0-dual closures for several classes of graphs

0-Dual Closures for Several Classes of Graphs

Ahmed Ainouche1 and Ingo Schiermeyer2

1 UAG-CEREGMIA-GRIMAAG, Campus de Schoelcher, B.P. 7209, 97275 SchoelcherCedex, Martinique (France). e-mail: [email protected] Department of Mathematics, Technical University of Freiberg, D-09596 Freiberg,Germany. e-mail: [email protected]

Abstract. We prove that for almost all sufficient conditions based on degree sums orneighborhood unions of 3-independent sets for a graph G to be hamiltonian imply that the0-dual closure of G is complete. The proofs are very short.

1. Introduction

We use Bondy and Murty for terminology and notation not defined here andconsider simple graphs onlyG ¼ ðV ;EÞ. By a and jwe denote the independence andthe vertex-connectivity number of G. If A;B are disjoint sets of V , we denote byEðA;BÞ the set of edges with an end inA and the other inB and by eðA;BÞ the numberof edges in EðA;BÞ: Also G A½ � is the subgraph induced by A and eðAÞ ¼ jEðG½A�Þj:

The open neighborhood, the closed neighborhood and the degree of a vertex uare denoted NðuÞ ¼ x 2 V jxu 2 Ef g; N ½u� ¼ fug [ NðuÞ and dðuÞ respectively. Avertex x is dominating if dðxÞ ¼ jV j � 1. For S � V and a 2 V nS; we denote byNSðaÞ (dSðaÞ resp.) the set (the number resp.) of neighbors of a in S. For 1 � k � a;we put Ik ¼ fY j Y is a k-independent setg. As in [1], for each pair ða; bÞ of non-adjacent vertices of a graph G we associate

rabðGÞ :¼ dGðaÞ þ dGðbÞcabðGÞ :¼ NGðaÞ [ NGðbÞj jkabðGÞ :¼ NGðaÞ \ NGðbÞj jTabðGÞ :¼ V n NG a½ � [ NG b½ �ð Þ; tab :¼ Tabj jaabðGÞ :¼ 2þ tab

If there is no confusion, we may omit G and/or the subscript ab. We also letdT1 � dT

2 � . . . � dTt be the degree (in G) sequence of vertices of T and denote by

jab the number of internally vertex disjoint paths from a to b. To allow particularconfigurations, we introduce the binary varible eab 2 0; 1f g where eab ¼ 0 if andonly if:

Graphs and Combinatorics (2003) 19:297–307Digital Object Identifier (DOI) 10.1007/s00373-002-0523-y Graphs and

Combinatorics� Springer-Verlag 2003

(i) T 6¼ ;; kab � t þ 1 and dTmaxð1;kab�1Þ ¼ dT

t

(ii) G T½ � ¼ Kr [ ðt � rÞK1 with 1 � r � t(iii) there exists W � NðaÞ [ NðbÞ with Wj j ¼ aab � r such that NðxÞnT ¼ W is

true for all x 2 Kr and dðyÞ � Wj j for all y 2 T nV ðKrÞ if r < t.

Finally we define qabðGÞ :¼ x 2 Tab j dGðxÞ � aab � eabf gj j if T 6¼ ;0 otherwise

�:

Note that dGðxÞ � aab � eab , dGðxÞ þ cab � n� eab:

2. Preliminary Results

In [6], Bondy and Chvatal introduced the concept of the k-closure for severalgraph properties. For hamiltonian graphs the n-closure generalises six earliersufficient degree conditions. In [1], Ainouche and Christofides introduced the0-dual closure C�0ðGÞ as an extension of the n-closure. In [14], the second authorshowed that C�0ðGÞ is complete whenever G satisfies four more sufficient con-ditions for hamiltonian graphs. In particular, the following two results areobtained.

Theorem 2.1. Let G be a 1-tough graph of order n � 3 and satisfying the conditionrab � n� 2 whenever ab =2EðGÞ: Then either c�0ðGÞ is complete or c�0ðGÞ ¼ðsþ 2ÞK2 _ K2s for some s � 2:

Note that the case c�0ðGÞ ¼ ðsþ 2ÞK2 _ K2s is missing in the statement given in[14].

