Gen. Math. Notes, Vol. 19, No. 1, November, 2013, pp.53-59
c
ISSN 2219-7184; Copyright ICSRS
Publication, 2013
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Some Results on Harmonic Graphs
A. Mahmoodi
Department of Mathematics
Payame Noor University
I. R. of Iran
E-mail: akmahmoodi@yahoo.com
(Received: 25-1-13 / Accepted: 20-9-13)
Abstract
A graph G is defined to be harmonic if there is a constant λ (necessarily
a natural number) such that, for every vertex v, the sum of the degrees of the
neighbors of v equals λdG (v) where dG (v) is the degree of v. In this paper, we
investigate the effects of some operations on harmonic graphs. Here, we present
some methods to construct new harmonic graphs from existing ones. Among
other results it is shown that complement of a harmonic graph is harmonic if
and only if it is a regular graph.
Keywords: Trees, eigenvalues, graph operations, harmonic graphs.
1
Introduction
We will consider only simple graphs, that is finite and undirected without loops
or multiple edges. If G is a graph with vertex set {1, . . . , n}, the adjacency
matrix of G is an n × n matrix A = (aij ), where aij = 1, if there is an edge
between the vertices i and j, and 0 otherwise. The eigenvalues of G are the
eigenvalues of A. The vertex set and edge set of graph G is denoted by V (G)
and E(G), respectively. Also, degree of a vertex v ∈ V (G) and the set of
neighbors of v are denoted by dG (v) and NG (v), respectively. A vertex of
degree k will be referred to as a k-vertex. In addition, vertices of degree one
are called pendent.
A graph G is defined to be harmonic if there is a constant λ (necessarily a
natural number) such that, for every vertex v, the sum of the degrees of the
54
A. Mahmoodi
neighbors of v equals λdG (v). It is easy to see that above definition are equivalent to Ad(G) = λd(G) i.e., the graph G is harmonic if and only if d(G) is one of
its eigenvectors, where d(G) denotes the column-vector (dG (v1 ), . . . , dG (vn ))t .
Clearly, λ is the eigenvalue corresponding to the eigenvector d(G). Suppose G
is a harmonic graph, then there is λ such that
X
λdG (vi ) =
dG (vj )
(1)
(vi ,vj )∈E(G)
for all i = 1, 2, . . . , n. Summing the expressions (1) over all i = 1, 2, . . . , n and
observing that each summand dG (vj ) is counted dG (vj ) times, we obtain
X
X
X
λ
dG (vi ) =
dG (vj )
(2)
1≤i≤n
1≤i≤n (vi ,vj )∈E(G)
So, we have
λ=
X
1
d2G (vi ).
2 | E(G) | 1≤i≤n
(3)
Clearly, a graph G is λ-harmonic if and only if A2 j = λAj holds. Where j is
column vector whose entry is one.
Harmonic graphs appeared for the first time in [1] where they were called
dual-degree-regular graphs. Almost ten years later, they reappeared in [4, 3]
under the present name and in connection with counting walks in a graph. All
(finite or infinite) harmonic trees were constructed in [6]. All finite harmonic
graphs with up to four independent cycles were characterized in [1] where it
was also shown that, while the number of finite harmonic trees is infinite, the
number of finite harmonic graphs with a fixed positive cyclomatic number is
finite.
The following elementary properties of harmonic graphs result from definition and the well-known spectral properties of graphs [2](cf. [1, pp. 18–20]).
Lemma 1. Let G be a connected λ-harmonic graph and | E(G) |= m.
(i) λ is the greatest eigenvalue of G and its multiplicity is one.
(ii) If m ≥ 1 then λ ≥ 1.
(iii) λ = 1 if and only if G = K2 .
2
Main Results
In view of Lemma1 it is reasonable to restrict our considerations to connected
non-regular graphs. In [6], all harmonic finite trees were determined.
55
Some Results on Harmonic Graphs
Let λ be a positive integer. Construct the tree Tλ in the following manner.
The tree Tλ has a total of λ3 − λ2 + λ + 1 vertices, of which one vertex is a
(λ2 − λ + 1)-vertex, λ2 − λ + 1 vertices are λ-vertices and (λ − 1)(λ2 − λ + 1)
vertices are pendent, i.e., each λ-vertex is connected to λ − 1 pendent vertices
and to the (λ2 − λ + 1)-vertex. The following theorem is proved in [6].
Theorem 1. For any positive integer λ there exists a unique λ-harmonic tree,
isomorphic to Tλ .
It is worth that one asks are complement of these trees harmonic? In the
following theorem, we provide negative answer to this question except in the
case λ = 1. To prove the theorem, we need next technical lemma.
Lemma 2. Let G be a λ-harmonic non-trivial graph with n P
vertex. Then its
dG (vi )−λdG (vj )
complement Ḡ is a µ-harmonic graph if and only if dḠ (vj ) = i6=j n−1−µ
for all 1 ≤ j ≤ n.
Proof. Suppose that Ḡ is a µ-harmonic graph and let Ā denote its adjacency matrix. Then we have Ā(n − 1 − dG (v1 ), . . . , n − 1 − dG (vn ))t =
µ(dḠ (v1 ), . . . , dḠ (vn ))t . Using A + Ā = J − I, one can obtain following equivalent equalities.




