Let G = (V, E) be a network with S and being a source and having a sync The maximum flow should be in f. Find an algorithm that determines whether a unique minimum reduction exists in G.
I have managed to get a similar question on this site:
A summary of the answer given here:
Find all the verticals arriving from S in the remaining graph and we got a minimum cut (S, T) in G.
Look at the same residual graph, starting with T
This group is also a minimum cut.
If this deduction is similar to your original cut, then there is only one, otherwise, you have just got 2 cuts, so the original one can not possibly be unique.
I do not understand if the deduction is similar to the original deduction, the deduction is unique, who can promise us that there is no other minimum deduction?
In advance, I do not consider that solution a lot. But in the original question, the second answer provided by Devin is absolutely correct.
I copy and paste it
With minimal ST reduction, (U, V) cut-edges e ', we make a simple observation: if it If the minimum deduction is not unique, then there are some other minimum deductible with the set of cut-outs E, such as E '!! = 'If yes, then we can repeat on each side of E, add to our capacity, re-calculate the maximum flow, and check whether it has increased. As a result of the above observation, an increase in E appears' When the increase, the maximum flow does not increase, if the original deduction is not unique. Some of your own explanations : Why do you really need to prove
There is an edge in E 'that there is no increase in the maximum flow when it increases & lt; = & Gt; The original deduction is not unique => You can increase the edge of e by 1, calculate the new maximum flow and it remains The same, which means that the e is not in the new low cut (if in e , then the minimum cut, according to the capability of F (E) = E, Since e is not in the new minute deduction, it is also a small cut of the original graph, which we know is the same volume with deduction is. Therefore, the original deduction is not unique. & lt; =:
The original deduction is not unique (let's call them E and E '), which means that you have the side e e is but there is no E in it. Then by increasing the capacity of only e 1. While calculating the minimum cut of the new graph, E ' already exists. Since e ' does not have an edge e , the maximum flow remains without any doubt. Hope you understand :)