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About a year and a half ago I asked this question regarding $\mathsf{P}=\mathsf{NP}$. The answers have helped me understand the problem tremendously and since then I've dabbled further into the topic.

With that stated, it is my understanding that $\mathsf{NP}$-complete problems are such that if a solution for $\mathsf{P}=\mathsf{NP}$ were found for that specific problem, then all $\mathsf{NP}$ problems could be solved using the same rules for resolving $\mathsf{NP}$.

With that stated, what is the simplest $\mathsf{P}=\mathsf{NP}$ problem outlined to date that is $\mathsf{NP}$-complete?

In other words, what is the most basic of problems that one could test a theoretical $\mathsf{P}=\mathsf{NP}$ solution against? I'm aware of many of the examples such as the Traveling Salesman or Knapsack problems but I assume there could be even simpler scenarios where all properties of the $\mathsf{P}=\mathsf{NP}$ or $\mathsf{P}\ne\mathsf{NP}$ dilemma are present.

xskxzr
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RLH
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    Sorry but this question doesn't make sense. You don't "test" a "P=NP solution". Your question suggests that people would say, "Great, Smith has proved that P=NP. I'll test if that's true by looking at such-and-such a problem" but that's not how it would work. A solution would be a proof that every problem in NP has a deterministic polynomial time algorithm. Most likely, that would be by exhibiting such an algorithm for some NP-compete problem. And that would give you an algorithm for whatever NP problem you're interested in; it wouldn't be a hypothesis that you'd test. – David Richerby Jun 10 '16 at 13:47
  • @DavidRicherby I see your point, but that's more a lack of my understanding of how to effectively communicate. My root question (regardless of reason) is what is considered the simplest, NP-Complete problem? Another way of looking at the reason is if I wanted to try to find an P=NP algorithm, what is the most bare bones problem that has been devised to date? – RLH Jun 10 '16 at 13:51
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    OK but "simplest" is purely a matter of opinion. Any search for the basics of NP-completeness will give you plenty of examples of fairly simple problems, such as SAT and CLIQUE. – David Richerby Jun 10 '16 at 14:09
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    "all NP problems could be solved using the same rules for resolving NP" -- what does that even mean? "what is the simplest P=NP problem" -- I have never heard of a class of such problems. Do you mean NP-complete problems? "simpler scenarios where all properties of the P=NP or P≠NP dilemma are present" -- yes, (almost) by definition in all NP-complete problems. Take your pick. – Raphael Jun 10 '16 at 14:16
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    https://en.wikipedia.org/wiki/Partition_problem is the simplest problem that I know of. It is simpler than Subset sum. – DSblizzard Sep 18 '17 at 08:15

2 Answers2

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Since all NP-complete problems are basically equivalent, it's hard to say which one is the easiest. SAT was one of the original problems and has important practical applications, so it is really well studied. Writing fast SAT-solvers seems like a reasonably interesting hobby to me and getting started is not terribly hard. Integer Linear Programming is similarly important and well studied.

The only objective way to discern NP-complete problems that I know of is their approximability. Some problems are hard to approximate, for others you can get almost arbitrarily good solutions in polynomial time.

Subset sum is both very simple to understand and "simple" to solve. Simple to solve means that it has an easy Pseudo-polynomial algorithm and is easy to approximate. However, I don't know about any research involving the problem.

adrianN
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Take the SAT problem. If you find a procedure which solves the SAT problem in polynomial time, then you proved it. But you should not take 2-SAT because it is already known to be in P.

Erhard Dinhobl
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  • Note that random SAT instances are easy. You can generate hard instances for example by encoding integer factorization as SAT. – adrianN Jun 10 '16 at 13:37
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    *shrug* Why is SAT "easier" than, say, clique or 3-colourability? Welcome to the site but, unfortunately, you've chosen to answer a question that doesn't really give you any sort of a chance to shine. – David Richerby Jun 10 '16 at 13:41
  • Since every NP-complete problem is reducible to SAT (or take any other problem) they have the same complexity! – Erhard Dinhobl Jun 10 '16 at 13:47
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    Sorry, I should have said "simpler", which is the word used in the question. Why is SAT "simpler" than, say, clique or 3-colourability. My point is that you're trying to answer a question that's so ill-formed that it's unanswerable. – David Richerby Jun 10 '16 at 13:49
  • I think the thread created wanted to know a problem which is simple to understand. It was not about the solution rather the problem itself. So I wanted to tell hin that SAT is an "easy" understandable problem, because I recognized 3-COL, CLIQUE, VERTEX COVER and what ever is not that easy to understand for people. (IMHO) – Erhard Dinhobl Jun 10 '16 at 14:01
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    OK, so this is purely opinion-based, which is exactly what we're not looking for in questions and answers. For example, my feeling is that clique is much simpler to explain. Consider all the people you know, and consider whether or not they know each other; is there a group of $k$ of them who all know each other? – David Richerby Jun 10 '16 at 14:08