GRAPHS AND PARTIAL ORDERINGS
BY
MARTIN AIGNER
University of Nort11 Carolina
Institute of Statistics Mimeo Series No.
October
600.1
1968
Research partially spo~,sored by the Air Force Office of
Scientific Research, Office of Aerospace Researcl1,
Uni ted States Air Force, under AFooR Grant No. 68-1406
DEPART~~~
OF STATISTICS
University of North Carolina
Chapel Hill, N. C.
27514
I.
I:ltroduction.
Every partial ordering of a set
G,
S gives rise to an m;directed graph
called its comparability graph, by identifying the vertices :)f G witll
the pJints of S and joining
points in
grap~ls
S are comparable.
hTO
vertices ::>f G iff tileir corresponding
The problem ::>f characterizLlg comparability
was solved by Walk [6J for the case of a tree, add later a charac-
terization in the general case in terms of subgraphs
G must not contain was
,
given by Gilmore-Hoffman [2) and independently by Gl1oui1a-Houri [1).
Before
we give their solution a fevT definitions are in order.
Definition:
Let
G be a fi:li te undirected graph witl10ut loops or IIDJ.1tip1e
edges 1), let V(G), E(G)
eralized path (of length
its vertex-and-edge-set, respectively, then a genk-l) is a progression of vertices
a i € V(G), (ai' a i +1 ) € E(G),
(a j , a j +1 )
once in
are the same.
ei~ler
direction).
a , a , ••• ,
1
2
~,
such that no two ordered pairs (ai' a i +1 ),
(i.e. an edge may be traversed tvrlce, but at most
We speak of a generalized cycle, if the vertex-
progression is a generalized path with
= ~.
1
then an edge
Definition:
Given the generalized path a , a , ••• , ~,
1 2
joining two vertices ai' a j with Ij - il > 1 is called a chord, in the
case
lj -
a
il= 2 we speak of a triangular chord.
Similar definitions hold
for generalized cycles with tile understanding that the definitions for chords
are given IOOdu10
k - 1.
The Gilmore-Hoffman theorem now reads:
G is a co:t::q)arabi1ity-graph iff
G contains no generalized cycles of odd length without triangular chords.
1)
For the remainder of this paper, a graph will always be assumed to be of
that kind.
2
In figures I and 2, we exllibit two examples
G, G
I
2
of graphs
G which do
not permit partial orderings; they will be seen to be basic for some of the
later developments 2)
f)(;----*b
e (7------4::1------_"0 c.
f,~, 1
In this paper we investigate some of the structural properties of comparability - and non - comparability - graphs.
In section II, we answer an extremal question by determining all NPOgraphs with a minimum number of lines and section III is devoted to the linegraph of a given graph and problems that arise in connection with partial
orderings.
There seems to be an analogy between "planar - non -
planar" and "com-
parability - non - comparability", as first the Kuratowski - type characterization in both instances, and then some theorems in this paper indicate.
For
example, theorems 3 and 4 have analogues for the planar - non - planar case
(Sedlacek [4 J), the extremal problem of section II has been solved by Wagner
[5 J for planar - non planar graphs for the point-version (the line-version
being trivial).
We hope to further develop this program in a sUbsequent
communication.
2)
G, G
are also two of the fundamental forbidden subgraphs for indifI
2
ference - systems, cf. Roberts [3].
3
Terminology:
1.
PO-graph stands for cor.rparability-graph, NPO-grapll
for noncompara-
bilitY-[7aph.
2.
He denote edges by
iff
(a, b)
€
3.
By the
~
S c V(G)
VIe
E(G),
t
iff
(a, b)
mean the subgraph which has
a, b
r/
are the tvlo end-points;
a..... b
E(G).
=a... , d=a.. ,
K-2
K-l
S as its vertex-set and includes
between any two vertices of S.
