In a DFA there is only one start state, so wouldn't u always be q0, which is the start state? That means O(n) runtime not including checking c(u', v') right?

I think I understand that, but I lose you in the grouping of pairs of states. What is the algorithm for finding these bad vertices, and what is the time complexity?

Sorry, I cannot connect how this is solving the bigger problem. I see that you are grouping (u,v) and (u',v') as vertices which are only adjacent by one directed edge if and only if there is a path from u' to v'. I just don't understand the bigger picture of grouping them. Is there anyway for you to clarify. I really want to understand. Thank you.

I am a little confused what you are doing by pairing states to form vertices in a graph.

Ahh so it doesn't. Thank you. Do you have any more insight?