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arXiv:2211.14178v1 [math.CO] 25 Nov 2022

Progress towards the two-thirds conjecture on locating-total

dominating sets∗

Dipayan Chakraborty†Florent Foucaud†Anni Hakanen†‡

Michael A. Henning§Annegret K. Wagler†

November 28, 2022

Abstract

We study upper bounds on the size of optimum locating-total dominating sets in graphs. A set

Sof vertices of a graph Gis a locating-total dominating set if every vertex of Ghas a neighbor in S,

and if any two vertices outside Shave distinct neighborhoods within S. The smallest size of such a

set is denoted by γL

t(G). It has been conjectured that γL

t(G)≤2n

3holds for every twin-free graph G

of order nwithout isolated vertices. We prove that the conjecture holds for cobipartite graphs, split

graphs, block graphs, subcubic graphs and outerplanar graphs.

1 Introduction

Our aim is to study upper bounds on the smallest size of locating-total dominating sets in graphs. This

notion is part of the extended research area of identiﬁcation problems in graphs and, more generally,

discrete structures like hypergraphs. In this type of problems, one seeks to select a small solution set,

generally vertices of a graph, in order to uniquely identify each vertex of the graph by its relationship

with the selected vertices. More precisely, given a set Dof vertices of a graph G, we say that two vertices

vand wof Gare located by Dif they have distinct sets of neighbors in D. Various notions based on

this location property have been studied, such as locating-dominating sets [33], identifying codes [27]

or separating sets [8], to name a few. We refer to the online bibliography on these topics maintained

by Lobstein [30] (almost 500 references by the time of writing, in 2022). This type of problems have

a wide range of applications, such as fault-detection in sensor or computer networks [27, 34], biological

testing [31], machine learning [10], or canonical representations of graphs [3, 29], to name a few.

In this paper, we study the notion of a locating-total dominating set. A set Dof vertices is a total

dominating set, abbreviated TD-set, of a graph Gif every vertex in Ghas a neighbor in D. Total

dominating sets are a natural and widely studied variant of the domination problem in graphs. We refer

to the book [26] for an overview on the topic. A locating-total dominating set of Gis a TD-set D⊂V(G)

such that any two vertices of Gnot in Dare located by D. The smallest size of such a locating-total

dominating set of a graph Gis called the locating-total domination number of Gand is denoted by γL

t(G).

A graph admits a (locating-)total dominating set if and only if it has no isolated vertex. We abbreviate

a locating-total dominating set by LTD-set, and we say that a γL

t-set of Gis a LTD-set of minimum

cardinality, γL

t(G), in G.

The concept of a locating-total dominating set was ﬁrst considered in [23], based on the similar concept

of a locating-dominating set (where total domination is replaced with usual domination) introduced by

Slater in the 1980s [33]. It was studied for example in [1, 2, 5, 6, 7, 13, 14, 15, 18, 19, 24, 25]. The

∗This work was sponsored by a public grant overseen by the French National Research Agency as part of the “Investisse-

ments d’Avenir” through the IMobS3 Laboratory of Excellence (ANR-10-LABX-0016) and the IDEX-ISITE initiative CAP

20-25 (ANR-16-IDEX-0001). We also acknowledge support of the ANR project GRALMECO (ANR-21-CE48-0004).

†Universit´e Clermont-Auvergne, CNRS, Mines de Saint-´

Etienne, Clermont-Auvergne-INP, LIMOS, 63000 Clermont-

Ferrand, France

‡Research supported by Jenny and Antti Wihuri Foundation

§Department of Mathematics and Applied Mathematics, University of Johannesburg

1

associated decision problem is NP-hard [28, 32]. The related concept where all vertices (not just the ones

outside of D) must be located was studied in [13, 21, 23].

It is known that any graph of order nwith every component of order at least 3 has a TD-set size at

most 2

3n[16], and this bound is tight only for the triangle, the 6-cycle and the family of 2-coronas of

graphs [11]. (The 2-corona H◦P2of a connected graph His the graph of order 3|V(H)|obtained from

Hby attaching a path of length 2 to each vertex of Hso that the resulting paths are vertex-disjoint.)

However, such a bound does not hold for locating-total dominating sets. Two vertices of a graph

are twins if they either have the same open neighborhood (open twins) or the same closed neighborhood

(closed twins). Consider a set Sof vertices that are pairwise twins of size at least 2 in a graph G(and S

forms either a clique or an independent set). Then, any locating-total dominating set Dneeds to contain

all vertices of Sexcept possibly one. Indeed, any two such vertices not in Dwould otherwise not be

located. For example, any complete graph of order at least 2 has only twins, and thus has its locating-

total domination number equal to its order minus one (while any two vertices form a total dominating

set). Other families of (twin-free) graphs with a total dominating set of size 2 and arbitrarily large

locating-total domination number have been described in [18].

Nevertheless, it seems that in the absence of twins, the locating-total domination number cannot be

as close to the graph’s order as in the general case. Towards such a fact, inspired by a similar problem

for (non-total) locating-dominating sets [17, 20, 22], two of the authors posed the following conjecture (a

graph is called twin-free if it contains no set of twins). A graph Gis isolate-free is it contains no isolated

vertex.

Conjecture 1 ([18]).Every twin-free isolate-free graph Gof order nsatisﬁes γL

t(G)≤2

3n.

Conjecture 1 was proved for graphs with no 4-cycle as a subgraph in [18]. It was also proved for line

graphs in [19]. It was also proved in [18] to hold for all graphs with minimum degree at least 26 for which

another related conjecture [20, 22] holds (which is the case for example for bipartite graphs and cubic

graphs). It was proved in a stronger form for claw-free cubic graphs in [24] (there, the 2

3factor in the

upper bound is in fact replaced with 1

2, and the authors conjectured that 1

2holds for all connected cubic

graphs, except K4and K3,3). An approximation of the conjecture was proved to hold for all twin-free

graphs in [18], where the 2

3factor in the upper bound is replaced with 3

4.

Note that, if true, the bound of Conjecture 1 is tight for the 6-cycle and 2-coronas, by the following.

Observation 2. If a graph Gof order nis a triangle, a 6-cycle or a 2-corona of any graph, then

γL

t(G) = 2

3n.

A graph is cobipartite if its vertex set can be partitioned into two cliques, and split if it can be

partitioned into a stable set (also called an independent set in the literature) and a clique. A graph is a

block graph if every 2-connected component forms a clique. A graph is subcubic if each vertex has degree

at most 3. A graph is outerplanar if it can be embedded in the plane without any edge-crossing, so that

all vertices lie on the same face of the embedding.

In this paper, we give further evidence towards Conjecture 1, by showing that it holds for cobipartite

graphs (Section 2), split graphs (Section 3), block graphs (Section 4), subcubic graphs (Section 5) and

outerplanar graphs (Section 6). We conclude in Section 7.

