Advances in algorithmic graph theorysummerschool2013.ccaba.upc.edu/.../Lecture6.2_Mertzios.pdfAdvances in algorithmic graph theory George B. Mertzios School of Engineering and Computing

Advances in algorithmic graph theory

George B. Mertzios

School of Engineering and Computing Sciences,Durham University, UK

Graph and routing dynamics: models and algorithms

Part I

EULER Summer School 2013

George Mertzios (Durham) Advances in algorithmic graph theory EULER, July 2013 1 / 75

Overview

Partial orders

Comparability graphs

Permutation graphs

Trapezoid graphs

Unless stated otherwise, definitions and theorems can be found in:[Golumbic, Alg. Graph Theory & Perfect Graphs, 2004]


Partial orders

Definition

A (strict) partial order is a pair P = (U,<P), where U is a finite set and<P is a transitive binary relation on U, i.e.:

whenever x , y , z ∈ U, if x <P y and y <P z , then x <P z .


Partial orders

Definition



A partial order is:

order: the elements of the “ground set” U are “ordered”

partial: not all elements of U are “ordered”


Partial orders

Definition



A partial order is:

order: the elements of the “ground set” U are “ordered”

partial: not all elements of U are “ordered”

Two elements x , y ∈ U are:

comparable in P, if x <P y or y <P x ,

incomparable in P, otherwise.

Element x is covered by element y (or y covers x) if:1 x <P y and

2 there exists no element z ∈ U with x <P z <P y .


Partial orders

For example, U = {a, b, c , d , e, f , g} and the transitive relation <P is:

a <P d , a <P e, a <P g ,b <P d , b <P e, b <P g , b <P c , b <P f ,d <P g , c <P f , c <P g

Do we need all this information for the relation <P ?

since we know that <P is transitive: a <P d and d <P g ⇒ a <P g

similarly: b <P d and d <P g ⇒ b <P g

similarly: b <P c and c <P f ⇒ b <P f

all other pairs of comparable elements are needed(i.e. they are not implied by others)

=⇒ all information we need is which element covers which other element !(since we know that <P is transitive)


Partial orders










Partial orders










Partial orders










Partial orders










Partial orders



An element x ∈ U is called:

minimal in <P , if there exists no element y ∈ U with y <P x ,

maximal in <P , if there exists no element y ∈ U with x <P y .


Partial orders






There may be many minimal/maximal elements !


Partial orders







In this example:

the minimal elements are: a, b,


Partial orders







In this example:

the minimal elements are: a, b,

the maximal elements are: e, f , g .


Partial orders



Consider a subset V ⊆ U of the ground set.

The restriction of the partial order P = (U,<P)on the elements of V is denoted by P(V ) = (V ,<P(V )).

the partial sub-order P(V ) = (V ,<P(V )) is:

a chain, if every two elements of V are comparable in <P ,

an antichain, if every two elements of V are incomparable in <P .

An (anti)chain is:

maximal, if it is not strictly included in another (anti)chain,

maximum, if it does not have less elements than any other (anti)chain.


Partial orders








An (anti)chain is:




Partial orders








An (anti)chain is:




Hasse diagrams



It is often useful to represent a partial order by a diagram:

succinct representation,

a better overview of the structure of the partial order,

useful to devise algorithms.


Hasse diagrams



This is done by the Hasse diagram of a partial order:

a layered diagram

first we place the maximal elements in the uppermost layer

in the next layer (below) we place the maximal elementsin the partial sub-order of the remaining elements

we continue recursively, until we place all elements in layers

Then:

two elements x , y of the same layer i are always incomparable

an element x in layer i is always covered by an element y in layer i + 1

the chains can be read top-down


Hasse diagrams



This is done by the Hasse diagram of a partial order:

a layered diagram

first we place the maximal elements in the uppermost layer

in the next layer (below) we place the maximal elementsin the partial sub-order of the remaining elements

we continue recursively, until we place all elements in layers

Then:

two elements x , y of the same layer i are always incomparable

an element x in layer i is always covered by an element y in layer i + 1

the chains can be read top-down


Hasse diagrams



The Hasse diagram of this partial order (edges ←→ comparable elements):

e f gmaximal elements −→

c damong {a, b, c, d}maximal elements −→

a bmaximal elements −→among {a, b}

maximal chains:{a, e}, {a, d, g}, {a, c, f}, . . .

maximum chains:{a, d, g}, {a, c, f}, . . .


Hasse diagrams










Hasse diagrams










Hasse diagrams










Hasse diagrams









not a maximal chain:{c,f}


Hasse diagrams









not a maximal chain:{c,f}


Linear orders and linear extensions

Definition

A linear order (or total order) is a partial order P = (U,<P),where every pair of elements in U is comparable.

Definition

Let P = (U,<P) be a partial order. A linear extension of P isa linear order L = (U,<L), such that:

if x <P y , then x <L y .

In other words, a linear extension L of a partial order P has:

all the comparabilities of P

and additional comparabilities to make L linear.



Lemma

Every partial order P has a linear extension L.



Lemma


Proof.

If P is a linear order ⇒ L = P

Otherwise we construct a linear extension L of P as follows:1 L← ∅2 In the Hasse diagram H of P, let the layers be H1, H2, . . . , Hk3 For i = 1 to k :

add to L the elements of the layer Hi in an arbitrary order



Lemma


Proof.

Since the elements within one layer Hi are all incomparable, thisprocess is equivalent to the following algorithm (Topological Sorting):

1 i = 12 While U 6= ∅:

ai ← an (arbitrary) minimal element of U(such an element always exists!)U ← U \ {ai}i ← i + 1

3 Return L← (a1, a2, . . . , an)

Correctness: since we always pick up as the next element a minimalelement ai in the remaining set, we do not violate in L anycomparability of P. (“smaller comes first”)



Lemma


Proof.

Since the elements within one layer Hi are all incomparable, thisprocess is equivalent to the following algorithm (Topological Sorting):

1 i = 12 While U 6= ∅:

ai ← an (arbitrary) minimal element of U(such an element always exists!)U ← U \ {ai}i ← i + 1

3 Return L← (a1, a2, . . . , an)

Correctness: since we always pick up as the next element a minimalelement ai in the remaining set, we do not violate in L anycomparability of P. (“smaller comes first”)


Interval orders

A generalization of linear orders:

Definition

An interval order is a partial order P = (U,<P), where every element xcan be assigned an interval Ix on the real line, such that x <P yif and only if Ix lies completely to the left of Iy .

Observation

A linear order is an interval order where all intervals are points.

Every interval representation of an interval graph correspondsbijectively to an interval order.

a

b c de


Interval orders


Definition


Observation



a

b c de


Interval orders


Definition


Observation



a

b c de


Dimension of partial orders

Definition

Let P1 = (U,<P1), P2 = (U,<P2), . . . , Pk = (U,<Pk) be k partial

orders on the same ground set U. The intersection of them is the partialorder P = P1 ∩ P2 ∩ . . . ∩ Pk , where:

x <P y if and only if x <Piy for every i = 1, 2, . . . , k.



Definition




Definition

Let P be a partial order and P be a class of partial orders(e.g. linear orders, interval orders, etc.).

The P-dimension of P is the smallest number kof orders P1, P2, . . . , Pk ∈ P , such that P = P1 ∩ P2 ∩ . . . ∩ Pk .

If P is the class of linear orders, then the P-dimension of P is calledthe dimension of P (denoted dim(P)).



Definition




Definition

Let P be a partial order and P be a class of partial orders(e.g. linear orders, interval orders, etc.).

The P-dimension of P is the smallest number kof orders P1, P2, . . . , Pk ∈ P , such that P = P1 ∩ P2 ∩ . . . ∩ Pk .

If P is the class of linear orders, then the P-dimension of P is calledthe dimension of P (denoted dim(P)).



Definition




Since every linear order is also a (trivial) interval order, it follows that:

Observation

interval-dimension of P ≤ dimension of P



Example of the dimension of a partial order P:

a′ b′ c′

a b c

Lemma

dim(P) = 3.

Proof.

(i) dim(P) ≤ 3: P = L1 ∩ L2 ∩ L3 (see figure)

L1

ab

c′cb′a′

L2 L3

b cc a

a′ b′a b

a′c′

c′b′



Example of the dimension of a partial order P:

a′ b′ c′

a b c

Lemma

dim(P) = 3.

Proof.

(ii) dim(P) ≥ 3: assume otherwise that dim(P) ≤ 2, i.e. P = L1 ∩ L2

Since a, b, c are incomparable in P⇒ a, b, c have opposite orderings in L1, L2

without loss of generality: a <L1 b <L1 c and c <L2 b <L2 a

Since c <P b′ ⇒ c <L1 b′ ⇒ b <L1 b′

Since a <P b′ ⇒ c <L2 b′ ⇒ b <L2 b′

b <L1 b′ and b <L2 b′ ⇒ b <P b′, contradiction


Overview

Partial orders


Permutation graphs

Trapezoid graphs



Given a partial order P = (U,<P), the comparability graph of Pis a graph G , where:

the vertices of G are the elements of U,there is an edge in G between any two comparable elements.

Definition

A graph G is a comparability graph if it is the comparability graphof a partial order P, i.e. if we can transitively orient the edges of G .

G is a cocomparability graph if its complement G is a comparability graph,i.e. if we can transitively orient the non-edges of G .

Observation

Interval graphs are cocomparability graphs.

a

b c de





Definition



Observation


a

b c de





Definition



Observation


a

b c de





Definition



Observation


a

b c de


Comparability graphsAn alternative definition using graph orientations

Given an undirected graph G = (V , E ), we can interpret Gas a directed graph (V , D(E )), where:

D(E ) is a set of directed arcs (instead of the edges of E ),for two vertices u, v ∈ V , uv ∈ E ⇔ 〈uv〉, 〈vu〉 ∈ D(E )(sometimes we may write E instead of D(E )).

For any subset F ⊆ D(E ):we call (V , F ) an orientation of G ,we denote by F−1 the inverse orientation of F ,where 〈uv〉 ∈ F−1 ⇔ 〈vu〉 ∈ F .we denote by F 2 = {〈uw〉 : 〈uv〉, 〈vw〉 ∈ F for some vertex v}.

(V,E) :






(V,E) : (V, F ) :






Definition

A graph G = (V , E ) is a comparability graph if there existsan orientation (V , F ) of G such that:

(i) F ∩ F−1 = ∅, (ii) F + F−1 = E , (iii) F 2 ⊆ F .


