On regular graphs with Šoltés vertices

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Introduction
All graphs under consideration in this paper are simple and undirected.The Wiener index of a graph G, denoted by W (G), is defined as where d G (u, v) is the distance between vertices u and v (i.e. the length of a shortest path between u and v).If the graph G is disconnected, we may take W (G) = ∞.The Wiener index was introduced in 1947 [17] and has been extensively studied ever since.For a recent survey on the Wiener index see [9].The transmission of a vertex v in a graph G, denoted by w G (v), is defined as w G (v) = u∈V (G) d G (u, v).Note that (1) can also be expressed as We say that a vertex v ∈ V (G) is a Šoltés vertex of G if W (G) = W (G − v), i.e. the Wiener index of G does not change if the vertex v is removed.If G is disconnected, with two non-trivial components or with at least three components, then every vertex is a Šoltés vertex.Therefore, it is natural to require that G is connected.Let and let 0 ≤ α ≤ 1.We say that a graph G is an α-Šoltés graph if |S(G)| ≥ α|V (G)|, i.e. the ratio between the number of Šoltés vertices of G and the order of G is at least α.For example, the graph in Figure 1 is the smallest cubic 1 3 -Šoltés graph.Note that G is an 1-Šoltés graph if every vertex in G is a Šoltés vertex.The only 1-Šoltés graph known to this day is the cycle on 11 vertices, C 11 .In this paper, Šoltés graph is simply the synonym for 1-Šoltés graph.
The Šoltés problem [15] was forgotten for nearly three decades.It was revived and popularised in 2018 by Knor et al. [7].They considered a relaxation of the original problem: one may look for graphs with a prescribed number of Šoltés vertices.They showed that there exists a unicyclic graph on n vertices with at least one Šoltés vertex for every n ≥ 9.They also showed that there exists a unicyclic graph with a cycle of length c and at least one Šoltés vertex for every c ≥ 5, and that every graph is an induced subgraph of some larger graph with a Šoltés vertex.They have further shown that a Šoltés vertex in a graph may have a prescribed degree [8].Namely, they proved that for any d ≥ 3 there exist infinitely many graphs with a Šoltés vertex of degree d.Necessary conditions for the existence of Šoltés vertices in Cartesian products of graphs were also considered [8].In 2021, Bok et al. [3] showed that for every k ≥ 1 there exist infinitely many cactus graphs with exactly k distinct Šoltés vertices.In 2021, Hu et al. [6] studied a variation of the problem and showed that there exist infinitely many graphs where the Wiener index remains the same even if r ≥ 2 distinct vertices are removed from the graph.
Akhmejanova et al. [1] considered another possible relaxation of the problem: Do there exist graphs with a given percentage of Šoltés vertices?They constructed two infinite families of graphs with a relatively high proportion of Šoltés vertices.Their first family comprises graphs B(k), k ≥ 2, where B(k) is a 2k 5k+6 -Šoltés graph on 5k +6 vertices.Two vertices of B(k) are of degree k + 1, while the remaining vertices are of degree 2. The percentage of Šoltés vertices is below 2  5 , but tends to 2 5 as k goes to infinity.They also introduced a two-parametric infinite family L(k, m), m ≥ 7 and k ≥ m−3 m−6 .Here, the percentage of Šoltés vertices is below 1  2 , but tends to 1 2 as k goes to infinity for a fixed m.These graphs contain at least one leaf, at least km vertices of degree 2, and a vertex of degree km + 1.
In the present paper we focus on regular graphs.Our intuition lead us to believe that the solutions to the original Šoltés problem should be graphs having all vertices of the same degree.
For a general regular graph G, the values W (G−u) and W (G−v) might be significantly different for two different vertices u, v ∈ V (G); it may happen that removal of one vertex increases the Wiener index, while removal of the other vertex descreases the Wiener index.However, W (G − u) and W (G − v) are equal if vertices u and v belong to the same vertex orbit.Therefore, we believe that a Šoltés graph is likely to be vertex transitive.It is not hard to obtain small examples of regular graphs with Šoltés vertices.We used the geng [11] software to generate small k-regular graphs (for k = 3, 4 and 5).Let R r denote the class of all r-regular graphs and let R r n denote the set of r-regular graphs on n vertices.Let N(G, k) be the number of graphs in the class G that contain exactly k Šoltés vertices.
Table 1 shows the numbers of (non-isomorphic) cubic graphs of orders n ≤ 24 that contain Šoltés vertices.We can see that cubic graphs of order 12 or less do not contain Šoltés vertices.There are plenty of examples with one Šoltés vertex.Cubic graphs with two Šoltés vertices first appear at order n = 14; there are three such graphs (see Figure 2(a)-(c)).Examples with three and four Šoltés vertices appear at order n = 16; there is one cubic graph with three and two cubic graphs with four Šoltés vertices (see Figure 2(d)-(f)).At order n = 18, there are no graphs with three Šoltés vertices, however there is only one graph with four Šoltés vertices (see Figure 2(g)).Numbers of 4-regular and 5-regular graphs with respect to their number of Šoltés vertices are given in Tables 2  and 3, respectively.gives the total number of cubic graphs of order n.Symbol '-' is a replacement for 0 (i.e.no such graph exists).Each of the next columns gives the numbers of graphs with k Šoltés vertices for k = 1, 2, . . ., 8. A blue-coloured number means that we have provided drawings of these graphs (see Figures 1 and 2).
Table 3: The numbers of (non-isomorphic) quintic graphs of order up to 14 with Šoltés vertices.Naming conventions used in Table 1 also apply here.In the next two sections we construct an infinite family of cubic 2-connected graphs with at least 2 r , r ≥ 1, Šoltés vertices.It recently came to our attention that Dobrynin independently found an infinite family of cubic graphs with four Šoltés vertices [4].However, our method is more general.

