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Consider the following statements: 
S1 : DFS of a directed graph always produces the same number of edges in the traversal, irrespective of the starting vertex.
S2 : If all of the back edges that are found while DFS traversal on directed graph are removed, the resulting graph is acyclic. 

Which of the following statements above are valid ?
  • a)
    Both S1 and S2 are valid
  • b)
    Only S1 is valid
  • c)
    Only S2 is valid
  • d)
      Neither s1 nor S2 is valid
Correct answer is option 'C'. Can you explain this answer?
Verified Answer
Consider the following statements:S1 :DFS of a directed graph always p...
Statement S1 : consider the graph
 
A (source vertex ) we will get 2 edges Starting with B will get only 1 edge
Starting with C we will get no edge
Therefore DFS on directed graph may not give same number of edges.
Statement S2 : Back edges are those edges (u,v) connecting a vertex u to an ancestor u in a depth-first tree. Self-loops are considered to be back edges. Back edges describe descendant-to-ancestor relations, as they lead from “high” to “low” nodes.
Suppose that there is a back edge (u, v). Then vertex v is an ancestor of vertex u in the depth-first forest.
There is thus a path from v to u in G, and the back edge (u,v) completes a cycle.
Removing the back edge will break the cycle. Therefore removing all the back edges will make the graph acyclic.
So the statement is true.
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Most Upvoted Answer
Consider the following statements:S1 :DFS of a directed graph always p...
Statement S1: DFS of a directed graph always produces the same number of edges in the traversal, irrespective of the starting vertex.




DFS (Depth First Search) is a graph traversal algorithm that explores as far as possible along each branch before backtracking. It visits all the vertices of the graph, and in the process, it may visit some edges multiple times.


The number of edges visited during a DFS traversal can vary depending on the starting vertex and the structure of the graph. Let's consider a simple example to understand this.


Suppose we have a directed graph with three vertices A, B, and C, and two edges from A to B and B to C. If we perform a DFS traversal starting from vertex A, we will visit all three vertices and two edges. However, if we start the DFS traversal from vertex B or C, we will only visit two vertices and one edge. Therefore, the number of edges visited during a DFS traversal can vary depending on the starting vertex, and statement S1 is not valid.




Statement S2: If all of the back edges that are found while DFS traversal on a directed graph are removed, the resulting graph is acyclic.




A back edge is an edge that is from a node to itself (self-loop) or one of its ancestors in the DFS traversal tree. These back edges create cycles in the graph. So, if we remove all the back edges from a directed graph, we are essentially removing all the cycles from the graph.


Removing back edges can be seen as removing all the edges that point from a node to one of its ancestors in the DFS traversal. By removing these edges, we break the cycles and make the graph acyclic.


Therefore, statement S2 is valid.




In summary, only statement S2 is valid. DFS of a directed graph does not always produce the same number of edges in the traversal, irrespective of the starting vertex. However, if all the back edges found during DFS traversal are removed, the resulting graph becomes acyclic.
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Consider the following statements:S1 :DFS of a directed graph always produces the same number of edges in the traversal, irrespective of the starting vertex.S2 :If all of the back edges that are found while DFS traversal on directed graph are removed, the resulting graph is acyclic.Which of the following statements above are valid ?a)Both S1 and S2 are validb)Only S1 is validc)Only S2 is validd) Neither s1 nor S2 is validCorrect answer is option 'C'. Can you explain this answer?
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