[Math] Traveling salesman problem: why visit each city only once

graph theorymath-historymotivationoptimization

According to wikipedia, the Traveling Salesman Problem (TSP) is:

Given a list of cities and the distances between each pair of cities,
what is the shortest possible route that visits each city exactly once
and returns to the origin city?

Okay, that's a cool problem, but the part about "visiting each city exactly once" makes little sense to me. If I were a traveling salesman, I would just want to minimize the length (time, cost, whatever) of my route, and if visiting the same city $17$ times achieves this (say, because that city has an especially "central" position in the graph) then so be it. There seems to be little sense in restricting attention to Hamiltonian cycles (i.e. cycles in which each vertex occurs precisely once); in particular, I would imagine that this restriction simultaneously makes the problem harder (computationally) and also less applicable (e.g. to problems "from the real world.")

Wikipedia goes on to say that:

The problem was first formulated in 1930 and is one of the most
intensively studied problems in optimization.

In light of my previous comments, I find this surprising.

Question. Why has the TSP been so intensively studied, while the variant (which I find more natural) has apparently received much less
attention?

In simple terms: why visit each city only once?

Let me just add that according to wikipedia, the general problem does not include the assumption that the triangle inequality holds; that special case is called the metric TSP. In this case, the restriction to Hamiltonian cycles is of course innocuous.

Best Answer

Let's look at an example. Suppose you have four cities, A, X, Y, and Z, and A is at distance $1$ from each of the others, while the distance between any two of X, Y, Z is $100$. Then of course a solution to TSP starting at A is given by A to X to A to Y to A to Z to A – let's write this as AXAYAZA, which has length $6$, and visits A multiple times.

Now let's carry out the construction in Rahul's comment. We form a new problem with the same cities but with the distance between cities replaced by the length of the shortest path between the two cities. So in the new problem, A is still at distance $1$ from each of the others, but the distance between any two of X, Y, Z is $2$. Now the tirangle inequality is satisfied (although the configuration is still unrealizable in any Euclidean space), and a solution is given by AXYZA, length $6$, no city visited twice (except for both starting and ending at A).

So we have taken a problem where the solution has multiple visits to a city, and replaced it with an equivalent problem which doesn't. I hope it's clear that you can carry out this construction for any (finite) set of cities and distances. So you can always replace a problem with an equivalent problem where the triangle inequality is satisfied and no city gets visited more than once. So, there is no loss of generality in assuming the triangle inequality holds and no city gets visited more than once. That's why we can restrict our attention to what looks like, but really isn't, a special case.

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