4 ms·
I think NP-hard is correct? You can reduce the halting problem and Kolmogorov complexity computation to each other [1], so the question is whether TSP can be re
by dataflow 5y ago
I think NP-hard is correct? You can reduce the halting problem and Kolmogorov complexity computation to each other [1], so the question is whether TSP can be reduced to the halting problem. Well, you can reduce SAT to the halting problem (they explain how here [2]) and you can reduce TSP to SAT given they're both NP-complete so... we're done?
[1] https://www.nearly42.org/cstheory/halting_to_kolmogorov/ https://www.nearly42.org/cstheory/halting_to_kolmogorov/
[2] https://en.wikipedia.org/wiki/NP-hardness#Examples https://en.wikipedia.org/wiki/NP-hardness#Examples
- hakuseki 5y agoThe reductions in your linked paper [1] are not polynomial-time reductions.
- bennofs 5y agoConstruct a TM which enumerates all possible variable assignments and checks if the SAT problem is satisfied then halts if so. You can construct this TM in polynomial time, and it halts exactly if the SAT problem is satisfiable. So this is a polynomial reduction from SAT to the halting problem.
- hakuseki 5y agoI do not dispute this. My comment was about the linked paper [1] regarding equivalence of the halting problem and Kolmogorov complexity, not the SAT problem.