3K2-decomposition of a graph

A c t a M a t h . A c a d . Sei. H u n g a r .

40 O---4), (1982), 201--208.

3K~-DECOMPOSITION OF A GRAPH A. BIALOSTOCKI and Y. RODITTY (Tel-Aviv)

1. Introduction

Graphs in this paper are finite, have no multiple edges and loops. DEFINITION 1. A graph G=G(V,E) is said to have an H-decomposition if it is the union of edge-disjoint isomorphic copies of the graph H. Necessary and sufficient conditions for H-decomposition have been determined mostly for the complete graph Kn, see [1, 2], but also for complete bipartite [2] and complete multipartite graphs [2, 5]. However only for particular graphs H. Recently Y. Caro and J. SchSnheim considered H-decomposition of a general graph where H is 2K~ or Pe (two-bars or a path of length 2). This problem was completely solved [3, 4]. This paper determines the graph G which have 3K~-decomposition. It is proved that the necessary conditions are also sufficient excluding a list of 26 graphs. 2. Preliminary results

The following two conditions for are obviously necessary: (1)

G=G(V, E) to have a 3K2-decomposition

E(G)=3k,

(2) deg

(v)--6. In the course of this paper we shall deal with the sufficiency problem. DEFINITION 2. If U is a subset of vertices of the graph will denote the degree of v in the graph induced by U U {v}.

G(V, E) then degv(v)

DEFINITION 3. Let G=G(V,E) satisfy (1) and (2) for a certain k. Denote VI={v[vEV(G), deg(v)=k}, and its cardinality by ~. DEFINITION 4. Let G=G(V, E) be a graph and H a subgraph of G. Denote by G',,,H the graph whose set of vertices is the same as that of G and its set of edges is the set E(G)\E(H). DEFINITION 5. Let G=G(V, E) satisfy (1) and (2) for a certain k. Define X = ~/~ degv~ (v), Y= ~, d e g v \ v I (v), where V1 is as in Definition 3. vEV a

vEV a A c t a M a t h e m a t i c a A c a c l e m i a e S e i e n ~ i a r u m H u n g a r i c a e 40, 1982

202

A. BIALSTOCKI A N D Y. RODITTY

THEOREM l. Let G=G(V, E) satisfy (l) and (2)for k. Then, (a)

X ~ 2(~-3)k

(b)

a=7.

~3,

PROOF. Summing degrees in G implies

(3)

X + Y = ~k.

Counting edges implies X ~ - + Y N 3k,

(4) or

(4')

X + 2Y ~ 6k.

Subtracting (4) from (3) implies (a). Subtracting (3) from

(5)

(4') implies

Y 3 then a, bEV(3K2). PROOF. The only graph that does not contain 2//2 are K3 and a star. But K3 and a star do not satisfy- (1) and (2) for any k. Hence our graph contains 2K2. If there are a, bEV(G) such that d e g ( a ) ~ 3 or deg(b)=>3 then there exist vertices x,yr b such that (x, a), (y, b)EE(G). In any case we shall take 2K2 as (x, a) and (y, b). Let zl, ..., z, (n~2) be the vertices left in G. If there is any edge (z~, zj), we are through. Otherwise all the vertices zl, i r . . . . , n are adjacent only to {a,b,x,y}. Let G2=G2(Vz,E~,) be a graph such that V2= {a, b, x, y} and E2={(s, t)EG]s, tEV2, (s, t);~(x, a), (y, b)}. Denote fl=]E(G2)]. Then obviously 0-