Theorem 2.2. Let G be a j-connected graph of order n � 3 and satisfying thecondition dðaÞ þ dðbÞ þ dðcÞ � nþ j for all triples fa; b; cg in G. Then c�0ðGÞ iscomplete.

The first author improved recently the closure condition given in [1]. Using thisnew version, we prove that for a larger spectra of sufficient conditions forhamiltonians graphs, the corresponding hamiltonian closure is either complete orbelongs to a well defined class of graphs.

The sufficient condition of Theorem 2.2 suggests considering a similar suffi-cient condition by Flandrin et al. [11], namely:

Theorem 2.3. Let G be a 2-connected graph of order n � 3 and satisfying thecondition dðaÞ þ dðbÞ þ dðcÞ � nþ NðaÞ \ NðbÞ \ NðcÞj j for all triples fa; b; cg inG. Then G is hamiltonian.

The two conditions are incomparable. For instance if G ¼ C6; the condition inTheorem 2.2 is not satisfied since dðaÞ þ dðbÞ þ dðcÞ ¼ 6 ¼ n < nþ j while byTheorem 2.3, G is hamiltonian since NðaÞ \ NðbÞ \ NðcÞj j ¼ 0. On the other hand,if G ¼ 2Km;m _ Km, m � 3 an opposite conclusion can be drawn sincedðaÞ þ dðbÞ þ dðcÞ ¼ nþ m ¼ nþ j for all fa; b; cg 2 I3ðGÞ and NðaÞ \ NðbÞ \jNðcÞj ¼ 2m for at least one triple.

298 A. Ainouche and I. Schiermeyer

We believe that c�0ðGÞ is complete if G satisfies the hypothesis of Theorem 2.3but we did not succeed in proving it. It is however easy to see thatdiamðc�0ðGÞÞ ¼ 2: In this paper, we prove that the 0-dual closures of the followingthree direct corollaries of Theorem 2.3 are complete.

Theorem 2.4. (Ainouche [2]). Let G be a 2-connected graph of order n � 3: Supposethat for all X 2 I3ðGÞ there is a vertex u 2 X such that

NðXnfugÞj j þ dðuÞ � n

Then G is hamiltonian.

Theorem 2.5. (Ainouche [2]). Let G be a 2-connected graph of order n � 3 and letX :¼ fx1; x2; x3g 2 I3. If

X3i¼1

NðXiÞj j > 2ðn� 2Þ; where Xi :¼ Xnfxig; i ¼ 1; 2; 3:

then G is hamiltonian.

Theorem 2.6. (Flandrin et al. [11]). Let G be a 2-connected graph of order n � 3:Suppose that for each set fa; bg 2 I2 we have

3cab þmax 2; kabf g > 2ðn� 1Þ:

Then G is hamiltonian.All these three last conditions are sharp (consider for instance the graph

3Kp þ K2 with p � 1). They were generalized in [2] to larger independent sets.Theorems 2.4, 2.5 and 2.6 are incomparable as we can see by the following smallgraphs. For instance if G ¼ C6, it is not possible to conclude that G is hamiltonianby Theorem 2.4. If G ¼ Km;m with m � 4; the condition of Theorem 2.5 is notsatisfied. Finally if G ¼ ðK2 [ P3Þ _ K2; it is easy to check that only the conditionof Theorem 2.6 is not satisfied.

3. Main Results

To prove our main results, we use the following strong closure condition forhamiltonian graphs, obtained as a relaxation of the main condition proved in [3].This condition improves the one given in [1].

Lemma 3.1 ([3]). Let G be a graph. If

jab � 2 and qab � minðtab; aab � kabÞ ðCCÞ

then G is hamiltonian if and only if ðGþ abÞ is hamiltonian.

