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为什么TSP NP难以在汉密尔顿路径NP完全？

[英]Why is TSP NP-hard while the Hamiltonian path NP-complete?

原文 2016-07-28 20:45:02 2 2 algorithm/ graph-theory/ np-complete/ np-hard

Why aren't these 2 problems, namely TSP and Hamiltonian path problem , both NP-complete? 为什么这两个问题，即TSP和哈密顿路径问题都不是NP完全？

They seem identical. 它们似乎相同。

2 个解决方案

For a problem X to be NP-complete , it has to satisfy: 对于问题X是NP完全的 ，它必须满足：

X is in NP , given a solution to X, the solution can be verified in polynomial time. X在NP中 ，给定X的解，可以在多项式时间内验证解。
X is in NP-hard , that is, every NP problem is reduceable to it in polynomial time (you can do this through a reduction from a known NP-hard problem (eg Hamiltonian Path)). X是NP难的 ，也就是说，每个NP问题在多项式时间内都可以减少（你可以通过减少已知的NP难问题（例如哈密顿路径）来实现）。

There are two versions of the The Travelling Salesman Problem (TSP) : 旅行商问题（TSP）有两个版本：

The optimization version (probably the one you are looking at), namely, find the optimum solution to the TSP. 优化版本（可能是您正在查看的版本），即找到TSP的最佳解决方案。 This is not a decision problem, and hence cannot be in NP, but it is however in NP-hard which can be proven via a Hamiltonian Path reduction . 这不是一个决策问题，因此不能在NP中，但它在NP-hard中可以通过哈密顿路径减少来证明。 Therefore this isn't an NP complete problem. 因此，这不是NP完全问题。
The decision version - given an integer K is there a path through every vertex in the graph of length < K? 决定版本 - 给定一个整数K是否有一条路径通过长度<K的图中的每个顶点？ This is a decision (yes/no) problem, and a solution can be verified in polynomial time (just traverse the path and see if it touches every vertex) and so it is in NP, but it is also in NP-hard (by an identical proof as above). 这是一个决定（是/否）问题，并且可以在多项式时间内验证解决方案（只是遍历路径并查看它是否接触到每个顶点），因此它在NP中，但它也在NP-hard（通过与上述相同的证据）。 Since it satisfies both requirements for NP-completeness, it is an NP-complete problem. 由于它满足NP完全性的两个要求，因此它是NP完全问题。

The definitions of NP-hardness and NP-completeness are related but different. NP-硬度和NP-完整性的定义是相关的但不同。 Specifically, a problem is NP-hard if every problem in NP reduces to it in polynomial time, and a problem is NP-complete if it's both NP-hard and itself in NP. 具体来说，如果NP中的每个问题在多项式时间内都减少，那么问题就是NP难，如果它既是NP难的又是NP本身的问题就是NP完全问题。

The class NP consists of decision problems, problems that have a yes/no answer. NP类由决策问题，有/无答案的问题组成。 As a result, TSP cannot be in NP because the expected answer is a number rather than yes or no. 因此，TSP不能在NP中，因为预期的答案是数字而不是是或否。 Therefore, TSP can be NP-hard, but it can't be NP-complete. 因此，TSP可以是NP难的，但它不能是NP完全的。

On the other hand, the Hamiltonian path problem asks for a yes/no answer, and it happens to be in NP. 另一方面，汉密尔顿路径问题要求是/否答案，它恰好在NP中。 Therefore, since it's NP-hard as well, it's NP-complete. 因此，既然它也是NP难的，它就是NP完全的。

Now, you can take TSP and convert it to a decision problem by changing the question from "what's the cheapest path?" 现在，您可以通过更改“最便宜的路径是什么？”中的问题来将TSP转换为决策问题。 to "is there a path that costs X or less?," and that latter formulation is in NP and also happens to be NP-complete. “是否存在成本为X或更低的路径？”，后一种配方在NP中也恰好是NP完全的。