Graphs with minimum fractional domatic number

The domatic number of a graph is the maximum number of vertex disjoint dominating sets that partition the vertex set of the graph. In this paper we consider the fractional variant of this notion. Graphs with fractional domatic number 1 are exactly the graphs that contain an isolated vertex. Furthermore, it is known that all other graphs have fractional domatic number at least 2. In this note we characterize graphs with fractional domatic number 2. More specifically, we show that a graph without isolated vertices has fractional domatic number 2 if and only if it has a vertex of degree 1 or a connected component isomorphic to a 4-cycle. We conjecture that if the fractional domatic number is more than 2, then it is at least 7/3.


Introduction
A set of vertices in a graph G is dominating if every vertex of the graph either belongs to the set or has a neighbour in the set.A domatic partition of G is a partition of its vertices into dominating sets.The maximum number of sets in a domatic partition of G is called the domatic number of G.This concept was introduced in 1975 by Cockayne and Hedetniemi [2], and since then appeared in many studies.It arises in several applications including facility location in networks [3] and lifetime maximization of sensor networks [5].
In sensor network applications, a typical scenario is that small battery-powered devices are deployed in a remote area where they need to continuously monitor environmental conditions (e.g.temperature, pressure, etc.) via sensors.The energy limitations of the devices and the remoteness of the network demand efficient power management.Dominating set based scheduling turned out to be a useful concept in this context.The redundancy graph consists of the vertices corresponding to the devices where two devices are connected by an edge if they can monitor the same area, i.e. one of them can be asleep when the other is active and vice versa.Thus, in order to monitor the entire area at any given moment it is sufficient that only nodes of a dominating set of the redundancy graph are active while the other nodes might be in a power saving mode.As an example let us consider a network that consists of five devices A, B, C, D, E each having one-month long battery and the redundancy graph is a 5- cycle (A, B, C, D, E, A).If no sleeping schedule is applied and all devices are always active, such a network could serve at most one month.A more efficient approach is to partition the vertices into two dominating sets, say {A, B, D} and {C, E}, and let only the devices in the first set to monitor the area for one month and then only the devices in the second set to do the job in the second month.Such a scheduling mechanism doubles the lifetime of the network.The domatic number of the redundancy graph is the maximum number of dominating sets that can successively monitor the network.
It turns out that one can achieve a longer lifetime by scheduling not necessarily disjoint dominating sets.In our example, we can attain a network lifetime of 2.5 months by activating devices in the following five dominating sets {A, C}, {B, D}, {C, E}, {A, D}, {B, E} in turn for half a month each.
It follows that the fractional domatic number of a graph is strictly greater than 2 if and only if the minimum degree of the graph is at least two and every connected component is different from a 4-cycle.
We prove the main result in Section 3 and conclude the paper in Section 4. In the next section we introduce necessary notions and auxiliary results.

Preliminaries
We consider simple graphs, i.e. undirected graphs, with no loops or multiple edges.We denote by V (G) and E(G) the sets of vertices and edges of a graph G, respectively.For a vertex v of a graph G = (V, E) we denote by N (v) the neighbourhood of v, i.e. the set of vertices adjacent to v, and by δ(G) the minimum degree of a vertex in G.As usual, C k and K p,q denote a k-vertex cycle and respectively a complete bipartite graph with p and q vertices in the parts.A cut vertex is a vertex in a connected graph that disconnects the graph upon deletion.Similarly, a cut edge is an edge in a connected graph that disconnects the graph upon deletion.A graph is 2-connected if it is connected and contains at least 3 vertices, but no cut vertex.A set Observation 2 (Ore [6]).If G is a graph without isolated vertices then the complement of a minimal dominating set of G is also a dominating set of G.
