Holes and Dominoes in Meyniel Graphs

International Journal of Foundations of Computer Science

World Scientic Publishing Company

HOLES AND DOMINOES IN MEYNIEL GRAPHS

FLORIAN ROUSSEL

LIFO Universite dOrleans bat IIIA BP

Orleans Cedex France

and

IRENA RUSU

LIFO Universite dOrleans bat IIIA BP

Orleans Cedex France

Received received date

Revised revised date

Communicated by Editors name

ABSTRACT

Many practical problems frequency assignement register allo cation timetables may b e

formulated as graph vertexcoloring problems but nding solutions for them is dicult

as long as they are treated in the most general case where the graph is arbitrary since

vertex coloring has b een proved to b e NPcomplete The problem b ecomes much easier

to solve if the graph resulting from the mo delisation of the practical application b elongs

to some particular class of graphs for which solutions to the problem are known Meyniel

graphs form such a class a fast coloring algorithm can b e found in for which an

ecient recognizing algorithm is therefore needed

A graph G V E is said to b e a Meyniel graph if every o dd cycle of G on at least ve

vertices contains at least two chords Meyniel graphs generalize b oth itriangulated and

parity graphs two well known classes of p erfect graphs that will b e present in our pap er

in Section

In Burlet and Fonlupt prop ose a characterization of Meyniel graphs which relies on

the following prop erty the class of Meyniel graphs may b e obtained from some basic

Meyniel graphs using a binary op eration called amalgam Besides the theoretical interest

of this result a practical interest arises b ecause of the p olynomial recognition algorithm

which can b e obtained Unfortunately it is quite exp ensive to verify if a given graph

is the amalgam of two graphs therefore the complexity of the whole algorithm is in

O n and this supp orts the idea that a new p oint of view is needed to nd a more

ecient algorithm

Our approach of Meyniel graphs will b e directed through the search of a general structure

Intuitively a Meyniel graph either will b e simple ie with no hole or domino or will

have a skeleton around which the rest of the graph will b e regularly organized As

suggested the rst typ e of Meyniel graphs is simple to identify For the second typ e a

deep er analysis is necessary it yields a characterization theorem which is used to deduce

the O m mn recognition algorithm

Intro duction

Meyniel proved that graphs whose o dd cycles of length ve or more have at

least two chords are p erfect that is for each of them induced subgraphs the clique

numb er equals the chromatic numb er Several years later Grotschel Lovasz and

Schirjvers showed that the NPproblems of computing the chromatic clique sta

bility and clique covering numb er b ecome p olynomial on p erfect graphs Together

these two results asserted the necessity to investigate the algorithmic asp ects of the

graphs now called Meyniel graphs

We are interested here in the recognition of these graphs for which we give

a characterization theorem and a recognition algorithm which improves a lot the

complexity of the previous one

The reader is supp osed familiar with the classical notions of elementary path

and cycle parity of a path or a cycle connected and connected graph connected

and connected comp onent For these notions as well as for nding linear algo

rithms to compute the connected comp onents of a graph see

i j of a path P x x is the path x x The subpath P

1 k +1 i i+1 x x

i j

x The similar notation will b e used for a subpath of a cycle Sp ecial attention

will b e payed to induced or chordless paths and cycles ie which have no edges

but the ones indicated b efore denoted resp ectively by P and C if they have k

k k

vertices

Several particular graphs will often b e used along the presentation the hole

chordless cycle of length at least equal to the domino cycle on six vertices with

a unique chord joining two vertices at distance three along the cycle and the house

cycle on ve vertices with a unique chord

We close this section by giving some notations All the vertices adjacent in G to

a xed vertex v will form the set N v or N v if no confusion is p ossible called

neighb ourho o d of v If either A V or A is an induced subgraph of G then G A

will denote the subgraph of G induced by all the vertices not in A Also if H is an

induced subgraph of G and either A V or A is an induced subgraph of G then

H A will b e the subgraph induced in G by the vertices in H and the vertices in

A One more simplication will b e made if A fxg we will write G x H x

instead of G fxg H fxg

Mo dular decomp osition

A set M V of vertices is said to b e a mo dule in G V E if every vertex in

V n M is either adjacent to all the vertices in M or to none of them A mo dule is

trivial if it is the empty set a singleton or V itself and a graph with more than

vertices is prime if all its mo dules are trivial

The mo dular decomp osition asso ciates with an arbitrary graph G a unique de

comp osition tree T G whose leaves are the vertices of G while the internal no des

are mo dules lab eled with P S and N for parallel serial and neighb ourho o d mo d

ule see or for details The ro ot of T G corresp onds to the entire G

which is uniquely decomp osed in mo dules represented in T G by the children of

the ro ot each mo dule is also decomp osed in submo dules and so on until only

singletons may b e found At the end if r is a no de of T G not a leaf with chil

dren r r r the graph with vertices r r r and edges r r such that

1 2 k 1 2 k i j

any vertex in M r is adjacent to any vertex in M r is called the representative

i j

graph of r then for a P no de an S no de an N no de the representative graph is

resp ectively a stable set a clique and a prime graph

The unicity of the decomp osition may b e found in As shown by McConnel

Spinrad and indep endently by Cournier Habib the mo dular decomp osition

tree of any graph is computable in linear in numb er of edges time

Preliminary asp ects

Consider a Meyniel graph G that we can supp ose connected and a hole C

obviously of even length of G Try to identify all the p ositions an arbitrary vertex

w out of C can o ccupy with resp ect to C

Lemma If G is a Meyniel graph and C is an induced cycle of G then for any

w V V C we have exactly one of the fol lowing statements

w is adjacent to no vertex of C

w is adjacent to exactly one vertex of C

C w is a bipartite connected graph

w is adjacent to every vertex of C

w is a copy of a certain z with respect to C ie w is exclusively adjacent

to z z z with the convention z z z z

i1 i i+1 0 k k +1 1

Pro of It is easy to see that any pair of statements cannot b e simultaneously

true Let C z z z z k and supp ose by contradiction that none

1 2 k 1

of the statements is veried Then w forms at least one o dd cycle with C of

length three or more therefore we can apply the following result to deduce that w

