Abstract. An oriented k-partite graph(multipartite graph) is the result of assigning a direction to each edge of a simple k-partite graph. Let D(V 1 , V 2 , · · · , V k ) be an oriented k-partite graph, and let d  as the imbalance of the vertex v ij . In this paper, we characterize the imbalances of oriented k-partite graphs and give a constructive and existence criteria for sequences of integers to be the imbalances of some oriented k-partite graph. Also, we show the existence of an oriented k-partite graph with the given imbalance set

Abdulaziz M Al-Assaf

Koko K Kayibi

S Pirzada

CiteSeerX

Acta Univ. Sapientiae, Mathematica, 3, 1 (2011) 34–42
On imbalances in oriented multipartite
graphs
S. Pirzada
King Fahd University of Petroleum and
Minerals
Dhahran, Saudi Arabia
email: sdpirzada@yahoo.co.in
Abdulaziz M. Al-Assaf
King Fahd University of Petroleum and
Minerals
Dhahran, Saudi Arabia
email: alassaf@kfupm@edu.sa
Koko K. Kayibi
King Fahd University of Petroleum and Minerals
Dhahran, Saudi Arabia
email: kayibi@kfupm.edu.sa
Abstract. An oriented k-partite graph(multipartite graph) is the result
of assigning a direction to each edge of a simple k-partite graph. Let
D(V1, V2, · · · , Vk) be an oriented k-partite graph, and let d
+
vij
and d−vij
be respectively the outdegree and indegree of a vertex vij in Vi. Define
bvij (or simply bij as bij = d
+
vij
−d−vij as the imbalance of the vertex vij.
In this paper, we characterize the imbalances of oriented k-partite graphs
and give a constructive and existence criteria for sequences of integers to
be the imbalances of some oriented k-partite graph. Also, we show the
existence of an oriented k-partite graph with the given imbalance set.
1 Introduction
A digraph without loops and without multi-arcs is called a simple digraph.
Mubayi et al. [1] defined the imbalance of a vertex vi in a digraph as bvi
(or simply bi) = d
+
vi
− d−vi , where d
+
vi
and d−vi are respectively the outdegree
and indegree of vi. The imbalance sequence of a simple digraph is formed by
2010 Mathematics Subject Classification: 05C20
Key words and phrases: digaph, imbalance, outdegree, indegree, oriented graph, oriented
multipartite graph, arc
34
On imbalances in oriented graphs 35
listing the vertex imbalances in non-increasing order. A sequence of integers
F = [f1, f2, · · · , fn] with f1 ≥ f2 ≥ · · · ≥ fn is feasible if it has sum zero and
satisfies
∑k
i=1 fi ≤ k(n − k), for 1 ≤ k < n.
The following result [1] provides a necessary and sufficient condition for a
sequence of integers to be the imbalance sequence of a simple digraph.
Theorem 1 A sequence is realizable as an imbalance sequence if and only if
it is feasible.
The above result is equivalent to saying that a sequence of integers B =
[b1, b2, · · · , bn] with b1 ≥ b2 ≥ · · · ≥ bn is an imbalance sequence of a simple
digraph if and only if for 1 ≤ k < n
k∑
i=1
bi ≤ k(n − k),
with equality when k = n.
On arranging the imbalance sequence in non-decreasing order, we have the
following observation.
Theorem 2 A sequence of integers B = [b1, b2, · · · , bn] with b1 ≤ b2 ≤ · · · ≤
bn is an imbalance sequence of a simple digraph if and only if for 1 ≤ k < n
k∑
i=1
bi ≥ k(n − k),
with equality when k = n.
Various results for imbalances in digraphs and oriented graphs can be found
in [2, 3, 4, 5].
2 Imbalance sequences in oriented multipartite
graphs
An oriented multipartite (k-partite) graph is the result of assigning a direction
to each edge of a simple multipartite (k-partite) graph, k ≥ 2. Throughout this
paper we denote an oriented k-partite graph by k-OG, unless otherwise stated.
