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Orthogonal [k − 1, k + 1]-factorizations in graphs
ELSEVIER
Journal of Statistical Planning and Inference 51 (1996) 195-200
joumalof statistical plan ' and r i n g
Orthogonal [ k - 1, k ÷ 1]-factorizations in graphs G u i z h e n L i u *, G u i y i n g Y a n Department of Mathematics, Shandong University, Jinan, Shandong 250100, People's Republic of China
Abstract Let G be a graph. Let F = {FI,F2 ..... Fd} be a factorization of G and H be a subgraph of G. If H has exactly one edge in common with Fi for all i = 1,2 ..... d, then we say that F is orthogonal to H. In this paper it is proved that for any d-matching M of a [kd - 1, kd + 1]-graph G, there is a [ k - l,k + l]-factorization of G orthogonal to M where k/> 2 is an integer.
1. Introduction The graphs considered in this paper are finite undirected graphs which have neither multiple edges nor loops. Let G be a graph with vertex set V(G) and edge set E(G). For a vertex x of G, the degree of x in G is denoted by de(x). Let g and f be two integer-valued functions defined on V(G) such that g(x) ~< f ( x ) for every x E V(G). Then a (g, f ) - f a c t o r of G is a spanning subgraph F o f G satisfying g(x) <~d e ( x ) <~f ( x ) for all x E V(G). If for all x E V(G), g(x) = a and f ( x ) = b, then a ( g , f ) - f a c t o r is called a [a,b]-factor. A (g,f)-factorization F = {F1,F2 . . . . . Fd} of a graph G is a partition o f E(G) into edge-disjoint (g,f)-factors FI,F2 . . . . . Fd. Similarly, we define an [a,b]-factorization. A subgraph H o f G is orthogonal to F if ] E ( H ) N E(Fi)1= 1, l <<,i <~d. Alspach et al. (1992) gave the following problem: Given a subgraph H o f G, does G have a factorization with certain prescribed properties which is orthogonal to H ? The purpose of this paper is to prove that for any d-matching M of a [kd - 1, kd + 1]-graph G, there is a [k - 1,k + 1]-factorization of G orthogonal to M. In particular, for any d-matching M o f a kd-regular graph there is a [ k - 1,k + 1]-factorization orthogonal to M. Further there is a kd-regular graph G with a d-matching M so that there is no k-factorization orthogonal to M. To prove the main results we need the following concepts and preliminary results. Theorem 1.1 (Ore (1960)). Let G be a graph with n >13 vertices. I f for any two nonadjacent vertices x and y, dG(x) + dG(y) >1n, then G has a Hamilton circuit. * Corresponding author 0378-3758/96/$15.00 (~) 1996--Elsevier Science B.V. All rights reserved SSDI 0378-3758(95)00084-4
196
G. Liu, G. YanlJournal of Statistical Planning and Inference 51 (1996) 195-200
For a subset S of V(G), we denote by G - S the subgraph obtained from G by deleting the vertices in S. For a subset E ' C E(G), G - E ' denotes the graph obtained from G by deleting the edges in E'. Let S and T be two disjoint subsets of V(G). We write E(S, T) = {xy : x E S and y C T} and e(S, T) =[ E(S, T) [. Let 9 and f be two integer-valued functions defined on V(G). Let C be a component of G - (S U T) such that 9(x) = f ( x ) for all x E V(C). We say that C is (S, T)-odd or even according to e(T, V(C)) + ~-~xcV(C)f ( x ) being odd or even. All other components of G - (S U T) are called (S, T)-neutral. Let h(S, T) denote the number of (S, T)-odd components of G - (S U T). For any function f , we write f ( X ) = ~x~X f ( x ) and f(O) = 0. Define 6(S, T) = dG(T) - e(S, T) - g(T) - h(S, T) + f ( S ) . In 1970 Lov~isz gave the following result. Theorem 1.2 (Lovhsz (1970)). Let G be a graph and 9 and f be two inteoer-valued functions defined on V(G) such that 9(x) <~f ( x ) for all x c V(G). Then G has a (9, f)-factor if and only if for all disjoint subsets S and T of V(G), 6(S, T) >~O. This result was generalized by Liu (1988). We define e(S, T) as follows. (a) e ( S , T ) = 2, if one of the following conditions holds: (1) S is not independent; (2) there is an (S, T)-even component C of G - ( S U T) such that either e(S, V(C)) >>.1 or there is a cut-edge e of C such that components C1 and C2 of C - e are (S, T)-even in G - e - ( S U T ) . (b) e(S, T ) = 1, if neither of (1) and (2) holds and there is an (S,T)-neutral component C of G - (S U T) such that either e(S, V(C))/> 1 or there is a cut-edge e of C such that one of the components C1 and C2 of C - e is (S, T)-even in G - e - (S U T). (c) e(S, T) = 0, otherwise. In 1988 Liu obtained the following result. Theorem 1.3 (Liu (1988)). Let G be a graph and 9 and f be &teoer-valued functions defined on V(G) such that 9(x) <~f ( x ) for all x E V(G). Then for any edge e o f G there is a (9,f)-factor containing e if and only if for all S, T C V ( G ) , S O T = 0, 6(S, T) >>.e(S, T). This result is essential for the results in this paper.
