Trivial approximation ratio of n/2 for 2-Opt for TSPI recently read that the 2-Opt algorithm...

Jovany Clayton

Jovany Clayton



Trivial approximation ratio of n/2 for 2-Opt for TSP
I recently read that the 2-Opt algorithm for the TSP yielded an approximation ratio of n/2 by trivial reasons. However, there was neither proof nor further context provided and I am curious how this bound can be obtained since I don't see these trivial reasons. I hope you can help me and give some intuition of why this ratio holds. (Note: Perhaps this can only obtained easily when considering metric TSP which would totally suffice for my concerns)

Answer & Explanation




2022-07-03Added 16 answers

Step 1
For all those that stumble upon this question: The answer actually is quite simple. Let me explain.
Given an Instance I = ( K n , w ) of metric TSP, consider the following algorithm:
1. Construct an arbitrary Tour T through V.
2. Return T.
I claim that this simple algorithm already achieves an approximation ratio of n/2: Let T denote an optimal Tour. Let e = ( v , w ) be an arbitrary edge of T. Note that T is the disjoint union of two paths P 1 and P 2 where one is a v-w-Path and the other is a w-v-Path. By the triangle inequality we have w ( e ) w ( P 1 ) = f E ( P 1 ) w ( f ) and similarily w ( e ) w ( P 2 ). Together with w ( T ) = f E ( T ) w ( f ) = w ( P 1 ) + w ( P 2 ) it follows that 2 w ( e ) w ( T ) .
Step 2
Therefore w ( T ) = e E ( T ) w ( e ) e E ( T ) 1 2 w ( T ) = n 2 w ( T ) ,, which proves the claim.
My initial confusion came from the misunderstanding this approximation ratio would have something to do with the specific 2-Opt algorithm. However, as argued above, this is not the case and this approximation ratio is quite trivial to achieve.

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