Graph Decomposition
with Constraints on
the Connectivity and
Minimum Degree
~
Ca rsten Tho mass en
MA THEMA TICAL INSTITUTE
THE TECHNICAL UNIVERSITY OF DENMARK
2800 LYNGBY, DENMARK
ABSTRACT
For each pairs,? of natural numbers there exist natural numbers f(s,t) andg(s,t)
such that the vertex set of each graph of connectivity at least f ( s , t )
(respectively minimum degree a t least g(s,t)) has a decomposition into sets
which induce subgraphs of connectivity (respectively minimum degree) at least
s and ?, respectively.
E. Gyoyi (B. Toft, private communication) has asked if, for each pair s,t of
natural numbers, there exists a (smallest) natural number f(s,t) satisfying the
assertion of the abstract.
We first establish the existence of g(s,t). The analogous problem for
maximum degrees was settled by Lovasz [ 1 I.
Theorem 1. If k is a natural number and G is a graph of minimum degree at
least 12k, then the vertex set V(G)of G can be decomposed into nonempty
sets S and T such that the subgraphs G(S) and G( 7 ) induced by S and T,
respectively, both have minimum degree at least k.
ProoJ: IfA and B are subsets of V(G)we denote by e(A,B) the number of
edges with one end in A and the other end in B and for short we write
e(A) = e(A,A).
We first claim that V(G) has a decomposition into nonempty subsets S and
T such that e(S) 2 kl SI and e ( r ) 1 kl TI. A well-known argument due to
Journal of Graph Theory, Vol. 7 (7 983)165-1 67
@ 1983 by John Wiley b Sons, Inc. CCCO364-9024/83/020165-03$01.30
166
JOURNAL OF GRAPH THEORY
Erdos shows that, if we partition V(G)into sets A and B such that e(A,B) is
maximal, then each vertex of the bipartite graph H containing all edges from
A to B has degree at least 6k. Assume ' AI I
!BI and let v,,v2, . . . be the
vertices of A such that the degrees in H of these vertices form a nonincreasing
sequence. Let A l = ( v I , v 3.,. . } and A 2 = (v2,v4,. . . ). Then
+ IB 1.
e ( A 2 , B )I
e ( A , , B )i e ( A 2 , B )
(1)
For i = 1,2, let Bj be the set of vertices of B which are joined to at least 2k
vertices of Ai. Then BI U B2 = B.
Assume first that jB, I 1'A IB I for i = 1,2. Then we choose subsets B:
B,suchthatB; U B i i s a p a r t i t i o n o f B a n d (B;I I
\B;i I
IB;I
l.then,
for i = 1,2,
+
which proves our claim.
Assume next that IB2I
have
<M
c
IBI. In this case we put B ; = B,\B2 and we
and
e ( A , , B ) 1 e ( A l ,B ; ) 2 ( 4 k
+ 1 ) IB; ! .
(3)
e ( A 2 ,B ) 1 e ( A l , B ) - IBI 1 ( 4 k + I ) l B ' l 1 - IBI 1 ( 4 k - l ) I B ; (
(4)
and by ( 2 ) and ( 4 ) we have
e ( A 2 ,B 2 ) L 2kj B; I 1 k j A 2 U B2 I .
(5)
Now ( 3 ) and ( 5 ) proves our claim with S = A l U B ; and T = A 2 U B2.
The case I B, I < 'A I BI is treated in a similar way.
Having proved our claim it is easy to complete the proof. We first consider
minimal nonempty subsets SI and TI of S and T,respectively, such that
e ( S I ) L k i S I Ia n d e ( T l ) l k]TII.ThenthesubgraphsG(Sl)andG(Tl)have
minimum degree greater than k. We next consider two disjoint supersets S2
and T2 of SI and T , , respectively, such that the subgraphs G(S2)and G(T2)
both have minimum degree at least k and such that S2 U T2is maximal under
these conditions. Then S2 U T2= V(G).For otherwise, G( V(G)\T2) would
have a vertex x of degree less than k and then the sets S2, T2 U (x) would
contradict the maximality of S2 U T2.This completes the proof.
167
GRAPH DECOMPOSITION
Mader [2] proved that, for each natural number k, there exists a natural
number n(k) such that each graph of minimum degree n(k) contains a kconnected subgraph. Combining this with Theorem 1 we answer Gyori’s
question in the affirmative.
+
Theorem 2. If s and t are natural numbers and G is an (s t - 1)-connected
graph of minumum degree at least g(n(s), n(t)), then the vertex set V(G)of G
can be decomposed into sets S and Tsuch that the subgraphs G(S) and G(r)
have connectivity at least s and t, respectively.
Proof. By Theorem 1, there exists a decomposition V(G)= S’ U T’ such
that G(S’) and G(T‘) have minimum degree at least n(s) and n(t),
respectively, and so by Mader’s theorem [2], G has two disjoint subgraphs of
connectivity at least s and c, respectively. We now choose disjoint vertex sets
S and T such that c(S) is s-connected, G(7) is f-connected and S U T is
maximal. We shall prove that S U T = V(G).
Suppose therefore (reductio ad adsurdum) that C = V(G)\(S U T ) is
nonempty. Then the maximality of S U T implies that G(S U C) is not sconnected, i.e., it contains a separating vertex set A of cardinality at most
s - 1. Since G(S) is s-connected, G(S U C) - A has a component whose
vertex set D is a subset of C. Again, the maximality of S U T implies that
G(T UD) has a separating set B of cardinality at most c - 1 and
G(A U 0)- B has a component whose vertex set is contained in D. But then
G - ( A U B ) is disconnected and A U B has cardinality at most s t - 2.
This contradiction proves the theorem.
Mader [2] proved that n(k) < 4k and sofand g are bounded above by a
linear function. The complete graph of order s t i- 1 shows that g(s,t) L
s t 1 andf(s,t) 2 s t 1. Maybe these inequalities are equalities for
all s and f . It is easy to see that g(s,2) = s 3 and in [4] it is shown that
f(s,2) = s 3. Recently, R Haggkvist [3] has found a very short proof of the
inequality g(s,t) I 2s t.
+
++
++
+
+
+
+
References
[ 11 L. Lovasz, On decomposition of graphs. Studia Sci. Math. Hungar. 1
(1966) 237-238.
[2] W. Mader, Existenz n-fach zusammenhangender Teilgraphen in
Graphen genugend grosser Kantendichte. Abh. Math. Sem. Univ.
Hamburg 37 (1972) 86-97.
[3] R Haggkvist, private communication.
[4] C. Thomassen, Nonseparating cycles in k-connected graphs. J. Graph
Theoly, 5 (1981) 351-354.