We say that a multiset every vertex of G belongs to (we shall also say is covered by) at most s dominating sets in D. The fractional domatic number F D(G) of G is the maximum of k/s over all natural numbers k and s such that G admits a (k, s)-configuration.
s .We will make use of the following known facts about the fractional domatic number Lemma 4 ( [4]).For any natural n, Lemma 5 ([4]).Let G and H be two graphs on disjoint vertex sets.Then A path in a graph G = (V, E) is a sequence (v 1 , v 2 , . . ., v k ) of pairwise distinct vertices, where v i v i+1 ∈ E for every i = 1, 2, . . ., k − 1.The first and the last vertices of a path are called the endvertices of the path, and all other vertices are the internal vertices of the path.A path in G is called binary if all its internal vertices have degree 2 in Proof.If G is 2-connected then we are done.Otherwise G contains a cut vertex.We consider two cases.
The first case is when G has a cut vertex of degree 2. Let v be such a vertex.The graph G − v has exactly two connected components.One of these components contains one neighbour of v and the other component contains the other neighbour.Note that v together with its two neighbours form a binary path in G.We extend this path to a maximal binary path P = (x, v 1 , v 2 , . . ., v k , y) in G, where v ∈ {v 1 , v 2 , . . ., v k }.Since P is maximal, the degree of each of x and y in G is different from 2. By assumption, their degree cannot be 1, so it is at least 3. Hence, the degree of x and y in G − {v 1 , v 2 , . . ., v k } is at least 2. Furthermore, the degree of every other vertex in G − {v 1 , v 2 , . . ., v k } is the same as in G. Consequently, G is a dumbbell with the handle P and the plates corresponding to the two connected components of G − {v 1 , v 2 , . . ., v k }.The minimum degree of the plates is at least 2.
Assume now that every cut vertex in G has degree at least 3. Let v be such a vertex and S 1 , S 2 , . . ., S t ⊆ V (G), t ≥ 2, be the connected components of G − v.Note that every connected component has at least two vertices, as otherwise the unique vertex in a connected component would have degree 1 in G.If in some connected component v has a unique neighbour u, then u is a cut vertex in G and hence, by assumption, it has degree at least 3. Then G is a dumbbell with the handle P = (u, v) and the plates that correspond to the two connected components of the graph obtained from G by deleting the cut edge uv.If v has more than two neighbours in each component, then G is a (H 1 , H 2 )-dumbbell with the handle P = (v) and the two plates each with minimum degree at least 2.

Fractional domatic number
In this section we prove our main Theorem 1.As discussed in the introduction, if a graph G has an isolated vertex, then its fractional domatic number is 1, and if the miminum degree of G as at least one, then its fractional domatic number is at least 2. Note that by Lemma 5, it is enough to prove Theorem 1 for connected graphs only.Furthermore, by Lemma 4, the fractional domatic number of a C 4 is 2. Hence, in this section, in order to prove Theorem 1 we will show Using Lemma 6, we will split our analysis into two parts.In Section 3.1, we will deal with dumbbells, and in Section 3.2 we will tackle 2-connected graphs.In Section 3.3 we will put everything together to prove Theorem 7. In the rest of this section we introduce necessary notation and prove some auxiliary results.We start with useful properties of configurations.
Observation 9.If a graph G has fractional domatic numer greater than 2, then it admits a (2k+1, k)configuration for any sufficiently large k.
Proof.By definition, G admits a (p, q)-configuration such that F D(G) = p/q > 2. It follows that p ≥ 2q + 1.By removing p − 2q + 1 dominating sets from the (p, q)-configuration, we obtain a (2q + 1, q)-configuration D of G. Since F D(G) > 2, G has no isolated vertices and therefore, by Observation 2, for any minimal dominating set D ⊆ V (G), its complement D = V (G) \ D is also a dominating set and thus they together form a (2, 1)-configuration D ′ .Now, for every k > q, from Observation 8, by adding k − q copies of D ′ to D we obtain a (2k + 1, k)-configuration of G, as required.
