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I am trying to find the minimum cut for a given graph. I am using QAOA and refering to the qiskit textbook: https://qiskit.org/textbook/ch-applications/qaoa.html. I understand I need to find the minimum eigenvalue.

I am confused on 2 main points:

  1. How the graph is connected to the circuit, we change it to a bit string and then parameterize it with the unitary Rx ...?
  2. how this maxcut def works so how can I change it to mincut?

Still working on understanding what max/min cut is.

def maxcut_obj(x, G): """ Given a bitstring as a solution, this function returns the number of edges shared between the two partitions of the graph.

Args:
    x: str
       solution bitstring
       
    G: networkx graph
    
Returns:
    obj: float
         Objective
"""
obj = 0
for i, j in G.edges():
    if x[i] != x[j]:
        obj -= 1
        
return obj
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1 Answer 1

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I think that it is certainly possible to formulate a QAOA instance for Minimum Cut, however I should mention that Minimum Cut is equivalent to max flow and there exist classical polynomial time algorithms to solve that problem (see here and here). Maximum Cut though is NP-Hard.

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  • $\begingroup$ It is possible, it's a matter of looking for the minimum eigenvalue instead of the maximum. I am actually trying to do a max flow however I don't think your answer is helpful here. I'm looking for how to switch from max to min, not if it's possible or applications $\endgroup$ Jan 18, 2023 at 12:19
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    $\begingroup$ @epsilonolispe, Elijah brought up a good point about min cut being solvable exactly in poly time by a classical algorithm. Doing this with QAOA just doesn't make any sense because it is a heuristic and can only give a probabilistic answer. Also, max cut is NP-complete and min-cut is in P, so it will not just be switching from max to min. $\endgroup$
    – MonteNero
    Jan 18, 2023 at 13:59
  • $\begingroup$ I'm not sure I agree with you on the point that it' switching from a max to a min, it's grabbing the minimum eigenvalue, no? Sure, perhaps it's not better but I'm still hoping to attempt it! $\endgroup$ Jan 18, 2023 at 17:32

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