Problem of parity - Can we draw a closed path made up of 20 line segments… [on hold]What am I getting for Christmas? Secret Santa and Graph theoryReturn of the lost ant 3DVariation of the opaque forest problem (a.k.a farmyard problem)A closed path is made up of 11 line segments. Can one line, not containing a vertex of the path, intersect each of its segments?Connecting $1997$ points in the plane- what am I missing?How many paths are there from point P to point Q if each step has to go closer to point Q.A problem involving divisibility , parity and extremely clever thinkingHow to go out from a circular forest if we are lost? Not the straight line?Does finding the line of tightest packing in a packing problem help?Cover the plane with closed disks
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Problem of parity - Can we draw a closed path made up of 20 line segments… [on hold]
What am I getting for Christmas? Secret Santa and Graph theoryReturn of the lost ant 3DVariation of the opaque forest problem (a.k.a farmyard problem)A closed path is made up of 11 line segments. Can one line, not containing a vertex of the path, intersect each of its segments?Connecting $1997$ points in the plane- what am I missing?How many paths are there from point P to point Q if each step has to go closer to point Q.A problem involving divisibility , parity and extremely clever thinkingHow to go out from a circular forest if we are lost? Not the straight line?Does finding the line of tightest packing in a packing problem help?Cover the plane with closed disks
$begingroup$
Can we draw a closed path made up of 20 line segments, each of which intersects exactly one of the other segments?
recreational-mathematics parity
New contributor
$endgroup$
put on hold as off-topic by José Carlos Santos, Javi, user21820, Xander Henderson, RRL 1 hour ago
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – José Carlos Santos, Javi, user21820, Xander Henderson, RRL
add a comment |
$begingroup$
Can we draw a closed path made up of 20 line segments, each of which intersects exactly one of the other segments?
recreational-mathematics parity
New contributor
$endgroup$
put on hold as off-topic by José Carlos Santos, Javi, user21820, Xander Henderson, RRL 1 hour ago
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – José Carlos Santos, Javi, user21820, Xander Henderson, RRL
add a comment |
$begingroup$
Can we draw a closed path made up of 20 line segments, each of which intersects exactly one of the other segments?
recreational-mathematics parity
New contributor
$endgroup$
Can we draw a closed path made up of 20 line segments, each of which intersects exactly one of the other segments?
recreational-mathematics parity
recreational-mathematics parity
New contributor
New contributor
New contributor
asked 20 hours ago
Luiz FariasLuiz Farias
281
281
New contributor
New contributor
put on hold as off-topic by José Carlos Santos, Javi, user21820, Xander Henderson, RRL 1 hour ago
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – José Carlos Santos, Javi, user21820, Xander Henderson, RRL
put on hold as off-topic by José Carlos Santos, Javi, user21820, Xander Henderson, RRL 1 hour ago
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – José Carlos Santos, Javi, user21820, Xander Henderson, RRL
add a comment |
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
David G. Stork's example with $18$ points and edges can easily be changed into an example with $10$ points and edges based on a pentagon inside another pentagon with alternating links. So take two of those $10$ solutions, one inside the other, and then join them appropriately to get something like this with $20$ points and edges.
$endgroup$
2
$begingroup$
Interesting that it has to "reverse direction"; I wonder if there's a winding-number argument to show something like this must be true...but I'm too groggy to work one out.
$endgroup$
– John Hughes
18 hours ago
1
$begingroup$
Bravo! (+1).... the key seems to be reversing chirality.
$endgroup$
– David G. Stork
18 hours ago
$begingroup$
It would be similarly possible to combine $6$ and $14$ solutions, and to have the sub-solutions next to each other rather than one inside the other
$endgroup$
– Henry
18 hours ago
$begingroup$
@Henry: Can you write code (Mathematica?) to generate a solution given $n = 2k$? That would be incredible. (I wrote code for my $n = 18$ "solution.")