Some of our results are actually slightly stronger. Indeed, the proved upper bound for cobipartite

graphs is in fact ⌈n

2⌉(which is tight). For twin-free split graphs, we show that the 2n

3bound of the

conjecture can never be reached. However, we construct inﬁnitely many connected split graphs that

come very close to the bound; this is interesting in its own right, showing that not only 2-coronas have

such large locating-total domination numbers. Moreover, the bound for subcubic graphs is proved to

hold for all subcubic graphs, even if they have twins.

We now introduce some of the notations used in the paper. The open and closed neighborhoods of a

vertex vin a graph Gare denoted NG(v) and NG[v], respectively (or N(v) and N[v] if Gis clear from the

context). We denote by degG(v) the degree of the vertex vin the graph G, that is, degG(v) = |NG(v)|.

The distance in Gbetween two vertices u, v is denoted dG(u, v). If two graphs Gand Hare isomorphic,

we note G∼

=H. For a graph Gwith a vertex or edge x, we denote by G−xthe subgraph of Gobtained

by removing x, and by G+xthe supergraph of Gobtained by adding x. Similarly, if Xis a set of vertices

and edges, we use the notations G−Xand G+Xfor the subgraph and supergraph of Gobtained by

2

deleting or adding all the elements of X. A leaf is a vertex of degree 1, and its unique neighbor is called

asupport vertex. We denote by δ(G) and ∆(G) the minimum and maximum degree, respectively, in the

graph G. We denote a path, a cycle, and a complete graph on nvertices by Pn,Cn, and Kn, respectively.

2 Cobipartite graphs

We now prove a stronger variant of the bound of Conjecture 1, whose proof is a reﬁnement of a similar

proof for the (non-total) locating-domination number from [20]. The bound is tight for complements

of half-graphs, which are graphs made from two cliques of the same size [20, Deﬁnition 5] and whose

locating(-total) domination number is equal to n

2.

Theorem 3. For any twin-free cobipartite graph Gof order n, we have γL

t(G)≤ ⌈n

2⌉.

Proof. Note that Gis connected (as a disconnected cobipartite graph is the disjoint union of two cliques,

and thus not twin-free: it either has closed twins if one of the cliques has order at least 2, or is a pair of

open twins if both cliques have order 1). Let C1and C2be two cliques of Gthat partition its vertex set.

Since Gis twin-free, both C1and C2have size at least 2 (assume that |C1| ≤ |C2|). Moreover, no two

vertices of C1have the same neighborhood in C2, and vice-versa. Thus, if any of C1, C2is a TD-set, then

it is also an LTD-set. Thus, if C1is a TD-set, we are done, as |C1| ≤ ⌊ n

2⌋. If however C1is not a TD-set,

it means that some vertex vof C2has no neighbor in C1; this vertex is unique since Gis twin-free.

If moreover, there is a vertex win C1with no neighbor in C2, we select D= (C1\ {w})∪ {x}as a

solution set, where x6=vis any vertex of C2other than v. This set is clearly a TD-set. Any two vertices

from C2are located, as vis only dominated by x, and any two other vertices from C2\ {x}have a distinct

and nonempty neighborhood within C1(and thus, within C1\ {w}). Moreover, wis the only vertex not

in Dnot dominated by x. Hence, Dis an LTD-set. Since |D|=|C1| ≤ ⌊ n

2⌋, we are done.

Otherwise, every vertex of C1has a neighbor in C2, and so C2is a TD-set and, in fact, as seen

previously, an LTD-set. Thus, if |C2| ≤ ⌈ n

2⌉, we are done. Otherwise, we have |C1|<⌊n

2⌋, and thus

|C1|+ 1 ≤ ⌈ n

2⌉. Moreover, by similar arguments as in the previous paragraph, the set C1together with

any vertex of C2other than vproduces a LTD-set of size |C1|+ 1, and we are done.

3 Split graphs

Consider a split graph G= (Q∪S, E ) where Qinduces a clique and Sa stable set. We suppose that G

is isolate-free to ensure the existence of an LTD-set in G, which further implies that Gis connected and

Qnon-empty (as every component not containing the clique Qneeds to be an isolated vertex from S).

Theorem 4. For any twin-free isolate-free split graph G= (Q∪S, E ) of order n, we have γL

t(G)<2

3n.

Proof. First, note that we have |Q|,|S| ≥ 2 as otherwise Gis a single vertex or not twin-free.

Observe next that Qis an LTD-set of Gsince Qis a TD-set and no two vertices in Shave the same

neighbors in Q(as Gis twin-free) showing that Qis also locating. Hence, the assertion is true if |Q|<2

3n.

Consider now a set Dconsisting of all vertices in Sand, for each s∈S, some arbitrary neighbor

qs∈Q(which exists since Gis connected). The set Dis an LTD-set of Gsince Dis a TD-set and no

two vertices in Q\Dhave the same neighbors in S(as Gis twin-free) implying that Dis also locating.

Hence, the assertion is true if |Q|>2

3nand, thus, |S|<1

3nand |D|= 2|S|<2

3n.

It is left to consider the case |Q|=2

3nand |S|=1

3n. We observe that any set Dconstructed as above

is an LTD-set of G, and we are going to show that it is not minimum. In fact, as Gis twin-free, there

are two vertices s, s′∈Sso that N(s)∩N(s′) is non-empty. To see this, observe that otherwise all sets

N(si) need to be pairwise disjoint, which would imply that

•either N(si) is composed of twins when |N(si)| ≥ 2 holds, or

•each set N(si) contains exactly one vertex and the 1

3nvertices in Qwithout a neighbor in Sare

twins.

As Gis twin-free, none of these cases can happen. Now, the set D′obtained from Dby replacing qsand

qs′by a common neighbor qs,s′of sand s′in Qis an LTD-set of G, indeed:

3

•each vertex from Shas a neighbor in D′, so does every vertex from Q(as all vertices in Q\ {qs,s′}

have qs,s′as a neighbor in D′and qs,s′has sand s′as neighbors in D′) which shows that D′is a

TD-set;

•no two vertices in Q\D′have the same neighbors in S(as Gis twin-free) which implies that D′is

also locating.

Thus, D′is an LTD-set of Gof size

|D′|=|D| − 1 = 2 · |S| − 1 = 2 ·1

3n−1<2

3n ,

which ﬁnally proves the assertion.

We next show that the bound of Theorem 4 cannot be improved.

Proposition 5. For each integer k≥3, there is a connected twin-free split graph Gkof order n= 3k

and γL

t(Gk)≥2k−1.

