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I an trying to learn about $DisNP$ Complexity class. I couldn't find this class in any book. But, regarding it, I found this definition in a paper:

Definition: A disjoint NP-pair (NP-pair for short) is a pair of nonempty sets $A$ and $B$ such that $A, B \in NP$ and $A \cap B = Empty$. $DisNP$ denote the class of all disjoint NP-pairs.

Now I am aware that when a problem like $SAT$ etc. is said to be in $NP$, it means that assuming we have a NDTM we can get the solution in polynomial time. Or in other words, if we can guess the solution, then the verification takes polynomial time.

But, what does it mean when we say a set $A \in NP$ or a set $B \in P$.

Does the elements of the sets represent the solution for which the corresponding $SAT$ instance for each set is true i.e. set $A$ and $B$ contain the solutions of some $SAT$ instances $SAT_A$ and $SAT_B$ such that $SAT_A$ is satisfiable iff its solution is in the set $A$?

I am not clear how we use sets (which I assume have binary strings) for defining $P, NP, NPC$?

J.Doe
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  • I don't understand how last 3 paragraphs correlate to question. Definitely, no NP-complete (or no pair of P-complete) problem is in DisNP. Just because these classes are not empty. – rus9384 Jul 10 '17 at 12:56
  • What I was trying to ask was in the definition, when we talk about 'two non empty sets $A$ and $B$', what does the elements of the set represent ? I am unclear about the meaning of it, thus the question. – J.Doe Jul 10 '17 at 12:59
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    https://en.wikipedia.org/wiki/Formal_language, https://en.wikipedia.org/wiki/Decision_problem, https://en.wikipedia.org/wiki/NP_(complexity) – D.W. Jul 10 '17 at 15:10
  • Thanks. I DO understand the definitions of $P$, $NP$ and $NPC$ in terms of Turing Machines. But, what I do not understand is the sets mentioned in $DisNP$. And I did check the link. It goes the Turing Machine way. If you could help me with this definition of $DisNP-pair$ i would be thankful. – J.Doe Jul 10 '17 at 15:18
  • Thank you. I am unsure why it was marked duplicate as definition of $P$ etc. Now as a followup as all problems in $NP$ can be reduced to an $NPC$ problem in polynomial time, so if we argue that no matter what language $A$ and $B$ have they can be converted to a $SAT$ problem pair (say $A'$ and $B'$) on same input variables, such that the same condition holds: i.e. $A'$ and $B'$ are satisfiable but they don't share a solution would that statement be correct ? – J.Doe Jul 10 '17 at 15:52
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    I marked it as a duplicate because the answer is found in https://cs.stackexchange.com/q/9556/755, where it says "We can identify a decision problem with..." It's also found in the Wikipedia links I gave; and in any good textbook which talks about formal languages. – D.W. Jul 10 '17 at 22:14
  • @J.Doe The root of your question is what it means for a set to be in NP. That's explained at all the links that D.W. posted. – David Richerby Jul 10 '17 at 23:14

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A disjoint NP pair consists of two NP languages $A,B$ which are disjoint. For example, $A$ could consist of all graphs which are $k(n)$-colorable (where $n$ is the number of vertices and $k$ is an arbitrary function), and $B$ could consist of all graphs having a clique of size $k(n)+1$.

Disjoint pairs of NP languages are used in proof complexity. For example, the disjoint pair discussed above (for $k(n) = n^c$) is used to prove exponential lower bounds on the size of proofs in the cutting planes proof system. See for example these notes. (For the state-of-the-art regarding cutting planes, see the recent paper Random CNFs are Hard for Cutting Planes.)

Yuval Filmus
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