$endgroup$
– David G. Stork
18 hours ago
$begingroup$
@DavidG.Stork - I am afraid no as I do not do Mathematica. But the answer should be realtively simple: if $k$ is odd (and at least $3$) use your solution, while if $k$ is even (and at least $6$) then split it into two odd numbers (each at least $3$) and use your solution on each, finally adjusting to join them. This means I do not have a solution for $k=4$, i.e. for $n=8$
$endgroup$
– Henry
17 hours ago
add a comment |
$begingroup$
(I assume there can be no crossings at vertices or corners.)
Here is one solution for $18$ (and @Henry, below, generalizes to $20$):
Since each segment is crossed by exactly one other segment, we can think of the problem as having 10 Xs that have to be linked without crossing.
$endgroup$
1
$begingroup$
Indeed - you seem to use $9$ being odd, though $10$ is not
$endgroup$
– Henry
19 hours ago
add a comment |
$begingroup$
You can certainly do it if your drawing is on a torus: draw a decagon that goes "through the hole"; then draw a zigzag (like the one in your picture) that crosses each edge of the decagon once. The two ends of the zigzag will end up on opposite "sides" of the original decagon, but can be joined "around the back". By converting the situation to one involving a "square donut" (akin to this one) you can probably do this all with straight lines, although that may be easier if the cross-section is a pentagon rather than a square...
$endgroup$
$begingroup$
I wonder if your "square donut" will force kinks in lines, thereby breaking the conditions of the problem. Possible... but not certain...
$endgroup$
– David G. Stork
18 hours ago
$begingroup$
You may well be right. Could be that there's a Z/2Z obstruction hiding in here somewhere.
$endgroup$
– John Hughes
18 hours ago
add a comment |
3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
David G. Stork's example with $18$ points and edges can easily be changed into an example with $10$ points and edges based on a pentagon inside another pentagon with alternating links. So take two of those $10$ solutions, one inside the other, and then join them appropriately to get something like this with $20$ points and edges.
$endgroup$
2
$begingroup$
Interesting that it has to "reverse direction"; I wonder if there's a winding-number argument to show something like this must be true...but I'm too groggy to work one out.
$endgroup$
– John Hughes
18 hours ago
1
$begingroup$
Bravo! (+1).... the key seems to be reversing chirality.
$endgroup$
– David G. Stork
18 hours ago
$begingroup$
It would be similarly possible to combine $6$ and $14$ solutions, and to have the sub-solutions next to each other rather than one inside the other
$endgroup$
– Henry
18 hours ago
$begingroup$
@Henry: Can you write code (Mathematica?) to generate a solution given $n = 2k$? That would be incredible. (I wrote code for my $n = 18$ "solution.")
$endgroup$
– David G. Stork
18 hours ago
$begingroup$
@DavidG.Stork - I am afraid no as I do not do Mathematica. But the answer should be realtively simple: if $k$ is odd (and at least $3$) use your solution, while if $k$ is even (and at least $6$) then split it into two odd numbers (each at least $3$) and use your solution on each, finally adjusting to join them. This means I do not have a solution for $k=4$, i.e. for $n=8$
$endgroup$
– Henry
17 hours ago
add a comment |
$begingroup$
David G. Stork's example with $18$ points and edges can easily be changed into an example with $10$ points and edges based on a pentagon inside another pentagon with alternating links. So take two of those $10$ solutions, one inside the other, and then join them appropriately to get something like this with $20$ points and edges.
$endgroup$
2
$begingroup$
Interesting that it has to "reverse direction"; I wonder if there's a winding-number argument to show something like this must be true...but I'm too groggy to work one out.
$endgroup$
– John Hughes
18 hours ago
1
$begingroup$
Bravo! (+1).... the key seems to be reversing chirality.
$endgroup$
– David G. Stork
18 hours ago
$begingroup$
It would be similarly possible to combine $6$ and $14$ solutions, and to have the sub-solutions next to each other rather than one inside the other
$endgroup$
– Henry
18 hours ago
$begingroup$
@Henry: Can you write code (Mathematica?) to generate a solution given $n = 2k$? That would be incredible. (I wrote code for my $n = 18$ "solution.")