Let Vi = {vi1, vi2, · · · , vini }, 1 ≤ i ≤ k, be k parts of k-OG D(V1, V2, · · · , Vk),
36 S. Pirzada, A. M. Al-Assaf, K. K. Kayibi
and let d+vij and d
−
vij
, 1 ≤ j ≤ ni, be respectively the outdegree and indegree of
a vertex vij in Vi. Define bvij (or simply bij as bij = d
+
vij
−d−vij as the imbalance
of the vertex vij. The sequences Bi = [bi1, bi2, · · · , bini ], 1 ≤ i ≤ k, in non-
decreasing order are called the imbalance sequences of D(V1, V2, · · · , Vk).
The k sequences of integers Bi = [bi1, bi2, · · · , bini ], 1 ≤ i ≤ k, in nonde-
creasing order are said to be realizable if there exists an k-OG with imbalance
sequences Bi, 1 ≤ i ≤ k. Various criterions for imbalance sequences in k-OG
can be found in [2].
For any two vertices vij in Vi and vlm in Vl (i 6= l, 1 ≤ i ≤ l ≤ k, 1 ≤
j ≤ ni, 1 ≤ m ≤ nl) of k-OG D(V1, V2, · · · , Vk), we have one of the following
possibilities.
(i). An arc directed from vij to vlm, denoted by vij(1 − 0)vlm.
(ii). An arc directed from vlm to vij, denoted by vij(0 − 1)vlm.
(iii). There is no arc from vij to vlm and there is no arc from vlm to vij and
this is denoted by vij(0 − 0)vlm.
A triple in k-OG is an induced suboriented graph of three vertices with
exactly one vertex from each part. For any three vertices vij, vlm and vpq in
k-OG D, the triples of the form vij(1 − 0)vlm(1 − 0)vpq(1 − 0)vij, or vij(1 −
0)vlm(1−0)vpq(0−0)vij are said to be oriented intransitive, while as the triples
of the form vij(1−0)vlm(1−0)vpq(0−1)vij, or vij(1−0)vlm(0−1)vpq(0−0)vij,
or vij(1 − 0)vlm(0 − 0)vpq(0 − 1)vij, or vij(1 − 0)vlm(0 − 0)vpq(0 − 0)vij, or
vij(0− 0)vlm(0− 0)vpq(0− 0)vij are said to be oriented transitive. An k-OG is
said to be oriented transitive if all its triples are oriented transitive, otherwise
oriented intransitive.
We have the following observation.
Theorem 3 Let D and D′be two k-OG with the same imbalance sequences.
Then D can be transformed to D′ by successively transforming appropriate
triples in one of the following ways. Either (a) by changing a cyclic triple
vij(1−0)vlm(1−0)vpq(1−0)vij to an oriented transitive triple vij(0−0)vlm(0−
0)vpq(0 − 0)vij which has the same imbalance sequences, or vice versa, or (b)
by changing an oriented intransitive triple vij(1 − 0)vlm(1 − 0)vpq(0 − 0)vij
to an oriented transitive triple vij(0 − 0)vlm(0 − 0)vpq(0 − 1)vij which has the
same imbalance sequences, or vice versa.
Proof. Let Bi be the imbalance sequences of k-OG D whose parts are Vi,
1 ≤ i ≤ k and |Vi| = ni. Let D
′ be k-OG with parts V ′i, 1 ≤ i ≤ k. To
prove the result, it is sufficent to show that D′ can be obtained from D by
successively transforming triples in any one of the ways as given in (a), or (b).
On imbalances in oriented graphs 37
We fix ni, 2 ≤ i ≤ k and use induction on n1. For n1 = 1, the result is
obvious. Assume that the result holds when there are fewer than n1 vertices
in the first part. Let j2, j3, · · · , jk be such that for l2 > j2, l3 > j3, · · · , lk > jk,
1 ≤ j2 < l2 ≤ n2, 1 ≤ j3 < l3 ≤ n3, · · · , 1 ≤ jk < lk ≤ nk, the corresponding
arcs have the same orientations in D and D′. For j2, j3, · · · , jk and 2 ≤ i, p, q ≤
k, p 6= q, we have three cases to consider.
(i). v1n1(1 − 0)vijp (1 − 0)vijq and v
′
1n1
(0 − 0)v′ijp (0 − 0)v
′
ijq
, (ii). v1n1(0 −
0)vijp (0−1)vijq and v
′
1n1
(1−0)v′ijp(0−0)v
′
ijq
and (iii). v1n1(1−0)vijp(0−0)vijq
and v′1n1(0 − 0)v
′
ijp
(0 − 1)v′ijq .