is adjacent to at least two consecutive vertices of C

Claim Let G be a Meyniel graph w a vertex of G and u v two nonadjacent

neighbours of w joined in G by an odd chord less path P Then V P N w

We nish the pro of of Lemma using this corollary of Prop osition in

Claim Let C z z z z be an induced cycle of length at least four of the

1 2 p 1

Meyniel graph G and w V V C a vertex adjacent to two consecutive vertices

say z z on C Then w is a copy of z or a copy of z or is adjacent to every

1 2 1 2

vertex of C

And now Lemma is proved 2

This lemma suggests a p ossibility to express the prop erty of the graph to b e

Meyniellike around a cycle by putting the vertices in sets with particular pro

p erties Unfortunately this approach is not really the b est one b ecause of the

instability of the partition in sets even a very small change of the cycle replacing

a vertex with one of its copies implies great changes in the sets therefore it is very

dicult to exploit the results obtained for one cycle when considering another one

That is why it is preferable to bring together all the copies of a given cycle C and

then dene all the other sets taking into account all these copies

Two more problems app ear here rstly the induced cycle on four vertices has a

particular b ehaviour a vertex w adjacent to all the vertices of a C is not necessarily

adjacent to the copies of the vertices while for a C k this is true secondly

identifying an induced cycle on at least six vertices ie a hole is not easy from an

algorithmical p oint of view A solution to these problems is to nd either a hole

or a domino in our graph since for domino es a statement equivalent to Lemma

holds see b elow

Now we have a rough representation of the Meyniel graph which we try to

rene following the two steps b elow details are found resp ectively in Sections

Step I augmenting the initial cycle or domino to a maximal connected

bipartite subgraph called skeleton of G for which Lemma is still

true putting together all the copies of and dening the other sets

with resp ect to all these copies

Step II improving using a rening op eration

Two kinds of Meyniel graphs are then found

the socalled HHDgraphs containing no hole no house and no domino

the Meyniel graphs which have an almost regular structure around a con

nected bipartite graph on at least six vertices the skeleton

We now start to give the details of the announced steps

Building a large skeleton

Let G V E b e a Meyniel graph containing a hole or a domino called C and

a bipartite connected subgraph of G containing C Any vertex z in can have

copies in G with resp ect to that is vertices w such that N w fz g N z In

the graph induced in G by the copies of z z included denote D z the connected

comp onent of z and notice that D z V fz g otherwise a vertex can b e

found in D z with a to o large neighb ourho o d in So z may b e seen as the

representative vertex of all its copies while the graph induced by D z

D z

may b e seen as the graph containing all the near copies of see Fig

The set of vertices adjacent to all the vertices in C recall that contains C is

denoted T In G T denote S all the vertices which are not in the same connected

comp onent as C

Till now we dened the sets of vertices which corresp ond to the statements

and a part of the set corresp onding to As indicated it is preferable to dene

the rest of the sets corresp onding to taking into account all the near

copies of ie all the graph At the b eginning they will b e mixed up but

progressively they will b e separated as needed For the moment dene see Fig

where z is an arbitrary vertex of

C z fw V V T j w is adjacent to all the vertices in D z g

W z fw V V C z T j w is adjacent to some vertex in D z g

L z fw V C z W z D z T j w is connected to C z or to W z

in G D z T g

Obviously for any connected bipartite subgraph of G this decomp osition of

G around is unique but the indicated sets do not partition the graph L Γ (z) CΓ (z) Γ

z T S DΓ (z) C

WΓ (z)

Fig Decomp osition of a Meyniel graph

With these denitions and with the notation T for the set of vertices adjacent

to all the vertices in a subgraph H we can now reformulate and slightly generalize

Claim in the following way

Claim Let H be either an induced cycle of length at least four or a domino of

the Meyniel graph G and w V V H a vertex which has two adjacent neighbours

say z z on H Then w D z D z T

H H H

Pro of If H is an induced cycle the result is immediate from Claim If it is

a domino we can apply Claim for one of the two C in H If w is adjacent to

every vertex of this C then again by Claim we must have w T otherwise it

4 H

is easy to see that w must b e the copy with resp ect to H of a vertex of H 2

Remark Notice that Claim is not true for any bipartite graph To see this it

is sucient to consider a cycle on six vertices with two chords joining vertices at

distance along the cycle There can be a vertex adjacent only to the vertices of

degree

Now a subgraph of G is called a skeleton of G with resp ect to the hole or

domino C if the following conditions are veried

is bipartite connected and contains C

for every z V D z is a mo dule in

for every z V W z is not connected to in G D z T

If these conditions hold in G T S the sets W z C z L z may b e seen

as the irregular parts of a quite regular structure In fact see the following lemma

as long as there exist some sets C z connected to in G D z T the regular

structure may b e augmented ie the skeleton may b e grown Finally no C z

will b e connected to in G D z T ie L z will contain no vertex of

D D

so the irregular parts will b e represented by some sets I z W z C z

L z attached to the corresp onding D z and adjacent to no other vertices but

p ossibly the ones in T In that case the sets D z I z T S will partition V

Lemma Let G be a Meyniel graph containing a hole or domino C Then G has

a skeleton with respect to C such that for every vertex z in the set C z is

not connected to in G D z T

Pro of Let C By induction on i we will prove that we can build a

sequence of skeletons f i r g of G such that every of them satises the

two auxiliary prop erties

A If C is an induced cycle of then T T

B If w V and z z V such that z z E w z E w z E then

1 2 1 2 1 2

i i

w D z D z T

1 2

and such that is the skeleton we are lo oking for

Remark Notice that because of the property of a skeleton of G every vertex

i i

z D z could be seen as the representative vertex of D z in instead of

z Therefore the properties A B here above wil l hold for any of the copies of

obtained by considering vertices in each D z

We will rstly prove that is a skeleton satisfying A B then supp osing that

i i+1

is built and satises the hyp othesis we will build and show it also veries

the hyp othesis

The statements A and B are easy to prove for using Claim Let us show

that is a skeleton of G To this end we give for and pro ofs as general as

p ossible which do not dep end on the bipartite graph but only on some of its

prop erties see the remarks b elow Exactly the same pro ofs will b e used later to

