Consider the following sequences of nodes for the undirected graph shown…

CONCEPT

A DFS preorder is controlled by the recursion stack. From the vertex at the top of the stack, DFS must visit an unvisited adjacent vertex whenever one exists; it may pop that vertex only after all of its neighbors have already been visited.

Therefore, for a proposed preorder, each next vertex must be reachable from the current stack top after popping only vertices that have no unvisited neighbors left.

APPLICATION

1. Sequence 1: after a → b → e → f, the stack top is f. Vertices c and g are still unvisited neighbors of f, so f cannot be popped to reach d next. This sequence violates the DFS stack rule.

2. Sequence 2: follow a → b → e → f → c. After c has no unvisited neighbor, return to f, continue to g, and then visit d from g. This realizes a → b → e → f → c → g → d.

3. Sequence 3: choose successive unvisited neighbors a → d → g → e → b → c → f. Every next vertex is available from the current stack top, so this preorder is realizable.

4. Sequence 4: after a → d → b → c, the stack top is c. Vertices e and f are still unvisited neighbors of c, so c cannot be popped to reach g next. This sequence violates the DFS stack rule.

CROSS-CHECK

The constructive traces certify sequences 2 and 3. The remaining-neighbor checks at f in sequence 1 and at c in sequence 4 independently show where their proposed preorder would require premature backtracking.

RESULT

Exactly sequences 2 and 3 are possible, so the answer is “2 and 3 only”.