Problem of parity - Can we draw a closed path made up of 20 line segments…What am I getting for Christmas?...
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Problem of parity - Can we draw a closed path made up of 20 line segments…
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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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
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$begingroup$
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
$endgroup$
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 1 hour ago
Luiz FariasLuiz Farias
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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.
$endgroup$
1
$begingroup$
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
42 mins ago
1
$begingroup$
Bravo! (+1).... the key seems to be reversing chirality.
$endgroup$
– David G. Stork
31 mins 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
9 mins 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
4 mins 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
1 hour 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
24 mins ago
$begingroup$
You may well be right. Could be that there's a Z/2Z obstruction hiding in here somewhere.
$endgroup$
– John Hughes
15 mins ago
add a comment |
Your Answer
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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.
$endgroup$
1
$begingroup$
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
42 mins ago
1
$begingroup$
Bravo! (+1).... the key seems to be reversing chirality.
$endgroup$
– David G. Stork
31 mins 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
9 mins 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
4 mins 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.
$endgroup$
1
$begingroup$
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
42 mins ago
1
$begingroup$
Bravo! (+1).... the key seems to be reversing chirality.
$endgroup$
– David G. Stork
31 mins 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
9 mins 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
4 mins 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 42 mins ago
answered 48 mins ago
HenryHenry
101k482170
101k482170
1
$begingroup$
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
42 mins ago
1
$begingroup$
Bravo! (+1).... the key seems to be reversing chirality.
$endgroup$
– David G. Stork
31 mins 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
9 mins 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
4 mins ago
add a comment |
1
$begingroup$
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
42 mins ago
1
$begingroup$
Bravo! (+1).... the key seems to be reversing chirality.
$endgroup$
– David G. Stork
31 mins 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
9 mins 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
4 mins ago
1
1
$begingroup$
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
42 mins 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
42 mins ago
1
1
$begingroup$
Bravo! (+1).... the key seems to be reversing chirality.
$endgroup$
– David G. Stork
31 mins ago
$begingroup$
Bravo! (+1).... the key seems to be reversing chirality.
$endgroup$
– David G. Stork
31 mins 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
9 mins 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
9 mins 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
4 mins 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
4 mins 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
1 hour 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
1 hour 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 28 mins ago
answered 1 hour ago
David G. StorkDavid G. Stork
12k41735
12k41735
1
$begingroup$
Indeed - you seem to use $9$ being odd, though $10$ is not
$endgroup$
– Henry
1 hour ago
add a comment |
1
$begingroup$
Indeed - you seem to use $9$ being odd, though $10$ is not
$endgroup$
– Henry
1 hour ago
1
1
$begingroup$
Indeed - you seem to use $9$ being odd, though $10$ is not
$endgroup$
– Henry
1 hour ago
$begingroup$
Indeed - you seem to use $9$ being odd, though $10$ is not
$endgroup$
– Henry
1 hour 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
24 mins ago
$begingroup$
You may well be right. Could be that there's a Z/2Z obstruction hiding in here somewhere.
$endgroup$
– John Hughes
15 mins 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
24 mins ago
$begingroup$
You may well be right. Could be that there's a Z/2Z obstruction hiding in here somewhere.
$endgroup$
– John Hughes
15 mins 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 51 mins 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
24 mins ago
$begingroup$
You may well be right. Could be that there's a Z/2Z obstruction hiding in here somewhere.
$endgroup$
– John Hughes
15 mins 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
24 mins ago
$begingroup$
You may well be right. Could be that there's a Z/2Z obstruction hiding in here somewhere.
$endgroup$
– John Hughes
15 mins 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
24 mins 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
24 mins ago
$begingroup$
You may well be right. Could be that there's a Z/2Z obstruction hiding in here somewhere.
$endgroup$
– John Hughes
15 mins ago
$begingroup$
You may well be right. Could be that there's a Z/2Z obstruction hiding in here somewhere.
$endgroup$
– John Hughes
15 mins ago
add a comment |
Luiz Farias is a new contributor. Be nice, and check out our Code of Conduct.
Luiz Farias is a new contributor. Be nice, and check out our Code of Conduct.
Luiz Farias is a new contributor. Be nice, and check out our Code of Conduct.
Luiz Farias is a new contributor. Be nice, and check out our Code of Conduct.
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