EUROPEAN JOURNAL OF PURE AND APPLIED MATHEMATICS
Vol. 14, No. 2, 2021, 358-365
ISSN 1307-5543 – ejpam.com
Published by New York Business Global

On Spectral-Equipartite Graphs and
Eccentricity-Equipartite Graphs

Arnel M. Yurfo1,∗, Joel G. Adanza2, Michael P. Baldado Jr.2

1 College of Arts and Sciences, Negros Oriental State University - Bayawan - Sta. Catalina
Campus, Bayawan City, Philippines
2 Mathematics Department, Negros Oriental State University - Main Campus, Dumaguete
City, Philippines

Abstract. Let G = (V,E) be a graph of order 2n. If A ⊆ V and 〈A〉 ∼= 〈V \A〉, then A is
said to be isospectral. If for every n-element subset A of V we have 〈A〉 ∼= 〈V \A〉, then we say
that G is spectral-equipartite. In [1], Igor Shparlinski communicated with Bibak et al., proposing
a full characterization of spectral-equipartite graphs. In this paper, we gave a characterization
of disconnected spectral-equipartite graphs. Moreover, we introduced the concept eccentricity-
equipartite graphs.

2020 Mathematics Subject Classifications: 05C50, 05C75

Key Words and Phrases: Spectral-equipartite graphs, eccentricity-equipartite graphs, isospec-
tral graphs, graph spectra

1. Introduction

Let G = (V,E) be a graph. The distance between vertices u and v in G, denoted by
d (u, v), is the length of the shortest path connecting u and v. If u and v is not connected,
then we define d (u, v) to be 0. The eccentricity of a vertex is its distance to a farthest
vertex. G is said to be k-regular if every vertex of G has the same degree which is k. G
is said to be weakly equipartite if every partition of V into two equal sets A and B, we
have 〈A〉 ∼= 〈B〉. In addition, if there is an automorphism mapping A onto B, then we say
that G is equipartite. The degree sequence of G is a non-decreasing sequence of degrees of
the vertices of G. G is degree-equipartite if for every n-element subset A of V , the degree
sequences of 〈A〉 and 〈V \A〉 are the same.

The adjacency matrix M = [aij ] of G is the square matrix of order 4n2 given by aij = 1
if vivj ∈ E (G), and aij = 0 otherwise.

∗Corresponding author.
DOI: https://doi.org/10.29020/nybg.ejpam.v14i2.3928

Email addresses: mathematicsrocks@yahoo.com (A. Yurfo),
joeladanza@yahoo.com (J. Adanza), michaelpbaldadojr@yahoo.com (M. Baldado Jr.)

http://www.ejpam.com 358 c© 2021 EJPAM All rights reserved.



A. Yurfo, J. Adanza, M. Baldado Jr. / Eur. J. Pure Appl. Math, 14 (2) (2021), 358-365 359

The spectrum of G is the collection of all eigenvalues of all its adjacency matrix. Two
graphs that have the same spectrum are said to be cospectral or isospectral. G is said to
be spectral-equipartite if for every n-element subset A of V , the induced subgraph of A
and V \A are isospectral.

Ferrero et al. in [5], mentioned the eccentricity sequence of a graph G as the non-
decreasing sequence of eccentricities of the vertices of G. A graph G = (V,E) of order
2n is said to be eccentricity-equipartite if for every n-element subset A of V , the induced
subgraph of A and V \A have the same eccentricity sequence.

Here after please refer to [6] for the other concepts.
Over the past years, various applications spectral graph theory in many fields were

discovered. In particular, spectral graph theory have important applications in chemistry,
physics, computer science, and common real world problems. For instance, in computer
science, the largest eigenvalue λ1 plays a significant role in simulating virus proliferation in
computer networks. Also mentioned in [3], Wang et al. in [10], claimed that the epidemic
threshold in spreading viruses is proportional to 1/λ1.

Furthermore, spectral graph theory was also applied in connection with the famous
’traveling salesman problem’. This is mentioned by Cvetković et al. in [4].

