12/14


Modules and common intervals

• Substituting a segment (vertex) by the realizer of apermutation graph (module) produces a new permutationgraph.

• A common interval of π1 and π2 forms a module in G• Strong modules correspond exactly to strong common

intervals [de Mongolfier 2003]

• A graph is a permutation graphs iff all prime nodes in themodular decomposition are permutation graphs.

13/14


Encoding realizers

Theorem ([Gallai ’67])

A prime permutation graph has aunique realizer, up to reversals.

The modular decomposition tree+ realizers of prime nodesencode all possible realizers of G ,cf. [Crespelle, Paul 2010].

3

5 6

1 2 4

��

��

��

��

��

��

1 4

series

parallel

prime

5 6

1

2 43

32

14/14


Conclusion

Summary

• Many optimization problems become polynomial onpermutation graphs

• Representations (intersection models) based on modulardecompositions

Some questions

• Is Bandwidth polynomial or NP-complete on permutationgraphs ?

• What about subgraph isomorphism from parametrized pointof view ?

Thank you ! Your questions ?

14/14


Conclusion

Summary

• Many optimization problems become polynomial onpermutation graphs

• Representations (intersection models) based on modulardecompositions

Some questions

• Is Bandwidth polynomial or NP-complete on permutationgraphs ?

• What about subgraph isomorphism from parametrized pointof view ?

Thank you ! Your questions ?

Education

AlgoPerm2012 - 05 Ioan Todinca