Assume all reductions are done in polynomial time
$P \subseteq NP \subseteq NPC \subseteq NPH$
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> and <math>C</math> is reduced to <math>A</math> and <math>B, C</math> $\in$ 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
- If problem <math>A</math> is reduced to a problem <math>B</math> and <math>B</math> $\in$ P, then <math>A</math> $\in$ <math>P</math>
- If problem <math>A</math> is reduced to a problem <math>B</math> and <math>B</math> $\in$ P, then <math>A</math> $\in$ <math>P</math>
- If problem <math>A</math> is reduced to a problem <math>B</math> and <math>B</math> $\in$ P, then <math>A</math> $\in$ <math>P</math>
- If problem <math>A</math> is reduced to a problem <math>B</math> and <math>B</math> $\in$ P, then <math>A</math> $\in$ <math>P</math>