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
asked Nov 27 '18 at 3:28
axplusbuaxplusbu
32
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add a comment |
$begingroup$
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
asked Nov 27 '18 at 3:28
axplusbuaxplusbu
32
$endgroup$
add a comment |
$begingroup$
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
asked Nov 27 '18 at 3:28
axplusbuaxplusbu
32
$endgroup$
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
asked Nov 27 '18 at 3:28
axplusbuaxplusbu
32
asked Nov 27 '18 at 3:28
axplusbuaxplusbu
32
edited Dec 1 '18 at 18:55
axplusbu
asked Nov 27 '18 at 3:28
axplusbuaxplusbu
32
asked Nov 27 '18 at 3:28
axplusbuaxplusbu
32
asked Nov 27 '18 at 3:28
axplusbuaxplusbu
32
32
add a comment |
add a comment |
1 Answer
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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.
answered Nov 28 '18 at 21:46
Puck RombachPuck Rombach
1266
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$begingroup$
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
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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!
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– Puck Rombach
Dec 1 '18 at 21:48
add a comment |
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1 Answer
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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.
answered Nov 28 '18 at 21:46
Puck RombachPuck Rombach
1266
$endgroup$
$begingroup$
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
add a comment |
$begingroup$
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.
answered Nov 28 '18 at 21:46
Puck RombachPuck Rombach
1266
$endgroup$
$begingroup$
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
add a comment |
$begingroup$
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.
answered Nov 28 '18 at 21:46
Puck RombachPuck Rombach
1266
$endgroup$
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.
answered Nov 28 '18 at 21:46
Puck RombachPuck Rombach
1266
answered Nov 28 '18 at 21:46
Puck RombachPuck Rombach
1266
answered Nov 28 '18 at 21:46
Puck RombachPuck Rombach
1266
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
add a comment |
$begingroup$
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
add a comment |
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