Minimal toughness in special graph classes

Let $t$ be a positive real number. A graph is called $t$-tough if the removal of any vertex set $S$ that disconnects the graph leaves at most $|S|/t$ components, and all graphs are considered 0-tough. The toughness of a graph is the largest $t$ for which the graph is $t$-tough, whereby the toughness of complete graphs is defined as infinity. A graph is minimally $t$-tough if the toughness of the graph is $t$, and the deletion of any edge from the graph decreases the toughness. In this paper, we investigate the minimum degree and the recognizability of minimally $t$-tough graphs in the classes of chordal graphs, split graphs, claw-free graphs, and $2K_2$-free graphs.


Introduction
All graphs considered in this paper are finite, simple and undirected.Let ω(G) denote the number of components, ∆(G) the maximum degree and κ(G) the connectivity number of the graph G. (Using ω(G) to denote the number of components might be confusing; most of the literature on toughness, however, uses this notation.)For a connected graph G, a vertex set S ⊆ V (G) is called a cutset if its removal disconnects the graph.
The notion of toughness was introduced by Chvátal (1973) to investigate Hamiltonicity.
Definition 1.1.Let t be a real number.A graph G is called t-tough if |S| ≥ t • ω(G − S) holds for any vertex set S ⊆ V (G) that disconnects the graph (i.e., for any S ⊆ V (G) with ω(G − S) > 1).The toughness of G, denoted by τ (G), is the largest t for which G is t-tough, taking τ (K n ) = ∞ for all n ≥ 1.
We say that a cutset Note that a graph is disconnected if and only if its toughness is 0. Obviously, the more edges a graph has, the larger its toughness can be.The main focus of this work is on graphs whose toughness decreases whenever any of their edges is deleted. 2 Gyula Y. Katona, Kitti Varga Definition 1.2.A graph G is said to be minimally t-tough if τ (G) = t and τ (G − e) < t for all e ∈ E(G).
For instance, every complete graph on at least two vertices is minimally ∞-tough.Note that since the toughness of any noncomplete graph is a rational number, there exist no minimally tough graphs with irrational toughness.
It follows directly from the definition that every t-tough, noncomplete graph is 2t-connected, where t is a nonnegative real number, implying κ(G) ≥ 2τ (G) for noncomplete graphs.Therefore, the minimum degree of any t-tough, noncomplete graph is at least ⌈2t⌉ for any nonnegative real number.
The following conjecture is motivated by a theorem of Mader (1971), which states that every minimally k-connected graph has a vertex of degree k, where k is a positive integer.
This conjecture can be naturally generalized to any positive real number t.
Conjecture 1.4 (Generalized Kriesell conjecture).Every minimally t-tough graph has a vertex of degree ⌈2t⌉, where t is a positive real number.
In the article by Katona et al. (2018), we proved that every minimally 1-tough graph has a vertex of degree at most n/3 + 1.Moreover, we also showed that Conjecture 1.3 holds for claw-free graphs by proving that every minimally 1-tough, claw-free graph is a cycle of length at least four.However, in general, determining whether a graph is minimally t-tough is hard: in the paper by Katona et al. (2021), we proved that this problem is DP-complete for any positive rational number t.We remark that DPcomplete problems are believed to be even harder than NP-complete ones; for more details about the complexity class DP, see the article by Papadimitriou and Yannakakis (1984).
Proving Conjecture 1.4 in general seems to be difficult, so the main motivation of this paper is to study the conjecture in some special graph classes.In the classes of chordal graphs, split graphs, and claw-free graphs, we can prove the conjecture for some values of t by giving a characterization of the minimally t-tough graphs in the special graph class, which easily implies the affirmative answer to the conjecture.In the case of 2K 2 -free graphs, the conjecture remains open, but we can at least show that these graphs can be recognized in polynomial time, which may give a chance to find a characterization later.The interesting property of these graph classes is that they are not closed under edge-deletion.
We prove that Conjecture 1.4 holds -for minimally t-tough, chordal graphs when t ≤ 1 is an arbitrary positive rational number, -for minimally t-tough, split graphs when t is an arbitrary positive rational number, -for minimally t-tough, claw-free graphs when t ≤ 1 is an arbitrary positive rational number.
In addition, we show that -minimally t-tough, split graphs for any positive rational number t, and -minimally t-tough, claw-free graphs for any positive rational number t ≤ 1, and -minimally t-tough, 2K 2 -free graphs for any positive rational number t can be recognized in polynomial time.Moreover, we give a characterization -of minimally t-tough, chordal graphs when t ∈ (1/2, 1] is an arbitrary rational number (more precisely, we show that there exist no such graphs), -of minimally t-tough, split graphs when t is an arbitrary positive rational number (more precisely, we show that there exist such graphs only when t is the reciprocal of an integer b ≥ 2), -of minimally t-tough, claw-free graphs when t ≤ 1 is an arbitrary positive rational number (by a theorem of Matthews and Sumner, we know that the toughness of a noncomplete, claw-free graph is either an integer or a half-integer, so we only need to study the cases when t = 1/2 or t = 1, and the latter case was already handled in our article Katona et al. (2018)).

