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$P \subseteq NP \subseteq NPC \subseteq NPH$
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Assume all reductions are done in polynomial time
 
Assume all reductions are done in polynomial time
 
$P \subseteq NP \subseteq NPC \subseteq NPH$
 
  
 
Consider problems $A$, $B$ and $C$
 
Consider problems $A$, $B$ and $C$

Revision as of 12:57, 30 December 2013

$P \subseteq NP \subseteq NPC \subseteq NPH$

Assume all reductions are done in polynomial time

Consider problems $A$, $B$ and $C$

  • If <math>A</math> is reduced to <math>B</math> and <math>B</math> $\in$ P, then <math>A</math> $\in$ <math>P</math>
  • If <math>A</math> is reduced to <math>B</math> and <math>B</math> $\in$ NP, then <math>A</math> $\in$ <math>NP</math>
    • (<math>A</math> may also be in <math>P</math>, but that cannot be inferred from the given statement)
  • If <math>A</math> is reduced to <math>B</math> and <math>B</math> $\in$ NPC, then <math>A</math> $\in$ <math>NP</math>
    • (<math>A</math> may also be in <math>NPC</math>, but that cannot be inferred from the given statement)
  • If <math>A</math> is reduced to <math>B</math>, <math>C</math> is reduced to <math>A</math> , <math>B \in NP </math> and $C \in$ NPC, then <math>A</math> $\in$ <math>NPC</math>
  • If <math>A</math> is reduced to <math>B</math> and <math>B</math> $\in$ NPH, then <math>A</math> $\in$ <math>?</math>
    • Here we can't say anything about A. It can be as hard as NPH, or as simple as P




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Assume all reductions are done in polynomial time

$P \subseteq NP \subseteq NPC \subseteq NPH$

Consider problems $A$, $B$ and $C$




blog comments powered by Disqus