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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).
Asymptotic notation Let us describe a few functions in terms of above asymptotic notation. Example: f(n) = 3n 3 + 2n 2 + 4n + 3 = 3n 3 + 2n 2 + O (n), as 4n + 3 is of
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include int choice, stack[10], top, element; void menu(); void push(); void pop(); void showelements(); void main() { choice=element=1; top=0; menu()
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