Preliminaries
In this section, we study some basic properties of minimally tough graphs.
The following proposition is a simple observation.
Proposition 2.1.Let t ≤ 1 be a positive rational number and G a graph with τ (G) = t.Then for any nonempty proper subset S of V (G). Proof: On the other hand, |S|/t ≥ 1 since S = ∅ and t ≤ 1.Therefore, ω(G − S) ≤ |S|/t holds in this case as well.
As is clear from its proof, the above proposition holds even if S is not a cutset.However, it does not necessarily hold if t > 1 and S is not a cutset: if t > 1, then the graph cannot contain a cut-vertex, therefore ω(G − S) = 1 for any subset S with |S| = 1, while |S|/t = 1/t < 1.
The following proposition describes the basic structure of minimally tough graphs.
Proposition 2.2.Let t be a positive rational number and G a minimally t-tough graph.For every edge e of G, -the edge e is a bridge in G, or -there exists a vertex set S = S(e) ⊆ V (G) with and the edge e is a bridge in G − S.
In the first case, we define S = S(e) = ∅.
Proof: Let e be an arbitrary edge of G which is not a bridge.Since G is minimally t-tough, τ (G − e) < t.Since e is not a bridge, G − e is still connected, so there exists a cutset So assume that t > 1 holds.Then there are two cases.

Gyula Y. Katona, Kitti Varga
Case 1: (t > 1 and) S is a cutset in G. Since τ (G) = t and S is a cutset, ω(G−S) ≤ |S|/t.This is only possible if e connects two components of (G − e) − S, i.e. if e is a bridge in G − S.
Case 2: (t > 1 and) S is not a cutset in G.
Then ω(G − S) = 1.Since S is a cutset in G − e, the edge e must connect two components of (G − e) − S. Thus e is a bridge in G − S and ω (G − e) − S = 2. Now we show that ω(G − S) ≤ |S|/t holds.Suppose to the contrary that ω(G − S) > |S|/t.Since ω(G − S) = 1, this implies |S| < t.Moreover, since τ (G) = t, the graph G is ⌈2t⌉-connected, thus it has at least 2t + 1 vertices.From this, it follows that S and one of the endpoints of e form a cutset in G: otherwise, G would only have |S| + 2 < t + 2 < 2t + 1 vertices (where the latter inequality is valid since which implies t < 1, and that is a contradiction.
So in both cases, hold, and e is a bridge in G − S.
The starting point of each of our proofs is showing that for each edge e of the investigated graphs, there exists a vertex set S(e) guaranteed by Proposition 2.2 with some nice properties.