$endgroup$
– David G. Stork
18 hours ago
$begingroup$
@DavidG.Stork - I am afraid no as I do not do Mathematica. But the answer should be realtively simple: if $k$ is odd (and at least $3$) use your solution, while if $k$ is even (and at least $6$) then split it into two odd numbers (each at least $3$) and use your solution on each, finally adjusting to join them. This means I do not have a solution for $k=4$, i.e. for $n=8$
$endgroup$
– Henry
17 hours ago
add a comment |
$begingroup$
David G. Stork's example with $18$ points and edges can easily be changed into an example with $10$ points and edges based on a pentagon inside another pentagon with alternating links. So take two of those $10$ solutions, one inside the other, and then join them appropriately to get something like this with $20$ points and edges.
$endgroup$
David G. Stork's example with $18$ points and edges can easily be changed into an example with $10$ points and edges based on a pentagon inside another pentagon with alternating links. So take two of those $10$ solutions, one inside the other, and then join them appropriately to get something like this with $20$ points and edges.
edited 18 hours ago
answered 19 hours ago
HenryHenry
101k482170
101k482170
2
$begingroup$
Interesting that it has to "reverse direction"; I wonder if there's a winding-number argument to show something like this must be true...but I'm too groggy to work one out.
$endgroup$
– John Hughes
18 hours ago
1
$begingroup$
Bravo! (+1).... the key seems to be reversing chirality.
$endgroup$
– David G. Stork
18 hours ago
$begingroup$
It would be similarly possible to combine $6$ and $14$ solutions, and to have the sub-solutions next to each other rather than one inside the other
$endgroup$
– Henry
18 hours ago
$begingroup$
@Henry: Can you write code (Mathematica?) to generate a solution given $n = 2k$? That would be incredible. (I wrote code for my $n = 18$ "solution.")
$endgroup$
– David G. Stork
18 hours ago
$begingroup$
@DavidG.Stork - I am afraid no as I do not do Mathematica. But the answer should be realtively simple: if $k$ is odd (and at least $3$) use your solution, while if $k$ is even (and at least $6$) then split it into two odd numbers (each at least $3$) and use your solution on each, finally adjusting to join them. This means I do not have a solution for $k=4$, i.e. for $n=8$
$endgroup$
– Henry
17 hours ago
add a comment |
2
$begingroup$
Interesting that it has to "reverse direction"; I wonder if there's a winding-number argument to show something like this must be true...but I'm too groggy to work one out.
$endgroup$
– John Hughes
18 hours ago
1
$begingroup$
Bravo! (+1).... the key seems to be reversing chirality.
$endgroup$
– David G. Stork
18 hours ago
$begingroup$
It would be similarly possible to combine $6$ and $14$ solutions, and to have the sub-solutions next to each other rather than one inside the other
$endgroup$
– Henry
18 hours ago
$begingroup$
@Henry: Can you write code (Mathematica?) to generate a solution given $n = 2k$? That would be incredible. (I wrote code for my $n = 18$ "solution.")
$endgroup$
– David G. Stork
18 hours ago
$begingroup$
@DavidG.Stork - I am afraid no as I do not do Mathematica. But the answer should be realtively simple: if $k$ is odd (and at least $3$) use your solution, while if $k$ is even (and at least $6$) then split it into two odd numbers (each at least $3$) and use your solution on each, finally adjusting to join them. This means I do not have a solution for $k=4$, i.e. for $n=8$
$endgroup$
– Henry
17 hours ago
2
2
$begingroup$
Interesting that it has to "reverse direction"; I wonder if there's a winding-number argument to show something like this must be true...but I'm too groggy to work one out.
$endgroup$
– John Hughes
18 hours ago
$begingroup$
Interesting that it has to "reverse direction"; I wonder if there's a winding-number argument to show something like this must be true...but I'm too groggy to work one out.