Proof. Let Q={q1,...,qk} ∪ {q′

1,...,q′

k}be a clique and S={s1,...,sk}a stable set, so that N(si) =

{qi, q′

i}for 1 ≤i < k and N(sk) = {q1,...,qk}. Note that q′

khas no neighbor in Sand that the sets

N(si) are disjoint for 1 ≤i < k. See Figure 1 for an illustration.

q1q′

1

s1

q2q′

2

s2

qkq′

k

sk

...

...

Q

S

Figure 1: The construction of graph Gkin the proof of Proposition 5, with an optimal LTD-set (black

vertices).

Let Cbe an LTD-set of Gk. Consider the k−1 closed neighborhoods N[si] for 1 ≤i < k. If we

have |N[si]∩C| ≥ 2 for all iwith 1 ≤i < k, then

S1≤i<k(N[si]∩C)

≥2k−2, and at least one of

the remaining vertices sk, qk, q′

kmust belong to C, as otherwise N(qk)∩C=N(q′

k)∩Cwould follow, a

contradiction. This implies |C| ≥ 2k−1.

If, however, for some iwith 1 ≤i < k, we have |N[si]∩C|= 1, then si/∈Csince otherwise siis not

totally dominated by the set C. If N[si]∩C={qi}, then N(q′

i)∩C=Q∩C. If N[si]∩C={q′

i}, then

N(qi)∩C= (Q∪ {sk})∩C. The two possibilities can occur at most once each. Assume that they both

occur once each, with N[sa]∩C={q′

a}and N[sb]∩C={qb}(with 1 ≤a < b < k). Note that sk∈C,

otherwise qaand q′

bare not located. Moreover, Cmust contain q′

k(otherwise q′

band q′

kare not located)

and qk(otherwise qaand qkare not located), and so |C| ≥ 2k−1, as claimed.

Similarly, if we have |N[si]∩C| ≥ 2 for all iwith 1 ≤i < k except that N[sa]∩C={q′

a}, if sk∈C,

then qk∈C, otherwise qaand qkare not located. If sk/∈C, then both qk, q′

kare in Cto locate the

vertices qa, qk, q′

k. Thus, again |C| ≥ 2k−1.

Finally, if we have |N[si]∩C| ≥ 2 for all iwith 1 ≤i < k except that N[sb]∩C={qb}, if sk∈C,

then q′

k∈C, otherwise q′

band q′

kare not located. If sk/∈C, then both qk, q′

kare in Cto locate the

vertices q′

b, qk, q′

k, and again |C| ≥ 2k−1.

Thus, in all the above cases, we have |C| ≥ 2k−1 and, together with the upper bound γL

t(Gk)<

2

3×3k= 2kfrom Theorem 4, we ﬁnally obtain γLT D (Gk) = 2k−1.

4 Block graphs

A block graph is a graph in which every maximal 2-connected subgraph (henceforth referred to as a block)

is complete. Equivalently, block graphs are diamond-free chordal graphs [4]. A cut-vertex vof a graph

4

Gis one such that the graph G−vhas more components than G. For any block graph G, a leaf block of

Gis a block that contains only a single cut-vertex of G. In this section, we show that our 2

3-conjecture

(namely, Conjecture 1), holds for block graphs. Trees are a subclass of block graphs in which every block

is of order 2. There are some concepts of trees which we use quite often in proving our result for block

graphs and we, therefore, deﬁne these concepts formally here.

Aroot of a tree is a ﬁxed vertex of the tree to which the name is designated. Having ﬁxed a root r

of a tree T, for any vertex uof T,

(1) a child of uis a vertex vof Tsuch that uv is an edge of Tand dT(v, r) = dT(u, r) + 1;

(2) a grandchild of uis a vertex wof Tsuch that uv and vw are edges of Tand dT(w, r) = dT(u, r) + 2;

and

(3) a great-grandchild of uis a vertex xof Tsuch that uv,vw and wx are edges of Tand dT(x, r) =

dT(u, r) + 3.

Conversely, the vertex uof Tis called the parent, the grandparent and the great-grandparent of v,w

and x, respectively. Given any two vertices uand vof a tree T, we say that uis above vin T(or vis below

uin T) if there exists a sequence of vertices x1, x2,...,xmof Tsuch that, for each i= 1,2,...,m−1,

xi+1 is a child of xiin T, where m≥2, x1=uand xm=v.

Theorem 6. If G∼

=P3or if Gis a twin-free isolate-free block graph of order n≥4, then γL

t(G)≤2

3n.

Proof. Since the locating-total domination number of a graph is the sum of the locating-total domination

numbers of each of the components of the graph, it is therefore enough to prove the theorem for a

connected twin-free block graph. Thus, let us assume that Gis either isomorphic to a 3-path or is a

connected twin-free block graph of order n≥4. The proof is by induction on n≥3. The base case

of the induction hypothesis is when n= 3, in which case G∼

=P3and γL

t(G) = 2 = 2

3n. Clearly, any

two consecutive vertices of P3constitute a minimum LTD-set of P3and hence, the result holds for the

base case of the induction hypothesis. We now assume, therefore, that n≥4 and that the induction

hypothesis is true for all connected twin-free block graphs of order at least 3 and at most n−1. Next, we

construct a new graph TGfrom Gin the following way (see Figure 2 for an example of the construction).

For every block Bof G, introduce a vertex uB∈V(TG) and for every cut-vertex c∈V(G), introduce

a vertex vc∈V(TG). Next, introduce edges uBvc∈E(TG) if and only if the cut-vertex cbelongs to the

block Bof G. By construction, therefore, TGis a tree. Thus, the vertices of the tree TGare of two types:

(1) u-type:uBintroduced in a one-to-one association with a block Bof G; and

(2) v-type:vcintroduced in a one-to-one association with a cut-vertex cof G.

Notice that any pair of vertices w, z of the tree TGsuch that wis the grandparent/grandchild of zin

TGare of the same vertex type. For a ﬁxed cut-vertex r∈V(G), designate vr∈V(TG) as the root of

TG(indeed, such a cut-vertex exists as n≥4 and the twin-free property of Gimplies that Ghas at least

two blocks). Notice that any leaf of the tree TGis a vertex of the type uBfor some leaf block Bof G.

By the twin-free nature of Gevery leaf block Bof Ghas order exactly 2. Now, ﬁx a leaf uFof TGthat

is at the farthest distance, in TG, from the root vrof TG. We now look at the great-grandparent of the

leaf uFin the tree TG(indeed, the great-grandparent of uFin TGexists because, on account of Gbeing

twin-free, at least one of the blocks of Gcontaining the vertex rhas a cut-vertex other than r). Notice

that the great-grandparent of uFin TGmust be a vertex of the type vpfor some cut-vertex pof G. We

next deﬁne the following.

Bp={B:Bis a block of Gand uBis either a child or a great-grandchild of vpin TG};

U=∪B∈BpV(B); and

A={x∈U:xis a cut-vertex of G}.

