Ordering tours in a Euclidean TSP according to (strictly) increasing length












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Let $H$ be the set of all Hamiltonian cycles on the complete graph $K_n$ associated with a set of $n geq 4$ points $P$ in the plane where edge weights are defined using the Euclidean distance between pairs of points in $P$. Suppose that $h_1 in H$ is an optimal tour with length $d(h_1)$ and that the tours are numbered in order of increasing length so that:
$d(h_1) leq d(h_2) leq cdots leq d(h_{|H|})$.



Question: Is there a set of conditions on the set of points in $P$ that would result in the ordering above being a sequence of strict inequalities? I am interested in finding a lower bound $lambda < 1$ that characterizes the ratio $d(h_1) / d(h_2) < lambda$ given some distribution of points, if possible. Intuitively it seems this may be related to the minimum distance between any pair of points in $P$, or the minimum difference in tour length resulting from an edge swap. Perhaps there are some other well known results on the distribution of tour lengths that might be related. I would be grateful for any references that come to mind on topics related to this question.






combinatorics graph-theory discrete-optimization







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    0












    $begingroup$


    Let $H$ be the set of all Hamiltonian cycles on the complete graph $K_n$ associated with a set of $n geq 4$ points $P$ in the plane where edge weights are defined using the Euclidean distance between pairs of points in $P$. Suppose that $h_1 in H$ is an optimal tour with length $d(h_1)$ and that the tours are numbered in order of increasing length so that:
    $d(h_1) leq d(h_2) leq cdots leq d(h_{|H|})$.



    Question: Is there a set of conditions on the set of points in $P$ that would result in the ordering above being a sequence of strict inequalities? I am interested in finding a lower bound $lambda < 1$ that characterizes the ratio $d(h_1) / d(h_2) < lambda$ given some distribution of points, if possible. Intuitively it seems this may be related to the minimum distance between any pair of points in $P$, or the minimum difference in tour length resulting from an edge swap. Perhaps there are some other well known results on the distribution of tour lengths that might be related. I would be grateful for any references that come to mind on topics related to this question.






    combinatorics graph-theory discrete-optimization







    share|cite|improve this question











    $endgroup$















      0












      0








      0





      $begingroup$


      Let $H$ be the set of all Hamiltonian cycles on the complete graph $K_n$ associated with a set of $n geq 4$ points $P$ in the plane where edge weights are defined using the Euclidean distance between pairs of points in $P$. Suppose that $h_1 in H$ is an optimal tour with length $d(h_1)$ and that the tours are numbered in order of increasing length so that:
      $d(h_1) leq d(h_2) leq cdots leq d(h_{|H|})$.



      Question: Is there a set of conditions on the set of points in $P$ that would result in the ordering above being a sequence of strict inequalities? I am interested in finding a lower bound $lambda < 1$ that characterizes the ratio $d(h_1) / d(h_2) < lambda$ given some distribution of points, if possible. Intuitively it seems this may be related to the minimum distance between any pair of points in $P$, or the minimum difference in tour length resulting from an edge swap. Perhaps there are some other well known results on the distribution of tour lengths that might be related. I would be grateful for any references that come to mind on topics related to this question.






      combinatorics graph-theory discrete-optimization







      share|cite|improve this question











      $endgroup$




      Let $H$ be the set of all Hamiltonian cycles on the complete graph $K_n$ associated with a set of $n geq 4$ points $P$ in the plane where edge weights are defined using the Euclidean distance between pairs of points in $P$. Suppose that $h_1 in H$ is an optimal tour with length $d(h_1)$ and that the tours are numbered in order of increasing length so that:
      $d(h_1) leq d(h_2) leq cdots leq d(h_{|H|})$.



