complexity theory - Question about NP-Completeness of the Independent Set Problem -

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I thought, when the problem is that pnp is full, then we reduce the problem to a known npc But, looking at the solution of the problem of independent set, it does not seem so.

To prove that the independent set is NP-complete, you take a graph, get reversed, and then calculate CLIQUE (G) but, doing it the other way It is taking a problem: PI does not know whether it is NPC and then it reduces NPC's problem.

'An example of the solution is. >

What am I missing here? This is not wrong because it is doing the other way?

To prove that P NP is complete, we need to show two things: / P>

  1. It is present in P.NP.
  2. There is a polytime reduction algorithm to reduce the number of NP-complete problems to the PO.

If we know that CLIQUE is in NPC, then we can easily prove that IS is in NPC.

  1. We are in the corner of the polyaitime firator, make sure that not everyone is in the solution of the candidate.
  2. We will now have to reduce the CIL to CIlqua. Looking at CLIQUE for a graph g and an integer n we want to see if there is a CLIQUE size n H Leave in the inverse of G . If you got an IS in H size n , then you have a CLIQUE size n with g In the same corner we reduced CLIQUE to IS.

If you want to reduce CLIQUE to IS, then you will not prove that NPC is not in reducing any other problem in NPC either.


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