www.elsevier.com/locate/disc
Note
A note on ascending subgraph decompositions of
complete multipartite graphs
Hung-Lin Fu
a; ∗, Wei-Hsin Hu
baDepartment of Applied Mathematics, National Chiao Tung University, Hsin Chu, Taiwan, ROC bDepartment of Accounting & Statistics, The Overseas Chinese College of Commerce, Taichung,
Taiwan, ROC
Received 23 December 1997; revised 2 November 1999; accepted 10 April 2000
Abstract
In this note, we prove that the ascending subgraph decomposition conjecture is true for com-plete multipartite graphs. c 2001 Elsevier Science B.V. All rights reserved.
MSC: 05C70
Keywords: Ascending subgraph decompositions; Multipartite graphs
The following conjecture about decomposing a graph G of size n+12 6|E(G)|¡ n+2
2
into n ascending subgraphs has been one of the most fascinating problems regularly mentioned by P. Erdos in his talks on ‘Unsolved Problems’.
Ascending subgraph decomposition conjecture (ASD conjecture). Let G be a graph of size n+1 2 6|E(G)| ¡ n+2 2
. Then, E(G) can be partitioned into n sets E1; E2; : : : ; En which induce subgraphs G1; G2; : : : ; Gn such that |E(Gi)| ¡ |E(Gi+1)| and Gi is isomor-phic to a subgraph of Gi+1 for i = 1; 2; : : : ; n − 1.
A graph G is said to have an ascending subgraph decomposition G1; G2; : : : ; Gn provided that the ASD conjecture holds for G. G1; G2; : : : ; Gn are called members of the decomposition.
In order to verify this conjecture, the following revised conjecture attracts more attention than the original one.
Research supported by National Science Council of the Republic of China (NSC-85-2121-M-009-010).
∗Corresponding author.
E-mail address: hifu@math.nctu.edu.tw (H.-L. Fu).
0012-365X/01/$ - see front matter c 2001 Elsevier Science B.V. All rights reserved. PII: S0012-365X(00)00171-0
Revised ASD conjecture. Let G be a graph of size 2 6q ¡ 2 . Then E(G) can be partitioned into n sets E1; E2; : : : ; En which induce subgraphs G1; G2; : : : ; Gn such that |E(Gi)| = i and Gi is isomorphic to a subgraph of Gi+1 for i = 1; 2; : : : ; n − 1, and E(Gn) = q − n2. So far, quite a few classes have been veri ed to satisfy this revised ASD conjecture, such as star forests [1,6], forests [4,10], graphs with bounded maximum degrees [5,7], split graphs [9], complete bipartite graphs [8], regular bipartite graphs [3], etc., but it is believed that to prove the conjecture in general is going to be very dicult.
In this note, we shall prove that every complete multipartite graph does have an ascending subgraph decomposition. In order to prove the main result, we need two de nitions and several lemmas.
A graph G is said to have an n-star decomposition if |E(G)|6 n+1 2
and G can be decomposed into n stars G1; G2; : : : ; Gn such that (i) all stars have di erent centers, (ii) |E(Gi)|6i for all i, and (iii) |E(Gi)|6|E(Gj)| for i ¡ j. And a graph G with size n+12 + t; 0 ¡ t, is said to have an (n; t)-star decomposition if G can be decom-posed into G1; G2; : : : ; Gn; T such that (i) all Gi’s are stars with di erent centers and (ii) |E(Gi)| = i for i = 1; 2; : : : ; n and |T| = t.
Lemma 1. Let G be a graph with |E(G)|6 n+1 2
. If V (G) = X ∪ Y and the subgraph of G induced by Y; G[Y ]; has an n0-star decomposition where n0¡ n; G[X ] is an empty
graph; |X | = n − n0; |Y | = n and G \ G[Y ] is a complete bipartite graph. Then G has
an n-star decomposition.
Proof. Clearly, G \ G[Y ] can be decomposed into n − n0 stars of size n and all the
centers are in X . Let those n−n0 stars be G0
n0+1; Gn00+2; : : : ; Gn0. Since G[Y ] has an n0-star
decomposition, let it be G0
1; G20; : : : ; G0n0. Now |E(Gj0)|−j=n−j for j=n0+1; n0+2; : : : ; n.
