domination in partitioned graphs with minimum degree two

Discrete Mathematics 307 (2007) 1115–1135www.elsevier.com/locate/disc

Domination in partitioned graphs with minimum degree two

Michael A. Henninga,1, Preben Dahl Vestergaardb

aSchool of Mathematical Sciences, University of KwaZulu-Natal, Pietermaritzburg 3209, South AfricabDepartment of Mathematics, Aalborg University, DK-9220 Aalborg East, Denmark

Received 20 February 2004; received in revised form 17 July 2006; accepted 20 July 2006Available online 10 October 2006

Abstract

Let V1, V2 be a partition of the vertex set in a graph G. For i = 1, 2, let �i denote the least number of vertices needed in G todominate Vi . It is known that if G has order n and minimum degree two, then �1 + �2 �2n/3. In this paper, we characterize thosegraphs of order n which are edge-minimal with respect to satisfying the conditions of connected, minimum degree at least two, and�1 + �2 = 2n/3.© 2006 Elsevier B.V. All rights reserved.

MSC: 05C69

Keywords: Domination; Minimum degree two; Partitioned graphs; 3-subdivision

1. Introduction

Let G = (V , E) be a graph with vertex set V = V (G) and edge set E = E(G), and let S ⊆ V . If U ⊆ V andevery vertex of U is in S or adjacent to a vertex in S, then S is said to dominate U . In particular, if U = V , then S is adominating set of G. The domination number of G, denoted by �(G), is the minimum cardinality of a dominating set.For U ⊆ V , we let �G(U) denote the minimum cardinality of a set of vertices in G that dominates U . If U = ∅, wedefine �G(U) = 0. Domination in graphs is now well studied in graph theory (see [3,4]).

By a partition of the vertices of a graph G, we shall mean two disjoint subsets V1, V2 of V (G) with V (G)=V1 ∪V2and V1 ∩ V2 = ∅; {V1, V2} = {∅, V (G)} is permitted. For i = 1, 2, if the graph G and the sets V1 and V2 are clear fromthe context, we shall denote �G(Vi) simply by �i , and �(G) by �.

A motivation given by Hartnell andVestergaard [2] for studying domination in partitioned graphs is that of minimizingthe number of file servers needed to serve workstations in a computer network. Some files (such as data and text files)are compatible with all computers on the network while other files (such as binary code files) are only compatiblewith a particular type of workstation (Mac, Sun, Sparc, for example). The file server ordinarily is shared among theworkstations and is located either at the workstation it caters to or one communication step away. Such a workstation issaid to be dominated by the file server. The computer network can be modeled by a graph where the vertices representthe workstations and the file servers and where the edges represent the connections between the workstations and theirlocal file servers. The requirement that the file server is, or is adjacent to, the workstation it serves corresponds in

E-mail addresses: [email protected] (M.A. Henning), [email protected] (P.D. Vestergaard).1 Research supported in part by the South African National Research Foundation and the University of KwaZulu-Natal.

0012-365X/$ - see front matter © 2006 Elsevier B.V. All rights reserved.doi:10.1016/j.disc.2006.07.024

http://www.elsevier.com/locate/disc

mailto:[email protected]

mailto:[email protected]

1116 M.A. Henning, P.D. Vestergaard / Discrete Mathematics 307 (2007) 1115–1135

the associated graph to demanding that some collection of text file servers dominate every vertex of the graph, sinceeach workstation must have access to common requirements of data, text, latex, emacs files, internet connection, openwindows, dos, etc. We are interested in workstations of two types, type 1 and 2, and for each i = 1, 2, some collectionof file servers must dominate every vertex of the graph of type i. We wish to minimize the number of file servers; thatis, we wish to minimize the sum � + �1 + �2 where � is the number of text file servers needed so that every workstationhas access to data, text, internet connection, latex, emacs files, and �i is the number of specialized file servers neededfor workstations of type i.

Hartnell and Vestergaard [2] established the following result.

Theorem 1 (Hartnell and Vestergaard [2]). If G is a graph of order n each component of which has order at least 3,then for any partition V1 and V2 of V (G), � + �1 + �2 �5n/4.

For a graph G, Tuza and Vestergaard [9] defined f (G) by

f (G) = max{�G(V1) + �G(V2) | V1, V2 is a partition of V (G)}.

Hence, we are interested in the sum of the least number of vertices needed in G to dominate V1 and the least numberof vertices needed in G to dominate V2. When one of the sets V1 or V2 is empty, f (G) is precisely the dominationnumber �(G) of the graph G. Hence, f (G)��(G) for every graph G. Tuza and Vestergaard [9] proved the followingresult.

Theorem 2 (Tuza and Vestergaard [9]). If G is a graph of order n each component of which has order at least 3, thenf (G)�4n/5 with equality if and only if G is obtained from a graph H by attaching two disjoint paths of length two toeach vertex of H.

If we restrict �(G), the minimum degree of G, to be at least two, then Seager [8] showed that the upper bound ofHartnell and Vestergaard in Theorem 1 can be improved:

Theorem 3 (Seager [8]). If G is a graph of order n with �(G)�2, then for any partition V1 and V2 of V (G),� + �1 + �2 �n.

As an immediate consequence of Theorem 3, Tuza and Vestergaard [9] established the following result.

Corollary 4 (Tuza and Vestergaard [9]). If G is a graph of order n with �(G)�2, then f (G)�2n/3. Furthermore, iff (G) = 2n/3, then �(G) = n/3 and for every partition V1, V2 of V (G) with �1 + �2 = 2n/3, �1 =�2 = n/3.

Proof. Let V1, V2 be a partition of V (G). If � > n/3, then, by Theorem 3, �1 + �2 �n − � < 2n/3. On the other hand,if ��n/3, then, since �i �� for i = 1, 2, �1 + �2 �2n/3. The desired result follows. �

As shown in Corollary 4, for every graph G with minimum degree at least two, f (G) is bounded above by two-thirdsthe order of the graph. In order to characterize the graphs achieving this bound, we define a graph G as a 2

3 -minimalgraph if G is edge-minimal with respect to satisfying the following three conditions:

(i) �(G)�2,(ii) G is connected, and

(iii) f (G) = 2n/3,

where n is the order of G. Our aim in this paper is to characterize 23 -minimal graphs. Similar results on the domination

and total domination numbers may be found in [5–7].

M.A. Henning, P.D. Vestergaard / Discrete Mathematics 307 (2007) 1115–1135 1117

(a) (b) (c)

Fig. 1. (a) A dumb-bell D(5, 5, 2); (b) A daisy D(5, 5); (c) A cycle C12.

Fig. 2. The graphs F1 and F2.

2. Notation

For notation and graph theory terminology we in general follow [1]. Specifically, let G = (V , E) be a graph withvertex set V of order n = |V | and edge set E of size |E|, and let v be a vertex in V . The open neighborhood of V

is N(v) = {u ∈ V | uv ∈ E} and the closed neighborhood of v is N [v] = {v} ∪ N(v). For a set S of vertices, theopen neighborhood of S is defined by N(S) = ∪v∈SN(v), and the closed neighborhood of S by N [S] = N(S) ∪ S. Theminimum degree among the vertices of G is denoted by �(G).

A 2-packing in a graph G is a set of vertices every two of which are at distance at least 3 apart in G. For notationalconvenience, we shall refer to a 2-packing simply as a packing throughout this paper. A maximum packing is a packingof maximum cardinality in the graph.

We define an elementary 3-subdivision of a nonempty graph G as a graph obtained from G by subdividing an edge ofG three times. A 3-subdivision of G is a graph obtained from G by a succession of elementary 3-subdivisions (includingthe possibility of none).

A cycle on n vertices is denoted by Cn and a path on n vertices by Pn.For n, m�3 and k�0, we define a dumb-bell D(n, m, k) to be the graph obtained from the disjoint union, Cn ∪Cm,

of Cn and Cm by joining a vertex of Cn to a vertex of Cm and subdividing this edge k times. Hence, D(n, m, k) hasorder n + m + k and size n + m + k + 1. A dumb-bell D(5, 5, 2) is shown in Fig. 1(a).

We define a daisy as a graph that can be constructed from two disjoint cycles by identifying a set of two vertices, onefrom each cycle, into one vertex. If the two cycles have lengths m and n, respectively, we denote the daisy by D(m, n).A daisy D(5, 5) is shown in Fig. 1(b).

3. The families F, G and H

In this section, we define three families of 23 -minimal graphs.

3.1. The family F

For k�1, letFk be the family of all graphs obtained from a cycle C3k+1 by joining two vertices at distance congruentto 2 modulo 3 apart on the cycle by a path of length 3. If F1 and F2 are the graphs shown in Fig. 2, then we observethat for k ∈ {1, 2}, Fi = {Fk}. Let F = {F | F ∈ Fk for some k�1}.


Fig. 3. A graph G in the family G with seven units.

3.2. The family G

We will refer to a graph G as a reduced graph if G has no induced path on six vertices, the internal vertices of whichhave degree 2 in G. Hence if u, v1, v2, v3, v4, v is a path in a reduced graph G, then degGvi �3 for at least one i,1� i�4 or uv ∈ E(G).

In this section, we define a family G of reduced 23 -minimal graphs. For this purpose, we define a unit to be a graph

that is isomorphic to a dumb-bell D(5, 5, k), where k ≡ 2 (mod 3) or to a daisy D(5, 5) or to a cycle Cn, wheren ≡ 0 (mod 3). There are three types of units. We call a unit type (a) or type (b) or type (c) according to whether it is adumb-bell or a daisy or a cycle, respectively.

In each unit, we define link vertices of the unit as follows. In a type (a) unit, if u and v denote the vertices of degree3 in the unit, then we call each vertex in a maximum packing of the u–v path a link vertex of the unit (and so eachvertex of the u–v path at distance congruent to 0 modulo 3 from u is a link vertex of the unit—in particular, both u

and v are link vertices). In a type (b) unit, we call the vertex of degree 4 in the unit the link vertex of the unit. In atype (c) unit, we select one of the three maximum packings in the unit (each such packing consists of a vertex v of thecycle and all vertices at distance congruent to 0 modulo 3 from v) and we call each vertex in this maximum packinga link vertex of the unit. Fig. 1 depicts the three types of units with the link vertices indicated by the large darkenedvertices.

Let G denote the family of all connected graphs that are obtained from the disjoint union of k�1 units (called theunits of G) by adding k − 1 edges joining link vertices of the units in such a way that

(i) every added edge is a bridge of G (which we call a link edge of G), and(ii) every link vertex, if any, of degree two in its unit is incident to at least one added edge.

In particular, we note that if G ∈ G, then the set of link vertices in G is precisely the set of vertices of degree at least3 in G. We call an edge of G that is not a link edge, a non-link edge of G.

A graph G in the family G with seven units (one of type (a), two of type (b), and four of type (c)), and therefore sixlink edges, is shown in Fig. 3 (the link vertices are indicated by the large darkened vertices).

3.3. The family H

Let H be the family consisting of all graphs H that can be obtained from a graph G in the family G by a successionof elementary 3-subdivisions of non-link edges of G, including the possibility of none. Hence if H ∈ H, then thereexists a sequence of graphs H0, . . . , Hk such that G = H0 and G ∈ G, H = Hk , and if k�1, then for i = 1, . . . , k,Hi is obtained from Hi−1 by an elementary 3-subdivision of an edge incident to a vertex of degree two in Hi−1, i.e., anon-link edge of G.


4. Main results

We shall prove:

Theorem 5. A graph G is a reduced 23 -minimal graph if and only if G ∈ G ∪ {F1}.