Given a multiset D of dominating sets of G and two distinct vertices x and y in G we define the following multisets We say that a (2r + 1, r)-configuration D of G is (x, y)-nice if each of x and y belongs to exactly r sets in D, and D x , D y , and D xy are all nonempty.
Lemma 10.Let G = (V, E) be a graph and let P = (x, v 1 , v 2 , . . ., v s , y) be a binary path in G with at least two vertices.If the graph We can assume that P has at least one internal vertex, i.e. {v 1 , v 2 , . . ., v s } = ∅, as otherwise H would coincide with G and the conclusion would trivially hold.Let P ′ denote the path consisting of the internal vertices of P , i.e.
R 3 are pairwise disjoint sets.We are now ready to define D ′ by extending every dominating set of H in D to a dominating set of G. Later we will show that D ′ is a (2r + 1, r)-configuration of G.We consider three cases.We are now going to show that D ′ is a (2r+1, r)-configuration in G.It is clear from the construction that every set in D ′ is a dominating set in G and |D ′ | = |D| = 2r + 1.Hence, it remains to show that every vertex in G belongs to at most r sets in D ′ .In fact, since D ′ \ D ⊆ {v 1 , v 2 , . . ., v s } for every D ∈ D, we only have to prove that every vertex in {v 1 , v 2 , . . ., v s } is covered by at most r sets in D ′ .
We will show the latter depending on the value of s modulo 3. The cases s ≡ 0 (mod 3) and s ≡ 1 (mod 3) are similar, so we consider only one of them.
Let s ≡ 0 (mod 3).By definition, R 0 , R 1 , and R 2 are pairwise disjoint, and therefore we can treat vertices v 1 , v 2 , . . ., v s depending on which of the three sets they belong to.By construction, all vertices in R 0 belong to |D where in the inequality we used (1).We define three pairwise disjoint sets R α = {d i : i ≡ α (mod 3), i ∈ [s]}, α ∈ {0, 1, 2}.It is straightforward to check that, depending on the value of s modulo 3, each of the three families of sets in Table 1 is a (7, 3)-configuration of G. s ≡ 0 (mod 3) We break down the analysis into three cases depending on the value of s modulo 3.As before we define three pairwise disjoint sets

Dumbbells
It is easy to check that all sets in F are dominating sets of G. Furthermore, notice that F is obtained by extending sets in D, so the two configurations have the same cardinality 2k + 1.It remains to verify that every vertex of G is covered by at most k sets in F. This is clearly the case for any vertex of H that is different from d s .For d s , we observe that it belongs to R 0 , and R 0 is added only to the sets from D ds .Hence d s is covered in F by the same number of sets as in D, i.e. by exactly k sets.Similarly, all the other vertices in R 0 are covered by exactly k sets.Vertex a and each of the vertices in ds ∪ D 2 ds be an arbitrary balanced partition of D ds .We define the desired (2k + 1, k)-configuration F of G as the union of the following five multisets: Following similar reasoning as in the previous case, on can check that |F| = |D|, the sets in F are dominating sets of G, and all vertices of H are covered by the same number of sets in F as in D. Furthermore, every vertex in R 1 is covered by exactly k sets, and every other vertex outside V (H) is covered by at most ⌈k/3⌉ + ⌈(k Finally, assume that s ≡ 2 (mod 3).Let D ds = D 1 ds ∪ D 2 ds ∪ D 3 ds be an arbitrary balanced partition of D ds .We define the desired (2k + 1, k)-configuration F of G as the union of the following four multisets: Using arguments similar to those used in the previous two cases, it is not hard to verify that F is indeed a (2k + 1, k)-configuration of G.
