C OMPOSITIO M ATHEMATICA
L OWELL W. B EINEKE G ARY C HARTRAND
The coarseness of a graph
Compositio Mathematica, tome 19, n
o4 (1968), p. 290-298
<http://www.numdam.org/item?id=CM_1968__19_4_290_0>
© Foundation Compositio Mathematica, 1968, tous droits réservés.
L’accès aux archives de la revue « Compositio Mathematica » (http:
//http://www.compositio.nl/) implique l’accord avec les conditions géné- rales d’utilisation (http://www.numdam.org/conditions). Toute utilisation commerciale ou impression systématique est constitutive d’une infrac- tion pénale. Toute copie ou impression de ce fichier doit contenir la présente mention de copyright.
Article numérisé dans le cadre du programme Numérisation de documents anciens mathématiques
http://www.numdam.org/
290
by
Lowell W. Beineke and
Gary
ChartrandIntroduction
A
graph
isplanar
if it can be drawn in theplane (or
on asphère)
so that no two of its
edges
intersect. The thicknesst(G)
of agraph G,
introducedby
Tutte[3],
is the minimum number ofedge- disjoint planar subgraphs
whose union is G. In contrast to the thickness is the coarseness 2c(G)
defined as the maximum number ofedge-disjoint nonplanar subgraphs
contained in G.Obviously,
G is
planar
if andonly
ifc(G)
= 0.We
present
some results hereconcerning
the coarseness of agraph.
Inparticular,
bounds on the coarseness ofcomplete graphs
are
given.
Basic results
A subdivision of a
graph
G is agraph
obtained from Gby
re-placing
anedge x
= uv of Gby
a new vertex and the two newedges
uw and wv. Twographs G,
andG2
arehomeomorphic
if thereexists a
graph G3
which can be obtained from each ofG,
andG2 by
a sequence of subdivisions.Following
Kuratowski[2],
we cal]a
graph
skew if it ishomeomorphic
to either thecomplete graph Ks
or thecomplete bigraph K3,3.
Thesegraphs
are shown inFigure
1. The classic result of Kuratowski then states that agraph
is
planar
if andonly
if it contains no skewsubgraph.
1 Research supported by grants from the U. S. Air Force Office of Scientific Research and the National Institute of Mental Health, Grant MH-10834.
2 This concept was suggested by Professor Paul Erdôs.
291
If a
graph
G has coarseness n, then the nedge-disjoint nonplanar subgraphs
contained in G may all be chosen to be skew, for ifone were not
skew,
thenby
Kuratowski’s theorem, some extra-neous
edges
could be deleted to obtain a skewgraph.
A block
of
agraph
G is a maximal connectedsubgraph
of Gwhich contains no eut vertices. It is clear that any skew
subgraph
of G must be a
subgraph
of some block ofG;
thus we arrive at thefollowing
result.THEOREM 1. The coarseness of a
graph
is the sum of the coarse-nesses of its blocks.
This
implies
that it is sufficient to determine the coarseness ofgraphs having
no eut vertices. Another observation which can be made concerns the number ofedges
in agraph.
The smallest number ofedges
which anonplanar graph
can possess is 9(and
then
only
if thegraph
isK3,3). Therefore,
itimmediately
followsthat if a
graph
Ghas q edges,
thenIn
[1]
it was shown that everygraph
ishomeomorphic
to agraph having
thickness 1 or 2. As we shall now see, however,coarseness is invariant under
homeomorphism.
THEOREM 2.
Homeomorphic graphs
have the same coarseness.PROOF. Let
Gi
andG2
behomeomorphic graphs.
Then thereexists a
graph G3
which can be obtained from each ofG,
andG2 by
a sequence of subdivisions. To show thatc(G1)
=c(G2),
it isclearly
sufficient to prove that the coarseness of one ofG,
andG2,
say
G1, equals
that ofG3 .
To provethis,
however, it is sufficient to show thatc(G1)
=c(G’1),
whereCi
is a subdivision ofGl. Thus,
there is an
edge x
= uv ofG,
which has beenreplaced by
a newvertex and the two new
edges
uw and wv to obtainGi .
Assumec(Gl) -
n. Consider a set of nedge-disjoint
skewsubgraphs
ofG1.
If none of thesesubgraphs contains x,
then this is a set ofsubgraphs
ofGi ,
soc(G’1) ~
n. Should one of thesesubgraphs
Hcontain x,
thenby replacing x by
the vertex w and theedges
uwand wv, a skew
subgraph
H’ ofG’
isproduced,
whichtogether
withthe other n -1
subgraphs
constitutes a collection of nedge- disjoint
skewsubgraphs
ofG’1.