Theorem 6. A graph H is a 23 -minimal graph if and only if H ∈ H ∪ F.

5. Preliminary results

Our aim in this section is to establish some preliminary results that we will need later when proving our main result.We begin by determining f (G) when G is a cycle or a path.

Proposition 7. For n�3, f (Cn) = �2n/3.

Proof. Let G be the cycle v1, v2, . . . , vn, v1. Let A = {vi | i ≡ 1 (mod 3)}. If n ≡ 0 (mod 3), then let V1 = A. Ifn ≡ 1 (mod 3), then let V1 = A − {vn}. If n ≡ 2 (mod 3), then let V1 = A − {vn−1}. Let V2 = V (G) − V1. Sinceany two vertices of V1 are at distance at least 3 apart in G, we need at least |V1| = �n/3 vertices to dominate V1.On the other hand, we need at least �n/3 vertices to dominate V2 if n /≡ 2 (mod 3) and at least �n/3 + 1 verticesto dominate V2 if n ≡ 2 (mod 3). Hence if n /≡ 2 (mod 3), then �1 + �2 �2�n/3 = �2n/3, while if n ≡ 2 (mod 3),then �1 + �2 �2�n/3 + 1 = �2n/3. In all cases, �1 + �2 ��2n/3, and so f (G)��2n/3. However, by Corollary 4,f (G)��2n/3. Consequently, f (G) = �2n/3. �

Proposition 8. For n�2, f (Pn) = 2n/3�.

Proof. Let G be the path v1, v2, . . . , vn. Let V1 = {vi | i ≡ 1 (mod 3)} and let V2 = V (G) − V1. Since any twovertices of V1 are at distance at least 3 apart in G, we need at least |V1| = n/3� vertices to dominate V1. On theother hand, we need at least �n/3 vertices to dominate V2 if n /≡ 2 (mod 3) and at least n/3� vertices to dominateV2 if n ≡ 2 (mod 3). Hence if n /≡ 2 (mod 3), then �1 + �2 �n/3� + �n/3 = 2n/3�, while if n ≡ 2 (mod 3), then�1 + �2 �2n/3� = 2n/3�. In all cases, �1 + �2 �2n/3�, and so f (G)�2n/3�.

To show that f (G)�2n/3�, let U1, U2 be an arbitrary partition of V (G). Let D = {vi | i ≡ 2 (mod 3)}. Ifn ≡ 0 (mod 3), then D dominates Ui for i = 1, 2. If n ≡ 1 (mod 3), then we may assume that vn ∈ U1, and so D ∪ {vn}dominates U1 while D dominate U2. If n ≡ 2 (mod 3), then D dominates Ui for i = 1, 2. In all cases, it follows that�G(U1) + �G(U2)�2n/3�, and so f (G)�2n/3�. Consequently, f (G) = 2n/3�. �

Lemma 9. If G is a 23 -minimal graph, then any edge joining two vertices of degree at least 3 in G is a bridge.

Proof. Suppose e=uv ∈ E(G) where degGu�3 and degGv�3. Then, �(G−e)�2, and so by Corollary 4, 23 |V (G)|=

f (G)�f (G − e)� 23 |V (G − e)| = 2

3 |V (G)|. Consequently, f (G − e) = 23 |V (G − e)|. Hence, by the minimality of

G, G − e must be disconnected. �

Lemma 10. If G′ be a connected nontrivial graph and G is obtained from G′ by an elementary 3-subdivision, thenf (G)�f (G′) + 2.

Proof. Suppose e = uv is the edge of G′ that is subdivided three times to produce G. Let u, a, b, c, v be the resultingu–v path of length 4 in G. Let V1, V2 be any partition of V (G) and let V ′

1 = V1 − {a, b, c} and V ′2 = V2 − {a, b, c}.

For i = 1, 2, let D′i be a smallest set of vertices of G′ that dominate V ′

i . Then, |D′1| + |D′

2|�f (G′). If u, v /∈ D′1, then

D1 = D′1 ∪ {b} dominates V1 in G, while if u or v is in D′

1, say v ∈ D′1, then D1 = D′

1 ∪ {a} dominates V1 in G. Inany event, D1 dominates V1 in G and |D1| = |D′

1| + 1. Similarly, we can add a vertex to D′2 to form a set D2 that

dominates V2 in G and such that |D2| = |D′2| + 1. Hence, �G(V1) + �G(V2)� |D1| + |D2|�f (G′) + 2. Since V1, V2

is an arbitrary partition of V (G), it follows that f (G)�f (G′) + 2, as desired. �

As a consequence of Lemma 10, we have the following result.


Lemma 11. Let G′ be a connected nontrivial graph with �(G′)�2 and let G be obtained from G′ by an elementary3-subdivision. If G is a 2

3 -minimal graph, then so too is G′.

Proof. Let G have order n, and so f (G) = 2n/3. Then, G′ has order n − 3. By Corollary 4, f (G′)�2(n − 3)/3.Thus, by Lemma 10, 2n/3 = f (G)�f (G′) + 2�2n/3. Consequently, we must have equality throughout this chain.In particular, f (G′) = 2

3 |V (G′)|, and so G′ is a 23 -minimal graph. �

We note that the converse of Lemma 11 is not necessarily true. For example, if G′ is a dumb-bell D(3, 3, 0), thenG′ is a 2

3 -minimal graph. However, the graph G = D(3, 3, 3) obtained from G′ by subdividing the bridge of G′ threetimes satisfies f (G) = 5 < 6 = 2

3 |V (G)|.

Lemma 12. Let H be a graph with �(H)�2 and let v ∈ V (H). Let G be the graph obtained by adding a path Pm

where 2�m�4 to H and joining its ends to v. Suppose f (G) = 23 |V (G)| and V1, V2 is a partition of V (G) with

�G(V1) + �G(V2) = f (G).

(a) If m ∈ {2, 3}, then each of V1 and V2 contains a vertex of the path Pm.(b) If m = 4, then V1 or V2 contains exactly one vertex of Pm and this vertex is at distance 2 from v.

Proof. Let H have order n and let Pm: v1, v2, . . . , vm be the path added to H. For i = 1, 2, let V ′i = Vi ∩ V (H) and let

D′i be a smallest set of vertices of H that dominate V ′

i . By Corollary 4, |D′1| + |D′

2|�f (H)�2n/3.Suppose m = 2 (and so G has order n + 2). If {v1, v2} ⊆ V1, then 2(n + 2)/3 = f (G) = �G(V1) + �G(V2)� |(D′

1 ∪{v})| + |D′

2|�(2n + 3)/3, which is impossible. Hence, V1 contains at most one vertex of Pm. Similarly, V2 containsat most one vertex of Pm. Consequently, each of V1 and V2 contains a vertex of Pm.

Suppose m = 3 (and so G has order n + 3). If {v1, v2, v3} ⊆ V1, then 2(n + 3)/3 = f (G)� |(D′1 ∪ {v2})| + |D′

2|�(2n + 3)/3, which is impossible. Hence, V1 contains at most two vertices of Pm. Similarly, V2 contains at most twovertices of Pm. Consequently, each of V1 and V2 contains a vertex of Pm.

Suppose m=4 (and so G has order n+4). Then V1 or V2, say V1, contains at least two vertices of Pm. If V1 containsall four vertices of Pm, then 2(n + 4)/3 = f (G)� |(D′

1 ∪ {v2, v3})| + |D′2|�2(n + 3)/3, which is impossible. If V1

contains exactly two vertices of Pm, then we can add one additional vertex to each of D′1 and D′

2 in order to dominate V1and V2, respectively, in G, and so f (G)� |D′

1|+|D′2|+2�f (H)+2�2(n+3)/3, a contradiction. Hence V1 contains

exactly three vertices of Pm. If v1 ∈ V2, then f (G)� |(D′1 ∪ {v3})| + |(D′

2 ∪ {v})|�2(n + 3)/3, a contradiction.Thus, v1 ∈ V1. Similarly, v4 ∈ V1. Hence, V2 contains exactly one vertex ofPm and this vertex is at distance 2from v. �

Note: Lemma 12 remains valid even if degH v = 1 and degH x�2 for all x in V (H) − {v}.

Lemma 13. Let H be a graph with �(H)�2 and let v ∈ V (H). Let G be obtained by adding a nontrivial path to Hand joining its ends to v. Let V1, V2 be a partition of V (G). For i = 1, 2, let Di be a smallest set of vertices of H thatdominate Vi ∩ V (H). If v ∈ D1 ∩ D2, then �G(V1) + �G(V2) < 2

3 |V (G)|.

Proof. Suppose H has order n and G is obtained from H by adding the path Pm: v1, v2, . . . , vm and joining its ends tov. By Lemma 10, it suffices to prove the result when 2�m�4. By Corollary 4, |D1| + |D2|�2n/3. For i = 1, 2, let�i = �G(Vi).

If m = 2, then, since v ∈ D1 ∩ D2 we have that Di dominates Vi for i = 1, 2, and so �1 + �2 � |D1| + |D2|�2n/3 <

2|V (G)|/3. If m = 3, then we may assume that v2 ∈ V1. Thus, D1 ∪ {v2} dominates V1 and D2 dominates V2, and so�1 + �2 � |D1| + |D2| + 1�2n/3 + 1 < 2|V (G)|/3. Finally, if m = 4, then for i = 1, 2, Di ∪ {v2} dominates Vi , andso �1 + �2 � |D1| + |D2| + 2�2n/3 + 2 < 2(n + 4)/3 = 2|V (G)|/3. �

Lemma 14. Let G be a graph of order n with �(G)�2 and f (G) = 2n/3. If u, x, y, v is an induced path in G withdegG x =degG y =2 and degG u�3 and degGv�3, and if V1, V2 is a partition of V (G) with �G(V1)+�G(V2)=f (G),then for i = 1, 2, |Vi ∩ {x, y}| = 1.


Proof. Suppose x, y ∈ V1. Let H =G−{x, y}. Then, �(H)�2 and so, by Corollary 4, f (H)�2|V (G′|/3=2(n−2)/3.For i = 1, 2, let Di be a smallest set of vertices of H that dominate Vi ∩ V (H). Then, D1 ∪ {x} dominates V1 in G andD2 dominates V2 in G, and so �G(V1) + �G(V2)� |D1| + |D2| + 1�2(n − 2)/3 + 1 < 2n/3, a contradiction. Hence,V1 contains at most one of x and y. Similarly, V2 contains at most one of x and y. The desired result follows. �

Lemma 15. Let G be a graph of order n with �(G)�2 and f (G)=2n/3. If C: v, x, w, y is an induced 4-cycle in G withdegG x=degG y=2 and degG v�3 and degG w�3, and if V1, V2 is a partition of V (G) with �G(V1)+�G(V2)=f (G),then for i = 1, 2, |Vi ∩ {x, y}| = 1.

Proof. Suppose x, y ∈ V1. Let F =G−x. Then, F is connected and �(F )�2. By Corollary 4, f (F )�2(n−1)/3. LetD1 be a smallest set of vertices of F that dominate V1 − {x} and let D2 be a smallest set of vertices of F that dominateV2. If |D1 ∩ {v, w}|�1, then D1 also dominates V1 in G, and so �G(V1) + �G(V2)�2(n − 1)/3, a contradiction.Hence we may assume that every such set D1 contains neither v nor w. But then in order for D1 to dominate y ∈ V1,necessarily y ∈ D1 and y must belong to every minimum set of vertices of F that dominates V1 − {x}. But this impliesthat v, w ∈ V1, and so each vertex of C is in V1.