Proof.We start by observing that since H 2 admits a (2k+1, k)-configuration, it also admits a (2r+1, r)configuration.Indeed, let D be a minimal dominating set in H 2 .As H 2 does not contain isolated vertices, by Observation 2, D = V (H 2 ) \ D is also a dominating set in H 2 .Hence, {D, D} is a (2, 1)configuration of H 2 .Therefore, Observation 8 implies that by extending a (2k + 1, k)-configuration of H 2 with r − k copies of {D, D} we obtain a (2r + 1, r)-configuration of H 2 . Let Let also P be the handle of G.Note that H 1 and H 2 either have exactly one vertex in common (if |V (P )| = 1), or they are vertex disjoint (if |V (P )| ≥ 2).Observe that in either case, for any Suppose first that H 1 and H 2 have one vertex in common, which we denote by x.Without loss of generality assume that the dominating sets in D are indexed in such a way that x belongs to D i if and only if i ∈ [t] for some t ≤ r.Similarly, assume that the sets in D ′ are indexed in such a way that x ∈ D ′ i if and only if i ∈ [t ′ ] for some t ′ ≤ r.Then it is easy to see that Suppose now that H 1 and H 2 are vertex disjoint and the handle P = (x, v 1 , v 2 , . . ., v s , y) has at least two vertices: x ∈ V (H 1 ) and y ∈ V (H 2 ).By definition, P is a binary path in G and In this case, the statement immediately follows from Lemma 10 if we show that H 1 ∪ H 2 admits a (x, y)-nice (2r + 1, r)-configuration.To prove the latter, we can assume that x belongs to exactly r sets in D. Indeed, otherwise we can add x to some sets in D that do not contain x to ensure the property.Similarly, we can assume that y belongs to exactly r sets in D ′ .Without loss of generality, suppose that x ∈ D i and y ∈ D ′ i if and only if i ∈ [r].Then is a (2r + 1, r)-configuration of H 1 ∪ H 2 and each of x and y belongs to exactly r sets in D ′′ , and D ′′ x , D ′′ y , and D ′′ xy are all nonempty, i.e.D ′′ is (x, y)-nice.This completes the proof.

2-connected graphs
The main goal of this subsection is to prove that the fractional domatic number of any 2-connected graph G, which is distinct from C 4 , is more than 2. To this end we will employ ear decompositions.An ear in a graph is either a cycle with at least 3 vertices or a path.
2. the first ear P 1 in the sequence is a cycle; and 3. for every i = 2, 3, . . ., t, P i is a path and exactly two vertices of P i , which are the endpoints of P i , belong to the earlier ears.
The following theorem is a classical result due to H. Whitney.
Theorem 15 (Whitney, [10]).A graph is 2-connected if and only if it has an open ear decomposition.
Our strategy is as follows.We will show that if the fractional domatic number of a graph G is more than 2, then after adding an open ear P = (x, . . ., y), to G the fractional domatic number of the resulting graph G ′ stays strictly above 2. To prove this, we will show (Lemma 19) that G admits a (x, y)-nice (2r + 1, r)-configuration.The result will then follow from Lemma 10 as P is a binary path in G ′ .
Using this result and Theorem 15 we will show (Theorem 20), by induction on the number of ears in an ear decomposition, that the fractional domatic number is above 2 as long as it is above 2 for the first ear in the ear decomposition.Since, by Lemma 4, the fractional domatic number of any cycle that is different from C 4 is more than 2 and any cycle can start an ear decomposition, the above would work for any 2-connected graph that has a cycle different from C 4 .As we show in the next two lemmas, in the case when all cycles in a 2-connected graph are C 4 s, the graph has very simple structure and its fractional domatic number is more than 2.
Lemma 16.Let G be a 2-connected graph with n ≥ 4 vertices that contains no (not necessarily induced) cycles of length other than 4. Then G = K 2,n−2 .