Comparability graphsExamples

Is this graph a comparability graph?a

b c

d




b c

d




b c

d




b c

d



Is this graph a comparability graph? YES.a

b c

d




b c

d


b

c

d e

f




b c

d


b

c

d e

f




b c

d


b

c

d e

f




b c

d


b

c

d e

f




b c

d


b

c

d e

f




b c

d


b

c

d e

f




b c

d

Is this graph a comparability graph? NO.a

b

c

d e

f


Comparability graphsThe forcing relation

What was common in these examples?

Some orientations were forcing some others !





Definition

The binary relation Γ on the edges of an undirected graph G = (V , E ) is:

〈ab〉 Γ 〈a′b′〉 if and only if:

either a = a′ and bb′ /∈ E

,or b = b′ and aa′ /∈ E .

In this case we say that 〈ab〉 directly forces 〈a′b′〉.

a = a′

b b′b = b′

a a′





Definition



either a = a′ and bb′ /∈ E ,or b = b′ and aa′ /∈ E .


a = a′

b b′b = b′

a a′





Definition





a = a′

b b′b = b′

a a′





Definition





Note that:

Γ is symmetric: 〈ab〉 Γ 〈a′b′〉 ⇔ 〈a′b′〉 Γ 〈ab〉





Definition





Note that:

Γ is symmetric: 〈ab〉 Γ 〈a′b′〉 ⇔ 〈a′b′〉 Γ 〈ab〉Γ is reflexive: 〈ab〉 Γ 〈ab〉 (since aa′ /∈ E and bb′ /∈ E )





Definition





Note that:

Γ is symmetric: 〈ab〉 Γ 〈a′b′〉 ⇔ 〈a′b′〉 Γ 〈ab〉Γ is reflexive: 〈ab〉 Γ 〈ab〉 (since aa′ /∈ E and bb′ /∈ E )

〈ab〉 6 Γ 〈ba〉 (since a 6= b)



To be more useful, we need a generalization of the forcing relation Γ:

Definition

The transitive closure Γ∗ of Γ is defined as follows:

1 〈ab〉 Γ 〈a′b′〉 =⇒ 〈ab〉 Γ∗ 〈a′b′〉

2 〈ab〉 Γ∗ 〈cd〉 and 〈cd〉 Γ∗ 〈xy〉 =⇒ 〈ab〉 Γ∗ 〈xy〉 (transitivity)(apply this rule as long as possible).

The transitive closure Γ∗ is an equivalence relation, since it is:

reflexive: 〈ab〉 Γ∗ 〈ab〉,symmetric: 〈ab〉 Γ∗ 〈cd〉 ⇐⇒ 〈cd〉 Γ∗ 〈ab〉,transitive: by condition 2 of the definition of Γ∗.

=⇒ Γ∗ partitions the arcs on E into equivalence classes:

the equivalence class of 〈ab〉 is: [〈ab〉] = {〈cd〉 : 〈ab〉 Γ∗ 〈cd〉},[〈ab〉] is called the implication class of 〈ab〉.




Definition


1 〈ab〉 Γ 〈a′b′〉 =⇒ 〈ab〉 Γ∗ 〈a′b′〉2 〈ab〉 Γ∗ 〈cd〉 and 〈cd〉 Γ∗ 〈xy〉 =⇒ 〈ab〉 Γ∗ 〈xy〉 (transitivity)

(apply this rule as long as possible).








Definition











Definition









Comparability graphsImplication classes

In other words:

〈ab〉 and 〈cd〉 are in the same implication class (i.e. [〈ab〉] = [〈cd〉])if and only if there exists a sequence of arcs:

〈ab〉 = 〈a0b0〉 Γ 〈a1b1〉 Γ . . . Γ 〈akbk〉 = 〈cd〉such a sequence is called a Γ-chain from 〈ab〉 to 〈cd〉in this case we say that 〈ab〉 eventually forces 〈cd〉.

Let A be an implication class. Then:

A = A∪ A−1 is the symmetric closure of A,

A corresponds to the undirected edges of the arcs in A,

A is called a color class of G .



In other words:

〈ab〉 and 〈cd〉 are in the same implication class (i.e. [〈ab〉] = [〈cd〉])if and only if there exists a sequence of arcs:

〈ab〉 = 〈a0b0〉 Γ 〈a1b1〉 Γ . . . Γ 〈akbk〉 = 〈cd〉such a sequence is called a Γ-chain from 〈ab〉 to 〈cd〉in this case we say that 〈ab〉 eventually forces 〈cd〉.

Let A be an implication class. Then:

A = A∪ A−1 is the symmetric closure of A,

A corresponds to the undirected edges of the arcs in A,

A is called a color class of G .



This graph has the following implication classes:

A1 = {〈ad〉, 〈ac〉, 〈ae〉} A−11 = {〈da〉, 〈ca〉, 〈ea〉}A2 = {〈bc〉, 〈bd〉, 〈be〉} A−12 = {〈cb〉, 〈db〉, 〈eb〉}A3 = {〈ab〉} A−13 = {〈ba〉}A4 = {〈cd〉} A−14 = {〈dc〉}

a

b

c d e

it is a comparability graph:




A1 = {〈ad〉, 〈ac〉, 〈ae〉} A−11 = {〈da〉, 〈ca〉, 〈ea〉}

A2 = {〈bc〉, 〈bd〉, 〈be〉} A−12 = {〈cb〉, 〈db〉, 〈eb〉}A3 = {〈ab〉} A−13 = {〈ba〉}A4 = {〈cd〉} A−14 = {〈dc〉}

a

b

c d e





A1 = {〈ad〉, 〈ac〉, 〈ae〉} A−11 = {〈da〉, 〈ca〉, 〈ea〉}A2 = {〈bc〉, 〈bd〉, 〈be〉} A−12 = {〈cb〉, 〈db〉, 〈eb〉}

A3 = {〈ab〉} A−13 = {〈ba〉}A4 = {〈cd〉} A−14 = {〈dc〉}

a

b

c d e





A1 = {〈ad〉, 〈ac〉, 〈ae〉} A−11 = {〈da〉, 〈ca〉, 〈ea〉}A2 = {〈bc〉, 〈bd〉, 〈be〉} A−12 = {〈cb〉, 〈db〉, 〈eb〉}A3 = {〈ab〉} A−13 = {〈ba〉}

A4 = {〈cd〉} A−14 = {〈dc〉}

a

b

c d e






a

b

c d e






a

b

c d e




This graph has one implication class:

A = {〈ab〉, 〈cb〉, 〈cd〉, 〈cf 〉, 〈ef 〉, 〈bf 〉, 〈ba〉, 〈bc〉, 〈dc〉, 〈fc〉, 〈fe〉, 〈fb〉}This is not a comparability graph:

a

b

c

d e

f




A = {〈ab〉, 〈cb〉, 〈cd〉, 〈cf 〉, 〈ef 〉, 〈bf 〉, 〈ba〉, 〈bc〉, 〈dc〉, 〈fc〉, 〈fe〉, 〈fb〉}

This is not a comparability graph:

a

b

c

d e

f




A = {〈ab〉, 〈cb〉, 〈cd〉, 〈cf 〉, 〈ef 〉, 〈bf 〉, 〈ba〉, 〈bc〉, 〈dc〉, 〈fc〉, 〈fe〉, 〈fb〉}This is not a comparability graph:

a

b

c

d e

f



Theorem

Let A be an implication class of G . If G has a transitive orientation F ,then either F ∩ A = A or F ∩ A = A−1. In both cases A∩ A−1 = ∅.



Theorem


Proof.

(i) A ⊆ F + F−1 (since F + F−1 = E as a transitive orientation)

(ii) F ∩ F−1 = ∅ (again, since F is a transitive orientation)

Furthermore: 〈ab〉 ∈ F and 〈ab〉 Γ∗ 〈a′b′〉 ⇒ 〈a′b′〉 ∈ F .

Thus, if F ∩ A 6= ∅, then A ⊆ F , i.e.: either F ∩ A = ∅ or A ⊆ F .



Theorem


Proof.







Theorem


Proof.







Theorem


Proof.



First case: F ∩ A = ∅. Then:

F ∩ A = ∅ ⇒ A ⊆ F−1 [by (i)]

⇒ A−1 ⊆ F

⇒ F ∩ A = F ∩(A∪ A−1

)= A−1

Furthermore A∩ A−1 = ∅ (since F ∩ A = ∅ and A−1 ⊆ F )



Theorem


Proof.



Second case: A ⊆ F . Then:

A ⊆ F ⇒ A−1 ⊆ F−1

⇒ F ∩ A−1 = ∅ [by (ii)]

⇒ F ∩ A = F ∩(A∪ A−1

)= A

Furthermore A∩ A−1 = ∅ (since A ⊆ F and F ∩ A−1 = ∅)



Theorem


The converse of the Theorem is also valid (we will prove it):

if A∩ A−1 = ∅ for every implication class of G ,then G has a transitive orientation F ,which includes exactly one of {A, A−1} for every A.

However:

if we choose for every A an arbitrary implication class from {A, A−1},do we always obtain a transitive orientation?

NO.

Example:



Theorem




However:


NO.

Example:



Theorem




However:


NO.

Example:

a

b c

a

b c



Theorem




However:


NO.

Example:

a

b c

a

b c



Let 〈ab〉 Γ∗ 〈cd〉, i.e. 〈ab〉 = 〈a0b0〉 Γ 〈a1b1〉 Γ . . . Γ 〈akbk〉 = 〈cd〉

Then for each i = 1, 2, . . . , k we have:

〈ai−1bi−1〉 Γ 〈aibi−1〉 Γ 〈aibi 〉

where the middle edge is equal to one of the other two,i.e. either ai−1 = ai or bi−1 = bi .

Therefore:

Lemma (canonical Γ-chain)

If 〈ab〉 Γ∗ 〈cd〉, then there exists a Γ-chain from 〈ab〉 to 〈cd〉 of the form

〈ab〉 = 〈a0b0〉 Γ 〈a1b0〉 Γ 〈a1b1〉 Γ 〈a2b1〉 Γ . . . Γ 〈akbk〉 = 〈cd〉.

Such a chain is called a canonical Γ-chain.