0-Dual Closures for Several Classes of Graphs 299

Condition ðCCÞ has two strong relaxations: a degree closure condition ðDCCÞinvolving the degree sum of fa; bg and a neighborhood closure condition ðNCCÞ,involving the neighborhood union of fa; bg: More precisely, we have:


qab � aab � kab ðor equivalently rab þ qab � nÞ ðDCCÞ

then G is hamiltonian if and only if ðGþ abÞ is hamiltonian.Note that ðDCCÞ reduces to Bondy-Chvatal’s closure condition if qab ¼ 0.


jab � 2 and qab ¼ tab ðor equivalently cab þminð2; jabÞ þ qab � nÞ ðNCCÞ

then G is hamiltonian if and only if ðGþ abÞ is hamiltonian.As usual, the 0-dual closure c�0ðGÞ is the graph obtained from G by successively

joining nonadjacent vertices satisfying one of the above conditions. The closurec�0ðGÞ is well defined. We shall write dc�0ðGÞ and nc�0ðGÞ if ðDCCÞ and ðNCCÞ arerespectively used instead of ðCCÞ for the construction of the closure. As a directconsequence of the above lemmas we have.

Theorem 3.4. Let G be a 2-connected graph. If nc�0ðGÞ is complete then G is ham-iltonian.

Computing the 0-dual closure nc�0ðGÞ, we obtain the following.

Theorem 3.5. If G satisfies the hypothesis of Theorem 2.4 for all X 2 I3 then nc�0ðGÞis complete and hence G is hamiltonian.



It happens that Theorems 2.4, 2.5 and 2.6 cover together a large spectra ofknown sufficient conditions for hamiltonian graphs. Thus Theorem 3.4 general-izes all the following sufficient conditions.

Theorem 3.8. Let G be a 2-connected graph of order n � 3: Then nc�0ðGÞ is completeif G satisfies one of the following conditions:

1. (Ainouche [2])P3

i¼1 NðXiÞj j þP

xi2X dðxiÞ > 3ðn� 1Þ for all X 2 I3ðGÞ2. (Ainouche [2]) NðX Þj j þ

Pxi2X dðxiÞ > 2ðn� 1Þ for all X 2 I3ðGÞ

3. (Tian [13])P3

i¼1 NðXiÞj j > 2ðn� 1Þ holds for all X 2 I3ðGÞ


4. (Bondy [5]) 2P

xi2X dðxiÞ > 3ðn� 1Þ holds for all X 2 I3ðGÞ5. (Flandrin et al., [11]) cab þmax dðaÞ; dðbÞf g � n holds for all fa; bg 2 I2ðGÞ6. (Ainouche [2]) 3cab > 2ðn� 2Þ holds for all fa; bg 2 I2ðGÞ7. (Chen, [7]) 2cab þ rab > 2ðn� 1Þ holds for all fa; bg 2 I2ðGÞ8. (Fraisse [12] and Faudree et al. [10] ) 3cab > 2ðn� 1Þ holds for all fa; bg 2 I2ðGÞ9. (Ainouche and Christofides [1]) cab þmin dðxÞ j x 2 Tabf g � n for allfa; bg 2 I2ðGÞ

10. (Ainouche and Christofides [1], Faudree et al. [9]) cab þ dðGÞ � n for allfa; bg 2 I2ðGÞ

11. (Ainouche and Christofides [1]) cab þ j � n for all fa; bg 2 I2ðGÞ12. (Ore [15]) rab � n for all fa; bg 2 I2ðGÞ

We would like to point out that the sufficient conditions of Theorems 2.4, 2.5, 2.6and 3.8 imply more than hamiltonicity. As proved in [3], the satisfaction ofcondition ðCCÞ implies that for any S � V with a; b 2 S; G is S-cyclable if and onlyif Gþ ab is S-cyclable. Moreover cðG; SÞ ¼ cðGþ ab; SÞ where cðG; SÞ denotes thelength of a longest cycle containing S.

In Fig. 1 is depicted the hierarchy among the different sufficient conditionsconsidered in this paper. The proofs of the different links are given within theproof of Theorem 3.8 in section 4.

Fig. 1


Allowing exceptional classes of graphs and considering the 0-dual closure weobtain a strong version of Theorem 2.1 as our last main result.

Theorem 3.9. Let G be a 2-connected graph of order n � 3: If r2 � n� 2 theneither H ¼ Kn; in which case G is hamiltonian or H is one of the following non-hamiltonian graphs:

1. H1 ¼ ðp þ rÞK1 þ Kp; p � 2 and 1 � r � 22. H2 ¼ pK1 [ K2ð Þ þ Kp; p � 33. H3 ¼ rK1 [ sK2ð Þ þ K2; 0 � r; 1 � s and r þ s ¼ 3

4. Proofs

For any set X 2 I3ðGÞ, we set si :¼ u =2X j NX ðuÞj j ¼ if gj j; i ¼ 0; 1; 2; 3: IfX ¼ x1; x2; x3f g, put Xi :¼ Xn xif g and for simplicity rX :¼

Px2X dðxÞ: The

following equalities are easy to derive and they will be useful for our proofs.