Proof.We prove the statement by induction on the minimum number of ears in an open ear decomposition.Let P 1 , P 2 , . . ., P k be an open ear decomposition of G with the minimum number of ears.If k = 1, then by the assumption G = C 4 = K 2,2 and the statement holds.Let now k ≥ 2 and let G ′ be the two connected graph with the ear decomposition P 1 , P 2 , . . ., P k−1 .Since G ′ is obtained from G by removing some edges and/or vertices (namely, the edges and the internal vertices of P k ), G ′ has no cycles of length other than 4. Hence, by the induction hypothesis, G ′ = K 2,p for some p ≥ 2. Now, if P k consists of a single edge xy, then x and y are non-adjacent in G ′ , and therefore they belong to the same part of the K 2,p .But then G contains a cycle on 3 vertices.Hence we assume that P k has at least one internal vertex and denote P k = (x, v 1 , v 2 , . . ., v t , y), where t ≥ 1.If the number t of the internal vertices in P k is at least 2, then it is easy to see that G contains a cycle of length more than 4. Hence t = 1.If x and y belong to the different parts of K 2,p or they are in the part with p ≥ 3 vertices, then it is easy to check that G contains a cycle on at least 5 vertices.This contradiction shows that x and y must belong to the part of K 2,p that contains 2 vertices, and hence G = K 2,p+1 .
Lemma 17.For every p ≥ 2, K 2,p admits a (3p − 2, p)-configuration, and therefore F D(K 2,p ) ≥ 3p−2 p .Proof.Let a 1 , a 2 and b 1 , b 2 , . . ., b p be the vertices of the K 2,p such that both a 1 and a 2 are adjacent to every vertex b i , i ∈ [p].It is straightforward to verify that the following family of dominating sets is a (3p − 2, p)-configuration: Proof.If G is equal to K 2,p for some p ≥ 3, then F D(G) ≥ 7  3 > 2 by Lemma 17.Hence, by Lemma 16, we assume that G contains a cycle C of length 3 or more than 4. It is known that for any cycle in a 2-connected graph there is an open ear decomposition that starts with this cycle (see e.g.[9, Theorem 4.2.8]).Let P 1 , P 2 , . . ., P t , t ≥ 1 be an open ear decomposition of G with P 1 = C.We will show by induction on i ∈ [t] that the graph G i with the open ear decomposition P 1 , P 2 , . . ., P i has fractional domatic number greater than 2. The theorem will then follow for i = t.
In the base case i = 1, the claim follows from Lemma 4. Assume therefore that i ≥ 2 and the statement holds for G i−1 .Let P i = (x, v 1 , v 2 , . . ., v q , y) and note that P i is a binary path in G i .Since F D(G i−1 ) > 2, by Observation 9 there exists a (2k + 1, k)-configuration of G i−1 , for some k ≥ 1, and hence, by Lemma 19, G i−1 admits a (x, y)-nice (2r + 1, r)-configuration for some r ≤ 2k + 1.Therefore, by Lemma 10, G i has a (2r + 1, r)-configuration implying F D(G) ≥ 2r+1 r > 2.

Putting everything together: proof of Theorem 7
In this section we combine the facts presented in the paper to prove our main result Theorem 7. Let G be a connected graph with δ(G) ≥ 2 that is different from C 4 .Then F D(G) > 2.
Proof.We will prove the statement by induction on the number of vertices in G. Clearly the statement holds for a single-vertex graph.Let n > 1 and let G be an arbitrary n-vertex graph.Suppose the statement hold for all graphs with less than |V (G)| vertices.
Without loss of generality we can assume that G is different from C 4 and δ(G) ≥ 2, as otherwise the fractional domatic numer of G is either 1 or 2. Since δ(G) ≥ 2, by Lemma 6, G is either a 2-connected graph or a (H 1 , H 2 )-dumbbell for some connected graphs H 1 and H 2 with δ(H 1 ) ≥ 2 and δ(H 2 ) ≥ 2. In the former case, the result holds by Theorem 20.Hence assume that G is a (H 1 , H 2 )-dumbbell.
