Consider the following setting. You have $n$ cities, and there is a cost to travel from a city $i$ to a city $j$ given by $c_{ij}>0$ where $c_{ij}\neq c_{ji}$. Moreover, if you are traveling to city $i$, you are allowed to stay exactly $d_i>0$ days. The problem is:

Find a travel schedule, this is: a number $m$ and a sequence of different cities $i_1,\dots,i_m\in\{1,\dots,n\}$, such that the cost $\sum_{k=1}^{m-1} c_{i_k,i_{k+1}}$ is minimal and that the whole trip lasts at least $L$, this is $\sum_{k=1}^m d_{i_k}\geq L$.

There are many similarities between this problem and the traveling salesman one. The key differences are the non symmetric costs $c_{ij}\neq c_{ji}$ and that instead of having to travel to all cities once, one just needs to travel to enough cities to make the trip last enough.

I understand that to show that this problem is NP-hard, I need to show that another NP-hard problem can be reduced to this one in polynomial time. I suspect that the original traveling salesman problem may work to do so through some clever trick I haven't found yet. Moreover, since there is plenty of literature on the traveling salesman problem, I suspect a similar problem to this may already be shown to be NP-hard.

My question is: Do you see how to show this problem to be NP-hard? Or, do you recognize this problem to be something standard in some piece of literature you can point me out?


1 Answer 1


I think if you choose all $d$'s to be exactly $1$ and choose $L$ to be exactly $n$ (number of cities), any solution must visit all cities, so the reduction will work out.

  • $\begingroup$ So this reduces this problem to the traveling salesman problem. But I need to reduce the traveling salesman problem to this one, and not the other way around right? $\endgroup$ May 27, 2021 at 14:16
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    $\begingroup$ The reduction is from the traveling salesman to this problem. I know its sometimes confusing how reductions work :) Take an instance of traveling salesman, and create the instance of this problem with $d=1$ and $L=n$. $\endgroup$
    – nir shahar
    May 27, 2021 at 14:17
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    $\begingroup$ Oh I see... I misunderstood the definition then. Yes, it is confusing. Let me think about this for a moment. Thanks! $\endgroup$ May 27, 2021 at 14:22
  • $\begingroup$ Can you point out where can I find a definition of "reduction" in this sense? I am using the book "Combinatorial optimization" by Korte and Vygen, I am not able to decipher this sense of reduction from there. Thanks!! $\endgroup$ May 27, 2021 at 14:33
  • $\begingroup$ Taken from cs.stackexchange.com/a/1249/126979: " To reduce problem X (the one you know is NP-hard) to problem Y (the one you're trying to prove is NP-hard, you need to describe an algorithm that transforms an arbitrary instance of X into a legal instance of Y. " is it that the "into a legal instance of Y" means "any legal instance of Y"? $\endgroup$ May 27, 2021 at 14:48

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