Notes on Hamiltonian threshold and chain graphs

We revisit results obtained in [F. Harary, U. Peled, Hamiltonian threshold graphs, Discrete Appl.~Math., 16 (1987), 11--15], where several necessary and necessary and sufficient conditions for a connected threshold graph to be Hamiltonian were obtained. We present these results in new forms, now stated in terms of structural parameters that uniquely define the threshold graph and we extend them to chain graphs. We also identify the chain graph with minimum number of Hamilton cycles within the class of Hamiltonian chain graphs of a given order.


Introduction
A threshold graph can be defined in many ways, as can be seen in [4]. Here we follow the definition via binary generating sequences. Accordingly, a threshold graph G(b) is obtained from its binary generating sequence of the form b = (b 1 b 2 · · · b n ) in the following way: (i) for i = 1, G 1 = G(b 1 ) = K 1 , i.e., a single vertex; is formed by adding an isolated vertex to G i−1 if b i = 0 (that is, a vertex nonadjacent to any vertex in G i−1 ) or by adding a dominating vertex to G i−1 if b i = 1 (that is, a vertex adjacent to all the vertices in G i−1 ).
Clearly, G(b) = G n . A schematic representation of a threshold graph is illustrated in Fig 1.1(a); its vertices are partitioned into cells U i , V i (1 ≤ i ≤ h).
The vertices in U = h i=1 U i induce a co-clique, while the vertices in V = h i=1 V i induce a clique.
A bipartite counterpart to the threshold graph is a chain graph which is generated by the same binary sequence in the following way: (i) for i = 1, G 1 = G(b 1 ) = K 1 , i.e., a single vertex belonging to one colour class, say white vertex; A schematic representation of a threshold and a chain graph.
written as (Naturally, t i 's and s i 's are lengths of maximum runs of consecutive zeros and ones, respectively.) In a so-called split graph the vertex set can be divided into two disjunct sets, say U and V , in such a way that U induces a co-clique and V induces a clique. Evidently, every threshold graph is a split graph and the neighbourhoods of the vertices are totally ordered by inclusion. By deleting all the edges that belong to the clique V = h i=1 V i (see Figure  1.1) of a threshold graph, we obtain the chain graph that is generated by the same binary sequence. Note, that if t 1 = 1, V is not necessarily the maximal clique in a threshold graph.
We say that a graph is Hamiltonian if it contains a cycle passing through all of its vertices. Every such a cycle is called a Hamilton cycle.
In this paper we revisit the results obtained in [1] on Hamiltonicity of threshold graphs. We give necessary and sufficient conditions for a threshold graph to be Hamiltonian in terms of its generating binary sequence.
The paper is organized as follows. In Section 2 we recall the results from [1]. In Section 3 we interpret these results in terms of the entries of the generating binary sequence of a threshold graph and give a criterion for Hamiltonicity of a threshold graph that can be deduced directly from its binary sequence. This criterion is implemented in the algorithm presented in Section 4, where we also include an algorithm that determines whether a given chain graph is Hamiltonian. In Section 5 we identify the chain graph of a given order that contains minimum number of Hamilton cycles.

Results obtained in [1]
Let G be a threshold graph with vertex set I ∪ J, where I (with |I| = r) induces a co-clique and J (with |J| = s) induces a maximal clique. Let further B denote the chain graph obtained from G by deleting all edges in the subgraph induced by J. For B, let d 1 ≤ d 2 ≤ · · · ≤ d r and e 1 ≤ e 2 ≤ · · · ≤ e s denote the degrees of the vertices x 1 , x 2 , . . . , x r ∈ I and y 1 , y 2 , . . . , y s ∈ J, respectively.
In order to determine whether a given threshold graph is Hamiltonian, the authors of [1] first showed that this problem can be reduced to the Hamiltonicity of the corresponding chain graph B with I and J of the same size, as shown in the sequel.
The first lemma gives sufficient conditions for a split graph to be non-Hamiltonian. In what follows we only consider threshold graphs with r ≥ 2, since for r = 0, G is Hamiltonian if and only if s ≥ 3, while for r = 1, G is Hamiltonian if and only if d 1 ≥ 2.
Since any threshold graph is a split graph, by the previous lemma, Hamiltonian threshold graphs satisfy 2 ≤ r ≤ s and e s−r > 0. The next lemma shows that the problem under the consideration can be reduced to the Hamiltonicity of threshold graphs with r = s. If r = s, then the edges in the clique cannot be used in any Hamiltonian cycle, and therefore can be dropped from G, yielding the chain graph B with r = s ≥ 2.
For q = 0, 1, . . . , r − 1 denote by S q the set of inequalities The next theorem gives two equivalent conditions for a chain graph B with r = s ≥ 2 to be Hamiltonian. To conclude, in order to determine whether an arbitrary threshold graph is Hamiltonian, all three results reported in Lemma 2.1, Lemma 2.2 and Theorem 2.3 should be employed. [1] In this section we restate the results from Section 2 in terms of the entries of the generating binary sequence of a given threshold graph G. Afterwards, we amalgamate them in order to obtain a unique result that gives necessary and sufficient conditions for a threshold graph G to be Hamiltonian.