Now let F = (G − {x, y}) ∪ {vw}. Then, F is connected and �(F )�2. By Corollary 4, f (F )�2(n − 2)/3. Let S1be a smallest set of vertices of F that dominate V1 − {x, y} and let S2 be a smallest set of vertices of F that dominateV2. Then, S2 dominates V2 in G. We can now extend S1 to a set that dominates V1 in G as follows: If v ∈ S1, thenadd w to S1, while if v /∈ S1, then add v to S1. Thus, �G(V1) + �G(V2)� |S1| + |S2| + 1�2(n − 2)/3 + 1 < 2n/3, acontradiction. Hence, V1 contains at most one of x and y. Similarly, V2 contains at most one of x and y. The desiredresult follows. �

6. Properties of graphs in F

Using the results established in Section 5, our aim in this section is to establish properties of graphs in the family F.

Lemma 16. Let u and v be the vertices of degree 3 in F1. Then, f (F1)=2n/3=4 and if V1, V2 is a partition of V (F1)

with �1 + �2 = 4, then V1 (say) consists of u, the vertex of degree 2 at distance 2 from u, and one common neighborof u and v. In particular, F1 is a reduced 2

3 -minimal graph. For i = 1, 2, {u, v} is a smallest set of vertices of F1 thatdominates Vi . Furthermore, if t ∈ V (F1), then f (F1 − t) = 3.

Proof. The graph F1 has order n = 6. The partition V1, V2 of V (F1) illustrated in Fig. 4 satisfies �1 + �2 = 4 = 2n/3,and so, by Corollary 4, f (F1)=2n/3. We show that this partition of V (F1) is unique (up to isomorphism). For i =1, 2,�i �� = 2. Let x and y denote the two common neighbors of u and v, and let u, w, z, v denote the u–v path of length2. By Lemma 14, |V1 ∩ {w, z}| = 1. We may assume z ∈ V1 and w ∈ V2. By Lemma 15, |V1 ∩ {x, y}| = 1. But thenu ∈ V1 and v ∈ V2 (for otherwise �1 + �2 �3). This then uniquely determines the sets V1 and V2. It follows that F1 isa reduced 2

3 -minimal graph. The rest of the proof is straightforward to verify. �

Lemma 17. Let G be a 3-subdivision of the graph F1. If G has order n and f (G) = 2n/3, then G ∈ F. Every graphG ∈ F has f (G) = 2|V (G)|/3. If V1, V2 is a partition of V (G) with �G(V1) + �G(V2) = f (G), and if w ∈ V (G),then for i = 1, 2, there exists a set of vertices of cardinality n/3 that contains w and that dominates Vi . Furthermore,there exists a set of vertices of cardinality n/3 that contains both vertices of degree 3 of G and that dominates Vi . Ifw ∈ V (G), then f (G − w)�2n/3 − 1. Up to isomorphism the partition V1, V2 is unique.

Fig. 4. A partition V1, V2 of V (F1) giving �1 + �2 = 4.


Fig. 5. The unique partition V1, V2 of V (G) with �G(V1) + �G(V2) = f (G).

Proof. Let u and v be the two vertices of degree 3 in G. Let x and y be the common neighbors of u and v in F1, andlet u, w, z, v be the u–v path of length 2 in F1.

We show first that no edge on the path u, w, z, v was subdivided when constructing G. Suppose, to the contrary, thatG is a 3-subdivision of F1 that contains an elementary 3-subdivision of at least one edge on the path u, w, z, v. Then,G is a 3-subdivision of the graph H, where H is the graph of order 9 obtained from F1 by an elementary 3-subdivisionof an edge on the path u, w, z, v. Let u, w1, w2, . . . , w5, v denote the u–v path in H of order 7. Let V1, V2 be a partitionof V (H) with �H (V1) + �H (V2) = f (H). For i = 1, 2, �H (Vi)��(H) = 3. By Lemma 15, we may assume x ∈ V1and y ∈ V2. Furthermore, it follows from the proofs of Lemmas 10 and 16, that we may assume u, w5 ∈ V1 andv, w1 ∈ V2. Now, {u, w4} dominates V1 − {w2} while {v, w2} dominates V2 − {w4}. Hence if w2 ∈ V2 or w4 ∈ V1,then �H (V1)�2 or �H (V2)�2, and so f (H)�5. On the other hand, if w2 ∈ V1 and w4 ∈ V2, then either w3 ∈ V1,in which case �H (V2) = 2, or w3 ∈ V2, in which case �H (V1) = 2. In any event, f (H)�5 < 2|V (H)|/3. Hence, byLemma 10, f (G) < 2n/3, a contradiction. Therefore, no edge on the path u, w, z, v was subdivided when constructingG. Hence, G ∈ F.

We show now that every graph G ∈ F of order n has f (G) = 2n/3. Let V1 contain all the vertices at distancecongruent to 0 or 1 modulo 3 from u on the u–v path of G containing x, all the vertices at distance congruent to 0modulo 3 from u on the u–v path of G containing y, and the vertex z. Let V2 contain all the remaining vertices of G.(Such a partition is illustrated in Fig. 5.) Then, it is straightforward to verify that �G(V1) = n/3 and �G(V2) = n/3, andso f (G) = 2n/3. Note that if w ∈ V (G), then there exists a dominating set of G of cardinality n/3 that contains w.Furthermore, there exists a dominating set of G of cardinality n/3 that contains both u and v. �

We show next that the partition V1, V2 is unique (up to isomorphism). For this purpose, it suffices to show that thispartition is unique for the graph F , where F is the graph of order 12 obtained from F1 by an elementary 3-subdivisionof one edge on the path u, x, v and of one edge on the path u, y, v. Let u, x1, x2, x3, x4, v denote the u–v path in F

that contains x, and let u, y1, y2, y3, y4, v denote the u–v path in F that contains y. Let V1, V2 be a partition of V (F)

with �F (V1) + �F (V2) = f (F ). For i = 1, 2, �F (Vi)��(F ) = 4. It follows from the proofs of Lemmas 10 and 16, thatwe may assume {u, x1, x4, z} ⊆ V1 and {v, y1, y4, w} ⊆ V2. Now, {v, x1, y2} dominates V1 − {x3} while {u, x2, y4}dominates V2 − {y2}. Hence if x3 ∈ V2 or y2 ∈ V1, then �F (V1)�3 or �F (V2)�3, and so f (F )�7. Suppose thenthat x3 ∈ V1 and y2 ∈ V2. If x2 ∈ V1 or if y3 ∈ V2, then �F (V2)�3 or �F (V1)�3, and so f (F )�7. On the otherhand, if x2 ∈ V2 and y3 ∈ V1, then for i = 1, 2, �F (Vi) = 4. Hence the partition V1 and V2 is unique. It follows fromthe proof of Lemma 10 that this partition extends (uniquely) to the graph G. Furthermore, by the above arguments, ifw ∈ V (F), then f (F − w)�2|V (F)|/3 − 1. It therefore follows from Lemma 10 that if w ∈ V (G), then f (G − w)

�2n/3 − 1. �

7. Properties of graphs in G

Using the results established in Section 5, our aim in this section is to establish properties of graphs in the familyG. We begin with an observation, the proof of which is immediate from the way in which graphs in the family G areconstructed.

Observation 18. If G ∈ G has order n, then the set consisting of all the link vertices of G and one vertex from each5-cycle in G, if any, that is not adjacent to a link vertex is a dominating set of G of cardinality n/3.


Fig. 6. A partition of the non-link vertices of a type (a) unit D(5, 5, 8).

Lemma 19. Let G be a type (a) unit of order n. Let u and v denote the two link vertices of degree 3 in G, and let Cu

and Cv denote the cycle containing u and v, respectively. Then,

(i) f (G) = 2n/3. Let V1, V2 be a partition of V (G) with �G(V1) + �G(V2) = f (G). Then,(ii) V1 (say) contains exactly one vertex from Cu −u and this vertex is at distance 2 from u, all vertices on the u–v path

at distance congruent to 1 modulo 3 from u, and exactly three vertices from Cv − v including the two neighbors ofv on Cv , while V2 contains all vertices on the u–v path at distance congruent to 2 modulo 3 from u (as illustratedin Fig. 6);

(iii) For i = 1, 2, we can choose a smallest set of vertices of G that dominates Vi to consist of the link vertices of G, avertex of Cu that is not adjacent to u, and a vertex of Cv that is not adjacent to v.

(iv) If w is any non-link vertex on the u.v path, any vertex on Cu at distance 2 from u or any vertex on Cv at distance2 from v, then f (G − w)�2n/3 − 1. Furthermore, for i = 1, 2, there exists a smallest set Di of vertices of G thatdominates Vi with w ∈ Di .

Proof. Let Cu: u, u1, u2, u3, u4, u and Cv: v, v1, v2, v3, v4, v, and let u=x0, x1, x2, . . . , x3k =v be the u–v path. Thus,n = 3(k + 3). Let X1 = {xi | i ≡ 1 (mod 3)}, X2 = {xi | i ≡ 2 (mod 3)} and S = {xi | i ≡ 0 (mod 3)}. Then, S is the setof link vertices of G, |S| = k + 1 and |X1| = |X2| = k. If V1, V2 is a partition of V (G) with X1 ∪ {u3, v1, v2, v4} ⊆ V1and X2 ∪ {u1, u2, u4, v3} ⊆ V2 (as illustrated in Fig. 6), then �1 + �2 = �G(V1) + �G(V2) = 2(k + 3) = 2n/3, and so,by Corollary 4, f (G) = 2n/3. This establishes (i).

We show next that if S denotes the set of link vertices of G, then this partition of V (G)−S is unique up to isomorphism.Let V1, V2 be a partition of V (G) with �1 + �2 = f (G) = 2(k + 3). By the note to Lemma 12, we may assume thatu3 ∈ V1 and {u1, u2, u4} ⊂ V2. Suppose V1 contains exactly one vertex of Cv − v. By Lemma 12, we may assumethat v3 ∈ V1. But then X2 ∪ {u4, v4} dominates V1 in G, and so �1 �k + 2, whence �1 + �2 �2k + 5 = 2n/3 − 1,a contradiction. Hence, by Lemma 12, V2 contains exactly one vertex of Cv − v and this vertex is at distance 2 from v.We may assume that v3 ∈ V2, and so {v1, v2, v4} ⊂ V1. Note now that for i =1, 2, S ∪{u3, v3}, for example, dominatesVi in G, and so �i �k + 3. For i ∈ {1, . . . , k − 1}, let Xi

2 = {x3j−1 | j � i} and Si+1 = {x3j | j � i + 1}. Then a set ofk + 2 vertices, namely for i = 0 the set S1 ∪ {u4, v2} and for i ∈ {1, . . . , k − 1} the set Xi

2 ∪ Si+1 ∪ {u4, v2}, dominatesV1 − {x3i+1} in G, and so if x3i+1 ∈ V2, then �1 �k + 2, whence �1 + �2 �2k + 5 = 2n/3 − 1, a contradiction. Thus,X1 ⊂ V1. Similarly, X2 ⊂ V2. This establishes (ii).

For i = 1, 2, S ∪{u′, v′} is a smallest set of vertices of G (of cardinality k + 3) that dominates Vi where u′ ∈ {u2, u3}and v′ ∈ {v2, v3}. This establishes (iii).

