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

Reduction[edit]

Reducing a problem <math>A</math> to problem <math>B</math> means converting an instance of problem <math>A</math> to an instance of problem <math>B</math>. Then, if we know the solution of problem <math>B</math>, we can solve problem <math>A</math> by using it.

Consider the following example:[edit]

You want to go from Bangalore to Delhi and you can get a ticket from Mumbai to Delhi. So, you ask your friend for a lift from Bangalore to Mumbai. So, the problem of going from Bangalore to Delhi got reduced to a problem of going from Mumbai to Delhi.

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

Assume all reductions are done in polynomial time

2 Important Points[edit]

1. If <math>A</math> is reduced to <math>B</math> and <math>B</math> $\in$ class $X$, then <math>A</math> cannot be harder than $X$, because we can always do a reduction from <math>A</math> to <math>B</math> and solve <math>B</math> instead of directly solving <math>A</math>. This reduction thus gives an upper bound for the complexity of <math>A</math>.

1. If <math>A</math> is reduced to <math>B</math> and <math>A</math> $\in$ class $X$, then <math>B</math> cannot be easier than $X$. The argument is exactly same as for the one above. This reduction is used to show if a problem belongs to NP-Hard - just reduce some known NP-Hard problem to the given problem. This reduction thus gives a lower bound for the complexity of <math>B</math>.

Somethings to take care of[edit]

• If <math>A</math> is reduced to <math>B</math> and <math>B</math> $\in$ NPC, then <math>A</math> $\in$ <math>NP</math>
  • Here, we cannot say <math>A</math> is <math>NPC</math>. All we can say is <math>A</math> cannot be harder than <math>NPC</math> and hence <math>NP</math> (all <math>NPC</math> problems are in <math>NP</math>). To belong to <math>NPC</math>, all <math>NP</math> problems must be reducible to <math>A</math>, which we cannot guarantee 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 $C \in$ <math>NPC</math>, then <math>A</math> $\in$ <math>NPC</math>
  • Here, the first reduction says that <math>A</math> is in <math>NP</math>. The second reduction says that all <math>NP</math> problems can be reduced to <math>A</math>. Hence, the two sufficient conditions for <math>NPC</math> are complete and hence <math>A</math> is in <math>NPC</math>
• If <math>A</math> is reduced to <math>B</math> and <math>B</math> $\in$ NP-Hard, then <math>A</math> $\in$ <math>?</math>
  • Here we can't say anything about <math>A</math>. It can be as hard as <math>NP-Hard</math>, or as simple as <math>P</math>