2. The main results Let a and b be integers and b>~a~>0. A graph G is called an [a,b]-graph if a <~de(x)<<, b for every x C V(G). An r-regular graph is an [r,r]-graph. Theorem 2.1. Any [kd - 1,kd + 1]-graph G with k >12 has a d-matching.
G. Liu, G. YanlJournal of Statistical Planning and Inference 51 (1996) 195-200
Proof. Let n
=[V(G)I. If
197
n <~2kd - 2, then for any x , y C V(G), we have
d o ( x ) + d o ( y ) >1 2(kd - 1) ~> n.
So G has a Hamilton circuit by Theorem 1.1. Since n >~kd >~2d, G has a d-matching. Suppose n > / 2 k d - 1 and G has no d-matching. Let M be a maximum matching of G. Then [ M [ ~ < d - 1 and any edge o f G - M is adjacent to an edge of M. Since d o ( x ) ~< kd - 1 for any x E V ( G ) , we have [E(G)[ ~< k d ( 2 d - 2) + d - 1 = ( d - 1)(2kd + 1). On the other hand, since d o ( x ) ~ k d - 1 and n >~2 k d - 1, we have [E(G)[ /> ½ ( k d - 1 ) ( 2 k d - 1) > ( d - 1)(2kd + 1), a contradiction. The theorem is proved. L e m m a 2.2. L e t G be a [kd - 1,kd + 1]-graph, k >>.2, a n d / e t M = {el,e2
. . . . .
ed-i }
be any ( d - 1)-matching o f G. Then given any edge e o f H = G - M, there is a [k - 1,k + 1]-factor F containing e so that G - E ( F ) is a [ k ( d - 1 ) - 1 , k ( d - 1) + 1]-graph.
clearly, the result is true. Assume d >~ 2. Then H = G - M is a [kd - 2,kd + 1]-graph. For each x E V ( H ) = V(G), define
Proof.
When
d = 1,
g(x) =
k k-
f (x ) =
k
1
for d o ( x ) = kd + 1, otherwise.
and k+ 1
for die(x) >>.kd, otherwise.
Clearly, for every x E V(H), we have k -
l~
1. Set A ( x ) =
die(x)/d - g(x) and A ' ( x ) = f ( x ) - die(x)/d. Thus
(d - 1)/d
for die(x) = kd + 1, for die(x) = kd, for die(x) = kd - 1,
(d - 2 )/d
for die(x) = kd - 2.
1/d
1 or 0 A(x) =
and
A'(x) :
(d - 1)/d
for die(x) = kd + 1,
1
for die(x) = kd, for die(x) = kd - 1, for die(x) = kd - 2.
1/d 2/d
Since d >/2, then for any x E V ( H ) , A ( x ) >>.0 and A ' ( x ) >~ lid. Let S, T C_ V ( H ) so that S N T = 0. Since g(x) ~ f ( x ) for each x E V ( H ) , H - ( S U T ) has neither (S,T)-odd nor even components. Hence we have 6 ( S , T ) = d i e ( T ) e(S, T ) - g ( T ) + f ( S ) .
Now we prove the lemma by Theorem 1.3.
198
G. Liu, G. YanlJournal of Statistical Plannin9 and Inference 51 (1996) 195-200
I f S # 0 and T = 0, then 3(S,T) = f ( S ) l>k>12. I f S = 0 and T # 0, then 6(S,T) = dH(T) - #(T)>>.kd - k - 1 ~> 1. Further, 3(0,0) = 0. In the following we assume that S # 0 and T # 0.
