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A depth-first search is performed on a directed acyclic graph. Let d[u] denote the time at which vertex u is visited for the first time and f[u] the time at which the dfs call to the vertex u terminates. Which of the following statements is always true for all edges (u, v) in the graph ?
- a)d[u] < d[v]
- b)d[u] < f[v]
- c)f[u] < f[v]
- d)f[u] > f[v]
Correct answer is option 'D'. Can you explain this answer?
Verified Answer
A depth-first search is performed on a directed acyclic graph. Let d[u...
I'm gonna disprove all wrong options here.
A) d[u] < d[v] , Counter Example => Well if we directly start DFS on V first. Then I call DFS on X which visits U.
B) d[u] < f[v] Counter example => Same as A)
C) f[u] < f[v] Counter example => Same as A) again
So answer is D)
A) d[u] < d[v] , Counter Example => Well if we directly start DFS on V first. Then I call DFS on X which visits U.
B) d[u] < f[v] Counter example => Same as A)
C) f[u] < f[v] Counter example => Same as A) again
So answer is D)
Most Upvoted Answer
A depth-first search is performed on a directed acyclic graph. Let d[u...
Explanation:
In a depth-first search (DFS) on a directed acyclic graph (DAG), we traverse the graph by exploring as far as possible along each branch before backtracking. During this process, we assign discovery and finish times to each vertex.
The discovery time, denoted as d[u], is the time at which vertex u is first visited during the DFS traversal. The finish time, denoted as f[u], is the time at which the DFS call to vertex u terminates.
Statement: f[u] ≤ f[v] for all edges (u, v) in the graph.
Proof:
To prove this statement, let's consider an edge (u, v) in the graph.
During the DFS traversal, the vertex u must be discovered before the vertex v because u is the starting point for the DFS call that eventually leads to v.
Therefore, we can conclude that d[u] < />
Now, let's consider two cases:
Case 1: d[u] < />
In this case, it means that the discovery time of vertex u is earlier than the discovery time of vertex v. As the DFS traversal progresses, vertex u will finish before vertex v because we explore each branch completely before backtracking.
Therefore, we can conclude that f[u] < />
Case 2: d[u] = d[v]
In this case, it means that the discovery time of vertex u is the same as the discovery time of vertex v. However, since the graph is acyclic, vertex u cannot be an ancestor of vertex v in the DFS traversal.
Therefore, we can conclude that f[u] < />
Conclusion:
In both cases, we have f[u] < f[v].="" hence,="" the="" statement="" f[u]="" ≤="" f[v]="" for="" all="" edges="" (u,="" v)="" in="" the="" graph="" is="" always="" true="" in="" a="" depth-first="" search="" of="" a="" directed="" acyclic="" graph.="" f[v].="" hence,="" the="" statement="" f[u]="" ≤="" f[v]="" for="" all="" edges="" (u,="" v)="" in="" the="" graph="" is="" always="" true="" in="" a="" depth-first="" search="" of="" a="" directed="" acyclic="" />
In a depth-first search (DFS) on a directed acyclic graph (DAG), we traverse the graph by exploring as far as possible along each branch before backtracking. During this process, we assign discovery and finish times to each vertex.
The discovery time, denoted as d[u], is the time at which vertex u is first visited during the DFS traversal. The finish time, denoted as f[u], is the time at which the DFS call to vertex u terminates.
Statement: f[u] ≤ f[v] for all edges (u, v) in the graph.
Proof:
To prove this statement, let's consider an edge (u, v) in the graph.
During the DFS traversal, the vertex u must be discovered before the vertex v because u is the starting point for the DFS call that eventually leads to v.
Therefore, we can conclude that d[u] < />
Now, let's consider two cases:
Case 1: d[u] < />
In this case, it means that the discovery time of vertex u is earlier than the discovery time of vertex v. As the DFS traversal progresses, vertex u will finish before vertex v because we explore each branch completely before backtracking.
Therefore, we can conclude that f[u] < />
Case 2: d[u] = d[v]
In this case, it means that the discovery time of vertex u is the same as the discovery time of vertex v. However, since the graph is acyclic, vertex u cannot be an ancestor of vertex v in the DFS traversal.