Grünbaum et al. in [7] characterized equipartite graphs. They also presented a problem
regarding the characterization of degree-equipartite graphs. Motivated by this problem,
Bibak and Haghighi [1] published a paper that contains the characterization of degree-
equipartite graphs. Moreover, they introduced a new type of graph called the spectral-
equipartite graph. This new type was suggested by Igor Shparlinski, who also asked for
its full characterization.

The latest study on equipartite graphs is by Shirdareh Haghighi et al. [9], which
characterizes equipartite graphs in terms of their Laplacian spectra.

2. Known Results

2.1. Weakly-Equipartite Graphs

Theorem 1 is due to Grünbaum et al. [7] in their study on equipartite graphs.

Theorem 1. ([7],Theorem 13) A graph G is weakly equipartite if and only if it is one of
the following graphs: 2nK1, nK2, 2C4, Kn,n\nK2, and 2Kn, or one of their complements:
K2n, K2n\nK2, K8\2C4, 2Kn + nK2, and Kn,n.

2.2. Degree-Equipartite Graphs

The following theorem is due to Bibak et al. [1] in their study on degree-equipartite
graphs.

Theorem 2. ([1],Theorem 10) A graph G of order 2n is degree-equipartite if and only if
it is one of the following graphs: 2nK1, nK2, 2C4, Kn,n\nK2, and 2Kn, or one of their
complements: K2n, K2n\nK2, K8\2C4, 2Kn + nK2, and Kn,n.



A. Yurfo, J. Adanza, M. Baldado Jr. / Eur. J. Pure Appl. Math, 14 (2) (2021), 358-365 360

2.3. Spectra of Graphs

The following theorems and lemmas are due to the different studies involving the
spectra of graphs.

Theorem 3. ([8], Theorem 2.1). Let G be a simple undirected graph and let A be its
adjacency matrix. Let H be a graph isomorphic to G and let B be the adjacency matrix
of H. Then, G and H have the same spectrum.

The next theorem is a consequence of Theorem 3 and the definition of isospectral
graphs.

Theorem 4. If two graphs are isomorphic, then they are isospectral.

Lemma 1. ([9], Lemma 2.5). If G is a non-complete regular graph such that every two
non-adjacent vertices of G form a vertex cut, then G is a cycle.

Lemma 2 can be verified easily as a direct consequence of [6] (F33, page 679).

Lemma 2. If H is a proper subgraph of G, then λ1 (H) < λ1 (G).

Lemma 3. ([6], F6, page 674). The spectrum of a graph is the union of the spectra of
its connected components.

Lemma 4. ([2], 1.4.1 and 1.4.2). Let m,n ∈ N. The spectrum of a complete graph Kn is
{−1n−1, n−1} and the spectrum of a complete bipartite graph Km,n is {±

√
mn, 0m+n−2}.

Lemma 5 is a direct consequence of Lemma 3 and Lemma 4.

Lemma 5. Let n ∈ N. The spectrum of an empty graph nK1 is {0n}.

Theorem 5. ([1], Problem 1, page 891). Every spectral-equipartite graph is regular.

3. Main Results

This section presents the main results of the study.

3.1. Characterization of disconnected Spectral-Equipartite Graphs

The following results lead to the characterization of disconnected spectral-equipartite
graphs.

This section also shows that the complement of a disconnected spectral-equipartite
graph is also spectral-equipartite.

Theorem 6. Every weakly-equipartite graph is spectral-equipartite.

Proof. Let G = (V,E) be a weakly-equipartite graph of order 2n and let A be an n
element subset of V . Then, 〈A〉 and 〈V \A〉 are isomorphic. Thus, by Theorem 4, 〈A〉 and
〈V \A〉 are isospectral. This shows that G is spectral-equipartite.



A. Yurfo, J. Adanza, M. Baldado Jr. / Eur. J. Pure Appl. Math, 14 (2) (2021), 358-365 361

Theorem 7. Every degree-equipartite graph is spectral-equipartite.

Proof. Let G be a degree-equipartite graph. By Theorem 1 and Theorem 2, ev-
ery weakly-equipartite graph is degree-equipartite, and every degree-equipartite graph is
weakly-equipartite. Hence, by Theorem 6, G is spectral-equipartite.

Remark 1. There exists a disconnected k-regular (with k > 1) spectral-equipartite graph.