Let 〈ab〉 Γ∗ 〈cd〉, i.e. 〈ab〉 = 〈a0b0〉 Γ 〈a1b1〉 Γ . . . Γ 〈akbk〉 = 〈cd〉

Then for each i = 1, 2, . . . , k we have:

〈ai−1bi−1〉 Γ 〈aibi−1〉 Γ 〈aibi 〉

where the middle edge is equal to one of the other two,i.e. either ai−1 = ai or bi−1 = bi .

Therefore:

Lemma (canonical Γ-chain)

If 〈ab〉 Γ∗ 〈cd〉, then there exists a Γ-chain from 〈ab〉 to 〈cd〉 of the form

〈ab〉 = 〈a0b0〉 Γ 〈a1b0〉 Γ 〈a1b1〉 Γ 〈a2b1〉 Γ . . . Γ 〈akbk〉 = 〈cd〉.

Such a chain is called a canonical Γ-chain.



Theorem

If A is an implication class of G , then exactly one of the following holds:

(i) A = A = A−1,

(ii) A∩ A−1 = ∅, A and A−1 are transitive, and they are the onlytransitive orientations of A.



Theorem


(i) A = A = A−1,


Proof.

(i) Assume A∩ A−1 6= ∅.

Let 〈ab〉 ∈ A∩ A−1 ⇒ 〈ba〉 ∈ A∩ A−1 ⇒ 〈ab〉 Γ∗ 〈ba〉.For every 〈cd〉 ∈ A, we have 〈cd〉 Γ∗ 〈ab〉 and 〈dc〉 Γ∗ 〈ba〉Thus, since Γ∗ is an equivalence relation and 〈ab〉 Γ∗ 〈ba〉:〈cd〉 Γ∗ 〈dc〉 ⇒ 〈dc〉 ∈ A.

Therefore A = A−1 ⇒ A = A = A−1.



Theorem


(i) A = A = A−1,


Proof.

(ii) Assume A∩ A−1 = ∅, and let 〈ab〉, 〈bc〉 ∈ A.

If ac /∈ E , then 〈ab〉 Γ∗ 〈cb〉 ⇒ 〈cb〉 ∈ A ⇒ 〈bc〉 ∈ A−1,which is a contradiction since 〈bc〉 ∈ A by assumption.Therefore ac ∈ E .

We will show that 〈ac〉 ∈ A, i.e. that A is transitive.



Theorem


(i) A = A = A−1,


Proof.

(ii) Assume A∩ A−1 = ∅, and let 〈ab〉, 〈bc〉 ∈ A.

If ac /∈ E , then 〈ab〉 Γ∗ 〈cb〉 ⇒ 〈cb〉 ∈ A ⇒ 〈bc〉 ∈ A−1,which is a contradiction since 〈bc〉 ∈ A by assumption.Therefore ac ∈ E .

We will show that 〈ac〉 ∈ A, i.e. that A is transitive.



Proof.

(ii) Assume A∩ A−1 = ∅, and let 〈ab〉, 〈bc〉 ∈ A. Then ac ∈ E .

Assume that 〈ac〉 ∈ B, where B 6= A.

Since 〈bc〉, 〈ab〉 ∈ A:

〈bc〉 = 〈b0c0〉 Γ 〈b1c0〉 Γ 〈b1c1〉 Γ 〈b2c1〉 Γ . . . Γ 〈bkck〉 = 〈ab〉We prove (by induction on i) that for every i = 0, 1, . . . , k :

a 6= bi , abi , aci ∈ E , and 〈abi 〉 ∈ A, 〈aci 〉 ∈ B

Induction basis (i = 0): true, since b0 = b and c0 = c .

a

b0 = b

A

A

B

c = c0b c



Proof.


Assume that 〈ac〉 ∈ B, where B 6= A. Since 〈bc〉, 〈ab〉 ∈ A:

〈bc〉 = 〈b0c0〉 Γ 〈b1c0〉 Γ 〈b1c1〉 Γ 〈b2c1〉 Γ . . . Γ 〈bkck〉 = 〈ab〉

We prove (by induction on i) that for every i = 0, 1, . . . , k :a 6= bi , abi , aci ∈ E , and 〈abi 〉 ∈ A, 〈aci 〉 ∈ B


a

b0 = b

A

A

B

c = c0



Proof.


Assume that 〈ac〉 ∈ B, where B 6= A. Since 〈bc〉, 〈ab〉 ∈ A:

〈bc〉 = 〈b0c0〉 Γ 〈b1c0〉 Γ 〈b1c1〉 Γ 〈b2c1〉 Γ . . . Γ 〈bkck〉 = 〈ab〉We prove (by induction on i) that for every i = 0, 1, . . . , k :



a

b0 = b

A

A

B

c = c0



Proof.


Induction hypothesis:assume that a 6= bi−1, abi−1, aci−1 ∈ E , 〈abi−1〉 ∈ A, 〈aci−1〉 ∈ B.

(i) To prove that a 6= bi , abi ∈ E and 〈abi 〉 ∈ A:

Since 〈aci−1〉 ∈ B and 〈bici−1〉 ∈ A ⇒ a 6= bi .

If abi /∈ E ⇒ 〈aci−1〉 Γ 〈bici−1〉⇒ A = B, contradiction.Therefore abi ∈ E

Since bi−1bi /∈ E ⇒ 〈abi−1〉 Γ 〈abi 〉 ⇒⇒ 〈abi 〉 ∈ A

a

A

A

B

bi−1 ci−1

bi



Proof.







a

A

A

B

bi−1 ci−1

A

bi



Proof.





If abi /∈ E ⇒ 〈aci−1〉 Γ 〈bici−1〉⇒ A = B, contradiction.

Therefore abi ∈ E


a

A

A

B

bi−1 ci−1

A

bi



Proof.







a

A

A

B

bi−1 ci−1

A

bi



Proof.







a

A

A

B

bi−1 ci−1

A

A

bi



Proof.



(ii) To prove that aci ∈ E and 〈aci 〉 ∈ B:

If aci /∈ E ⇒ 〈bici 〉 Γ 〈bia〉⇒ A = A−1, contradiction.Therefore aci ∈ E

Since ci−1ci /∈ E ⇒ 〈aci−1〉 Γ 〈aci 〉 ⇒⇒ 〈aci 〉 ∈ B

a

A

A

B

ci−1bi

ci



Proof.




If aci /∈ E ⇒ 〈bici 〉 Γ 〈bia〉⇒ A = A−1, contradiction.

Therefore aci ∈ E


a

A

A

B

ci−1bi

A

ci



Proof.






a

A

A

B

ci−1bi

A

ci



Proof.






a

A

A

B

ci−1bi

A

B

ci



Proof.

Summarizing:

〈bc〉 = 〈b0c0〉 Γ 〈b1c0〉 Γ 〈b1c1〉 Γ 〈b2c1〉 Γ . . . Γ 〈bkck〉 = 〈ab〉for every i = 0, 1, . . . , k:


This is a contradiction, since 〈bkck〉 = 〈ab〉 and a 6= bk .



Proof.

Summarizing:

〈bc〉 = 〈b0c0〉 Γ 〈b1c0〉 Γ 〈b1c1〉 Γ 〈b2c1〉 Γ . . . Γ 〈bkck〉 = 〈ab〉for every i = 0, 1, . . . , k:


This is a contradiction, since 〈bkck〉 = 〈ab〉 and a 6= bk .

Corollary

Every color class A of a graph G :

either has exactly two transitive orientations A and A−1,one being the reversal of the other (case A∩ A−1 = ∅),

or has no transitive orientation (case A = A = A−1)(in this case G is not a comparability graph).


Comparability graphsA transitive orientation algorithm

So far we studied transitive orientations, forcing, implication classes.

How can we decide whether a graph G is a comparability graph?

We need some extra notions:

Definition

Let G = (V , E ) be a graph. A partition E = B1 + B2 + . . . + Bk

of the edges is a G -decomposition of E if for every i = 1, 2, . . . , k:

Bi is an implication class of Bi + Bi+1 + . . . + Bk .

A sequence of arcs [〈x1y1〉, . . . , 〈xkyk〉] is a decomposition scheme for Gif there exists a G -decomposition E = B1 + B2 + . . . + Bk such that:

〈xiyi 〉 ∈ Bi , for every i = 1, 2, . . . , k.

For any G -decomposition: many decomposition schemes(any representative for each Bi )

For any decomposition scheme: exactly one G -decomposition






Definition













Definition













Definition










How to find a G -decomposition?

Decomposition Algorithm

1 i ← 1; E1 ← E

2 While Ei 6= ∅:

3 Arbitrarily pick an edge xiyi ∈ Ei

4 Enumerate the implication class Bi of Ei that contains 〈xiyi 〉5 Ei+1 ← Ei − Bi ; i ← i + 1

Do all graphs have a G -decomposition?

YES: the algorithm returns a scheme [〈x1y1〉, 〈x2y2〉, . . . , 〈xkyk〉]and a G -decomposition E = B1 + B2 + . . . + Bk for any graph G .

If we choose 〈yixi 〉 instead of 〈xiyi 〉 for some i=⇒ we include B−1i instead of Bi in the G -decomposition.



How to find a G -decomposition?

Decomposition Algorithm

1 i ← 1; E1 ← E

2 While Ei 6= ∅:


4 Enumerate the implication class Bi of Ei that contains 〈xiyi 〉5 Ei+1 ← Ei − Bi ; i ← i + 1

Do all graphs have a G -decomposition?

YES: the algorithm returns a scheme [〈x1y1〉, 〈x2y2〉, . . . , 〈xkyk〉]and a G -decomposition E = B1 + B2 + . . . + Bk for any graph G .

If we choose 〈yixi 〉 instead of 〈xiyi 〉 for some i=⇒ we include B−1i instead of Bi in the G -decomposition.