Furthermore, as shown earlier, if w ∈ X1 ∪ X2, then in G − w, �1 + �2 �2n/3 − 1. If w = u3, then X1 ∪ {v, v2}is a set of vertices of G − w (of cardinality k + 2) that dominate V1 − {w}, and so in G − w, �1 + �2 �2n/3 − 1. Ifw = u2, then S ∪ {v2} is a set of vertices of G − w (of cardinality k + 2) that dominate V2 − {w}, and so in G − w,�1 +�2 �2n/3−1. Similarly, if w ∈ {v2, v3}, then in G−w, �1 +�2 �2n/3−1. Finally, X1 ∪{u3, v, v2} dominates V1in G and X1 ∪{u2, u, v4} dominates V2 in G, while X2 ∪{u4, v, v2} dominates V1 in G and X2 ∪{u, u2, v3} dominatesV2 in G. This establishes (iv). �

The proof of the following lemma follows readily from Lemma 12.

Lemma 20. Let G be a type (b) unit of order n = 9 and let v be its link vertex. Then f (G) = 6 (=2n/3), and if V1, V2is a partition of V (G) with �G(V1) + �G(V2) = f (G), then V1 − {v} contains exactly one vertex from the one 5-cycleand this vertex is at distance 2 from v while V2 − {v} contains exactly one vertex from the other 5-cycle and this vertexis at distance 2 from v. Furthermore, for i = 1, 2, we can choose a smallest set of vertices of G (of cardinality 3) that


(b) (c)

Fig. 7. A partition of the non-link vertices of a type (b) or type (c) unit.

dominates Vi to contain v and from each cycle a vertex that is not adjacent to v. If w is a vertex in V (G) − {v} notadjacent to v, then f (G − w) < 6, i.e., f (G − w) < 2n/3.

Lemma 21. Let G ∈ G contain a type (c) unit F . Let S denotes the set of link vertices of G. If V1, V2 is a partition ofV (G) with �G(V1) + �G(V2) = f (G), then for i = 1, 2 the set (Vi ∩ V (F)) − S is a maximum packing in F and theset of link vertices of F is a smallest set of vertices of F that dominate Vi in F .

Proof. By Lemma 14, if x and y are adjacent vertices of F that are not link vertices, then for i = 1, 2, |Vi ∩ {x, y}|= 1.Suppose that the set (V1 ∩ V (F)) − S is not a maximum packing in F . Then there must exist vertices u and w ofV1 ∩ V (F) that are adjacent to a common link vertex v. By construction of graphs in the family G, every neighbor of v

different from u and w, is a link vertex of some unit of G (different from F ). By Observation 18, the set D consistingof S and one vertex from each 5-cycle in G, if any, that is not adjacent to a link vertex is a dominating set of G ofcardinality n/3, where n = |V (G)|. But then since v is adjacent to a vertex of S in some unit different from F , D − {v}still dominates V2 in G, and so �G(V1) + �G(V2)�2n/3 − 1, a contradiction. Hence, (V1 ∩ V (F)) − S is a maximumpacking in F . Similarly, (V2 ∩ V (F)) − S is a maximum packing in F . It follows that, for i = 1, 2, the set of linkvertices of F is a smallest set of vertices of F (of cardinality |V (F)|/3) that dominates Vi in F . �

As an immediate consequence of Observation 18 and Lemmas 19, 20, and 21, we have the following result (Fig. 7).

Lemma 22. If G ∈ G, then G is a reduced 23 -minimal graph. Furthermore, if G has order n and V1, V2 is a partition

of V (G) with �G(V1) + �G(V2) = f (G), then for i = 1, 2, the set consisting of all the link vertices of G and one vertexfrom each 5-cycle in G that is not adjacent to a link vertex is a smallest set of vertices of G (of cardinality n/3) thatdominates Vi . Furthermore, if v is any vertex of G that is not a vertex of a 5-cycle that is adjacent to a link vertex, thenfor i = 1, 2, there exists a smallest set Di of vertices of G that dominates Vi with v ∈ Di .

8. Properties of graphs in H

Our aim in this section is to establish properties of graphs in the family H. We begin with an observation.

Observation 23. Let H be a 3-subdivision of the non-link edges of a graph G ∈ G. Then there is a dominating set ofH of cardinality |V (H)|/3 that contains all the link vertices of G and no vertex of degree 2 in H that is adjacent to alink vertex.

Proof. Let D be a dominating set of H constructed as follows: for each subgraph F of H that is a 3-subdivision of atype (a) unit of G (and so H is a dumb-bell D(3k + 2, 3�+ 2, 3m− 1) of order 3(k + �+m+ 1) for some k, �, m�1),let D ∩ V (F) be a dominating set of F consisting of the maximum packing (of cardinality m + 1) on the path of F

that joins the two link vertices of degree 3 in F and any remaining k + � vertices of D ∩ V (F) that do not belong tothis path (necessarily, such vertices are not adjacent to a link vertex). For each subgraph F of H that is a 3-subdivisionof a type (b) unit of G (and so F is a daisy D(3k + 2, 3� + 2) of order 3(k + � + 1) for some k, ��1), let D ∩ V (F)

consist of any dominating set of F of cardinality k + � + 1 of F that contains the link vertex of G (and therefore noneighbor of the link vertex). For each subgraph F of H that is a 3-subdivision of a type (c) unit of G (and so F is a


Fig. 8. A partition of the vertices of a 3-subdivision of a type (a) unit.

cycle of length congruent to 0 modulo 3), let D ∩ V (F) be a maximum packing of F that consists of the link verticesof G. �

Lemma 24. If H is a graph of order n that is a 3-subdivision of a graph G ∈ G that contains an elementary 3-subdivisionof at least one link edge of G, then f (H)�2n/3 − 1.

Proof. Assume precisely one link edge uv of G is subdivided (a multiple of three times) in constructing H andlet u = u1, u2, . . . , u3k+2 = v denote the resulting path in H. By Observation 23, there is a dominating set D ofH − {u2, u3, . . . , u3k+1} that contains all the link vertices of G and in particular {u1, u3k+2} ⊆ D. Let P denote theu3–u3k path (of order 3k − 2) and let F = H − V (P ). Then, F has order n − 3k + 2 and D ∩ V (F) is a dominatingset of F of cardinality n/3 − k. It follows that f (F )�2n/3 − 2k. By Proposition 8, f (P ) = 2(3k − 2)/3� = 2k − 1.Hence, f (H)�f (F ) + f (P )�2n/3 − 1, as claimed. If several link edges are subdivided, then, by Lemma 10,f (H)�2n/3 − 1. �

The proof of the following lemma is routine (though tedious) and is similar to the proof of Lemma 19, and wetherefore omit the details.

Lemma 25. Let H ∈ H be a graph of order n that is a 3-subdivision of a graph G ∈ G (by definition of H no linkedge is subdivided) and let F be a subgraph of H that is a 3-subdivision of a type (a) unit of G. Let u and v denote thetwo link vertices of degree 3 in F , and let Cu: u, u1, u2, . . . , u3k+1, u and Cv: v, v1, v2, . . . , v3�+1, v denote the cyclecontaining u and v, respectively, in F . Then,

(i) f (F ) = 2|V (F)|/3,(ii) If V1, V2 is a partition of V (H) with �H (V1)+ �H (V2)=f (H), then V1 (say) satisfies V1 ∩ (V (Cu)−{u})={u3i |

i = 1, . . . , k} and V2 satisfies V2 ∩ (V (Cv) − {v}) = {v3i | i = 1, . . . , �}. Furthermore, V1 contains all thosevertices on the u–v path at distance congruent to 1 modulo 3 from u, while V2 contains all vertices on the u–v

path at distance congruent to 2 modulo 3 from u (as illustrated in Fig. 8).(iii) For i = 1, 2, we can choose a smallest set of vertices of H that dominates Vi ∩ V (F) to contain all those vertices

(including u and v) on the u–v path at distance congruent to 0 modulo 3 from u.(iv) If w is any vertex of degree 2 different from the two neighbors of u on Cu and the two neighbors of v on Cv , and

if w is not a vertex on the u–v path at distance congruent to 0 modulo 3 from u, then f (H − w)�2n/3 − 1.Furthermore, for i = 1, 2, there exists a smallest set Di of vertices of H that dominates Vi ∩ V (F) with w ∈ Di .

The proofs of the following two lemmas are routine and we therefore omit the details.

Lemma 26. Let H ∈ H be a graph of order n that is a 3-subdivision of (non-link edges of) a graph G ∈ G andlet F be a subgraph of H that is a 3-subdivision of a type (b) unit of G. Let v denote the link vertex of G in F ,and let C1: v, u1, u2, . . . , u3k+1, v and C2: v, v1, v2, . . . , v3�+1, v denote the two cycles containing v in F . Then,f (F ) = 2|V (F)|/3, and if V1, V2 is a partition of V (H) with �H (V1) + �H (V2) = f (H), then V1 (say) satisfiesV1 ∩ (V (C1) − {v}) = {u3i | i = 1, . . . , k} and V2 satisfies V2 ∩ (V (C2) − {v}) = {v3i | i = 1, . . . , �}. For i = 1, 2,if w is any vertex in V (F) − {u1, u3k+1, v1, v3k+1}, then there exists a smallest set Di of vertices of H that dominatesVi ∩ V (F) with w ∈ Di . Furthermore if w ∈ V (F) − v, then f (H − w)�2n/3 − 1.


Lemma 27. Let H ∈ H be a graph of order n that is a 3-subdivision of (non-link edges of) a graph G ∈ G and let F

be a subgraph of H that is a 3-subdivision of a type (c) unit of G. Let v denote a link vertex of G in F , and let F be thecycle v = v0, v1, v2, . . . , v3k−1, v0. Let X = {vi | i ≡ 0 (mod 3)} (and so X contains all the link vertices of G in F ). IfV1, V2 is a partition of V (H) with �H (V1) + �H (V2) = f (H), then for i = 1, 2 the set (Vi ∩ V (F)) − X is a maximumpacking in F and the set X is a smallest set of vertices of H (and of F ) that dominates Vi in F . For i = 1, 2, if Di is anydominating set of Vi in H, then |Di ∩ (Vi ∩ V (F))|� |V (F)|/3. In particular, f (F ) = 2|V (F)|/3. If w ∈ V (F) − X,then f (H − w)�2n/3 − 1.

As an immediate consequence of Lemmas 25–27, it follows that if H ∈ H has order n, then there exists a partitionV1, V2 of V (H) such that �H (V1)+ �H (V2)�2n/3. Consequently, f (H)=2n/3. Therefore, by construction of graphsin the family H, and by Observation 23, we have the following result.

Lemma 28. If H ∈ H, then H is a 23 -minimal graph. Furthermore, if H has order n and V1, V2 is a partition of V (H)

with �H (V1) + �H (V2) = f (H), then for i = 1, 2, there exists a set of vertices of cardinality n/3 which dominates Vi

and contains all the link vertices of H and no vertex of degree 2 in H that is adjacent to a link vertex. Furthermore, if v

is any vertex of H except a vertex of a cycle with length congruent to 2 modulo 3 that is adjacent to a link vertex, thenfor i = 1, 2, there exists a smallest set Di of vertices of H that dominates Vi with v ∈ Di .

As an immediate consequence of Lemmas 11, 17, 24, and 28 we have the following result.

Lemma 29. If Theorem 5 is true, then so too is Theorem 6.

9. Proof of Theorem 5

The sufficiency follows from Lemmas 16 and 22. To prove the necessary condition of Theorem 5, we proceed byinduction on the order n�3 of a reduced 2

3 -minimal graph. Necessarily, n ≡ 0 (mod 3). If n = 3, then G = C3 ∈ G.Let n�6 and assume the result is true for all reduced 2

3 -minimal graphs of order less than n. Hence, by Lemma 29,if H is a 2

3 -minimal graph of order less than n, then H ∈ F ∪ H. Let G = (V , E) be a reduced 23 -minimal graph of

order n. Further, let V1, V2 be a partition of V (G) such that �G(V1) + �G(V2) = f (G) = 2n/3. If e is an edge of G,then f (G − e)�f (G). Hence, by the minimality of G, we have the following observation.