$endgroup$
– John Hughes
18 hours ago
1
1
$begingroup$
Bravo! (+1).... the key seems to be reversing chirality.
$endgroup$
– David G. Stork
18 hours ago
$begingroup$
Bravo! (+1).... the key seems to be reversing chirality.
$endgroup$
– David G. Stork
18 hours ago
$begingroup$
It would be similarly possible to combine $6$ and $14$ solutions, and to have the sub-solutions next to each other rather than one inside the other
$endgroup$
– Henry
18 hours ago
$begingroup$
It would be similarly possible to combine $6$ and $14$ solutions, and to have the sub-solutions next to each other rather than one inside the other
$endgroup$
– Henry
18 hours ago
$begingroup$
@Henry: Can you write code (Mathematica?) to generate a solution given $n = 2k$? That would be incredible. (I wrote code for my $n = 18$ "solution.")
$endgroup$
– David G. Stork
18 hours ago
$begingroup$
@Henry: Can you write code (Mathematica?) to generate a solution given $n = 2k$? That would be incredible. (I wrote code for my $n = 18$ "solution.")
$endgroup$
– David G. Stork
18 hours ago
$begingroup$
@DavidG.Stork - I am afraid no as I do not do Mathematica. But the answer should be realtively simple: if $k$ is odd (and at least $3$) use your solution, while if $k$ is even (and at least $6$) then split it into two odd numbers (each at least $3$) and use your solution on each, finally adjusting to join them. This means I do not have a solution for $k=4$, i.e. for $n=8$
$endgroup$
– Henry
17 hours ago
$begingroup$
@DavidG.Stork - I am afraid no as I do not do Mathematica. But the answer should be realtively simple: if $k$ is odd (and at least $3$) use your solution, while if $k$ is even (and at least $6$) then split it into two odd numbers (each at least $3$) and use your solution on each, finally adjusting to join them. This means I do not have a solution for $k=4$, i.e. for $n=8$
$endgroup$
– Henry
17 hours ago
add a comment |
$begingroup$
(I assume there can be no crossings at vertices or corners.)
Here is one solution for $18$ (and @Henry, below, generalizes to $20$):
Since each segment is crossed by exactly one other segment, we can think of the problem as having 10 Xs that have to be linked without crossing.
$endgroup$
1
$begingroup$
Indeed - you seem to use $9$ being odd, though $10$ is not
$endgroup$
– Henry
19 hours ago
add a comment |
$begingroup$
(I assume there can be no crossings at vertices or corners.)
Here is one solution for $18$ (and @Henry, below, generalizes to $20$):
Since each segment is crossed by exactly one other segment, we can think of the problem as having 10 Xs that have to be linked without crossing.
$endgroup$
1
$begingroup$
Indeed - you seem to use $9$ being odd, though $10$ is not
$endgroup$
– Henry
19 hours ago
add a comment |
$begingroup$
(I assume there can be no crossings at vertices or corners.)
Here is one solution for $18$ (and @Henry, below, generalizes to $20$):
Since each segment is crossed by exactly one other segment, we can think of the problem as having 10 Xs that have to be linked without crossing.
$endgroup$
(I assume there can be no crossings at vertices or corners.)
Here is one solution for $18$ (and @Henry, below, generalizes to $20$):
Since each segment is crossed by exactly one other segment, we can think of the problem as having 10 Xs that have to be linked without crossing.
edited 18 hours ago
answered 19 hours ago
David G. StorkDavid G. Stork
12k41735
12k41735
1
$begingroup$
Indeed - you seem to use $9$ being odd, though $10$ is not
$endgroup$
– Henry
19 hours ago
add a comment |
1
$begingroup$
Indeed - you seem to use $9$ being odd, though $10$ is not
$endgroup$
– Henry
19 hours ago
1
1
$begingroup$
Indeed - you seem to use $9$ being odd, though $10$ is not
$endgroup$
– Henry
19 hours ago
$begingroup$
Indeed - you seem to use $9$ being odd, though $10$ is not
$endgroup$
– Henry
19 hours ago
add a comment |
$begingroup$
You can certainly do it if your drawing is on a torus: draw a decagon that goes "through the hole"; then draw a zigzag (like the one in your picture) that crosses each edge of the decagon once. The two ends of the zigzag will end up on opposite "sides" of the original decagon, but can be joined "around the back". By converting the situation to one involving a "square donut" (akin to this one) you can probably do this all with straight lines, although that may be easier if the cross-section is a pentagon rather than a square...