Case (i). Since v1n1 and v
′
1n1
have equal imbalances, we have v1n1(0−1)vijq
and v′1n1(0−0)v
′
ijq
, or v1n1(0−0)vijq and v
′
1n1
(1−0)v′ijq . Thus there is a triple
v1n1(1 − 0)vijp (1 − 0)vijq (1 − 0)v1n1 , or v1n1(1 − 0)vijp (1 − 0)vijq (0 − 0)v1n1
in D, and corresponding to these v′1n1(0 − 0)v
′
ijp
(0 − 0)v′ijq (0 − 0)v
′
1n1
, or
v′1n1(0 − 0)v
′
ijp
(0 − 0)v′ijq (0 − 1)v
′
1n1
respectively is a triple in D′.
Case (ii). Since v1n1 and v
′
1n1
have equal imbalances, we have v1n1(1−0)vijq
and v′1n1(0 − 0)v
′
ijq
. Thus there is a triple v1n1(0 − 0)vijp (0 − 1)vijq (0 − 1)v1n1
in D and corresponding to this v′1n1(1 − 0)v
′
ijp
(0 − 0)v′ijq (0 − 0)v
′
1n1
is a triple
in D′.
Case (iii). Since v1n1 and v
′
1n1
have equal imbalances, therefore we have
v1n1(0 − 1)vijq and v
′
1n1
(0 − 0)v′ijq . Thus v1n1(1 − 0)vijp (0 − 0)vijq (1 − 0)v1n1
is a triple in D, and corresponding to this v′1n1(0− 0)v
′
ijp
(0− 1)v′ijq (0− 0)v
′
1n1
is a triple in D′.
Therefore from (i), (ii) and (iii) it follows that there is an k-OG that can
be obtained from D by any one of the transformations (a) or (b) with the
imbalances remaining unchanged. Hence the result follows by induction. 
Corollary 1 Among all k-OG with given imbalance sequences, those with the
fewest arcs are oriented transitive.
A transmitter is a vertex with indegree zero. In a transitive oriented k-OG
with imbalance sequences Bi = [bi1, bi2, · · · , bini ], 1 ≤ i ≤ k, any of the
vertices with imbalances bini , can act as a transmitter.
The next result provides a useful recursive test of checking whether the
sequences of integers are the imbalance sequences of k-OG.
Theorem 4 Let Bi = [bi1, bi2, · · · , bini ], 1 ≤ i ≤ k, be k sequences of integers
in non-decreasing order with b1n1 > 0 and bjnj ≤
∑k
r=1,r6=j nr, for all j,
2 ≤ i ≤ k. Let B′1 be obtained from B1 by deleting one entry b1n1 , and let
38 S. Pirzada, A. M. Al-Assaf, K. K. Kayibi
B′2, B
′
3, · · · , B
′
k, be obtained from B2, B3, · · · , Bk by increasing b1n1 smallest
entries of B2, B3, · · · , Bk by one each. Then Bi are imbalance sequences of
some k-OG if and only if B′i are imbalance sequences.
Proof. Suppose B′i be the imbalance sequences of some k-OG D
′ with parts
V ′i, 1 ≤ i ≤ k. Then k-OG D with imbalance sequences Bi can be obtained by
adding a vertex v1n1 in V
′
1 such that v1n1(1 − 0)vij for those vertices vij in V
′
i,
i 6= 1 whose imbalances are increased by one in going from Bi to B
′
i.
Conversely, let Bi be the imbalance sequences of k-OG D with parts Vi,
1 ≤ i ≤ k. By Corollary 4, any of the vertices vini in Vi with imbalances
bini , 1 ≤ i ≤ k can be a transmitter. Assume that the vertex v1n1 in V1
with imbalance b1n1 be a transmitter. Clearly, d
+
v1n1
> 0 and d−v1n1
= 0 so
that b1n1 = d
+
v1n1
− d−v1n1
> 0. Also, d+vjnj
≤
∑k
r=1,r6=j nr and d
−
vjnj
≥ 0 for
2 ≤ i ≤ k so that bjnj = d
+
vjnj
− d−vjnj
≤
∑k
r=1,r6=j nr.