      Question: Is there a set of conditions on the set of points in $P$ that would result in the ordering above being a sequence of strict inequalities? I am interested in finding a lower bound $lambda < 1$ that characterizes the ratio $d(h_1) / d(h_2) < lambda$ given some distribution of points, if possible. Intuitively it seems this may be related to the minimum distance between any pair of points in $P$, or the minimum difference in tour length resulting from an edge swap. Perhaps there are some other well known results on the distribution of tour lengths that might be related. I would be grateful for any references that come to mind on topics related to this question.






      combinatorics graph-theory discrete-optimization




      combinatorics graph-theory discrete-optimization






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      share|cite|improve this question













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      edited Dec 1 '18 at 18:55







      axplusbu

















      asked Nov 27 '18 at 3:28









      axplusbuaxplusbu

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          This is not a complete answer, but since there are no other answers: if we have 4 points arranged in a square, this gives $d(h_2)/d(h_1)=frac{2}{1+sqrt{2}} $, and I have a feeling this is the best you can do.



          I don't think that the minimum distance between any pair is the right place to look: it seems intuitive two swap the two edges adjacent to the shortest edge on $h_1$, but the increase in total length depends also on the angles between those edges, so there may be a "smallest" swap that does not involve the shortest edge.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Thanks, puck! I understand your idea about the square graph. However, I am still looking for a more general condition.
            $endgroup$
            – axplusbu
            Dec 1 '18 at 18:58










          • $begingroup$
            Yes, I understand. What I meant is that I think that this is an extremal example, and $2/(1+sqrt{2})$ is going to be the best upper bound in general. It also seems like $h_2$ will differ from $h_1$ by a swap of one pair of edges. I can't think of an example involving a swap with more than 2 edges. This is all just conjecturing, though. I don't see a straightforward way to prove this. It's a fun problem!
            $endgroup$
            – Puck Rombach
            Dec 1 '18 at 21:48











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          1 Answer
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          $begingroup$

          This is not a complete answer, but since there are no other answers: if we have 4 points arranged in a square, this gives $d(h_2)/d(h_1)=frac{2}{1+sqrt{2}} $, and I have a feeling this is the best you can do.



          I don't think that the minimum distance between any pair is the right place to look: it seems intuitive two swap the two edges adjacent to the shortest edge on $h_1$, but the increase in total length depends also on the angles between those edges, so there may be a "smallest" swap that does not involve the shortest edge.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Thanks, puck! I understand your idea about the square graph. However, I am still looking for a more general condition.
            $endgroup$
            – axplusbu
            Dec 1 '18 at 18:58










          • $begingroup$
            Yes, I understand. What I meant is that I think that this is an extremal example, and $2/(1+sqrt{2})$ is going to be the best upper bound in general. It also seems like $h_2$ will differ from $h_1$ by a swap of one pair of edges. I can't think of an example involving a swap with more than 2 edges. This is all just conjecturing, though. I don't see a straightforward way to prove this. It's a fun problem!
            $endgroup$
            – Puck Rombach
            Dec 1 '18 at 21:48
















          0












          $begingroup$

          This is not a complete answer, but since there are no other answers: if we have 4 points arranged in a square, this gives $d(h_2)/d(h_1)=frac{2}{1+sqrt{2}} $, and I have a feeling this is the best you can do.



          I don't think that the minimum distance between any pair is the right place to look: it seems intuitive two swap the two edges adjacent to the shortest edge on $h_1$, but the increase in total length depends also on the angles between those edges, so there may be a "smallest" swap that does not involve the shortest edge.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Thanks, puck! I understand your idea about the square graph. However, I am still looking for a more general condition.
            $endgroup$
            – axplusbu
            Dec 1 '18 at 18:58










          • $begingroup$
            Yes, I understand. What I meant is that I think that this is an extremal example, and $2/(1+sqrt{2})$ is going to be the best upper bound in general. It also seems like $h_2$ will differ from $h_1$ by a swap of one pair of edges. I can't think of an example involving a swap with more than 2 edges. This is all just conjecturing, though. I don't see a straightforward way to prove this. It's a fun problem!
            $endgroup$
            – Puck Rombach
            Dec 1 '18 at 21:48














          0












          0








          0





          $begingroup$

          This is not a complete answer, but since there are no other answers: if we have 4 points arranged in a square, this gives $d(h_2)/d(h_1)=frac{2}{1+sqrt{2}} $, and I have a feeling this is the best you can do.