NðX Þj j ¼ s1 þ s2 þ s3rX ¼ s1 þ 2s2 þ 3s3

kx1x2 þ kx2x3 þ kx3x1 ¼ s2 þ 3s3P3i¼1 NðXiÞj j ¼ 2rX � ðkx1x2 þ kx2x3 þ kx3x1Þ ¼ 2s1 þ 3s2 þ 3s3

8>>><>>>:

ð1Þ

For the proofs of Theorems 3.5, 3.6 and 3.6, we assume H ¼ nc�0ðGÞ is notcomplete and qab < t must be true whenever ab =2EðHÞ. Moreover there must existat least one triple X 2 I3ðGÞ.

Proof of Theorem 3.5. Choose X :¼ fa; b; ug 2 I3 such that rX is minimum amongall possible elements of I3. Clearly NðXn uf gÞj j þ dðuÞ � n, cab þ dðuÞ �n, dðuÞ � aab: By the choice of X , we have dðuÞ ¼ min dðxÞ j x 2 Tabf g: Thisimplies that ðNCCÞ holds and hence ab 2 EðHÞ. This is a contradiction to ourassumption. (

Proof of Theorem 3.6. Choose X :¼ x1; x2; x3f g 2 I3ðHÞ such that rX is minimum.By hypothesis,

Pxi2X N Xið Þj j � 2n� 3: Moreover NðX Þj j � n� 3: Using (1) we

easily establish

s1 þ s2 þ s3 � n� 3 ð2Þ

2s1 þ 3s2 þ 3s3 � 2n� 3 ð3Þ

Moreover, since X 2 I3ðHÞ; NðXhÞj j þ dðxhÞ � n� 1 is true for all h 2 f1; 2; 3g:Summing these inequalities we get

Xi2f1;2;3g

N Xið Þj j þ rX ¼ 3s1 þ 5s2 þ 6s3 � 3ðn� 1Þ ð4Þ


From (3) and (4) we derive s1 þ 2s2 þ 3s3 ¼ rX � n: From (2) and (3) we derives1 þ 2s2 þ 3s3 � nþ s3: Both inequalities imply s1 þ 2s2 þ 3s3 ¼ n and s3 ¼ 0:With these equalities in hand and using (3) and (4), we get s2 ¼ 3: MoreoverNðX Þj j ¼ s1 þ s2 þ s3 ¼ n� 3; that is

NðX Þ [ X ¼ V : ð5Þ

Next, suppose for instance Nðx1Þ \ Nðx2Þ ¼ ;: Then N X3ð Þj j þ dðx3Þ ¼ rX ¼ n: Asx1x2 =2EðHÞ; we have N X3ð Þj j þ dðx3Þ < n: This is a contradiction. Therefore andby symmetry, we must have

kx1x2 ¼ kx2x3 ¼ kx3x1 ¼ 1 ð6Þ

Set Nðx2Þ \ Nðx3Þ ¼ af g; Nðx3Þ \ Nðx1Þ ¼ bf g and Nðx1Þ \ Nðx2Þ ¼ cf g anddefine Sx1 :¼ Nðx1Þn b; cf g; Sx2 :¼ Nðx2Þn c; af g; Sx3 :¼ Nðx3Þn a; bf g: Clearlyax1x2 ¼ dðx3Þ þ 1 and hence ex1x2 ¼ 0 for otherwise ðNCCÞ holds for x1; x2f g.Therefore x3f g [ Sx3 ¼ Tx1x2 must be a clique and N uð ÞnTx1x2 ¼ a; bf g for allu 2 Tx1x2 : Similarly N uð ÞnTx2x3 ¼ b; cf g for all u 2 Tx2x3 and N uð ÞnTx3x1 ¼ c; af g forall u 2 Tx3x1 . But now, we have in particular Nðx1Þ [ NðaÞ ¼ V n x1; af g, that isaax1 ¼ 2 and hence ax1 2 EðHÞ by ðNCCÞ: This is a contradiction sincea 2 Nðx1Þ \ Nðx2Þ \ Nðx3Þ and hence s3 > 0. (

