Confusion about NP-hard and NP-Complete in Traveling Salesman problems

Question

Traveling Salesman Optimization(TSP-OPT) is a NP-hard problem and Traveling Salesman Search(TSP) is NP-complete. However, TSP-OPT can be reduced to TSP since if TSP can be solved in polynomial time, then so can TSP-OPT(1). I thought for A to be reduced to B, B has to be as hard if not harder than A. As I can see in the below references, TSP-OPT can be reduced to TSP. TSP-OPT is supposed to be harder than TSP. I am confused...

References: (1)Algorithm, Dasgupta, Papadimitriou, Vazirani Exercise 8.1 http://algorithmics.lsi.upc.edu/docs/Dasgupta-Papadimitriou-Vazirani.pdf https://cseweb.ucsd.edu/classes/sp08/cse101/hw/hw6soln.pdf

http://cs170.org/assets/disc/dis10-sol.pdf

Answer 1

I took a quick look at the references you gave, and I must admit there's one thing I really dislike in your textbook (1st pdf) : they address NP-completeness while barely mentioning decision problems. The provided definition of an NP-complete problem also somewhat deviates from what I'd expect from a textbook. I assume that was a conscious decision to make the introduction more appealing...

I'll provide a short answer, followed by a more detailed explanation about related notions.

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Short version

Intuitively (and informally), a problem is in NP if it is easy to verify its solutions.

On the other hand, a problem is NP-hard if it is difficult to solve, or find a solution.

Now, a problem is NP-complete if it is both in NP, and NP-hard. Therefore you have two key, intuitive properties to NP-completeness. Easy to verify, but hard to find solutions.

Although they may seem similar, verifying and finding solutions are two different things. When you use reduction arguments, you're looking at whether you can find a solution. In that regard, both TSP and TSP-OPT are NP-hard, and it is difficult to find their solutions. In fact, the third pdf provides a solution to excercise 8.1 of your textbook, which directly shows that TSP and TSP-OPT are equivalently hard to solve.

Now, one major distinction between TSP and TSP-OPT is that the former is (what your textbook call) a search problem, whereas the latter is an optimization problem. The notion of verifying the solution of a search problem makes sense, and in the case of TSP, it is easy to do, therefore it is NP-complete. For optimization problems, the notion of verifying a solution becomes weird, because I can't think of any way to do that without first computing the size of an optimal solution, which is roughly equivalent to solving the problem itself. Since we do not know an efficient way to verify a solution for TSP-OPT, we cannot say that it is in NP, thus we cannot say that it is NP-complete. (More on this topic in the detailed explanation.)

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The tl;dr is that for TSP-OPT, it is both hard to verify and hard to find solutions, while for TSP it is easy to verify and hard to find solutions. Reductions arguments only help when it comes to finding solutions, so you need other arguments to distinguish them when it comes to verifying solutions.

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More details

One thing your textbook is very brief about is what a decision problem is. Formally, the notion of NP-completeness, NP-hardness, NP, P, etc, were developed in the context of decision problems, and not optimization or search problems.

Here's a quick example of the differences between these different types of problems.

A decision problem is a problem whose answer is either YES or NO.

TSP decision problem

Input: a graph G, a budget b

Output: Does G admit a tour of weight at most b ? (YES/NO)

Here is the search version

TSP search problem

Input: a graph G, a budget b

Output: Find a tour of G of weight at most b, if it exists.

And now the optimization version

TSP optimization problem

Input: a graph G

Output: Find a tour of G with minimum weight.

Out of context, the TSP problem could refer to any of these. What I personally refer to as the TSP is the decision version. Here your textbook use TSP for the search version, and TSP-OPT for the optimization version.

The problem here is that those various problems are strictly distinct. The decision version only ask for existence, while the search version asks for more, it needs one example of such a solution. In practice, we often want to have the actual solution, so more practical approaches may omit to mention decision problems.

Now what about it? The definition of an NP-complete problem was meant for decision problems, so it technically does not apply directly to search or optimization problems. But because the theory behind it is well defined and useful, it is handy to still apply the term NP-complete/NP-hard to search/optimization problem, so that you have an idea of how hard these problems are to solve. So when someone says the travelling salesman problem is NP-complete, formally it should be the decision problem version of the problem.

Obviously, many notions can be extended so that they also cover search problems, and that is how it is presented in your textbook. The differences between decision, search, and optimization, are precisely what the exercises 8.1 and 8.2 try to cover in your textbook. Those exercises are probably meant to get you interested in the relationship between these different types of problems, and how they relate to one another.

Answer 2

Short Version

The decision problem is NP-complete because you can both have a polynomial time verifier for the solution, as well as the fact that the hamiltonian cycle problem is reducible to TSP_DECIDE in polynomial time.

However, the optimization problem is strictly NP-hard, because even though TSP_OPTIMIZE is reducible from the hamiltonian (HAM) cycle problem in polynomial time, you don't have a poly time verifier for a claimed hamiltonian cycle C, whether it is the shortest or not, because you simply have to enumerate all possibilities (which consumes the factorial order space & time).

What the given reference define is, bottleneck TSP

The Bottleneck traveling salesman problem (bottleneck TSP) is a problem in discrete or combinatorial optimization. The problem is to find the Hamiltonian cycle in a weighted graph which minimizes the weight of the most weighty edge of the cycle.

The problem is known to be NP-hard. The decision problem version of this, "for a given length x is there a Hamiltonian cycle in a graph G with no edge longer than x?", is NP-complete. NP-completeness follows immediately by a reduction from the problem of finding a Hamiltonian cycle.

This problem can be solved by performing a binary search or sequential search for the smallest x such that the subgraph of edges of weight at most x has a Hamiltonian cycle. This method leads to solutions whose running time is only a logarithmic factor larger than the time to find a Hamiltonian cycle.

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Long Version

The mistake is to say that the TSP is NP complete. Truth is that TSP is NP hard. Let me explain a bit:

The TSP is a problem defined by a set of cities and the distances between each city pair. The problem is to find a circuit that goes through each city once and that ends where it starts. This in itself isn't difficult. What makes the problem interesting is to find the shortest circuit among all those that are possible.

Solving this problem is quite simple. One merely need to compute the length of all possible circuits, then keep the shortest one. Issue is that the number of such circuits grows very quickly with the number of cities. If there are n cities then this number is factorial of n-1 = (n-1)(n-2)...3.2.

A problem is NP if one can easily (in polynomial time) check that a proposed solution is indeed a solution.

Here is the trick.

In order to check that a proposed tour is a solution of the TSP we need to check two things, namely

1. That each city is is visited only once
2. That there is no shorter tour than the one we are checking

We didn't check the second condition! The second condition is what makes the problem difficult to solve. As of today, no one has found a way to check condition 2 in polynomial time. It means that the TSP isn't in NP, as far as we know.

Therefore, TSP isn't NP complete as far as we know. We can only say that TSP is NP hard.

When they write that TSP is NP complete, they mean that the following decision problem (yes/no question) is NP complete:

TSP_DECISION : Given a number L, a set of cities, and distance between all city pairs, is there a tour visiting each city exactly once of length less than L?

This problem is indeed NP complete, as it is easy (polynomial time) to check that a given tour leads to a yes answer to TSPDECISION.