Let U be the set of v1n1 vertices of smallest imbalances in Vj, 2 ≤ i ≤ k and
let W = V2 ∪ V3 ∪ · · · ∪ Vk − U. Now construct D such that v1n1(1 − 0)u for
all u in U. Clearly D − {v1n1 } realizes V
′
i, 1 ≤ i ≤ k. 
Theorem 5 provides an algorithm for determining whether or not the se-
quences Bi, 1 ≤ i ≤ k of integers in non-decreasing order are the imbalance
sequences and for constructing a corresponding k-OG.
Suppose Bi = [bi1, bi2, · · · , bini ], 1 ≤ i ≤ k, be imbalance sequences of k-OG
with parts Vi = {vi1, vi2, · · · , vini }, where b1n1 > 0 and bjnj ≤
∑k
r=1,r6=j nr,
2 ≤ i ≤ k. Deleting b1n1 and increasing b1n1 smallest entries of B2, B3, · · · , Bk
by 1 each to form B′2, B
′
3, · · · , B
′
k. Then arcs are defined by v1n1(1 − 0)vij,
for which b′vij = bvij + 1, where i 6= 1. If at least one of the conditions
b1n1 > 0, or bjnj ≤
∑k
r=1,r6=j nr does not hold, then we delete bini for that
i for which the conditions get satisfied and the same argument is used for
defining arcs. If this method is applied recursively, then (i) it tests whether Bi
are the imbalance sequences, and if Bi are the imbalance sequences (ii) k-OG
△(Bi) with imbalance sequences Bi is constructed.
We illustrate this reduction and resulting construction as follows.
Consider the four sequences B1 = [1, 3, 4], B2 = [−3, 2, 2], B3 = [−4,−3] and
B4 = [−3, 1].
(i) [1, 3, 4], [−3, 2, 2], [−4,−3], [−3, 1]
(ii) [1, 3], [−2, 2, 2], [−3,−2], [−2, 1], v13(1−0)v21, v13(1−0)v31, v13(1−0)v32,
v13(1 − 0)v41
(iii) [1], [−1, 2, 2], [−2,−1], [−2, 1], v12(1−0)v21, v12(1−0)v31, v12(1−0)v32
(iv) ∅, [−1, 2, 2], [−2,−1], [−2, 1], v11(1 − 0)v31
On imbalances in oriented graphs 39
(v) ∅, [−1, 2], [0,−1], [−1, 1], v23(1−0)v31, v23(1−0)v41 or, ∅, [−1, 2], [−1, 0],
[−1, 1]
(vi)∅, [−1], [0, 0], [0, 1], v22(1 − 0)v32, v22(1 − 0)v41
(vii)∅, [0], [0, 0], [0, 0], v42(1 − 0)v21.
Clearly 4-OG with parts V1 = {v11, v12, v13}, V2 = {v21, v22, v23}, V3 =
{v31, v32} and V4 = {v41, v42} in which v13(1 − 0)v21, v13(1 − 0)v31, v13(1 −
0)v32, v13(1 − 0)v41, v12(1 − 0)v21, v12(1 − 0)v31, v12(1 − 0)v32, v11(1 − 0)v31,
v23(1 − 0)v31, v23(1 − 0)v41, v22(1 − 0)v32, v22(1 − 0)v41, v42(1 − 0)v21 are arcs
has imbalance sequences [1, 3, 4], [−3, 2, 2], [−4,−3] and [−3,−1].
The next result gives a combinatorial criterion for determining whether k
sequences of integers are realizable as imbalances.
Theorem 5 Let Bi = [bi1, bi2, · · · , bini ], 1 ≤ i ≤ k, be k sequences of integers
in non-decreasing order. Then Bi are the imbalance sequences of some k-OG
if and only if
k∑
i=1
mi∑
j=1
bij ≥ 2
k−1∑
i=1
k∑
j=i+1
mimj −
k∑
i=1
ni
k∑
j=1
mj −
k∑
i=1
mini, (1)
for all sets of k integers mi, 0 ≤ mi ≤ ni with equality when mi = ni.