Proof of Theorem 3.7. We assume that 3cab þmax 2; kabf g > 2ðn� 1Þ holds for allpairs ða; bÞ of nonadjacent vertices in H . Clearly aab � 3 since ab =2EðHÞ and setX :¼ a; b; cf g:We may assume that dðaÞ � dðbÞ � dðcÞ and X is chosen so that rX

is minimum. Moreover we may assume kab ¼ max kab; kbc; kcaf g if dðaÞ ¼dðbÞ ¼ dðcÞ: Suppose first kab � 2: Then 3cab þ kab ¼ 2cab þ rab >2ðn� 1Þ ) 2cab þ 2dðcÞ > 2ðn� 1Þ and hence cab þ dðcÞ � n: By the choice of X ,we have cab þ dT

1 � n and hence ab 2 EðHÞ: This is a contradiction to ourassumption. Next, suppose kab � 1 and set h ¼ 2dðcÞ � dðaÞ þ dðbÞð Þ: Now itis easy to check that cab þ dðcÞ ¼ cab þ dT

1 > ðn� 2Þ þ hþkab2 : If hþ kab � 2; a

contradiction arises as above. So we assume hþ kab � 1: If h ¼ 0 thendðaÞ ¼ dðbÞ ¼ dðcÞ and by assumption, kab ¼ max kab; kbc; kcaf g: In this case wehave 3min cab; cbc; ccaf g > 2ðn� 2Þ and hence, setting a; b; cf g ¼ x1; x2; x3f g; weget

P3i¼1 NðXiÞj j > 2ðn� 2Þ and the proof follows that of Theorem 3.6. Finally

we consider the case where h > 0 and hence kab ¼ 0: In [11], it is proved thatrX � nþ s3 whenever G satisfies the condition of Theorem 3.7. As kab ¼ 0 we getrX ¼ cab þ dðcÞ � n. Therefore cab þ dT

1 � n; implying ab 2 EðHÞ: (

Proof of Theorem 3.8. We shall write i) j to mean that the truth of the state-ment i of Theorem 3.8 implies that of statement j. We sometimes write i)Theorem 3.x to mean that the truth of the statement i of Theorem 3.8 implies thatconditions of Theorem 3.x are satisfied. As a consequence of Theorems 3.5, 3.6and 3.7 we have

Claim 1. (Theorem 2.4) _ (Theorem 2.5) _ (Theorem 2.6) ) Theorem 3.4


As already mentioned, Theorem 2.3 is stronger than Theorems 2.4, 2.5 and 2.6as it is proved below.

Claim 2. (Theorem 2.4) _ (Theorem 2.5) _ (Theorem 2.6) ) Theorem 2.3

Let us consider any X 2 I3ðGÞ:We first note that the condition of Theorem 2.3is equivalent to rX � nþ s3: Considering Theorem 2.4, we observe thatNðXn uf gÞj j þ dðuÞ � n, rX � kXn uf g � n and hence rX � nþ s3: Thus Theo-rem 2.4) Theorem 2.3. Considering Theorem 2.5, suppose by contradiction thatP3

i¼1 NðXiÞj j ¼ 2s1 þ 3s2 þ 3s3 > 2ðn� 2Þ but rX ¼ s1 þ 2s2 þ 3s3 � n� 1þ s3:Then s2 þ s3 � 1: On the other hand, from NðX Þj j ¼ s1 þ s2 þ s3 � n� 3 and2s1 þ 3s2 þ 3s3 > 2ðn� 2Þ we derive s2 þ s3 � 3; a contradiction. ThereforeTheorem 2.5 ) Theorem 2.3. It is already known ([11]) that Theorem 2.6 )Theorem 2.3.

For the remaining we shall prove that all sufficient conditions of Theorem 3.8are corollaries of one or more of the Theorems 2.4, 2.5 and 2.6. It is then obviousthat their neighborhood 0-dual closure is complete.