Observation 30. If e ∈ E, then either e is a bridge of G or �(G − e) = 1.

The next result is a consequence of the inductive hypothesis and Lemma 29.

Observation 31. If G′ is a connected subgraph of G of order n′ < n with �(G′)�2, then either f (G′) < 2n′/3 orG′ ∈ F ∪ H.

Since G is a reduced 23 -minimal graph and n�6, G is not a cycle. Thus, G contains at least one vertex of degree

at least 3. Let S = {v ∈ V | deg v�3}. Each vertex of V − S therefore has degree 2. For each v ∈ S, we define the2-graph of v to be the component of G − (S − {v}) that contains v. So each vertex of the 2-graph of v has degree 2in G, except for v. Furthermore, the 2-graph of v consists of edge-disjoint cycles through v, which we call 2-graphcycles, and paths emanating from v, which we call 2-graph paths.

Since G is a reduced 23 -minimal graph, every 2-graph cycle has length at most five, while every 2-graph path has

length either one or two.Using the inductive hypothesis, and the structure and properties of graphs in the families F and H established in

Sections 6 and 8, we shall prove the following lemma, a proof of which is given in Subsection 9.1.

Lemma 32. If S is not an independent set, then G ∈ G.

By Lemma 32, we may assume that S is an independent set, for otherwise G ∈ G.In the proofs of the lemmas that follow, we shall use the following notation to simply the proofs. If v ∈ S, we let

Gv be the component of G − (S − {v}) that contains v. Thus, Gv consists of 2-graph cycles of v, if any, and 2-graphspaths of v (which have length 1 or 2), if any. If |S|�2, we let Hv = G − V (Gv). For i = 1, 2, let Ri be a smallest setof vertices of Hv that dominates Vi ∩ V (Hv). Then, |R1| + |R2|�f (Hv).


With the assumption that S is an independent set, we shall prove the following lemma, a proof of which is given inSubsection 9.2.

Lemma 33. If v ∈ S has at least two 2-graph cycles, then G = D(5, 5).

As an immediate consequence of Lemma 33, we note that if G is a daisy, then G = D(5, 5). In particular, if |S| = 1,then G = D(5, 5) ∈ G. Hence we may assume that each vertex of S has at most one 2-graph cycle and that |S|�2, forotherwise G ∈ G. With these assumptions (and still n�6) we shall prove the following three lemmas, proofs of whichare given in Subsections 9.3, 9.4 and 9.5, respectively.

Lemma 34. No 2-graph cycle in G is a 3-cycle.

Lemma 35. No 2-graph cycle in G is a 4-cycle.

Lemma 36. If G has a 2-graph cycle, then G = D(5, 5, 2).

By Lemma 36, we may assume that there is no 2-graph cycle in G, for otherwise G = D(5, 5, 2) ∈ G. With thisassumption we shall prove the following lemma, a proof of which is given in Subsection 9.6.

Lemma 37. G = F1.

This completes the proof of Theorem 5. �

9.1. Proof of Lemma 32

Let e = uv be an edge, where u, v ∈ S. By Observation 30, e must be a bridge of G. Let G1 and G2 be thetwo components of G − e where u ∈ V (G1). By Corollary 4, for i = 1, 2, f (Gi)�2|V (Gi)|/3. For i = 1, 2 andj = 1, 2, let Vij = Vj ∩ V (Gi) and let Dij be a smallest set of vertices of Gi that dominates Vij . Then, 2n/3 =f (G)�(|D11|+ |D12|)+ (|D21|+ |D22|)�f (G1)+f (G2)�2n/3. Consequently, we must have equality throughoutthis inequality chain. If follows that for i = 1, 2, Gi is a 2

3 -minimal graph and Vi1 and Vi2 is a partition of V (Gi) with�Gi

(Vi1) + �Gi(Vi2) = 2|V (Gi)|/3. By Observation 31, for i = 1, 2, Gi ∈ F ∪ H.

Suppose G1, G2 ∈ F. By Lemma 17, we can choose D11 and D12 to contain u. It follows that, since f (G2 − v)

�2|V (G2)|/3 − 1 = f (G2) − 1 by Lemma 17, f (G)�f (G1) + f (G2) − 1�2n/3 − 1, a contradiction.Suppose G1 ∈ H and G2 ∈ F. By Lemma 9, u is not a vertex of a cycle congruent to 2 modulo 3 in G1 that is

adjacent to a link vertex of G1 (for otherwise G had a cycle edge joining two vertices of S). Hence, by Lemma 28, wecan choose D11 and D12 to contain u. Thus, as before, we can show that f (G)�2n/3 − 1, a contradiction. Similarly,if G1 ∈ F and G2 ∈ H, we reach a contradiction. Hence we must have G1, G2 ∈ H.

It now follows from Lemmas 25–27, and the fact that f (G) = 2n/3, that u (respectively, v) (i) has degG1u�3

(resp., degG2v�3), i.e., is a link vertex of G1 (resp., G2) or (ii) belongs to a subgraph F of G1 (resp., G2) that is a

3-subdivision of a type (a) unit, lies on the path in F joining the two link vertices of degree 3 in F and is at distancecongruent to 0 modulo 3 from these two link vertices in F or (iii) belongs to a subgraph of G1 (resp., G2) that is a3-subdivision of a type (c) unit and is at distance congruent to 0 modulo 3 from a link vertex in that unit. It follows thatG ∈ H. Since G is a reduced 2

3 -minimal graph, we must have G ∈ G. �Henceforth we assume that S is independent.


Suppose v ∈ S has at least two 2-graph cycles. We proceed with a proof of Lemma 33 by establishing five claims.

Claim 38. No 2-graph cycle of v is a 3-cycle.

Proof. Suppose that v has a 2-graph cycle that is a 3-cycle, say v, v1, v2, v. By Lemma 12, we may assume that v1 ∈ V1and v2 ∈ V2. Let C be a 2-graph cycle of v different from v, v1, v2, v and let F = G − (V (C) − {v}). Then F is a


connected graph with �(F )�2. For i = 1, 2, let Di be a smallest set of vertices of F that dominates Vi ∩ V (F). Inorder to dominate v1, we may assume v ∈ D1 (if vi is in D1, then replace vi in D1 with v). In order to dominate v2,we may assume v ∈ D2. Hence, v ∈ D1 ∩ D2. Thus, by Lemma 13, �G(V1) + �G(V2) < 2n/3, a contradiction. Hence,no 2-graph cycle of v is a 3-cycle. �

Claim 39. If v has two 2-graph cycles, then at most one of them is a 4-cycle.

Proof. Suppose that v has two 2-graph cycles both of which are 4-cycles, say v, v1, v2, v3, v and v, u1, u2, u3, v.By Lemma 12, we may assume that v2 ∈ V1 and v3 ∈ V2. Now either v1 ∈ V1 or v1 ∈ V2. We treat the casev1 ∈ V1 as v1 ∈ V2 is analogous. Let F = G − {u1, u2, u3}. Then F is a connected graph with �(F )�2. ByCorollary 4, f (F )�2(n − 3)/3. For i = 1, 2, let Di be a smallest set of vertices of F that dominate Vi ∩ V (F). Then,|D1| + |D2|�f (F )�2(n − 3)/3. In order to dominate v3, we may assume v ∈ D2.

If u2 ∈ V1, then D1 ∪ {u2} dominates V1 in G and D2 dominates V2 in G, and so �G(V1) + �G(V2)� |D1| + |D2| +1�2n/3 − 1, a contradiction. Hence, u2 ∈ V2. By Lemma 12, we may assume that u1 ∈ V1.

Suppose v has a third 2-graph cycle C. For i = 1, 2, let Ti be a smallest dominating set of Vi ∩ (G − (V (C) − {v})).To dominate u1 and v3 we may assume v ∈ T1 ∩ T2. Then by Lemma 13, f (G) < 2|V (G)|/3, a contradiction. Hence,v, v1, v2, v3, v and v, u1, u2, u3, v are the only 2-graph cycles of v.

If n = 7, then {v1, v} dominates V1 and {u2, v} dominates V2, implying that �G(V1) + �G(V2) = 4 < 2n/3. Hence,n�8 and so degG v�5.

Suppose degG v�6. Let K ′ = G − {v, v1, v2, v3, u1, u2, u3} and let K be the graph obtained from K ′ by adding alledges between neighbors of v in K ′. Then K is a connected graph with �(K)�2. Hence, by Corollary 4, f (K)�2|V (K)|/3 = 2(n − 7)/3. For i = 1, 2, let Si be a smallest set of vertices of K that dominate Vi ∩ V (K) in K. Then,|S1| + |S2|�f (K)�2(n − 7)/3. Now S1 ∪ {v, v1} dominates V1 in G and S2 ∪ {v, u2} dominates V2 in G, and so�G(V1)+�G(V2)� |S1|+|S2|+4�2(n−7)/3+4 < 2n/3, a contradiction. Hence, degG v=5. Since S is independent,v therefore has a 2-graph path P (of length 1 or 2).

Using the notation introduced after the proof of Lemma 32, Hv is a connected graph with �(Hv)�2 (note thatHv =G−{v1, v2, v3, u1, u2, u3}−V (P )). By Corollary 4, |R1|+|R2|�2|V (Hv)|/3. If P has length 1, then R1∪{v, v1}dominates V1 in G and R2 ∪ {v, u2} dominates V2 in G, and so �G(V1) + �G(V2)� |R1| + |R2| + 4�f (Hv) +4�2(n − 8)/3 + 4 < 2n/3, a contradiction. Hence P has length 2. Let P be the path v, w, x. If x ∈ V1, then R1 ∪{v, v1, x} dominates V1 in G and R2 ∪{v, u2} dominates V2 in G, and so �G(V1)+�G(V2)� |R1|+|R2|+5�f (Hv)+5�2(n − 9)/3 + 5 = 2n/3 − 1, a contradiction. If x ∈ V2, then R1 ∪ {v, v1} dominates V1 in G and R2 ∪ {v, u2, x}dominates V2 in G, once again producing a contradiction. We deduce, therefore, that v cannot have two 2-graph cyclesboth of which are 4-cycles. �

Claim 40. Every 2-graph cycle of v is a 5-cycle.

Proof. Suppose v has a 2-graph cycle that is a 4-cycle, say v, v1, v2, v3, v. By Claims 38 and 39, all other 2-graphcycles of v must be 5-cycles. By Lemma 12, we may assume that v2 ∈ V1 and v3 ∈ V2. Let v, w1, w2, w3, w4, v

be a 2-graph cycle of v of length 5. By Lemma 12, we may assume that w2 ∈ V1 and w3 ∈ V2. Furthermore, either{w1, w4} ⊂ V1 or {w1, w4} ⊂ V2.

Suppose {w1, w4} ⊂ V2. Let F = G − {w1, w2, w3, w4}. Then F is a connected graph with �(F )�2. By Corollary4, f (F )�2(n − 4)/3. For i = 1, 2, let Di be a smallest set of vertices of F that dominates Vi ∩ V (F). Then,|D1| + |D2|�f (F )�2(n − 4)/3. In order to dominate v2, we may assume v ∈ D2. But then for i = 1, 2, Di ∪ {w2}dominates Vi in G, and so �G(V1) + �G(V2)� |D1| + |D2| + 2�2(n − 4)/3 + 2 < 2n/3, a contradiction. Hence,{w1, w4} ⊂ V1.