Proof. The necessity of the condition follows from the fact that the k-OG
induced by mi vertices for 1 ≤ i ≤ k, 1 ≤ mi ≤ ni has a sum of imbalances
2
∑k−1
i=1
∑k
j=i+1 mimj −
∑k
i=1 ni
∑k
j=1 mj −
∑k
i=1 mini.
For sufficiency, assume that Bi = [bi1, bi2, · · · , bini ], 1 ≤ i ≤ k be the
sequences of integers in non-decreasing order satisfying conditions (1) but are
not the imbalance sequences of any k-OG. Let these sequences be chosen in
such a way that ni, 1 ≤ i ≤ k are the smallest possible and b11 is the least for
the choice of ni. We consider the following two cases.
Case (i). Suppose equality in (1) holds for some mj ≤ nj, 1 ≤ i ≤ k − 1,
mk ≤ nk, so that
k∑
i=1
mi∑
j=1
bij = 2
k−1∑
i=1
k∑
j=i+1
mimj −
k∑
i=1
ni
k∑
j=1
mj −
k∑
i=1
mini.
By the minimality of ni, 1 ≤ i ≤ k the sequences B
′
i = [bi1, bi2, · · · , bimi ]
are the imbalance sequences of some k-OG D′(V ′1, V
′
2, · · · , V
′
k).
Define B′′i = [bi(mi+1), bi(mi+2), · · · , bi(ni)], 1 ≤ i ≤ k.
40 S. Pirzada, A. M. Al-Assaf, K. K. Kayibi
Consider the sum
k∑
i=1
fi∑
j=1
bi(mi+j) =
k∑
i=1
mi+fi∑
j=1
bij −
k∑
i=1
mi∑
j=1
bij
≥ 2
k−1∑
i=1
k∑
j=i+1
(mi + fi)(mj + fj) −
k∑
i=1
ni
k∑
j=1
(mj + fj)
−
k∑
i=1
(mi + fi)ni − 2
k−1∑
i=1
k∑
j=i+1
mimj +
k∑
i=1
ni
k∑
j=1
mj +
k∑
i=1
mini
= 2
k−1∑
i=1
k∑
j=i+1
mimj + 2
k−1∑
i=1
k∑
j=i+1
(mifj + fimj + fifj) −
k∑
i=1
ni
k∑
j=1
fj
−
k∑
i=1
mini −
k∑
i=1
fini − 2
k−1∑
i=1
k∑
j=i+1
mimj +
k∑
i=1
ni
k∑
j=1
mj+
+
k∑
i=1
mini ≥ 2
k−1∑
i=1
k∑
j=i+1
fifj −
k∑
i=1
ni
k∑
j=1
fj −
k∑
i=1
fini,
for 1 ≤ fi ≤ ni − mi, with equality when fi = ni − mi for all i, 1 ≤ i ≤ k.
So by the minimality of ni, 1 ≤ i ≤ k, the sequences B
′′
i form the imbalance
sequence of some k-OG D′′(V ′′1 , V
′′
2 , · · · , V
′′
k).
Construct a new k-OG D(V1, V2, · · · , Vk) as follows. Let V1 = V
′
1 ∪
V ′′1 , V2 = V
′
2 ∪ V
′′
2 , · · ·Vk = V
′
k ∪ V
′′
k with V
′
i ∩ V
′′
i = ∅ and the arc set con-
taining those arcs which are among V ′1, V
′
2, · · · , V
′
k and among V
′′
1 , V
′′
2 , · · · , V
′′
k.
Then D(V1, V2, · · · , Vk) has imbalance sequences Bi, 1 ≤ i ≤ k, which is a
contradiction.