To see this, the following are disconnected k-regular (with k > 1) spectral-equipartite
graph: 2nK1; nK2; 2C4; and 2Kn.

Lemma 6. If a1, a2, . . . , an ∈ N with a1 ≥ a2 ≥ . . . ≥ an ≥ 4, then (a1 + a2 + · · ·+ an−1)−
2 (n− 1) ≥ an for all positive integer n ≥ 3.

Proof. We use induction. For n = 3, we have a1 ≥ a2 ≥ a3 ≥ 4, that is a1 ≥ a3 and
a2 ≥ a3. Since a3 ≥ 4, a2 − 4 ≥ 0. Thus, a1 + a2 − 4 ≥ a3, that is a1 + a2 − 2(3− 1) ≥ a3.
Hence the assertion holds for n = 3. Now, let k ≥ 3 and assume that the assertion holds
for k. Then, (a1+a2+· · ·+ak−1)−2 (k − 1) ≥ ak ≥ ak+1. Since ak ≥ ak+1 ≥ 4, ak−2 ≥ 0.
Thus, (a1 + a2 + · · ·+ ak) − 2k ≥ ak+1. Thus, the assertion also holds for k + 1. This
shows the lemma.

Lemma 7. If a1, a2, . . . , an ∈ N with a1 ≥ a2 ≥ . . . ≥ an ≥ 4, then there exists r ∈ N such
that (a1+a2+· · ·+ar)−2 (r) ≥ ar+1+ar+2+· · ·+an and (a1+a2+· · ·+ar−1)−2 (r − 1) <
ar + ar+1 + ar+2 · · ·+ an.

Proof. Let S = {k ∈ N : (a1 + a2 + · · ·+ ak)− 2 (k) ≥ ak+1 + ak+2 + · · ·+ an}. By
Lemma 6, n − 1 ∈ S, that is, S 6= ∅. By the Well-ordering Principle, S contains a least
element, say r. If r ∈ S, then (a1 + a2 + · · · + ar) − 2 (r) ≥ ar+1 + ar+2 + · · · + an.
Since r is the least element of S, r − 1 /∈ S. Hence, (a1 + a2 + · · · + ar−1) − 2 (r − 1) <
ar + ar+1 + ar+2 · · ·+ an.

The following remark follows from Lemma 7.

Remark 2. Let n ≥ 3 and A = {G1, G2, . . . , Gn} be the set of all components of a
disconnected k-regular (k > 1) graph. Then there exists a set B = {Gi1 , Gi2 , . . . , Gir}
(r < n) subset of A such that |V (Gi1)\{u1} ∪ V (Gi2)\{u2} ∪ · · · ∪ V (Gir)\{ur}| ≥
|V (Gir+1)∪V (Gir+2)∪ · · ·∪V (Gin)∪{u1, u2, . . . , ur} |, and |V (Gi1)\{u1}∪V (Gi2)\{u2}∪
· · ·∪V (Gir−1)\{ur−1}| < |V (Gir)∪V (Gir+1)∪V (Gir+2)∪· · ·∪V (Gin)∪{u1, u2, . . . , ur−1} |.

Lemma 8. A disconnected k-regular (k > 1) spectral-equipartite graph cannot have more
than two components.

Proof. Supposed G = (V,E) has more than two components, say A = G1 ∪ G2 ∪
. . . ∪ Gn (n > 2), where Gi is a component for i = 1, 2, . . . , n. By Remark 2, Then
there exists a set B = {Gi1 , Gi2 , . . . , Gir} (r < n) subset of A such that |V (Gi1)\{u1} ∪



A. Yurfo, J. Adanza, M. Baldado Jr. / Eur. J. Pure Appl. Math, 14 (2) (2021), 358-365 362

V (Gi2)\{u2}∪· · ·∪V (Gir)\{ur}| ≥ |V (Gir+1)∪V (Gir+2)∪· · ·∪V (Gin)∪{u1, u2, . . . , ur} |,
and |V (Gi1)\{u1}∪V (Gi2)\{u2}∪· · ·∪V (Gir−1)\{ur−1}| < |V (Gir)∪V (Gir+1)∪V (Gir+2)∪
· · · ∪V (Gin)∪{u1, u2, . . . , ur−1} |. Partition V as follows: (1) Remove from V (Gij ) a non-
empty set of vertices Aj from V (Gij ) to form V ′j = V (Gij )\Aj for j = 1, 2, . . . , r such that
|V ′1∪V ′2∪· · ·∪V ′r | = |V (Gir+1)∪V (Gir+2)∪· · · (Gin)∪A1∪A2∪· · ·∪Ar|. (2) LetH1 =

