Zrjydtj

Test whether all array elements are factors of a number

How could an uplifted falcon's brain work?

"to be prejudice towards/against someone" vs "to be prejudiced against/towards someone"

What do the dots in this tr command do: tr .............A-Z A-ZA-Z <<< "JVPQBOV" (with 13 dots)

Watching something be written to a file live with tail

Has the BBC provided arguments for saying Brexit being cancelled is unlikely?

Is it unprofessional to ask if a job posting on GlassDoor is real?

To string or not to string

What do you call a Matrix-like slowdown and camera movement effect?

Can a Warlock become Neutral Good?

Smoothness of finite-dimensional functional calculus

Why Is Death Allowed In the Matrix?

tikz: show 0 at the axis origin

Is it important to consider tone, melody, and musical form while writing a song?

Is it legal for company to use my work email to pretend I still work there?

Which models of the Boeing 737 are still in production?

A newer friend of my brother's gave him a load of baseball cards that are supposedly extremely valuable. Is this a scam?

Why does Kotter return in Welcome Back Kotter?

What does "Puller Prush Person" mean?

Arthur Somervell: 1000 Exercises - Meaning of this notation

Risk of getting Chronic Wasting Disease (CWD) in the United States?

Have astronauts in space suits ever taken selfies? If so, how?

What would happen to a modern skyscraper if it rains micro blackholes?

I’m planning on buying a laser printer but concerned about the life cycle of toner in the machine

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

2

$begingroup$

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 1 hour ago

Luiz FariasLuiz Farias

111

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.

$endgroup$

add a comment | 

2

$begingroup$

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 1 hour ago

Luiz FariasLuiz Farias

111

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.

$endgroup$

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 1 hour ago

Luiz FariasLuiz Farias

111

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 1 hour ago

Luiz FariasLuiz Farias

111

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 1 hour ago

Luiz FariasLuiz Farias

111

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 1 hour ago

Luiz FariasLuiz Farias

111

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 1 hour ago

Luiz FariasLuiz Farias

111

3 Answers
3

active

oldest

votes

3

$begingroup$

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 30 mins ago

answered 37 mins 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
  30 mins ago
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

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

  
  
  
  

add a comment | 

1

$begingroup$

(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 17 mins ago

answered 1 hour ago

David G. StorkDavid G. Stork

12k41735

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Indeed - you seem to use $9$ being odd, though $10$ is not
  $endgroup$
  – Henry
  1 hour 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 39 mins 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
  12 mins ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

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

  
  
  
  

add a comment | 

Your Answer

StackExchange.ifUsing("editor", function ()
return StackExchange.using("mathjaxEditing", function ()
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix)
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
);
);
, "mathjax-editing");

StackExchange.ready(function()
var channelOptions =
tags: "".split(" "),
id: "69"
;
initTagRenderer("".split(" "), "".split(" "), channelOptions);

StackExchange.using("externalEditor", function()
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled)
StackExchange.using("snippets", function()
createEditor();
);

else
createEditor();

);

function createEditor()
StackExchange.prepareEditor(
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader:
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
,
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
);



);

Luiz Farias is a new contributor. Be nice, and check out our Code of Conduct.

draft saved

draft discarded

Sign up or log in

StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);

Sign up using Google

Sign up using Facebook

Sign up using Email and Password

Post as a guest

Name

Email

Required, but never shown

StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3177522%2fproblem-of-parity-can-we-draw-a-closed-path-made-up-of-20-line-segments%23new-answer', 'question_page');

);

Post as a guest

Name

Email

Required, but never shown

3

$begingroup$

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 30 mins ago

answered 37 mins 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
  30 mins ago
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

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

  
  
  
  

add a comment | 

3

$begingroup$

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 30 mins ago

answered 37 mins 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
  30 mins ago
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Bravo! (+1).... the key seems to be reversing chirality.
  $endgroup$
  – David G. Stork
  19 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.

share|cite|improve this answer

edited 30 mins ago

answered 37 mins ago

HenryHenry

101k482170

$endgroup$

share|cite|improve this answer

edited 30 mins ago

answered 37 mins ago

HenryHenry

101k482170

answered 37 mins ago

HenryHenry

101k482170

answered 37 mins ago

HenryHenry

101k482170

1

$begingroup$

(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 17 mins ago

answered 1 hour ago

David G. StorkDavid G. Stork

12k41735

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

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

  
  
  
  

add a comment | 

1

$begingroup$

(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 17 mins ago

answered 1 hour ago

David G. StorkDavid G. Stork

12k41735

$endgroup$

• 
  
  
  

  
  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.

share|cite|improve this answer

edited 17 mins ago

answered 1 hour ago

David G. StorkDavid G. Stork

12k41735

$endgroup$

share|cite|improve this answer

edited 17 mins ago

answered 1 hour ago

David G. StorkDavid G. Stork

12k41735

answered 1 hour ago

David G. StorkDavid G. Stork

12k41735

answered 1 hour 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 39 mins 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
  12 mins ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  You may well be right. Could be that there's a Z/2Z obstruction hiding in here somewhere.
  $endgroup$
  – John Hughes
  4 mins 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 39 mins 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
  12 mins ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  You may well be right. Could be that there's a Z/2Z obstruction hiding in here somewhere.
  $endgroup$
  – John Hughes
  4 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...

share|cite|improve this answer

answered 39 mins ago

John HughesJohn Hughes

65.2k24293

$endgroup$

share|cite|improve this answer

answered 39 mins ago

John HughesJohn Hughes

65.2k24293

answered 39 mins ago

John HughesJohn Hughes

65.2k24293

answered 39 mins ago

John HughesJohn Hughes

65.2k24293