Hence, in any case,c(Gi )
> n.To show that
c(Gi )
> n is notpossible,
suppose thatG"
containsn+1
edge-disjoint
skewsubgraphs.
If neither of theedges
uv norze» is contained in any of these
subgraphs,
then these skewgraphs
are also
subgraphs
ofGl,
but this cannothappen
sincec(G1)
= n.Thus, one of these
edges,
say uw, mustbelong
to some skewgraph
S. Because no skewgraph
contains a vertex ofdegree
1,the
edge
zew is alsonecessarily
in S. Butthen,
a skewgraph
S’can be obtained
by replacing
theedges
uw and zew and the vertexw
by
theedge
x. Thegraph
S’along
with the other n skewgraphs
form a set of n+1
edge-disjoint
skewsubgraphs
ofGl,
but thisis also a contradiction.
By taking
asufficiently large
number of subdivisions of agraph
G with
high
coarseness, agraph
G’ can be obtained whose thickness is 2.By
theprevious theorem,
however,c(G) == c(G’).
We statethis as a
corollary.
COROLLARY 2a. For every
positive integer
n, there exists agraph
G such thatc(G) - t(G)
= n.More
specifically,
consider thenonplanar complete bigraph K3,3.’
· It iseasily
observed thatK3, 3n
is theedge-disjoint
unionof
K1,3n
andK2,3n,
each of which isplanar.
Thust(K3,3n)
= 2.One also sees that
K3, 3n
contains nedge-disjoint copies
ofK3, 3
so that
c(K3,3n)
~ n, but since thisgraph
has 9nedges, by
in-equality (1), c(K3,3n)
= n. These facts haveanother, perhaps
somewhat
unexpected interpretation:
For anypositive integer
n,it is
possible
to find twoplanar graphs
which can be combined into agraph G,
which then in turn can bedecomposed
into nedge-disjoint nonplanar graphs.
On the coarseness of
complète graphs
The
complete graph K p
with p vertices hasp(p-1)/2 edges;
therefore
by (1),
we haveIn this section we also
present
lower bounds forc(K,).
The deter-mination of a formula for
c(Kp)
appears to be a very difficultproblem.
The known values forc(Kp)
aregiven
in Table 1. Thusthe smallest value of p for which
c(Kp)
is unknown is 13. Also unknown is the smallestpositive integer n
for which there existsno p such that
c(Kf))
= n.For several of the entries in Table 1, a
simple
construction anda
degree argument provides
the answer. We illustrate this for p = 9.By dividing
the 9 vertices into 3 sets of 3 verticeseach,
one sees that every two of these sets induces a copy of
K3,3
so that293
TABLE 1
The known coarsenesses of complete graphs
c(K9) ~
3. SinceK9
has 36edges,
itfollows, by (2),
thatc(K9) ~
4.However,
forc(K9)
to have the value 4, it wouldnecessarily
haveto contain 4
edge-disjoint copies
ofK3,3,
but each vertex ofK9
has
degree
8 so at least twoedges meeting
at each vertex cannot be in any of thesecopies
ofK3,3 implying
that 3 is the maximumnumber of
edge-disjoint nonplanar
skewgraphs
contained inK9.
For p
=11and p
=12 anargument
similar to thatfor p
= 9 canbe
given,
but for p = 10 a differenttechnique
isemployed.
We labelthe vertices of
Kio
as v1, v2, ···, v10. Thenc(K10) ~ 4
sinceK1o
contains 4
edge-disjoint
skewgraphs,
e.g., those shown inFigure
2.Figure 2
By (2), however, c(K10)
ç 5. Ifc(K10)
= 5, then, sinceKlo
has45
edges, K10
must contain 5edge disjoint copies Gi (1 i 5)
ofthe
graph K3, 3 . Necessarily,
each vertex appears in 3 of thesecopies.
We now relabel the vertices ofK10 by
ul , U2’ ..., U10 sothat ul appears in
G1, G2 ,
andG3
and so that the vertices to which ul isadjacent
are as indicated inFigure
3.Figure 3
The two unlabeled vertices in
Gi clearly belong
to the set{u5,
U6’ .. -,u10}.
If both these vertices are in the set{u5,
u6,u7}
or in
{u8,
U95U1015
then these two vertices must appeartogether
in some copy of
K3,3
since theedge
determinedby
them is insome
Gi .
However, no suchgraph
canexist,
for these vertices must bemutually adjacent
to two other vertices in thisGi . Hence,
of the two unlabeled vertices inG1,
one must be in{u5,
U6’u7}
and the other in{u8,
u9,u10},
say U5 and u8.Similarly,
of the two unlabeled vertices in
G2,
one mustbelong
to the set{u2,
u3,u4}
and the other to{u8,
u9 ,u10}.