i+1

show and for

There is nothing to do to prove

To prove consider z z V and supp ose by contradiction that there

0 0

exists z D z adjacent to z D z but nonadjacent to z in the same set

1 2

Since the graph induced by D z is connected we may supp ose that z z E

1 2

0 0

We will prove that N z N z To this end let if it exists u b e a

0 0

neighb our of z on such that z u E and consider C a smallest cycle of

containing z u and z Then C contains at least one subpath U with extremities z

and z containing u Then U fz z g fz z z g is a cycle whose chords but one

1 2

have an extremity z otherwise C was not a smallest cycle It is easy now to see

that z must b e adjacent to every vertex on U fz z g and the same is valid for

z since U is included in and z z have the same b ehaviour with resp ect to the

0 0

vertices in We deduce that is not bipartite a contradiction

0 0

A similar argument proves the other inclusion so N z N z therefore

by the denition of the set D w ie the largest connected set containing copies

of w z z should have b een in the same set D w with w V they have

the same neighb ourho o d in and are in the same connected comp onent in the

graph of copies thanks to the edge z z a contradiction So is proved

Remark Notice that the proof of does not depend on the graph that we

use but only on the fact that is bipartite and connected

To prove supp ose the contrary holds for z V and let P b e a shortest

0 0

path joining W z to in G D z T denote resp ectively by w z its

extremities Notice that the only vertex on this path which could have other

neighb ours on D z is the one preceding z that we call s otherwise P

is not a shortest path Since w W z there must exist two adjacent vertices

0 0

z z D z such that w z E w z E The graph induced by D z is

connected so we can cho ose z z such that z z E Moreover since is

proved we can see Remark supp ose that z z by changing the representative

vertex for D z if necessary

from z to z and denote z the vertex which Take P a chordless path in

0 1

precedes z on this path Without loss of generality we can supp ose that s has no

neighb our on P fz z g otherwise we suitably change z therefore in P P

0 0

the only p ossible chords have an extremity z and the other one r V P such

that r C z otherwise P is not a shortest path as dened Consider if it

exists r w the neighb our of z the closest to w on P w Because of z r E

and r C z we also have z r E and therefore r w E otherwise z z P

w r

induce a bad cycle since z cannot have other neighb ours on P they would b e

r w

in W z and P wouldnt b e as short as p ossible Then z must also b e adjacent

to all the vertices on P P apply Claim for z with the neighb ours z and the

appropriate vertex among r w dep ending on the parity of the path consequently

0 0

z z Now the statement B for s z z implies s D z D z T a

0 1 0 0

contradiction

We deduce that z has no neighb our but w and p ossibly z if z z on P

0 0 1

Then P P is an induced cycle on three or more vertices and P P z

contradicts the hyp othesis that G is a Meyniel graph except if P P has exactly

three vertices ie s w z z But then the statement B for s z z and

0 1 0

gives another contradiction Thus is proved

Remark Notice that the proof of does not depend on the graph that we

use for G but only on the fact that B and are true

Now we have that is a skeleton which satises A and B By induction

supp ose that G has a skeleton which satises A and B If no set C z is

i i

connected to in G D z T then we denote and Lemma is proved

i+1 i

Otherwise we build another skeleton strictly containing and also satisfying

A and B

i i

Since there exists some z V such that C z is connected to in

i i

G D z T let P b e a shortest path joining a vertex of C z to in

G D z T denote its extremities resp ectively by c and z Again only s the

vertex preceding z on P can have other neighb ours on D z b ecause of

the minimality of P Once more take P a chordless path joining z and z in

and supp ose that s has no neighb our on P fz z g Notice that P P has no

chord otherwise such a chord would have the extremities z r with r V P and

i i i

r could b e neither in W z since W z is not connected to nor in C z

the path P wouldnt b e the shortest with the indicated prop erties

i+1 i

Dene P obviously this is a bipartite connected graph and let

i+1

us prove that is a skeleton of G satisfying A and B

i+1

A is true for

i+1 i

Let C b e an induced cycle of If it is also a subgraph of then by

induction we have the desired conclusion Otherwise V C V P implies

that s V C As s can have more than one neighb our on two p ossibilities

may app ear

First case The neighb ours of s on C are z z V

1 2

If the path included in C fz z sg has more than one Let t T

1 2 1 C

vertex then by applying B in for t and two adjacent vertices on this path we

deduce that t T the other two p ossibilities are not valid since t has two many

1 1

neighb ours on If this path has exactly one vertex z then by applying B for

and t z z we deduce that t D z T If t T then we are done in

1 1 3 1 3 1

the other case s is adjacent to z t for which the prop erty B may b e applied

1 1

i i

thanks to Remark and we have s D z D z T a contradiction

1 3

Second case Only one neighb our of s on C is in

Then C contains the whole path P and the rest of C is a path in with

an extremity z the neighb our of c in and the other one denoted z If this

we have path has more than three vertices then by B for and any t T

1 C

t T If it has exactly three vertices say z z z then by B we have t

1 1 2 1

i i

D z T but t D z is imp ossible since then c t z would give by B a

2 1 2 1

contradiction If it has exactly two vertices z and z then again by B we must

have t D i z D i z T The only p ossible case is t T the other two

1 1 1

imply that either c or s has two adjacent neighb ours on

i+1

Then A is proved for

i+1

B is true for

Take w z z as indicated We will prove that if w T then w D i+1 z

1 2 1

i+1

D z To this end take H z z z z z a chordless cycle of

2 1 1 2 3 k 1

containing z z Claim for this cycle implies that either w is adjacent exclusively

