5

Let's have a complete undirected metric graph with k nodes; metric graph is a graph that satisfy the triangle inequality, so being w the weight function the for all nodes a, b, c it is true that w(a, c) is less or equal to w(a,b) + w(b,c).

Wlog let's say that the cycle: <1, 2, 3, ..., k, 1> is the optimal TSP solution for that graph.

My question is: if I remove one node from the graph (for example the n-th) and I shortcut the cycle just skipping n is the resulting cycle still an optimal TSP solution?

n.b., The cycle would become <1, 2, ..., n-1, n+1, ..., k, 1>

Paolo.Bolzoni
  • 2,416
  • 1
  • 18
  • 29
  • What would be more interesting (since the answer to this is clearly no): would any graph satisfying the triangle inequality have this property? – Paul Draper Apr 10 '14 at 10:27
  • @PaulDraper doesn't a metric graph by definition satisfy the triangle inequality? I couldn't find a decent definition of metric graphs online. (For completeness: my interpretation of a metric graph is a plane graph for which the weight of an edge `(u, v)` is the Euclidean distance between `u` and `v`. Maybe that's totally wrong.) – Vincent van der Weele Apr 10 '14 at 10:38
  • @Heuster, you might be right. In that case, timrau's answer is incorrect, as 1->2->3 is shorter than 1->3. Paulo, could you clarify what a "metric graph" is? – Paul Draper Apr 10 '14 at 10:44
  • Metric graph is a graph such that satisfy triangle inequality. So being w the weight function for every node a, b, c you have that w(a,b) + w(b,c) is more or equal than w(a,c). Edited the question. – Paolo.Bolzoni Apr 10 '14 at 10:52

1 Answers1

4

No, this does not hold. A rather handwaving counterexample is given below. I trust you can add the numbers, do the math, and formally verify this (I used this online solver to verify my claims).

Consider these points:

enter image description here

The top point is clearly far away, so it has to be connected to the closest points. The other links then follow, as shown here:

enter image description here

If we exclude the top point, it is more optimal to have the two top points connect to the center point, as shown below. So just short-cutting is not optimal:

enter image description here

Vincent van der Weele
  • 12,927
  • 1
  • 33
  • 61