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In an unweighted, undirected connected graph, the shortest path from a node S to every other node is computed most efficiently, in terms of time complexity, by
- a)Dijkstra’s algorithm starting from S.
- b)Warshall’s algorithm.
- c)Performing a DFS starting from S.
- d)Performing a BFS starting from S.
Correct answer is option 'D'. Can you explain this answer?
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Verified Answer
In an unweighted, undirected connected graph, the shortest path from a...
Dijkastra and warshall 's algo only used for weighted graph.
Both DFS and BFS can be used for finding path between 2 vertices in undirected and unweighted graph but BFS can only give the shortest path as concern in given question so BFS is Ans.
Note :- finding only path(DFS) and finding shortest path(BFS) ..Matters a lot
Most Upvoted Answer
In an unweighted, undirected connected graph, the shortest path from a...
Explanation:
- In an unweighted, undirected connected graph, all edges have the same weight, hence there is no need for Dijkstra's algorithm, which is designed for weighted graphs.
- Warshall's algorithm is used for computing the transitive closure of a graph, which is not the same as finding the shortest path from a node to every other node.
- DFS (depth-first search) and BFS (breadth-first search) are both graph traversal algorithms that can be used to find paths in a graph, but BFS is more suitable for finding the shortest path in an unweighted graph.
Reasons why performing a BFS starting from S is the most efficient way to find the shortest path from a node S to every other node in an unweighted, undirected connected graph:
- BFS explores the graph in layers, starting from the source node S and visiting all nodes at distance 1, then all nodes at distance 2, and so on, until all reachable nodes have been visited.
- Since all edges have the same weight, the distance from S to any node can be found by counting the number of edges in the shortest path from S to that node.
- BFS guarantees that when a node is visited, its distance from S is the shortest possible distance, because all nodes at the same or shorter distance have already been visited.
- BFS has a time complexity of O(V+E), where V is the number of vertices and E is the number of edges, which is the best possible time complexity for finding the shortest path in an unweighted graph, because every edge must be considered at least once.
- In an unweighted, undirected connected graph, all edges have the same weight, hence there is no need for Dijkstra's algorithm, which is designed for weighted graphs.
- Warshall's algorithm is used for computing the transitive closure of a graph, which is not the same as finding the shortest path from a node to every other node.
- DFS (depth-first search) and BFS (breadth-first search) are both graph traversal algorithms that can be used to find paths in a graph, but BFS is more suitable for finding the shortest path in an unweighted graph.
Reasons why performing a BFS starting from S is the most efficient way to find the shortest path from a node S to every other node in an unweighted, undirected connected graph:
- BFS explores the graph in layers, starting from the source node S and visiting all nodes at distance 1, then all nodes at distance 2, and so on, until all reachable nodes have been visited.
- Since all edges have the same weight, the distance from S to any node can be found by counting the number of edges in the shortest path from S to that node.
- BFS guarantees that when a node is visited, its distance from S is the shortest possible distance, because all nodes at the same or shorter distance have already been visited.
- BFS has a time complexity of O(V+E), where V is the number of vertices and E is the number of edges, which is the best possible time complexity for finding the shortest path in an unweighted graph, because every edge must be considered at least once.
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In an unweighted, undirected connected graph, the shortest path from a node S to every other node is computed most efficiently, in terms of time complexity, bya)Dijkstra’s algorithm starting from S.b)Warshall’s algorithm.c)Performing a DFS starting from S.d)Performing a BFS starting from S.Correct answer is option 'D'. Can you explain this answer?
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