Recall our previous example:(with eight implication classes A ←→ four color classes A)

A1 = {〈ad〉, 〈ac〉, 〈ae〉} A−11 = {〈da〉, 〈ca〉, 〈ea〉}A2 = {〈bc〉, 〈bd〉, 〈be〉} A−12 = {〈cb〉, 〈db〉, 〈eb〉}A3 = {〈ab〉} A−13 = {〈ba〉}A4 = {〈cd〉} A−14 = {〈dc〉}a

b

c d e




The Decomposition Algorithm:

a

b

c e

(V,Ei)i 〈xiyi〉 (V,Bi)

i = 1 〈ac〉

a

b

c d e

a

b

c d e

a

b

c d ei = 2 〈bc〉

a

b

c d e

a

b

c d ei = 3 〈dc〉

=

E B2 B3B1

+ +

d




a

b

c e


i = 1 〈ac〉

a

b

c d e

a

b

c d e

a

b

c d ei = 2 〈bc〉

a

b

c d e

a

b

c d ei = 3 〈dc〉

=

E B2 B3B1

+ +

d




a

b

c e


i = 1 〈ac〉

a

b

c d e

a

b

c d e

a

b

c d ei = 2 〈bc〉

a

b

c d e

a

b

c d ei = 3 〈dc〉

=

E B2 B3B1

+ +

d




a

b

c e


i = 1 〈ac〉

a

b

c d e

a

b

c d e

a

b

c d ei = 2 〈bc〉

a

b

c d e

a

b

c d ei = 3 〈dc〉

=

E B2 B3B1

+ +

d




decomposition scheme: [〈ac〉, 〈bc〉, 〈dc〉]G -decomposition: B1 = A1, B2 = A2 + A−13 , B3 = A−14

Note that:

Although 〈ba〉 and 〈bc〉 are not Γ-related in the original graph:〈ba〉 Γ 〈bc〉 once B1 is removed from G .

In general: each implication class of Ei+1

is the union of some implication classes of Ei .

———————————————————————————————A1 = {〈ad〉, 〈ac〉, 〈ae〉} A−11 = {〈da〉, 〈ca〉, 〈ea〉}A2 = {〈bc〉, 〈bd〉, 〈be〉} A−12 = {〈cb〉, 〈db〉, 〈eb〉}A3 = {〈ab〉} A−13 = {〈ba〉}A4 = {〈cd〉} A−14 = {〈dc〉}

a

b

c d e





Note that:





a

b

c d e





Note that:





a

b

c d e



Theorem (Transitive Orientation Theorem)

Let E = B1 + B2 + . . . + Bk be a G -decomposition of a graph G = (V , E ).The following statements are equivalent:

(i) G is a comparability graph,

(ii) A∩ A−1 = ∅, for every implication class A of G ,

(iii) Bi ∩ B−1i = ∅, for every i = 1, 2, . . . , k.








Proof.

(i) ⇒ (ii): it follows by the first theorem.








Proof.

(ii) ⇒ (iii):

Let E1 = E .

For i ≥ 2 let Ei = Bi + Bi+1 + . . . + Bk .

It suffices to prove by induction on i that:for every implication class Ci of Ei , we have Ci ∩ C−1i = ∅.



Proof.

Induction basis (i = 1): it follows by (ii), since E1 = E .

Induction hypothesis (i ≥ 1):for every implication class Ci of Ei , we have Ci ∩ C−1i = ∅Bi is an implication class of Ei (by the Decomposition Algorithm)

Let Ci+1 be an implication class of Ei+1 = Ei − Bi

If Ci+1 is an implication class of Ei ⇒ Ci+1 ∩C−1i+1 = ∅ (by ind. hyp.)

Otherwise Ci+1 = D1 + D2 + . . . + Dx , where D1, D2, . . . , Dx , x ≥ 2,are disjoint implication classes of Ei .

By induction hypothesis: Dj ∩D−1j = ∅, 1 ≤ j ≤ x . Thus:

Ci+1 ∩ C−1i+1 = (D1 + D2 + . . . + Dx ) ∩ (D−11 + D−12 + . . . + D−1x )

= ∑n

j ,`=1(Dj ∩D−1` )

= (D1 ∩D−11 ) + . . . + (Dx ∩D−1x ) = ∅



Proof.







Ci+1 ∩ C−1i+1 = (D1 + D2 + . . . + Dx ) ∩ (D−11 + D−12 + . . . + D−1x )

= ∑n

j ,`=1(Dj ∩D−1` )

= (D1 ∩D−11 ) + . . . + (Dx ∩D−1x ) = ∅



Proof.







Ci+1 ∩ C−1i+1 = (D1 + D2 + . . . + Dx ) ∩ (D−11 + D−12 + . . . + D−1x )

= ∑n

j ,`=1(Dj ∩D−1` )

= (D1 ∩D−11 ) + . . . + (Dx ∩D−1x ) = ∅



Proof.







Ci+1 ∩ C−1i+1 = (D1 + D2 + . . . + Dx ) ∩ (D−11 + D−12 + . . . + D−1x )

= ∑n

j ,`=1(Dj ∩D−1` )

= (D1 ∩D−11 ) + . . . + (Dx ∩D−1x ) = ∅



Proof.







Ci+1 ∩ C−1i+1 = (D1 + D2 + . . . + Dx ) ∩ (D−11 + D−12 + . . . + D−1x )

= ∑n

j ,`=1(Dj ∩D−1` )

= (D1 ∩D−11 ) + . . . + (Dx ∩D−1x ) = ∅



Proof.

(iii) ⇒ (i):

Let E = B1 + B2 + . . . + Bk be a G -decomposition of E ,where Bi ∩ B−1i = ∅ for every i = 1, 2, . . . , k⇒ in particular, B1 is transitive (by the previous theorem)

We prove that B1 + B2 + . . . + Bk is a transitive orientation of G

Proof by induction on the length k of the G -decomposition.

Induction basis (k = 1):if E = Bi ⇒ B1 is clearly a transitive orientation of G

Induction hypothesis: the statement is true for all G -decompositionsof graphs of length ≤ k − 1

⇒ F = B2 + . . . + Bk is a transitive orientation of E − B1

We will show that B1 + F is transitive



Proof.

(iii) ⇒ (i):










Proof.

(iii) ⇒ (i):










Proof.

Let 〈ab〉, 〈bc〉 ∈ B1 + F

For transitivity of B1 + F we need to prove that: 〈ac〉 ∈ B1 + F

Case 1: 〈ab〉, 〈bc〉 ∈ B1 ⇒ 〈ac〉 ∈ B1 ⊆ B1 + F (B1 is transitive)

Case 2: 〈ab〉, 〈bc〉 ∈ F ⇒ 〈ac〉 ∈ F ⊆ B1 + F (F is transitive)

Case 3: 〈ab〉 ∈ B1 and 〈bc〉 ∈ F

If ac /∈ E ⇒ 〈ab〉 Γ 〈cb〉 ⇒ B1 ∩ F 6= ∅, contradiction ⇒ ac ∈ E

Assume that 〈ac〉 /∈ B1 + F ⇒ 〈ca〉 ∈ B1 + F . Then:

assume that 〈ca〉 ∈ B1. Since also 〈ab〉 ∈ B1 ⇒ 〈cb〉 ∈ B1,(contradiction, since 〈bc〉 ∈ F in Case 3)

assume that 〈ca〉 ∈ F . Since also 〈bc〉 ∈ F ⇒ 〈ba〉 ∈ F ,(contradiction, since 〈ab〉 ∈ B1 in Case 3)

Case 4: 〈ab〉 ∈ F and 〈bc〉 ∈ B1: symmetric to Case 3



Proof.

Let 〈ab〉, 〈bc〉 ∈ B1 + F












Proof.

Let 〈ab〉, 〈bc〉 ∈ B1 + F












Proof.

Let 〈ab〉, 〈bc〉 ∈ B1 + F












Proof.

Let 〈ab〉, 〈bc〉 ∈ B1 + F












Proof.

Let 〈ab〉, 〈bc〉 ∈ B1 + F












Proof.

Let 〈ab〉, 〈bc〉 ∈ B1 + F












The Transitive Orientation Theorem implies the next algorithm:

Transitive Orientation Algorithm

1 i ← 1; E1 ← E ; F ← ∅2 While Ei 6= ∅:


4 Enumerate the implication class Bi of Ei that contains 〈xiyi 〉5 If Bi ∩ B−1i = ∅ then add Bi to F

else Return “G is not a comparability graph”

6 Ei+1 ← Ei − Bi ; i ← i + 1

7 Return F



Running time of the algorithm for a graph G = (V , E ):O(δ|E |) = O(|V ||E |) = O(|V |3) in worst case(see details in [Golumbic, Alg. Graph Theory & Perfect Graphs, 2004])

Several later algorithms for transitive orientation

The currently fastest algorithm:[McCornell, Spinrad, Discrete Mathematics, 1999]

Given a graph G = (V , E ), it produces in linear O(|V |+ |E |) timean orientation F of G such that:

F is transitive ⇐⇒ G is a comparability graph

F is given implicitly through a linear order L, which serves asa linear extension of F (topological sort)

main tool: a fast “modular decomposition” of a graph

But, in order to check whether F is transitive:matrix multiplication (currently in O(|V |2.376) time)






















Comparability graphsColoring and clique

Recall that for every graph G = (V , E ):

a clique U ⊆ V is a maximal clique if there exists no other cliqueU ′ ⊆ V such that U ⊂ U ′,a clique U ⊆ V is a maximum clique if there exists no other cliqueU ′ ⊆ V such that |U | < |U ′|,the clique number ω(G ) is the size of a maximum clique of G ,

a proper k-coloring of G is:an assignment of k colors to the vertices of G such that,if uv ∈ E then u and v have different colors,a partition V = V1 ∪ V2 ∪ . . . ∪ Vk , where each Vi isan independent set (“color class”),

the chromatic number χ(G ) is the smallest k for which G hasa proper k-coloring.

Theorem

For any graph G , χ(G ) ≥ ω(G ).




a clique U ⊆ V is a maximal clique if there exists no other cliqueU ′ ⊆ V such that U ⊂ U ′,a clique U ⊆ V is a maximum clique if there exists no other cliqueU ′ ⊆ V such that |U | < |U ′|,the clique number ω(G ) is the size of a maximum clique of G ,a proper k-coloring of G is:

an assignment of k colors to the vertices of G such that,if uv ∈ E then u and v have different colors,a partition V = V1 ∪ V2 ∪ . . . ∪ Vk , where each Vi isan independent set (“color class”),


Theorem





a clique U ⊆ V is a maximal clique if there exists no other cliqueU ′ ⊆ V such that U ⊂ U ′,a clique U ⊆ V is a maximum clique if there exists no other cliqueU ′ ⊆ V such that |U | < |U ′|,the clique number ω(G ) is the size of a maximum clique of G ,a proper k-coloring of G is:

an assignment of k colors to the vertices of G such that,if uv ∈ E then u and v have different colors,a partition V = V1 ∪ V2 ∪ . . . ∪ Vk , where each Vi isan independent set (“color class”),


Theorem




Definition

A graph G is perfect if χ(GA) = ω(GA) for every induced subgraph GA.

Is every comparability graph G also a perfect graph?

