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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).
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any simple algoritham questions with answers
Normally a potential y satisfies y r = 0 and 0 ³ y w - c vw -y v . Given an integer K³0, define a K-potential to be an array y that satisfies yr = 0 and K ³ y w - c vw -y v
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