We now establish the following two claims related to the sets deﬁned above.

Claim A. The set Ais an LTD-set of G[U].

5

r= 1

3

p= 2

B2

B1

6

5

B3

F=B4

B5

4

B6

B7

(a)

uB1

uB2

uB6

uB7

uB3

uF=uB4uB5

vr=v1

vp=v2v3

v4

v5v6

(b)

Figure 2: Figure (a) represents a twin-free block graph Gand Figure (b) represents TG. The vertices

underneath the dashed curve represent those deleted from Gto obtain G′. The black vertices represent

vertices in the set A. (All notations are as in the proof of Theorem 6.)

Proof of claim. That Ais a TD-set of G[U] is clear from the structure of G. We show that Ais also a

locating set of G[U]. So, let us assume that vertices w, z ∈U\A. Since Gis twin-free, notice that wand

zbelong to distinct blocks of Bp, say Band B′, respectively.

First, assume that uBand uB′are both children of vpin TG. Then, at least one of uBand uB′must

have a child in TG(or equivalently, at least one of Band B′must contain a cut-vertex of G). This is

because if none of uBand uB′had a child in TG, it would mean that Band B′are leaf blocks of Gand

so, |B|=|B′|= 2 (since Gis twin-free). That, in turn, would mean that Ghas open twins (with the

leaves of Gin the blocks Band B′sharing the common support vertex vp), contrary to our assumption.

Thus, if qis a cut-vertex of Gin V(B)∪V(B′), then q(∈A) locates wand z. Next, we assume that uB

and uB′are both great-grandchildren of vpin TG(or equivalently, Band B′are both leaf blocks of G).

Then, uBand uB′have parents vqand vq′, say, respectively, in TG, where q6=q′due to the fact that G

is twin-free. This implies that wand zare located in Gby {q, q′}. Finally, we assume that uBis a child

of vpand that uB′is a great-grandchild of vpin TG. Then, wand zare located in Gby p. This proves

the claim. (✷)

Claim B. |U| ≥ 2|A| − 1.

Proof of claim. Let uB1, uB2, . . . , uBmbe m≥1 children of vpin TGand let each block Biof Gbe of

order ni. Now, due to the twin-free nature of G, each vertex uBiof TGhas at least ni−2 and at most

ni−1 children (and hence at least ni−2 and at most ni−1 grandchildren as well). To be more precise,

assume that, for 0 ≤s≤m, the vertices uB1, uB2,...,uBshave exactly n1−2, n2−2,...,ns−2 children,

respectively, in TG; and that the vertices uBs+1 , uBs+2 ,...,uBmhave exactly ns+1 −1, ns+2 −1,...,nm−1

children, respectively, in TG. This implies that we have the following.

|U|= 1 −s+ 2 X

1≤i≤m

(ni−1) = 1 −2m−s+ 2 X

1≤i≤m

ni.

Moreover, we have

|A|= 1 −s+X

1≤i≤m

(ni−1) = 1 −m−s+X

1≤i≤m

ni.

6

Combining the above two equations, therefore, we have

|U| − (2|A| − 1) = s≥0

which proves the claim. (✷)

Now, let G′=G−U, that is, G′is the graph obtained by deleting from Gall vertices (and edges

incident with them) in the blocks B∈ Bp. Notice that G′is still a connected block graph; and assume

that the order of G′is n′(which is strictly less than n). We next divide the proof according to whether

G′is twin-free, has twins or is isomorphic to a 3-path.

Case 1 (G′is either twin-free or is isomorphic to a 3-path).We further subdivide this case into the

following.

Subcase 1.1 (n′≤2).In this subcase, if n′= 2, then the two vertices of G′form an edge of G′(since Gis

connected). Hence, the two vertices of G′are closed twins of degree 1, contrary to our initial assumption

in this case. Therefore, let us assume that n′≤1. If n′= 1, then vphas no grandparent in TG. In other

words, there is no vertex of v-type above vpin TG. This implies that vpmust itself be the root vertex

of TG. However, this, in turn, implies that G′is an empty graph which contradicts the fact that n′= 1.

Thus, we must have n′= 0. In this case too, vpmust itself be the root vertex of of TGand so, G=G[U]

and |U|=n. Therefore, by Claim A, the set Ais an LTD-set of G[U] = G. Moreover, by Claim B, we

have

|A| ≤ 1

2(|U|+ 1) = 1

2(n+ 1) ≤2

3n,

where the last inequality is true since n≥4.

Subcase 1.2 (n′≥3).In this subcase, by the induction hypothesis, we have γL

t(G′)≤2

3n′. Suppose

now that S′⊂V(G′) is a minimum LTD-set of G′, that is with |S′|=γL

t(G′). We then claim that the

set S=S′∪Ais an LTD-set of G. To prove so, we ﬁrst see that the set Sis a TD-set of G, since S′is a

TD-set of G′and Ais a TD-set of G[U]. Moreover, Sis also a locating set of Gdue to the following two

reasons.

(1) Any two distinct vertices w∈V(G′)\S′and z∈V(G)\Sare located by S′.

(2) By Claim A, the set Ais a locating set of G[U].

Using Claim B, therefore, the two-thirds bound on γL

t(G) in this subcase is established by the following

inequality.

γL

t(G)≤ |S|=|S′|+|A| ≤ γL

t(G′) + 2

32|A| − 1[since |A| ≥ 2]

≤2

3n′+ 2|A| − 1≤2

3n′+|U|=2

3n.

We next turn to the case that G′has twins.

Case 2 (G′is neither twin-free nor is isomorphic to a 3-path).Assume that xand yare two vertices of

G′which are twins in G′. Then, without loss of generality, there exists an edge in Gbetween the vertices

pand xand there is no edge in Gbetween pand y. This implies that xand pbelong to the same block

X, say, of Gto which ydoes not belong. Let the block of Gto which ybelongs be called Y. Next, we

prove the following claim.

Claim C. The vertex xalso belongs to the block Yof G.

Proof of claim. Toward a contradiction, let us assume that x /∈V(Y). If the blocks Xand Yof Ghad

no common vertex in G′, by the connectedness of G′, it would mean that the symmetric diﬀerence of

the sets NG′(x) and NG′(y) is non-empty, and so, xand ywould not be twins in G′, a contradiction.

So, let V(X)∩V(Y) = {v}(note that two blocks of a block graph can intersect at not more than a

single vertex). Now, if either of xand ywere cut-vertices of G, or if V(X)∪V(Y) contained any vertex

7

x

p

y

z

x′′

X

Y

X′′

(a) The vertices to the left of the dashed curve

represent those deleted from Gto obtain G′′.

