# The minimum number of Hamilton cycles in a hamiltonian threshold graph of a prescribed order^{1}

^{1}

## Abstract

We prove that the minimum number of Hamilton cycles in a hamiltonian threshold graph of order is and this minimum number is attained uniquely by the graph with degree sequence of distinct degrees. This graph is also the unique graph of minimum size among all hamiltonian threshold graphs of order

Key words. Threshold graph; Hamilton cycle; minimum size

## 1 Introduction

There are few results concerning the precise value of the minimum or maximum number of Hamilton cycles of graphs in a special class with a prescribed order. For example, it is known that the minimum number of Hamilton cycles in a simple hamiltonian cubic graph of order is which follows from Smith’s theorem [1, p.493] and an easy construction [10, p.479], but the maximum number of Hamilton cycles is not known; even the conjectured upper bound [2, p.312] has not been proved. Another example is Sheehan’s conjecture that every simple hamiltonian 4-regular graph has at least two Hamilton cycles [10] (see also [1, p.494 and p.590]), which is still unsolved.

In this paper we will determine the minimum number of Hamilton cycles in a hamiltonian threshold graph of order and the unique minimizing graph. Threshold graphs were introduced by Chvtal and Hammer [3] in 1973. Besides the original definition, seven equivalent characterizations are given in the book [7].

Definition 1. A finite simple graph is called a threshold graph if there exists a nonnegative real-valued function defined on the vertex set of and a nonnegative real number such that for any two distinct vertices and and are adjacent if and only if

The class of threshold graphs play a special role for many reasons, some of which are the following: 1) They have geometrical significance. Let be the convex hull of all degree sequences of the simple graphs of order Then the extreme points of the polytope are exactly the degree sequences of threshold graphs of order [6] (for another proof see [9]). 2) A nonnegative integer sequence is graphical if and only if it is majorized by the degree sequence of some threshold graph [9]. 3) A graphical sequence has a unique labeled realization if and only if it is the degree sequence of a threshold graph [7, p.72].

For terminology and notations we follow the textbooks [1,11]. The order of a graph is its number of vertices, and the size its number of edges. We regard isomorphic graphs as the same graph. Thus for two graphs and means that and are isomorphic. and denote the neighborhood and closed neighborhood of a vertex respectively. For a real number denotes the largest integer less than or equal to and denotes the least integer larger than or equal to The notation denotes the cardinality of a set

## 2 Main Results

Let be a graph whose distinct positive vertex-degrees are and let Denote for The sequence is called the degree partition of Each is called a degree set. Sometimes when is empty it may be omitted. These notations will be used throughout. We will need the following characterization [7, p.11] which describes the basic structure of a threshold graph.

Lemma 1. is a threshold graph if and only if for each

In other words, for and is adjacent to if and only if

Clearly, Lemma 1 not only implies another characterization that the vicinal preorder of a threshold graph is a total preorder, but also indicates that every threshold graph is determined uniquely by its degree sequence [7, p.72].

The following lemma can be found in [7, pp.11-13].

Lemma 2. For any threshold graph,

For two subsets and of the vertex set of a graph the notation denotes the set of edges of with one end-vertex in and the other end-vertex in Here and need not be disjoint. In the case is just the edge set of the subgraph of induced by Next we define a new concept which will be used in the proofs.

Definition 2. An edge of a threshold graph with degree partition is called a key edge of if it lies in for some with

Thus when is even we have only one type of key edges, and when is odd () we have two types of key edges. For example, if then the set of key edges is while if then the set of key edges is We will need the following two lemmas concerning properties of key edges.

Lemma 3. If is a key edge of a threshold graph then is a threshold graph.

Proof. Denote and let be the number of distinct positive vertex-degrees of Let First suppose that and for some We write TPO for the conditions in Lemma 1 (suggesting total preorder). To prove that is a threshold graph, by Lemma 1 it suffices to show that the degree sets of satisfy TPO. The structural change of the degree partitions depends on the sizes of the two sets and We distinguish four cases.

Case 1. and The condition implies that is possible only if is odd, since if is even then Hence implying that and are two distinct sets. By Lemma 2,

After deleting the two sets and become empty, and they disappear in goes to and goes to Now and the adjacency relations among the vertices of still satisfy TPO.

Case 2. and As in case 1, and are two distinct sets. By Lemma 2, we have

When deleting stays in and goes to Thus and satisfies TPO.

Case 3. and We have When deleting goes to If is even, and then When deleting goes to and the set disappears. Thus In all other cases, we have In fact, if is odd or is even and we have while if is even, and we have When deleting remains in Thus In each case, satisfies TPO.

Case 4. and We have If is even, and then When deleting remains in (but with degree ) and a new degree set appears. Now In all other cases, two new degree sets appear, one containing only and the other containing only so that In either case, satisfies TPO and hence it is a threshold graph.

