Question

Show that we can use a depth-first search of an undirected graph G to identify the connected components of G, and that the depth-first forest contains as many trees as G has connected components. More precisely, show how to modify depth-first search so that it assigns to each vertex an integer label v.cc between 1 and k, where k is the number of connected components of G, such that u.cc = v.cc if and only if u and are in the same connected component.

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Take a look at the DFS\textsf{DFS} algorithm. The key observation here is that we are in an undirected graph so each time the function DFS-Visit\textsf{DFS-Visit} is called in DFS\textsf{DFS}, it will discover all connected vertices, that is, it will discover one connected component \textbf{one connected component }of the graph. That is where we can sneak in our .cc.cc label that identifies all the components of the graph. The algorithm is as follows:

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