We need to exploit the Hasse diagram of the partial order for G !



Definition

A graph G is perfect if χ(GA) = ω(GA) for every induced subgraph GA.

Is every comparability graph G also a perfect graph?

We need to exploit the Hasse diagram of the partial order for G !

maximal elements −→


maximal elements −→ q

r s

t x y

p

among {p,q,r,s}

among {p,q}

d(t) = d(x) = d(y) = 1

d(r) = d(s) = 2

d(p) = d(q) = 3



Let G = (V , E ) be the comparability graph of the partial order P.

We define a depth function d : V −→N, such that for every v ∈ V :if v is a maximal element in P, then d(v) = 1,otherwise d(v) = 1 + max{d(w) : v <P w}.

If the Hasse diagram has k layers, then:the longest top-down path in the diagram has k vertices, i.e.the maximum chain has k vertices.

Every chain in P corresponds to a clique in G ⇒ ω(G ) ≥ k.




r s

t x y

p

among {p,q,r,s}

among {p,q}

d(t) = d(x) = d(y) = 1

d(r) = d(s) = 2

d(p) = d(q) = 3










r s

t x y

p

among {p,q,r,s}

among {p,q}

d(t) = d(x) = d(y) = 1

d(r) = d(s) = 2

d(p) = d(q) = 3










r s

t x y

p

among {p,q,r,s}

among {p,q}

d(t) = d(x) = d(y) = 1

d(r) = d(s) = 2

d(p) = d(q) = 3










r s

t x y

p

among {p,q,r,s}

among {p,q}

d(t) = d(x) = d(y) = 1

d(r) = d(s) = 2

d(p) = d(q) = 3










r s

t x y

p

among {p,q,r,s}

among {p,q}

d(t) = d(x) = d(y) = 1

d(r) = d(s) = 2

d(p) = d(q) = 3










r s

t x y

p

among {p,q,r,s}

among {p,q}

d(t) = d(x) = d(y) = 1

d(r) = d(s) = 2

d(p) = d(q) = 3



On the other hand:

Every layer in the Hasse diagram is an independent set

If the diagram has k layers, then:every layer is a color class of G ⇒ χ(G ) ≤ k

Therefore ω(G ) ≥ k ≥ χ(G ) ≥ ω(G ) ⇒ χ(G ) = ω(G )

Every induced subgraph GA of G is a comparability graph⇒ χ(GA) = ω(GA) for every induced subgraph GA

⇒ G is a perfect graph.




r s

t x y

p

among {p,q,r,s}

among {p,q}

d(t) = d(x) = d(y) = 1

d(r) = d(s) = 2

d(p) = d(q) = 3



On the other hand:









r s

t x y

p

among {p,q,r,s}

among {p,q}

d(t) = d(x) = d(y) = 1

d(r) = d(s) = 2

d(p) = d(q) = 3



On the other hand:






Theorem

Every comparability graph G is a perfect graph.

Theorem

Given the Hasse diagram of a comparability graph G = (V , E ), we cancompute a minimum coloring and a maximum clique in O(|V |+ |E |) time.


Overview

Partial orders


Permutation graphs

Trapezoid graphs


Permutation graphs

Graphs can be defined using mathematical structures:

we saw: partial orders −→ comparability graphswhat about permutations?

Consider the numbers 1, 2, . . . , n

Consider a permutation π = [π1, π2, . . . , πn] of these numbers


Permutation graphs





For every i = 1, 2, . . . , n:

πi is the ith number in the permutation π


Permutation graphs





For every i = 1, 2, . . . , n:


π−1i is the position of number i in the the permutation π


Permutation graphs





For every i = 1, 2, . . . , n:



Example: n = 6

π = [4, 3, 6, 1, 5, 2]


Permutation graphs





For every i = 1, 2, . . . , n:



Example: n = 6

π = [4, 3, 6, 1, 5, 2]

π1 = 4


Permutation graphs





For every i = 1, 2, . . . , n:



Example: n = 6

π = [4, 3, 6, 1, 5, 2]

π1 = 4, π2 = 3


Permutation graphs





For every i = 1, 2, . . . , n:



Example: n = 6

π = [4, 3, 6, 1, 5, 2]

π1 = 4, π2 = 3, π3 = 6


Permutation graphs





For every i = 1, 2, . . . , n:



Example: n = 6

π = [4, 3, 6, 1, 5, 2]

π1 = 4, π2 = 3, π3 = 6, π4 = 1


Permutation graphs





For every i = 1, 2, . . . , n:



Example: n = 6

π = [4, 3, 6, 1, 5, 2]

π1 = 4, π2 = 3, π3 = 6, π4 = 1, π5 = 5


Permutation graphs





For every i = 1, 2, . . . , n:



Example: n = 6

π = [4, 3, 6, 1, 5, 2]

π1 = 4, π2 = 3, π3 = 6, π4 = 1, π5 = 5, π6 = 2


Permutation graphs





For every i = 1, 2, . . . , n:



Example: n = 6

π = [4, 3, 6, 1, 5, 2]

π1 = 4, π2 = 3, π3 = 6, π4 = 1, π5 = 5, π6 = 2,

π−11 = 4


Permutation graphs





For every i = 1, 2, . . . , n:



Example: n = 6

π = [4, 3, 6, 1, 5, 2]

π1 = 4, π2 = 3, π3 = 6, π4 = 1, π5 = 5, π6 = 2,

π−11 = 4, π−12 = 6


Permutation graphs





For every i = 1, 2, . . . , n:



Example: n = 6

π = [4, 3, 6, 1, 5, 2]

π1 = 4, π2 = 3, π3 = 6, π4 = 1, π5 = 5, π6 = 2,

π−11 = 4, π−12 = 6, π−13 = 2


Permutation graphs





For every i = 1, 2, . . . , n:



Example: n = 6

π = [4, 3, 6, 1, 5, 2]

π1 = 4, π2 = 3, π3 = 6, π4 = 1, π5 = 5, π6 = 2,

π−11 = 4, π−12 = 6, π−13 = 2, π−14 = 1


Permutation graphs





For every i = 1, 2, . . . , n:



Example: n = 6

π = [4, 3, 6, 1, 5, 2]

π1 = 4, π2 = 3, π3 = 6, π4 = 1, π5 = 5, π6 = 2,

π−11 = 4, π−12 = 6, π−13 = 2, π−14 = 1, π−15 = 5


Permutation graphs





For every i = 1, 2, . . . , n:



Example: n = 6

π = [4, 3, 6, 1, 5, 2]

π1 = 4, π2 = 3, π3 = 6, π4 = 1, π5 = 5, π6 = 2,

π−11 = 4, π−12 = 6, π−13 = 2, π−14 = 1, π−15 = 5, π−16 = 3


Permutation graphs

For every permutation π we can define the graph G [π] as follows:

the vertices of G [π] are {1, 2, . . . , n},there is an edge between vertex i and vertex j (where i < j)when j appears to the left of i in π i.e. π−1j < π−1i


Permutation graphs



Example: π = [4, 3, 6, 1, 5, 2]

1 2

4

3

5

6


Permutation graphs



Equivalent definition of G [π] = (V , E ):

V = {1, 2, . . . , n},ij ∈ E ⇔ (i − j)(π−1i − π−1j ) < 0.

Definition

A graph G is a permutation graph if there exists a permutation πsuch that G is isomorphic to G [π].


Permutation graphs

For a permutation π = [π1, π2, . . . , πn], the reverse permutation π is:

π = [π1, π2, . . . , πn] = [πn, πn−1, . . . , π1]


Permutation graphs


π = [π1, π2, . . . , πn] = [πn, πn−1, . . . , π1]

Note that for every i = 1, 2, . . . , n:

πi = πn−i


Permutation graphs


π = [π1, π2, . . . , πn] = [πn, πn−1, . . . , π1]

Note that for every i = 1, 2, . . . , n:

πi = πn−iπ−1i = n− π−1i


Permutation graphs


π = [π1, π2, . . . , πn] = [πn, πn−1, . . . , π1]

Lemma

For every permutation π: G [π] = G [π].


Permutation graphs


π = [π1, π2, . . . , πn] = [πn, πn−1, . . . , π1]

Lemma


Proof.

Denote G [π] = (V , E ) and G [π] = (V , E ).

Then: ij ∈ E ⇔ (i − j)(

π−1i − π−1j

)< 0

⇔ (i − j)((

n− π−1i

)−(

n− π−1j

))< 0

⇔ (i − j)(

π−1i − π−1j

)> 0

⇔ ij /∈ E


Permutation graphs


π = [π1, π2, . . . , πn] = [πn, πn−1, . . . , π1]

Lemma


Proof.

Denote G [π] = (V , E ) and G [π] = (V , E ).

Then: ij ∈ E ⇔ (i − j)(

π−1i − π−1j

)< 0

⇔ (i − j)((

n− π−1i

)−(

n− π−1j

))< 0

⇔ (i − j)(

π−1i − π−1j

)> 0

⇔ ij /∈ E


Permutation graphs


π = [π1, π2, . . . , πn] = [πn, πn−1, . . . , π1]

Lemma


Corollary

The complement of a permutation graph is also a permutation graph.


Permutation graphs

Lemma

Every permutation graph G is a comparability graph(i.e. G is transitively orientable).

Proof.

Denote G = G [π] = (V , E ).

Define an orientation F of G such that, whenever ij ∈ E :

〈ij〉 ∈ F ⇐⇒ i < j

We show that F is transitive. Assume that 〈ij〉, 〈jk〉 ∈ F

Then: i < j < k (by definition of F )

Furthermore: π−1i > π−1j and π−1j > π−1k (by definition of G [π])

⇒ π−1i > π−1k ⇒ ik ∈ E and 〈ik〉 ∈ F

⇒ F is transitive

⇒ G is a comparability graph


Permutation graphs

Lemma


Proof.

Denote G = G [π] = (V , E ).


〈ij〉 ∈ F ⇐⇒ i < j




⇒ π−1i > π−1k ⇒ ik ∈ E and 〈ik〉 ∈ F

⇒ F is transitive



Permutation graphs

Lemma


Proof.

Denote G = G [π] = (V , E ).


〈ij〉 ∈ F ⇐⇒ i < j




⇒ π−1i > π−1k ⇒ ik ∈ E and 〈ik〉 ∈ F

⇒ F is transitive



Permutation graphs

Lemma


Proof.

Denote G = G [π] = (V , E ).