New versions of results obtained in
Let G and B be a threshold graph and a chain graph generated by a binary sequence (1.1). If t 1 = 1, then the degrees of vertices in B, corresponding to the colour classes U and V , are: Otherwise, if t 1 = 1, then G is a split graph in which the subgraph induced by V ∪ U 1 gives a maximal clique. Then the degrees of vertices in B corresponding to the colour classes h i=2 U i and V ∪ U 1 are: From the previous observations, it follows that the size of the maximal clique s and the size of the corresponding co-clique r satisfy We first state, without proof, the following lemma that determines when e s−r = 0 may occur.
Lemma 3.1. Let G be a threshold graph generated by (1.1), such that s > r. Then e s−r = 0 holds if and only if t 1 = 1 and s − r ∈ {1, · · · , s 1 + 1}. Now, Lemma 2.1 applied to threshold graphs states the following.
In the sequel we consider only threshold graphs, with s ≥ r if t 1 = 1 and with s ≥ r + s 1 + 2 if t 1 = 1. Let the integer be defined in the following way: if s − r < s 1 , then such an integer exists). Observing that the threshold graph G * obtained by deleting the vertices y 1 , y 2 , . . . , y s−r from G is generated by the binary sequence we get a reformulation of Lemma 2.2.
, the degrees of vertices in U * and V * in the corresponding bipartite graph B * of G * given in non-decreasing order are Next, we consider the system of inequalities S q , q ∈ {0, 1, . . . , r − 1}, for s = r. Proof. Recall from Section 2 that a threshold graph G, with s = r, is Hamiltonian if and only if the corresponding chain graph B is Hamiltonian. Next, by Theorem 2.3, the chain graph B is Hamiltonian if and only S q holds for q = r − 1. On the other hand, for (3.2) and each repeated vertex degree, we have

Now, it is easy to see that S r−1 holds if and only if (3.2) holds. Note that the inequality
is not included, since (by the assumption that s = r) it holds as equality. Gathering all the previous results, we arrive at our main result, the criterion for the Hamiltonicity of a threshold graph based on its generating binary sequence. Theorem 3.5. Let G be a threshold graph generated by (1.1), such that r ≥ 2. If either s < r for t 1 = 1 or s < r + s 1 + 1 for t 1 = 1, then G is not Hamiltonian. Otherwise, G is Hamiltonian if and only if where for s−r < s 1 , = 0, and otherwise is the least integer such that i=1 s i ≤ s−r < As a corollary we state a necessary and sufficient condition for a chain graph to be Hamiltonian. Note that Hamiltonian chain graph have the colour classes of the same size (see [2]) and can not have any pendant edges. Moreover, in any Hamiltonian chain graph generated by (1.1), the inequalities t 1 ≥ s 1 + 1 and s h ≥ t h + 1 must hold.

Algorithms
In this section we present algorithms for recognition of Hamiltonian threshold graph and Hamiltonian chain graph. The input is a binary generating sequence, and in return we obtain the decision whether the corresponding threshold (resp. chain) graph is Hamiltonian or not.
(1) Calculate r and s. r = h i=1 t i and s = h i=1 s i . (2) If r = s or r = s and t 1 < s 1 + 1 or s h < t h + 1, then G is not Hamiltonian.
RETURN. (Otherwise, go to the next step.) (3) If the inequalities h i=j s i ≥ h i=j t i + 1 hold for all j = 2, 3, . . . , h then G is Hamiltonian. Otherwise, is not.
We give some examples illustrating the application of the previous algorithms.
Example 4.1. Let G be a complete split graph, i.e, a threshold graph generated by (0 t 1 1 s 1 ). If t 1 = 1, then G is Hamiltonian if and only if s 1 ≥ 2. Otherwise, for t 1 ≥ 2, if s 1 < t 1 , then G is not Hamiltonian. If s 1 ≥ t 1 , then G is Hamiltonian, since in this case we have = 0 and + 1 = h = 1. Therefore, we conclude that a complete split graph is Hamiltonian if and only if the size of the clique is greater than or equal to the size of the co-clique, except for the case where both are equal to 1.
Example 4.2. Let G be a threshold graph generated by (0 t 1 1 s 1 0 t 2 1 s 2 ). If either t 1 + t 2 > s 1 + s 2 , t 1 = 1 or s 2 < t 1 + t 2 , t 1 = 1, then G is not Hamiltonian. Otherwise, if s − r < s 1 , then G is Hamiltonian if and only if s 2 ≥ t 2 + 1, while if s − r ≥ s 1 , then G is necessarily Hamiltonian.