(a, b, ••• , c, d)
a, b, ••• ,
b
where
subgraph on S of G or sUbgraph induced by the set
all edges of E(G)
4.
a
(a, b),
with
denotes the generalized cycle
~
a
= 0.1,
b
= a2,
••• , c
= a l = a. We speak of the sequence of vertices
to indicate the direction in which we run through the cycle.
(e.g. the sequence a, b
is different from the edge
(a, b)).
Subcycles of
a given generalized cycle are again indicated by their first and last vertices
as long as there is no danger of ambiguity.
(e.g. subcycle
exartrple above DEans the given cycle minus the first vertex a
two vertices
c, d,
(b,
(b, ••• )
in the
and the last
of course, must be adjacent to the vertex imme-
diately preceding c)).
5.
Cycles of length k are denoted by C ;
k
no vertex occurs more than once.
we speak of a simple cycle if
4
II.
~Uninal
Configurations.
Accordine to the Gilmore-Hoffr:lal1 theorem, an NPO-graph nust contain a generalized cycle of odd length lri tl10ut triangular chords llhich ire will call a
GH-cycle for the remainder of the paper.
In this section the properties of a
shortest such cycle in a NPO-eraph will be investigated and in theorem 2 all
the minimal graphs (in the sense that the deletion of any edge results in a
PO-graph) will be determined.
Theorem 1:
An arbitrary NPO-graph G must contain a block which together
with its outgoing edges does not admit a partial ordering.
Proof:
The algorithm designed by Gilmore and Hoffman allows us to start
with any particular edge (or for that matter, with any PO-subgraph of G)
order to construct a partial ordering of the points.
in
Hence if all the blocks
plus their outgoing edges are PO-graphs, we rray start with anyone of them and
then keep on orienting the edges.
Since by the definition of a block, we
never return to the same block once we left it, the algorithm clearly yields
a PO-graph.
Remark 1:
outgoing edges"
Remarl~
2:
In theorem 1, we cannot dispose of the condition "with its
as the graph G illustrates.
l
A trivial corollary of theorem 1 is the fact that all
forests can be partially ordered.
In view of theorem 1, we, henceforth, confine ourselves to blocks plus pos-
In lemma.s 1 - 4, we will study a. shortest GH-cycle
sible outgoing edges.
of an In?O-graph G.
We run through C in one of the two possible directions,
but keep this direction fixed once we have chosen it.
follows the vertex a,
of a,
C
we call a
If the vertex b
the predecessor of b,
and indicate this fact by a, b.
b
the successor
5
Lemma 1:
Suppose the vertex a
appears more than once in
e,
say,
e = (a, b, ••••• , c, a, d, ••• , e). Suppose w.l.o.g. that
e' = (a, b, ••• , c) is of odd length, then
(1)
b - c,
(2)
b - d,
c - e,
unless
(3)
d - e,
unless
e" = (a, d, ••• , e)
Proof:
e' is a generalized cycle of odd length, "mich together with the
hypothesis on
e implies (1).
e = (a, b, ••• , c, a, d = e),
If
b
f
at least 4, then the generalized cycle
has length h.
d or
c
f
e
and
e" is of length
(a, b, ••• , c, a, d}3)
or
(a, b, ••• , c, a, e}3), respectively, would contradict the minimality of
To prove (3) we assume
generalized cycle
e" has length greater than 4, then by looking at the
(a, b, ••• , c, a, d, a, e)3)
If for two vertices
and b, a,
a, b, a
Proof:
that C'
we infer
d - e,
making
e.
again use of the hypothesis on
Lemma 2:
e.
a
and b,
e contains both sequences a, b
then they must be consecutive sequences, i.e.
e contains
or b, a, b.
Let
=
e
= ( ••• a, b, ••• , b, a ••• ),
(b, ••• )
and hence
then we may assume w.l.o.g.
e" = (a, b, ••• , b) are of odd length. But
this lvould clearly contradict (1) in lemma 1, thus the conclusion follows.
Lemma 3:
Let G be a minimal I'iPO-graph and
of G t1USt appear in
Proof:
e
as before, then every edge
e at least once.
