Abstract
A graph $G$ is said to be determined by its generalized spectra (DGS for short) if, for any graph $H$, graphs $H$ and $G$ are cospectral with cospectral complements imply that $H$ is isomorphic to $G$. In Wang (J. Combin. Theory, Ser. B, 122 (2017) 438-451), the author gave a simple method for a graph to be DGS. However, the method does not apply to Eulerian graphs. In this paper, we gave a simple method for a large family of Eulerian graphs to be DGS. Numerical experiments are also presented to illustrate the effectiveness of the proposed method.
Highlights
The spectrum of a graph encodes a lot of combinatorial information about the given graph and has long been a useful tool in spectral graph theory.A fundamental question in this area is: “Which graphs are determined by their spectra (DS for short)?”
The above theorem fails for Eulerian graphs. (Recall that a graph is Eulerian if it admits an Eulerian tour, which traverses each edge exactly once; or equivalently, if it is connected and the degree of every vertex is even)
We show that for an Eulerian graph G with det W (G)/2
Summary
The spectrum of a graph encodes a lot of combinatorial information about the given graph and has long been a useful tool in spectral graph theory. If det W (G)/2 n/2 (which is always an integer) is odd and square-free, G is DGS It is noticed, the above theorem fails for Eulerian graphs. (Recall that a graph is Eulerian if it admits an Eulerian tour, which traverses each edge exactly once; or equivalently, if it is connected and the degree of every vertex is even) This is because for an Eulerian graph, every entry (except for the ones in the first column) of the walk-matrix is divisible by 2, and 2n−1 divides det W and det W/2 n/2 can never be odd and square-free.
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