Bigraph-factorization of symmetric complete
bipartite multi-digraphs
Kazuhiko Ushio
Department of Industrial Engineering
Faculty of Science and Technology
Kinki University, Osaka 577-8502, JAPAN
E-mail:[email protected]
Abstract
We show that a necessary and sufficient condition for the existence of
a Kp,q - factorization of the symmetric complete bipartite multi-digraph
)"K:n n is (i) m = n == 0 (mod p) for p = q and (ii) m = n == a (mod
d(p' q')p'q'/e) for p =f. q, where d = (p,q), p' = p/d, q' = q/d, e =
()..,p'q').
+
1. Introduction
The symmetric complete bipartite multi-digraph )..K:n,n is the symmetric complete
bipartite digraph K:n,n in which every arc is taken).. times. Let Kp,q denote the
complete bipartite di&raph in which all arcs are directed away from p start-vertices
to q end-vertices. A spanning subgraph F of )..K:n,n is called a Kp,q - factor if each
component of F is isomorphic to Kp,q. If )..K:n,n is expressed as an arc-disjoint sum
of Kp,q - factors, then this sum is called a Kp,q - factorization of )..K:n,n. In this
paper, it is shown that a necessary and sufficient condition for the existence of such
a factorization is (i) m = n == 0 (mod p) for p = q and (ii) m = n == a (mod
d(p' + q')p'q'/e) for p -I- q, where d = (p, q), p' = p/d, q' = q/d, e = ()"'p'q').
Let Knl ,nz, K~l ,nz' Knl ,n2,na' and K~l ,n2,n3 denote the complete bipartite graph, the
symmetric complete bipartite digraph, the complete tripartite graph, and the symmetric complete tripartite digraph, respectively. Let Ck , fh, Fk , and Kp,q denote
the cycle or the directed cycle, the star or the directed star, the path or the directed path, and the complete bipartite graph or the complete bipartite digraph,
respectively, on two partite sets Vi and Vj. Then the problems of giving the necessary and sufficient conditions of Ck - factorization of Knl ,n2' K~l ,n2' and K~l ,n2 ,na
have been completely solved by Enomoto, Miyamoto and Ushio[3] and Ushio[12].
fh - factorization of Knl ,n2' K~l ,n2' and K~l ,n2 ,n3 have been studied by DU[2], U shio
and Tsuruno[9], Ushio[14], and Wang[18]. Recently, Martin[5,6] and Ushio[ll] gave
Australasian Journal of Combinatorics 20(1999), pp.261-265
-A-
necessary and sufficient conditions for Sk - factorization of K n1 ,n2 and K~l ,n2'
factorization of K n1 ,n2 and K~l ,n2 have been studied by U shio and Tsuruno[8], and
Ushio[7,10). Kp,q - factorization of K n1 ,n2 has been studied by Martin[5]. Ushio[13]
gave necessary and sufficient conditions for Kp,q - factorization of K~l ,n2' For graph
theoretical terms, see [1,4].
2. Kp,q - factor of )"K:n,n
The following theorem is on the existence of a Kp,q - factor of
)'Kr~,n'
Theorem 1. ),K:n,n has a Kp,q - factor if and only if for p = q (i) m = n == 0 (mod
+ n == 0 (mod p + q), (iii) pm - qn == 0 (mod p2 _ q2),
q2), (v) pm 2:: qn and (vi) pn 2:: qm.
p), and for p -I q (ii) m
(iv) pn - qm == 0 (mod p2
Proof. (Necessity) Suppose that ),K:n,n has a Kp,q - factor F. Let t be the number
of components of F. Then t = (m + n)/(p + q). Among these t components, let tl
and t2 be the number of components whose start-vertices are in \Ii and 112, respectively. Then, since F is a spanning subgraph of )"K:n,n, we have ptl + qt2 = m and
qtl + pt2 = n. When p = q, we have pil + pt2 = m and ptl + pt2 = n. Therefore,
(pm - qn) / (p2 - q2) and
Condi tion (i) is necessary. When p # q, we have t 1
t2 = (pn
qm)/(p2 - q2). ~From 0 ::; tl ::; m and 0 ::; t2 ::; n, we must have pm 2:: qn
and pn 2:: qm. Condition (ii)-(vi) are, therefore, necessary.
