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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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













3












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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?










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    3












    $begingroup$


    Can we draw a closed path made up of 20 line segments, each of which intersects exactly one of the other segments?










    share|cite|improve this question







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    Luiz Farias is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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      3












      3








      3





      $begingroup$


      Can we draw a closed path made up of 20 line segments, each of which intersects exactly one of the other segments?










      share|cite|improve this question







      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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      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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      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







      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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      asked 2 hours ago









      Luiz FariasLuiz Farias

      161




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      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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      Check out our Code of Conduct.




















          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.



          enter image description here






          share|cite|improve this answer











          $endgroup$








          • 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
            1 hour ago






          • 1




            $begingroup$
            Bravo! (+1).... the key seems to be reversing chirality.
            $endgroup$
            – David G. Stork
            59 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
            37 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
            32 mins 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
            11 mins ago


















          1












          $begingroup$

          (I assume there can be no crossings at vertices or corners.)



          Here is one solution for $18$ (and @Henry, below, generalizes to $20$):



          enter image description here



          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











          $endgroup$








          • 1




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


















          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









          $endgroup$












          • $begingroup$
            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
            52 mins ago










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











          Your Answer





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          3 Answers
          3






          active

          oldest

          votes








          3 Answers
          3






          active

          oldest

          votes









          active

          oldest

          votes






          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.



          enter image description here






          share|cite|improve this answer











          $endgroup$








          • 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
            1 hour ago






          • 1




            $begingroup$
            Bravo! (+1).... the key seems to be reversing chirality.
            $endgroup$
            – David G. Stork
            59 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
            37 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
            32 mins 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
            11 mins ago















          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.



          enter image description here






          share|cite|improve this answer











          $endgroup$








          • 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
            1 hour ago






          • 1




            $begingroup$
            Bravo! (+1).... the key seems to be reversing chirality.
            $endgroup$
            – David G. Stork
            59 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
            37 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
            32 mins 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
            11 mins ago













          3












          3








          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.



          enter image description here






          share|cite|improve this answer











          $endgroup$



          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.



          enter image description here







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited 1 hour ago

























          answered 1 hour 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
            1 hour ago






          • 1




            $begingroup$
            Bravo! (+1).... the key seems to be reversing chirality.
            $endgroup$
            – David G. Stork
            59 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
            37 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
            32 mins 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
            11 mins ago












          • 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
            1 hour ago






          • 1




            $begingroup$
            Bravo! (+1).... the key seems to be reversing chirality.
            $endgroup$
            – David G. Stork
            59 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
            37 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
            32 mins 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
            11 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
          1 hour 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
          1 hour ago




          1




          1




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





          $begingroup$
          Bravo! (+1).... the key seems to be reversing chirality.
          $endgroup$
          – David G. Stork
          59 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
          37 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
          37 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
          32 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
          32 mins 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
          11 mins 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
          11 mins ago











          1












          $begingroup$

          (I assume there can be no crossings at vertices or corners.)



          Here is one solution for $18$ (and @Henry, below, generalizes to $20$):



          enter image description here



          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











          $endgroup$








          • 1




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















          1












          $begingroup$

          (I assume there can be no crossings at vertices or corners.)



          Here is one solution for $18$ (and @Henry, below, generalizes to $20$):



          enter image description here



          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











          $endgroup$








          • 1




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













          1












          1








          1





          $begingroup$

          (I assume there can be no crossings at vertices or corners.)



          Here is one solution for $18$ (and @Henry, below, generalizes to $20$):



          enter image description here



          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











          $endgroup$



          (I assume there can be no crossings at vertices or corners.)



          Here is one solution for $18$ (and @Henry, below, generalizes to $20$):



          enter image description here



          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














          share|cite|improve this answer



          share|cite|improve this answer








          edited 57 mins ago

























          answered 2 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
            1 hour ago












          • 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











          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









          $endgroup$












          • $begingroup$
            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
            52 mins ago










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















          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









          $endgroup$












          • $begingroup$
            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
            52 mins ago










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













          0












          0








          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









          $endgroup$



          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...







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered 1 hour 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
            52 mins ago










          • $begingroup$
            You may well be right. Could be that there's a Z/2Z obstruction hiding in here somewhere.
            $endgroup$
            – John Hughes
            43 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
            52 mins ago










          • $begingroup$
            You may well be right. Could be that there's a Z/2Z obstruction hiding in here somewhere.
            $endgroup$
            – John Hughes
            43 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
          52 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
          52 mins ago












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




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










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