If
is a permutation of 
, the graph 
has vertices 
where xy is an edge of if and only if (x, y) or (y, x) is an inversion of . Any graph isomorphic to is called a permutation graph.
In 1967 Gallai characterized permutation graphs in terms of forbidden induced subgraphs. In 1971 Pnueli, Lempel, and Even showed that a graph is a permutation graph if and only if both the graph and its complement have transitive orientations. In 2010 Limouzy characterized permutation graphs in terms of forbidden Seidel minors.
In this paper, we characterize permutation graphs in terms of a cohesive order of its vertices. We show that only the caterpillars are permutation graphs among the trees. A simple method of constructing permutation graphs is also presented here.
of 
to itself is called a permutation of order
. We shall write 
for
. We shall denote by 
the set of all permutations of
.
is an ordered pair 
where
but
. Equivalently, 
is an inversion if and only if 
and
.
. The graph of inversions of
, denoted by
, is the graph with vertices 
where 
is an edge of 
if and only if 
or 
is an inversion of
.
for some permutation 
will be called a permutation graph. There is an implementation PermutationGraph[p] in the Combinatorica package of Mathematica [
.
is a permutation graph if and only if both 
and its complement 
have transitive orientations. Recently in 2010 Limouzy [
will be denoted by 
while the edge-set will be denoted by
. An edge with end-vertices 
and 
will be denoted by 
or
. For graph theoretic terms used here without definition, the book by Harary [
. The inversions of 
are
, 
, 
, 
, 
, and
. It is convenient to draw the graph 
with the vertices in a line following their arrangement in
. A drawing of 
is shown in 
and 
are two edges of the graph where 
lies between 
and 
in the drawing, then 
is also an edge.
is an edge and 
is a vertex that lies between 
and 
in the drawing, then either 
is an edge or 
is an edge.
be a graph of order
. An arrangement 
of the vertices of 
is called a cohesive vertex-set order of 
(or simply cohesive order
) if the following conditions are satisfied:
and
, 
, then 
.
and
, then 
or
.
, denoted by 
has the same vertex-set as 
and two distinct vertices 
and 
form the edge 
in 
if and only if 
is not an edge in
.
be a graph. Then 
is a cohesive order of 
if and only if 
is a cohesive order of
.
be a cohesive order of
. We claim that the same is a cohesive order of
. To prove 
for
, let 
and 
be vertices of 
such that
. Then 
and 
are not edges in
. By property 
of a cohesive order, the edge 
is not in
. Hence, 
is an edge of
. To prove 
for
, let 
be an edge of 
with
. Let 
be an integer such that
. Since 
is in
, then it is not in
. By property 
of a cohesive order (for
) the edges 
and 
cannot be both in
. Hence at least one of them is in
.
.
. Then
.
is a cohesive order of a graph 
if and only if 
is a cohesive order of
.
of a graph 
means to change 
to either the ordered pair 
or the ordered pair
.
is the digraph obtained by assigning a direction to each edge of
. The directed edges, which are ordered pairs, are called arcs.
is said to be transitive if 
is an arc of 
whenever 
and 
are arcs in
.
, the out-degree of a vertex
, denoted by 
or simply 
is the numnber of vertices 
in 
such that 
is an arc in
. The in-degree of
, denoted by 
or 
is the number of vertices 
in 
such that 
is an arc in
.
in a tournament, denoted by 
is defined by
. The score sequence of a tournament is the sequence of scores arranged in non-decreasing order.
be a tournament of order
. The following statements are equivalent:
is transitive.
and 
in
, if 
is an arc of
, then
.
and 
in
, if
, then 
is an arc of
.
is
.
is a permutation graph if and only if it has a cohesive order.
is a permutation graph, then 
is isomorphic to 
for some permutation
. By Theorem 2.1, 
is a a cohesive order of
. Let 
be an isomorphism of 
to
. Then 
is a cohesive order of
.
be a graph with a cohesive order
. Orient 
to obtain a digraph 
as follows: For each edge 
in
, assign the direction 
if
; otherwise assign the direction
.
of a cohesive order, it follows that 
is transitive. Extend 
to a tournament by orienting the complement 
of 
as follows: If 
but 
is not in
, assign the direction 
to the edge 
in
. By Lemma 2.1 
is a cohesive order of
. So likewise, the orientation of 
obtained in this manner is also transitive. Let us denote this digraph by
.
and 
is an orientation of
. Since 
is complete, then 
is a tournament. We claim that 
is a transitive tournament. Let 
and 
be arcs of
. If both arcs belong to 
or to
, then 
is in 
because both 
and 
are transitive. So let us assume that one of the arcs belong to 
and the other arc belong to
. Without loss of generality, assume that 
is an arc in
, and 
is an arc in
. If 
is in
, we are done. If 
is not in
, then 
is in
. Since 
is transitive and
, 
are in
, then 
is in
. This is a contradiction because 
is in
.
is 
. Let 
be the permutation defined by
, where 
is the score of 
in
. We claim that the mapping 
is an isomorphism of 
to
.
is bijective since the scores of the vertices are distinct. It remains to show that 
preserves adjacency. Let 
be an edge of
, where
. In 
we have the arc
. Since the tournament 
is transitive, then by Theorem 2.2,
. Hence, 
is an inversion of
. Therefore, 
and 
are adjacent in
. Conversely, let 
be an edge in
. Then either 
or 
is an inversion. Without loss of generality, assume that 
is an inversion. Let 
and
, where
. Since 
is an inversion, we have
, or
. Therefore, the arc 
is in
. Since
, the arc 
must be in
. Consequently, the edge 
is in
.
shown in 
.
.