The deletion of any edge not in
e
would not alter the character of
G as to partial orderings, in contradiction to the minimality of G.
3)
It may, of course, happen that these cycles are not in conformity with the
definition for generalized cycles any more, i.e. the sequences d, a or a, e
may occur twice, but these cases are easily seen to yield the same conclusions.
6
Lemma l~:
e
that
b
Let
contains the sequence
a, b, c, d
a shortest GH-cycle.
with
a .... d,
then
Suppose
=c
a
or
= d.
Proof:
e
e
G be a minimal NPO-graph and
We proceed to prove the assertion by contradiction.
contains the edge
(a, d).
Assume first
e = {a,
or even length.
e
fact that
a .... d,
b, c, d, ••• , a, d, ••• ).
e' =
We have to consider two cases depending on whether
Since
In the first case, (2) would imply
{d, ••• , a)
a .... c,
has odd
contradicting the
In the latter case, we
does not possess any triangular chords.
e = {a, d, ••• , a, b, c, d, ... ) with e" = {a, d, ••• ) being a gen-
have
eralized cycle of odd length.
a contradiction.
By appealing to (2) again, we conclude
The case iihere
e
contains
d
as predecessor of
be settled in an analogous manner, thus the edge
the sequence
a, b, c, d,
(a, d)
b .... d,
a
can
must appear within
and the lennna follows.
Every simple odd cycle without chords of length at least 5 obviously is
a rninir.1al NPO-graph, as are the graphs
G and
l
G
2
in section 1.
The fol-
lovr.u1g theorem makes the converse assertion that these are all the minimal
graphs.
Theorem 2:
~
The only minimal NPO-graphs are the simple odd cycles of length
5 . vrithout chords and the graphs Gl
Proof:
cycle.
Let
and G •
2
G be an arbitrary minimal NPO-graph and
If no vertex of
G appears more than once in
represents a simple cycle of odd length.
all the edges of G,
e,
Since by lemma
there can be no chords in
e,
e
and
a shortest GHthen
e
clearly
3, e must contain
vTe
obtain the first
class of the above rentioned graphs.
x
Suppose now there are points that occur at least twice in
e,
times, we can think of
e
is such a vertex appearing, say,
union of
k
k
cycles, each starting and terminating at
x.
Since
then if
as the
e
is of odd
7
lenGth, at least one of these cycles must also have odd length; let us denote
by e(x)
one of these cycles of shortest odd length.
appearing at least twice, choose the point a
cycle of shortest length among all e (x) ,
In the set of points
such that e(a)
e = (a, b, c, ••• , d, e, a, f, h, ••• h', g)
=
and e"
(a, b, c, ... , d, e)
=
e'
is a
call the complementary cycle
and let
e'
=
(a, f, ... , g).
e" ,
with
If Ire can show that G
contains either G or G as a full sUbgraph, then the theorem will follow.
l
2
By the construction of e t i t is clear that whenever a vertex appears more
than once in e', the number of edges between the two occurrences is even, a
fact vThich will be used extensively in the sequel.
(by (1», and by the minima.lity of e t
wise
e(c)
would be shorter
tl~
In
e t,
we have b .... e
= e(a) we infer c" d, since other-
e(a).
Thus
a, b, c, d, e
are 5 dis-
tinct vertices.
Case (A):
f = g.
minimality of e t
notice
Here lemma 2 implies that e" = (a, f),
= e (a)
f., b, f ., e
again we he.ve
f
f
c,
and clearly f'" b, f .., e,
f
f
d.
and by the
Furthermore, we
and so the following situ-
ation results:
where the dotted lines indicate that these edges are missing.
(a, b, c, d, e, f)
induce G ,
I
To show that
we have to demonstrate the absence of the 3
8
edees
(h, d), (c, e), (c, d).
= C(a) (they would violate the minimality of C'),
because they cannot be in C'
and C"
The first two of these edges are missing
only consists of a, f, a.
Finally, if
were in G, we could
(c, d)
delete it and still retain a lrPO-graph, namely G , herewith contradicting the
l
hypothesis on G.