〈ij〉 ∈ F ⇐⇒ i < j




⇒ π−1i > π−1k ⇒ ik ∈ E and 〈ik〉 ∈ F

⇒ F is transitive



Permutation graphs

Lemma


Proof.

Denote G = G [π] = (V , E ).


〈ij〉 ∈ F ⇐⇒ i < j




⇒ π−1i > π−1k ⇒ ik ∈ E and 〈ik〉 ∈ F

⇒ F is transitive



Permutation graphs

Theorem (Pnueli, Lempel, Even, 1971)

A graph G = (V , E ) is a permutation graph ⇔ G and G arecomparability graphs.


Permutation graphs



Proof.

(⇒): Let G = G [π] for some permutation π. Then:

G is a comparability graph (last theorem)

G = G [π] is a permutation graph ⇒ G is a comparability graph


Permutation graphs



Proof.

(⇐): Let G and G be comparability graphs.

Let (V , F1) be a transitive orientation of G


We show that F1 + F2 is a transitive orientation of the clique⇒ Assume otherwise that 〈ij〉, 〈jk〉, 〈ki〉 ∈ F1 + F2

⇒ at least two of them belong to the same Fi , wlog. 〈ij〉, 〈jk〉 ∈ F1

⇒ by transitivity of F1: 〈ik〉 ∈ F1, contradiction⇒ F1 + F2 is transitive

Similarly: F−11 + F2 is transitive


Permutation graphs



Proof.









Permutation graphs



Proof.









Permutation graphs

Proof.

For every transitive orientation of the clique:

a unique linear ordering of its vertices

We now construct in 3 steps a permutation π such that G = G [π]:

1 Label the vertices according to the linear ordering of F1 + F2:the ith vertex v in this ordering gets label L(v) = i

2 Re-label the vertices according to the linear ordering of F−11 + F2:the ith vertex v in this ordering gets label L′(v) = i

Let u, v be two vertices of G . Then:

uv ∈ E ⇔ (by definition of F1)

u and v have different relative position in the labelings L(·) and L′(·)⇔ (L(u)− L(v)) (L′(u)− L′(v)) < 0


Permutation graphs

Proof.










Permutation graphs

Proof.










Permutation graphs

Proof.










Permutation graphs

Proof.




3 Define π as follows:

For every vertex v , if L(v) = i then define π−1i = L′(v).

Then: uv ∈ E ⇔ (L(u)− L(v)) (L′(u)− L′(v)) < 0 ⇔⇔ (i − j)

(π−1i − π−1j

)< 0 ⇔ G = G [π]


Permutation graphsThe recognition problem

Therefore, given a graph G :

We can decide efficiently whether G is a permutation graph(recognition problem)

If both G and G are comparability graphs then:

G is a permutation graphconstruct a permutation π (using the theorem) such that G = G [π]

Otherwise G is not a permutation graph

Time complexity:

using the Transitive Orientation Algorithm: O(n3)

the currently fastest algorithm: O(n + m)[McCornell, Spinrad, Discrete Mathematics, 1999]

using modular decomposition of G


Permutation graphsThe recognition problem

Therefore, given a graph G :

We can decide efficiently whether G is a permutation graph(recognition problem)

If both G and G are comparability graphs then:

G is a permutation graphconstruct a permutation π (using the theorem) such that G = G [π]

Otherwise G is not a permutation graph

Time complexity:

using the Transitive Orientation Algorithm: O(n3)

the currently fastest algorithm: O(n + m)[McCornell, Spinrad, Discrete Mathematics, 1999]

using modular decomposition of G


Permutation graphsAn intersection model

Definition

An undirected graph G = (V , E ) is called an intersection graph, if eachvertex v ∈ V can be assigned to a set Sv , such that two vertices of G areadjacent if and only if the corresponding sets have a nonemptyintersection, i.e. E = {uv | Su ∩ Sv 6= ∅}.Then, F = {Sv | v ∈ V } is the intersection model of G .



Definition

An undirected graph G = (V , E ) is called an intersection graph, if eachvertex v ∈ V can be assigned to a set Sv , such that two vertices of G areadjacent if and only if the corresponding sets have a nonemptyintersection, i.e. E = {uv | Su ∩ Sv 6= ∅}.Then, F = {Sv | v ∈ V } is the intersection model of G .

Definition

A graph G is called an interval graph, if G is the intersection graph of aset of intervals on the real line.

a

b c de

a

b

c d

e⇔



Lemma

Permutation graphs are exactly the intersection graphs of line segmentsbetween two parallel lines.

L1

L2

a

c de b

a

b

c d

e⇔

a

c d eb

Proof idea:

for every i = 1, 2, . . . , n, the ith line segment has endpoints i and πi

the ith line segment intersects the jth line segment

⇔ i and j appear in reversed order in π

⇔ i and j are adjacent in G [π]



Lemma

Permutation graphs are exactly the intersection graphs of line segmentsbetween two parallel lines.

L1

L2

a

c de b

a

b

c d

e⇔

a

c d eb

Proof idea:

for every i = 1, 2, . . . , n, the ith line segment has endpoints i and πi

the ith line segment intersects the jth line segment

⇔ i and j appear in reversed order in π

⇔ i and j are adjacent in G [π]


Partial orders of dimension 2

Theorem (see Spinrad, Efficient graph representations, 2003)

The complements of permutation graphs are exactly the comparabilitygraphs of partial orders with dimension at most 2.





Proof idea:

Every line segment ù is defined by two points pu and qu

Define a partial order P from the line segments:ù <P `v ⇔ ù lies to the left of `v

Define two linear orders P1 and P2 from the points on L1 and L2

ù <P `v ⇔ pu <P1 pv and qu <P2 qv , i.e. P = P1 ∩ P2

L1

L2

ù `v

qu qv

pu pv





Therefore:


For k ≤ 2, we can decide in polynomial time whether the dimension of apartial order is at most k.


Overview

Partial orders


Permutation graphs

Trapezoid graphs


Trapezoid graphs

A generalization of permutation graphs:

Definition (Dagan, Golumbic, Pinter, 1988; Corneil, Kamula, 1987)

A graph G is called a trapezoid graph, if G is the intersection graph of aset of trapezoids between two parallel lines.

L1

L2

⇔Tb

Tc

a

b

c d

eTa Td Te


Trapezoid graphs




L1

L2

⇔Tb

Tc

a

b

c d

eTa Td Te

Special cases of trapezoid graphs:

Permutation graphs: when trapezoids are line segments

Interval graphs: when trapezoids are rectangles

Parallelogram graphs: when trapezoids are parallelograms


Trapezoid graphs




L1

L2

⇔Tb

Tc

a

b

c d

eTa Td Te






Trapezoid graphs




L1

L2

⇔Tb

Tc

a

b

c d

eTa Td Te






Trapezoid graphs

Theorem (Dagan, Golumbic, Pinter, Discr. Applied Math., 1988)

The complements of trapezoid graphs are exactly the comparabilitygraphs of partial orders with interval dimension at most 2.


Trapezoid graphs



Proof idea:

every trapezoid Tu is defined by two intervals Iu and JuDefine a partial order P from the trapezoids:Tu <P Tv ⇔ Tu lies to the left of Tv

Define two interval orders P1 and P2 from the intervals on L1 and L2

Tu <P Tv ⇔ Iu <P1 Iv and Ju <P2 Jv , i.e. P = P1 ∩ P2

L1

L2

Tu Tv

Iu Iv

Ju JvGeorge Mertzios (Durham) Advances in algorithmic graph theory EULER, July 2013 58 / 75

Trapezoid graphs



Therefore:

Observation

Trapezoid graphs are cocomparability graphs.

L1

L2

TbTc

Ta Td Te


Trapezoid graphs

Question: Is every trapezoid graph also a comparability graph?

Equivalently: Is every trapezoid graph also a permutation graph?(i.e. can we always “rewrite” a trapezoid graph as a permutation graph?)

Answer: No.

Example:

G is not a comparability graph ⇒ not a permutation graph !If d becomes a line segment ⇒ f intersects b. (contradiction)


Trapezoid graphs



Answer: No.

Example: a b

c

d

e

f

g

L1

L2

b c a g ed d f

bca ge dd f

G :



Trapezoid graphs



Answer: No.

Example: a b

c

d

e

f

g

L1

L2

b c a g ed d f

bca ge dd f

G :



Trapezoid graphs



Answer: No.

Example: a b

c

d

e

f

g

L1

L2

b c a g ed d f

bca ge dd f

G :



Trapezoid graphs



Answer: No.

Example: a b

c

d

e

f

g

L1

L2

b c a g ed d f

bca ge dd f

G :



Trapezoid graphs



Answer: No.

Example: a b

c

d

e

f

g

L1

L2

b c a g ed d f

bca ge dd f

G :



Trapezoid graphs



Answer: No.

Example: a b

c

d

e

f

g

L1

L2

b c a g ed d f

bca ge dd f

G :



Trapezoid graphs



Answer: No.

Example: a b

c

d

e

f

g

L1

L2

b c a g ed d f

bca ge dd f

G :



Trapezoid graphs



Answer: No.

Example: a b

c

d

e

f

g

L1

L2

b c a g ed d f

bca ge dd f

G :

G is not a comparability graph ⇒ not a permutation graph !

If d becomes a line segment ⇒ f intersects b. (contradiction)


Trapezoid graphs



Answer: No.

Example: a b

c

d

e

f

g

L1

L2

b c a g ed d f

bca ge dd f

G :




Trapezoid graphs



Answer: No.

Example: a b

c

d

e

f

g

L1

L2

b c a g ed d f

bca ge dd f

G :




Trapezoid graphs



Answer: No.

Example: a b

c

d

e

f

g

L1

L2

b c a g ed d f

bca ge dd f

G :




Trapezoid graphs



Answer: No.

Example: a b

c

d

e

f

g

L1

L2

b c a g ed d f

bca ge dd f

G :




Trapezoid graphs



Answer: No.

Example: a b

c

d

e

f

g

L1

L2

b c a g ed d f

bca ge dd f

G :




Trapezoid graphs



Answer: No.

Example: a b

c

d

e

f

g

L1

L2

b c a g ed d f

bca ge dd f

G :




Trapezoid graphs



Answer: No.

Example: a b

c

d

e

f

g

L1

L2

b c a g ed d f

bca ge dd f

G :




Trapezoid graphs



Answer: No.