1 2

otherwise b ecause of the to z on H fz z g or exclusively to z w T

k 1 1 2 3 H

i+1

statement A for we should have w T Without loss of generality supp ose

that the rst p ossibility is valid in the second one the reasoning is similar and let

i+1 i+1 i+1

us prove that w D z that is N w fz g N z

1 1 1

We prove

i+1

By contradiction supp ose rstly that there exists a neighb our z V of w

which is not a neighb our of z Take H z z u u z a shortest cycle of

1 2 1 2 1 p 1

i+1

containing z z and z and denote U a chordless subpath of the graph induced

1 2

by z z u u joining z and u z Notice that u u is chordless and

1 2 1 i 1 i 1 i

all the chords of the path z z u u contain z

1 2 1 i 1

if z V U ie z has a neighb our on u u then Claim for w

2 1 2 i1

and U or for w and U z dep ending on the parity implies that w is adjacent to

every vertex on U Since z z E U z has at least four vertices so w is adjacent

1 2

i+1

to a P of the graph on four vertices cannot b e a C since by A we would

4 4

obtain w T Let us show that in the second case see here b elow we have the

same conclusion

if z V U then Claim for w and U or for w and U z dep ending on

2 1

the parity implies again that w is adjacent to every vertex on U Now the case

where z is not the other neighb our of z on U easily implies that w is adjacent to

i+1

every vertex of a P of The contrary case ie z z E gives that z see

4 2 k

the reasoning ab ove z z z form another P whose vertices are adjacent to w

1 2 4

i+1

We have to prove now that if w is adjacent to a P abcd of it must b e

i+1 i+1

adjacent to the whole Let H b e a shortest cycle in containing ab and cd

If the four vertices are in the order a b c d on H then notice that bc is an

edge of the cycle and the only p ossible chords have an extremity b or c otherwise H

is not as short as p ossible Since H is an even cycle H fb cg is an o dd induced

path and implies with w and by Claim that w is adjacent to every vertex in H

By the statement A for a chordless sub cycle of H we have the desired conclusion

If the four vertices are in the order a c d b on H then call U the subpath of

H joining a c not containing d and U the subpath of H joining b d not containing

a It is clear that the paths U and U have no chord otherwise H is not as short

1 2

as p ossible Moreover if there exists a chord xy on H joining a vertex from U c

to a vertex from U b then we can found a cycle H containing xy and such that

a b c d are in this order on H H must contain all the vertices of H since H is

as short as p ossible moreover for H we can p erform the preceding reasoning to

obtain the desired conclusion Thus we may supp ose that all the chords of H have

an extremity b or c and the other one resp ectively on U a or U d Therefore

1 2

w a and U imply by Claim that w is adjacent to all the vertices on U Then

2 2

w is adjacent to all the vertices of a chordless sub cycle of U c and by A we have

w T a contradiction

We prove

i+1

Supp ose now that there exists a neighb our z V of z which is not a

i+1

neighb our of w Then there exists in a cycle H containing z z and z z whose

1 1 2

only chords have an extremity z Call U H z Among all the neighb ours of

1 1

z on U z and z included there exists at least one namely z which is non

1 2

adjacent to w Consider such a neighb our z which is the closest to z along U

3 2

z the closest to z Then z w E and let z b e the neighb our of z on U

3 3 4 4 1 z z

3 2

is o dd and has only one chord b ecause z w E The cycle induced by w z U

3 1 z z

3 4

i+1 i+1

of N z a contradiction w N z w cannot have neighb ours on U

4 1 z z

3 4

The statement B is proved

i+1

Now to show that is a skeleton of G we should prove the conditions

hold But is obviously true and immediately follow from

Remarks

Lemma is proved as long as the indicated condition on the sets C z is

i+1

not veried a new graph may b e built with the same prop erties When it is

veried the last graph we have built is the one we were lo oking for call it 2

If we call I z the set W z C z L z for any z then we have the

announced prop erty ie I z may b e consider as attached in G T S to

D z since it has no other connections to the rest of G T S

Remark There is no diculty to see that every vertex t T is adjacent in fact

to every vertex in D z Indeed take arbitrary vertices z C z D z

z C

and simply apply Claim for t and the graph C z z isomorphic to C

On the contrary while considering other vertices z of C T may b e adjacent

to only a part of D z This why we will try to improve this structure in order to

have an homogeneous b ehaviour of T with resp ect to the skeleton

Improving the skeleton

Let us limit to its induced subgraph containing all the vertices z such

that every vertex of D z is adjacent to every vertex of T Obviously is a

bipartite graph containing C Moreover we have

Lemma The graph is connected

Pro of If this is not the case then consider K the connected comp onent of

containing C and z V K a vertex such that z is not connected Therefore

there exists a connected comp onent R of which contains z and has no other

connections with K

Since is connected there must exist a path joining R and K in z

Consider U a shortest path joining in z a vertex in R to a vertex in K denote

u V R and v V K its extremities I a chordless path joining z u in R and J

a chordless path joining z v in K Without loss of generality we may supp ose that

no chord exists joining vertices from I z or J z to vertices of U Then the only

p ossible chords of U I J have an extremity z and the other one on U

Claim If r is the neighbour of z on I and w V K is a neighbour of v out of

J then r u and v z w u E

Pro of It is quite easy to see that the path I J contains at least three vertices

and U R K at least one situated in We have two p ossibilities

If the path I J contains at least four vertices then I or J contains more that

two vertices We supp ose that I has at least three vertices otherwise by replacing