⋃r
j=1 V

′
j

and H2 =
(⋃n

j=r+1 V (Gij )
)
∪
(⋃r

j=1Ai

)
. Then 〈H2〉 has k-regular components while 〈H1〉

does not have. Thus, by Lemma 2 and Lemma 3 spec(〈H1〉) 6= spec(〈H2〉). This shows
the lemma.

Lemma 9. Let G be a disconnected k-regular (with k > 1) graph of order 2n. If G is a
spectral-equipartite graph, then it has only two components which are both of order n.

Proof. Let G be a disconnected k-regular (k > 1) graph of order 2n and V be the
vertex set of G. Suppose G is a spectral-equipartite graph. By Lemma 8 and by the
definition of disconnected graphs, G has exactly two components. Next, we will prove
that the two components of G are both of order n. Let G1 and G2 be the components of
G. Suppose to the contrary |V (G1)| 6= |V (G2)|. Without loss of generality, assume that
|V (G1)| < |V (G2)|. Let |V (G2)| − |V (G1)| = m. Partition V into two sets A and B with
〈A〉 = 〈V (G2) \V (mK1)〉 and 〈B〉 = G1∪mK1. By Lemma 2 and Lemma 3, the spectrum
of 〈A〉 does not contain k while the spectrum of 〈B〉 does. Hence, Spec (〈A〉) 6= Spec (〈B〉).
Thus, 〈A〉 and 〈B〉 are not isospectral, and G is not spectral-equipartite. This proves that
the two components of G have an equal number of vertices which is n. This shows the
lemma.

Theorem 8. Let G be a disconnected k-regular (with k > 1) graph of order 2n. If G is
a spectral-equipartite graph, then G = 2Kn or G = 2C4.

Proof. By Lemma 9, the two components of G, say G1 and G2, has n vertices each.
Suppose to the contrary that G1 is not a complete graph nor a cycle. By Lemma 1, we
can find two non-adjacent vertices, say x and y, in G1 where G1\ {x, y} is connected. Let
pq be an in G2. We can partition V (G) into two sets, A and B, with n vertices each such
that 〈A〉 = 〈V (G1) \ {x, y}〉∪(p, q) and 〈B〉 = 〈V (G2) \ {p, q}〉∪〈{x}〉∪〈{y}〉. By Lemma
3, Lemma 4, and Lemma 5, we have Spec (〈A〉) = Spec (〈V (G1) \ {x, y}〉) ∪ {1,−1} and
Spec (〈B〉) = Spec (〈V (G2) \ {p, q}〉) ∪ {0} ∪ {0}.
Consider the following cases:
Case 1. Both G1 and G2 are not bipartite
If both G1 and G2 are not bipartite, then Spec (〈V (G2) \ {p, q}〉) cannot have 1 and −1
at the same time as elements. Hence, Spec (〈A〉) 6= Spec (〈B〉). Thus, 〈A〉 and 〈B〉 are
not isospectral, and G is not spectral-equipartite.
Case 2. Only one between G1 and G2 is bipartite
Suppose that G1 is bipartite and G2 is not. Hence, using the same argument in Case 1
above, if G2 is not bipartite, then Spec (〈V (G2) \ {p, q}〉) cannot have 1 and −1 at the
same time as elements. Hence, Spec (〈A〉) 6= Spec (〈B〉). Thus, 〈A〉 and 〈B〉 are not
isospectral, and G is not spectral-equipartite.



