Vol. 10 No 6 - Pi Mu Epsilon

450 PI MU EPSILON JOURNAL
INDIANA JONES AND THE QUEST FOR ANTICONNECTED DIGRAPHS 451
produces an anti connected digraph.
Henry Jones continued, "We must first show that ar(D) exists for every
connected digraph D and then present a bound on this parameter. Before
continuing, we 'II need the following definitions. "
A tree is a connected graph containing no cycles. A bipartite graph is
a graph G whose vertex set can be partitioned into two sets U and V such
that every edge joins a vertex of U to a vertex of V. It is well-known (see
[1], for example) that a tree of order n has size n - 1 and that every tree is
bipartite. A spanning tree of a connected graph G is a tree that is a subgraph
of G and has the same vertex set as G. Every connected graph has a
spanning tree.
Theorem 2. For a connected digraph D of order n, the (anticonnected)
reversal number ar(D) ~ L (n - 1 l /21.
Proof. Let D be a connected digraph of order p and let The a spanning
directed tree of D. Then Tis bipartite with partite sets U and V. Suppose
that T has m 1 arcs from U to V and ~ arcs from V to U. Then m =
min{m 1
,~} ~ L(n-1) /21. Reversing the direction of these m arcs will
produce an anticonnected directed tree and hence an anticonnected digraph.
Thus, ar(D) ~ L (n - 1) /21. 0
"Now notice Dad, that the bound in Theorem 2 is sharp in the sense that
for every positive integer n, the directed path of order n has (anticonnected)
reversal number L(n - 1) /2J."
For a connected graph G, the (anticonnected) reversal number ar(G) is
defined as ar(G) = max {ar(D)}, where the maximum is taken over all
orientations D of G. So, by Theorem 2, for a connected graph G of order
n, the reversal number ar(G) ~ L(n - 1) /2J. It was already noted that the
directed path of order n bas reversal number L (n - 1) /21. Thus, for the path
Pn of order n, it follows that ar(Pn) = L(n - 1) /21. It was also noted that if
G is a graph of order n ~ 3 such that deg v ~ (3n - 1)/4 for every vertex
v of G, then every orientation of Gin anticonnected. Hence ar(G) = 0 for
each such graph G.
"With what we know now we should be able to determine the reversal
IUII7Iber of a tree. "Henry Jones continued to scrawl out theoretical. results.
Theorem 3. For a tree T of order n ~ 3 , the reversal number
ar(1) = L(n - 1) /21.
Proof. Let T be a tree of order n. Then T is bipartite with partite sets
U and V. Obtain an orientation D of Thy directing r