Cubic 2-connected graphs with two Šoltés vertices
In the present section we prove the following result.Theorem 4.There exist infinitely many cubic 2-connected graphs G which contain at least two Šoltés vertices.
We prove Theorem 4 by a sequence of lemmas.We start by giving several definitions.First, we define a graph G t on 8t + 8 vertices, where t ≥ 1.Take 2t copies of the diamond graph (i.e.K 4 − e) and connect their degree-2 vertices, so that a ring of 2t copies of K 4 − e is formed.Add a disjoint 4-cycle to that graph.Then subdivide one of the edges that connects two consecutive diamonds by two vertices, denote them by z 1 and z 2 , and connect z 1 and z 2 with two opposite vertices of the 4-cycle.Add a leaf to each remaning degree-2 vertex of the 4-cycle.Denote the resulting graph by G t .For an illustration, see G 3 in Figure 3.Note that G t has exactly two leaves, denote them by v 1 and v 2 , and all the remaining vertices have degree 3. Denote by u 1 and u 2 the two vertices at the longest distance from v 1 .This distance is Observe that G t has an automorphism (a symmetry) fixing both v 1 and v 2 , while interchanging u 1 with u 2 .
We compute f (t) by summing the contributions of all vertices (first the contribution of quadruples of vertices of all copies of K 4 − e, and then the contribution of the vertices which are not in any copy of K 4 − e).As the calculation is long and tedious, we present just the result which was checked by a computer.Actually, the exact value of f (t) is not important here.The crucial property is that for t big enough, f (t) is positive.In fact, lim t→∞ f (t) = ∞; see also Section 3, where a lower bound for f (t) is given.To motivate the above definition, we briefly describe the main idea of the proof.We attach trees T 1 and T 2 to vertices v 1 and v 2 of G t , and then we add edges to them so that the resulting graph H will be cubic and 2-connected.See Figure 4 for an example.This will be done in three phases.P1) In the first phase, we will construct trees T 1 and T 2 to guarantee that vertices u 1 and u 2 are indeed Šoltés vertices.The vertices of T 1 and T 2 can be partitioned into several layers based on their distance to v 1 and v 2 , respectively.The resulting graph will be denoted by Q.
P2) In the second phase, we will add the 'red' edges, whose endpoints are in two consecutive layers, to graph Q.The resulting graph will be denoted by R. The purpose of the second phase is to make sure that the sum of free valencies is even within each layer, making the next phase possible.Note that although the final graph H is cubic, the intermediate graphs, Q and R, are subcubic.The free valency of a vertex P3) In the last phase, we will add blue edges to R in order to to obtain cycles and paths, so that the resulting graph H will be cubic and 2-connected.The endpoints of 'blue' edges will reside in the same layer of the forest T 1 ∪ T 2 .Adding 'red' and 'blue' edges has no influence on Šoltésness of u 1 and u 2 .Let us consider the resulting graph We need to find suitable trees T 1 and T 2 rooted at v 1 and v 2 , respectively.Each of these trees will have q vertices and their depth, say d, will be determined later.What next properties do T 1 and T 2 need to have?Let ℓ i be the number of vertices of the forest Since the resulting graph will be cubic and . .and for the last value ℓ d we have 1 ≤ ℓ d ≤ 2 d+1 .The trees attached to v 1 and v 2 may be paths in which case we get (since each of T 1 and T 2 have exactly q vertices).In this case the transmission of v j in T j is biggest possible, 1 ≤ j ≤ 2. In the other extremal situation the transmission of v j in T j is smallest possible, 1 ≤ j ≤ 2, which results in

