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# Progress towards the two-thirds conjecture on locating-total dominating sets

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We study upper bounds on the size of optimum locating-total dominating sets in graphs. A set $S$ of vertices of a graph $G$ is a locating-total dominating set if every vertex of $G$ has a neighbor in $S$, and if any two vertices outside $S$ have distinct neighborhoods within $S$. The smallest size of such a set is denoted by $\gamma^L_t(G)$. It has been conjectured that $\gamma^L_t(G)\leq\frac{2n}{3}$ holds for every twin-free graph $G$ of order $n$ without isolated vertices. We prove that the conjecture holds for cobipartite graphs, split graphs, block graphs, subcubic graphs and outerplanar graphs.
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arXiv:2211.14178v1 [math.CO] 25 Nov 2022
Progress towards the two-thirds conjecture on locating-total
dominating sets
Dipayan ChakrabortyFlorent FoucaudAnni 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 DV(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 HP2of 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 Gxthe 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 GXand 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= (QS, 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= (QS, 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 sS, some arbitrary neighbor
qsQ(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, sSso 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 Dobtained from Dby replacing qsand
qsby a common neighbor qs,sof sand sin 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,sas a neighbor in Dand qs,shas sand sas neighbors in D) which shows that Dis a
TD-set;
no two vertices in Q\Dhave the same neighbors in S(as Gis twin-free) which implies that Dis
also locating.
Thus, Dis an LTD-set of Gof size
|D|=|D| 1 = 2 · |S| 1 = 2 ·1
3n1<2
3n ,
which ﬁnally proves the assertion.
We next show that the bound of Theorem 4 cannot be improved.
Proposition 5. For each integer k3, there is a connected twin-free split graph Gkof order n= 3k
and γL
t(Gk)2k1.
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 k1 closed neighborhoods N[si] for 1 i < k. If we
have |N[si]C| 2 for all iwith 1 i < k, then
S1i<k(N[si]C)
2k2, 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
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=QC. 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 skC,
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| 2k1, as claimed.
Similarly, if we have |N[si]C| 2 for all iwith 1 i < k except that N[sa]C={q
a}, if skC,
then qkC, 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| 2k1.
Finally, if we have |N[si]C| 2 for all iwith 1 i < k except that N[sb]C={qb}, if skC,
then q
kC, 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| 2k1.
Thus, in all the above cases, we have |C| 2k1 and, together with the upper bound γL
t(Gk)<
2
3×3k= 2kfrom Theorem 4, we ﬁnally obtain γLT D (Gk) = 2k1.
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 Gvhas 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,...,m1,
xi+1 is a child of xiin T, where m2, x1=uand xm=v.
Theorem 6. If G
=P3or if Gis a twin-free isolate-free block graph of order n4, 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 n4. The proof is by induction on n3. 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 n4 and that the induction
hypothesis is true for all connected twin-free block graphs of order at least 3 and at most n1. 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 uBV(TG) and for every cut-vertex cV(G), introduce
a vertex vcV(TG). Next, introduce edges uBvcE(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 rV(G), designate vrV(TG) as the root of
TG(indeed, such a cut-vertex exists as n4 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={xU: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 uBare both children of vpin TG. Then, at least one of uBand uBmust
have a child in TG(or equivalently, at least one of Band Bmust contain a cut-vertex of G). This is
because if none of uBand uBhad a child in TG, it would mean that Band Bare 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 Bsharing 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 uBare both great-grandchildren of vpin TG(or equivalently, Band Bare both leaf blocks of G).
Then, uBand uBhave parents vqand vq, say, respectively, in TG, where q6=qdue 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 uBis 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 m1 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 ni2 and at most
ni1 children (and hence at least ni2 and at most ni1 grandchildren as well). To be more precise,
assume that, for 0 sm, the vertices uB1, uB2,...,uBshave exactly n12, n22,...,ns2 children,
respectively, in TG; and that the vertices uBs+1 , uBs+2 ,...,uBmhave exactly ns+1 1, ns+2 1,...,nm1
children, respectively, in TG. This implies that we have the following.
|U|= 1 s+ 2 X
1im
(ni1) = 1 2ms+ 2 X
1im
ni.