(Sufficiency) When p = q, put m = 8p and n = 8p. Then obviously )'K:n n has
a Kp,p - factor formed by 8 Kp,p's. When p # q, for those parameters m ~nd n
satisfying (ii)-(vi), let tl = (pm - qn)/(p2 q2) and t2 = (pn - qm)/(p2 - q2). Then
tl and t2 are integers such that 0 ::; tl ::; m and 0 ::; t2 S; n. Hence, ptl + qi2 = m and
qil +pt2 = n. Using ptl vertices in 1!i and qtl vertices in 112, consider i l Kp,q's whose
start-vertices are in 1!i. Using the remaining qi2 vertices in 1!i and the remaining pt2
vertices in 112, conyider t2 Kp,q's whose start-vertices are in 112. Then these tl + t2
Kp,q's are arc-disjoint and they form a Kp,q factor of )..K:n,n'
Corollary 2. )"K~,n has a Kp,q - factor if and only if n == 0 (mod p) for p
== 0 (mod p + q) for p -I q.
= q and
n
3. Kp,q - factorization of )"K:n,n
We use the following notation.
Notation. Given a Kp,q - factorization of )"K:n,n, let
r be the number of factors
t be the number of components of each factor
b be the total number of components.
Among t components of each factor, let i l and t2 be the numbers of components
whose start-vertices are in 1!i and 112, respectively.
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Among T components having vertex x in Vi, let
whose start-vertices are in Vj.
Tij
be the numbers of components
We give the following necessary condition for the existence of a Kp,q - factorization
of )"K:n,n.
Theorem 3. Let d = (p,q), p'
- factorization, then (i) m = n
d(p' + q')p' q'/ e) for p i- q.
= p/d, q' = q/d, e = ()..,p'q'). If )"K:n,n has a Kp,q
== 0 (mod p) for p = q and (ii) m = n == 0 (mod
Proof. Suppose that )"K:n,n has a Kp,q - factorization. Then b = 2)..mn/pq,
t = (m+n)/(p+q), T = bit = 2)..mn(p+q)/(m+n)pq. And tl = (pm_qn)/(p2_q2),
t2 = (pn - qm)/(p2 - q2) for p i- q. Moreover, qTl1 = )..n, PT12 = )..n, PT21 = )..m,
and qT22 = )..m. Thus we have T = Tl1 + T12 = )..n(p + q)/pq and T = T21 + T22 =
)..m(p + q)/pq. Therefore, m = n is necessary.
Moreover, when m = nand p = q, we have b = 2)..n 2/p2, t = nip, r = 2)..n/p, TU =
T12 = T21 = r22 = )..n/p. Therefore, n == 0 (mod p) is also necessary. When m = n
and Pi- q, we have b = 2)..n 2/d 2p'q', t = 2n/d(p' + q'), T = )..n(p' + q')/dp'q', TU =
T22 = )..n/dq', T12 = T21 = )..n/dp', tl = t2 =n/d(p' + q'). Thus we have n == 0 (mod
d(p' + q')) and )..n == 0 (mod dp'q'). Therefore, n == 0 (mod d(p' + q')p'q'/e) is also
necessary.
Lemma 4. Let G, Hand K be digraphs. If G has an H - factorization and H has
a K - factorization, then G has a K - factorization.
Proof. Let E( G) = Ur=l E(Fi) be an H - factorization of G. Let HY) (1 :S j :S t)
be the components of Fi . And let E(H?)) = Uk=l E(Kki,j)) be a K - factorization of
HJi). Then E( G)
= Ui=l Uk=l E(U~=l Kki,j))
is a K - factorization of G.
We prove the following extension theorems, which we use later in this paper.
Theorem 5. If )"K~,n has a Kp,q - factorization, then
tion for every positive integer s.
