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某些NP完全问题又如何成为NP-Hard？

[英]How can some NP-Complete problems be also NP-Hard?

原文 2014-01-08 20:18:39 6 2 complexity-theory/ computer-science/ np-complete/ np-hard

I'm trying wrap my heard around P, NP, NP-Complete and NP-Hard in an intuitive way so that I don't have to remember their definitions. 我正在尝试以一种直观的方式围绕P，NP，NP-Complete和NP-Hard听到我的声音，这样我就不必记住它们的定义了。

In the following image (the left hand scenario, P != NP), there's an overlapping area between NP-Complete and NP-Hard. 在下图中（左手的情况，P！= NP），在NP-Complete和NP-Hard之间存在重叠区域。 Does it mean that some problems are both NP-Complete and NP-Hard? 这是否意味着NP-Complete和NP-Hard都存在一些问题？ I find that contradictory, according to this particular answer: What are the differences between NP, NP-Complete and NP-Hard? 根据这个特定答案，我发现这是矛盾的： NP，NP-Complete和NP-Hard有什么区别？ . 。

The table in the above link says an NP-Complete problem is verifiable in polynomial time and an NP-Hard problem is not. 上面链接中的表格说，NP完全问题可以在多项式时间内验证，而NP硬问题则不能。 So how can there be an overlap? 那么怎么会有重叠呢？

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2 个解决方案

Part of the definition of NP-completeness is being NP hard. NP完整性的部分定义是NP困难。 Therefore, every NP-complete problem is NP-hard. 因此，每个NP完全问题都是NP难的。 This is also reflected by both of your graphs. 这两个图也反映了这一点。

The table you linked to was wrong until I fixed it a few hours ago. 您链接到的表是错误的，直到几小时前我将其修复。 NP-Complete problems are a subset of NP problems, and all NP problems are verifiable in polynomial time by definition. NP-完全问题是NP问题的子集，并且所有NP问题根据定义都可以在多项式时间内验证。 NP-hard problems are those problems which are at least as hard as any other NP problem (which is sort of unintuitive, because problems not in NP can be NP-hard). NP难题是至少与任何其他NP问题一样困难的问题（这是不直观的，因为NP之外的问题可能是NP难题）。

To be NP-Complete a problem must be 要成为NP-Complete，必须解决一个问题

in NP 在NP中
NP-hard NP硬

to be complete in a specific complexity class, a problem must be 要在特定的复杂性类别中完整，必须解决一个问题

in that complexity class 在那个复杂度等级
at least as hard as any other problem in that complexity class 至少与该复杂性类别中的任何其他问题一样困难

We have to define "at least as hard". 我们必须定义“至少一样困难”。 Suppose we have a problem A in NP. 假设我们在NP中有一个问题A。 To prove it is NP-hard (and therefore NP-Complete), we show that all problems in NP can be converted to A in polynomial time. 为了证明它是NP难的（因此是NP完全的），我们证明NP中的所有问题都可以在多项式时间内转换为A。 Because A takes at least polynomial time to solve, and polynomials are closed under addition, the conversion is now negligible, and the runtime is the same as the runtime of A (in terms of it being polynomial or not). 因为A至少需要多项式时间来求解，并且多项式在加法之后被关闭，所以转换现在可以忽略不计，并且运行时间与A的运行时间相同（就其是否为多项式而言）。

Once you have one NP-Complete problem, you can prove a problem A in NP is NP-hard (and therefore NP-Complete) by taking another NP-Complete problem B and converting it to A in polynomial time. 一旦遇到一个NP完全问题，就可以通过采用另一个NP完全问题B并在多项式时间内将其转换为A来证明NP中的问题A是NP难的（因此是NP完全的）。

I hope this makes it clear that NP-Complete is a subset of NP-hard (and that the table you linked to was wrong). 我希望这可以清楚地表明NP-Complete是NP-hard的子集（并且链接到的表是错误的）。