dG (v1 )
dḠ (v1 )




..
..
(n − 1)Āj − (J − I − A) 
 = µ

.
.
dG (vn )
dḠ (vn )




dḠ (v1 )
dG (v1 )
dḠ (v1 )
i6=1 dG (vi )






 
..
..
..
..
⇐⇒ (n − 1) 
 + λ
 = µ

−
.
.
.
.
P
dG (vn )
dḠ (vn )
dḠ (vn )
i6=n dG (vi )

  P

dḠ (v1 )
i6=1 dG (vi ) − λdG (v1 )




.
..
..
⇐⇒ (n − 1 − µ) 
=
.
.
P
dḠ (vn )
i6=n dG (vi ) − λdG (vn )


 P

On the other hand, µ 6= n − 1 since G is a non-trivial graph. Therefor, we are
done.
Theorem 2. Let G be a λ-harmonic non-trivial graph with n vertex. Then
its complement Ḡ is a µ-harmonic graph if and only if G is a regular graph.
Proof. The implication “⇐=” is clear. To prove the other, suppose that Ḡ
is a µ-harmonic graph. By Lemma2, for 1 ≤ j ≤ n and 1 ≤ k ≤ n, we have
2 | E(G) | −(1 + λ)dG (vj )
n−1−µ
2 | E(G) | −(1 + λ)dG (vk )
dḠ (vk ) =
.
n−1−µ
dḠ (vj ) =
(4)
(5)
56
A. Mahmoodi
These equalities imply that
(−1 +
1+λ
1+λ
)dG (vj ) = (−1 +
)dG (vk ).
n−1−µ
n−1−µ
Now, one can deduce that dG (vj ) = dG (vk ) provided that µ 6= n − 2 − λ.
Then, we may assume that µ = n − 2 − λ. Using this together with (4), we
have (n − 1)(λ + 1) = 2 | E(G) |. So, application [2, Theorem3.8], one can
obtain 2 | E(G) |≥ n(n − 1). Therefor, since G is a simple graph, this implies
that it is a complete graph.
As said before, using Theorem2, we can characterize the µ-harmonicity of
the complement of a λ-harmonic tree.
Corollary 1. Let G be a λ-harmonic tree. Then its complement is µ-harmonic
if and only if λ = 1 and µ = 0.
In continue, we give some methods to construct new harmonic graphs from
existing ones.
The product G×H of graphs G and H is a graph such that the vertex set
of G×H is the Cartesian product V (G) × V (H); and any two vertices (u, u0 )
and (v, v 0 ) are adjacent in G×H if and only if u is adjacent with v and u0 is
adjacent with v 0 . Define the Kronecker product A ⊗ B of two matrices A and
B to be the matrix we get by replacing the ij-entry of A by Aij B, for all i and
j.
Theorem 3. Let G be a λ-harmonic graph and H be a λ0 -harmonic graph.
Then G×H is a λλ0 -harmonic graph.
Proof. Suppose that A and B are adjacency matrices of G and H, respectively. So, by [5, Section 9.7] adjacency matrices of G×H is A ⊗ B and we
have
(A ⊗ B)2 (jn ⊗ jm ) = (A ⊗ B)(Ajn ⊗ Bjm ) = A2 jn ⊗ B 2 jm
= λAjn ⊗ λ0 Bjm = λλ0 (A ⊗ B)(jn ⊗ jm ).
This completes the proof.
The Cartesian product GH of graphs G and H is a graph such that the
vertex set of GH is the Cartesian product V (G)×V (H); and any two vertices
(u, u0 ) and (v, v 0 ) are adjacent in GH if and only if either u = v and u0 is
adjacent with v 0 in H, or u0 = v 0 and u is adjacent with v in G. Suppose
that G and H are two r and s-regular graphs, respectively. In this case we
know GH is (r + s)-regular and harmonic graph. There are many examples
57
Some Results on Harmonic Graphs
show that regularity is necessary to GH be harmonic. For example consider
that K2 T2 which is not harmonic. In this case K2 and T2 are 1-harmonic
and 2-harmonic, respectively. Also, one can see it is possible to choose both