Denote by D the sum of distances from all vertices of
Observe that we need to find a finite sequence (ℓ 1 , ℓ 2 , . . ., ℓ d ) so that f (t) = D.As the resulting graph H has to be cubic, we need to add an even number of vertices, 2q, to the graph G t .What are the bounds for D? First, we determine the lower bound; let us denote it by D m .This bound will be obtained when ℓ i attains the maximum possible value of 2 i+1 for every 1 ≤ i ≤ d − 1.In other words, we are attaching complete binary trees to v 1 and v 2 .Let a = ⌊log 2 (2q + 3)⌋.
Recall that the depth of a complete binary tree with n vertices is ⌊log 2 (n)⌋ and note that in our case a − 1 = d.Then The above sequence will be called the short sequence and denoted by L m .Using the formula a i=1 ix i−1 = (ax a+1 − (a + 1)x a + 1)/(x − 1) 2 , we get Now we find the upper bound D m for D. In this case ℓ 1 = ℓ 2 = • • • = ℓ q = 2; this sequence will be called the long sequence and denoted by L m .Therefore, Note that D m and D m are functions of q and t.  4: The minimum and maximum value of q which satisfy the condition of Lemma 5.
Lemma 5. Let t ≥ 3. Then there exists q, such that For small values of t we computed the minimum and maximum value of q that satisfies the condition of the lemma; see Table 4.
Note that the value q in Lemma 5 is uniquely determined only for t = 3.For larger values of t we get a range of options.Any q between q min and q max can be used.Moreover, different values of q lead to non-isomorphic graphs H. Now, we show that for every integer D, D m ≤ D ≤ D m , there exists a finite sequence (ℓ 1 , ℓ 2 , . . ., ℓ d ) which realises D.Moreover, the graph G t can be extended to H by attaching trees T 1 and T 2 that have Let us define an operation M that modifies one such sequence L = (ℓ 1 , ℓ 2 , . . ., ℓ d ).
Every other D can be realised by a sequence that is obtained from L m by iteratively applying operation M. Assume that after (D − 1) − D m steps we obtained the sequence L = (ℓ 1 , ℓ 2 , . . ., ℓ d ) which realises D − 1. Obviously, M(L) realises D. It is easy to check that conditions (i) to (iii) are satisfied for M(L).
Next, we prove two additional properties that hold when operation M is iteratively applied on L m .Lemma 8. Let t ≥ 1 and q ≥ 0. Let L = (ℓ 1 , ℓ 2 , . . ., ℓ d ) be obtained from L m by iteratively applying operation M. Then the following holds: (i) If (ii) of Definition 6 applies, then ℓ j = 2 for all j < i.
Proof.(i): Observe that if part (ii) of Definition 6 applies and i ≥ 2, then ℓ i−1 = since otherwise i − 1 satisfies the assumption of the definition, thus i is not the smallest such value.Moreover, if there is k, k < i, with ℓ k ≥ 3, then choose the largest possible k with this property.Then Hence, k satisfies the assumption of the definition, a contradiction.It means that if part (ii) of the definition applies, then (ii): If ℓ i+1 = 0 then clearly ℓ i ≥ 3. Suppose that ℓ i = 3.Then ℓ i−1 = 2, since otherwise the assumptions apply to i − 1.As 2q is even, there must be k < i such that ℓ k ≥ 3. Let k be the largest possible value with this property.Then the assumptions apply to k, a contradiction.Thus, if ℓ i+1 = 0 then ℓ i ≥ 4.
Note that part (ii) of Lemma 8 means that in the sequence L ′ = M(L), ℓ ′ i ≥ 3 and ℓ ′ i+1 = 1.The corresponding graph H will thus have a single vertex of degree 3 in the final layer.
Example 9.For an illustration, the sequence of sequences for 2q = 20 is Lemma 10.Let t ≥ 3.There exist rooted trees T 1 and T 2 such that vertices u 1 and u 2 are Šoltés vertices in the graph Q obtained from G t by attaching T 1 and T 2 to vertices v 1 and v 2 .
Proof.By Lemma 5, there exists q such that D m ≤ f (t) ≤ D m .From Lemma 7, we obtain the sequence L, which gives us the appropriate number of vertices in every layer of the forest T = T 1 ∪ T 2 .This ensures that v 1 and v 2 are Šoltés vertices in Q.
Now, we construct a graph Q containing G t and realising L. Let T j be the tree rooted at v j , 1 ≤ j ≤ 2. Then T 1 will have ⌈ℓ i /2⌉ vertices at distance i from v 1 and T 2 will have ⌊ℓ i /2⌋ vertices at distance i from v 2 .Observe that it is possible to construct both T 1 and T 2 .We have two possibilities.
Case 1: ℓ i = 2ℓ i−1 .Then either i = 2 i+1 , ℓ i−1 = 2 i , . . ., ℓ 1 = 4 and on the first i levels both T 1 and T 2 are complete binary trees of height i; or , which means that both T 1 and T 2 contain one vertex at levels 1, 2, . . ., i − 1 and two vertices at level i.
