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I suspect such a reduction would significantly inflate the size of your problem. I say this because there exist algorithms such as Christofides algorithm which

I suspect such a reduction would significantly inflate the size of your problem. I say this because there exist algorithms such as Christofides algorithm which explicitly work on Metric TSP. If there were such a trivial transformation that preserved the complexity class, why wouldn't Christofides algorithm apply to the general case?

jsprogrammer 11y ago
>I suspect such a reduction would significantly inflate the size of your problem. If Metric TSP and a more general TSP are both NP-complete then there is an algorithm in P to transform a general TSP to a Metric TSP. So, a reduction could inflate the size of the problem, but that inflation will not be significant relative to the most naive approach. Here is a SE algorithm: http://cstheory.stackexchange.com/a/14049 http://cstheory.stackexchange.com/a/14049
- zzazzdsa 11y ago
  There's a more fundamental issue here-- when talking about approximation algorithms reductions don't necessarily preserve approximation ratios. If you take a standard TSP instance with optimal tour length C and add the largest distance between two points M to every edge, you get a new (metric) TSP with optimal tour length C + nM, where n is the number of nodes of the graph. Applying the Christofides algorithm to this new metric TSP instance can only guarantee a tour of length 1.5C + 1.5nM, and when you subtract M from every edge again to get a tour in the original TSP instance, this scheme will return a tour of length at most 1.5C + .5nM, a bound which is provably tight. This is a terrible approximate solution, since if the largest two-node distance is O(C) (for example), the approximation obtained has value O(n)C-- a monstrously bad O(n) approximation!
btilly 11y ago
Well yes..and no. Let M be the largest distance between any two points in your graph. Add M to the length of every edge. Now the triangle inequality is trivially satisfied and you've got a Metric TSP. However the "you find a path within a factor of 3/2 of the best possible" guarantee of the Christofides algorithm is much less useful than you might hope because all paths just got a lot longer.

jsprogrammer 11y ago

>I suspect such a reduction would significantly inflate the size of your problem. If Metric TSP and a more general TSP are both NP-complete then there is an algorithm in P to transform a general TSP to a Metric TSP. So, a reduction could inflate the size of the problem, but that inflation will not be significant relative to the most naive approach. Here is a SE algorithm: http://cstheory.stackexchange.com/a/14049 http://cstheory.stackexchange.com/a/14049

zzazzdsa 11y ago
There's a more fundamental issue here-- when talking about approximation algorithms reductions don't necessarily preserve approximation ratios. If you take a standard TSP instance with optimal tour length C and add the largest distance between two points M to every edge, you get a new (metric) TSP with optimal tour length C + nM, where n is the number of nodes of the graph. Applying the Christofides algorithm to this new metric TSP instance can only guarantee a tour of length 1.5C + 1.5nM, and when you subtract M from every edge again to get a tour in the original TSP instance, this scheme will return a tour of length at most 1.5C + .5nM, a bound which is provably tight. This is a terrible approximate solution, since if the largest two-node distance is O(C) (for example), the approximation obtained has value O(n)C-- a monstrously bad O(n) approximation!

btilly 11y ago

Well yes..and no. Let M be the largest distance between any two points in your graph. Add M to the length of every edge. Now the triangle inequality is trivially satisfied and you've got a Metric TSP. However the "you find a path within a factor of 3/2 of the best possible" guarantee of the Christofides algorithm is much less useful than you might hope because all paths just got a lot longer.