          I don't think that the minimum distance between any pair is the right place to look: it seems intuitive two swap the two edges adjacent to the shortest edge on $h_1$, but the increase in total length depends also on the angles between those edges, so there may be a "smallest" swap that does not involve the shortest edge.






          share|cite|improve this answer









          $endgroup$



          This is not a complete answer, but since there are no other answers: if we have 4 points arranged in a square, this gives $d(h_2)/d(h_1)=frac{2}{1+sqrt{2}} $, and I have a feeling this is the best you can do.



          I don't think that the minimum distance between any pair is the right place to look: it seems intuitive two swap the two edges adjacent to the shortest edge on $h_1$, but the increase in total length depends also on the angles between those edges, so there may be a "smallest" swap that does not involve the shortest edge.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Nov 28 '18 at 21:46









          Puck RombachPuck Rombach

          1266




          1266












          • $begingroup$
            Thanks, puck! I understand your idea about the square graph. However, I am still looking for a more general condition.
            $endgroup$
            – axplusbu
            Dec 1 '18 at 18:58










          • $begingroup$
            Yes, I understand. What I meant is that I think that this is an extremal example, and $2/(1+sqrt{2})$ is going to be the best upper bound in general. It also seems like $h_2$ will differ from $h_1$ by a swap of one pair of edges. I can't think of an example involving a swap with more than 2 edges. This is all just conjecturing, though. I don't see a straightforward way to prove this. It's a fun problem!
            $endgroup$
            – Puck Rombach
            Dec 1 '18 at 21:48


















          • $begingroup$
            Thanks, puck! I understand your idea about the square graph. However, I am still looking for a more general condition.
            $endgroup$
            – axplusbu
            Dec 1 '18 at 18:58










          • $begingroup$
            Yes, I understand. What I meant is that I think that this is an extremal example, and $2/(1+sqrt{2})$ is going to be the best upper bound in general. It also seems like $h_2$ will differ from $h_1$ by a swap of one pair of edges. I can't think of an example involving a swap with more than 2 edges. This is all just conjecturing, though. I don't see a straightforward way to prove this. It's a fun problem!
            $endgroup$
            – Puck Rombach
            Dec 1 '18 at 21:48
















          $begingroup$
          Thanks, puck! I understand your idea about the square graph. However, I am still looking for a more general condition.
          $endgroup$
          – axplusbu
          Dec 1 '18 at 18:58




          $begingroup$
          Thanks, puck! I understand your idea about the square graph. However, I am still looking for a more general condition.
          $endgroup$
          – axplusbu
          Dec 1 '18 at 18:58












          $begingroup$
          Yes, I understand. What I meant is that I think that this is an extremal example, and $2/(1+sqrt{2})$ is going to be the best upper bound in general. It also seems like $h_2$ will differ from $h_1$ by a swap of one pair of edges. I can't think of an example involving a swap with more than 2 edges. This is all just conjecturing, though. I don't see a straightforward way to prove this. It's a fun problem!
          $endgroup$
          – Puck Rombach
          Dec 1 '18 at 21:48




          $begingroup$
          Yes, I understand. What I meant is that I think that this is an extremal example, and $2/(1+sqrt{2})$ is going to be the best upper bound in general. It also seems like $h_2$ will differ from $h_1$ by a swap of one pair of edges. I can't think of an example involving a swap with more than 2 edges. This is all just conjecturing, though. I don't see a straightforward way to prove this. It's a fun problem!
          $endgroup$
          – Puck Rombach
          Dec 1 '18 at 21:48


















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