G′′ ∼

=P3. The black vertices constitute the

set Aand the grey vertices constitute an LTD-

set S′′ of G′′.

x

p

y

z

x′′

X

Y

X′′

(b) The vertices to the left of the dashed curve

represent those deleted from Gto obtain G⋆.

G′′ has twins x′′ and y; and G⋆is a twin-free

block graph. The black vertices constitute the

set A∪ {x}and the grey vertices constitute

an LTD-set S⋆of G⋆.

Figure 3: Twin-free block graph G. The dotted boxes mark the blocks X,X′′ and Yof Gas in the proof

of Theorem 6.

other than p, x, v and y, then again, xand ywould not be twins in G′, the same contradiction as before.

This implies that V(G′) = {x, v, y}and that G′is isomorphic to a 3-path, again a contradiction to our

assumption in this case. Hence, the claim holds. (✷)

Now, clearly, X6=Y, or else, py would be an edge in G, contradicting our earlier observation. Thus,

xis a cut-vertex of Gbelonging to the distinct blocks Xand Yof G. Moreover, |X|= 2, or else, again,

xand ywould not be twins in G′, a contradiction. More precisely, V(X) = {x, p}. We also observe here

that ycannot be a cut-vertex of G, or else, xand ywould not be twins in G′, again the contradiction

as before. Therefore, since Gis twin-free, every vertex other than yof the block Ymust be a cut-vertex

of G.

Now, look at the block graph G′′ =G′−xon, say, n′′ vertices (the graph induced by the vertices

on the right of the dashed curve in Figure 3a). Notice that, in the tree TG, the vertex vxcannot have

any children other than uX, or else, xand ycannot be twins in G′, contrary to our assumption for this

case. This implies that the block graph G′′ is also connected. We also have y∈V(G′′). Thus, n′′ ≥1.

However, we claim that n′′ 6= 2.

Claim D. The order n′′ of the block graph G′′ cannot be 2.

Proof of claim. Toward a contradiction, let us assume n′′ = 2 such that zis the vertex of G′′ other than

y. That is, V(G′′ ) = {y, z }and that yz is an edge of G, by the connectedness of G′′ . Therefore, yand z

belong to the same block Z, say, of G. If Z=Y, then by our observation preceding this claim, the vertex

z, being diﬀerent from y, must be a cut-vertex of Gmaking n′′ ≥3, contradicting our assumption. So,

Zand Yare distinct blocks of Gto both of which the vertex ybelongs. This makes ya cut-vertex of

G, again contradicting our earlier observation that ycannot be a cut-vertex of G. This proves the claim.

(✷)

Thus, we next divide this case into the following two subcases according to the order of G′′ .

Subcase 2.1 (n′′ = 1).In this subcase, we claim that S=A∪ {x}is an LTD-set of G. It is clear that

Sis a TD-set of G; by Claim A, Ais an LTD-set of G[U]. The vertex yis located from any vertex in

8

U\Aby the vertex x, and thus the set Sis also locating. Therefore, in this case, we have

γL

t(G)≤ |S|=|A|+ 1 <2

32|A| − 1+4

3[since |A| ≥ 2]

≤2

3|U|+ 2=2

3n. [by Claim B]

Subcase 2.2 (n′′ ≥3 and G′′ is either twin-free or is isomorphic to a 3-path).Since n′′ is at least 3

and is strictly less than n, by the induction hypothesis, we have γL

t(G′′)≤2

3n′′. Moreover, let S′′ be a

minimum LTD-set of G′′ , that is with |S′′|=γL

t(G′′). We next claim the following.

Claim E. The set S=S′′ ∪Ais an LTD-set of G.

Proof of claim. Since S′′ is a TD-set of G′′ and Ais a TD-set of G[U∪ {x}], Sis therefore a TD-set of

G. Next we show that Sis also a locating set of G. To begin with, we note that Y′′ =Y−xis a block of

G′′ containing the vertex y. Now, since yis not a cut-vertex of G, we have S′′ ∩Y′′ 6=∅(or else, yis not

dominated by S′′). This implies that xis located by S′′ from all vertices in U\A. Moreover, xis also

located by pfrom all vertices in V(G′′ )\S′′. Next, any pair w, z of distinct vertices with w∈V(G′′)\S′′

and z∈V(G)\(S∪ {x}) are located by S′′. Finally, any distinct pair of vertices w, z ∈U\Aare located

by A, since the latter is an LTD-set of G[U] by Claim A. (✷)

Therefore, in this subcase, using Claim B again, the theorem follows from the following inequality.

γL

t(G)≤ |S|=|S′′|+|A| ≤ γL

t(G′′) + 2

32|A| − 1[since |A| ≥ 2]

≤2

3n′′ + 2|A| − 1≤2

3n′′ +|U|<2

3n.

Subcase 2.3 (n′′ ≥3 and G′′ is neither twin-free nor is isomorphic to a 3-path).Assume that x′′ and

y′′ are a pair of twins of G′′ . Moreover, for x′′ and y′′ to be twins in G′′, at least one of them must be in

the block Y. Let us, without loss of generality, assume that y′′ ∈V(Y).

We next observe that the vertices yand y′′ are the same. To prove so, by contradiction, let us assume

that y′′ 6=y. Then y′′ is a cut-vertex of Gand so, for x′′ and y′′ to be twins in G′′,x′′ must not belong

to the block Yof G. However, this, in turn, implies that yis a neighbor of y′′ but not of x′′ and so, x′′

and y′′ are not twins in G′′ , a contradiction all the same. This, therefore, proves the observation.

Again, the vertex x′′ /∈Y, since otherwise, x′′ 6=y′′ =yimplies that x′′ is a cut-vertex of G, thus

forcing x′′ and y′′ to not be twins, contrary to our assumption. Let x′′ belong to the block X′′ (6=Y) of

G′′ (and of G). We now try to establish the structure of the block Yof G. Notice that, by the structure

of a block graph, the twins x′′ and yin G′′ must have a single common neighbor z, say, in G′′ such that

zis a cut-vertex of Gbelonging to both the blocks Yand X′′ of G. Furthermore, if the block Ycontains

any vertex of Gother than the vertices x, y and z, then x′′ and yare not twins in G′′ , a contradiction.

Thus, we have V(Y) = {x, y, z}.

Next, to understand the structure of the block X′′ of G′′, we see that neither can X′′ contain any

vertex other than zand x′′ , nor can x′′ be a cut-vertex of G; or else, we again have the contradiction

that x′′ and yare not twins in G′′. Therefore, this implies that V(X′′) = {x′′ , z}, that is, X′′ is a leaf

block of G′′ (and of G). See Figure 3 for the structure of the blocks X′′ and Y.