Thus, there are at least n − j G0
i’s for i = 1; 2; : : : ; n0 such that i − |E(Gi0)| ¿ 0. Starting from j=n0+1, we delete n−n0−1 edges from G0
n0+1 in which these edges are incident
to the centers of G0
i’s where i −|E(Gi0)| ¿ 0. Then, add these edges to G0i, respectively. Note that if there are more than n − n0− 1 G0
i’s with i − |E(Gi0)| ¿ 0, we shall add the edge to G0
i which has larger i − |E(G0i)|. By repeating this process, delete n − n0− 2 edges from G0
n0+2; n − n0− 3 edges from Gn00+3; etc., we conclude the proof.
Lemma 2. Let G be a graph with |E(G)|6 n+12 . If V (G) = X ∪ Y; G[Y ] has an n0-star decomposition where n0¡ n; G[X ] is an empty graph; |X | = n − n0 and all
the vertices in X have degree not greater than n0; then G has an n-star
decomposi-tion.
Proof. Since the bipartite graph obtained from (X; Y ) can be decomposed into n − n0
stars with centers in X and all of these stars are of size not greater than n0, an
n-star decomposition of G can be obtained by rearranging the members in the n0-star
Fig. 1. Pregnant stars.
Lemma 3. Let G be a graph with |E(G)| = n+12 + t; 0 ¡ t. If V (G) = X ∪ Y; G[Y ] has an n0-star decomposition where n0¡ n; G[X ] is an empty graph; |X | = n − n0; |Y |
= n and G \ G[Y ] is a complete bipartite graph. Then G has an (n; t)-star decom-position.
Proof. Let G0
1; G20; : : : ; G0n0 be the n0stars obtained by the n0-star decomposition of G[Y ]
and G0
n0+1; Gn00+2; : : : ; Gn0 be the stars of size n obtained from the decomposition of the
complete bipartite graph (X; Y ). Consider i6n0, where m = i − |E(G0
i)| is maximum. Then there are at least m of G0
n0+1; Gn00+2; : : : ; Gn0 satisfying |E(Gj0)| ¿ j for j ∈ {n0+
1; n0+2; : : : ; n}. (Choose the ones with larger |E(G0
j)|−j.) Therefore, we can delete one edge from each of the above members and add them to G0
i. (Note that the center of G0i is adjacent to all the centers of G0
j where j=n0+1; n0+2; : : : ; n.) By repeating the above process, we have the members G1; G2; : : : ; Gn0 where |E(Gi)| = i for i = 1; 2; : : : ; n0. As
the larger members, Gn0+1; : : : ; Gn, we can delete t edges from them suitably and the
set of t edges gives the T we need.
Since the proof of the main result is quite complicated, we believe that an explanation of how we do it will be helpful in going through the details of the proof.
Our goal of decomposition is to obtain G1; G2; : : : ; Gn−1; Gn∪ T where |E(Gi)| = i and |T| = t. For the smaller members, we shall use stars. Although, it is quite possible that we have a decomposition in which every member is a star, but if this is not so, we shall mainly use pregnant stars (small star hiding in a large star) for the larger Gi’s, see Fig. 1, and the smaller members remain as stars.
In order to obtain the decomposition, we shall rst decompose the complete mul-tipartite graph Km1;m2;:::;mk; m1¿m2¿ · · · ¿mk, into k − 1 complete bipartite graphs
Hi= Kmi;mi−1+mi−2+···+m1; i = 2; 3; : : : ; k. Therefore |E(Hi)| = mi(m1+ m2+ · · · + mi−1);
i = 2; 3; : : : ; k. Then, Km1;m2;:::;mk can be depicted as Fig. 2.