Case (ii). Assume that the strict inequality holds in (1) for some mi 6= ni,
1 ≤ i ≤ k. Let B′1 = [b11 − 1, b12, · · · , b1n1−1, b1n1 ] and let B
′
j = [bj1, bj2, · · · ,
bjnj ] for all j, 2 ≤ j ≤ k. Clearly the sequences B
′
i, 1 ≤ i ≤ k satisfy conditions
(1). Therefore, by the minimality of b11, the sequences B
′
i, 1 ≤ i ≤ k are the
imbalance sequences of some k-OG D′(V ′1, V
′
2, · · · , V
′
k). Let bv11 = b11−1 and
bv1n1
= b1n1 + 1. Since bv1n1 > bv11 + 1, there exists a vertex vij either in Vi,
1 ≤ i ≤ k, 1 ≤ j ≤ ni, such that v1n1(0 − 0)vij(1 − 0)v11, or v1n1(1 − 0)vij(0 −
0)v11, or v1n1(1−0)vij(1−0)v11, or v1n1(0−0)vij(0−0)v11 in D
′(V ′1, V
′
2, · · · , V
′
k),
and if these are changed to v1n1(0−1)vij(0−0)v11, or v1n1(0−0)vij(0−1)v11,
or v1n1(0−0)vij(0−0)v11, or v1n1(0−1)vij(0−1)v11 respectively, the result is
k-OG with imbalance sequences Bi, which is a contradiction. This completes
the proof. 
On imbalances in oriented graphs 41
3 Imbalance sets in oriented multipartite graphs
The set of distinct imbalances of the vertices in k-OG is called its imbalance
set. Now we give the existence of k-OG with a given imbalance set.
Theorem 6 Let S = {s1, s2, · · · , sn} and T = {−t1, −t2, · · · , −tn}, where
s1, s2, · · · , sn, t1, t2, · · · , tn are positive integers with s1 < s2 < · · · < sn and
t1 < t2 < · · · < tn. Then there exists k-OG with imbalance set S ∪ T .
Proof. First assume that k ≥ 2 is even. Construct k-OG D(V1, V2, · · · , Vk) as
follows. Let
V1 = V11 ∪ V12 ∪ · · · ∪ V1n,
V2 = V21 ∪ V22 ∪ · · · ∪ V2n,
. . .
Vk = Vk1 ∪ Vk2 ∪ · · · ∪ Vkn,
with Vij ∩ Vlm = ∅, |Vij| = ti for all odd i, 1 ≤ i ≤ k − 1, 1 ≤ j ≤ n and
|Vij| = si for all even i, 2 ≤ i ≤ k, 1 ≤ j ≤ n. Let there be an arc from each
vertex of Vij to every vertex of V(i+1)j for all odd i, 1 ≤ i ≤ k − 1, 1 ≤ j ≤ n
so that we obtain k-OG with imbalance of vertices as follows.
For odd i, 1 ≤ i ≤ k − 1 and 1 ≤ j ≤ n
bvij = |V(i+1)j| − 0 = si,
for all vij ∈ Vij; and for even i, 2 ≤ i ≤ k and 1 ≤ j ≤ n
bvij = 0 − |V(i+1)j| = −ti,
for all vij ∈ Vij
Therefore imbalance set of D(V1, V2, · · · , Vk) is S ∪ T .
Now assume k ≥ 3 is odd. Construct k-OG D(V1, V2, · · · , Vk) as below. Let
V1 = V11 ∪ V
′
11 ∪ V12 ∪ V
′
12 ∪ . . . ∪ V1n ∪ V
′
1n,
V2 = V21 ∪ V22 ∪ . . . ∪ V2n,
. . .
Vk−1 = V(k−1)1 ∪ V(k−1)2 ∪ . . . ∪ V(k−1)n,
Vk = V
′
k1 ∪ V
′
k2 ∪ . . . ∪ V
′
kn,
42 S. Pirzada, A. M. Al-Assaf, K. K. Kayibi
with Vij∩Vlm = ∅, V
′
ij∩V
′
lm = ∅, Vij∩V
′
lm = ∅, |Vij| = ti for all i, 1 ≤ i ≤ k−2,
1 ≤ j ≤ n, |Vij| = si for all even i, 2 ≤ i ≤ k − 1, 1 ≤ j ≤ n, |V
′
ij| = ti for
all j, 1 ≤ j ≤ n and |V ′kj| = sj for all j, 1 ≤ j ≤ n. Let there be an arc from
each vertex of Vij to every vertex of V(i+1)j for all i, 1 ≤ i ≤ k − 2, 1 ≤ j ≤ n
and let there be an arc from each vertex of V ′1j to every vertex of V
′
kj for all j,
1 ≤ j ≤ n, so that we obtain k-OG with imbalance set S ∪ T , as above. 
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Received: October 20, 2010


On imbalances in oriented multipartite graphs

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On imbalances in oriented multipartite graphs

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