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The title pretty much says it all.

I am curious if the TSP is dependent only upon the number of cities, $c$, involved, or if the dimensionality, $d$, matters to the problem.

For example, if $d>>c$ does the problem shift in some manner, making the time to solve the problem less dependent on the number of cities?

chase
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2 Answers2

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The traveling salesman problem is, in its basic form, a graph theory problem. All vertices (the cities to visit) are connected, and all edges are assigned a "distance", and you need to answer: "what is the shortest route that starts at a given vertex, visits all other vertices, and comes back to the start point?"

The only thing that changes in a multi-dimensional format is that calculating these distances takes more variables, but this is calculated in linear time (I'm assuming Euclidean distances). But once you get calculate them all and get the graph described above, the problem is exactly the same. So no change in complexity.

Jeffery
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    Graph theory does have a notion of dimensionality to it, in that a graph may or may not be embeddable into $\mathbb{R}^n$ for certain $n$ (where "embedabble" means it can be drawn without intersecting edges). The smallest $n$ for which this can be done can be interpreted as the dimension of the graph itself. This dimensionality enters into certain problems (for example, coloring problems). It doesn't really enter into the TSP in any significant way, however. – Ian Dec 01 '18 at 20:45
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    What does affect TSP is whether the edge costs are metric: whether they satisfy the triangle inequality $d(u,w) \le d(u,v) + d(v,w)$. If the graph is embeddable in $\mathbb R^n$ for any $n$, then the edge costs (Euclidean distances) are metric, which allows for some approximation algorithms that do not work in general. – Misha Lavrov Dec 01 '18 at 22:29
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    Except of course for $d=1$: If the cities are points of $\Bbb R$ with the euclidean distance then the TSP is pretty easy... – David C. Ullrich Dec 02 '18 at 18:49
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Please allow me 2 comments on the average and expected tour length in higher dimensions:

Comment 1: What changes with dimension, is how the expected tour length 'scales' with $N$:
When you consider the Euclidean traveling salesman problem for $N$ cities randomly distributed in the unit d-dimensional hypercube, you will find that the optimal tour length for large $N$ will scale as $\beta_d N^{1-1/d} (1-O(1/N))$. The constant $\beta_2$ is the famous 'TSP constant'.

For large $d$, you expect that the so-called mean-field approximation will be very good and allow you to provide good estimates of the constant and the expected tour length.

One standard reference is 'Finite Size and Dimensional Dependence in the Euclidean Traveling Salesman Problem' by Allon G. Percus and Olivier C. Martin

Comment 2: If the dimensionality of space is large compared to the number of cities, then you expect that the distances are statistically mutually independent and the order of the tour does not make much of a difference any longer in most cases - your situation is the similar to the TSP on the complete graph (see: Can the 'TSP polynomial for simple graphs' be calculated for standard families of graphs (paths, cycles, grid graphs)?), where you just need to know the number of points/ cities.
In a sense that is similar to comment 1: for large $d$ the expected tour length will scale close to $N$.

Of course, in a multi-dimensional space you still can pick cities that all lie in a plane and you are effectively back to 2d but those configurations are statistically, when you pick city locations randomly, irrelevant.

Let me re-phrase: The time to find the optimum solution does not go down (unless you add further constraints on the structure of space as with the complete graph); however, the cost of missing the optimum strongly reduces with higher $d$ on average.

Michael T
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