3(S, T) = de(T) - [ d e ( S ) - de-r(S)] - 9(T) + f ( S ) _ [d~T)
9(T)]+[f(S)
= A(T) + A'(S)+ ( 1 - 1 )
d~S)]+(1-1)[dH(T)-dH(S)]+ae_T(S de(T) - (1-1)
[e(S,T) + dH-r(S)] + de-T(S)
= A(T) + A'(S) +
1-
[ d e ( T ) - e(S, r)] +
= A ( T ) + A'(S)+
1-
dH_s(T)+
Thus 1
3(S,T) >1At(S) >~ ~ > O. Since 3(S, T) is an integer, we have
3(s, T) >11. When S is not independent, we have dH-T(S)/d >t 2/d. Hence, ifde_s(T)~> 1, then
6(S,T) >>. 1 -
de_s(r)+ 1
2
>~l---d+- d =1+~.
1
6(S, T) is an integer, so 6(S, T) i> 2. If dH-s(T) = 0, then since dH(X) ~> kd - 2 and T # 0, we have I S [ /> kd - 2 and therefore we have 3(S, T) >~ A ' ( S ) + d e - r ( S ) d
>>-
+ d~e - r ( S )
kd - 2 2 >.---d--+-d =k>~2. Hence we have 3(S, T)/> e(S, T). By Theorem 1.3, for any edge e of H there is a ( # , f ) - f a c t o r F containing e and F is a [k - 1,k + 1J-factor.
)
G. Liu, G. YanlJournal of Statistical Planning and Inference 51 (1996) 195-200
199
By the definition of g(x) and f ( x ) , it follows that H' = H - E ( F ) is a [k(d - 1) - 2 , k ( d - 1 ) + 1]-graph and dn,(x) = k ( d - 1 ) - 2 implies dt4(x) = k d - 2, or de(x) --- k d - 1. Thus G - E ( F ) is a [ k ( d - 1 ) - l , k ( d - 1 ) + 1]-graph as required.
Theorem 2.3. Let k >12 and d be positive &tegers and let G be a [kd - 1, kd + 1]graph. Then for any d-matching M of G, there is a [ k - 1,k + l]-factorization F orthogonal to M. Proof. We employ induction on d. The theorem holds trivially for d = 1. Assume now that d = 2 and the theorem is true for d - 1. Let M = { e l , e 2 . . . . . ed} and Ml = { e l , e 2 . . . . . e d - l } . By Lemma 2.2 G has a [k1,k + 1]-factor Fd containing the edge ed and excluding M1 such that G' = G - E(Fd) is a [ k ( d - 1 ) - 1 , k ( d - 1 ) + 1]-graph. Clearly, Ml is a ( d - 1) -matching of G'. By the induction hypothesis G' has a [k - 1, k + 1]-factorization F' = {F1,F2 . . . . . Fd-1} orthogonal to M1. Thus G has a [k - l,k + 1]-factorization F = { F 1 , F2 . . . . . Fd } orthogonal to M. A kd-regular graph is a [kd - 1, kd + 1]-graph, hence we have the following result.
Corollary 2.4. Let G be any kd-regular graph. Then for any d-matching M, G has a [ k - 1,k + 1]-factorization F orthogonal to M. Remark. If k = 1, a [kd - 1, kd + 1]-graph may not have any d-matching. There are kd-regular graphs such that for some d-matching M there is no k-factorization orthogonal to M. If k is odd, then a kd-graph may have no k-factorization. If k is even, we can construct a kd-regular graph G with a d-matching which has no orthogonal k-factorization. We illustrate with the following example. Let k/> 2 be a even integer and let G be a kd-regular graph with V(G) = {ui, vi: 1 <~i <~kd} and E(G) = { u i v j : 1 <~i <~kd, 1 <~j <. kd}\{ulvkd, VlUkd} 0 {uluka, v, Vkd}. Let M be a d-matching of G such that el = UlUkd E M and e2 = VlVkd E M. Since G - {et,e2} is a bipartite graph, clearly, the graph G has no 2-factor containing el and excluding e2 and hence has no k-factorization orthogonal to M. Therefore, the assertions o f Theorem 2.3 and Corollary 2.4 are best possible in the following sense: if k-factorization replaces [k - 1,k + 1]-factorization, they do not hold again.
Acknowledgements The authors wish to thank the referees for their helpful suggestions.