Chordal graphs
First, we study chordal graphs.Unlike for the other graph classes considered in this paper, the complexity of determining the toughness of chordal graphs is open.
A classical theorem of Dirac (1961) states that every chordal graph has a simplicial vertex; moreover, every noncomplete, chordal graph contains at least two nonadjacent simplicial vertices.
The main idea of the following proof is that in a minimally t-tough, chordal graph with t ≤ 1, if e is an edge in the neighborhood of a simplicial vertex v, then the vertex set S(e) guaranteed by Proposition 2.2 should, but cannot, contain v.
Proof: If t ≤ 1/2, then let G be a minimally t-tough, chordal graph, and if 1/2 < t ≤ 1, then suppose to the contrary that G is a minimally t-tough, chordal graph.
In both cases, let v be a simplicial vertex of G.If t ≤ 1/2, then suppose to the contrary that v has degree at least 2. If 1/2 < t ≤ 1, then every vertex of G has degree at least 2t > 1, so v has degree at least 2.
In both cases, let u and w be two neighbors of v. Since v is simplicial, u and w are adjacent.Let e = uw.Obviously, e is not a bridge, so by Proposition 2.2, there exists a vertex set and e is a bridge in G − S. Clearly, S must contain v.
Case 1: |S| ≤ 1, i.e. S = {v}.Since v is simplicial, (G − e) − {v} has exactly two components, and v has exactly one neighbor in both of them, so the degree of v is exactly 2. Hence, where the inequality 1 − 1/t ≤ 0 is valid since t ≤ 1, thus we obtained a contradiction.
Thus, Conjecture 1.4 is true for minimally t-tough, chordal graphs with t ≤ 1.Now it is natural to ask if there exist minimally t-tough, chordal graphs with t ≤ 1/2.Katona and Khan (2023) proved that the answer is affirmative if and only if t is a reciprocal of an integer b ≥ 2.Moreover, they gave a complete characterization of minimally t-tough, chordal graphs with t ≤ 1/2.Theorem 3.4 (Katona and Khan (2023)).A chordal graph is minimally t-tough for some positive rational number t ≤ 1/2 if and only if it can be obtained from a tree with maximum degree ∆ ≥ 3 by removing an independent set W of vertices of degree 3 and connecting the three neighbors of each vertex of W by a triangle, where W satisfies the following.
-Either ∆ = 3 and W is the set of all vertices of degree 3 such that every neighbor of any vertex in W has degree 2, Gyula Y. Katona, Kitti Varga -or ∆ ≥ 3 and W is some subset of vertices of degree 3 such that every neighbor of any vertex in W has degree ∆ ≥ 3.
The above theorem implies that for any positive rational number t ≤ 1/2, the class of minimally t-tough, chordal graphs can be recognized in polynomial time.

