Consider the following statements regarding directed acyclic graph:

• I. If a directed graph G is cyclic but can be made acyclic by removing one edge, then a depth-first search in G will encounter exactly one back edge.
• II. If a depth-first search on a directed graph G = (V,E) produces exactly one back edge, then it is possible to choose an edge e E such that the graph G′ = (V,E − { e }) is acyclic.

Which of the following option is correct?
(A) Statement I is false.
(B) Only statement II is true.
(C) Both (A) and (B)
(D) None


Answer: (C)

Explanation:

• I. False: you can have multiple back edges, yet it can be possible to remove one edge that destroys all cycles. For example, in graph G = (V,E) = ({a,b,c},{(a,b), (b,c),(b,a),(c,a)}), there are two cycles ([a,b,a] and [a,b,c,a]) and a DFS from a in G returns two back edges ((b,a) and (c,a)), but a single removal of edge (a,b) can disrupt both cycles, making the resulting graph acyclic.
• II. True: Removing the back edge will result in a graph with no back edges, and thus a graph with no cycles (as every graph with at least one cycle has at least one back edge). Notice that a graph can have two cycles but a single back edge, thus removing some edge that disrupts that cycle is insufficient, you have to remove specifically the back edge. For example, in graph G = (V,E) = ({a,b,c}, {(a,b), (b,c), (a,c), (c,a)}), there are two cycles [a,b,c,a] and [a,c,a], but only one back edge (c,a) Removing edge (b,c) disrupts one of the cycles that gave rise to the back edge ([a,b,c,a]), but another cycle remains, [a,c,a].

Since statement I is false and II is true. So, both option (A) and (B) are true, hence option (C) is correct choice.

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