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G. Liu, G. Yan/Journal o f Statistical Plannino and Inference 51 (1996) 195-200
References [1] Alspach, B., K. Heinrich and G. Liu (1992). Orthooonal Factorizations, Contemporary Design Theory - A Collection o f Surveys. Wiley, New York, 13-37. [2] Liu, G. (1988). On (9,f)-covered graphs, Math. Acta. Sei. 8, 181-184. [3] Lov~isz, L. (1970). Subgraphs with prescribed valencies, J. Combin. Theory 8, 319~116. [4] Ore, O. (1960). Note on Hamilton circuits, Amer. Math. Monthy 67, 55.
Journal of Statistical Planning and Inference 51 (1996) 195-200
joumalof statistical plan ' and r i n g
Orthogonal [ k - 1, k ÷ 1]-factorizations in graphs G u i z h e n L i u *, G u i y i n g Y a n Department of Mathematics, Shandong University, Jinan, Shandong 250100, People's Republic of China
Abstract Let G be a graph. Let F = {FI,F2 ..... Fd} be a factorization of G and H be a subgraph of G. If H has exactly one edge in common with Fi for all i = 1,2 ..... d, then we say that F is orthogonal to H. In this paper it is proved that for any d-matching M of a [kd - 1, kd + 1]-graph G, there is a [ k - l,k + l]-factorization of G orthogonal to M where k/> 2 is an integer.
1. Introduction The graphs considered in this paper are finite undirected graphs which have neither multiple edges nor loops. Let G be a graph with vertex set V(G) and edge set E(G). For a vertex x of G, the degree of x in G is denoted by de(x). Let g and f be two integer-valued functions defined on V(G) such that g(x) ~< f ( x ) for every x E V(G). Then a (g, f ) - f a c t o r of G is a spanning subgraph F o f G satisfying g(x) <~d e ( x ) <~f ( x ) for all x E V(G). If for all x E V(G), g(x) = a and f ( x ) = b, then a ( g , f ) - f a c t o r is called a [a,b]-factor. A (g,f)-factorization F = {F1,F2 . . . . . Fd} of a graph G is a partition o f E(G) into edge-disjoint (g,f)-factors FI,F2 . . . . . Fd. Similarly, we define an [a,b]-factorization. A subgraph H o f G is orthogonal to F if ] E ( H ) N E(Fi)1= 1, l <<,i <~d. Alspach et al. (1992) gave the following problem: Given a subgraph H o f G, does G have a factorization with certain prescribed properties which is orthogonal to H ? The purpose of this paper is to prove that for any d-matching M of a [kd - 1, kd + 1]-graph G, there is a [k - 1,k + 1]-factorization of G orthogonal to M. In particular, for any d-matching M o f a kd-regular graph there is a [ k - 1,k + 1]-factorization orthogonal to M. Further there is a kd-regular graph G with a d-matching M so that there is no k-factorization orthogonal to M. To prove the main results we need the following concepts and preliminary results. Theorem 1.1 (Ore (1960)). Let G be a graph with n >13 vertices. I f for any two nonadjacent vertices x and y, dG(x) + dG(y) >1n, then G has a Hamilton circuit. * Corresponding author 0378-3758/96/$15.00 (~) 1996--Elsevier Science B.V. All rights reserved SSDI 0378-3758(95)00084-4
196
G. Liu, G. YanlJournal of Statistical Planning and Inference 51 (1996) 195-200
For a subset S of V(G), we denote by G - S the subgraph obtained from G by deleting the vertices in S. For a subset E ' C E(G), G - E ' denotes the graph obtained from G by deleting the edges in E'. Let S and T be two disjoint subsets of V(G). We write E(S, T) = {xy : x E S and y C T} and e(S, T) =[ E(S, T) [. Let 9 and f be two integer-valued functions defined on V(G). Let C be a component of G - (S U T) such that 9(x) = f ( x ) for all x E V(C). We say that C is (S, T)-odd or even according to e(T, V(C)) + ~-~xcV(C)f ( x ) being odd or even. All other components of G - (S U T) are called (S, T)-neutral. Let h(S, T) denote the number of (S, T)-odd components of G - (S U T). For any function f , we write f ( X ) = ~x~X f ( x ) and f(O) = 0. Define 6(S, T) = dG(T) - e(S, T) - g(T) - h(S, T) + f ( S ) . In 1970 Lov~isz gave the following result. Theorem 1.2 (Lovhsz (1970)). Let G be a graph and 9 and f be two inteoer-valued functions defined on V(G) such that 9(x) <~f ( x ) for all x c V(G). Then G has a (9, f)-factor if and only if for all disjoint subsets S and T of V(G), 6(S, T) >~O. This result was generalized by Liu (1988). We define e(S, T) as follows. (a) e ( S , T ) = 2, if one of the following conditions holds: (1) S is not independent; (2) there is an (S, T)-even component C of G - ( S U T) such that either e(S, V(C)) >>.1 or there is a cut-edge e of C such that components C1 and C2 of C - e are (S, T)-even in G - e - ( S U T ) . (b) e(S, T ) = 1, if neither of (1) and (2) holds and there is an (S,T)-neutral component C of G - (S U T) such that either e(S, V(C))/> 1 or there is a cut-edge e of C such that one of the components C1 and C2 of C - e is (S, T)-even in G - e - (S U T). (c) e(S, T) = 0, otherwise. In 1988 Liu obtained the following result. Theorem 1.3 (Liu (1988)). Let G be a graph and 9 and f be &teoer-valued functions defined on V(G) such that 9(x) <~f ( x ) for all x E V(G). Then for any edge e o f G there is a (9,f)-factor containing e if and only if for all S, T C V ( G ) , S O T = 0, 6(S, T) >>.e(S, T). This result is essential for the results in this paper.
