Depth First Search Graph

DepthFirstSearch (DFS) : parent node; discovery/finishing time; parenthesis theorem.

Requirements: implement/update specific methods for the DFS of a graph; for at least 2 graphs (1 being the provided one), show the DFS order of vertices in the graph, and for each node, specify its parent node in the search (the node from which the currect node was reached). Moreover, display for each node the discovery and finishing time, to check that the Parenthesis Theorem holds true.

Approach:You may start from the DFT in the textbook and do the appropriate changes. For counting the dfs time, you should set a global counter (lets name it time), which is set to 0 in the moment the search starts with the root vertex. At any moment the search starts with a new vertex (that is, justafter entering a new dfs) the counter has to be incremented by one; also it has to be incremented by one when the search from the current node ends (just before exit a dfs). For each node, calculate also the discovery and finishing time; for doing this, you may use two arrays (lets name them d[] for discovery and f[] for finishing). The discovery time of a given vertex v receives the value of the counter just when enter to the dfs for that vertex (that is d[v]=time, before incrementing time), and the finishing just before exit from that dfs (that is f[v]=time, before decrementing time).

Parentheses Theorem: In any depth first search, for any 2 vertices u and v, one of the following 3 conditions holds true: The intervals [d[u], f[u]] and [d[v], f[v]] are completely disjoint;

The [d[u], f[u]] interval is completely contained within interval [d[v], f[v]], and u is a descendant of v in the dfs tree;

The [d[v], f[v]] interval is completely contained within interval [d[u], f[u]], and v is a descendant of u in the dfs tree;

Design and implement a driver to show the following (check for 2 graphs; 1 is provided, including the starting vertex):

• Display the dfs starting from a specified vertex;
• Display the discovery/finishing time for each node in the graph;
• Display the parent node for each node in the dfs;
• Show the Parentheses Theorem holds true, by mentioning the specific condition in each case (this has to me manually calculated and added in the documentation).

Input data: You should test your application and include the tests in your documentation for at least two graphs; one is mandatory to be this one provided below. It is represented in the G = (V, E) representation, where V is the vertices set, and E is the edges set. Please note that our graph is a directed one (that is edges have directions, thus, the presence of an (u, v) edge does not imply (v, u) is also present in the graph). Nevertheless, this has no impact on the algorithm and its implementation. The dfs should start with vertex 1.

V= {1, 2, 3, 4, 5, 6, 7}

E= {1, 2}, {1, 6}, {2, 3}, {2, 4}, {2, 5}, {3, 5}, {4, 5}, {5, 1},{6, 4}, {6, 7}