Abstract
Spectral techniques are used for the study of several network properties: community detection, bipartition, clustering, design of highly synchronizable networks, and so forth. In this paper, we investigate which kinds of bicyclic networks are determined by their per-spectra. We find that the permanental spectra cannot determine sandglass graphs in general. When we restrict our consideration to connected graphs or quadrangle-free graphs, sandglass graphs are determined by their permanental spectra. Furthermore, we construct countless pairs of per-cospectra bicyclic networks.
1. Introduction
It was recognized in about last decade that graph spectra have several important applications in computer science. Graph spectra appear in internet technologies, pattern recognition, computer vision, data mining, multiprocessor systems, statistical databases, and many other areas. For example, spectral filtering is applied in the study of Internet structure [1]. This method uses the eigenvectors of the adjacency and other graph matrices and some clusters in data sets represented by graphs. For more information about the applications of graph spectra in computer science see [2–6], among others.
The permanent of matrix is defined as where the sum is taken over all permutations of . Valiant [7] showed that computing the permanent is #P-complete even when restricted to (0, 1)-matrices.
Let be a graph with vertices, and let be (0,1)-adjacency matrix of . The permanental polynomial of , denoted by , is defined as . The permanental spectrum (per-spectrum for short) of graph , denoted by , is the set of all roots (together with their multiplicities) of .
Two graphs are per-cospectral if they share the same per-spectrum. A graph is said to be determined by its per-spectrum (DPS for short) if there is no other nonisomorphic graph with the same per-spectrum.
Which graphs are determined by their adjacency spectra is an old problem in graph spectra theory. van Dam and Haemers [8, 9] gave an excellent survey of answers to the question of which graphs are determined by the spectra of some graph polynomials. Merris et al. [10] first considered the problem which graph is DPS. And they showed that the five pairs adjacency cospectral graphs (see [11]) are DPS. Based on the result, they formulated that the per-spectrum seems a little better than the adjacency spectrum when it comes to distinguishing graphs which are not trees. In fact, characterizing what kinds of graphs are determined by the per-spectra is generally a very hard problem. Up to now, only a few types of graphs are proved to be DPS; see [12–17].
A bicyclic network is a simple connected graph in which the number of edges equals the number of vertices plus one [18]. The sandglass graph is a bicyclic network, denoted by , obtained by appending a triangle to each pendant vertex of the path . Lu et al. [19] proved that sandglass graphs are determined by their adjacency spectra. Motivated by the statement of Merris et al., a natural problem is whether sandglass graphs are determined by their per-spectra. In this paper, we give a solution of this question.
In what follows, we begin with some definitions and notions. Let be the union of two graphs and which have no common vertices. For any positive integer , let denote the union of disjoint copies of graph . The path and cycle on vertices are denoted by and , respectively. Let denote the number of -cycles in .
Let and be two vertex-disjoint cycles. Suppose that is a vertex of and is a vertex of . Joining and by a path of length , where and means identifying with , the resulting graph (see Figure 1), denoted by , is called -graph. Let , , and be three vertex-disjoint paths, where and at most one of them is . Identifying the three initial vertices and terminal vertices of them, respectively, the resulting graph (see Figure 1), denoted by , is called -graph. Then bicyclic networks can be partitioned into two classes: the class of graphs which contain -graph as its induced subgraph and the class of graphs which contain -graph as its induced subgraph.
A subgraph of is a Sachs subgraph if each component of is a single edge or a cycle. Merris et al. [10] gave a modified Sachs formula to compute the coefficients of the permanental polynomials of graphs.
Lemma 1 (see [10]). Let be a graph with . Then where the sum is taken over all Sachs subgraphs of on vertices, and is the number of cycles in .
Lemma 2 (see [13]). Let be a graph with vertices and edges, and let be the degree sequence of . Then
Lemma 3 (see [17]). Let be a graph with edges, and let denote the degree sum of the three vertices on th triangle in . Then
Lemma 4 (see [13]). The following can be deduced from the permanental polynomial of a graph :(i)The number of vertices(ii)The number of edges(iii)The number of triangles(iv)The length of the shortest odd cycle(v)The number of the shortest odd cycles(vi)Whether is bipartite
2. Sandglass Graphs Are DPS
In this section, we will give the solutions of the problem which sandglass graphs are DPS?
Checking graph depicted in Figure 2, direct computation yields + . This implies that the permanental spectra cannot determine sandglass graphs in general. Examining graph again, we know that is not connected and contains a quadrangle. It is natural to consider the problem which sandglass graphs are DPS when we restrict our consideration to connected graphs or quadrangle-free graphs, where the quadrangle-free graph is one which contains no quadrangles (i.e., cycles of length 4). We will answer these questions one by one in the following.
First, we give some lemmas which will play an important role in the proof of main theorems.
Lemma 5. Let be a graph with vertices. Then .
Proof. By Lemma 3, the proof is trivial.