graphs λ-harmonic but their Cartesian product is not harmonic. Next simple
example shows this.
Example 1. Let G = T2 and H = C3 . They are 2-harmonic graphs, but
T2 C3 is not harmonic.
Theorem 4. Let G and H are two λ-harmonic graphs. Then GH is µharmonic graph if and only if G and H are regular graphs.
Proof. As said before, one implication is obvious. To prove the other, suppose that A and B are adjacency matrices of G and H, respectively. Then by
[5, Section 9.7] adjacency matrices of GH is In ⊗ B + A ⊗ Im where n and m
denote the order of G and H, respectively. Since GH is a µ-harmonic graph,
we have (In ⊗ B + A ⊗ Im )2 (jn ⊗ jm ) = µ(In ⊗ B + A ⊗ Im )(jn ⊗ jm ). Let
(d1 , . . . , dn ) and (d01 , . . . , d0n ) denote degree vertices of G and H, respectively.
So, we will have
(In ⊗ B + A ⊗ Im )2 (jn ⊗ jm )
= (In ⊗ B + A ⊗ Im )(jn ⊗ Bjm + Ajn ⊗ jm )
= jn ⊗ B 2 jm + 2(Ajn ⊗ Bjm ) + A2 jn ⊗ jm
t
= jn ⊗ λBjm + 2( d1 · · · dn ⊗ d01 · · ·
=µ
d01 + d1 · · ·
d0m + d1 · · ·
d0m
d01 + dn · · ·
t
) + λAjn ⊗ jm
t
d0m + dn .
Now, for (1 ≤ i ≤ n) and (1 ≤ j ≤ m), we get mn equations
λd0j + λdi + 2di d0j = µ(d0j + di ).
For i = 1 and 1 ≤ j ≤ m, one can obtain from m first equations d01 = d02 =
· · · = d0m . Therefore H is a regular graph. We note that two graphs GH and
HG are isomorphic, so by similar method, G is a regular graph.
Let G and H be two graphs with disjoint vertex sets V1 and V2 . The join
graph G ∨ H is the graph with vertex set V1 ∪ V2 and edge set E(G) ∪ E(H) by
adding all of the edges from vertices in G to those in H. It seems that if both
of G and H are regular, then G ∨ H is harmonic. But this is not the case for
example consider K2 ∨ Cn , where n ≥ 4.
Theorem 5. Let G be a r-regular graph of order n and H be a s-regular graph
of order m. Then G ∨ H is µ-harmonic graph if and only if r + m = s + n.
58
A. Mahmoodi
Proof. It is suffices to prove sufficiency. Suppose that G ∨ H is µ-harmonic
graph and v ∈ V (G). So, one can obtain
X
µ(dG (v) + m) =
dG∨H (u)
u∈NG∨H (v)
=
X
u∈NG (v)
dG∨H (u) +
X
dG∨H (u)
u∈NH (v)
= r(r + m) + m(s + n).
Similarly, for v 0 ∈ H we have µ(dH (v) + n) = s(s + n) + n(r + m). Using above
relations, we get
µ=
s(s + n) + n(r + m)
r(r + m) + m(s + n)
=
.
r+m
s+n
Then one can deduced that (r + m)(rs − nm) = (s + n)(sr − mn). On the
other hand, we know that r < n and s < m. Therefore r + m = s + n.
Corollary 2. Let G and H be two r-regular graphs of order n and m, respectively. Then G ∨ H is µ-harmonic graph if and only if m = n.
Regularity for both graphs G and H in Theorem5 is necessary. There are
many examples which show that although one of them is regular but their join
is not.
Example 2. Let G = k2 and H = T2 . Then k2 ∨ T2 is not harmonic.
Acknowledgements:
This research was supported by grants from Payame Noor University. The
author is indebted to Payame Noor University for support.
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Some Results on Harmonic Graphs
59
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