Split graphs
In this section, we study split graphs.It is not difficult to see that every split graph is chordal, thus the results of the previous section can be applied here.
Definition 4.1.A graph is a split graph if its vertex set can be partitioned into a clique and an independent set.
The toughness of split graphs can be computed in polynomial time; first, this was shown for t = 1 by Kratsch et al. (1996), then by Woeginger (1998) for all positive rational numbers t.
Theorem 4.2 (Woeginger (1998)).For any positive rational number t, the class of t-tough, split graphs can be recognized in polynomial time.
Note that if an edge goes between the clique and the independent set of the split graph, then after the removal of this edge, the graph is still a split graph, so we can compute in polynomial time whether the toughness decreased.The following lemma says that the vertex sets showing that the removal of any edge in the clique decreases the toughness can be uniquely determined.By Proposition 2.2, the edge e must be a bridge in G − S, so u, v / ∈ S and S 0 ⊆ S. Suppose to the contrary that S 0 S. Note that in (G − e) − S 0 , every vertex except u and v has degree at most 1, so since u, v / ∈ S, we obtain where the last inequality holds by the assumption S 0 S. Then by Proposition 2.2, which is a contradiction.
The next theorem states that except for the complete graphs, there exist no minimally t-tough, split graphs with t > 1/2.The main idea of the proof is to show that if such a graph existed with a partitioning Q ∪ I of its vertex set, where Q spans a clique and I an independent set, then every vertex of Q would have at most one neighbor in I, but these split graphs cannot be minimally t-tough.
Proof: Since every split graph is chordal, we can assume by Theorem 3.3 that t > 1 holds.Suppose to the contrary that there exists a minimally t-tough, split graph G. Obviously, G is noncomplete, otherwise its toughness would be infinity and not t.Let G be partitioned into a clique Q and an independent set I. Since G is noncomplete, I = ∅ and since G is connected and noncomplete, Q = ∅ also holds.
Now we show that we can assume |I| ≥ 2. We already know that I = ∅, so let us consider the case when |I| = 1 holds.Since G is noncomplete, we can pick a vertex w ∈ Q which is not adjacent to the vertex of I. Then we can consider for some b ≥ 2, and so τ (G) = 1/b ≤ 1/2, which would be a contradiction.So let e = uv be an edge in Q and S = S(e) ⊆ V (G) a vertex set guaranteed by Proposition 2.2.By Lemma 4.3, Then every component of G − S has size 1 except for the component of the edge e.Let Since By Proposition 2.2, Since t is positive and x is an integer, this implies x ≤ 1.Now we show l u ≤ 1.Since S is a cutset in G and u is not an isolated vertex in G − S, it follows that S ∪ {u} is also a cutset in G. Thus, On the other hand, by Proposition 2.2, Similarly, l v ≤ 1.Now we show that u has at most one neighbor in I, i.e., l u + x ≤ 1. Suppose to the contrary that u has more neighbors in I.By the above observations, this is only possible if x = 1 and l u = 1.Then the vertex set S ′ = (S \ I) ∪ {u} is a cutset in G and ω(G − S ′ ) = ω(G − S) + 1.In addition, since G is t-tough, ω(G − S ′ ) ≤ |S ′ |/t holds.Then by Proposition 2.2, which is a contradiction.Since e is an arbitrary edge in Q, it follows that every vertex of Q has at most one neighbor in I.By a previous assumption, |I| ≥ 2, and note that G is connected (otherwise its toughness would be 0).Therefore, there exists an edge f ∈ E(Q) both of whose endpoints have a neighbor in I but not a common one.Then by Lemma 4.3, and by Proposition 2.2, On the other hand, let a ∈ I be fixed, and let see Figure 2. Since κ(G) ≥ 2t, the vertex a has degree at least 2t, so which is a contradiction since t > 1.
We have just shown that for any rational number t > 1/2, there exist no minimally t-tough, split graphs.However, note that complete graphs are minimally ∞-tough, split graphs.Now we characterize minimally t-tough, split graphs for t ≤ 1/2: in the following theorem, we show that t must be a reciprocal of an integer b ≥ 2, and the class of minimally 1/b-tough, split graphs can be seen in Figure 3.  Then {v} is a cutset, and its removal leaves exactly 2 components, so it is a tough set having all the properties described in Proposition 2.2.This lemma has the following straightforward consequence.
Corollary 5.7.The class of minimally 1/2-tough, claw-free graphs can be recognized in polynomial time.
The following lemma describes another interesting property of minimally 1/2-tough, claw-free graphs.
Lemma 5.8.If G is a minimally 1/2-tough, claw-free graph, then all of its cycles have length 3.