With that, we look at the block graph G⋆=G′′ −y(the graph induced by the vertices on the right

of the dashed curve in Figure 3b). Then, G⋆is again a connected graph, since yis not a cut-vertex of

G. Moreover, the order n⋆of G⋆is at least 2 (since x′′ , z ∈V(G⋆)). If, however, n⋆= 2, then we have

V(G′′) = {x′′ , y, z }and thus, G′′ is isomorphic to a 3-path, contrary to our assumption in this subcase.

Therefore, we have n⋆≥3. We next show the following claim.

Claim F. The graph G⋆is twin-free.

Proof of claim. Toward a contradiction, let us assume that the block graph G⋆has a pair of twins. Then

one of them must be the cut-vertex zof G. Let x⋆be the other vertex of G⋆such that x⋆and zare twins

in G⋆. Since x′′ is a neighbor of zalone in G⋆, therefore zcannot be a twin in G⋆of any vertex other

9

than x′′. In other words, x⋆=x′′. However, since degG⋆(x′′) = 1, we have degG⋆(z) = 1 and, hence, the

graph G⋆is simply the edge x′′ zof G. This however, contradicts the fact that n⋆≥3. Hence, this proves

that G⋆is twin-free. (✷)

Since n⋆is at least 3 and is strictly less than n, by the induction hypothesis, we have γL

t(G⋆)≤2

3n⋆.

Moreover, let S⋆be a minimum LTD-set of G⋆, that is with |S⋆|=γL

t(G⋆). We next claim the following.

Claim G. The set S=S⋆∪A∪ {x}is an LTD-set of G.

Proof of claim. Since S⋆is a TD-set of G⋆and A∪ {x}is a TD-set of G[U∪ {x, y}], Sis therefore a

TD-set of G. Next we show that Sis also a locating set of G. To begin with, we note that, since x′′

is a leaf in G⋆, its support vertex zmust be in the LTD-set S⋆of G⋆. Thus, the vertex yis located

from every other vertex in V(G)\Sby the set {x, z}. Next, any pair w1, w2of distinct vertices with

w1∈V(G⋆)\S⋆and w2∈V(G)\S, respectively, are located by S⋆. Finally, by Claim A, any pair of

distinct vertices w1, w2∈U\Aare located by the set A.(✷)

Therefore, again using Claim B, in this subcase, the theorem follows from the following inequality.

γL

t(G)≤ |S|=|S⋆|+|A|+ 1 < γL

t(G⋆) + 2

32|A|+ 1[since |A| ≥ 2]

≤2

3n⋆+ 2|A|+ 1≤2

3n⋆+|U|+ 2=2

3n.

This completes the proof.

For any block graph Hof order k≥2, the 2-corona G=H◦P2is a twin-free block graph of

order n= 3kand by Observation 2, it has locating-total domination number equal to its total domination

number, that is, γL

t(G) = γt(G) = 2k=2

3n. See Figure 4 for an illustration with Ha complete graph.

Thus, we obtain the following.

Proposition 7. There are inﬁnitely many connected twin-free block graphs Gof order nwith γL

t(G) =

2

3n.

Figure 4: The 2-corona K6◦P2of a complete graph of order 6.

5 Subcubic graphs

In this section, we establish a tight upper bound on the locating-total domination number of a subcubic

graph, where a subcubic graph is a graph with maximum degree at most 3. For this purpose, let Ftdom

be the family consisting of the three complete graphs K1,K2, and K4, and a star K1,3, that is,

Ftdom ={K1, K2, K4, K1,3}.

We denote a path, a cycle, and a complete graph on nvertices by Pn,Cn, and Kn, respectively. A

diamond is the graph K4−ewhere eis an arbitrary edge of the K4. A paw is the graph obtained from a

10

triangle K3by adding a new vertex and joining it with an edge to one vertex of the triangle. Equivalently,

a paw is obtained from K1,3by adding an edge between two leaves.

For k≥2, we say a graph Gcontains a (d1, d2,...,dk)-sequence if there exists a path v1v2...vksuch

that degG(vi) = difor all i∈[k]. We are now in a position to prove the following upper bound on the

locating-total domination number of a subcubic graph.

Theorem 8. If G /∈ Ftdom is a connected subcubic graph of order n≥3, then γL

t(G)≤2

3n.

Proof. Suppose, to the contrary, that the theorem is false. Among all counterexamples, let Gbe one of

minimum order n. If n= 3, then G∼

=P3or G∼

=K3, and in both cases γL

t(G) = 2 = 2

3n, a contradiction.

Hence, n≥4. Suppose n= 4. By assumption, G /∈ {K4, K1,3}. If Gis a diamond or a paw, then let

Sconsist of one vertex of degree 2 and one vertex of degree 3, and if Gis a path or a cycle, then let S

consist of two adjacent vertices of degree 2. In all cases, Sis a LTD-set of Gof cardinality 2, and so

γL

t(G)≤2<2

3n, a contradiction. Hence, n≥5.

Suppose that n= 5. If Gis a path P5or a cycle C5, then γL

t(G) = 3 <2

3n(choose three consecutive

vertices of degree 2), a contradiction. Hence, ∆(G) = 3. Let vbe a vertex of degree 3 in Gwith neighbors

v1, v2, v3. Let v4be the remaining vertex of G. Since Gis connected, we may assume, renaming vertices

if necessary, that v1v4is an edge. The set {v, v1, v2}is a LTD-set of G, and so γL

t(G)≤3<2

3n, a

contradiction. Hence, n≥6.

Suppose that n= 6. If Gis a path P6or a cycle C6, then γL

t(G) = 4 = 2

3n(choose four consecutive

vertices of degree 2), a contradiction. Hence, ∆(G) = 3. Let vbe a vertex of degree 3 in Gwith neighbors

v1, v2, v3, and let v4and v5be the two remaining vertices of G. Since Gis connected, we may assume,

renaming vertices if necessary, that v1v4is an edge. One of the sets {v, v1, v2, v4}and {v, v1, v3, v4}is a

LTD-set of G, and so γL

t(G)≤4 = 2

3n, a contradiction. Hence, n≥7.

In what follows, we adopt the notation that if there is a (d1, d2,...,dk)-sequence in G, then P:v1v2...vk

denotes a path in Gassociated with such a sequence, where degG(vi) = difor all i∈[k]. Further, we let

G′=G−V(P) and let G′have order n′, and so n′=n−k. Recall that n≥7.

We show ﬁrstly that there is no vertex of degree 1.

Claim H. δ(G)≥2.

Proof of claim. Suppose, to the contrary, that δ(G) = 1. We proceed further with a series of structural

properties of the graph Gthat show that certain (d1, d2,...,dk)-sequences are forbidden.

Subclaim H.1. The following properties hold in the graph G.

(a) There is no (1,3,1)-sequence.

(b) There is no (1,2,2)-sequence.