$endgroup$
$begingroup$
I wonder if your "square donut" will force kinks in lines, thereby breaking the conditions of the problem. Possible... but not certain...
$endgroup$
– David G. Stork
18 hours ago
$begingroup$
You may well be right. Could be that there's a Z/2Z obstruction hiding in here somewhere.
$endgroup$
– John Hughes
18 hours ago
add a comment |
$begingroup$
You can certainly do it if your drawing is on a torus: draw a decagon that goes "through the hole"; then draw a zigzag (like the one in your picture) that crosses each edge of the decagon once. The two ends of the zigzag will end up on opposite "sides" of the original decagon, but can be joined "around the back". By converting the situation to one involving a "square donut" (akin to this one) you can probably do this all with straight lines, although that may be easier if the cross-section is a pentagon rather than a square...
$endgroup$
$begingroup$
I wonder if your "square donut" will force kinks in lines, thereby breaking the conditions of the problem. Possible... but not certain...
$endgroup$
– David G. Stork
18 hours ago
$begingroup$
You may well be right. Could be that there's a Z/2Z obstruction hiding in here somewhere.
$endgroup$
– John Hughes
18 hours ago
add a comment |
$begingroup$
You can certainly do it if your drawing is on a torus: draw a decagon that goes "through the hole"; then draw a zigzag (like the one in your picture) that crosses each edge of the decagon once. The two ends of the zigzag will end up on opposite "sides" of the original decagon, but can be joined "around the back". By converting the situation to one involving a "square donut" (akin to this one) you can probably do this all with straight lines, although that may be easier if the cross-section is a pentagon rather than a square...
$endgroup$
You can certainly do it if your drawing is on a torus: draw a decagon that goes "through the hole"; then draw a zigzag (like the one in your picture) that crosses each edge of the decagon once. The two ends of the zigzag will end up on opposite "sides" of the original decagon, but can be joined "around the back". By converting the situation to one involving a "square donut" (akin to this one) you can probably do this all with straight lines, although that may be easier if the cross-section is a pentagon rather than a square...
answered 19 hours ago
John HughesJohn Hughes
65.2k24293
65.2k24293
$begingroup$
I wonder if your "square donut" will force kinks in lines, thereby breaking the conditions of the problem. Possible... but not certain...
$endgroup$
– David G. Stork
18 hours ago
$begingroup$
You may well be right. Could be that there's a Z/2Z obstruction hiding in here somewhere.
$endgroup$
– John Hughes
18 hours ago
add a comment |
$begingroup$
I wonder if your "square donut" will force kinks in lines, thereby breaking the conditions of the problem. Possible... but not certain...
$endgroup$
– David G. Stork
18 hours ago
$begingroup$
You may well be right. Could be that there's a Z/2Z obstruction hiding in here somewhere.
$endgroup$
– John Hughes
18 hours ago
$begingroup$
I wonder if your "square donut" will force kinks in lines, thereby breaking the conditions of the problem. Possible... but not certain...
$endgroup$
– David G. Stork
18 hours ago
$begingroup$
I wonder if your "square donut" will force kinks in lines, thereby breaking the conditions of the problem. Possible... but not certain...
$endgroup$
– David G. Stork
18 hours ago
$begingroup$
You may well be right. Could be that there's a Z/2Z obstruction hiding in here somewhere.
$endgroup$
– John Hughes
18 hours ago
$begingroup$
You may well be right. Could be that there's a Z/2Z obstruction hiding in here somewhere.
$endgroup$
– John Hughes
18 hours ago
add a comment |