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asked in Algorithms by Boss (5.1k points)   | 83 views

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Regarding Bellman's Ford algorithm , which is used for finding single source shortest paths where Dijkstra's Algo fails i.e. in case of negative edge weight cycle , it gives shortest path from a source to  a particular node provided it is reachable from the source .So considering this , the statement 1 is false actually since it says that between s and t we get shortest path always which is not the case [Fails in case it is not reachable from the source]

Statement 2 is true as it has mentioned the path between s and t is well defined meaning reachability is there and even if negative edge weight cycle exists , it will be handled by Bellman Ford Algo correctly.

 

Hence statement 1 is false and statement 2 is true.
answered by Veteran (59.6k points)  
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@habib...But the algorithm returns False if it finds a single negative weight cycle.

But Bellman Ford will at least conclude whether there is a feasible shortest path between s and t after (v-1) iterations where v is the number of vertices..

Contrary to this Dijkstra's Algo will be stuck in loop there is negative edge weight cycle.
I have a doubt here...what do you mean by stuck in loop?
Dijkstra ends up with wrong solution if there is a negative weight cycle.
So in Bellman-ford.... is it like, it will say that cycle is present but also give correct distance for the vertices which are not present in the cycle?
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I) If there is no path from S to T then Bellman Ford Algo will not return shortest path from S to T.

II) A well defined path doesn't contain a negative edge weight cycle so definitely a shortest path value (-ve or +ve) will be returned by the Bellman Ford Algorithm for the source vertex V for which shortest path is well defined. Graph G contains a negative edge weight cycle acc to question but it will not be reachable from the well defined shortest path for vertex V !!!  :) HENCE STATEMENT 2 IS TRUE
answered by (103 points)  
edited by
Statement 2 is not talking about two specific vertices s and t as in first statement hence no issue with second statement
I got your point, and edited the ans.

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