I with J and by a similar pro of we obtain the same conclusion By applying Claim

for any t T and the path I U J fr z g we obtain that t is adjacent to any

vertex on U I J

Now since any vertex q on U I J is situated on an induced subgraph of

U I J isomorphic to a hole or a domino this is a consequence of the fact that

is bipartite and all the chords in U I J have an extremity z as in Remark we

deduce that any t T is adjacent to any vertex in D q for an arbitrary q U

But then every vertex in U should b e in a contradiction

If the path I J contains exactly three vertices then b oth I J are paths of

length one whose vertices will b e z u in this case u r resp ectively z v

Take w z a neighb our of v on K and supp ose by contradiction that w u E

Let q b e the neighb our of v on U We can see that every vertex of U v is

nonadjacent to w this is true for q b ecause is bipartite and for each vertex of

U fq v g b ecause U is as short as p ossible Then we apply Claim for any t T

and the path U w to deduce that t is adjacent to all the vertices of U

Now we deduce as ab ove that V U V a contradiction if U z has

at least six vertices On the contrary if U z has exactly four vertices the same

reasoning do esnt work but if q D q where q is the unique vertex on U I J

the cycle given by t u q v w implies tq E Claim is proved 2

We can easily conclude the pro of of Lemma the Claim b efore implies that u

and K are in fact in the same connected comp onent of a contradiction 2

So is bipartite and connected but it is not necessarily a skeleton of G

t t

satisfying Lemma since for two vertices z z V the sets C z C z

1 2 1 2

t t

may b e connected in G if N z N z see b elow But then z z may

1 2 1 2

b e seen as copies of the same vertex since their role in the graph is the same

in other words they will b e a part of a mo dule of

As recalled in Section the mo dular decomp osition asso ciates to any graph

a representative graph which is either a clique or a stable set or a prime graph

ie a graph with no mo dules such that the initial graph may b e obtained from

the representative graph by replacing any of its vertices with a mo dule Apply the

t t

mo dular decomp osition to our graph call the asso ciated representative graph

which is a prime graph and notice that necessarily contains C Indeed if this

is not the case then C should b e in a mo dule C is prime so all its vertices will

always b e in the same mo dule Since is connected all the vertices of this mo dule

should b e adjacent to all the vertices of another mo dule But then wouldnt b e

bipartite

t t t

So contains C and is obtained from by replacing any vertex with a

r r

set of vertices which in fact should induce stable sets otherwise wouldnt b e

bipartite Obviously is bipartite but it is not necessarily connected Then

let b e the connected comp onent of containing C and z an arbitrary vertex

t t

the vertices z z z of We denote of representing as a vertex of

1 2 p

t t

D z D z where in fact D z D z

i i i

i=1

the subgraph induced by the vertices in D z

z V ( )

W z fw V V T j w is adjacent to some vertices in D z g

Lz fw V W z D z T j w is connected to W z in G D z T g

As it can b e easily seen these notations closely follow the initial ones the only two

dierences b eing the following ones

the sets D z are not necessarily connected while D z were

the vertices w V V T adjacent to some vertices in D z are no

more separated in W z and C z but put together in the set W z Indeed as

long as these vertices do not have dierent b ehaviours with resp ect to the rest of

the graph there is no reason to treat them dierently This was necessary to b e

done in the inductive pro cessing for the skeletons but it is no more necessary

Now we prove that is a prop er skeleton of G with resp ect to C ie

is bipartite connected and contains C

for every z V D z is a mo dule in

for every z V W z is not connected to in G D z T

every vertex in is adjacent to all the vertices of T

Once more these prop erties are similar to the one of a skeleton except for the

fourth one

Lemma Let G be a Meyniel graph containing an induced hole or domino C

Then the graph built as described is a proper skeleton of G with respect to C

Pro of Prop erties and are obvious Prop erty follows immediately

from the construction of and prop erty of We only have to prove

By contradiction supp ose there exists a vertex z V such that W z is

in G D z T Take U a shortest path joining W z to connected to

D D

in G D z T whose extremities are denoted r and v resp ectively We may

supp ose that z is adjacent to r otherwise by we can suitably change the

representative vertex of D z in Clearly any two vertices on U are situated in

dierent sets D z D z of otherwise U is not as short as p ossible and

1 2 D

then we can supp ose that V U V if there existed a vertex of U b elonging to

some W x C x L x then U would necessarily contain at least two vertices

of D x Let J b e a chordless path joining v and z in and assume without loss

of generality that the vertex s preceding v on U has no other neighb our on J Then

the cycle U J has no chords otherwise U wouldnt b e as short as p ossible

Remark The path U contains at least one vertex q of ie such that not

every t T is adjacent to al l q D q otherwise U should be included in

Consider q chosen as close as possible to r

Two p ossibilities exist for r

If r V then denote u the vertex preceding q on U Now by changing

r q

the notation of U z in I of U in U of in K and taking w a neighb our

ur q v

of v on K J we may apply Claim to deduce that w r E so r is in the same

connected comp onent of as a contradiction

If r V there exists r D r and t T such that tr E Denote

U U r r clearly U J has no chord

First case J has at least four vertices Then by Claim we have that t is

adjacent to all the cycle U J and this contradicts the assumption tr E

Second case J has exactly three vertices Then in b oth cases jV U J j

t t

and jV U J j we can deduce that N z N v To this end let us

rstly show that in b oth cases we can assume that t has no neighb our on U v

If jV U J j then the contrary should imply by Claim that tr E a

contradiction If jV U J j then V U n fv g fr g and r t E

t t t

In b oth cases we prove now that N v N z Indeed if w N v

N z then Claim for the cycle given by t z U and the vertex w implies since

is a bipartite that w z E notice that w is not in U since t has no other

t t

neighb our on U v as proved b efore Conversely if w N z N v then

again by Claim for the cycle given by t z U and the vertex w implies again

since is bipartite w v E So the neighb ourho o ds of v and z in are identical

Since is a prime graph this is not p ossible

Third case J has exactly two vertices Then recall that s is the neighb our of v

on U Claim for any cycle U J s s with s D s and any t T implies

that ts E so s V Now if we still call J the path J s U the path U s

and we apply the reasoning in the case b efore where J has exactly three vertices