Case (B):
f
f
g.
The edge
otherl·dse the deletion of
d, e),
must be contained in C',
Now we may assume w.l.o.g. that
since if we have the sequence
in C',
(a, b, y', ••• d, e)
= b or y'
or 1 .... y, 1 .... y'
= e,
in which cases
with both edges in C".
cation of lemma 1 easily yields y .... y',
=e
c = x and. d
C'
= (a,
to prove that C'
e
again)
is the successor
In the latter case, appli-
which in turn implies
contains the sequence
b, c, b, e, d, e).
y
=b
or
x, b, e, x'.
It is our goal
To this end it suffices to show
= x', as lemma 2 clearly indicates. We assume therefore
c 1= x
(the case
d
f
x'
can be dealt with in an analogous fashion)
and proceed to derive a contradiction.
(i)
x 1= e,
Xl
1= d,
(ii)
x = e,
Xl
f
(iii)
x 1= e,
Xl
= d,
(iv)
x = e,
Xl
= d.
(i).
(a, b, c,
by lemma 4.
Hence, assume
w.l.o.g.
and
e
y, e, b, y'
are both of odd length (because of the minimality of C(a)
and we either have y
y'
since
would yield the GH-cycle (a, b, c, ••• ,
then the generalized cycles
••• , y, e)
of b,
(b, e)
hence G would not be minimal.
is the successor of b
in C',
(b, e)
Since
c
Applying lemma 1 to
f
We are faced with four possibilities:
d,
x and x
(a, x)
f
and
e,
we infer a .... x with
(e, a)
(a, x)
and recalling x.., e,
€
C".
we conclude
9
= f.
x
next
x'
f
d
implies
x' = b.
by lern.18. h
a"" x',
a"" x",
= f,
G.
'vhich yields the sequence
(ii).
As before
f
x'
d
c, ••• , x ", x = e ),
g
f, b, e, b, f,
x''''' a,
= x",
e, b, e, x'
(iii).
=g
In this case,
minimallty of
c
f
x
x, b, e, x' = d, e, a, e),
in
and
the predecessor of
for we could delete
c'
a
= b,
c'.
If
and
(e, d»,
tv
d
tv
c
x.
the edges
tv
e
where
c'
is
f = x
(b, c)
or
c
would
successor
c's
(e, c'), if it were not in
e"" c I
C' ,
The fact that
=b
cI
or
x'
= d.
If
d
= c I,
then
"Ie
d
... ,
x"
=b,
e
tv
(e, c)
b,
a
first establish
x
x"
= b,
= f),
follows by means of an application of lemma 1, part (3) to
and
(x = f, b).
As
b
Fc ' ,
this cycle
l1aS
cI
=d =
x'
tv
Next it is easily seen that
(c, c',
and
without
is easily seen (by lemma 1 applied to
by the same argument as in (i).
c
By the
It clearly follows that
and vre consider the odd generalized cycle
vThere
e"" c.
and another application of lemma 1 would yield
contradiction to the hypothesis
x = f
As in (i),
•
C' , ;. cf. the remark at the
and hence by lemma 4 we have
c I,
then
CI
(a, b,
The resulting sequence
destroying the above cycle, thus obtaining a GH-cycle.
in turn implies
in
(e, c) ¢ C I (otherwise either
..., x",
C' ,
x
and (2) of lemma 1 imply
(c, c ' ,
€
by an
g
Now vre consider the odd generalized cycle
beginning of the proof).
(e, c')
x' =
presents the desired contradiction to lemma 2.
p, the edge
x"
in violation of lemma 2.
and hence
= g.
have to be contained in an even subcycle of
CI
In this cycle
'''auld produce a GH-cycle,
is the predecessor of
a ,... x" , which in turn yields x"
we infer
and
C'
nOvT consider the odd generalized cycle
x"
where
in
hence
By the same arguIl1£mt as above, we conclude
ir.tplies
analogous argument as in (i).
x'
(a, b = x I, x", "., d, e).