(c) There is no (1,2,3,1)-sequence.

(d) There is no (1,2,3,2,1)-sequence.

(e) There is no (1,2,3)-sequence.

(f) There is no (1,2)-sequence.

(g) There is no (1,3,2)-sequence.

(h) There is no (1,3,3,1)-sequence.

Proof of subclaim. (a) Suppose that there is a (1,3,1)-sequence in G. In this case, n′=n−3≥4.

Since Gis connected, so too is the graph G′. Thus, G′is not a counterexample to our theorem, except

possibly when n= 4 and G′∈ Ftdom . Let v′be the third neighbor of v2in Gnot on the path P. Suppose

G′∈ Ftdom, implying that G′∼

=K1,3with v′as a leaf in G′. The graph Gis therefore determined, and

has order n= 7. In this case, choosing Sto consist of the two support vertices (of degree 3) and a leaf

neighbor of each support vertex produces a LTD-set of Gof cardinality 4, and so γL

t(G′)≤4<2

3n.

Hence, G′/∈ Ftdom. Since G′is not a counterexample, it holds that γL

t(G′)≤2

3n′=2

3n−2. Every

11

γL

t-set of G′can be extended to a LTD-set of Gby adding to it the vertices v2and v3, implying that

γL

t(G)≤γL

t(G′) + 2 ≤2

3n, a contradiction.

(b) Suppose that there is a (1,2,2)-sequence in G. Let v′be the second neighbor of v3. As in the

previous case, n′=n−3≥4 and G′is connected. By part (a), there is no (1,3,1)-sequence, implying

that G′/∈ Ftdom and γL

t(G′)≤2

3n′=2

3n−2. As before every γL

t-set of G′can be extended to a LTD-set

of Gby adding to it the vertices v2and v3, implying that γL

t(G)≤2

3n, a contradiction.

(c) Suppose that there is a (1,2,3,1)-sequence in G. In this case, n′=n−4≥3 and G′is connected.

By part (a), there is no (1,3,1)-sequence, implying that G′/∈ Ftdom and γL

t(G′)≤2

3n′=2

3(n−4) <2

3n−2.

Every γL

t-set of G′can be extended to a LTD-set of Gby adding to it the vertices v2and v3, implying

that γL

t(G)≤2

3n, a contradiction.

(d) Suppose that there is a (1,2,3,2,1)-sequence in G. In this case, G′is connected and n′=n−5≥2.

If G′∈ Ftdom, then G′∼

=K2by the fact that there is no (1,3,1)-sequence in Gby part (a). The graph G

is therefore determined, and is obtained from a star K1,3by subdividing every edge once. We note that G

has order n= 7 and the set N[v3] (of non-leaves of G) is a LTD-set of G, implying that γL

t(G)≤4<2

3n,

a contradiction. Hence, G′/∈ Ftdom. Thus, γL

t(G′)≤2

3n′=2

3(n−5) <2

3n−3. Every γL

t-set of G′can

be extended to a LTD-set of Gby adding to it the vertices v2, v3, and v4, implying that γL

t(G)<2

3n, a

contradiction.

(e) Suppose that there is a (1,2,3)-sequence in G. In this case, n′=n−3≥4 and G′contains at most

two components. Let v4and v′

4be the two neighbors of v3diﬀerent from v2. By our earlier observations,

each of v4and v′

4has degree at least 2 in G, and therefore degree at least 1 in G′.

Suppose that G′is disconnected. In this case, since there is no (1,3,1)-sequence, no (1,2,3,1)-

sequence, and no (1,2,3,2,1)-sequence in G, neither component of G′belongs to Ftdom. By linearity, we

therefore have that γL

t(G′)≤2

3n′=2

3n−2. Every γL

t-set of G′can be extended to a LTD-set of Gby

adding to it the vertices v2and v3, implying that γL

t(G)≤2

3n, a contradiction. Hence, G′is connected.

Recall that n′≥4.

Suppose now that G′is connected. If G′∈ Ftdom, then G′∼

=K1,3. Let v5be the central vertex of

G′, and so each of v4and v′

4is a leaf neighbor of v5in G′. The graph Gis therefore determined and

n= 7. The set {v2, v3, v4, v5}is a LTD-set of G, implying that γL

t(G)≤4<2

3n, a contradiction. Hence,

G′/∈ Ftdom. Thus, γL

t(G′)≤2

3n′=2

3n−2. Every γL

t-set of G′can be extended to a LTD-set of Gby

adding to it the vertices v2and v3, implying that γL

t(G)≤2

3n, a contradiction.

(f) Since there is no (1,2,1)-sequence (since n≥7), no (1,2,2)-sequence by (b) and no (1,2,3)-

sequence by (e), there can be no (1,2)-sequence in G. Hence, part (f) follows immediately from parts (b)

and (e).

(g) Suppose that there is a (1,3,2)-sequence in G. In this case, n′=n−3≥4. If G′is disconnected,

then by parts (a)–(f), neither component of G′belongs to Ftdom. By linearity, we therefore have that

γL

t(G′)≤2

3n′=2

3n−2. Every γL

t-set of G′can be extended to a LTD-set of Gby adding to it the

vertices v2and v3, implying that γL

t(G)≤2

3n, a contradiction. Hence, G′is connected. If G′∈ Ftdom,

then G′∼

=K1,3. In this case, the graph Ghas order n= 7 and is obtained from a 5-cycle by selecting two

non-adjacent vertices on the cycle and adding a pendant edge to these two vertices. In this case, the set

consisting of the two vertices of degree 3 and any two vertices of degree 2 is a LTD-set of G, implying that

γL

t(G)≤4<2

3n, a contradiction. Hence, G′/∈ Ftdom . Thus, γL

t(G′)≤2

3n′=2

3n−2. Every γL

t-set of G′

can be extended to a LTD-set of Gby adding to it the vertices v2and v3, implying that γL

t(G)≤2

3n, a

contradiction.

(h) Suppose that there is a (1,3,3,1)-sequence in G. In this case, n′=n−4≥3 and G′contains

at most two components. Let uibe the neighbor of vinot on Pfor i∈ {2,3}. Possibly, u2=u3. By

parts (a) and (g), the vertex uihas degree 3 in Gfor i∈ {2,3}. Suppose that G′is disconnected. In this

case, by parts (a)–(g), neither component of G′belongs to Ftdom. By linearity, we therefore have that

γL

t(G′)≤2

3n′=2

3(n−4) <2

3n−2. Every γL

t-set of G′can be extended to a LTD-set of Gby adding

to it the vertices v2and v3, implying that γL

t(G)<2

3n, a contradiction. Hence, G′is connected. Recall

that n′≥3. By parts (a)–(g), we note that G′/∈ Ftdom , implying that γL

t(G′)≤2

3n′<2

3n−2. Every

γL

t-set of G′can be extended to a LTD-set of Gby adding to it the vertices v2and v3, implying that

γL

t(G)<2

3n, a contradiction.