t t

we obtain that N s N z This is a contradiction with the hyp othesis on

t t

s should b e in the mo dule of represented by z in so is proved 2

t t

This metho d for nding with its three steps building will b e

called the rening op eration

Characterization Theorem

One more simple structural result is necessary b efore proving the main theorem

Obviously the statement A which was true for any of the graphs in the inductive

pro cess see Lemma is also true in The statement B and a new one are

proved here b elow

Claim The fol lowing two statements hold for G with the proper skeleton

B If w V and z z V such that z z E w z E w z E

1 2 1 2 1 2

then w D z D z T

1 2

C If t t T such that t t E then N t N t

1 2 1 2 GT S 1 GT S 2

Pro of B is true

Using the statement B for we obtain that w D z D z T Since

1 2

D z D z and D z D z by denition we are done

1 1 2 2

C is true

Obviously if this is not the case there must exist z V and r W z Lz

such that r t E but r t E Take U a chordless path from r to z in the

1 2

subgraph induced by D z W z Lz if it do esnt exist since D z could b e

nonconnected we suitably change the representative vertex z of the set D z in

and supp ose without loss of generality that r is the rst vertex on this path

while going from z to r such that r is adjacent to exactly one of the two vertices

t t Supp ose also that z is the only vertex of D z on U otherwise we can again

1 2

change the representative vertex of D z on Let z b e a neighb our of z on

and notice that b oth z and z are adjacent to t t Then take the cycle induced

1 2

by z t U If it is o dd then Claim for t and U z implies that t is adjacent

1 1 1

to every vertex on U if it is even the same result is obtained by applying Claim

for t and U In b oth cases we must have by the choice of U that t is adjacent

1 2

to any vertex on U except r Now take s the neighb our of r on U and z V

such that z z E Then the cycle given by r s t t z is o dd and has only one

1 2

chord a contradiction by for r z E 2

While p erforming the preceding reasoning the condition for G to b e a Meyniel

graph has b een used lo cally around the skeleton but not globally parts of

W z Lz may have not b een tested Therefore it is not sucient for a graph

to have a prop er skeleton to deduce that it is a Meyniel graph However the graph

may b e reduced by removing some edges of to a smaller with resp ect to the

numb er of edges graph which has the same nature as G Meyniel graph or not

but a smaller numb er of holes and domino es This is done as follows

Consider S a spanning tree of and let G b e the graph obtained from G

by removing all the edges in b etween two sets D z D z with z z V

1 2 1 2

z z and z z E S In other words the structure identied around

1 2 1 2

remains unchanged but is reduced to a spanning tree by removing some of its

edges Nothing is changed inside the sets D z W z Lz T S for any z V

Notice that in this way C is destroyed and p ossibly some other holes or domino es

of G It remains to prove that the nature of the graph do esnt change

Theorem Characterization Theorem The graph G V E is a Meyniel graph

if and only if either it is HHDfree or for any even hole or domino C the conditions

below are simultaneously veried

the graph obtained by the rening operation is a proper skeleton of G

is a Meyniel the reduced graph G with respect to any spanning tree S

graph

Pro of The pro of of has already b een done so we will prove To

this end supp ose the contrary holds and take C v v v v k o dd at

1 2 k 1

least equal to a cycle of G with at most one chord Now since G is a Meyniel

graph at least one chord of the cycle induced by V C exists in G but is removed

in G Say this chord is v v and supp ose without loss of generality that i the

1 i

and case v v E is similar to the case v v E Hence v v is an edge of

1 3 1 k 1 1 i

we can without loss of generality supp ose that v v V Let us distinguish

1 i

two cases

such that x T S There exists x V C

v v

2 i1

If x T then we have x v or x v otherwise xv and xv are two chords

2 i1 1 i

of C in G Supp ose without loss of generality that x v so v v is a chord

2 2 i

and let y V C If y V then y v is the second chord of C and we

v v

i+1

have a contradiction Supp ose that y T Then y v E this would b e a second

chord in C so by the prop erty C N v N y we deduce that v

GT S 2 GT S 3

is not in G T S v y would b e the second chord of C in G therefore v T

3 3

would b e in T and another chord will b e if v S then another vertex on C

v v

2 i1

found But now v v E and we have again a contradiction Since y T we also

3 1

which b elongs to have y S otherwise there necessarily exists a vertex of C

v v

i+1

has all its vertices in W v W v but no T and we can deduce that C

1 i

v v

i+1

vertex of W v is adjacent to a vertex of W v in G This is a contradiction with

1 i

is connected the fact that C

v v

i+1

which b elong to T and we If x S then there exists some vertices of C

v v

2 i1

obtain as previously a contradiction

is completely contained in G T S C

v v

2 i1

V we can nd on V and V C In b oth cases V C

v v v v

D D

2 i1 2 i1

at least two vertices of In the rst case this is obvious since v v C

2 i1

v v

2 i1

W z In the second one at least one z V exists such that V C

v v

2 i1

D z j since any path from W z to Then jV C must have vertices

v v

1 i

then we are done if for instance from D z If b oth these vertices are on C

v v

2 i1

v D z the second vertex v D z j i is not the neighb our of v on C

1 j i

since then v v should have b een deleted We deduce that at least one vertex of

i j

W z v D z since v D z so V C exists on C

i 1

v v v v

j +1 i1 j +1 i1

is a subgraph of G T S otherwise there exists a vertex Thus C

v v

i+1

T which forms two chords with two vertices of V C of V C

v v v v

2 i1 i+1

a contradiction Therefore C is also a subgraph of G T S Notice that at

least two vertices v v of C must b e situated in the same set D z Indeed if

a b

V C V then C or a copy of it cannot b e induced exclusively by vertices

at of since is bipartite the indicated conclusion follows If V C V

least one vertex z V exists such that V C W z This implies as

b efore that jV C D z j

We can supp ose without loss of generality that z v and a b We cho ose a as

small as p ossible so v V D z but take more precautions for cho osing

D z it is b If z v then we necessarily can nd b i such that v V

i b+1

sucient to take the largest b smaller than i such that z D z If z v then

b i

we p ossibly cannot nd such a b smaller than i Then we take b as small as p ossible