even then we can construct i-th level of both T 1 and T 2 , and at least one vertex of level i − 1 of T 2 will have degree less than 3. On the other hand, if ℓ i−1 is odd then T 2 has only (ℓ i−1 − 1)/2 vertices at level i − 1.However, it has ⌊ℓ i /2⌋ vertices at level i and , so in T 2 , the number of vertices at level i is at most twice the number of vertices at level i − 1.In this case, at least one vertex at level i − 1 in T 1 will have degree less than 3.This concludes Case 2.
We construct the trees T 1 and T 2 so that at each level we minimise the number of vertices of degree 3. Observe that then there is no level in which there are vertices of degree 1 and also vertices of degree 3.
Consider v 1 and v 2 as vertices of level 0, and set ℓ 0 = 2.We plan to add edges within levels (i.e.'blue' edges) to create a cubic graph, but sometimes we must also add edges between consecutive levels (i.e.'red' edges).First, we add necessary edges connecting vertices of different levels.For every i ≥ 1, if i j=0 ℓ j is odd then add an edge joining a vertex (of degree ≤ 2) of (i − 1)-th level with a vertex of i-th level.Lemma 11.It is possible to add edges to Q as described above, so that the resulting graph has no parallel edges and it is subcubic.
Proof.We add the red edges step by step starting with level 1, together with creating the trees T 1 and T 2 .And we show that at each level it is possible to add a required edge.We distinguish two cases: Case 1: There is no red edge between levels i − 2 and i − 1.
for all j ≤ i.In both subcases i j=0 ℓ j is even, and no red edge is added between levels i − 1 and i. Suppose that 2ℓ i−1 > ℓ i .Then there is a vertex at (i − 1)-st level, say x, whose degree is less than 3.If 2ℓ i > ℓ i+1 , then there is a vertex at i-th level, say y, whose degree is also less than 3, and we can add the edge xy.(Observe that if we add these additional edges together with the construction of trees T 1 and T 2 , then we do not create parallel edges.The only problem occures when x is the unique vertex at level i − 1 in T j and y is also in T j .But then either ℓ i−1 = 3 and j = 2, in which case x can be chosen in T 1 , or ℓ i−1 = 2 and ℓ i = 3 in which case x can be chosen in T 2 and y in T 1 .)On the other hand, if 2ℓ i = ℓ i+1 then (recall that and ℓ i+1 = 4, so i j=0 ℓ j is even, and no red edge is added between levels i − 1 and i. Case 2: There is a red edge between levels i − 2 and i − 1.Then i−1 j=0 ℓ j is odd.Assume that we also have to add a red edge between levels i − 1 and i.Then i j=0 ℓ j is also odd which means that ℓ i is even and that 2ℓ i−1 > ℓ i , as shown in Case 1.Hence, 2ℓ i−1 > ℓ i + 1, so there is a vertex at (i − 1)-st level, say x, whose degree is less than 3.As 2ℓ i > ℓ i+1 , there is a vertex at i-th level, say y, whose degree is also less than 3, and we can add the edge xy.(Multiple edges can be avoided analogously as in Case 1.) We remark that we did not precisely specify how to choose the vertices x and y, when a red edge is add between levels i − 1 and i in case there are several possibilities.Here are a few simple rules to follow when choosing x or y at level i: (i) if there are at least three leaves (in T 1 ∪ T 2 ) at level i, then do not choose these leaves; (ii) if there is eactly one leaf w at level i, w ∈ V (T j ), then there must be a protected degree-2 vertex at level i in T 3−j (i.e. the protected vertex shall not be chosen); (iii) if there are two leaves w and w ′ at level i (one of them is in T 1 and the other in T 2 , as we will prove later), then one degree-2 vertex from T 1 and one degree-2 vertex from T 2 has to be protected; (iv) if there are no leaves at level i, we have no constraints.
The above rules will be fully justified later, when we will consider 2-connectivity of the resulting graph H. Denote by R the graph obtained after adding red edges, as described above.We have the following statement.
Lemma 12.In each level of R, the sum of free valencies is even.
Proof.Let 1 ≤ i ≤ d.We prove the statement for level i.So denote by a 1 , a 2 , . . ., a ℓ i the vertices at i-th level.Our task is to show that ℓ i j=1 (3 − deg R (a j )) is even.We distinguish two cases, with two subcases each.

