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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).
Explain the term - Branching There are two common ways of branching: case of ..... otherwise ...... endcase if ..... then ..... else ..... endif case of
Write a program that uses the radix sort to sort 1000 random digits. Print the data before and after the sort. Each sort bucket should be a linked list. At the end of the sort, the
memory address of any element of lower left triangular sparse matrix
What is an Algorithm? An algorithm is a sequence of unambiguous instructions for solving a problem, i.e., for getting a needed output for any legitimate input in a finite amoun
For the following graph find the adjacency matrix and adjacency list representation of the graph.
Q. Define a sparse matrix. Explain different types of sparse matrices? Show how a triangular array is stored in memory. Evaluate the method to calculate address of any element ajk
Thus far, we have seen the demonstration of a single queue, but several practical applications in computer science needs several queues. Multi queue is data structure in which mult
Explain Space Complexity Space Complexity :- The space complexity of an algorithm is the amount of memory it requires to run to completion. Some of the reasons to study space
Q. Illustrate the steps for converting the infix expression into the postfix expression for the given expression (a + b)∗ (c + d)/(e + f ) ↑ g .
* Initialise d & pi* for each vertex v within V( g ) g.d[v] := infinity g.pi[v] := nil g.d[s] := 0; * Set S to empty * S := { 0 } Q := V(g) * While (V-S)
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