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Directed Acyclic Graphs
7.5.1 Test for Acyclicity
Result:
A digraph is acyclic if and only if its first search does not have back
arcs.
Proof:
First we prove that

backarc
cycle.

Let (u w)
be a backarc. This means that w is an ancestor of v. Thus (w,...,
v, w) will be a cycle in the digraph.

Next we show that

cycle
backarc.
Suppose G is cyclic. Consider a cycle and let v be the vertex with the
lowest dfnumber on the cycle. See Figure 7.13.
Because v is on a cycle, there is a vertex u such that (uv)
is an edge. Since v has the lowest dfnumber among all vertices on the cycle,
umust be an descendant of v.

it can not be a tree arc since dfnumber(v) dfnumber(u)

it can not be a forward arc for the same reason

it can not be a cross arc since v and u are on the same cycle.
Note that the above test for acyclicity has worst case complexity O(e).
Next:7.5.2
Topological SortUp:7.5
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Directed Acyclic Graphs
eEL,CSA_Dept,IISc,Bangalore