Case 1:
i j=0 ℓ j is odd.Then there are ℓ i + 1 edges between levels ℓ i−1 and ℓ i in R. If ℓ i+1 is odd, then i+1 j=0 ℓ j is even, so there are ℓ i+1 edges between levels i and i + 1.
On the other hand, if ℓ i+1 is even, then i+1 j=0 ℓ j is odd, so there are ℓ i+1 + 1 edges between levels i and i + 1.Hence,

Case 2:
i j=0 ℓ j is even.Then there are ℓ i edges between levels ℓ i−1 and ℓ i in R. If ℓ i+1 is odd, then i+1 j=0 ℓ j is odd, so there are ℓ i+1 + 1 edges between levels i and i + 1.
On the other hand, if ℓ i+1 is even, then i+1 j=0 ℓ j is also even, so there are ℓ i+1 edges between levels i and i + 1.Hence, By Lemma 12, the sum of free valencies is even at each level of R.This means that, in general, after we add some edges connecting vertices within level i, and when we do that for all i, 1 ≤ i ≤ d, the resulting graph H will be cubic.We now describe how to add these 'blue' edges, so that H will be 2-connected, and how to resolve the cases when a level has small number of remaining degree-2 vertices.
Observation.H will be 2-connected if for every leaf x of T , say x ∈ V (T k ), where 1 ≤ k ≤ 2 and x is a vertex at level i, there is a path, say P , containing only the vertices of level i and connecting x with a vertex, say y, of T 3−k .
We refer to the above as the 2-connectivity condition.The reason is that P can be completed to a cycle using a (v k , x)-path in T k and a (v 3−k , y)-path in T 3−k .Since all cycles constructed in this way contain three vertices of G t , the resulting graph H will be 2-connected.We remark that in one special case the path P will contain vertices of levels i and i + 1, but it will still be possible to complete P to a cycle containing three vertices of G t .Thus, our attention will be focused on the leaves of T .Lemma 13.It is possible to add edges to R so that the resulting graph H will be cubic and 2-connected.
Proof.First, we consider the d-th (i.e., the last) level.Note that all the vertices at level d are leaves of T .Since d j=1 ℓ j = 2q is even, each vertex at level d has degree 1 in R. We distinguish three cases.Case 2: ℓ d = 1.Then ℓ d−1 ≥ 3, as already shown.In this case, we replace the sequence L = (ℓ 1 , ℓ 2 , . . ., ℓ d−1 , 1) by L * = (ℓ 1 , ℓ 2 , . . ., ℓ d−1 , 3) and we find a 2-connected cubic graph H * realizing L * .In this graph, the pendant vertices of level d are connected to three different vertices of level d − 1 in T , since at each level we minimised the number of degree-3 vertices.Since d j=1 ℓ j is even, there are no other edges connecting vertices of level d − 1 with those of level d in R. Hence, we add a 3-cycle as described in Case 1, and then contract the three vertices at level d to a single vertex.If H * is cubic and 2-connected, then so is the resulting graph H.If ℓ d−1 ≥ 3, then pick a vertex of degree 1 at level d − 1, say x, and join it to both vertices of level d.Then x has degree 3, but since it is a leaf of T , it does not satisfy the 2-connectivity condition in the strict sense.Nevertheless, there is a cycle in H which contains edges of G t , a path connecting v 1 with a vertex of level d in T 1 , a path connecting v 2 with a vertex of level d in T 2 , and the two edges connecting x with the vertices of level d, which is sufficient.Then add to H the edge connecting vertices of level d and add a path passing through all vertices of level d − 1 except x, and starting/ending in the two degree-2 vertices.This resolves the problem for levels d and d − 1.This concludes Case 3.
We now turn to level i, 1 ≤ i < d.In case ℓ d = 2, we assume i < d − 1.Then vertices of level i are connected to vertices of levels i − 1 and i + 1 using only the edges of R, and we now add only edges connecting vertices within level i, i.e. the blue edges.
In some cases, we specify positions of red edges that were added to T to form R, to justify the four rules for choosing vertices x and y in the process of creating R.
If there is no leaf at level i, then all vertices of this level have degrees 2 and 3 in R. By Lemma 12, there is an even number of degree-2 vertices.Thus, we can add a collection of independent edges so that all vertices of level i will have degree 3. Since there were no leaves, the vertices of level i satisfy the 2-connectivity condition.Now suppose that there are leaves at level i.Since we minimised the number of vertices of degree 3 when constructing T , there are no vertices of degree 3 in level i.Consequently, each vertex of level i is connected to at most one vertex in level i + 1 in T .Hence, T 1 and This means that the numbers of leaves at level i in T 1 and T 2 differ by at most one, and also that there are no degree-3 vertices at level i in T .Moreover, since i < d, level i contains two vertices, say b We distinguish three cases.