A. Yurfo, J. Adanza, M. Baldado Jr. / Eur. J. Pure Appl. Math, 14 (2) (2021), 358-365 363

Suppose that G2 is bipartite and G1 is not. By Lemma 2,
λ1 (〈V (G2) \ {p, q}〉) > 1 since K2 ⊂ 〈V (G2) \ {p, q}〉. Since G1 is not bipartite, only one
between λ1 and −λ1 may exist as an eigenvalue of 〈V (G1) \ {x, y}〉. Thus, Spec (〈A〉) 6=
Spec (〈B〉). Thus, 〈A〉 and 〈B〉 are not isospectral, and G is not spectral-equipartite.
Case 3. Both G1 and G2 are bipartite

If both G1 and G2 are bipartite, then we have the following subcases:
Subcase 1. Both G1 and G2 are not complete bipartite graphs.
If both G1 and G2 are not complete bipartite graphs, then we will partition G into two
sets A and B with n vertices each such that 〈A〉 = 〈V (G1) \ {x, y}〉 ∪ (p, q) and 〈B〉 =
〈V (G2) \ {p, q}〉 ∪ 〈{x}〉 ∪ 〈{y}〉. Since G1 is not a complete bipartite graph, we can
have two non-adjacent vertices, x, and y, to belong to different partite sets. Hence,
they do not have a common neighbor. Thus, for 〈V (G1) \ {x, y}〉, there are 2k vertices
with degree k − 1 and (n− 2) − 2k vertices of degree k. On the other hand, since pq is
an edge, p and q must belong to different partite sets, so they do not have a common
neighbor. Hence, for 〈V (G2) \ {p, q}〉, there are 2k − 2 vertices of degree k − 1 and
(n− 2) − (2k − 2) = n − 2k vertices with degree k. Clearly, 〈V (G2) \ {p, q}〉 contains
one more edge when compared to 〈V (G1) \ {x, y}〉. With 2k − 2 < 2k for vertices with
degree k − 1 and n − 2k > (n− 2) − 2k for vertices with degree k, we could say that
〈V (G1) \ {x, y}〉 is isomorphic to some proper subgraph of 〈V (G2) \ {p, q}〉. Thus, by
Lemma 2 λ1 (〈V (G2) \ {p, q}〉) > λ1 (〈V (G1) \ {x, y}〉). Since K2 is a proper subset of
〈V (G2) \ {p, q}〉, then λ1 (〈V (G2) \ {p, q}〉) > 1. Hence, Spec (〈A〉) 6= Spec (〈B〉). Thus,
〈A〉 and 〈B〉 are not isospectral, so G is not spectral-equipartite.
Subcase 2. Both G1 and G2 are complete bipartite graphs.
If both G1 and G2 are complete bipartite graphs, then we will use a different partitioning
of the graph G into two sets, A and B, with n vertices each and consider the following
subsubcases:
Subsubcase 1. n/2 is even.
If n/2 is even, then we will partition V (G) into two sets, A and B, with n vertices each
such that 〈A〉 =

(
n
2 − 1

)
K1 ∪ Kn

4
,n+4

4
, and 〈B〉 = K1,n

2
∪ Kn

4
,n+4

4
. Thus,by Lemma 3

and Lemma 4, we have spec 〈(A)〉 =
{

0
n
2
−1
}
∪
{
±
√

n
4

(
n
4 + 1

)
, 0

n
2
−1
}

and Spec (〈B〉) ={
±
√

n
2 , 0

n
2
−1
}
∪
{
±
√

n
4

(
n
4 − 1

)
, 0

n
2
−3
}

. Hence, Spec (〈A〉) 6= Spec (〈B〉). Thus, 〈A〉 and

〈B〉 are not isospectral, and so G is not spectral-equipartite.
Subsubcase 2. n/2 is odd.
If n/2 is odd, then we will partition V (G) into two sets, A and B, with n vertices each
such that 〈A〉 =

(
n
2 − 1

)
K1 ∪Kn+2

4
,n+2

4
, and 〈B〉 = K1,n

2
∪Kn−2

4
,n−2

4
. Thus, by Lemma

3, Lemma 4, and Lemma 5, spec 〈(A)〉 =
{

0
n
2
−1
}
∪
{
±n+2

4 , 0
n
2
−1
}

and spec 〈(B)〉 ={
±
√

n
2 , 0

n
2
−1
}
∪
{
±n−2

4 , 0
n
2
−3
}

. Hence, Spec (〈A〉) 6= Spec (〈B〉). Thus, 〈A〉 and 〈B〉 are

not isospectral, and so G is not spectral-equipartite.
By Cases 1,2, and 3, we have shown that if G1 is neither a complete graph nor a cycle,

then G will not be a spectral-equipartite graph. Thus, G1 must either be a complete



A. Yurfo, J. Adanza, M. Baldado Jr. / Eur. J. Pure Appl. Math, 14 (2) (2021), 358-365 364

graph or a cycle. If G1 is complete, then G2 having the same order and size as G1 is also
complete. Whence, G = 2Kn.