If v has a third 2-graph cycle (necessarily a 5-cycle), then it is readily shown that we may assume v ∈ D1 ∩ D2and, by Lemma 13, we can reach a contradiction. Hence, v, v1, v2, v3, v and v, w1, w2, w3, w4, v are the only 2-graphcycles of v.

If n=8, then {v, v1, w2} dominates V1 and {v, w3} dominates V2, implying that �G(V1)+�G(V2)�5 < 2n/3. Hence,n�9 and so degG v�5.

Suppose degGv�6. Let K ′ =G−{v, v1, v2, v3, w1, w2, w3, w4} and let K be the graph obtained from K ′ by addingall edges between neighbors of v in K ′. Then �(K)�2 and it is easily seen (using a similar proof to that of Claim 39)


that �G(V1)+�G(V2)�f (K)+5�2(n−8)/3+5 < 2n/3, a contradiction. Hence, degG v=5. Since S is independent,v therefore has a 2-graph path P (of length 1 or 2).

Using the notation introduced after the proof of Lemma 32, �(Hv)�2 and it is easily seen (using a similar proof tothat of Claim 39) that either P has length 1 and �G(V1)+ �G(V2)�f (Hv)+ 5�2(n− 9)/3 + 5 < 2n/3 or P has length2 and �G(V1) + �G(V2)�f (Hv) + 6�2(n − 10)/3 + 6 < 2n/3. Both cases produce a contradiction. Hence, v has no2-graph cycle that is a 4-cycle. The desired result now follows from Claim 38. �

Claim 41. There are exactly two 2-graph cycles of v each of which is a 5-cycle.

Proof. Suppose v has at least three 2-graph cycles. By Claim 40, each such cycle is a 5-cycle. By Lemma 12, wemay assume that v has two 2-graph cycles, say v, v1, v2, v3, v4, v and v, w1, w2, w3, w4, v, such that v2 and w2 arethe only vertices of these two cycles in V2 − {v}. Let F = G − {w1, w2, w3, w4}. Then F is a connected graph with�(F )�2. By Corollary 4, f (F )�2(n − 4)/3. For i = 1, 2, let Di be a smallest set of vertices of F that dominatesVi ∩ V (F). Then, |D1| + |D2|�f (F )�2(n − 4)/3. In order to dominate v1 and v3 we may assume v ∈ D1. Butthen for i = 1, 2, Di ∪ {w2} dominates Vi in G, and so �G(V1) + �G(V2)� |D1| + |D2| + 2�2(n − 4)/3 + 2 < 2n/3,a contradiction. �

By Claim 41, there are exactly two 2-graph cycles of v each of which is a 5-cycle. Let v, v1, v2, v3, v4, v andv, w1, w2, w3, w4, v be these two 2-graph cycles of v. Furthermore, it follows from Lemma 12 and the proof of Claim41 that V1 − {v} contains exactly one vertex from one of these 5-cycles and this vertex is at distance 2 from v andV2 − {v} contains exactly one vertex from the other 5-cycle and this vertex is at distance 2 from v. We may assumethat {v1, v3, v4, w2} ⊆ V1 and that {v2, w1, w3, w4} ⊆ V2.

Claim 42. degG v = 4.

Proof. Suppose degG v�5. By assumption, v is not adjacent to a vertex of S.Suppose degG v = 5. Using the notation introduced after the proof of Lemma 32, Hv is a connected graph with

�(Hv)�2. By Corollary 4, f (Hv)�2|V (Hv)|/3. If the 2-graph path of v has length 1, then for i =1, 2, Ri ∪{v, v2, w2}dominates Vi in G, and so �G(V1)+ �G(V2)� |R1| + |R2| + 6�2(n− 10)/3 + 6 < 2n/3, a contradiction. On the otherhand, if the 2-graph path of v has length 2, then let this path be v, x, y. If y ∈ V1, then R1 ∪ {v, v2, w2, y} dominatesV1 in G and R2 ∪{v, v2, w2} dominates V2 in G, and so �G(V1)+ �G(V2)� |R1|+ |R2|+ 7�2(n− 11)/3 + 7 < 2n/3,a contradiction. Similarly, if y ∈ V2 we reach a contradiction. Hence, degGv�6.

Suppose there is a 2-graph path v, x of v of length 1. Let F = G − x. Then, �(F )�2. By Corollary 4, f (F )�2|V (F)|/3 = 2(n − 1)/3. For i = 1, 2, let Di be a smallest set of vertices of F that dominates Vi ∩ V (F). Then,|D1| + |D2|�f (F ). For i = 1, 2, we may choose Di to contain {v, v2, w2}. But then for i = 1, 2, Di also dominatesVi in G, and so �G(V1) + �G(V2)� |D1| + |D2|�2(n − 1)/3 < 2n/3, a contradiction. Hence every 2-graph path of v

has length 2.Suppose �(Hv)�2. By Corollary 4, f (Hv)�2|V (Hv)|/3. For i = 1, 2, we now extend Ri to a set that dominates Vi

in G as follows: add v, v2, w2 to Ri , and for each end-vertex u of a 2-graph path of v, if u ∈ Vi , then add u to Ri . Henceif v has k 2-graph paths (k�2 since degGv�6), then �G(V1) + �G(V2)� |R1| + |R2| + 6 + k�2(n − 9 − 2k)/3 + 6 +k = (2n − k)/3 < 2n/3, a contradiction. Hence, �(Hv)�1.

Let w be a vertex of Hv of degree at most 1. Necessarily, w ∈ S and w is adjacent to the two end-vertices of at leasttwo 2-graph paths of v. Let v, x1, x2 and v, y1, y2 be two 2-graph paths of v where x2 and y2 are adjacent to w. Let G′be obtained from G by deleting the edges vx1 and vy1 and adding the edge x1y1, i.e., G′ = (G − {vx1, vy1}) ∪ {x1y1}.Then, �(G′)�2 and w, x2, x1, y1, y2, w is a 2-graph cycle of w in G′. (Possibly, G′ is disconnected.) By Corollary 4,f (G′)�2|V (G′)|/3=2n/3. For i =1, 2, let Si be a smallest set of vertices of G′ that dominates Vi in G′. As before, fori=1, 2, we may choose Si to contain {v, v2, w2}. Furthermore, by Lemma 12, we may assume that {x1, x2, y2} ⊂ V1andy1 ∈ V2 and therefore that {x1, w} ⊂ S1 (and y2 ∈ S2). Since v ∈ S1 ∩ S2, it follows that S1 − {x1} dominates V1 in G

and S2 dominates V2 in G. Hence �G(V1) + �G(V2)� |S1| + |S2| − 1�f (G′) − 1�2n/3 − 1, a contradiction. Hence,degG v = 4.

By Claim 42, degG v = 4, and so G = D(5, 5). This completes the proof of Lemma 33. �



Suppose that v ∈ S has a 2-graph cycle that is a 3-cycle, say v, v1, v2, v. By assumption this is the only 2-graphcycle of v. By Lemma 12, we may assume that v1 ∈ V1 and v2 ∈ V2. Since n�6, degG v�3. By assumption, v is notadjacent to a vertex of S, and so each neighbor of v distinct from v1 and v2 belongs to a 2-graph path of v.

We show first that there is no 2-graph path of v of length 1. Suppose, to the contrary, that v, x is a 2-graph pathof v of length 1. Thus the neighbor of x different from v belongs to S. Let F = G − x. Then, �(F )�2 (possibly,F is disconnected). By Corollary 4, f (F )�2(n − 1)/3. For i = 1, 2, let Di be a smallest set of vertices of F thatdominate Vi ∩ V (F). We may assume that v ∈ D1 ∩ D2. But then for i = 1, 2, Di dominates Vi in G, and so�G(V1) + �G(V2)� |D1| + |D2|�2(n − 1)/3, a contradiction. Hence every 2-graph path of v has length 2.

Let v, x, y be a 2-graph path of v of length 2. Thus the neighbor of y different from x belongs to S. Let K =G−{x, y}.Then, �(K)�2 (possibly, K is disconnected). By Corollary 4, f (K)�2(n − 2)/3. For i = 1, 2, let Ti be a smallest setof vertices of K that dominate Vi ∩ V (K). We may assume that v ∈ T1 ∩ T2. If y ∈ V1, then T1 ∪ {y} dominates V1 inG while T2 dominates V2 in G, and so �G(V1) + �G(V2)� |T1| + |T2| + 1�2(n − 2)/3 + 1 < 2n/3, a contradiction.Similarly, if y ∈ V2, then adding y to T2 shows that �G(V1) + �G(V2)� |T1| + |T2| + 1 < 2n/3. We deduce, therefore,that no 2-graph cycle of v is a 3-cycle, as desired. �


Suppose v ∈ S has a 2-graph cycle that is a 4-cycle, say v, v1, v2, v3, v. By assumption this is the only 2-graphcycle of v. By Lemma 12, we may assume that v2 ∈ V1 and v3 ∈ V2. Since n�6, degG v�3. By assumption, v is notadjacent to a vertex of S, and so each neighbor of v distinct from v1 and v3 belongs to a 2-graph path of v.

Claim 43. degG v�4.

Proof. Suppose that degG v = 3. Then the graph Hv is connected and �(Hv)�2. If the 2-graph path of v has length1, then R1 ∪ {v, v2} dominates V1 in G and R2 ∪ {v} dominates V2 in G, and so �G(V1) + �G(V2)� |R1| + |R2| +3�2(n − 5)/3 + 3 < 2n/3, a contradiction. On the other hand, if the 2-graph path of v has length 2, then let this pathbe given by v, x, y and let w be the vertex of S adjacent to y. Now, |V (Hv)| = n − 6 and either |R1| + |R2| = 2n/3 − 4or |R1| + |R2|�2n/3 − 5. If |R1| + |R2|�2n/3 − 5, then R1 ∪ {v1, x} dominates V1, while R2 ∪ {v, y} dominatesV2, and so �G(V1) + �G(V2)� |R1| + |R2| + 4�2n/3 − 1, a contradiction. If |R1| + |R2| = 2n/3 − 4, then, byObservation 31, Hv ∈ F ∪ H. Since the set S is independent, the vertex w cannot be a vertex of degree 2 in Hv

adjacent to a vertex of degree at least 3 in Hv . Hence, by Lemmas 17 and 28, we can choose R1 and R2 so thatw ∈ R1 ∩ R2. Thus, R1 ∪ {v1, v} dominates V1 in G, while R2 ∪ {v} dominates V2 in G (since y is dominated by w),and so �G(V1) + �G(V2)� |R1| + |R2| + 3�2n/3 − 1, a contradiction. Hence, degGv�4. �

Claim 44. Every 2-graph path of v has length 2.

