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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

4

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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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asked 7 hours ago

Luiz FariasLuiz Farias

241

New contributor

Luiz Farias is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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4

$begingroup$

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

share|cite|improve this question

asked 7 hours ago

Luiz FariasLuiz Farias

241

New contributor

Luiz Farias is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

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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

share|cite|improve this question

asked 7 hours ago

Luiz FariasLuiz Farias

241

New contributor

Luiz Farias is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

$endgroup$

share|cite|improve this question

asked 7 hours ago

Luiz FariasLuiz Farias

241

New contributor

Luiz Farias is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

share|cite|improve this question

asked 7 hours ago

Luiz FariasLuiz Farias

241

New contributor

Luiz Farias is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

asked 7 hours ago

Luiz FariasLuiz Farias

241

New contributor

Luiz Farias is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

asked 7 hours ago

Luiz FariasLuiz Farias

241

3 Answers
3

active

oldest

votes

5

$begingroup$

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.

share|cite|improve this answer

edited 6 hours ago

answered 6 hours ago

HenryHenry

101k482170

$endgroup$

• 
  
  
  

  
  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
  6 hours ago
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Bravo! (+1).... the key seems to be reversing chirality.
  $endgroup$
  – David G. Stork
  6 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
  5 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
  5 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
  5 hours ago
  

  
  
  
  

add a comment | 

2

$begingroup$

(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.

share|cite|improve this answer

edited 6 hours ago

answered 7 hours ago

David G. StorkDavid G. Stork

12k41735

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Indeed - you seem to use $9$ being odd, though $10$ is not
  $endgroup$
  – Henry
  6 hours ago
  

  
  
  
  

add a comment | 

0

$begingroup$

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...

share|cite|improve this answer

answered 6 hours ago

John HughesJohn Hughes

65.2k24293

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  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
  5 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  You may well be right. Could be that there's a Z/2Z obstruction hiding in here somewhere.
  $endgroup$
  – John Hughes
  5 hours ago
  

  
  
  
  

add a comment | 

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5

$begingroup$

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.

share|cite|improve this answer

edited 6 hours ago

answered 6 hours ago

HenryHenry

101k482170

$endgroup$

• 
  
  
  

  
  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
  6 hours ago
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Bravo! (+1).... the key seems to be reversing chirality.
  $endgroup$
  – David G. Stork
  6 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
  5 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
  5 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
  5 hours ago
  

  
  
  
  

add a comment | 

5

$begingroup$

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.

share|cite|improve this answer

edited 6 hours ago

answered 6 hours ago

HenryHenry

101k482170

$endgroup$

• 
  
  
  

  
  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
  6 hours ago
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Bravo! (+1).... the key seems to be reversing chirality.
  $endgroup$
  – David G. Stork
  6 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
  5 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
  5 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
  5 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.

share|cite|improve this answer

edited 6 hours ago

answered 6 hours ago

HenryHenry

101k482170

$endgroup$

share|cite|improve this answer

edited 6 hours ago

answered 6 hours ago

HenryHenry

101k482170

answered 6 hours ago

HenryHenry

101k482170

answered 6 hours ago

HenryHenry

101k482170

2

$begingroup$

(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.

share|cite|improve this answer

edited 6 hours ago

answered 7 hours ago

David G. StorkDavid G. Stork

12k41735

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Indeed - you seem to use $9$ being odd, though $10$ is not
  $endgroup$
  – Henry
  6 hours ago
  

  
  
  
  

add a comment | 

2

$begingroup$

(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.

share|cite|improve this answer

edited 6 hours ago

answered 7 hours ago

David G. StorkDavid G. Stork

12k41735

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Indeed - you seem to use $9$ being odd, though $10$ is not
  $endgroup$
  – Henry
  6 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.

share|cite|improve this answer

edited 6 hours ago

answered 7 hours ago

David G. StorkDavid G. Stork

12k41735

$endgroup$

share|cite|improve this answer

edited 6 hours ago

answered 7 hours ago

David G. StorkDavid G. Stork

12k41735

answered 7 hours ago

David G. StorkDavid G. Stork

12k41735

answered 7 hours ago

David G. StorkDavid G. Stork

12k41735

0

$begingroup$

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...

share|cite|improve this answer

answered 6 hours ago

John HughesJohn Hughes

65.2k24293

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  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
  5 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  You may well be right. Could be that there's a Z/2Z obstruction hiding in here somewhere.
  $endgroup$
  – John Hughes
  5 hours ago
  

  
  
  
  

add a comment | 

0

$begingroup$

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...

share|cite|improve this answer

answered 6 hours ago

John HughesJohn Hughes

65.2k24293

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  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
  5 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  You may well be right. Could be that there's a Z/2Z obstruction hiding in here somewhere.
  $endgroup$
  – John Hughes
  5 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...

share|cite|improve this answer

answered 6 hours ago

John HughesJohn Hughes

65.2k24293

$endgroup$

share|cite|improve this answer

answered 6 hours ago

John HughesJohn Hughes

65.2k24293

answered 6 hours ago

John HughesJohn Hughes

65.2k24293

answered 6 hours ago

John HughesJohn Hughes

65.2k24293