Example: a b

c

d

e

f

g

L1

L2

b c a g ed d f

bca ge dd f

G :



Trapezoid graphs

How can we recognize whether a graph G is a trapezoid graph?

We can first check whether G is a comparability graph

necessary but not sufficient condition !

Several algorithms are known:

[Felsner, Habib, Mohring, SIAM Journal on Discrete Math., 1994]

[Ma, Spinrad, Journal of Algorithms, 1994]

[Langley, Discrete Applied Mathematics, 1995]

[Mertzios, Corneil, Discrete Applied Mathematics, 2011]

We sketch the algorithm of [Ma, Spinrad, 1994]

Running time: O(n2)

Main tools of this algorithm:

trapezoid graphs ←→ partial orders of interval dimension ≤ 2

cover of a bipartite graph with “chain graphs”


Trapezoid graphs










Running time: O(n2)





Trapezoid graphs










Running time: O(n2)





Trapezoid graphs

A graph G is bipartite ⇔ χ(G ) ≤ 2⇔ its vertices can be partitioned into two independent sets U, V

We denote a bipartite graph as G = (U, V , E ),where U, V are its color classes

Definition

A bipartite graph G = (U, V , E ) is a chain graph if N(a) ⊆ N(b) orN(b) ⊆ N(a), for any two vertices a, b in the same color class.

Therefore, in a chain graph G = (U, V , E ):

we can enumerate the vertices of U = {u1, u2, . . . , un} such that

N(u1) ⊆ N(u2) ⊆ . . . ⊆ N(un)

and the vertices of V = {v1, v2, . . . , vm} such that

N(v1) ⊆ N(v2) ⊆ . . . ⊆ N(vm)


Trapezoid graphs

A graph G is bipartite ⇔ χ(G ) ≤ 2⇔ its vertices can be partitioned into two independent sets U, V

We denote a bipartite graph as G = (U, V , E ),where U, V are its color classes

Definition

A bipartite graph G = (U, V , E ) is a chain graph if N(a) ⊆ N(b) orN(b) ⊆ N(a), for any two vertices a, b in the same color class.

Therefore, in a chain graph G = (U, V , E ):

we can enumerate the vertices of U = {u1, u2, . . . , un} such that

N(u1) ⊆ N(u2) ⊆ . . . ⊆ N(un)

and the vertices of V = {v1, v2, . . . , vm} such that

N(v1) ⊆ N(v2) ⊆ . . . ⊆ N(vm)


Trapezoid graphs

Definition

Let G = (U, V , E ) be a bipartite graph. The bipartite complementG = (U, V , E ) of G is the bipartite graph such that for u ∈ U, v ∈ V :

uv ∈ E ⇔ uv /∈ E


Trapezoid graphs

Definition


uv ∈ E ⇔ uv /∈ E

Lemma

Chain graphs are exactly the bipartite graphs with no induced 2K2.

2K2 :

u1 u2

v1 v2

Proof.

G = (U, V , E ) includes an induced 2K2

⇔ N(u1) * N(u2) and N(u2) * N(u1) for two vertices u1, u2

⇔ G is not a chain graph.


Trapezoid graphs

Definition


uv ∈ E ⇔ uv /∈ E

Lemma

Chain graphs are exactly the bipartite graphs with no induced 2K2.

Corollary

G = (U, V , E ) is a chain graph ⇔ G = (U, V , E ) is a chain graph.

That is:

chain graphs are closed under bipartite complementation.


Trapezoid graphs

Definition

Let G = (U, V , E ) be a bipartite graph. Then:

the chain cover number ch(G ) of G is the smallest number k of chaingraphs Gi = (U, V , Ei ), 1 ≤ i ≤ k , such that E =

⋃ki=1 Ei .

the chain dimension chdim(G ) of G is the smallest number k of chaingraphs Gi = (U, V , Ei ), 1 ≤ i ≤ k , such that E =

⋂ki=1 Ei .

Lemma

For any bipartite graph G = (U, V , E ): ch(G ) = chdim(G ).

Proof idea: chain graphs are closed under bipartite complementation

Therefore:

The problems of computing ch(G ) and computing chdim(G ) forbipartite graphs G are polynomially equivalent to each other.


Trapezoid graphs

Definition



⋃ki=1 Ei .


⋂ki=1 Ei .

Lemma



Therefore:



Trapezoid graphs

Definition



⋃ki=1 Ei .


⋂ki=1 Ei .

Lemma



Therefore:



Trapezoid graphs

Definition



⋃ki=1 Ei .


⋂ki=1 Ei .

Lemma



Therefore:



Trapezoid graphs

How can we efficiently recognize trapezoid graphs?

We use the chain cover number ch(G ) for bipartite graphs G

Definition (Ma, Spinrad, Journal of Algorithms, 1994)

Let P = (U,<P) be a partial order, where U = {u1, u2, . . . , un}. DefineV = {v1, v2, . . . , vn}. The bipartite graph C (P) = (U, V , E ) is defined as:

uivj ∈ E ⇔ ui <P uj .

“C” stands for “Comparable”


Trapezoid graphs







Example:u1 u2 u3 u4

v1 v2 v3 v4

C(P ) :

P :u1 <P u3 u3 <P u4u1 <P u4 u2 <P u4


Trapezoid graphs







Example:u1 u2 u3 u4

v1 v2 v3 v4

C(P ) :

P :u1 <P u3 u3 <P u4u1 <P u4 u2 <P u4


Trapezoid graphs







Example:u1 u2 u3 u4

v1 v2 v3 v4

C(P ) :

P :u1 <P u3 u3 <P u4u1 <P u4 u2 <P u4


Trapezoid graphs







Example:u1 u2 u3 u4

v1 v2 v3 v4

C(P ) :

P :u1 <P u3 u3 <P u4u1 <P u4 u2 <P u4


Trapezoid graphs







Lemma (Ma, Spinrad, Journal of Algorithms, 1994)

A partial order P is an interval order ⇔ C (P) is a chain graph.


Trapezoid graphs

Moreover something more general is true:


For every partial order P = (U,<P): ch(C (P)) is equal to the intervaldimension of P.

Proof.

(≥): Assume C (P) is covered by k chain graphs H1, H2, . . . , Hk

Construct a partial order Q` = (U,<Q`) for every H` such that:

ui <Qùj ⇔ uivj ∈ H`

Every Q` is an interval order, since C (Q`) = H` is a chain graph

We prove that P = Q1 ∩Q2 ∩ . . . ∩Qk :


Trapezoid graphs




Proof.







Trapezoid graphs




Proof.







Trapezoid graphs




Proof.






ui <P uj ⇔ uivj ∈ C (P) ⇔ uivj /∈ C (P)

⇔ uivj /∈ H` for every ` ⇔ uivj ∈ H` for every `⇔ ui <Q`

uj for every `


Trapezoid graphs




Proof.






Therefore ch(C (P)) ≥ Idim(P).


Trapezoid graphs




Proof.

(≤): Assume P = Q1 ∩Q2 ∩ . . . ∩Qk , where all Qi are interval orders

We prove that C (P) is covered by C (Q1), C (Q2), . . . , C (Qk)

if uivj ∈ C (P) ⇔ uivj /∈ C (P) ⇔ ui ≮P uj⇔ ui ≮Q`

uj for some ` ⇔ uivj /∈ C (Q`) for some `

⇔ uivj ∈ C (Q`) for some `



Trapezoid graphs




Proof.

(≤): Assume P = Q1 ∩Q2 ∩ . . . ∩Qk , where all Qi are interval orders

We prove that C (P) is covered by C (Q1), C (Q2), . . . , C (Qk)

if uivj ∈ C (P) ⇔ uivj /∈ C (P) ⇔ ui ≮P uj⇔ ui ≮Q`

uj for some ` ⇔ uivj /∈ C (Q`) for some `

⇔ uivj ∈ C (Q`) for some `



Trapezoid graphs




Recall that:




Trapezoid graphs




Recall that:



Therefore, to recognize trapezoid graphs:

it suffices to recognize bipartite graphs G with ch(G ) ≤ 2 !


Trapezoid graphs




Recall that:



Therefore, to recognize trapezoid graphs:

it suffices to recognize bipartite graphs G with ch(G ) ≤ 2 !

Theorem (Yannakakis, SIAM J. Alg. Discr. Methods, 1982)

For k ≥ 3, it is NP-complete to decide whether ch(G ) ≤ k.


Trapezoid graphs

Recognition of bipartite graphs G with ch(G ) ≤ 2:[Ma, Spinrad, Journal of Algorithms, 1994]

Notation

Let P = (A,<P) be a partial order and a ∈ A. The upper set of ais the set U(a) = {b : a <P b}.

Let G = (X , Y , E ) be a bipartite graph

We compute a partial order P = (X ∪ Y ,<P) with these rules:(initially P = ∅)

1 If N(a) = N(b) for two vertices a, b ∈ X ∪ Y ⇒ remove a from G

2 If N(x) ⊂ N(x ′), where x , x ′ ∈ X : add x ′ <P x to P

3 If N(y) ⊂ N(y ′), where y , y ′ ∈ Y : add y <P y ′ to P

4 If xy /∈ E , where x ∈ X , y ∈ Y : add y <P x to P

5 If U(y) ∩ X ⊆ U(x) ∩ X , where x ∈ X , y ∈ Y : add x <P y to P


Trapezoid graphs

Recognition of bipartite graphs G with ch(G ) ≤ 2:[Ma, Spinrad, Journal of Algorithms, 1994]

Notation

Let P = (A,<P) be a partial order and a ∈ A. The upper set of ais the set U(a) = {b : a <P b}.

Let G = (X , Y , E ) be a bipartite graph

We compute a partial order P = (X ∪ Y ,<P) with these rules:(initially P = ∅)







Trapezoid graphs

Theorem (Ma, Spinrad, Journal of Algorithms, 1994)

For the constructed partial order P: dim(P) = ch(G ).