Proof. Suppose that v, x is a 2-graph path of v has length 1. Suppose degG v�5. Let F ′ = G − {v, v1, v2, v3, x} andlet F be the graph obtained from F ′ by adding edges between neighbors of v in F ′. Then, �(F )�2 (possibly, F isdisconnected). By Corollary 4, f (F )�2(n − 5)/3. For i = 1, 2, let Di be a smallest set of vertices of F that dominateVi ∩ V (F). Then, D1 ∪ {v2, v} dominates V1 in G and D2 ∪ {v} dominates V2 in G, and so �G(V1) + �G(V2)� |D1| +|D2| + 3�2(n − 5)/3 + 3 < 2n/3, a contradiction. Hence, by Claim 43, degG v = 4. Let P be the 2-graph path of v

different from v, x. Then, Hv = G − (V (P ) ∪ {x1, v1, v2, v3}).If �(Hv)�2, then, by Corollary 4, |R1| + |R2|�2|V (Hv)|/3. If P has length 1, then R1 ∪ {v2, v} dominates V1 in G

and R2 ∪ {v} dominates V2 in G, and so �G(V1) + �G(V2)� |R1| + |R2| + 3�2(n − 6)/3 + 3 < 2n/3, a contradiction.Hence P has length 2. Let P be the path v, y, z. If z ∈ V1, then R1 ∪ {v2, v, z} dominates V1 in G and R2 ∪ {v}dominates V2 in G, while if z ∈ V2, then R1 ∪ {v2, v} dominates V1 in G and R2 ∪ {v, z} dominates V2 in G. In anyevent, �G(V1) + �G(V2)� |R1| + |R2| + 4�2(n − 7)/3 + 4 < 2n/3, a contradiction. Hence, �(Hv) = 1.

Let w be a vertex of Hv of degree 1. Necessarily, w ∈ S and w is adjacent to the two end-vertices of the two 2-graphpaths of v. Let K = G − {v1, v2, v3}. Then, K is a connected graph with �(K)�2 and degKw = 3. For i = 1, 2, let Si

be a smallest set of vertices of K that dominate Vi ∩ V (K).


If both 2-graph paths of v have length 1, then v belongs to a 2-graph cycle of w of length 4 in K. Hence, K /∈F∪H,and so, by Observation 31, f (K)�2|V (K)|/3 − 1 = 2n/3 − 3. For i = 1, 2, Si ∪ {v2} dominates Vi in G, and so�G(V1) + �G(V2)�f (K) + 2 < 2n/3, a contradiction. Hence P has length 2 and so v belongs to a 2-graph cycle of w

of length 5 in K. Let P be the path v, y, z.If |S1| + |S2|�2|V (K)|/3 − 1, then as before we get a contradiction. Hence, f (K) = |S1| + |S2| = 2|V (K)|/3 =

2(n − 3)/3. Since there is a 2-graph cycle in K, the inductive hypothesis implies that K ∈ H. However, since G is areduced 2

3 -minimal graph, so too is K, and so K ∈ G. By Lemma 22, we can choose S1 and S2 to contain v. Hence,S1 ∪{v2} dominates V1 in G and S2 dominates V2 in G, and so �G(V1)+�G(V2)�f (K)+1=2n/3−1, a contradiction.Hence there can be no 2-graph path of v of length 1. �

By Claim 43, degG v�4. By Claim 44, every 2-graph path of v has length 2.

Claim 45. �(Hv)�1.

Proof. Suppose �(Hv)�2. By Corollary 4, f (Hv)�2|V (Hv)|/3. For i=1, 2, we now extend Ri to a set that dominatesVi in G as follows: add v and v1 to R1, add v to R2, and for each end-vertex u of a 2-graph path of v, if u ∈ Vi , thenadd u to Ri . Hence if v has k 2-graph paths (where k�2 since degG v�4), then �G(V1) + �G(V2)� |R1| + |R2| + 3 +k�2(n − 4 − 2k)/3 + 3 + k = (2n + 1 − k)/3 < 2n/3, a contradiction. �

By Claim 45, �(Hv)�1. Let w be a vertex of Hv of degree at most 1. Necessarily, w ∈ S and w is adjacent to thetwo end-vertices of at least two 2-graph paths of v. Let v, x1, x2 and v, y1, y2 be two 2-graph paths of v where x2 andy2 are adjacent to w. Let G′ be obtained from G − {v1, v2, v3} by deleting the edges x1x2 and y1y2 and adding theedges x1y1 and x2y2. Then, �(G′)�2 and in G′, w, x2, y2, w is a 2-graph cycle of w while v, x1, y1, v is a 2-graphcycle of v. Since degG′ w�3 and (by assumption) every neighbor of w has degree 2, it follows that G′ /∈F ∪ H.Hence, f (G′)�2|V (G′)|/3 − 1 = 2n/3 − 3. For i = 1, 2, let Si be a smallest set of vertices of G′ that dominatesVi in G′. For i = 1, 2, we may assume that x1 /∈ Si (otherwise we replace x1 in Si by v) and x2 /∈ Si (otherwise wereplace x2 in Si by w). Similarly, we may assume y1, y2 /∈ Si . But then for i = 1, 2, Si ∪ {v2} dominates Vi in G, andso �G(V1) + �G(V2)� |S1| + |S2| + 2�f (G′) + 2�2n/3 − 1, a contradiction. Hence there can be no 2-graph cycleof length 4 in G. This completes the proof of Lemma 35.


Suppose v ∈ S has a 2-graph cycle. By Lemmas 34 and 35, such a 2-graph cycle is a 5-cycle, say v, v1, v2, v3, v4, v.By assumption this is the only 2-graph cycle of v. By Lemma 12, we may assume that {v1, v2, v4} ⊆ V1 and v3 ∈ V2.Since n�6, degG v�3. By assumption, v is not adjacent to a vertex of S, and so each neighbor of v distinct from v1and v4 belongs to a 2-graph path of v.

Claim 46. If v, x, y is a 2-graph path of v of length 2, then x ∈ V2 and y ∈ V1.

Proof. Consider the graph F = G − {x, y}. Then, �(F )�2. For i = 1, 2, let Si be a smallest set of vertices of F

that dominates Vi ∩ V (F). By Corollary 4, |S1| + |S2|�2|V (F)|/3. We may assume that {v1, v} ⊆ S1 and v4 ∈ S2.If x ∈ V1, then adding y to Vi if y ∈ Vi , shows that �G(V1) + �G(V2)� |S1| + |S2| + 1�2(n − 2)/3 + 1 < 2n/3,a contradiction. Hence, x ∈ V2. Thus, by Lemma 14, y ∈ V1. �

Claim 47. If v, x is a 2-graph path of v of length 1, then x ∈ V2.

Proof. Consider the graph F = G − x. Then, �(F )�2. Letting Si be defined as in the proof of Claim 46, we onceagain may assume {v1, v} ⊆ S1 and v4 ∈ S2. Hence if x ∈ V1, then for i = 1, 2, Si dominates Vi in G, and so�G(V1) + �G(V2)� |S1| + |S2|�2(n − 1)/3 < 2n/3, a contradiction. Hence, x ∈ V2. �

Claim 48. If degG v�4, then �(Hv)�1.


Proof. Suppose �(Hv)�2. Then, by Corollary 4, f (Hv)�2|V (Hv)|/3. Since degG v�4, there are at least two 2-graphpaths (each of length 1 or 2) that contain the vertex v. Suppose there are � such 2-graph paths of length 1 and k such2-graph paths of length 2. Since degG v�4, k + ��2. Adding v and v1 to R1, v and v4 to R2, and for each end-vertexy of a 2-graph path of length 2 of v, if any, adding y to Ri if y ∈ Vi shows that �G(V1) + �G(V2)� |R1| + |R2| + 4 +k�2(n − 5 − � − 2k)/3 + 4 + k = (2n + 2 − 2� − k)/3�2n/3 with strict inequality if � �= 0 or k �= 2. Hence wemust have � = 0 and k = 2.

Let v, x, y and v, u, z be the two 2-graph paths of v, and let N(y) = {x, w}. Then, w ∈ S. By Claim 46, x, u ∈ V2and y, z ∈ V1. If |R1| + |R2| < 2|V (Hv)|/3, then �G(V1) + �G(V2) < 2n/3, a contradiction. Hence, |R1| + |R2| =2|V (Hv)|/3 = 2n/3 − 6. By Observation 31, each component of Hv is in F ∪ H. Since the set S is independent, thevertex w is not adjacent to any vertex of degree at least 3 in Hv . Hence, by Lemmas 17 and 28, we can choose R1 andR2 to contain w. Thus, R1 ∪ {v, v1, z} dominates V1 in G, while R2 ∪ {v, v4} dominates V2 in G (since y is dominatedby w), and so �G(V1) + �G(V2)� |R1| + |R2| + 5 = 2n/3 − 1, a contradiction. Hence we must have �(Hv)�1. �

Claim 49. degG v = 3.

Proof. Suppose that degG v�4. By Claim 48, �(Hv)�1. Let w be a vertex of Hv of degree at most 1. Necessarily,w ∈ S and w is adjacent to end-vertices of at least two 2-graph paths of v. Suppose y and z are end-vertices of two2-graph paths of v that are adjacent to w. If each of the 2-graph paths v–y and v–z has length 1, then Lemma 15 andClaim 47 produce a contradiction. Hence we may assume that the 2-graph path v–y has length 2 and is given by v, x, y.By Claim 46, x ∈ V2 and y ∈ V1. Let P denote the 2-graph path v–z.

Suppose P has length 2. Let P be the path v, u, z. Then, by Claim 46, u ∈ V2 and z ∈ V1. Let F = G − {x, y}. Then,�(F )�2, and so f (F )�2|V (F)|/3. For i = 1, 2, let Si be a smallest set of vertices of F that dominates Vi ∩ V (F).We may assume that {v1, v, w} ⊆ S1 (and v4 ∈ S2). Hence, S1 dominates V1 in G and S2 ∪ {v} dominates V2 in G,and so �G(V1) + �G(V2)� |S1| + |S2| + 1 = 2(n − 4)/3 + 1 < 2n/3, a contradiction. Hence, P has length 1, i.e., P isthe path v, z. By Claim 47, z ∈ V2.

If degG w = 3, then let F = (G−{x, y, z}) ∪ {vw}, while if degG w�4, then let F =G−{x, y, z}. Then, �(F )�2,and so f (F )�2|V (F)|/3. For i = 1, 2, let Ti be a smallest set of vertices of F that dominates Vi ∩ V (F). Wemay assume that {v1, v} ⊆ T1 and v4 ∈ T2. If |T1| + |T2|�2n/3 − 3, then adding w to T1 and v to T2 shows that�G(V1) + �G(V2)�2n/3 − 1, a contradiction. Hence, |T1| + |T2| = 2|V (F)|/3 = 2n/3 − 2. By Observation 31, eachcomponent of Hv is in F∪H. It follows from Lemmas 17 and 28 that we can choose T1 and T2 to contain w. Hence, T1dominates V1 in G and T2 ∪{v}dominates V2 in G, and so �G(V1)+�G(V2)� |T1|+|T2|+1=2(n−3)/3+1=2n/3−1,a contradiction. Hence, degGv = 3 as claimed. �

The proof of Lemma 36 now follows from Claim 50.

Claim 50. If degG v = 3, then G = D(5, 5, 2).

Proof. The graph Hv is connected with �(Hv)�2. By Corollary 4, f (Hv)�2|V (Hv)|/3.Suppose the 2-graph path of v has length 1 and is given by v, x. Let N(x)={v, w}. Then, w ∈ S and degHv

w�2. If|R1|+|R2|�2|V (Hv)|/3−1=2n/3−5, then R1∪{v, v1} dominates V1 in G and R2∪{v, v4} dominates V2 in G, and so�G(V1)+ �G(V2)� |R1|+ |R2|+4=2n/3−1, a contradiction. Hence |R1|+ |R2|=f (Hv)=2|V (Hv)|/3=2n/3−4.By Observation 31, Hv ∈ F ∪ H. Since the set S is independent, the vertex w cannot be a vertex of degree 2 inHv adjacent to a vertex of degree at least 3 in Hv . Hence, by Lemmas 17 and 28, we can choose R1 and R2 so thatw ∈ R1 ∩ R2. Thus, R1 ∪ {v1, v} dominates V1 in G, while R2 ∪ {v4} dominates V2 in G (since x is dominated by w),and so �G(V1) + �G(V2)� |R1| + |R2| + 3�2n/3 − 1, a contradiction. Hence the 2-graph path of v has length 2.

