Arjun Suresh (talk | contribs) |
Arjun Suresh (talk | contribs) |
||
(27 intermediate revisions by the same user not shown) | |||
Line 1: | Line 1: | ||
+ | <metadesc>Reduction to|from P, NP, NPC and NP-Hard problems</metadesc> | ||
+ | |||
+ | [[NP,_NP_Complete,_NP_Hard | P, NP, NPC, NPH]] | ||
+ | |||
+ | *<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> | ||
+ | |||
+ | ==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. | ||
+ | |||
+ | ===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. | ||
+ | |||
+ | ===Some Inferences=== | ||
+ | Consider problems $A$, $B$ and $C$. | ||
+ | |||
Assume all reductions are done in polynomial time | 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>. | ||
− | + | ===Somethings to take care of=== | |
− | + | *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. | |
− | *If <math>A</math> is reduced to <math>B</math> and <math>B</math> $\in$ | + | *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> (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> | + | *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>. | |
− | *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 | ||
Line 15: | Line 33: | ||
+ | [[Category: Automata Theory Notes]] | ||
− | [[Category: | + | [[Category: Compact Notes for Reference of Understanding]] |
− | |||
− |
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.
Consider problems $A$, $B$ and $C$.
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 $C$