reason for computing topological ordering for shortest path in a weighted DAG











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What is the reason for having to first compute the topological ordering for finding the shortest path in a DAG? For example, if we are given a DAG in an adjacency list format, can't we just iterate BFS or DFS style to find the shortest path? (starting with the start node of course) Why does there have to be a unique ordering? I've been doing examples, and it seems that finding a topological ordering does not make a difference.










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    What is the reason for having to first compute the topological ordering for finding the shortest path in a DAG? For example, if we are given a DAG in an adjacency list format, can't we just iterate BFS or DFS style to find the shortest path? (starting with the start node of course) Why does there have to be a unique ordering? I've been doing examples, and it seems that finding a topological ordering does not make a difference.










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      What is the reason for having to first compute the topological ordering for finding the shortest path in a DAG? For example, if we are given a DAG in an adjacency list format, can't we just iterate BFS or DFS style to find the shortest path? (starting with the start node of course) Why does there have to be a unique ordering? I've been doing examples, and it seems that finding a topological ordering does not make a difference.










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      What is the reason for having to first compute the topological ordering for finding the shortest path in a DAG? For example, if we are given a DAG in an adjacency list format, can't we just iterate BFS or DFS style to find the shortest path? (starting with the start node of course) Why does there have to be a unique ordering? I've been doing examples, and it seems that finding a topological ordering does not make a difference.







      directed-acyclic-graphs graph-traversal topological-sort weighted-graph






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      asked Nov 8 at 4:32









      JohnnyD27

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