(b, e)
= x,
x''''' f
the successor of
otheruisc deletion of
contrary to the hypothesis on
x"
x"
Let us de:1ote by
consiCer the odd generalizeci. cycle
we r:ust 11ave
lemma 1 then gives
length at
10
least 5 and we plainly obtain
x =
r,
b, e,
(iv).
Xl
=d
= cl
Lemma 4 applied to the sequence
nmf presents the desired contradiction.
x"
By exactly the saI"ile argument, we conclude
cessor of
x
=e
cI = b
either
in
or
CI ) ,
xI
odd eeneralized cycle
e .... c l
c .... c,
= d = C I.
and hence the contradiction
lenrna
= f.
c I.... x
(c, c
(c l
=G
(x"
is prede-
is successor of
c)
In the first case, we again obtain
b ....
Xl
I, ...,
d .... c
In the latter case, we consider the
= d.
x"
and
= g,
e).
Clearly then
c .... g
and by
4 applied to the sequence g, a, b, c we would have e = b, which is
x" = g, e, b, e.
incompatible with the sequence
up to a contradiction,
hence
v1e
conclude
c
e'l
,"
=x
and similarly
d = xI
The
•
4.
situation at this stage is indicated in fig.
- - ....
All four possibilities lead
.....
l~
I}
b
r g, C"
is of length at least 4.
Since
f
g .... e
and if either one of the edges
Using lemma 1
(f, b), (g, e)
conclude
i'Te
f .... b,
CI, we
were outside
clearly would obtain a GH-cycle after deleting this edge, tlnls contradicting
the minimality of G.
g
r d).
and
hi
Next we notice
f:
c
Hence we infer
h
r a,
b, e
f
and
=e
hi
because of lemma 2, and lastly h
and
f:
g
=b
a, b, e,
f:
c
and
(as
f
f:
c
furthermore
h'
f
and
h
f:
d, since the
d
11
opposite would contradict what ,re just proved about a shortest odd cycle
e(x),
amone all
Lemma 1 applied to the cycle
yields
h - d,
h - b,
Similarly we obtain
sider the edge
that either
conclude
and application to the cycle
(h', e)
a - h'
h = h',
or
€
h' - c
e",
h = h',
and
(f
(f
h' - e,
= e,
= e,
h, " ' , c, b, e, d)
h, " ' , c, b)
gives
Now let us finally con-
Using once again lemma. 1 it is easily seen
As the first possibility cannot occur, we
and the vertices
a, b, c, d, e, h
induce
G ,
2
12
III.
n1e Linegraph.
In this section, the linegraph
L(G)
of a graph
G shall be investigated
with respect to partial orderinGs of the vertex-set.
We begin 't·Tith two easy le:r.unas:
LeI:lr.la 5:
G contains an odd cycle of length at, least 5 (with or without
If
chords), then
Let
Proof:
C
L(G)
= (al ,
does not admit a partial ordering.
C be such an odd cycle in G, say,
a , ••• , a + ),
2k l
2
then the edges
(a , a::;), ••• , (a2k , a 2k + ),
l
2
lell{;th
L(G)
2k+l
(a2k + , a )
l
l
(a , a ),
l
2
plain~ induce an odd cycle of
without triangular chords in
L(G),
i.e. a GH-cycle, hence
is an IU'O-graph.
Ler.1r.l8. 6:
Let
G be one of the minimal graphs of theorem 2 and
corresponding linegraph, then
Proof:
i'1e
on~
L(G)
have to verify the cases
G= G
l
and G= G ,
2
G
2
Theorem::;:
If G is an NPO-graph, then
L(G)
Vle consider a shortest GH-cycle in
it fixed throughout the argument.
C,
i.e.
assertion.
G,
If no edge of
denote it by
and
clear~
Suppose now there are edges which appear in
>1
a, b, c ' b, c ' ••• , c t ' b, d
l
2
for
appealing to lemma. 1 we obtain
a - c t ' d - c l ' a - d,
(a, ct' b, c ' d)
l
results.
t
C,
and keep
G appears more than once
C is an ordinary cycle, then lemma. 5
such sequences.