If both H 1 and H 2 are C 4 s, then the result follows from Lemma 11.If exactly one of the graphs, say H 1 , is C 4 , and graph H 2 is different from C 4 and δ(H 2 ) ≥ 2, then F D(H 2 ) > 2 by the induction hypothesis, and hence H 2 admits a (2k+1, k)-configuration for some k ≥ 3. Then the statement follows from Lemma 12. Finally, if both H 1 and H 2 are different from C 4 , then by the induction hypothesis, both of them have fractional domatic number more than 2 and therefore they admit (2r + 1, r)-and (2k + 1, k)-configurations respectively for some positive r and k.The statement then follows from Lemma 13.

Conclusion
In this paper we characterized graphs with fractional domatic number 2. In order to do this, we showed that any connected graph of minimum degree at least two that is different from C 4 has fractional domatic number more than 2. While our proof does not bound the fractional domatic number away from 2, we did not find any graph with this parameter being strictly between 2 and 7/3.In fact we believe that there are no such graphs Conjecture 21.If the fractional domatic number of a graph is greater than 2, then it is at least 7/3.
In our approach the only obstacle to proving this conjecture is Lemma 19 which guarantees that given a (2k +1, k)-configuration one can always find a (x, y)-nice (2r+1, r)-configuration for some r ≤ 2k +1, but does not guarantee that one can always find such a configuration with r = k, which would be enough to settle the conjecture.
and |D xy | ≥ 1.The latter together with the fact that D x ∪ D y ∪ D xy ∪ D xy is a partition of D implies a simple but important inequality that we will use later

1 .
If s ≡ 0 (mod 3), then (a) R 0 dominates all vertices of P ′ except v 1 , and for every D ∈ D x we let D ′ = D ∪ R 0 ; (b) R 1 dominates all vertices of P ′ except v s , and for every D ∈ D y we let D ′ = D ∪ R 1 ; (c) R 2 dominates all vertices of P ′ , and for every D ∈ D xy we let D ′ = D ∪ R 2 ; (d) for every D ∈ D xy we define D ′ = D ∪ R 0 .2. If s ≡ 1 (mod 3), then (a) R 0 dominates all vertices of P ′ except v 1 , and for every D ∈ D x we let D ′ = D ∪ R 0 ; (b) R 1 dominates all vertices of P ′ , and for every D ∈ D xy we let D ′ = D ∪ R 1 ; (c) R 2 dominates all vertices of P ′ , except v s , and for every D ∈ D y we let D ′ = D ∪ R 2 ; (d) for every D ∈ D xy we define D ′ = D ∪ R 0 .3. If s ≡ 2 (mod 3), then (a) R 0 dominates all vertices of P ′ except v 1 and v s , and for every D ∈ D xy we let D ′ = D ∪ R 0 ; (b) R 1 dominates all vertices of P ′ , and for every D ∈ D x we let D ′ = D ∪ R 1 ; (c) R 2 dominates all vertices of P ′ , and for every D ∈ D y we let D ′ = D ∪ R 2 ; (d) first, we partition D xy arbitrary into two parts D 1 xy and D 2 xy , with |D 1 xy | = min{r − |D x |, |D xy |}; second, for every D ∈ D 1 xy we define D ′ = D ∪ R 1 , and for every D ∈ D 2 xy we define D ′ = D ∪ R 2 .

1 .> 2 .