However, no vertex of{u2,
u3,u4}
can be inC2
since each of these vertices isalready adjacent to u5
inCI.
This is acontradiction;
thusK1o
does notcontain 5
edge-disjoint copies of K3,3
so thatc (Klo)
= 4. This factwill be useful in
obtaining
a lower bound forc(Kp).
THEOREM 3. For the
complete graph K p ,
PROOF.
For p
= 3r, the vertices ofKp
can be divided into rsets of 3 vertices each.
Every
two of thèse setsclearly
déterminea copy of
K3,3’
Since there are2
suchpairs
of sets,K3r
contains2 copies of K3,3
and thusc(K3r) ~ (r2).
Assume p
= 3r+1. Let v be one vertex ofKp
and divide theremaining
vertices into r sets of 3 vertices each. Asbefore,
one295
gets (r2) copies of K3,3.
However, if one takes 3triples
of verticestogether
with v, 4 skewgraphs
can be obtained as in thedecompo-
sition of
K1o .
These canreplace
the 3copies
ofK3,3
which hadbeen obtained
originally.
There are[r 3] disjoint
collections of 3triples,
and it ispossible
toproduce
an additional skewgraph
foreach such collection.
Hence, c(K3r+1) ~ (r2) + [r 3].
For p
= 3r + 2, weagain
take r sets of 3 vertices each and denotethe
remaining
verticesby u
and v. On one of these setstogether
with u
and v,
a copy ofKs
can be formed. Let C denote the col- lection of3 [r+24]
vertices in[r+2 4]
of theremaining
r-1 sets.A skew
graph homeomorphic
toK5
can then be constructed for each of the r-1-[r+2 4]
stillremaining
setsby using
such a setwith u and and
assigning
a vertex of C as a vertex ofdegree
2between u and v. Since
there are
enough
vertices toaccomplish
this. Therefore,In terms of p, the bounds
provided by (2)
and Theorem 3 can be stated as follows:The
inequalities (3) immediately
establish an estimate ofc(Kp)
for
large
p.COROLLARY 3a. The coarseness of the
complete graph K p
isasymptotic
top2/18.
The lower bound
a(p)
and the upper boundb(p) given by (3)
are
presented
in Table 2 for 1 çp ~
40.TABLE 2
Lower Bounds a(p) and Upper Bounds b(p) on c(Kp), 1 ~ p ~ 40
The amount
by
whicha(p)
andb(p)
can differ is now considered.THEOREM 4. The lower bound
a(p)
and the upper boundb(p)
of
c(Kp), given
in(3), satisfy
theinequality:
PROOF. We consider the 3 cases p = 3r, 3r+1, and 3r+2
individually.
First,
assume p = 3r. ThenNext,
suppose p = 3r+1. ThenHowever, for p
= 3r + 1,297 This
inequality
is satisfiedonly
when r - 1(mod 3), i. e.,
whenr = 3s+1. But
then p
= 9s+4, in which caseFinally,
assume p = 3r+2. Thenand
Because
also,
theinequality
is satisfied for all p = 3r+2.Since the difference
b (p ) -a (p ) is always
aninteger,
the theoremfollows.
In
conclusion,
wepresent
bounds for the numberc(Km,n).
Since
K
has mnedges,
wehave, by (1), c(Km,n) ~ [mn 9].
We now obtain a lower bound for
c(Km,n).
Assume the vertex set V’ ofKm,n
ispartitioned
asVl
UV2,
where[V1]
- m,[V2]
= n,and every
edge joins
a vertex ofVI
with a vertex ofV2.
The setV1
has[m3]
distincttriples
of vertices whileV2
has[n 3]
suchtriples.
Thus,Km, n
has at least[m 3] [n 3] copies
ofK3,3.
Theseobservations are summarized below.
THEOREM 5. For the
complete bigraph Km, n,
COROLLARY 5a. If m = n = 0 mod 3, then
c(Km,n)
=mn/9.
REFERENCES
L. W. BEINEKE
[1] On the decomposition of complete graphs into planar and other subgraphs. Doctoral thesis, The University of Michigan, 1965.
K. KURATOWSKI
[2] Sur le problème des courbes gauches en topologie. Fund. Math. 16 (1930),
2712014283.
W. T. TUTTE
[3] The thickness of a graph. Indag. Math. 25 (1963), 5672014577.
(Oblatum 2-8-1967) University College,
London and Purdue University,
Fort Wayne
The University of Michigan
and Western Michigan University