D z if b k we denote v v larger than i such that v V

b+1 1 b+1

If v v in C then v v and v v are two chords of C in G and we

a1 b+1 a b+1 b a1

have a contradiction v v v v are not removed in G since v v v v are

a b+1 b a1 a1 a b b+1

not Otherwise that is if v v v we necessarily have v z and then

a1 b+1 1 i

v v v v should have b een removed in G as well as v v since v v v D z

a 1 b 1 1 i i a b

a contradiction

Supp ose now that was nd by the rening op eration and that G is a

Meyniel graph but G is not a Meyniel graph Once more there must exists a cycle

C v v v k o dd at least equal to of G containing at most one chord

1 k 1

Since G is a Meyniel graph C cannot induce a cycle in G it would have at most

one chord to o therefore at least one edge of C say v v is missing in G

1 2

There is no Hence v v corresp onds to an edge z z of not contained in S

1 2 1 2

loss of generality if we supp ose that v v Two cases can o ccur

1 2

There exists x V C such that x T S

v v

If x T then we have x v or x v otherwise xv and xv are two chords

3 k 1 2

of C in G Supp ose without loss of generality that x v so v v is a chord

3 1 3

Now no vertex y of C is in T V votherwise v v and y v resp ectively

1 3 2

v v

W v Lv S In fact no y v are two chords of C in G Then V C

1 1 3

v v

is in S since in this case any path joining it to v should contain a vertex of C

v v

W v Lv and then v D v T Lv In vertex in T Hence V C

1 1 4 1 1

v v

the rst two cases v v and v v are two chords of C in G a contradiction In the

3 1 4 2

Then C v y third one recall that there exists y V such that y v S

2 1

is a cycle of G with only one chord a contradiction

If x S then there exists some vertices of C which b elong to T and we

v v

obtain as previously a contradiction

C is completely contained in G T S

v v

Once more at least two vertices v v of C must b e situated in the same set

a b

D z Indeed if V C V then C or a copy of it cannot b e induced

exclusively by vertices of since is bipartite If V C V at least one

vertex z V exists such that V C W z This implies as b efore that

jV C D z j call z z two vertices in the intersection

a b

We can supp ose without loss of generality that z v so v v and v v

1 a 1 b 1

and a b We cho ose b as large as p ossible if b k we denote v v so v

b+1 1 b+1

D z We cho ose a the largest p ossible smaller than b such that v V

D z If v v or v v then we have a contradiction v v v v V

a 2 b k a b+1 b a1

D v are two chords of C in G Otherwise we have z v and V C

2 2

v v

but v has at least one neighb our W v Lv The edge v v is not an edge of S

2 2 2 1 2

0 0

is not a spanning otherwise S y V such that the edge v y is an edge of S

tree But then C v y is isomorphic to C and is an induced subgraph of G

so G is not a Meyniel graph This contradicts the hyp othesis 2

The Algorithm

The characterization theorem gives us now the p ossibility to indicate a recogni

tion algorithm for Meyniel graphs To this end we successively reduce the numb er

of edges in the initial graph G V E with jV j n jE j m without changing

its nature Finally it is sucient to see if the last graph which will have no holes

or domino es otherwise it could b e simplied again is a Meyniel graph or not This

will also b e the answer G

The complexity of this algorithm will b e of O m mn In order to simplify

the notations in this section the graphs will b e identied with their set of vertices

sometimes represented as a list of disjoint subsets this is the case of

Closely following the theoretical asp ects presented in the previous section we

can describ e the main steps of the algorithm as b elow details on each step are given

afterwards

Recognition Algorithm

A E

while A do

let xy A A A xy

lo ok for an even hole or domino C containing xy

if not found then return G is not a Meyniel graph or go to

nd T

C compute D z for every z C fD z j z C g

with a BFSlike algorithm

grow and as much as p ossible

verify that is a skeleton if not return G is not a Meyniel graph

t t

or return G is not a Meyniel graph compute

not containing xy and G G G A A E G compute S

endwhile

return G is a Meyniel graph

Obviously the complexity is given by the lo op while whose b o dy is executed at

most m times since the cardinality of A reduces at every execution The numb er

of op erations required by the steps strongly dep ends on the data structures

that are used they are supp osed to b e suitable chosen to supp ort easy access For

instance G is represented b oth using the adjacency matrix and the adjacency lists

of neighb ours sorted in ascending order of vertices that can b e done in linear time

using bucketssort see This allows us on one hand to know in constant time

if a given pair of vertices is an edge or not and on the other hand to quickly nd

all the neighb ours of a given vertex For the same reason for any of the sets C T

etc a double representation may b e used when needed

With these general but imp ortant remarks we describ e the realisation of the

steps contained in the lo op in order to have a complexity in O m n for each of

them The most frequently used algorithm is the breathrstsearch that we call

BFS or BFSz when we want to sp ecify the departure vertex z of the traversal

It is well known that BFS needs O m n op erations for every graph H with

H H

m edges and n vertices

H H

Steps and Take initially xy the rst edge in the list Remove the com

mon neighb ours of x y all the edges joining a private neighb our of x to a private

neighb our of y and the edge xy itself In the remaining graph nd if it exists and

using the BFS algorithm a chordless path P x z z z y and notice

1 2 k

that the graph induced by P in G is a cycle of length at least containing xy and

at most one chord If it is o dd G is not a Meyniel graph If it is even either it is a

hole and we are done or a domino and we are done once more or contains the

chordless cycle z z z z of length at least In this last case we forget

2 3 k 1 2

the initial edge xy and recall xy the edge z z The former xy is no more removed

2 k 1

from A but the new one do es All these op erations are easily done in O m n

time

Step Obviously T N x Then it is sucient to verify for every w

N x how many of its neighb ours are in C and then to compare this numb er

with the cardinality of C

Step To compute D z for every z C we rstly nd G T Then for every

z C we p erform the BFS algorithm in this graph starting with z and considering

only vertices which have not yet b een placed in some D z If the current vertex

u is adjacent to the two or three neighb ours of z on C then u is added to D z

and the BFS algorithm continues otherwise u is ignored by BFSz as if it was

not met It is easy to see that any edge uv in G is used at most two times and

the maximum is realized when u and v are in D z D z with z z in this

case b oth the algorithms BFSz and BFSz use this edge But the complexity

remains in O m n

Remark In this way we put in D z the vertices u such that N u N z

C C

while by the denition of D z we should have put the vertices for which the equality

