I am helping a friend with a work related project where, he needs to calculate the maximum capacity from a node a to a node b, where the edge has a capacity. However the max
The above answer has been very well explained. Just in case anyone needs an explanation of the correctness of the algorithm, here you go:
Proof:
At any point in the algorithm, there will be 2 sets of vertices A and B. The vertices in A will be the vertices to which the correct maximum minimum capacity path has been found. And set B has vertices to which we haven't found the answer.
Inductive Hypothesis: At any step, all vertices in set A have the correct values of maximum minimum capacity path to them. ie., all previous iterations are correct.
Correctness of base case: When the set A has the vertex S only. Then the value to S is infinity, which is correct.
In current iteration, we set
val[W] = max(val[W], min(val[V], width_between(V-W)))
Inductive step: Suppose, W is the vertex in set B with the largest val[W]. And W is dequeued from the queue and W has been set the answer val[W].
Now, we need to show that every other S-W path has a width <= val[W]. This will be always true because all other ways of reaching W will go through some other vertex (call it X) in the set B.
And for all other vertices X in set B, val[X] <= val[W]
Thus any other path to W will be constrained by val[X], which is never greater than val[W].
Thus the current estimate of val[W] is optimum and hence algorithm computes the correct values for all the vertices.