Problem of parity - Can we draw a closed path made up of 20 line segments…












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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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    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







    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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      4












      4








      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







      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




      recreational-mathematics parity






      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.











      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.









      share|cite|improve this question




      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.
      Check out our Code of Conduct.









      asked 7 hours ago









      Luiz FariasLuiz Farias

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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.






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






















          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.



          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
            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
            6 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



















          2












          $begingroup$

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



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



          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
            7 hours 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
            6 hours ago










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









          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.



          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
            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
            6 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
















          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.



          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
            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
            6 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














          5












          5








          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.



          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 6 hours ago

























          answered 6 hours 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
            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
            6 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














          • 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
            6 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








          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
          6 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
          6 hours ago




          1




          1




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






          $begingroup$
          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
          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
          6 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






          $begingroup$
          @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




          $begingroup$
          @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











          2












          $begingroup$

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



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



          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
            7 hours ago
















          2












          $begingroup$

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



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



          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
            7 hours ago














          2












          2








          2





          $begingroup$

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



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



          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 6 hours ago

























          answered 7 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
            7 hours ago














          • 1




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








          1




          1




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




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










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










          • $begingroup$
            You may well be right. Could be that there's a Z/2Z obstruction hiding in here somewhere.
            $endgroup$
            – John Hughes
            6 hours 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 6 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
            6 hours ago










          • $begingroup$
            You may well be right. Could be that there's a Z/2Z obstruction hiding in here somewhere.
            $endgroup$
            – John Hughes
            6 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
            6 hours ago










          • $begingroup$
            You may well be right. Could be that there's a Z/2Z obstruction hiding in here somewhere.
            $endgroup$
            – John Hughes
            6 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
          6 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
          6 hours ago












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




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










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










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          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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