Claim 3. ð11) 10) 9) 1Þ ) Theorem 2.4

Obviously 11) 10) 9) 1: MoreoverP3

i¼1 NðXiÞj j þ rX > 3ðn� 1Þ )maxi2 1;...;3½ � NðXiÞj j þ dðxiÞf g � n and hence 1) Theorem 2.4.

Claim 4. 1 _ 5ð Þ ) Theorem 2.4

By Claim 3, 1) Theorem 2.4. Next set X :¼ a; b; cf g and assumedðaÞ � dðbÞ � dðcÞ: Then cab þ dðbÞ � n) NðXn cf gÞj j þ dðcÞ � n and hence 5)Theorem 2.4.


The implication is immediate.


Obviously 6 _ 7ð Þ ) Theorem 2.6 since 3cab þ 2 > 2ðn� 1Þ , 3cab > 2ðn� 2Þand 3cab þ kab > 2ðn� 1Þ , 2cab þ rab > 2ðn� 1Þ are both relaxations of3cab þmaxð2; kabÞ > 2ðn� 1Þ.

Claim 7. ð7) 1Þ and ð7) 5Þ

For all X 2 I3ðGÞ; if 2cab þ dðaÞ þ dðbÞ > 2ðn� 1Þ holds for all pairs ða; bÞ ofvertices of X then 2

P3i¼1 NðXiÞj j þ 2rX > 6ðn� 1Þ and hence 7) 1: For all

a; bf g 2 I2ðGÞ, 2cab þ rab > 2ðn� 1Þ ) 2cab þ 2max dðaÞ; dðbÞf g > 2ðn� 1Þ;that is cab þmax dðaÞ; dðbÞf g � n: Thus 7) 5:

Claim 8. 3 _ 4 _ 9ð Þ ) 2) 1


Using(1),wegetP3

i¼1 NðXiÞj j ¼ 2s1 þ 3s2 þ 3s3 � 2s1 þ 3s2 þ 4s3 ¼ NðX Þj jþrX

and hence 3) 2: Next we prove 4) 2: Indeed 2rX > 3ðn� 1Þ ) NðX Þj j þ rX

> 2ðn� 1Þ since 12 2s1 þ 3s2 þ 4s3ð Þ � 2

3 ðs1 þ 2s2 þ 3s3Þ: The implication 9) 2 isobvious. Finally NðX Þj j þ rX ¼ 2s1 þ 3s2 þ 4s3 > 2ðn� 2Þ implies

P3i¼1 NðXiÞj jþ

rX ¼ 3s1 þ 5s2 þ 6s3 > 3ðn� 2Þ; that is 2) 1:

Claim 9. 8 _ 10 _ 12ð Þ ) 7

For all a; bf g 2 I2ðGÞ; 3cab > 2ðn� 1Þ ) 2cab þ rab > 2ðn� 1Þ and hence8) 7: Similarly cab þ dab � n) 2ðcab þ dabÞ > 2ðn� 1Þ ) 2cab þ rab > 2ðn� 1Þand hence 10) 7: Finally rab � n) 2rab > 2ðn� 1Þ ) 2cab þ rab > 2ðn� 1Þand hence 12) 7:

Claim 10. 8) 3ð Þ, 8) 6ð Þ and 12) 4ð Þ

The implications are obvious. (

Proof of Theorem 3.9. For the proof of this Theorem we assume that H is con-structed under ðCCÞ. If H ¼ Kn then G is hamiltonian. Suppose now H notcomplete and choose any pair ða; bÞ of nonadjacent vertices. Set D :¼ NðaÞ \ NðbÞ(and hence Dj j ¼ kabÞ; A :¼ NðaÞnNðbÞ; B :¼ NðbÞnNðaÞ and assume that thevertices x1; x2; . . . ; xtf g of T ¼ Tab are labelled so that dðxiÞ ¼ dT

i for alli ¼ 1; . . . ; t: We note that T 6¼ ; since G is 2-connected.

Claim 1. Aj j � 1 and Bj j � 1.

As ab =2EðHÞ, by ðNCCÞ we have cab þ dðx1Þ ¼ dðaÞ þ Bj j þ dðx1Þ < n: On theother hand and by hypothesis we have dðx1Þ þ dðaÞ � n� 2: It follows thatBj j � 1: Similarly Aj j � 1:

Claim 2. D is a set of dominating vertices.

