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| − | $P \subseteq NP \subseteq NPC \subseteq NPH$
| + | <metadesc>Reduction to|from P, NP, NPC and NP-Hard problems</metadesc> |
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| − | Assume all reductions are done in polynomial time
| + | [[NP,_NP_Complete,_NP_Hard | P, NP, NPC, NPH]] |
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| | + | *<math>P</math> is a subset of <math>NP</math>, but whether <math>P = NP</math> is still not known |
| | + | *<math>NPC</math> is a subset of <math>NP</math> (only if <math>P \neq NP</math>, <math>NPC</math> is a proper subset of <math>NP</math>) |
| | + | *<math>NPC</math> is a proper subset of <math>NPH</math> |
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| | + | ==Reduction== |
| | + | 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. |
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| | + | ===Consider the following example:=== |
| | + | 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. |
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| − | Consider problems $A$, $B$ and $C$ | + | ===Some Inferences=== |
| | + | Consider problems $A$, $B$ and $C$. |
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| − | *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>.
| + | Assume all reductions are done in polynomial time |
| | + | ===2 Important Points=== |
| | + | #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>. |
| | + | #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 <math>NPH</math> - just reduce some known <math>NPH</math> problem to the given problem. This reduction thus gives a lower bound for the complexity of <math>B</math>. |
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| | ===Somethings to take care of=== | | ===Somethings to take care of=== |
| − | *If <math>A</math> is reduced to <math>B</math> and <math>B</math> $\in$ NPC, then <math>A</math> $\in$ <math>NP</math> | + | *If <math>A</math> is reduced to <math>B</math> and <math>B</math> $\in$ <math>NPC</math>, 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. | | **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> | | *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> | + | **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> (because <math>C</math> is in <math>NPC</math>, by definition of <math>NPC</math>, all problems in <math>NP</math> are reducible to <math>C</math> and now <math>C</math> is reduced to <math>A</math>). Hence, the two necessary and sufficient conditions for <math>NPC</math> are satisfied and <math>A</math> is 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> | + | *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 <math>A</math>. It can be as hard as <math>NPH</math>, or as simple as <math>P</math> | + | **Here we can't say anything about <math>A</math>. It can be as hard as <math>NPH</math>, or as simple as <math>P</math>. |
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| | + | [[Category: Automata Theory Notes]] |
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| − | [[Category: Automata Theory]] | + | [[Category: Compact Notes for Reference of Understanding]] |
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| − | [[Category:Notes & Ebooks for GATE Preparation]]
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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.
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.
GATECSE