, we assign the arc that goes from right to left. Then the result is a transitive tournament. It is not difficult to get the score of any vertex in this tournament. We simply count the eastbound arcs and the westbound arcs with a fixed tail. Consider for example,
. The number of eastbound arcs with tail at 
is 3. The number of westbound arcs is simply the number of vertices to its left that are not adjacent to to
. The table below summarizes the scores of the vertices.
defined by
. Then
. The permutation graph 
is shown in 
be a graph. The following are equivalent:
is a permutation graph.
is a permutation graph.
is a permutation graph.
is a permutation graph.
has a cohesive order if and only if 
has a cohesive order. Therefore, (a) and (b) are equivalent.
is a cohesive order of
, then the subgraph of 
induced by a set of vertices
, where 
has cohesive order 
and therefore is a permutation graph. Hence, (a) and (c) are equivalent.
is an induced subgraph of
.
have connected components 
and let 
be the order of
. Let
. Then
. Therefore 
is a permutation graph.
and stars 
are permutation graphs because they have cohesive orders as illustrated in 
, we have
, etc.
, and 
such that
. Note for example that 
is an arc and the vertices 1 and 4 are between 2 and 3 in the drawing. We have 1 adjacent to 2 and 4 adjacent to 3. This illustrates condition (b).
we see that for every arc 
where 
all vertices 
with 
are between 
and
. Moreover, the vertex 
is adjacent to 0. Therefore condition (b) is satisfied. Condition (a) is satisfied vacuously.
formed by subdividing each edge of the star 
into two edges, as shown in 
has no cohesive order. Therefore this is not a permutation graph. This result is also established by Limouzy [
for
.
.
as a subgraph.
is not a permutation
obtained by subdividing the edges of
.
is not a permutation graph. Therefore, all we need to show is that a caterpillar is a permutation graph. Let 
be a caterpillar and let 
be the path obtained from 
by removing the pendant vertices. If
, then 
is either the trivial graph or the star 
for some
. Since the trivial graph and the stars are permutation graphs, we assume that
.
as shown in 
be a set of pendant vertices of 
all adjacent to the same vertex 
of
. If 
is odd, we insert the vertices in 
immediately to the left of the vertex 
of the path (see 
is even we insert the vertices in 
between 
and
. The result is a cohesive order of
. Therefore 
is a permutation graph.
be a graph with vertices 
and let 
be a collection of arbitrary graphs. The composition by 
of 
, denoted by 
is the graph formed by taking the disjoint union of the graphs 
and then adding all edges of the form 
where 
is in
, 
is in 
whenever 
is an edge of
.
is equal to a fixed graph
, we use the symbol 
to denote the composition.
and 
, denoted by 
is formed by taking the disjoint union of 
and 
and then adding all edges of the form 
where 
and
. Thus, the composition 
is formed by taking the disjoint union of the graphs 
and then forming the sum 
if the associated vertices 
and 
of 
are adjacent.
be a graph of order 
and let 
be arbitrary graphs. Then
, 
are permutation graphs.
is a permutation graph. Each graph 
is an induced subgraph of
. Therefore, each 
is a permutation graph. If we take a vertex 
from each
, then the subgraph induced by these vertices is isomorphic to
. Therefore 
is a permutation graph.
, 
are all permutation graphs. Then there is a cohesive order 
of
. Let 
be the order of
. Then the vertices of 
has a cohesive order
.
is a cohesive order of
.
be the star 
with central vertex 
and pendant vertices
, then
is a permutation graph.
of order 
may be written as
and
.
can be written as a composition, then we say that 
is prime.
and 
are prime permutation graphs.
, 
, and 
are prime permutation graphs. These are all caterpillars that do not have two pendant vertices adjacent to a common vertex. Note that 
which is excluded from the list is a caterpillar with two pendant vertices having a common neighbor.
be a tree of order
. Assume that 
is a prime permutation graph. By Theorem 3.2 T is a caterpillar. Suppose that 
and 
are pendant vertices with a common neighbor
. Let 
be the tree obtained from 
by identifying 
and
. Let 
be the vertices of 
Without loss of generality, assume that 
is the vertex resulting from the identification of 
and
. Let 
be the graph with two vertices but without an edge, and let 
be the trivial graph for
. Then
.
is prime.
is a caterpillar with no two pendant vertices having a common neighbor. Suppose that 
is a not a prime permutation graph. Then for some non-trivial graph 
with vertices
,
.
contains at least two vertices. Now, 
must be connected for otherwise, 
is disconnected. Let 
be adjacent to 
without loss of generality. Then
. If 
has at least two vertices, then there will be a cycle in
. Therefore, 
has only one vertex. In
, 
cannot be adjacent anymore to any other vertex for otherwise, we would also create a cycle of length 4. Now consider
. There cannot be adjacent vertices in 
for otherwise we will create a cycle of length 3. But then all vertices in 
are pendant vertices of 
and they have a common neighbor, the vertex in
. This is a contradiction. 
be a decomposable permutation graph. Then there exists a non-trivial prime permutation graph 
and permutation graphs
such that
.
, where 
is non-trivial. If we take one vertex 
from each
, then the subgraph induced by these vertices is isomorphic to
. Hence, 
must be a permutation graph. Each 
is an induced subgraph of
. Therefore, each 
is a permutation graph. Assume that 
has smallest order among all such decompositions of
. We claim that 
is a prime permutation graph. Suppose that 
is not prime. Let
, where 
is non-trivial. Since 
is a decomposition of 
and vertices of 
are the induced subgraphs 
then each 
is a associated with subset of
. We may assume that 
is an induced subgraph of
. Hence 
. But this contradicts the choice of
. Therefore, 
must be a prime permutation graph. 
in the decomposition