L(G ) = G
2
l
is an NPO-graph.
then by lemma. 2 they must be consecutive sequences in
cycle
since simple
contains a 5 - cycle now prove the assertion.
the fact that
in
its
is an NPO-graph.
cycles 'tvithout chords are isom::>rphic to their linegraphs.
Proof:
L(G)
verifies the
C in both directions,
C.
Let
such a sequence.
If
t
~
2, then
and hence the 5 -
Let us assume therefore
t
=1
for all
13
Case (A):
b, c, b, d,
Every sequence
e,
x
with
d
t
b, c, b
x,
i.e.
is embedded in the larger sequence
a, ••• , x
are distinct vertices of
In this case we construct the following sequence of edges in
(b, c), (a, b), (a, d), (d, e), (e, x),
ence is assured by lemma 1.
C
L(G), we clearly arrive at a GH-cycle
'\-There
There exists a sequence
a, b, c, d, e
t a,
e, b, d.
5 - cycle
If
t a,
f
induce
= a,
then
Assume now g"" b,
struct the GH-cycle
ima.lity of
then
C,
( h, b, d, a, g)
g
t
a
a"" d, b ,.., f,
in
hence
(since otherwise we would obtain the
and furthermore,
which in turn implies
b, c, d, e, f, g
g ,.., b.
By lemma. 1, we infer
(a, b, d, f, g}),
(f, d, a, b, c)),
Finally if
f
a, b, c, b, d, c, d, f, g, h, i
are distinct vertices as we have shown in the para-
graph preceding case (A).
g
whose exist-
L{G), thus proving the theorem.
Case (B):
G,
(a, d)
G.
(a, b),
Replacing every such sequence of vertices in
by trle indicated sequence of edges in
in
using the edge
L{G):
a,
g
t
f
c.
f
(otherwise 5 - cycle
c
But now the six vertices
5), and lemma. 6 completes the proof.
G
l
(fig.
fig.
5 again represents Gl
then in the case where
h
with f
t
a, h
f
= a,
b,
unless
we con-
(••• a, b, c, b, g, h ••• ), thus contradicting the min-
and in the case
h
t
a, h ,.., b,
and appeal to lemma. 5.
we find the
5 - cycle
As to the only remaining possibility
14
h = a,
,·re apply to the sequence
b, d, e, d,
have just shOvln above (note that because of
longer than
(a, b, c, b, d, e, d, a, g})
possible since the sequence
a, b
f = a,
g"" b,
C;,
11
= a,
the GH-cycle
and conclude
already occurred in
b
= i,
L(G),
C must be
which is in-
G whose line-
too, admit a partial ordering of the vertex-set by some ori-
entation of the edges.
nected components of
G.
what we
C.
In vie'" of theorem 3, we now vn.sh to determine all PO-eraphs
graphs
i
Since connected components of
G correspond to con-
L(G), vTe may confine ourselves to cormected PO-graphs
He observe first that in the graph
add the edges joining the outer points
G,
l
as displayed in fig. 1, we may
d, e, f
(or some of these edges) and
the resulting graph will still be a NPO-graph; in other words, in order to
preserve the NPO-character,
adjacent to an inner point.
Definition:
iff
a
l
A graph
r bl r cl r a l
inner point
only have to make sure that no outer point is
This gives rise to the following definition:
T as in fig. 6 is called a big triple centered at
and no outer point
a , b , c •
l
l
l
Since the linegraph
'\e
L(T)
G with
L(G)
coincides with an
(outer points, however, may coincide).
of any big triple
possible edges joining outer points in
dition for
a , b , c
2
2
2
x
G ,
l
T
is isomorphic to
G
l
plus
we at once have a necessary con-
a PO-graph, namely, that it contain no big triple.