Therefore, we can conclude that f[u] < />
Conclusion:
In both cases, we have f[u] < f[v].="" hence,="" the="" statement="" f[u]="" ≤="" f[v]="" for="" all="" edges="" (u,="" v)="" in="" the="" graph="" is="" always="" true="" in="" a="" depth-first="" search="" of="" a="" directed="" acyclic="" graph.="" f[v].="" hence,="" the="" statement="" f[u]="" ≤="" f[v]="" for="" all="" edges="" (u,="" v)="" in="" the="" graph="" is="" always="" true="" in="" a="" depth-first="" search="" of="" a="" directed="" acyclic="" />
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A depth-first search is performed on a directed acyclic graph. Let d[u] denote the time at which vertex u is visited for the first time and f[u] the time at which the dfs call to the vertex u terminates. Which of the following statements is always true for all edges (u, v) in the graph ?a)d[u] < d[v]b)d[u] < f[v]c)f[u] < f[v]d)f[u] > f[v]Correct answer is option 'D'. Can you explain this answer? for Computer Science Engineering (CSE) 2025 is part of Computer Science Engineering (CSE) preparation. The Question and answers have been prepared according to the Computer Science Engineering (CSE) exam syllabus. Information about A depth-first search is performed on a directed acyclic graph. Let d[u] denote the time at which vertex u is visited for the first time and f[u] the time at which the dfs call to the vertex u terminates. Which of the following statements is always true for all edges (u, v) in the graph ?a)d[u] < d[v]b)d[u] < f[v]c)f[u] < f[v]d)f[u] > f[v]Correct answer is option 'D'. Can you explain this answer? covers all topics & solutions for Computer Science Engineering (CSE) 2025 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A depth-first search is performed on a directed acyclic graph. Let d[u] denote the time at which vertex u is visited for the first time and f[u] the time at which the dfs call to the vertex u terminates. Which of the following statements is always true for all edges (u, v) in the graph ?a)d[u] < d[v]b)d[u] < f[v]c)f[u] < f[v]d)f[u] > f[v]Correct answer is option 'D'. Can you explain this answer?.
A depth-first search is performed on a directed acyclic graph. Let d[u] denote the time at which vertex u is visited for the first time and f[u] the time at which the dfs call to the vertex u terminates. Which of the following statements is always true for all edges (u, v) in the graph ?a)d[u] < d[v]b)d[u] < f[v]c)f[u] < f[v]d)f[u] > f[v]Correct answer is option 'D'. Can you explain this answer? for Computer Science Engineering (CSE) 2025 is part of Computer Science Engineering (CSE) preparation. The Question and answers have been prepared according to the Computer Science Engineering (CSE) exam syllabus. Information about A depth-first search is performed on a directed acyclic graph. Let d[u] denote the time at which vertex u is visited for the first time and f[u] the time at which the dfs call to the vertex u terminates. Which of the following statements is always true for all edges (u, v) in the graph ?a)d[u] < d[v]b)d[u] < f[v]c)f[u] < f[v]d)f[u] > f[v]Correct answer is option 'D'. Can you explain this answer? covers all topics & solutions for Computer Science Engineering (CSE) 2025 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A depth-first search is performed on a directed acyclic graph. Let d[u] denote the time at which vertex u is visited for the first time and f[u] the time at which the dfs call to the vertex u terminates. Which of the following statements is always true for all edges (u, v) in the graph ?a)d[u] < d[v]b)d[u] < f[v]c)f[u] < f[v]d)f[u] > f[v]Correct answer is option 'D'. Can you explain this answer?.
Solutions for A depth-first search is performed on a directed acyclic graph. Let d[u] denote the time at which vertex u is visited for the first time and f[u] the time at which the dfs call to the vertex u terminates. Which of the following statements is always true for all edges (u, v) in the graph ?a)d[u] < d[v]b)d[u] < f[v]c)f[u] < f[v]d)f[u] > f[v]Correct answer is option 'D'. Can you explain this answer? in English & in Hindi are available as part of our courses for Computer Science Engineering (CSE).
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A depth-first search is performed on a directed acyclic graph. Let d[u] denote the time at which vertex u is visited for the first time and f[u] the time at which the dfs call to the vertex u terminates. Which of the following statements is always true for all edges (u, v) in the graph ?a)d[u] < d[v]b)d[u] < f[v]c)f[u] < f[v]d)f[u] > f[v]Correct answer is option 'D'. Can you explain this answer?, a detailed solution for A depth-first search is performed on a directed acyclic graph. Let d[u] denote the time at which vertex u is visited for the first time and f[u] the time at which the dfs call to the vertex u terminates. Which of the following statements is always true for all edges (u, v) in the graph ?a)d[u] < d[v]b)d[u] < f[v]c)f[u] < f[v]d)f[u] > f[v]Correct answer is option 'D'. Can you explain this answer? has been provided alongside types of A depth-first search is performed on a directed acyclic graph. Let d[u] denote the time at which vertex u is visited for the first time and f[u] the time at which the dfs call to the vertex u terminates. Which of the following statements is always true for all edges (u, v) in the graph ?a)d[u] < d[v]b)d[u] < f[v]c)f[u] < f[v]d)f[u] > f[v]Correct answer is option 'D'. Can you explain this answer? theory, EduRev gives you an
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