Moreover, we have
|A|= 1 s+X
1im
(ni1) = 1 ms+X
1im
ni.
6
Combining the above two equations, therefore, we have
|U| (2|A| 1) = s0
which proves the claim. ()
Now, let G=GU, that is, Gis the graph obtained by deleting from Gall vertices (and edges
incident with them) in the blocks B Bp. Notice that Gis still a connected block graph; and assume
that the order of Gis n(which is strictly less than n). We next divide the proof according to whether
Gis twin-free, has twins or is isomorphic to a 3-path.
Case 1 (Gis either twin-free or is isomorphic to a 3-path).We further subdivide this case into the
following.
Subcase 1.1 (n2).In this subcase, if n= 2, then the two vertices of Gform an edge of G(since Gis
connected). Hence, the two vertices of Gare closed twins of degree 1, contrary to our initial assumption
in this case. Therefore, let us assume that n1. 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 Gis 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 n4.
Subcase 1.2 (n3).In this subcase, by the induction hypothesis, we have γL
t(G)2
3n. Suppose
now that SV(G) is a minimum LTD-set of G, that is with |S|=γL
t(G). We then claim that the
set S=SAis an LTD-set of G. To prove so, we ﬁrst see that the set Sis a TD-set of G, since Sis a
TD-set of Gand 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 wV(G)\Sand zV(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| 12
3n+|U|=2
3n.
We next turn to the case that Ghas twins.
Case 2 (Gis neither twin-free nor is isomorphic to a 3-path).Assume that xand yare two vertices of
Gwhich 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 Gis a twin-free
block graph. The black vertices constitute the
set A {x}and the grey vertices constitute
an LTD-set Sof 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 Gis 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′′ =Gxon, 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 yV(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′′ =Yxis 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 wV(G′′)\S′′
and zV(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| 12
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, Gis again a connected graph, since yis not a cut-vertex of
G. Moreover, the order nof Gis 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 n3. We next show the following claim.
Claim F. The graph Gis twin-free.
Proof of claim. Toward a contradiction, let us assume that the block graph Ghas a pair of twins. Then
one of them must be the cut-vertex zof G. Let xbe the other vertex of Gsuch that xand zare twins
in G. Since x′′ is a neighbor of zalone in G, therefore zcannot be a twin in Gof 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 Gis simply the edge x′′ zof G. This however, contradicts the fact that n3. Hence, this proves
that Gis twin-free. ()
Since nis at least 3 and is strictly less than n, by the induction hypothesis, we have γL
t(G)2
3n.
Moreover, let Sbe a minimum LTD-set of G, that is with |S|=γL
t(G). We next claim the following.
Claim G. The set S=SA {x}is an LTD-set of G.
Proof of claim. Since Sis a TD-set of Gand 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 Sof 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
w1V(G)\Sand w2V(G)\S, respectively, are located by S. Finally, by Claim A, any pair of
distinct vertices w1, w2U\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|+ 12
3n+|U|+ 2=2
3n.
This completes the proof.
For any block graph Hof order k2, the 2-corona G=HP2is 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 K6P2of 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 K4ewhere 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 k2, 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 n3, 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
Hence, n4. 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
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
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
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=GV(P) and let Ghave order n, and so n=nk. Recall that n7.
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=n34.
Since Gis connected, so too is the graph G. Thus, Gis not a counterexample to our theorem, except
possibly when n= 4 and G Ftdom . Let vbe the third neighbor of v2in Gnot on the path P. Suppose
G Ftdom, implying that G
=K1,3with vas 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 Gis not a counterexample, it holds that γL
t(G)2
3n=2
3n2. Every
11
γL
t-set of Gcan 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
(b) Suppose that there is a (1,2,2)-sequence in G. Let vbe the second neighbor of v3. As in the
previous case, n=n34 and Gis connected. By part (a), there is no (1,3,1)-sequence, implying
that G/ Ftdom and γL
t(G)2
3n=2
3n2. As before every γL
t-set of Gcan be extended to a LTD-set
of Gby adding to it the vertices v2and v3, implying that γL
t(G)2
(c) Suppose that there is a (1,2,3,1)-sequence in G. In this case, n=n43 and Gis connected.
By part (a), there is no (1,3,1)-sequence, implying that G/ Ftdom and γL
t(G)2
3n=2
3(n4) <2
3n2.