Now suppose that is odd and where Apply Lemma 2. If when deleting both and go to and the degree set disappears. Then and satisfies TPO. Otherwise When deleting a new degree set appears, where and are nonadjacent. In this case and again satisfies TPO.

Lemma 4. Every key edge of a hamiltonian threshold graph lies in at least one Hamilton cycle.

Proof. Let be a hamiltonian threshold graph with degree partition Let be a key edge of with and for some Choose any Hamilton cycle of If lies in C, we are done. Otherwise let Then and are adjacent, and and are adjacent. Applying Lemma 1 we deduce that and are adjacent. Now the classical cycle exchange [1, p.485] with and yields a new Hamilton cycle containing the edge

Different necessary and sufficient conditions for a threshold graph to be hamiltonian are given by Golumbic [4], Harary and Peled [5], and Mahadev and Peled [8]. What we need is the following one by Golumbic [4, p.231] whose proof can be found in [7, p.25].

Lemma 5. Let be a threshold graph of order at least with the degree partition Then is hamiltonian if and only if

and if is even, then

Definition 3. For every integer we denote by the graph with degree sequence of distinct degrees.

is a hamiltonian threshold graph. is depicted in Figure 1.

Now we are ready to prove the main results.

Theorem 6. The minimum number of Hamilton cycles in a hamiltonian threshold graph of order is and this minimum number is attained uniquely by the graph

Proof. We first determine the minimizing graph and then count its number of Hamilton cycles. Let be a hamiltonian threshold graph of order having the minimum number of Hamilton cycles. Let be the degree partition of Note that for any threshold graph with we have This follows from

by Lemma 2.

The theorem holds trivially for the case Next suppose means that is a complete graph, which is impossible. Thus We claim that Lemma 5 with implies Hence it suffices to prove To the contrary suppose Let be any edge in Then is a key edge by definition. By Lemma 3, is a threshold graph. Since is a hamiltonian threshold graph, its degree sets satisfy the inequalities in Lemma 5. Analyzing the change of degree partitions from to as in the proof of Lemma 3, we see that the degree sets of also satisfy the inequalities in Lemma 5. Hence by Lemma 5, is hamiltonian. Deleting any edge cannot increase the number of Hamilton cycles. By Lemma 4, the key edge lies in at least one Hamilton cycle of It follows that has fewer Hamilton cycles than contradicting the minimum property of This proves

If then by Lemma 5 we have Since we must have and Then applying Lemma 1 we deduce that has the degree sequence so that Next suppose By Lemma 5, Hence We first consider the case (The case will be treated later). We claim that Otherwise, as argued above, deleting any key edge in would reduce the number of Hamilton cycles such that is still a hamiltonian threshold graph, a contradiction. Then using the fact that and and applying Lemma 5 we deduce that if is odd or if is even and then and if then Continuing in this way, by successively deleting a key edge in for if we conclude that for each Then using the fact that and and applying Lemma 5, we conclude that for each and that if is odd then and if is even then Thus, if is even then is even, has the degree sequence and hence

If and is odd, we assert that As remarked at the beginning, we always have Thus it suffices to show To the contrary suppose By Lemma 4, any key edge in lies in at least one Hamilton cycle. With the assumption that applying Lemma 3 and Lemma 5 we see that is also a hamiltonian threshold graph with fewer Hamilton cycles than a contradiction. This shows Now is odd. Combining all the above information about we deduce that has the degree sequence and hence

Denote the number of Hamilton cycles of by Since to prove it suffices to show the following

Claim. For every integer

In let and By Lemma 5, neither nor is hamiltonian. Thus the path must lie in every Hamilton cycle of Deleting the vertex and adding the edge we obtain a graph which is isomorphic to and has the same number of Hamilton cycles as Hence

In , let Then the edge lies in every Hamilton cycle of Denote Clearly is isomorphic to and hence has Hamilton cycles. Since in from each Hamilton cycle of we can obtain two distinct Hamilton cycles of by replacing the vertex by the edge in two ways. More precisely, a Hamilton cycle of yields two Hamilton cycles and of Conversely every Hamilton cycle of can be obtained in such a vertex-to-edge expansion from a Hamilton cycle of Hence This shows the claim and completes the proof.

The above proof of Theorem 6 also proves that is the unique graph that has the minimum size among all hamiltonian threshold graphs of order To see this, just replace the assumption that has the minimum number of Hamilton cycles by the one that has the minimum size. Also note that the size of a threshold graph is easy to count, since it is a split graph with the clique and the independent set Thus we have the following result.

Theorem 7. The minimum size of a hamiltonian threshold graph of order is

and this minimum size is attained uniquely by the graph

### Footnotes

- E-mail addresses: 235711gm@sina.com(P.Qiao), zhan@math.ecnu.edu.cn(X.Zhan). This research was supported by Science and Technology Commission of Shanghai Municipality (STCSM) grant 13dz2260400 and the NSFC grant 11671148.

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