Proof: If every edge of G is a bridge, then G is a tree, so it does not contain any cycles.
Let us assume that there exists an edge e = uv, which is not a bridge.Then by Lemma 5.6, there exists a vertex w which is a cut-vertex in G such that e is a bridge in G − {w}.Let L 1 and L 2 be the components of G − {w} (note that since G is 1/2-tough, G − {w} has exactly 2 components).Without loss of generality, we can assume u, v ∈ L 1 .Let L 1,1 and L 1,2 denote the components of (G − e) − {w} containing u and v, respectively.Since G is connected and e is not a bridge in G, the vertex w has neighbors in L 1,1 , L 1,2 , and L 2 .Since G is claw-free and w has neighbors in L 2 , the neighbors of w in L 1 must span a clique.Since e is a bridge in L 1 , and w has neighbors both in L 1,1 and L 1,2 , it follows that w has exactly one neighbor in L 1,1 , namely u, and has exactly one neighbor in L 1,2 , namely v. Since e is not a bridge in G, there exists a cycle containing the edge e, but then this cycle must also contain the vertex w.By the previous observations, this cycle must be {u, v, w}.
This means that in G every cycle has length 3. Now we are ready to characterize minimally 1/2-tough, claw-free graphs.
Theorem 5.9.The class of minimally 1/2-tough, claw-free graphs consists of exactly those graphs that can be built up in the following way.
1. Take a tree T with maximum degree at most 3 where the set of vertices of degree 1 and 3 together form an independent set.
2. Now delete every vertex of degree 3, but connect its 3 neighbors with a triangle.
Proof: Let G be a graph that can be obtained as described in the theorem.It is easy to see that G is claw-free.If G does not contain a triangle, then it must be a path on at least 3 vertices, which is clearly minimally 1/2-tough.If G contains some triangles, then by the construction of G, when removing any set S of vertices, the removal of each vertex of S creates at most one more new component, so G is 1/2-tough.On the other hand, if an edge of a triangle is deleted, then by removing the third vertex of this triangle, we obtain 3 components since the vertices of the triangle were not leaves in the original tree T .
All the other edges of the graph are bridges, so the graph is minimally 1/2-tough.Now we show that if a graph G is claw-free and minimally 1/2-tough, then it can be obtained as described in the theorem.
Case 1: G is a tree.Since G is a claw-free tree, it cannot have a vertex of degree at least 3, so G must be a path.A path on 2 vertices is K 2 , whose toughness is infinity, so it is not minimally 1/2-tough.A path on at least 3 vertices is obtained by setting T to be exactly this path.(In this case, Step 2 of the construction does not change T .) Case 2: G is not a tree.By Lemma 5.6, every vertex of every triangle is a cut-vertex.By Lemma 5.8, two triangles cannot share an edge.Since G is claw-free, any vertex not contained in any triangle has degree 1 or 2. By Lemma 5.8, if a vertex is not contained in any triangle, then it is not contained in any cycle; thus it is either of degree 1 or is a cut-vertex.Now apply the reverse of the operation given in Step 2 (i.e., for each triangle, remove its edges, add a new vertex and connect it with the vertices of the triangle).Let us call the newly added vertices red, the vertices of the triangles green, and the other vertices blue.Since G does not contain any cycles except for the triangles, the resulting graph must be a tree.Clearly, the red vertices have degree 3 and the blue ones have degree 1 or 2 in the tree.Now we show that the green vertices also have degrees 1 or 2 in the tree.Since the green vertices are contained in a triangle of G, they are cut-vertices.Since the toughness of G is 1/2, it is not difficult to see that the green vertices have degree 1 or 2 in the tree.So the red vertices have degree 3, and all the other vertices have degree 1 or 2 in the tree.
To complete the proof, we need to show that the set of vertices of degrees 1 and 3 in the tree together form an independent set.Two leaves of the tree cannot be adjacent since the graph is connected and has at least 3 vertices.Two vertices of degree 3 of the tree (i.e.red vertices) cannot be adjacent because of the way they were created.Two vertices of degrees 1 and 3 of the tree cannot be adjacent since that would mean that this leaf of the tree is contained in a triangle in G and has no neighbors outside the triangle, which contradicts the fact that it is a cut-vertex in G.
For an example to obtain a minimally 1/2-tough, claw-free graph described in Theorem 5.9, see Figure 5.With the help of this characterization, we can easily determine the minimum degree of minimally 1/2tough, claw-free graphs.