holds In fact by Claim and recal ling that we are working in G T this equality

should hold It wil l be veried later after the nal and al l D z are found by

verifying that al l D z are modules in

Step To nd and all the sets D z we use the following algorithm in

G G T S ie the connected comp onent of C in G T It corresp onds to

the construction we p erformed in the pro of of Lemma Recall that C at the

b eginning of step

aux

while aux do

take z aux aux aux z

compute W z and C z

using BFS nd the set D el of vertices not in D z accessible from W z

if D el then return G is not a Meyniel graph

else remove D el from G

1 2 p

compute the connected comp onents K K K of C z

i i

for i to p do arbitrarily take vertices x K call D x K

i i

fD x i pg fx j i pg

D D i i

aux aux fx j i pg

endwhile

if is not bipartite then return G is not a Meyniel graph

for every z verify if D z is a mo dule in if not return G is not a

Meyniel graph

Remark Some of the K added to and x added to by the algorithm below

D i

may in fact correspond to the sets C z which in Lemma are not connected

to the skeleton This is due to our construction which doesnt test the connection to

the skeleton but only the adjacency to al l D z Therefore we need to nd here

the connected component of containing C and to cal l it again in order to

obtain the skeleton concerned by Lemma This is done in O m n see

An update of is also needed

The algorithm BFS requires O m n op erations for every z recall that

D el D el

the subgraphs are identied in this section with their set of vertices Also the

connected comp onents may b e found again using BFS in O m n since

C (z ) C (z )

all the sets involved here b efore are disjoint the total complexity is in O m n for

time where R D z D el n these op erations Removing D el needs O m

z R R

z z

for every z Once more we obtain a global complexity of O m n for this op eration

The test if is bipartite uses the BFS algorithm whose complexity is linear

Verifying if the sets D z in are mo dules is again an easy task it is sucient

z for any z D z This is easily done for a z to any N to compare N

D D

xed z in O jN z j if the memb ership of a vertex to a set D u is easy to test

suitable data structures exist and if the adjacency lists of neighb ours are sorted

in ascending order of vertices

Step To compute we consider every vertex z and for every z D z

calculate the numb er of its neighb ours in T It must b e exactly jT j for every z to

deduce that z

the O m n algorithm in may b e applied The algorithm in To nd

is used again once to verify whether is connected or not and return G is

not a Meyniel graph if necessary and once to nd the connected comp onent

of containing C In the same time the sets D z are built by putting together

the necessary D z

Step The spanning tree S is easily built using the BFS algorithm the edge

xy will b e ignored since we want S not to contain xy this is not really necessary

but it is more natural to destroy a cycle by removing the edge which generated

it The actualisation of A is done by considering every edge uv in A nding if

they exist z z such that u D z v D z and verifying if z z is an edge

1 2 1 2 1 2

of A or not For a xed uv this is done in constant time

Remark It is easy to see that in step of the algorithm G must be a Meyniel

graph the remaining graph G has no chord less cycle on at least six vertices and

the induced C are not contained in a house To see this let by contradiction

u u u u be a C included in the house H also containing the vertex u adjacent

1 2 3 4 4 5

to u u Then while considering u u A at step we should have obtained G

1 2 3 4

is not a Meyniel graph in step a contradiction

Thus the steps need O m n op erations therefore the whole algorithm

for the Meyniel graphs recognition is done in O m mn

Concluding remarks

Following and we dene the parity graphs resp itriangulated graphs

as the graphs in which every o dd cycle has at least two crossing resp noncrossing

chords It is easily seen that all these graphs are Meyniel graphs therefore the

general structure established in Section is valid for these graphs but it will have

some particular asp ects in any of the two cases

for parity graphs the set T and therefore S must b e empty otherwise any

vertex in t and any P of form an o dd cycle with no crossing chords

for itriangulated graphs the sets D z must have no edges otherwise two

adjacent vertices in this set and a chordless cycle of containing z form an o dd

cycle with no noncrossing chords moreover the graph induced by T must b e a

clique ie every pair of vertices forms an edge

Statements similar to Theorem are valid for b oth the parity and the itrian

gulated graphs yielding

In the case of parity graphs the pro of of Theorem is still valid taking into

account the fact that T is an empty set the two chords found at every step crosses

each other

In the case of itriangulated graphs we can notice that for all z in we have

D z fz g so W z Then and b ecause of Lemma which

guarantees that the sets C z are not connected to in G D z T we

deduce that

P for every pair x y of vertices in all the paths joining them in G T S

are contained in

Now the pro of for the cases where some x T exists on the cycle C is more

simple since T is a clique In the case where C is included in G T S resp ectively

G T S by the prop erty P and b ecause at least two vertices of C are in

we deduce that V C V and this contradicts the fact that is bipartite

As a consequence we can imagine to follow the same reasoning in order to obtain

O m mn recognition algorithms for parity and itriangulated graphs but we

can notice that in step of the algorithm we cannot decide directly that the graph

is a parity graph resp ectively an itriangulated graph Indeed small bad structures

may still b e present a cycle of length ve with two noncrossing edges in the case

of parity graphs and a cycle of length ve with two crossing edges in the case of

itriangulated graphs Therefore in step of the algorithm we should verify that

the remaining graph G do esnt contain resp ectively these two structures The rst

one is easy to nd if it exists in O mn n by considering every vertex and

verifying if the graph induced by its neighb ours is P free or not see For the

second one it is sucient to take any pair of nonadjacent vertices x y of G and to

P or not ie test if the graph induced by their common neighb ourho o d contains a

if this graph is a multipartite graph the mo dular decomp osition algorithm may b e

applied this time O mn op erations are needed

In this way we can approach the recognition of these sub classes using the same

to ols as for the recognition of Meyniel graphs see for dierent approaches

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