Case 1: T has at least 3 leaves at level i.If E(R) \ E(T ) contains an edge connecting levels i − 1 and i, then this edge will terminate at b 1 , and if E(R) \ E(T ) contains an edge connecting levels i and i + 1, then this edge will start at b 2 .(Note that if we create T and R simultaneously, level by level, then we can form b 1 and b 2 , so that we do not get parallel edges.In the worst case we relabel b 1 and b 2 , so that b 1 ∈ V (T 2 ) and b 2 ∈ V (T 1 ).)This leaves the leaves untouched.Then we add a cycle passing through all leaves of level i and add a collection of independent edges so that all vertices of level i become degree-3 vertices.Since, at level i, at least one leaf is in T 1 and at least one is in T 2 , the vertices at level i satisfy the 2-connectivity condition.
Case 2: T has exactly two leaves at level i. Denote these vertices by a 1 and a 2 .As mentioned above, we may assume that a 1 ∈ V (T 1 ) and a 2 ∈ V (T 2 ).If E(R) \ E(T ) contains an edge connecting levels i − 1 and i, then this edge will terminate at a 1 , and if E(R) \ E(T ) contains an edge connecting levels i and i + 1, then this edge will start at a 2 .(Again, not to create parallel edges, the red edge between levels i − 1 and i may be connected to a 2 instead of a 1 , and then possible red edge between levels i and i + 1 will start at a 1 .)Then add edges a 1 b 2 , a 2 b 1 , and a collection of independent edges so that all vertices of level i become degree-3 vertices.Due to the presence of edges a 1 b 2 and a 2 b 1 , the vertices at level i satisfy the 2-connectivity condition.
Case 3: T has exactly one leaf at level i. Denote this vertex by a. Without loss of generality, assume that a ∈ V (T 1 ).If E(R) \ E(T ) contains an edge connecting levels i − 1 and i, then this edge will terminate at b 1 , and if E(R) \ E(T ) contains an edge connecting levels i and i + 1, then this edge will start at a. (Not to create parallel edges, the red edge between levels i − 1 and i may be connected to a instead of b 1 , and then possible red edge between levels i and i + 1 will start at b 1 .)Then add the edge ab 2 , and a collection of independent edges so that all vertices of level i become degree-3 vertices.Due to the presence of edge ab 2 , vertices at level i satisfy the 2-connectivity condition.
3 Cubic 2-connected graphs with 2 r Šoltés vertices Now we generalise Theorem 4 to higher amount of Šoltés vertices.
Theorem 14.Let r ≥ 1.There exist infinitely many cubic 2-connected graphs G which contain at least 2 r Šoltés vertices.
Proof.We reconsider the graph G t from the proof of Theorem 4. This graph consists of a chain of 2t diamonds attached to vertices z 1 and z 2 of a graph on 8 vertices.Denote this graph on 8 vertices by F .We construct G t,r .Take a binary tree B of depth r − 1.This tree has 2 r − 1 vertices, out of which 2 r−1 are leaves.Denote these leaves by a 1 , a 2 , . . ., a 2 r−1 .Let B ′ be a copy of B. To distinguish endvertices of B ′ from those of B, put to the endvertices of B ′ dashes.Now take 2 r−1 chains of 2t diamonds and identify the ends (the vertices of degree 2) of k-th chain with a k and a ′ k , respectively.Finally, join the roots of B and B ′ (i.e., the vertices of degree 2) by edges to z 1 and z 2 .Denote by G t,r the resulting graph, see Figure 5 for G 3,2 .Then G t,r has 8t2 r−1 +2(2 r−1 − 1)+8 vertices.Moreover, all central vertices of 2 r−1 chains of diamonds belong to the same orbit of G t,r .Observe that there are 2 r such vertices.Let u 1 and u 2 be central vertices of one of the chains of diamonds.If we show that lim t→∞ (W (G t,r − u 1 ) − W (G t,r )) = ∞, we can complete G t,r analogously as G t was completed to H in the proof of Theorem 4, to obtain a cubic 2−connected graph with at least 2 r Šoltés vertices.
Thus, it remains to show that W (G t,r − u 1 ) − W (G t,r ) tends to infinity as t → ∞.
We first estimate w Gt,r (u 1 ) from above.For small i, there are at most 4 vertices at distance i from u 1 .For bigger i the amount of vertices at distance i grows, but it cannot exceed 4 • 2 r + 8 since there are 2 r−1 chains attached to F and F itself has 8 vertices.Thus, w Gt,r (u 1 ) ≤ (2 r+2 + 8) 1+3t+2r+3t+3 i=1 i.And if we held r constant, w Gt,r (u 1 ) can be bounded from above by a quadratic polynomial in t.Now we estimate x,y∈V (Gt,r)\{u 1 } d Gt,r−u 1 (x, y) − d Gt,r (x, y) from below.For every x, y ∈ V (G t,r ) \ {u 1 } we have d Gt,r−u 1 (x, y) ≥ d Gt,r (x, y), since G t,r has all paths which exist in G t,r − u 1 .However, it suffices to consider only x, y being in the same chain of diamonds as u 1 .Observe that the distance from u 2 to a neighbour of u The distance from u 2 to the second neighbour of u 1 is increased at least by 6t − 4, etc.However, we should consider also a neighbour of u 2 ( = u 1 ).For this vertex the distances are increased at least by 6t − 4, 6t − 6, . . .Summing up, Consequently, W (G t,r −u 1 )−W (G t,r ) is bounded from below by a cubic polynomial (in t) with leading coefficient 9. Thus, lim t→∞ (W (G t,r − u 1 ) − W (G t,r )) = ∞ as required.