If G1 is a cycle Cn, then k = 2 and G2 is also Cn. Suppose n > 4. Then, we choose two
non-adjacent vertices, a and b, from G1 that have a common neighbor and two adjacent
vertices, c and d, from G2. Now, we partition G into two sets, A and B, with n vertices
each such that 〈A〉 = 〈V (G2) \ {c, d}〉 ∪ 〈{a, b}〉 and 〈B〉 = 〈V (G1) \ {a, b}〉 ∪ 〈{c, d}〉.
Then 〈A〉 = Pn−2 ∪K1 ∪K1 and 〈B〉 = Pn−3 ∪K1 ∪ P2. Clearly, Pn−3, P2, and K1 are
proper subgraphs of Pn−2. By Lemma 2, λ1 (Pn−2) > λ1 (Pn−3) > λ1 (P2) > 0. Hence, by
Lemma 3, Spec (〈A〉) 6= Spec (〈B〉). Therefore, n ≤ 4. For n = 1, 2 or 3, we will have K1,
K2, and K3, respectively, and for n = 4, we have C4. Thus, we have G = 2K1, 2K2, 2K3,
and 2C4. Therefore, if G is a disconnected k-regular spectral-equipartite graph of order
2n with k > 1, then G = 2Kn or G = 2C4.

Theorem 9. Let G be a disconnected graph of order 2n. G is spectral-equipartite if and
only if it is one of the following graphs: 2nK1, nK2, 2Kn, and 2C4.

Proof. Suppose G is one of the graphs 2nK1, nK2, 2Kn, and 2C4. By Theorem 2, G
is weakly-equipartite, and by Theorem 6, it must be spectral-equipartite.

Now suppose that G is a disconnected spectral-equipartite graph. By Theorem 5, G
must be a k-regular graph of order 2n. If k = 0, then the 2n vertices of G are isolated.
Hence, we have G = 2nK1. If k = 1, then G is just the union of the n copies of K2. Hence,
we have G = nK2. If k > 1, then by Theorem 8, G = 2Kn or G = 2C4.

Theorem 10. If G is a disconnected spectral-equipartite graph, then its complement is
also spectral-equipartite.

Proof. Let G be a disconnected spectral-equipartite graph. By Theorem 9, G is one of
the graphs 2nK1, nK2, 2C4, and 2Kn. Observe that the complements of these graphs are
K2n, K2n\nK2, K8\2C4, and Kn,n. By Theorem 2, these graphs are weakly equipartite,
and by Theorem 6, they must be spectral-equipartite.

3.2. Eccentricity-Equipartite Graphs

This section gives some eccentricity-equipartite graphs.

Theorem 11. Every weakly-equipartite graph is eccentricity-equipartite.

Proof. Let G be a weakly-equipartite graph. Thus, every partition of V (G) into two
sets, A and B, with n vertices each, 〈A〉 and 〈B〉 are isomorphic. Hence, 〈A〉 and 〈B〉
have the same eccentricity sequence. Thus, G is eccentricity-equipartite.

Corollary 1. Let n ∈ N. The following graphs are eccentricity-equipartite: 2nK1; nK2;
2C4; Kn,n\nK2; 2Kn; K2n; K2n\nK2; K8\2C4; 2Kn + nK2 and Kn,n.

Proof. The statement immediately follows from Theorem 1 and Theorem 11.



REFERENCES 365

Theorem 12. Every degree-equipartite graph is eccentricity-equipartite.

Proof. The proof for this theorem will immediately follow from Theorem 2 and Corol-
lary 1.

Acknowledgements

The authors would like to thank the Rural Engineering and Technology Center of
Negros Oriental State University for partially supporting this research.

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