Proof. Obvious. Construct a Kp,q:" factorization of
we have a Kp,q - factorization of s)"K~,n.
s)"K~,n
)"K~,n
has a Kp,q - factoriza-
repeatly s times. Then
Theorem 6. If )"K~,n has a Kp,q - factorization, then )"K:n,sn has a Kp,q - factorization for every positive integer s.
Proof. Since )"K~,n has a Kp,q - factorization, )"K:n,sn has a Ksp,sq - factorization.
Obviously, Ksp,sq has a Kp,q - factorization. Therefore, by Lemma 4 )"K:n,sn has a
Kp,q - factorization.
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We use the following notation for a Kp,q.
Notation. Let U = {UI,U2, ... ,Up} and V = {VI,V2, ... ,Vq}. For a Kp,q whose startvertex set is U and end-vertex set is V, we denote (UI,U2, ... ,Up;VI,V2, ... ,Vq) or
(U;1l).
We give the following sufficient conditions for the existence of a, Kp,q - factorization
of )"K~,n'
Theorem 7. When n
== 0 (mod p),
)"K~,n
has a Kp,p - factorization.
Proof. Put n = sp. Let Vi = {I, 2, .",p} and
(1;2; Vi) are Kp,p - factors and they comprise a
Theorem 5 and Theorem 6, we see that )"K~,n
V2 = {I', 2', ... ,p'}. Then (VI; 1;2) and
Kp,p - factorization of K;,p. Applying
has a Kp,p - factorization.
Theorem 8. For p #- q, let d = (p,q), p' = pld, q'
n = d(p' + q')p' q' I A, >"K~,n has a Kp;q - factorization,
= q/d.
When p'q'
= x)..
and
{1,2, ... ,n}, V2 = {I',2', .. "n'}, For i = 1,2, ... ,p'+q' and
+ q')2 Kp,q factors Fij as following:
Fij = { ((A + 1, .. " A + p);(B + g + 1, ... , B + g+ q)')
((A + p + 1, .. " A + 2p);(B + g + q + 1, , .. , B + g + 2q)')
Proof. Let
j = 1,2, .. "p'
Vi
=
+ q',
construct (p'
((A + (f - l)p + 1, ... , A + fp);(B + g + (J - l)q + 1, .. " B + g + fq)')
((B + 1, .. " B + p)';(A + g + 1, ... , A + g + q))
((B + p + 1, ... , B + 2p)';(A + g + q + 1, ,.. , A + g + 2q))
((B + (f - l)p + 1, , .. , B + fp)';(A + g + (J - l)q + 1, ... , A + g + fq)) },
where A = (i - l)dp'q', B = (j - l)dp'q', f = p'q'lA, g = fp, and the additions are
taken modulo n with residues 1,2, ... , n.
Then they comprise a Kp,q - factorization of )"K~,n'
Theorem 9. For p #- q, let d = (p, q), p' = p/d, q' = q/d, e
(mod d(p' + q')p'q'le), >"K~,n has a Kp,q - factorization.
Proof.
d(p'
Let x
+ q')p'q's/e
= (>",p'q').
When n
== 0
= p'q'le and y = >../e. Then p'q' = xe and>" = yeo Put n =
and N = d(p' + q')p'q'le. By Theorem 8, eK'N,N has a Kp,q - fac-
torization. Applying Theorem 4 and Theorem 5, we see that
factorization.
>"K~,n
has a Kp,q -
We have the following main theorem and its corollary.
Main Theorem. Let d = (p,q), p' = p/d, q' = qld, e = (A,p'q'). >..K:n,n has a Kp,q
- factorization if and only if (i) m = n == 0 (mod p) for p = q and (ii) m = n == 0
(mod d(p' + q')p' q' Ie) for p =I- q.
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Corollary[13]. Let d = (p, q), p' = p/d, q' = q/d. K:n,n has a Kp,q - factorization if
and only if (i) m = n == 0 (mod p) for p = q and (ii) m = n == 0 (mod d(p' + q')p'q')
for p =I- q.
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(Received 5/3/99)
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