Example:


Trapezoid graphs



A

Example:

G :

Xx1 x2 x3

y1 y2 y3Y

after Step 1: P = ∅

after Step 2: N(x1) ⊂ N(x2)⇒ x2 <P x1

N(x3) ⊂ N(x2)⇒ x2 <P x3

after Step 3: N(y3) ⊂ N(y1)⇒ y3 <P y1

N(y3) ⊂ N(y2)⇒ y3 <P y2

after Step 4 (from the non-adjacent vertices):

y1 <P x3, y2 <P x1y3 <P x1, y3 <P x3

after Step 5:

U(y1) ∩X = {x3}U(y2) ∩X = {x1}U(y3) ∩X = {x1, x3}

U(x1) ∩X = ∅

U(x3) ∩X = ∅U(x2) ∩X = {x1, x3}

x2 <P y1, x2 <P y2,x2 <P y3


Trapezoid graphs




Example:

G :

Xx1 x2 x3

y1 y2 y3Y



N(x3) ⊂ N(x2)⇒ x2 <P x3


N(y3) ⊂ N(y2)⇒ y3 <P y2


y1 <P x3, y2 <P x1y3 <P x1, y3 <P x3

after Step 5:

U(y1) ∩X = {x3}U(y2) ∩X = {x1}U(y3) ∩X = {x1, x3}

U(x1) ∩X = ∅

U(x3) ∩X = ∅U(x2) ∩X = {x1, x3}

x2 <P y1, x2 <P y2,x2 <P y3


Trapezoid graphs




Example:

G :

Xx1 x2 x3

y1 y2 y3Y



N(x3) ⊂ N(x2)⇒ x2 <P x3


N(y3) ⊂ N(y2)⇒ y3 <P y2


y1 <P x3, y2 <P x1y3 <P x1, y3 <P x3

after Step 5:

U(y1) ∩X = {x3}U(y2) ∩X = {x1}U(y3) ∩X = {x1, x3}

U(x1) ∩X = ∅

U(x3) ∩X = ∅U(x2) ∩X = {x1, x3}

x2 <P y1, x2 <P y2,x2 <P y3


Trapezoid graphs




Example:

G :

Xx1 x2 x3

y1 y2 y3Y



N(x3) ⊂ N(x2)⇒ x2 <P x3


N(y3) ⊂ N(y2)⇒ y3 <P y2


y1 <P x3, y2 <P x1y3 <P x1, y3 <P x3

after Step 5:

U(y1) ∩X = {x3}U(y2) ∩X = {x1}U(y3) ∩X = {x1, x3}

U(x1) ∩X = ∅

U(x3) ∩X = ∅U(x2) ∩X = {x1, x3}

x2 <P y1, x2 <P y2,x2 <P y3


Trapezoid graphs




Example:

G :

Xx1 x2 x3

y1 y2 y3Y



N(x3) ⊂ N(x2)⇒ x2 <P x3


N(y3) ⊂ N(y2)⇒ y3 <P y2


y1 <P x3, y2 <P x1y3 <P x1, y3 <P x3

after Step 5:

U(y1) ∩X = {x3}U(y2) ∩X = {x1}U(y3) ∩X = {x1, x3}

U(x1) ∩X = ∅

U(x3) ∩X = ∅U(x2) ∩X = {x1, x3}

x2 <P y1, x2 <P y2,x2 <P y3


Trapezoid graphs




Example:

G :

Xx1 x2 x3

y1 y2 y3Y



N(x3) ⊂ N(x2)⇒ x2 <P x3


N(y3) ⊂ N(y2)⇒ y3 <P y2


y1 <P x3, y2 <P x1y3 <P x1, y3 <P x3

after Step 5:

U(y1) ∩X = {x3}U(y2) ∩X = {x1}U(y3) ∩X = {x1, x3}

U(x1) ∩X = ∅

U(x3) ∩X = ∅U(x2) ∩X = {x1, x3}

x2 <P y1, x2 <P y2,x2 <P y3


Trapezoid graphs




Example:

G :

Xx1 x2 x3

y1 y2 y3Y



N(x3) ⊂ N(x2)⇒ x2 <P x3


N(y3) ⊂ N(y2)⇒ y3 <P y2


y1 <P x3, y2 <P x1y3 <P x1, y3 <P x3

after Step 5:

U(y1) ∩X = {x3}U(y2) ∩X = {x1}U(y3) ∩X = {x1, x3}

U(x1) ∩X = ∅

U(x3) ∩X = ∅U(x2) ∩X = {x1, x3}

x2 <P y1, x2 <P y2,x2 <P y3


Trapezoid graphs




Example:

G :

Xx1 x2 x3

y1 y2 y3Y



N(x3) ⊂ N(x2)⇒ x2 <P x3


N(y3) ⊂ N(y2)⇒ y3 <P y2


y1 <P x3, y2 <P x1y3 <P x1, y3 <P x3

after Step 5:

U(y1) ∩X = {x3}U(y2) ∩X = {x1}U(y3) ∩X = {x1, x3}

U(x1) ∩X = ∅

U(x3) ∩X = ∅U(x2) ∩X = {x1, x3}

x2 <P y1, x2 <P y2,x2 <P y3

Hasse diagram of P :

x2

y3

y2

x1

y1

x3

ch(G) = 2

dim(P ) = 2

P1 = {x2, y3, y1, x3, y2, x1}P2 = {x2, y3, y2, x1, y1, x3}


Trapezoid graphs




Example:

G :

Xx1 x2 x3

y1 y2 y3Y



N(x3) ⊂ N(x2)⇒ x2 <P x3


N(y3) ⊂ N(y2)⇒ y3 <P y2


y1 <P x3, y2 <P x1y3 <P x1, y3 <P x3

after Step 5:

U(y1) ∩X = {x3}U(y2) ∩X = {x1}U(y3) ∩X = {x1, x3}

U(x1) ∩X = ∅

U(x3) ∩X = ∅U(x2) ∩X = {x1, x3}

x2 <P y1, x2 <P y2,x2 <P y3


x2

y3

y2

x1

y1

x3

ch(G) = 2G = G1 ∪G2

dim(P ) = 2

P1 = {x2, y3, y1, x3, y2, x1}P2 = {x2, y3, y2, x1, y1, x3}


Trapezoid graphs




Example:

G :

Xx1 x2 x3

y1 y2 y3Y



N(x3) ⊂ N(x2)⇒ x2 <P x3


N(y3) ⊂ N(y2)⇒ y3 <P y2


y1 <P x3, y2 <P x1y3 <P x1, y3 <P x3

after Step 5:

U(y1) ∩X = {x3}U(y2) ∩X = {x1}U(y3) ∩X = {x1, x3}

U(x1) ∩X = ∅

U(x3) ∩X = ∅U(x2) ∩X = {x1, x3}

x2 <P y1, x2 <P y2,x2 <P y3


x2

y3

y2

x1

y1

x3

ch(G) = 2G = G1 ∪G2

dim(P ) = 2

P = P1 ∩ P2

P1 = {x2, y3, y1, x3, y2, x1}P2 = {x2, y3, y2, x1, y1, x3}


Trapezoid graphs



Therefore:

We can we can efficiently check whether ch(G ) ≤ 2

Summarizing the algorithm: [Ma, Spinrad, Journal of Algorithms, 1994]

Let G be the input cocomparability graph

Let P be a partial order of the complement G

We compute the bipartite graph C (P)

If ch(C (P)) ≤ 2 then:

P has interval dimension at most 2we compute two interval orders P1, P2 such that P = P1 ∩ P2

we compute from P1, P2 a trapezoid representation of G

else return “G is not a trapezoid graph”


Trapezoid graphs



Therefore:

We can we can efficiently check whether ch(G ) ≤ 2

Summarizing the algorithm: [Ma, Spinrad, Journal of Algorithms, 1994]

Let G be the input cocomparability graph

Let P be a partial order of the complement G

We compute the bipartite graph C (P)

If ch(C (P)) ≤ 2 then:

P has interval dimension at most 2we compute two interval orders P1, P2 such that P = P1 ∩ P2

we compute from P1, P2 a trapezoid representation of G

else return “G is not a trapezoid graph”


Trapezoid graphsSpecial subclasses

Definition

Parallelogram graphs are the intersection graphs of parallelograms betweentwo parallel lines.

Very similar structure with trapezoid graphs (subclass)

Can we recognize parallelogram graphs efficiently?

Theorem (Mertzios, Sau, Zaks, SIAM Journal on Computing, 2011)

Given a graph G , it is NP-complete to decide whether G is a parallelogramgraph.



Definition








Definition








Definition (Corneil, Kamula, 1987)

A trapezoid graph G is called a triangle (or PI∗) graph, if it admitsa trapezoid representation, in which every trapezoid is a triangle.

L1

L2


A trapezoid graph G is called a simple-triangle (or PI) graph, if it admitsa trapezoid representation, in which every trapezoid is a triangle withone point on L1 and the other two points (i.e. interval) on L2.

L1

L2




A trapezoid graph G is called a triangle (or PI∗) graph, if it admitsa trapezoid representation, in which every trapezoid is a triangle.

L1

L2


A trapezoid graph G is called a simple-triangle (or PI) graph, if it admitsa trapezoid representation, in which every trapezoid is a triangle withone point on L1 and the other two points (i.e. interval) on L2.

L1

L2



Permutation ⊂ Simple-Triangle ⊂ Triangle ⊂ Trapezoid

Triangle and simple-triangle graphs:

again very similar structure with trapezoid graphs (subclass)

can we recognize parallelogram graphs efficiently?

these problems remained open since 1987 (until recently)

Theorem (Mertzios, Theoretical Computer Science, 2012)

The recognition of triangle (i.e. PI∗) graphs is NP-complete.

Theorem (Mertzios, ESA, 2013)

Given a graph G with n vertices and m non-edges, we can decidein O(n2m) time whether G is a simple-triangle (i.e. PI) graph.



































Simple-Triangle (PI) graphs

Definition

A partial order P is a linear-interval order if P = P1 ∩ P2, where:

P1 is a linear order,

P2 is an interval order.



Definition




Theorem

The complements of simple-triangle graphs are exactly the comparabilitygraphs of linear-interval orders.

Proof idea:

L1

L2

Ju Jv

Tu Tv

pu pv



Definition




Theorem

The complements of simple-triangle graphs are exactly the comparabilitygraphs of linear-interval orders.

How to recognize linear-interval orders efficiently?

We can not apply directly the previous results for permutationand trapezoid graphs !

new techniques are needed





Main idea of the algorithm:

Construct a Boolean formula φ = φ1 ∧ φ2 such that:

φ1 is a 3SAT formula

φ2 is a 2SAT formula

φ is satisfiable ⇔ such G is a PI graph

φ has a special structure:satisfiability on φ can be solved in linear time


Thank you for your attention!


Documents

Advances in algorithmic graph theorysummerschool2013.ccaba.upc.edu/.../Lecture6.2_Mertzios.pdfAdvances in algorithmic graph theory George B. Mertzios School of Engineering and Computing