Concluding remarks and further work
We believe that if there exists another Šoltés graph in addition to C 11 , it is likely to be vertex-transitive or has a low number of vertex orbits.Vertices of the same orbit are either all Šoltés vertices or none of them is.
Holt and Royle [5] have constructed a census of all vertex-transitive graphs with less than 48 vertices; these graphs can be obtained from their Zenodo repository [14] in the graph6 format [10].The repository contains 100 720 391 graphs in total, 100 716 591 of which are connected [12].The computer search revealed that the only Šoltés graph among them is the well-known C 11 .
We also examined the census of cubic vertex-transitive graphs by Potočnik, Spiga and Verret [13].Their census contains all (connected) cubic vertex-transitive graphs on up to 1280 vertices; there are 111 360 such graphs.CVT(n, i) denotes the i-th graph of order n in the census.No Šoltés graph has been found, but the search revealed that there exist graphs that are 1  3 -Šoltés, i.e. 1  3 of all vertices are Šoltés vertices.We found 7 cubic 1 3 -Šoltés graphs; all of them are trunctations of certain cubic vertex-transitive graphs.In this paper, the truncation of a graph G is denoted by Tr(G).Note that the truncation of a vertex-transitive graph is not necessarily a vertex-transitive graph; in the case of cubic graphs, there may be up to 3 vertex orbits.When doing the computer search, we have to check the Šoltés property for one vertex from each orbit only.Here is the list of cubic vertex-transitive graphs G, such that Tr(G) is a We were able to identify seven such graphs.However, we believe that there could exist many more.
Problem 15.Find an infinite family of cubic vertex-trainsitive graphs {G i } ∞ i=1 , such that Tr(G i ) is a 1  3 -Šoltés graph for all i ≥ 1.Moreover, we also found an example of a 4-regular 1  3 -Šoltés graph, namely the graph L(CVT(324, 104)).It has order 486 and is the line graph of CVT(324, 104), which is a Cayley graph.More data can be found in the Appendix.
Of course, 1 3 -Šoltés is still a long way from being Šoltés.The next conjecture is additionally reinforced by the fact that there are no Šoltés graphs among vertex-transitive graphs with less than 48 vertices.
Conjecture 16.The cycle on eleven vertices, C 11 , is the only Šoltés graph.