2. The main results Let a and b be integers and b>~a~>0. A graph G is called an [a,b]-graph if a <~de(x)<<, b for every x C V(G). An r-regular graph is an [r,r]-graph. Theorem 2.1. Any [kd - 1,kd + 1]-graph G with k >12 has a d-matching.
G. Liu, G. YanlJournal of Statistical Planning and Inference 51 (1996) 195-200
Proof. Let n
=[V(G)I. If
197
n <~2kd - 2, then for any x , y C V(G), we have
d o ( x ) + d o ( y ) >1 2(kd - 1) ~> n.
So G has a Hamilton circuit by Theorem 1.1. Since n >~kd >~2d, G has a d-matching. Suppose n > / 2 k d - 1 and G has no d-matching. Let M be a maximum matching of G. Then [ M [ ~ < d - 1 and any edge o f G - M is adjacent to an edge of M. Since d o ( x ) ~< kd - 1 for any x E V ( G ) , we have [E(G)[ ~< k d ( 2 d - 2) + d - 1 = ( d - 1)(2kd + 1). On the other hand, since d o ( x ) ~ k d - 1 and n >~2 k d - 1, we have [E(G)[ /> ½ ( k d - 1 ) ( 2 k d - 1) > ( d - 1)(2kd + 1), a contradiction. The theorem is proved. L e m m a 2.2. L e t G be a [kd - 1,kd + 1]-graph, k >>.2, a n d / e t M = {el,e2
. . . . .
ed-i }
be any ( d - 1)-matching o f G. Then given any edge e o f H = G - M, there is a [k - 1,k + 1]-factor F containing e so that G - E ( F ) is a [ k ( d - 1 ) - 1 , k ( d - 1) + 1]-graph.
clearly, the result is true. Assume d >~ 2. Then H = G - M is a [kd - 2,kd + 1]-graph. For each x E V ( H ) = V(G), define
Proof.
When
d = 1,
g(x) =
k k-
f (x ) =
k
1
for d o ( x ) = kd + 1, otherwise.
and k+ 1
for die(x) >>.kd, otherwise.
Clearly, for every x E V(H), we have k -
l~
1. Set A ( x ) =
die(x)/d - g(x) and A ' ( x ) = f ( x ) - die(x)/d. Thus
(d - 1)/d
for die(x) = kd + 1, for die(x) = kd, for die(x) = kd - 1,
(d - 2 )/d
for die(x) = kd - 2.
1/d
1 or 0 A(x) =
and
A'(x) :
(d - 1)/d
for die(x) = kd + 1,
1
for die(x) = kd, for die(x) = kd - 1, for die(x) = kd - 2.
1/d 2/d
Since d >/2, then for any x E V ( H ) , A ( x ) >>.0 and A ' ( x ) >~ lid. Let S, T C_ V ( H ) so that S N T = 0. Since g(x) ~ f ( x ) for each x E V ( H ) , H - ( S U T ) has neither (S,T)-odd nor even components. Hence we have 6 ( S , T ) = d i e ( T ) e(S, T ) - g ( T ) + f ( S ) .
Now we prove the lemma by Theorem 1.3.
198
G. Liu, G. YanlJournal of Statistical Plannin9 and Inference 51 (1996) 195-200
I f S # 0 and T = 0, then 3(S,T) = f ( S ) l>k>12. I f S = 0 and T # 0, then 6(S,T) = dH(T) - #(T)>>.kd - k - 1 ~> 1. Further, 3(0,0) = 0. In the following we assume that S # 0 and T # 0.
