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posted by martyb on Tuesday October 10 2017, @07:49AM   Printer-friendly
from the problem-of-looking-for-an-asymmetric-traveling-salesman dept.

The real-world version of the famous traveling salesman problem finally gets a good-enough solution.

From the abstract on arXiv https://arxiv.org/abs/1708.04215 (full article is available):

We give a constant-factor approximation algorithm for the asymmetric traveling salesman problem. Our approximation guarantee is analyzed with respect to the standard LP relaxation, and thus our result confirms the conjectured constant integrality gap of that relaxation. Our techniques build upon the constant-factor approximation algorithm for the special case of node-weighted metrics. Specifically, we give a generic reduction to structured instances that resemble but are more general than those arising from node-weighted metrics. For those instances, we then solve Local-Connectivity ATSP, a problem known to be equivalent (in terms of constant-factor approximation) to the asymmetric traveling salesman problem.

[1] LP == Linear Programming
[2] ATSP == Asymmetric Traveling Salesman Problem

Any Soylentils able to explain what, if any, impact this research has on the class of NP-complete problems?

source:
https://www.quantamagazine.org/one-way-salesman-finds-fast-path-home-20171005/


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  • (Score: 5, Informative) by bradley13 on Tuesday October 10 2017, @08:22AM (2 children)

    by bradley13 (3053) on Tuesday October 10 2017, @08:22AM (#579695) Homepage Journal

    I didn't read TFA, because this is Soylent. However, from TFS:

    "We give a constant-factor approximation algorithm..." The word "approximation" means that there is no guarantee of an optimal solution. There are lots of algorithms for getting "pretty good" solutions, or even optimal solutions "most of the time" for NP-complete problems in polynomial time. So: this result presumably offers nothing new, except maybe a different way of doing the approximation.

    --
    Everyone is somebody else's weirdo.
    • (Score: 2) by TheRaven on Tuesday October 10 2017, @10:38AM (1 child)

      by TheRaven (270) on Tuesday October 10 2017, @10:38AM (#579720) Journal
      To add to that: this is true in the specific case of the travelling salesman problem. There are a number of algorithms that give good solutions in varying complexity classes. I seem to recall (though, checking Wikipedia, it looks as if I may have made this up) that the travelling salesman problem is much easier to solve if you place some constraints on the structure of your graph that happen to map quite nicely to the kinds of roads that people build, making the practical version a lot easier in the majority of cases than the graph theoretic problem.
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      • (Score: 0) by Anonymous Coward on Thursday October 12 2017, @07:07PM

        by Anonymous Coward on Thursday October 12 2017, @07:07PM (#581293)

        > the travelling salesman problem is much easier to solve if you place some constraints

        Goodness me yes! It's an absolutely tractable problem under lots of conditions. I forget what boundaries are big-O (complexity) better and which are just empirically useful, but some which have been investigated include: directed graphs, trees, planar graphs, maximum degree (# of edges/node), bounded edge length...

        There's some pretty amazing properties where, like including or excluding '0' from the domain of a function, suddenly the problem can have a vastly different meaning.

        Any graph maths or CS types around here able to link to an understandable overview of modified travelling salesman?

  • (Score: 5, Informative) by stormwyrm on Tuesday October 10 2017, @08:43AM

    by stormwyrm (717) on Tuesday October 10 2017, @08:43AM (#579699) Journal
    What we have here is a heuristic that they have proved is at most 1.5 times longer than the most optimal route. However, the NP-completeness of the travelling salesman problem is based on the exact solution, the actual shortest path that the travelling salesman could take. There exists a polynomial-time algorithm capable of transforming an instance of any NP-complete problem into any other (polynomial time reduction), so if one can solve any one NP-complete problem in polynomial time, you can solve any other NP-complete problem in polynomial time too. Say, if you had the shortest path for an instance of the travelling salesman problem, you could transform that travelling salesman problem into a Hamiltonian cycle problem (also NP-complete) and your shortest path in the TSP problem would also be reducible to the answer of the Hamiltonian cycle. However, if you had only an approximate answer to the travelling salesman problem, even if it were good enough for practical purposes such as if you actually needed to travel to all those places yourself, it would not provide you with the solution to the equivalent Hamiltonian cycle problem, or any other NP-complete problem you could reduce it to.
    --
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  • (Score: 2) by turgid on Tuesday October 10 2017, @10:37AM (3 children)

    by turgid (4318) Subscriber Badge on Tuesday October 10 2017, @10:37AM (#579719) Journal

    I usually find I get home fastest when I'm on my beer scooter.

    • (Score: 1, Touché) by Anonymous Coward on Tuesday October 10 2017, @12:02PM (2 children)

      by Anonymous Coward on Tuesday October 10 2017, @12:02PM (#579740)

      But, how many pubs do you stop at on the way home? And do you visit them in an optimal sequence??

  • (Score: 2, Insightful) by All Your Lawn Are Belong To Us on Tuesday October 10 2017, @02:56PM (1 child)

    by All Your Lawn Are Belong To Us (6553) on Tuesday October 10 2017, @02:56PM (#579833) Journal

    As in, chess programs make sequences of legal moves and then make numeric calculations to evaluate what the board looks like in that condition. Best condition is the played next move. They don't "play" chess the way humans do necessarily (humans think about what the board looks like and try to find optimal next moves but do not assign centipawn values to conditional factors or use hash tabling - at least not that I'm aware of,) but the approximation is now good enough to hold or beat grandmaster level players. But it's only an approximation of the solution.

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    • (Score: 0) by Anonymous Coward on Thursday October 12 2017, @07:14PM

      by Anonymous Coward on Thursday October 12 2017, @07:14PM (#581296)

      > centipawn

      Well, sure, but we don't call it that. Though, internally, those silicon neural nets don't 'call' it anything either!

      Proof: "it's good to control the middle of the board. But most of the time, not so good that it's worth losing a piece or a pawn over."

      Heck, any time two positions look preeeeeetty similarly good, that's a decipawn or centipawn difference. The non-piece-determining conditionals X moves ahead under Y circumstances, admittedly, is not something common to elementary school chess clubs, but certainly the better human chess players consider more than just "can I get a piece through this sequence", and unless the position advantage is "worth" a pawn+, it's definitially pawn-fractional.

      Pawn-fractional. Well, thank you soylentil, for expanding my lexicon. Much wordy love!

  • (Score: 2) by wonkey_monkey on Wednesday October 11 2017, @04:47PM (1 child)

    by wonkey_monkey (279) on Wednesday October 11 2017, @04:47PM (#580564) Homepage

    finally gets a good-enough solution.

    So, not an actual solution then?

    --
    systemd is Roko's Basilisk
    • (Score: 0) by Anonymous Coward on Thursday October 12 2017, @11:16PM

      by Anonymous Coward on Thursday October 12 2017, @11:16PM (#581422)

      Right. Exactly. This is why it won't transpose to other problems in its class, though the approximation bound *might* morph, for carefully constructed parallel problems.

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