Lemma 6. Let be a graph with vertices. Then -th coefficient of is
Proof. Suppose that is odd. Then the Sachs subgraph of order in is only . By Lemma 1, we have . Otherwise, if is even, then the Sachs subgraph of order in is and . By Lemma 1, we have .
Lemma 7. Let be a quadrangle-free graph with vertices. If , then the degree sequence of is , where means .
Proof. Suppose that the degree sequence is , where , and are integers. Since and have the same number of edges, thenBy Lemma 2, we haveSince , we have . By a simple calculation, we obtainChecking (8), it is easy to see that if or , then , a contradiction. Hence,Furthermore, if except , then , a contradiction. ThusSolving simultaneous equations (6)–(10), we have Thus the degree sequence of is possible , , or . It is not difficult to check that only satisfies the well-known hand-shaking theorem. So, the degree sequence of is .
Theorem 8. Restricting consideration on quadrangle-free graphs, sandglass graphs are determined by their per-spectra.
Proof. Let be a quadrangle-free graph with vertices per-cospectral with . By Lemma 7, we know that the degree sequence of is . Then is isomorphic to or , where denotes the union of disjoint cycles of length . By the above, it implies that . It is not difficult to see that if , then is isomorphic to . So, assume and consider the following two cases.
Case 1. Assume that is isomorphic to . We further discuss the following three subcases.
Subcase 1.1. Exact one of the two triangles belongs to the bicyclic component. Then for , and . By Lemma 3, we obtain that . By Lemma 5, we have . This contradicts the assumption that and are per-cospectral.
Subcase 1.2. Both of the two triangles belong to the bicyclic component. Then . If , then is isomorphic to . Next we suppose that . By the structure of and Lemma 1, we can obtain that when is even, and when is odd. By Lemma 6, this is a contradiction.
Subcase 1.3. Neither of the two triangles belongs to the bicyclic component. Then for . By Lemma 3, we have . So, . This contradicts the assumption that and are per-cospectral.
Case 2. Assume that is isomorphic to . We further consider the following three subcases.
Subcase 2.1. Exact one of the two triangles belongs to the bicyclic component. Then for and . By the structure of and Lemma 1, we can obtain that if is even, then ; and if is odd, then . By Lemma 6, this is a contradiction.
Subcase 2.2. Both of the two triangles belong to the bicyclic component. Then , which is a contradiction to being a quadrual-free graph.
Subcase 2.3. Neither of the two triangles belongs to the bicyclic component. Then for , or , . By Lemma 3, we have . By Lemma 6, we have a contradiction, and the theorem follows.
Theorem 9. Restricting consideration on connected graphs, sandglass graphs are determined by their per-spectra.
Proof. Let be a connected graph, where and , and let be per-cospectral with . By Lemma 4, is a bicyclic graph with two triangles and must be isomorphic to a graph containing a sandglass graph as its induced subgraph or (isomorphic to -graph when and ) as its induced subgraph.
Suppose that is isomorphic to a graph which contains a sandglass graph as its induced subgraph. Then contains no quadrangles. By Lemma 7, must be isomorphic to the sandglass graph .
In the following, we will prove that is isomorphic to a graph containing as its induced subgraph.
Suppose that is even. By Lemma 6, we know that . This implies, by Lemma 1, that must have odd perfect matchings. Examining the structure of , we see that has at most two perfect matchings. So, only has uniquely one perfect matching. This implies that all triangles or 4-cycle in are not a component of some Sachs subgraph of order . Thus, the perfect matching of is a unique Sachs subgraph of order . By Lemma 1, , which contradicts the fact that .
Assume that is odd. By Lemma 1 and examining the structure of , we know that the Sachs subgraphs of order in is only the union of a triangle and a perfect matching of deleting all edges on the triangle. Then . This contradicts .
This completes the proof.
For any bicyclic network, it is difficult to discuss which is determined by its per-spectrum. We can construct countless pairs per-cospectral bicyclic networks. Let be an arbitrary graph with a fixed vertex and let denote the coalescence of and with respect to and , which is the graph obtained from by identifying and . Similarly, we define . Borowiecki [20] showed that if both and are per-cospectral, then both and are also per-cospectral. As an example, let be the bicyclic network depicted in Figure 3. As and are isomorphic, they are per-cospectral. By the above-mentioned result of Borowiecki [20], for any graph , both and are per-cospectral.
3. Summary
Per-spectra is an important part of graph spectra. In this paper, we discuss properties of permanental spectra of bicyclic networks. We show that without some limitations bicyclic networks are not DPS. Particularly, we find a pair of per-cospectral graphs. Combining the result of Lu et al. [19], our results (Theorems 8 and 9) are beyond Merris et al.’s imagination. Finally, we pose the following conjecture.
Conjecture 10. Sandglass graphs with a perfect matching are DPS.
Conflicts of Interest
The authors declare that they have no conflicts of interest.
Acknowledgments
The authors thank Dr. Shunyi Liu for providing Figure 2. The authors are supported by NSF of Qinghai (2016-ZJ-947Q) and High-level personnel of scientific research projects of QHMU(2016XJG07).