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NETS 150 Homework 2. This is true. Based on the description, in regards to DFS, a given node would only finish if and only if every node that are discoverable from that particular node have been finished. Consider the following graph: For a to finish, b, c, and d must have already finished. Let’s say for example that we begin a DFS on the above graph at node c. First, c is discovered, then d is discovered; d then finishes, and then c finishes. If we next move to node a and conduct a DFS from there, a and then b are discovered. Following this, b and then a finish. Finally, we move to discover f and e and then finish e and f. Node: A B C D E F Discovery 5 6 1 2 10 9 Finish 8 7 4 3 11 12 This upholds the idea that if there is a path from node A to B, the start time of B must be <= the end time of A. based on the above graph this is true in the graph notwithstanding of starting position. The only way this could be broken is if we understand “path” to take account of going against ...
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