the p sets {a 1 , b i }, i ∈ [p]; 2. the p sets {a 2 , b i }, i ∈ [p]; and 3. p − 2 copies of {b 1 , b 2 , . . ., b p }.Observation 18.Let G = (V, E) be a graph, let S be a dominating set in G, and let D be a (2k+1, k)configuration of G. Then the multiset D ′ that consists of the set S and two copies of every element in D is a (2r + 1, r)-configuration of G, where r = 2k + 1.Proof.By construction, |D ′ | = 2|D| + 1 = 2r + 1.Moreover, every vertex in G belongs to at most 2k sets in D ′ \ {S} and therefore every vertex belongs to at most r = 2k + 1 sets of D ′ .Hence the claim.Lemma 19.Let G = (V, E) be a graph and let x, y be two distinct vertices in G.If G has a (2k +1, k)configuration D, then, for some r ≤ 2k + 1, G has a (x, y)-nice (2r + 1, r)-configuration D ′ .Proof.First, let us assume that 1 ≤ |D xy | < k.In this case, the desired (2r + 1, r)-configuration D ′ , where r = k, is obtained from D by extending some of the sets in D as follows: 1. add x to k − |D x | − |D xy | sets of D xy ; 2. add y to some other k − |D y | − |D xy | sets of D xy .Notice that such a modification can always be done, because (k − |D x | − |D xy |) + (k − |D y | − |D xy |) ≤ (k − |D x | − |D xy |) + (k − |D y |) + 1 = |D xy |.Clearly, every new set remains dominating in G, and each of x and y belongs to exactly k sets of D ′ .Moreover, since |D xy | < k, we have that both D ′ x and D ′ y are nonempty.Hence, D ′ is (x, y)-nice.Suppose now |D xy | = 0. Then we define D ′ to be the multiset obtained from D by duplicating every element in D and adding the dominating set V .By Observation 18, the multiset D ′ is a (2r + 1, r)configuration of G, where r = 2k + 1.Moreover, |D ′ xy | = 1 < r, and therefore, by the above argument, D ′ can be turned into the desired (x, y)-nice (2r + 1, r)-configuration.Similarly, if |D xy | = k, we define D ′ to be the multiset obtained from D by duplicating every element in D and adding the set V \ {x, y}, which is dominating because every vertex in G has degree at least two as F D(G) ≥ 2k+1 k As before, D ′ is a (2r + 1, r)-configuration with r = 2k + 1.Moreover, 1 ≤ |D ′ xy | = 2k < r, and hence, by the above argument the desired (x, y)-nice (2r + 1, r)-configuration can be obtained from D ′ .Theorem 20.Let G be a 2-connected graph, which is not a C 4 .Then F D(G) > 2.
Given a path or a cycle P in G we denote by E(P ) the set of edges of P .Let H 1 and H 2 be two graphs that have at most one vertex in common.A graph G is called a (H 1 , H 2 )-dumbbell if G is obtained from H 1 and H 2 by connecting them with a path.More formally, G is the union of H 1 , H 2 , and P , where 1. P is a binary path in G with |V (P )| ≥ 1; 2. for each i ∈ {1, 2}, the set V (H i ) ∩ V (P ) has exactly one element, which is an end-vertex of P ; 3. if |V (P )| ≥ 2, then H 1 and H 2 are vertex disjoint.The path P is called the handle of the dumbbell and the graphs H 1 and H 2 are its plates.A graph is a dumbbell if it is a (H 1 , H 2 )-dumbbell for some H 1 and H 2 .

Table 1 :
} (7, 3)-configurations of G depending on the value of s modulo 3.3.1.2(C4 , H)-dumbbellsLet S be an n-element set.A partition of S into k ≥ n sets is balanced if the cardinalities of any two parts in the partition differ at most by one.Clearly, from the definition, the cardinality of any part is either ⌈n/k⌉ or ⌊n/k⌋.
Lemma 12. Let G be a (C 4 , H)-dumbbell.If H admits a (2k + 1, k)-configuration, for k ≥ 3, then G does too.Proof.Let (a, b, c, d 1 , a) be the C 4 plate of the dumbbell and P = (d 1 , d 2 , . . ., d s ), s ≥ 1 be the handle of the dumbbell.Let D be a (2k + 1, k)-configuration of H, D ds be the family of dominating sets in D that contain d s , and D ds be the family of dominating sets that do not contain d s , i.e.D ds = D \ D ds .Without loss of generality we assume that |D ds | = k.