Appendix
There are 7 cubic vertex-transitive graphs G on up to 1280 vertices, such that Tr(G) is a 1  3 -Šoltés graph.Since all these graph are Cayley graphs, we give the generating set for the Cayley graph.Note that the group itself (its permutation representation) is given implicitly by these generators; however, we also give the group's ID from GAP's library of small groups [2].SmallGroup(n, k) is the k-th group of order n from that library.We also calculated girth, diameter and tested all graphs for bipartiteness.

Conjecture 2 .
If G is a Šoltés graph, then G is vertex transitive.Among truncations of cubic vertex-transitive graphs we found several 1 3 -Šoltés graphs; see Section 4. Interestingly, all our examples are in fact Cayley graphs and this leads us to pose the following conjecture.Conjecture 3. If G is a Šoltés graph, then G is a Cayley graph.

Figure 2 :
Figure 2: Examples of small regular graphs with two or more Šoltés vertices: (a) to (c) are the three cubic graphs of order 14 with two Šoltés vertices; (d) and (e) are the two cubic graphs of order 16 with four Šoltés vertices; (f) is the only cubic graph of order 16 with three Šoltés vertices; (g) is the only cubic graph of order 18 with four Šoltés vertices; (h) is the only quartic graph of order 13 with two Šoltés vertices.Šoltés vertices are coloured red.

Figure 4 :
Figure 4: A graph H with two Šoltés vertices, namely u 1 and u 2 , that contains G 3 .

Case 1 :
ℓ d ≥ 3.In this case, we add to graph R a cycle passing through all the vertices of level d.Then the vertices of level d will have degree 3 and they will satisfy the 2connectivity condition, since ⌈ℓ d /2⌉ vertices of level d are in T 1 and ⌊ℓ d /d⌋ of them are in T 2 .

Case 3 :
ℓ d = 2.In this case, we simultaneously resolve the problem for levels d and d −1.Since d−1 j=1 ℓ j is even, all vertices of level d − 1 have degree 1 except for two, which have degree 2. (Recall that T was constructed so that at each level the number of vertices of degree 3 was minimised.)If ℓ d−1 = 2, then connect both vertices of level d − 1 with both vertices of level d and add an edge connecting the vertices of level d.Then the vertices of levels d − 1 and d have degree 3 and they satisfy the 2-connectivity condition.

1 and b 2 ,
such that b 1 ∈ V (T 1 ), b 2 ∈ V (T 2 ) and b 1 , b 2 are not leaves in T .(Recall that if i = d − 1 and ℓ d = 1, then then we solve this case for L * where ℓ d = 3, and afterwards we provide the contraction of vertices at level d, see Case 2 above.)Then deg T (b 1 ) = deg T (b 2 ) = 2.

Figure 5 :
Figure 5: The graph G 3,2 .Edges of binary trees of depth 1 are red.

Table 1 :
The numbers of (non-isomorphic) cubic graphs with Šoltés vertices.There are no graphs with Šoltés vertices for orders up to 12.The column labeled |R3n |

Table 2 :
The numbers of (non-isomorphic) quartic graphs of orders up to 17 with Šoltés vertices.Naming conventions used in Table1also apply here.
Interestingly, all these graphs are Cayley graphs.Several properties of these graphs are listed in the Appendix.The graph CVT(768, 3650) is the only non-bipartite example, while the rest are bipartite.Girths of these graph are values from the set {4, 6, 8, 10, 12}.