15
Theorel'";l 4:
The linegrapl1
L(G)
of a graph
G ac1r.lits n partial ordering
iff
~
5,
(i)
G contains no oud cycle of length
(ii)
G contains no triangle with edges leading from the three vertices
into distinct blocks, or vnth two paths of length at least two leading from
t"TO of the vertices into distinct blocks, or with an edge leading from one of
the three vertices into a different block, if the triangle is embedded in a
complete graph on four points,
(iii)
G contains no big triple4).
Proof:
The necessity of (i) and (iii) is clear.
contains a triangle
~dth
As to (ii), suppose
G
tluree edges attached to it as specified in the
theorem, then these six edges clearly induce
G in L(G). If the triangle
2
has two paths of length at least two attached to it, then these seven edges
are readily seen to induce a 5 - cycle with two chords anu t"TO outgoing edges
opposite the two chords, which is an :NPO-graph.
Finally it is easily verified
that the linegraph of a complete graph on 4 points plus an edge does not admit
a partial ordering, thus completing the proof of the necessity-part.
Suppose then G satisfies the conditions of the theorem, and let
C be a longest cycle in G.
k
Case (A):
k ~ 6, even. By (i) and (iii)
again by (iii) there can only be
ClaWS
C
k
contains no chords, and
5 ) attached to C ,
k
centered at the
4) • It is interesting to note the analogy between the above conditions and
the conditions in Sedlacek [4] for a graph to have a planar linegraph.
5). A claw or stargraph consists of a vertex, called the center, plUS outer
points and all the edges joining the center with every outer point.
16
ver-::'iccs of
implies.
C ' and any t",TO cla"rs JiUlst be point-disjoint, as (i) clearly
k
The line graph of such a graph is easily seen to be a PO-graph.
Case (D):
k
= 4.
Assur:Je first there are no chords in
readily provides the answer.
Let
C4
=
(a, b, c, d)
C,.
4
"lith the chord
(a, c).
or at
Again applying (iii), we
C4' centered at the vertices
(ii) now implies that we either have claws centered at
l~
b
and
d.
The line6l'aphs of such graphs are PO-graphs.
we have both chords in
C4,
then (iii)
Next assume there is exactly one chord in C •
conclude there can only be cla."rs attached to
of
C4'
a
and
c.,
Finally if
ire appeal to (ii) again, noting that the line-
graph of the cOIlI.Plete graph on 4 points admits a partial ordering.
Case (C):
k = 3.
By (i) two triangles cannot be in the sane block unless
they have an edge in common.
The hypothesis
k
=3
and (iii)
now imply
that there are at most two triangles, joined by a path (possibly of length
zero, in which case the triangles have a common vertex).
(ii) and (iii)
applied. -::'0 the triangles and the connecting path now settle this case.
Case (D):
k
readily recognize
= 1,
i.e.
G is a tree.
Making use of (iii) once IOOre, we
G to be a path with point-disjoint cla"Ts attached to it,
the linegraph of which is a PO-graph.
17
Refere:-ice s.
[lJ.
,
/
~
A. Ghouila - Houri, Caracterisation des graphes nonorientes dont
on peut orienter les aretes de ma.ni~re
'a.
obtenir le graphe d tune
relation d'ordre, C. R. Acad. Sci. Paris, 25h(19G2), 1370 - 1371[2J.
P. C. Gilmore and A. J. Hoffman.. A characterization of cQIll)arability graphs and of Interval/graphS, Can. J. l.fath ... 16 (1964),
539 - 548.
[3 J.
F. S. Roberts, Representations of Indifference Relations, Ph. D.
Thesis (1968), Dept. of Math... Stanford University.
[4).
J. Sedlacek, Some Properties of Interchange graphs, Proc. Symp.
Smolenice (1963), lh5 - 150.
[5).
K. Wagner.. FastpJiittbare Graphen.. J. Combinatorial Theory.. 3
[6).
E. S. Wolk, The Comparability Graph of a Tree, Prec. Am. Math.
Soc. 13 (1962) 789 - 795.