Fig. 2 will give us a rough idea of the decomposition we are looking for. First, we claim that Pki=1mi¿ n¿Pki=2mi= (G). The left-hand inequality is easy to see. Assume that n ¡Pki=2mi. Then (G)¿n + 1. This implies that |V (G)|¿n + 2 and therefore |E(G)|¿1
2(n+1)(n+2). This is a contradiction. Hence, in Fig. 2, there exists an R in (P0; P1] such that RQ = n. Now, we can draw a dashed line RR0 such that “R0RQ = 45◦
and this dashed line provides some information for the decomposition. For example, 4B2 tells us how many edges can be removed from H2 in order to have members which are stars, and 4A1 shows the de ciency we have in order to construct
Fig. 2. Km1;m2;:::;mk.
stars as small members. Also, by Fig. 2 and geometric arguments, since (the height of Hi+1) − (the height of Hi)
= (m1+ m2+ · · · + mi) − (m1+ m2+ · · · + mi−1) = mi
= the width of Hi; and “R0RQ = 45◦;
the height (vertical) of 4B2; 4B3; : : : are equal and the height (vertical) of 4A1; 4A2;
4A3; : : : are not increasing.
Now, the decomposition will be obtained recursively, starting from the small members.
Theorem 4. Let G = Km1;m2;:::;mk with m1¿m2¿ · · · ¿mk¿1 and |E(G)| = n+12
+ t; 06t6n. Then G has an ASD.
Proof. Let V (G)=Ski=1Vi where |Vi|=mi. Clearly, G can be decomposed into n stars S0
1; S20; S30; : : : ; Sn0 such that the rst n − (Pki=2mi) stars have zero edges, then there are m2 stars with m1 edges, m3 stars with m1+ m2 edges, etc., and mk stars with Pk−1i=1 mi edges.
For i = 1; 2; : : : ; k, let Ai denote the sum of j − |Sj0| for all j where the center of Sj0 is in Vi and j − |Sj0| ¿ 0, and let Bi denote the sum of |Sj0| − j for all j where the center of S0
Case 1: 4(G) = m =Pk−1i=1 mi6n. By the argument following Fig. 2, we con-clude that B1= 0; B2= B3= · · · = Bk and A1¿A2¿ · · · ¿Ak. Since Pk−1i=1 mi+ mk= Pk
i=1mi¿ n, hence m2¿mk¿ n − m. Delete n − m stars Xn−m; Xn−m−1; : : : ; X1 with n − m; n − m − 1; : : : ; 1 edges, respectively, starting from the left-hand side of P1P2, and then add these n − m stars to S0
n; Sn−10 ; : : : ; Sm+10 to obtain Gn; Gn−1; : : : ; Gm+1. By the reason that S0
j; j = m + 1; : : : ; n, has center in Vk and Sj0 is incident to each vertex of Sk−1
i=1 Vi; Gm+1; Gm+2; : : : ; Gn are pregnant stars. Now, we construct the small members recursively. First, it is clear that H2\Sn−mi=1 Xi has an (n−Pki=3mi)-star decomposition. By Lemma 1, H =(H2\Sn−mi=1 Xi)∪H30 (see Fig. 2) has an n0=(n−Pki=4mi−P20P3)-star decomposition in case that the above graph has at most n0+1
2
edges. On the other hand, if the above graph has more than n0+1
2
edges, then by Lemma 3, we have an (n0; t0)-star decomposition for some t0. Here H0
3 is a part of H3 with height n3=m1+m2 and base =|{j | |Sj| ¿ j and the center of Sj is in V3}|. By Lemma 2, the n0-star (or (n0; t0)-star) decomposition of H can be extended to H ∪ (H
3\ H30). Continuing the above processes, we have an (n; t)-star decomposition for G \ (Sni=m+1Gi). Then the proof follows by adding T to Gn.
Case 2: 4(G) ¿ n and A2= A3= · · · = Ak= 0. First, if m1¿ n then each star Si0 of positive size has at least n edges. By Theorem in [11], we have the desired ASD with each member a star. On the other hand, let m16n. Let i be the largest integer such that G[Sij=1Vj] contains edges not greater than n02+1 edges where n0= n −Pkj=i+1mj. By Lemma 1, we are able to obtain an n0-star decomposition of G[Si
j=1Vj] following a way similar to what we have in Case 1. Since Aj = 0 for each j¿i, for each l ¿ i; G[Slj=1Vj] contains more than n
0+Pl j=i+1mj
2
edges. By Lemma 3, G[Plj=1Vj] has an (nl; tl)-star decomposition where nl= n0+Plj=i+1mj and some tl¿ 0. This implies that G(l = k) has an ASD by adding T to the largest member Gn where T is obtained in an (n; t)-star decomposition of G. This concludes the proof of Case 2.
