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Step 1: Choose a vertex in the graph and make it the source vertex & mark it visited.
Step 2: Determine a vertex which is adjacent to the source vertex and begun a new search if it is not already visited.
Step 3: Repeat step 2 via a new source vertex. While all adjacent vertices are visited, return to earlier source vertex and continue search from there.
If n refer to the number of vertices in the graph & the graph is represented through an adjacency matrix, then the total time taken to carry out DFS is O(n2). If G is revel by an adjacency list and the number of edges of G are e, then the time taken to carry out DFS is O(e).
A list item stores pointers and an element to predecessor and successor. We call a pointer to a list item a handle . This looks simple enough, but pointers are so powerful tha
In the last section, we discussed regarding shortest path algorithm that starts with a single source and determines shortest path to all vertices in the graph. In this section, we
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This notation gives an upper bound for a function to within a constant factor. Given Figure illustrates the plot of f(n) = O(g(n)) depend on big O notation. We write f(n) = O(g(n))
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