Choose any vertex y of D. Then xy 2 EðHÞ is true for all x 2 T for otherwiseNðyÞnNðxÞ � a; bf g; contradicting Claim 1 (by setting ða; bÞ ¼ ðx; yÞ). ThusNðyÞ T 8y 2 D: Furthermore, if u 2 A then ux =2EðHÞ for otherwise NðuÞnNðbÞ � a; xf g: This in turn implies NðyÞ A 8y 2 D for otherwise NðyÞnNðuÞ � b; xf g: Similarly NðyÞ B 8y 2 D: As Dj j < n

2 for otherwise dðaÞþdðbÞ � n; it is clear that D must be a clique.

Claim 3. Dj j ¼ kab � 2 and EðA;BÞ ¼ ;.

Since G is 2-connected and NðT ÞnT � D by Claim 2, it follows thatDj j ¼ kab � 2: Suppose now, by contradiction uv 2 EðHÞ with u 2 A and v 2 B:Then NðuÞnNðxÞ � b; vf g; a contradiction to Claim 1.

Claim 4. eðT Þ � 1.

We first observe that dT ðxÞ � 1 must be true for all x 2 T for otherwise, settingða; bÞ ¼ a; xð Þ we contradict Claim 1. As ab =2EðHÞ; we have qab þ kab � aab � 1by ðDCCÞ. By hypothesis, dðaÞ þ dðbÞ � n� 2, aab � kab þ 2: Therefore


qab � 1: ð7Þ

Suppose first eðT Þ � 1: Then for all x 2 T such that dT ðxÞ ¼ 1 we havedðxÞ ¼ 1þ Dj j ¼ 1þ kab � 1þ t since by hypothesis, dðaÞ þ dðbÞ � n� 2,aab � kab þ 2 and hence t � kab: Therefore x 2 T j dðxÞ � 1þ t ¼ aab � 1f gj j ¼

x 2 T j dT ðxÞ ¼ 1f gj j � 2eðT Þ > 1: This is a contradiction to (7) if eab ¼ 1:Therefore eab ¼ 0 and hence qab ¼ 0: But then G T½ � ¼ Kr [ ðt � rÞK1; r � 1: Thisimplies r ¼ 1; 2 and eðT Þ � 1:

For the remaining, we assume that a; b are chosen so that Aj j þ Bj j is mini-mum.

Claim 5. H ¼ H1 ¼ ðp þ rÞK1_ Kp, p � 2; 1 � r � 2 if T is an independent set

Suppose that T is an independent set. By the choice of ða; bÞ we deduceA ¼ B ¼ ;; that is NðaÞ ¼ NðbÞ ¼ D. It is then easy to see thatH ¼ ðp þ rÞK1 _ Kp; with p ¼ Dj j � 2; p þ r ¼ T [ a; bf gj j; where t � 1:

Claim 6. H ¼ H2 ¼ ðpK1 [ K2Þ_ Kp, p � 3 if eðT Þ ¼ 1 and t � 3.

Let xy be the only edge of G T½ � and v be any vertex of T n x; yf g. We firstobserve that NðaÞ ¼ NðbÞ ¼ D: Otherwise if NðaÞ 6¼ D and NðbÞ 6¼ D then settingða; bÞ ¼ ðx; vÞ; we contradict the choice of ða; bÞ: If, for instance NðaÞ 6¼ D butNðbÞ ¼ D then setting ða; bÞ ¼ ðv; bÞ; we get the same contradiction. MoreoverNðuÞ ¼ D is true for all u 2 T n x; yf g by Claim 2. It is now obvious thatH ¼ pK1 [ K2ð Þ _ Kp; where p ¼ a; bf g [ T n x; yf gð Þj j � 3:

Claim 7. H ¼ H3 ¼ ðrK1 [ sK2Þ_ K2, 0 � r; 1 � s; r þ s ¼ 3 if t ¼ 2.

Now eðT Þ ¼ 1 and by the choice of ða; bÞ we must have Aj j � 1 and Bj j � 1:Then clearly s ¼ eðT Þ þ Aj jþ Bj j and r ¼ s� 3:

The proof of the Theorem is now complete. (

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Received: February 14, 2002Final version received: November 7, 2002


0-dual closures for several classes of graphs

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