01
List direct neighbours before tracing longer routes
From A, the direct outgoing neighbours are B and D. C is not a direct neighbour because there is no arrow from A to C. From B, the only outgoing neighbour is C, and from D it is also C. C has no outgoing neighbours in the supplied graph. Write these four adjacency lists before tracing routes: A lists B and D; B lists C; D lists C; C lists none. This separates the actual edges from connections that may become possible by following more than one edge in sequence.
02
Count destinations separately from the routes that reach them
Starting at A, B and D are reachable in one step. C is reachable in two steps through either A to B to C or A to D to C. There are therefore three distinct reachable destinations from A, even though C has two different routes leading to it. The shortest route from A to C has two steps. This exercise treats every edge as one step; it assigns no time, cost or capacity to any edge. Equal step counts do not establish that two real-world routes would take equal time or perform equally well.
03
Check the reverse question against the actual arrows
Starting at C, none of A, B or D is reachable because C has no outgoing edges. The existence of a route from A to C does not create a reverse route from C to A. A reusable worksheet records the node list, directed edge list, start node, direct neighbours, reachable destinations and one explicit route for each claimed destination. If a new arrow from C to A is added, create a revised edge list before recalculating. That change creates cycles and alters reachability, so conclusions from the original four-edge graph should not be carried forward without review.