3(S, T) = de(T) - [ d e ( S ) - de-r(S)] - 9(T) + f ( S ) _ [d~T)
9(T)]+[f(S)
= A(T) + A'(S)+ ( 1 - 1 )
d~S)]+(1-1)[dH(T)-dH(S)]+ae_T(S de(T) - (1-1)
[e(S,T) + dH-r(S)] + de-T(S)
= A(T) + A'(S) +
1-
[ d e ( T ) - e(S, r)] +
= A ( T ) + A'(S)+
1-
dH_s(T)+
Thus 1
3(S,T) >1At(S) >~ ~ > O. Since 3(S, T) is an integer, we have
3(s, T) >11. When S is not independent, we have dH-T(S)/d >t 2/d. Hence, ifde_s(T)~> 1, then
6(S,T) >>. 1 -
de_s(r)+ 1
2
>~l---d+- d =1+~.
1
6(S, T) is an integer, so 6(S, T) i> 2. If dH-s(T) = 0, then since dH(X) ~> kd - 2 and T # 0, we have I S [ /> kd - 2 and therefore we have 3(S, T) >~ A ' ( S ) + d e - r ( S ) d
>>-
+ d~e - r ( S )
kd - 2 2 >.---d--+-d =k>~2. Hence we have 3(S, T)/> e(S, T). By Theorem 1.3, for any edge e of H there is a ( # , f ) - f a c t o r F containing e and F is a [k - 1,k + 1J-factor.
)
G. Liu, G. YanlJournal of Statistical Planning and Inference 51 (1996) 195-200
199
By the definition of g(x) and f ( x ) , it follows that H' = H - E ( F ) is a [k(d - 1) - 2 , k ( d - 1 ) + 1]-graph and dn,(x) = k ( d - 1 ) - 2 implies dt4(x) = k d - 2, or de(x) --- k d - 1. Thus G - E ( F ) is a [ k ( d - 1 ) - l , k ( d - 1 ) + 1]-graph as required.
Theorem 2.3. Let k >12 and d be positive &tegers and let G be a [kd - 1, kd + 1]graph. Then for any d-matching M of G, there is a [ k - 1,k + l]-factorization F orthogonal to M. Proof. We employ induction on d. The theorem holds trivially for d = 1. Assume now that d = 2 and the theorem is true for d - 1. Let M = { e l , e 2 . . . . . ed} and Ml = { e l , e 2 . . . . . e d - l } . By Lemma 2.2 G has a [k1,k + 1]-factor Fd containing the edge ed and excluding M1 such that G' = G - E(Fd) is a [ k ( d - 1 ) - 1 , k ( d - 1 ) + 1]-graph. Clearly, Ml is a ( d - 1) -matching of G'. By the induction hypothesis G' has a [k - 1, k + 1]-factorization F' = {F1,F2 . . . . . Fd-1} orthogonal to M1. Thus G has a [k - l,k + 1]-factorization F = { F 1 , F2 . . . . . Fd } orthogonal to M. A kd-regular graph is a [kd - 1, kd + 1]-graph, hence we have the following result.
Corollary 2.4. Let G be any kd-regular graph. Then for any d-matching M, G has a [ k - 1,k + 1]-factorization F orthogonal to M. Remark. If k = 1, a [kd - 1, kd + 1]-graph may not have any d-matching. There are kd-regular graphs such that for some d-matching M there is no k-factorization orthogonal to M. If k is odd, then a kd-graph may have no k-factorization. If k is even, we can construct a kd-regular graph G with a d-matching which has no orthogonal k-factorization. We illustrate with the following example. Let k/> 2 be a even integer and let G be a kd-regular graph with V(G) = {ui, vi: 1 <~i <~kd} and E(G) = { u i v j : 1 <~i <~kd, 1 <~j <. kd}\{ulvkd, VlUkd} 0 {uluka, v, Vkd}. Let M be a d-matching of G such that el = UlUkd E M and e2 = VlVkd E M. Since G - {et,e2} is a bipartite graph, clearly, the graph G has no 2-factor containing el and excluding e2 and hence has no k-factorization orthogonal to M. Therefore, the assertions o f Theorem 2.3 and Corollary 2.4 are best possible in the following sense: if k-factorization replaces [k - 1,k + 1]-factorization, they do not hold again.
Acknowledgements The authors wish to thank the referees for their helpful suggestions.
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