Gyula Y. Katona, Kitti Varga
Let L u and L v denote the two components of (G − e) − S for which u ∈ L u and v ∈ L v .By Claim 6.3, all the components of G − S are isolated vertices except for the component of the edge e.Since G is 2K 2free, these isolated vertices cannot be adjacent to w.We can assume that w has neighbors both in L u and L v , otherwise we can consider S = S \ {w} instead of S, see Figure 6.Since w is not adjacent to either u or v, both L u and L v must have size at least two.Since G is 2K 2 -free, both L u and L v are stars.(If there were any other edge in L u , then this edge and any edge induced by L v would be independent.)Clearly, S ∪ {u} is a cutset in G, and thus Obviously, the same holds for S ∪ {v}.By Proposition 2.2, we obtain Case 1: S is a cutset in G.
Then S \ {w} is also a cutset in G (since w has neighbors only in that component of G − S which contains the edge e), so By Proposition 2.2, we obtain {w} spans a cycle of length 5, namely wu 1 uvv 1 w.Therefore, S ′ = S \ {w} ∪ {u 1 } has the required properties.
Case 2: S is not a cutset in G.
Case 2.1: (S is not a cutset in G and) t > 1.
Case 2.2: (S is not a cutset in G and) t ≤ 1.Then |S| = 1 must hold, otherwise S \ {w} = ∅, so by Proposition 2.1, and so by Proposition 2.2, which would imply t > 1, a contradiction.
Theorem 6.5.For any positive rational number t, the class of minimally t-tough, 2K 2 -free graphs can be recognized in polynomial time.
Proof: By Theorem 6.2, we can compute the toughness of a 2K 2 -free graph in polynomial time.We only need to examine whether the removal of any edge from the graph decreases the toughness.Let e = uv be an arbitrary edge of the graph.If e is a bridge, then its removal from G obviously decreases the toughness.Let us assume that e is not a bridge.If G is minimally t-tough, then by Proposition 2.2, there exists S = S(e) ⊆ V (G) for which ω (G − e) − S > |S|/t, and by Theorem 6.4, we can assume that |S| = 1 or that S is contained in the open neighborhood of {u, v}.First, check whether there exists a cut-vertex in G − e whose removal from G − e leaves more than 1/t components (this can be clearly done in polynomial time).If it does, then the removal of e from G obviously decreases the toughness.If it does not, then start a BFS algorithm in G at u and v simultaneously (this can be also done in polynomial time).Since G is 2K 2 -free, the BFS tree has at most two levels (not counting the zeroth level containing u and v) and inside the second level there are no edges.If G is minimally t-tough, then the above mentioned set S exists and (since |S| = 1) it is contained in the open neighborhood of {u, v}, i.e. in the first level of this BFS tree.In addition, by Claim 6.3, all the components of G − S are isolated vertices except for the component of the edge e, thus every vertex in the first level belongs either to S or to the component of e.Therefore, if we remove the edge e from G and expand the first level into a clique by adding all necessary edges, then the toughness of the obtained split graph is equal to the toughness of G − e.By Theorem 4.2, we can compute the toughness of this split graph in polynomial time and check if it is less than t.
So we can decide in polynomial time whether a given 2K 2 -free graph is minimally tough.
Lemma 4.3.Let t be a positive rational number and G a minimally t-tough, split graph partitioned into a clique Q and an independent set I. Let e = uv be an edge between two vertices of Q and S = S(e) ⊆ V (G) a vertex set guaranteed by Proposition 2.2.ThenS = Q \ {u, v} ∪ w ∈ I uw, vw ∈ E(G) .Proof: Let S = S(e) ⊆ V (G) be an arbitrary vertex set guaranteed by Proposition 2.2 (such a vertex set exists since G is minimally t-tough), and let S 0 = Q \ {u, v} ∪ w ∈ I uw, vw ∈ E(G)}.Now we show that S = S 0 holds.

Fig. 1 :
Fig. 1: The set S = S(e) consisting of the vertices of Q \ {u, v} and the common neighbors of u and v.

Fig. 2 :
Fig. 2: The set R = R(a) consisting of those vertices of Q that are not adjacent to a.

Theorem 4. 5 .
Fig.4: A minimally 3/2-tough, claw-free graph and its edge e for which there exist no vertex sets S = S(e) guaranteed by Proposition 2.2 with |S| ≤ 3.

Fig. 6 :
Fig.6: The set S = S(e) guaranteed by Proposition 2.2 in a minimally 2K2-free graph with a vertex w ∈ S which is not adjacent to any of the endpoints of e.