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My question is - if we consider the Travelling salesman problem without the returning to the starting point part does the greedy (nearest neighbor) algorithm solves this problem correctly ? In a previous thread someone stated that this problem is equivalent to the shortest Hamiltonian path, but I think we dont need the restriction on visiting every node exactly once, so I would like a better explenation.

Thank you,

Dannynis
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  • Have you tried to prove or disprove it? There is a trivial counter-example. Also this belongs on cs.stackexchange.com, stackoverflow is for coding questions – BlueRaja - Danny Pflughoeft Feb 11 '18 at 15:08
  • I started to think about it because - https://stackoverflow.com/questions/6733999/what-is-the-problem-name-for-traveling-salesman-problemtsp-without-considering?rq=1 where no clear conclusion was made – Dannynis Feb 11 '18 at 20:12

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