The minimum number of Hamilton cycles in a Hamiltonian chain graph of a prescribed order
In this section we give some observations on Hamiltonian chain graphs and we also determine chain graphs with minimum number of Hamilton cycles. The problem on the value of the minimum number of Hamilton cycles in a given graph has been considered for some special graph classes. For existing literature and recent results related to threshold graphs, we refer reader to [3]. An edge of a chain graph G generated by (1.1) is called a key edge of G if it joins a vertex in U i to a vertex in V i for some 1 ≤ i ≤ h (see 1.1 (b)). As it will be shown in the sequel, key edges play a significant role in determining Hamiltonian chain graphs. We proceed by the following two lemmas.
Lemma 5.1. Let e be a key edge of a chain graph G generated by (1.1), then G − e is a chain graph.
Proof. Let e = uv, where u ∈ U i and v ∈ V i . We consider the following cases. Case 1. If t i > 1, s i > 1, then G − e is a chain graph generated by Case 2. If t i > 1, s i = 1, i.e., if G is generated by then G − e is a chain graph generated by then G − e is a chain graph generated by Case 4. If t i = s i = 1, i.e., if G is generated by then G − e is a chain graph generated by Lemma 5.2. Every key edge of a Hamiltonian chain graph lies in at least one Hamilton cycle.
Proof. Let G be Hamiltonian chain graph, e = uv be a key edge of G, with u ∈ U i and v ∈ V i , and let C be a Hamilton cycle. If e ∈ C, there is nothing to prove. Otherwise, let C has the form (u, s, . . . , v, t . . .). Then u ∼ s and v ∼ t which implies s ∈ V j for some j ≥ i and v ∈ U k for some k ≤ i and consequently s ∼ t. So, both uv and st are the edges of G. The cycle C obtained by adding these two edges to C and deleting us and vt from C is Hamilton and contains e.
We are ready for the main result of this section.
Theorem 5.3. The minimum number of Hamilton cycles in a Hamiltonian chain graph of order n = 2h, h ≥ 2, is 2 h−2 and this number is attained uniquely by the chain graph G n generated by .
Proof. If G is Hamiltonian chain graph, then G has colour classes of the same order, say h (see [2]). If G is generated by 0 t 1 1 s 1 · · · 0 t k 1 s k , then t 1 = 1 and s k = 1. The chain graph with minimum number of Hamilton cycles is defined with minimum values of t i 's, s i 's that, according to Corollary 3.6, are t 1 = s k = 2, t i = 1, i = 1, s j = 1, j = k. The graph under consideration, i.e., the graph generated by (5.1) is Hamiltonian, which is an easy exercise to prove. If any of t i 's, s i 's takes a greater value than the given one, then by deleting any key edge (which by Lemma 5.2 belongs to at least one Hamilton cycle), we would obtain a graph that, in case that it is Hamiltonian, would have fewer number of Hamilton cycles (as the deletion of an edge cannot increase the number of Hamilton cycles).
It remains to compute the number of Hamilton cycles, say c 2h , which can be performed by induction on h. If h = 2, then G 4 = C 4 and c 4 = 2 2−2 = 1.
Let n = 2(h + 1). If U 1 = {x, y} and V 1 = {z}, then by Lemma 5.2, neither G 2(h+1) − xz nor G 2(h+1) − yz is Hamiltonian. Thus the path xzy must lie in every Hamilton cycle of G 2(h+1) and so every Hamilton cycle of G 2(h+1) must go through z. And from z it may continue either through x or y. Assume, without loss of generality, that z is followed by x. The remaining part of the Hamilton cycle must continue through a vertex a / ∈ {x, y, z} and before it returns to z it should go through y. Since G 2(h+1) \ {xz} is isomorphic to G 2h , together with 2 possible choices starting from z we obtain c 2(h+1) = 2c 2h = 2 h−1 . This completes the proof.