Let the 2-graph path of v be given by v, x, y, and let N(y) = {x, w}. By Claim 46, x ∈ V2 and y ∈ V1. Let F beobtained from G − {v, x, y} by adding the edges wv1 and wv4. Then, F is a connected graph of order n − 3 with�(F )�2, and so f (F )�2|V (F)|/3 = 2n/3 − 2. For i = 1, 2, let Si be a smallest set of vertices of F that dominatesVi ∩ V (F). We may assume that {v1, w} ⊆ S1 and v4 ∈ S2. Then, S1 ∪ {v} dominates V1 in G and S2 ∪ {y} dominatesV2 in G. Hence if |S1| + |S2|�2n/3 − 3, then �G(V1) + �G(V2)� |S1| + |S2| + 2�2n/3 − 1, a contradiction. Thus|S1| + |S2| = 2|V (F)|/3, and so, since F has a 2-graph cycle, it follows from Observation 31 that F ∈ H. However,since G is a reduced 2

3 -minimal graph, so too is F , and so F ∈ G. Note that w, v1, v2, v3, v4, w is a 2-graph cycle of


w in F and degF w�4. By our assumption that S is independent, every neighbor of w in F has degree 2. By the wayin which the family G is defined, this is only possible if F = D(5, 5) (with w the vertex of degree 4 in F ). But thenG = D(5, 5, 2). �


We proceed with the Proof of Lemma 37 by establishing four claims.

Claim 51. If |S| = 2, then G = F1.

Proof. Let u and v be the two vertices of S. Then, u and v are joined by at least three paths each of length 2 or 3.In particular, n�5. If n�7, then for i = 1, 2, {u, v} dominates Vi in G, and so �G(V1) + �G(V2)�4 < 2n/3. Hence,n ∈ {5, 6}. If G = K2,3 or if G = K2,4, then we may assume that v ∈ V1, and so {u, v} dominates V1 in G and {u}dominates V2 in G. Thus, �G(V1) + �G(V2)�3 < 2n/3, a contradiction. It follows that G = F1, as desired. �

By Claim 51, we may assume |S|�3, for otherwise the desired result follows.

Claim 52. There is no 4-cycle in G.

Proof. Suppose that C: v, x, w, y is a 4-cycle in G. By assumption, there is no 2-graph cycle in G and the set S isindependent. Hence C contains exactly two vertices of S. We may assume that v, w ∈ S. By Lemma 15, we may assumethat x ∈ V1 and y ∈ V2. Let u and z be neighbors of v and w, respectively, that are not on C. Since S is independent,degG u = degG z = 2.

Suppose degG v = degG w = 3. Since |S|�3, u �= z. Let F be obtained from G by deleting the vertices of C andadding the edge uz, i.e., F = (G − V (C)) ∪ {uz}. Then, �(F )�2 and so, by Corollary 4, f (F )�2(n − 4)/3. Fori = 1, 2, let Di be a smallest set of vertices of F that dominates Vi ∩ V (F). We can now extend D1 to a set thatdominates V1 in G as follows: If u ∈ D1, then add w to D1; if u /∈ D1 and z ∈ D1, then add v to D1; otherwise addx to D1. Similarly, we can extend D2 to a set that dominates V2 in G by adding one further vertex to D2. Hence,�G(V1) + �G(V2)� |D1| + |D2| + 2�f (F ) + 2�2(n − 4)/3 + 2 < 2n/3, a contradiction. Hence we may assumedegG v�4 and u �= z. Further, we may assume u ∈ V1.

Suppose that v, u is a 2-graph path of v of length 1. Let F ′ = G − {u, x}. Then, �(F ′)�2 and so, by Corollary 4,f (F ′)�2(n − 2)/3. Now any minimum set of vertices of F ′ that dominates V2 in F ′ also dominates V2 in G, whileany minimum set of vertices of F ′ that dominates V1 in F ′ can be extended to a set that dominates V1 in G by addingthe vertex v. Hence, �G(V1) + �G(V2)�f (F ′) + 1�2(n − 2)/3 + 1 < 2n/3, a contradiction. Hence, u must be theinternal vertex of a 2-graph path of v of length 2, say v, u, t . Since |S|�3, we may assume that t is not adjacent to w.

Let F ∗ =G−{t, u, x}. Then, �(F ∗)�2. By Observation 31, f (F ∗) < 2(n− 3)/3 or each component of F ∗ belongsto F ∪ H.

Suppose f (F ∗) < 2(n − 3)/3. If t ∈ V1, then any minimum set of vertices of F ∗ that dominates V2 in F ∗ alsodominates V2 in G, while any minimum set of vertices of F ∗ that dominates V1 in F ∗ can be extended to a setthat dominates V1 in G by adding u and v. On the other hand, if t ∈ V2, then any minimum set of vertices of F ∗that dominates V1 in F ∗ can be extended to a set that dominates V1 in G by adding v while any minimum set ofvertices of F ∗ that dominates V2 in F ∗ can be extended to a set that dominates V2 in G by adding u. In any event,�G(V1) + �G(V2)�f (F ∗) + 2 < 2(n − 3)/3 + 2 = 2n/3, a contradiction. Hence, each component of F ∗ belongs toF ∪ H.

By Lemmas 17 and 28, for i = 1, 2, we can choose a smallest set Di of vertices of F ∗ that dominates Vi ∩ V (F ∗)to contain the neighbor of t different from u. But then any minimum set of vertices of F ∗ that dominates V2 in F ∗also dominates V2 in G, while any minimum set of vertices of F ∗ that dominates V1 in F ∗ can be extended to a setthat dominates V1 in G by adding v. Hence, �G(V1) + �G(V2)�f (F ∗) + 1 = 2(n − 3)/3 + 1 < 2n/3, a contradiction.Similarly, if u ∈ V2, we reach a contradiction. The desired result follows. �

Claim 53. If v ∈ S, then degG v = 3 and v has exactly two 2-graph paths of length 1 and one 2-graph path oflength 2.


Fig. 9. A subgraph of G when d �= z.

Proof. We show first that v has at least one 2-graph path of length 2. Suppose, to the contrary, that each 2-graph path ofv has length 1. Then, Gv = K1,k where k�3. If �(Hv) = 1, then G contains a 4-cycle, contradicting Claim 52. Hence,�(Hv)�2. Thus, by Corollary 4, f (Hv)�2|V (Hv)|/3 = 2(n− k − 1)/3�2(n− 4)/3. For i = 1, 2, Ri ∪{v} dominatesVi in G. Hence, �G(V1) + �G(V2)� |R1| + |R2| + 2�2(n − 4)/3 + 2 < 2n/3, a contradiction. Hence, v has at leastone 2-graph path of length 2.

For each vertex v ∈ S, let �v denote the sum of the number of vertices different from v that belong to a 2-graph pathof v. We have shown that each vertex of S has at least one 2-graph path of length 2, and so �v �4. Since each vertex ofV − S belongs to a 2-graph path that contains exactly two vertices of S, each vertex of V − S is counted twice in thesum

∑v∈S�v , and so

n − |S| = |V − S| = 1

2

∑

v∈S

�v � 1

2

∑

v∈S

4 = 2|S|.

Consequently, |S|�n/3 with equality if and only if �v = 4 for each v ∈ S. Equivalently, |S|�n/3 with equality if andonly if each v ∈ S has exactly two 2-graph paths of length 1 and one 2-graph path of length 2. For i = 1, 2, the set Sdominates Vi in G, and so 2n/3 = �G(V1) + �G(V2)�2|S|�2n/3. Consequently, we must have equality throughoutthis inequality chain. In particular, |S| = n/3, and so each v ∈ S has exactly two 2-graph paths of length 1 and one2-graph path of length 2.

Let v ∈ S. By Claim 53, v has exactly two 2-graph paths of length 1, say v, x and v, y, and one 2-graph path oflength 2, say v, a, b. Thus, Gv is obtained from K1,3 by subdividing one edge exactly once.

Suppose �(Hv)�2. Thus, by Corollary 4, f (Hv)�2|V (Hv)|/3 = 2(n − 5)/3. For i = 1, 2, we can now extendRi to a set that dominates Vi in G as follows: add v to both R1 and R2, and add b to Ri if b ∈ Vi . Thus, �G(V1) +�G(V2)� |R1| + |R2| + 3�2(n − 5)/3 + 3 < 2n/3, a contradiction. Hence, �(Hv)�1.

Let w be a vertex of Hv of degree at most 1. Necessarily, w ∈ S and, by Claim 52, w is adjacent either to x or y (butnot both) and to b. We may assume w is adjacent to y. By Claim 53, degG w = 3 and w has exactly two 2-graph pathsof length 1 and one 2-graph path of length 2. Let w, c be the 2-graph path of w different from w, y, and let d be thevertex of S − {w} adjacent to c. Let z be the vertex of S − {v} adjacent to x.

Suppose d �= z (see Fig. 9). Let F = G − N [v] − N [w]. Then, �(F )�2 and F has order n − 7. For i = 1, 2, let Si

be a smallest set of vertices of F that dominates Vi ∩ V (F). Then, for i = 1, 2, Si ∪ {v, w} dominates Vi in G, and so�G(V1) + �G(V2)� |S1| + |S2| + 4�2(n − 7)/3 + 4 < 2n/3, a contradiction. Hence, d = z.

Let K be obtained from G−{a, b, v, w, y} by adding the edge cx. Then, �(K)�2 and K has order n−5. For i =1, 2,let Ti be a smallest set of vertices of K that dominates Vi ∩ V (K). Since z, x, c, z is a 3-cycle in K with x and c havingdegree two in K, we may assume that x, c /∈ T1 ∪ T2 (if x ∈ T1, for example, then simply replace x in T1 with z). Itfollows that �G(V1)+�G(V2)� |S1|+ |S2|+f (C5)�2(n−5)/3+3 < 2n/3, a contradiction. This completes the proofof Lemma 37. �

References

[1] G. Chartrand, L. Lesniak, Graphs and Digraphs, Third ed., Chapman & Hall, London, 1996.

[2] B.L. Hartnell, P.D. Vestergaard, Partitions and dominations in a graph, J. Combin. Math. Combin. Comput. 46 (2003) 113–128.

[3] T.W. Haynes, S.T. Hedetniemi, P.J. Slater, Fundamentals of Domination in Graphs, Marcel Dekker, New York, 1998.


[4] T.W. Haynes, S.T. Hedetniemi, P.J. Slater (Eds), Domination in Graphs: Advanced Topics, Marcel Dekker, New York, 1998.[5] M.A. Henning, Graphs with large total domination number, J. Graph Theory 35 (1) (2000) 21–45.[6] W. McCuaig, B. Shepherd, Domination in graphs with minimum degree two, J. Graph Theory 13 (1989) 749–762.[7] L.A. Sanchis, Domination related to domination in graphs with minimum degree two, J. Graph Theory 25 (1997) 139–152.[8] S.M. Seager, Partition dominations of graphs of minimum degree 2, Congress. Numer. 132 (1998) 85–91.[9] Z. Tuza, P.D. Vestergaard, Domination in partitioned graph, Discuss. Math. Graph Theory 22 (1) (2002) 199–210.

domination in partitioned graphs with minimum degree two

Documents