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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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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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