Case 3: 4(G) ¿ n and A2¿ 0. Let s be the integer such that As−1¿ 0 and As= 0. (Ak= 0 since 4(G) ¿ n and A2¿ 0.) There are two situations to consider:
(i) Bs−16As−1. Since A1¿A2¿ · · · ¿As−1¿Bs−1= Bs−2= · · · = B2 and B1= 0, Pj−1
i=1Ai¿Pji=2Bi for j6s. Hence G[Ssi=1Vi] contains at most n
0+1
2
edges for some n0=n−Pk
j=s+1mj. Then by Lemmas 1 and 2, G[Ssi=1Vi] has an n0-star decomposition. Thus, for k¿x ¿ s; G[Sxi=1Vi] has an nx-star decomposition or (nx; tx)-star decompo-sition. By the time x = k, we have the ASD.
(ii) Bs−1¿ As−1. First, if s−1¿3, then rearrange V1; V2; : : : ; Vkto the order V1; V2; : : : ; Vs−2, Vs; Vs+1; : : : ; Vk; Vs−1. Now, the proof can be obtained by a similar idea as that of Case 1. Therefore, it is left to consider s−1=2. It is easy to see that there are some S0
j’s with centers in V2 and are of size less than their index. Let the number of such stars be u and clearly |V2|¿2u+1. Let G0 be the graph obtained by deleting 2u+1 vertices from V2 such that the number of edges deleted is 2u+12 +(n−2u)(2u+1)=(2u+1)(n−u). Let n0= n − (2u + 1), then G0 has n0(n0 + 1)=2 + t edges. As in Case 2, G0 has an
and their neighbors form a complete bipartite graph (X; Y; E) where |X | = 2u + 1 and
|Y | = n − u. Therefore, (X; Y; E0) can be decomposed into G
n0+1; Gn0+2; : : : ; Gn such that
Gi is a star if i6n − u and if i ¿ n − u then Gi is a union of two stars with size n − u and i − (n − u) such that it is a double star with common leaves (of small star). Now, combining the two decompositions and adding T to Gn we have the desired ASD. For further reading
The following reference is also of interest to the reader: [2]. Acknowledgements
The authors would like to express their appreciation to all the referees for their helpful comments.
References
[1] Y. Alavi, A.J. Boals, G. Chartrand, P. Erdos, O. Oellerman, The ascending subgraph decomposition problem, Congr. Numer. 58 (1987) 7–14.
[2] Fu-Long Chen, Hung-Lin Fu, Partition integral set into subsets with prescribed sums, in preprints. [3] H. Chen, K. Ma, On the ascending subgraph decompositions of regular graphs, Appl. Math. J. Chinese
Univ. Ser. B 13 (2) (1998) 165–170.
[4] R.J. Faudree, R.J. Gould, Ascending subgraph decomposition for forests, Congr. Numer. 70 (1990) 221–229.
[5] R.J. Faudree, R.J. Gould, M.S. Jacobson, L. Lesniak, Graphs with an ascending subgraph decomposition, Congr. Numer. 65 (1988) 33–41.
[6] R.J. Faudree, A. Gyarfas, R.H. Schelp, Graphs which have an ascending subgraph decomposition, Congr. Numer. 59 (1987) 49–54.
[7] H.L. Fu, A note on the ascending subgraph decomposition problem, Discrete Math. 84 (1990) 315–318. [8] H.L. Fu, Some results on the ascending subgraph decomposition, Bull. Inst. Math. Acad. Sin. 16 (4)
(1988) 341–345.
[9] H.L. Fu, W.H. Hu, Two classes of graphs which have ascending subgraph decompositions, Arc Combin. 35-A (1993) 65–70.
[10] Wei-Hsin Hu, On ascending subgraph decomposition and its related problems, Ph.D. Thesis, National Chiao Tung University, Hsin Chu, Taiwan, June 1992.
[11] K. Ma, H. Zhou, J. Zhou, On the ascending star subgraph decomposition of star forest, Combinatorica 14 (3) (1994) 307–320.