Thus, the proof of the subclaim is complete. (⋄)

12

We now return to the proof of Claim H. By Subclaim H.1(f), the neighbor of every vertex of degree 1

has degree 3 in G. Further by Subclaim H.1(g), such a vertex of degree 3 has both its other two neighbors

of degree 3. Therefore the existence of a vertex of degree 1 implies that there is a (1,3,3)-sequence in

G. In this case, n′=n−3≥4. Let u2be the neighbor of v2not on P, and let u3and w3be the

two neighbors of v3not on P. By our earlier observations, the vertex u2has degree 3 in G, and, by

Subclaim H.1(h), both vertices u3and w3have degree at least 2 in G.

Suppose that G′contains a component that belongs to Ftdom . By Subclaim H.1, this is only possible

if u3and w3are adjacent and both vertices have degree 2 in G. In this case, G[{v3, u3, w3}] is a triangle

in G. We now consider the connected graph G∗=G−{v1, v2, v3, u3, w3}of order n∗=n−5. Since u2has

degree 2 in G∗, we note that n∗≥3 and G∗/∈ Ftd om. Hence, γL

t(G∗)≤2

3n∗=2

3(n−5) <2

3n−3. Every

γL

t-set of G∗can be extended to a LTD-set of Gby adding to it the vertices v2,v3, and u3, implying that

γL

t(G)≤γL

t(G∗) + 3 <2

3n, a contradiction. Hence, no component of G′belongs to the family Ftdom.

By linearity, we therefore have that γL

t(G′)≤2

3n′=2

3n−2. Every γL

t-set of G′can be extended to a

LTD-set of Gby adding to it the vertices v2and v3, implying that γL

t(G)<2

3n, a contradiction. This

completes the proof of Claim H. (✷)

By Claim H, every vertex in Ghas degree 2 or 3.

Claim I. The graph Gis triangle-free.

Proof of claim. Suppose that Gcontains a triangle K3. Among all triangles in G, let Tcontain the

maximum number of vertices of degree 2 in G. Let V(T) = {v1, v2, v3}, where 2 ≤degG(v1)≤degG(v2)≤

degG(v3)≤3. Since n≥7, the triangle Tcontains at most two vertices of degree 2, and so degG(v3) = 3.

Let G′=G−V(T) and let G′have order n′, and so n′=n−3≥4.

Suppose that degG(v1) = 2. We note that degG(v2) = 2 or degG(v2) = 3. Since every vertex in G

has degree 2 or 3, no component of G′belongs to Ftdom . Hence by linearity, γL

t(G′)≤2

3n′=2

3n−2.

Every γL

t-set of G′can be extended to a LTD-set of Gby adding to it the vertices v2and v3, implying

that γL

t(G)<2

3n, a contradiction. Hence, degG(v1) = 3, implying that every vertex in Thas degree 3 in

G. Hence by our choice of the triangle T, no vertex of degree 2 in Gbelongs to a triangle.

Let uibe the neighbor of vinot in the triangle Tfor i∈[3]. We note that the vertices u1,u2, and u3

are not necessarily distinct. Suppose that G′contains no component that belongs to Ftdom. By linearity,

this yields γL

t(G′)≤2

3n′=2

3n−2. Every γL

t-set of G′can be extended to a LTD-set of Gby adding to it

the vertices v2and v3, implying that γL

t(G)≤2

3n, a contradiction. Hence, G′contains a component that

belongs to Ftdom . Since n≥7 and no vertex of degree 2 in Gbelongs to a triangle, this is only possible

if either G′∼

=K1,3or if G′contains a K2-component.

On the one hand, if G′∼

=K1,3, then the three vertices u1,u2, and u3are leaves in G′that are adjacent

to a common neighbor (of degree 3) in G′. In this case, the graph Gis determined and n= 7, and the

set V(T)∪ {u1}is a LTD-set of G, implying that γL

t(G)≤4<2

3n, a contradiction.

On the other hand, if G′contains a K2-component, then renaming vertices if necessary, we may

assume that u1and u2belong to such a component. We note that u1and u2both have degree 2 in G,

and u1v1v2u2u1is a 4-cycle in G. Further we note that in this case, G′contains two components, where

the second component contains the vertex u3. We now consider the graph G∗=G− {v1, v2, v3, u1, u2}.

Let G∗have order n∗=n−5. By the fact that δ(G)≥2 by Claim H, the graph G∗/∈ Ftdom, implying

that γL

t(G∗)≤2

3n∗=2

3(n−5) <2

3n−3. Every γL

t-set of G∗can be extended to a LTD-set of Gby

adding to it, for example, the vertices u2, v2and v3, implying that γL

t(G)<2

3n, a contradiction. (✷)

By Claim I, the graph Gis triangle-free. We show next that there is no vertex of degree 2.

Claim J. The graph Gis a cubic graph.

Proof of claim. Suppose, to the contrary, that δ(G) = 2. As before, we obtain a series of structural

properties of the graph Gthat show that certain (d1, d2,...,dk)-sequences are forbidden. These forbidden

sequences will enable us to deduce the desired result of the claim that Gmust be a cubic graph.

Subclaim J.1. The following properties hold in the graph G.

13

(a) There is no (2,2,2)-sequence.

(b) There is no (2,3,2)-sequence.

(c) There is no (2,2,3)-sequence.

(d) There is no (2,2)-sequence.

(e) There is no (2,3,3)-sequence.

Proof of subclaim. (a) Suppose that there is a (2,2,2)-sequence in G. In this case, n′=n−3≥4. Since

n≥7, δ(G) = 2, and Gcontains no triangle, no component of G′belongs to Ftdom. Hence by linearity,

γL

t(G′)≤2

3n′=2

3n−2. Every γL

t-set of G′can be extended to a LTD-set of Gby adding to it the

vertices v2and v3, implying that γL

t(G)≤2

3n, a contradiction.

(b) Suppose that there is a (2,3,2)-sequence in G. As before, n′=n−3≥4. Suppose that G′contains

a component that belongs to Ftdom. Since there is no (2,2,2)-sequence and n≥7, and since δ(G)≥2

and Gcontains no triangle, this is only possible if G′∼

=K1,3. But then the graph Gis determined and

n= 7, and γL

t(G′) = 4 <2

3n(by considering the set N[v2]), a contradiction. Hence, no component of G′

belongs to Ftdom . By linearity, this yields γL

t(G′)≤2

3n′=2

3n−2. Every γL

t-set of G′can be extended

to a LTD-set of Gby adding to it the vertices v2and v3, implying that γL

t(G)≤γL

t(G′) + 2 ≤2