bellman ford without -ve edge weight , is simply dijkastra

if we update value one by one in directed graph , total updation will be 100

if we update value one by one in directed graph , total updation will be 100

4 votes

We have a Directed Graph with 100 vertexes. v1 --> v2 --> ... v100 and all edges weights is equal to 1. we want to used bellman-ford for finding all shortest paths from v1 to other vertexes. this algorithm in each step check all edges in arbitrary order. if in each step the shortest distance v1 to all others vertexes is not changed, this algorithm halt. the number of steps is related to checking order of edges. what is the minimum and maximum of steps in this problem?

ُSolution says 2 and 100.

anyone can say how the min and max steps is calculated?

ُSolution says 2 and 100.

anyone can say how the min and max steps is calculated?

0

bellman ford without -ve edge weight , is simply dijkastra

if we update value one by one in directed graph , total updation will be 100

if we update value one by one in directed graph , total updation will be 100

0

We will get the final result in 2 steps if our order is v1v2,v2v3,v3v4,...v99v100? of we get 100 in this order? would you please submit as an answer?

0

v1 --> v2 --> v3 --> v4 --> v5 --> ... v100

If your graph is like this, then maximum steps will be 100..

For 2 steps, it should be modelled differently..

0

@Arjun sir,

I think that if the graph is like this ==>

v1 --> v2 --> v3 --> v4 --> v5 --> v6 --> ... v100 and all the edge weights are equal to 1.

so, i think graph is topologically sorted by bellman ford and so now we process the edges

in order (v1,v2) (v2,v3) (v3,v4) (v4,v5) (v5,v6) as the edges in the path in one more than the preceding one.

so, we can find distance from source to any vertex by path relaxation property of bellman ford ....

I think that if the graph is like this ==>

v1 --> v2 --> v3 --> v4 --> v5 --> v6 --> ... v100 and all the edge weights are equal to 1.

so, i think graph is topologically sorted by bellman ford and so now we process the edges

in order (v1,v2) (v2,v3) (v3,v4) (v4,v5) (v5,v6) as the edges in the path in one more than the preceding one.

so, we can find distance from source to any vertex by path relaxation property of bellman ford ....

4 votes

1) Minimum number of steps for the given problem is 2.

if the given graph is like

v1 --> v2

v1 ---> v3

:

:

v1 --> v100

in this case, the first iteration will find the distance from source to each edge and second iteration won't change any weight and algorithm will stop.

2) Maximum number of steps for the given problem is 100

if the given graph is like

v1 --> v2 --> v3 --> v4 ............--> v100

i.e., then all are in stright line. then we need 99 iterations to update the distance to 100 th vertex and the last iteration won't change the distance.

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