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Suppose that A, B, and C are decision problems. Suppose also that A is polynomial-time reducible to B and that B is polynomial-time reducible to C. If both A and C are NP-complete, then does it imply that B is also NP-complete?

I know that, if A is NP-complete and it is polynomial-time reducible to B, then B is NP-hard. However, in order for a problem to be NP-complete, it must meet (1) it's in NP, and (2) it's NP-hard.

I have no idea how to prove the first requirement of NP-complete.

nbro
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hugoinperson
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2 Answers2

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If A is NP-complete and it is polynomial time reducible to B, then B is NP-hard.

If B is polynomial time reducible to C and C is NP-complete, then B is in NP.

A problem in NP which is in NP-hard is NP-complete.

Another way to show B is NP-complete is to notice that any two NP-complete problems (e.g A and C) are polynomially reducible to each other, and thus B is equivalent (two-way polynomially reducible) to any NP-complete problem.

n. m. could be an AI
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Les try Out:- (REC= Recursive lang, REL=Recursive Enumerable lang, UD= Undecidable, D= Decidable)  

if P < Q than

  UD-->UD

  D<--D

  P<--P

  P,NP<--NP

  NPC-->NPH

  P,NP--> we can't anything it may be (NP,NPH,REC,REL)

  REC<-- REC

  REL<--REL

  D--> Can't say anything.

   ?<--UD


we know that P is Proper Subset of NP. (as P != Np)

and All NPC is NPH.

to prove NPC:-
""
if NP reducible to X problem than that X is NPH.

if X reducible to any NPC than that X is NPC.""

p^NPC=0