Every γL
t-set of Gcan be extended to a LTD-set of Gby adding to it the vertices v2and v3, implying
that γL
t(G)2
(d) Suppose that there is a (1,2,3,2,1)-sequence in G. In this case, Gis connected and n=n52.
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(n5) <2
3n3. Every γL
t-set of Gcan
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
(e) Suppose that there is a (1,2,3)-sequence in G. In this case, n=n34 and Gcontains 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 Gis 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 Gbelongs to Ftdom. By linearity, we
therefore have that γL
t(G)2
3n=2
3n2. Every γL
t-set of Gcan 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, Gis connected.
Recall that n4.
Suppose now that Gis 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
G/ Ftdom. Thus, γL
t(G)2
3n=2
3n2. Every γL
t-set of Gcan be extended to a LTD-set of Gby
adding to it the vertices v2and v3, implying that γL
t(G)2
(f) Since there is no (1,2,1)-sequence (since n7), 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=n34. If Gis disconnected,
then by parts (a)–(f), neither component of Gbelongs to Ftdom. By linearity, we therefore have that
γL
t(G)2
3n=2
3n2. Every γL
t-set of Gcan 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, Gis 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
3n2. 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
(h) Suppose that there is a (1,3,3,1)-sequence in G. In this case, n=n43 and Gcontains
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 Gis disconnected. In this
case, by parts (a)–(g), neither component of Gbelongs to Ftdom. By linearity, we therefore have that
γL
t(G)2
3n=2
3(n4) <2
3n2. Every γL
t-set of Gcan 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, Gis connected. Recall
that n3. By parts (a)–(g), we note that G/ Ftdom , implying that γL
t(G)2
3n<2
3n2. Every
γL
t-set of Gcan be extended to a LTD-set of Gby adding to it the vertices v2and v3, implying that
γL
t(G)<2
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=n34. 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 Gcontains 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=n5. Since u2has
degree 2 in G, we note that n3 and G/ Ftd om. Hence, γL
t(G)2
3n=2
3(n5) <2
3n3. Every
γL
t-set of Gcan 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 Gbelongs to the family Ftdom.
By linearity, we therefore have that γL
t(G)2
3n=2
3n2. Every γL
t-set of Gcan be extended to a
LTD-set of Gby adding to it the vertices v2and v3, implying that γL
t(G)<2
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 n7, the triangle Tcontains at most two vertices of degree 2, and so degG(v3) = 3.
Let G=GV(T) and let Ghave order n, and so n=n34.
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 Gbelongs to Ftdom . Hence by linearity, γL
t(G)2
3n=2
3n2.
Every γL
t-set of Gcan 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 Gcontains no component that belongs to Ftdom. By linearity,
this yields γL
t(G)2
3n=2
3n2. Every γL
t-set of Gcan 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, Gcontains a component that
belongs to Ftdom . Since n7 and no vertex of degree 2 in Gbelongs to a triangle, this is only possible
if either G
=K1,3or if Gcontains a K2-component.
On the one hand, if G
=K1,3, then the three vertices u1,u2, and u3are leaves in Gthat 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
On the other hand, if Gcontains 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, Gcontains two components, where
the second component contains the vertex u3. We now consider the graph G=G {v1, v2, v3, u1, u2}.
Let Ghave order n=n5. By the fact that δ(G)2 by Claim H, the graph G/ Ftdom, implying
that γL
t(G)2
3n=2
3(n5) <2
3n3. Every γL
t-set of Gcan be extended to a LTD-set of Gby
adding to it, for example, the vertices u2, v2and v3, implying that γL
t(G)<2
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=n34. Since
n7, δ(G) = 2, and Gcontains no triangle, no component of Gbelongs to Ftdom. Hence by linearity,
γL
t(G)2
3n=2
3n2. Every γL
t-set of Gcan be extended to a LTD-set of Gby adding to it the
vertices v2and v3, implying that γL
t(G)2
(b) Suppose that there is a (2,3,2)-sequence in G. As before, n=n34. Suppose that Gcontains
a component that belongs to Ftdom. Since there is no (2,2,2)-sequence and n7, 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
3n2. Every γL
t-set of Gcan 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