graph algorithm - Does a Given Network has a Unique Min-Cut? -

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 :)