Data Structures and Algorithms in Java[1].pdf - Fulvio Frisone

ordering such that any directed path in G traverses vertices in increasing order. (See
Figure 13.11.) Note that a digraph may have more than one topological ordering.
Figure 13.11: Two topological orderings of the same
acyclic digraph.
Proposition 13.21:
has a topological ordering if and only if it is acyclic.
Justification: The necessity (the "only if" part of the statement) is easy to
demonstrate. Suppose
is topologically ordered. Assume, for the sake of a
contradiction, that has a cycle consisting of edges (vi 0 , vi 1 ), (vi 1 , vi 2 ),…, (vik−
1, vi 0 ). Because of the topological ordering, we must have i 0 < i 1 ... < ik− 1 < i 0 ,
which is clearly impossible. Thus,
must be acyclic.
We now argue the sufficiency of the condition (the "if" part). Suppose is
acyclic. We will give an algorithmic description of how to build a topological
ordering for . Since is acyclic, must have a vertex with no incoming edges
(that is, with in-degree 0). Let v 1 be such a vertex. Indeed, if v 1 did not exist, then
in tracing a directed path from an arbitrary start vertex we would eventually
encounter a previously visited vertex, thus contradicting the acyclicity of
. If we
remove v 1 from , together with its outgoing edges, the resulting digraph is still
acyclic. Hence, the resulting digraph also has a vertex with no incoming edges, and
we let v2 be such a vertex. By repeating this process until the digraph becomes
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