Exercise 2: Detecting a Cycle in A-B, B-C, C-D, D-B — Possible Solution
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GIVEN
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Edges: A-B, B-C, C-D, D-B

STEP 1: DFS WITH PARENT TRACKING, STARTING FROM A
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Visit A (parent: none). Go to neighbor B.

Visit B (parent: A). Neighbors: A (parent, skip), C, D. Go to C first.

Visit C (parent: B). Neighbors: B (parent, skip), D. Go to D.

Visit D (parent: C). Neighbors: C (parent, skip), B.
  B has already been visited, AND B is NOT D's parent (D's parent is
  C, not B).

STEP 2: THE CYCLE
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Per this chapter's own parent-tracking rule, reaching an already-
visited vertex that isn't the current parent means a genuine cycle
exists. The edge D-B is exactly this case - it connects back to B,
which was visited earlier via a different path (A-B-C-D), not via the
edge D just arrived on.

RESULT
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Yes, this graph contains a cycle: B - C - D - B (a triangle-shaped
cycle involving B, C, and D). Vertex A sits outside the cycle,
connected only as an extra branch off of B.

WHY THIS WORKS AS AN ANSWER
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The DFS trace explicitly tracks each vertex's parent throughout,
correctly skips the trivial "came from" edges (B's edge back to A, C's
edge back to B, D's edge back to C), and identifies the specific edge
(D-B) that violates the parent-tracking rule, naming the exact cycle
it closes rather than just answering yes/no.