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













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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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    3












    $begingroup$


    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







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      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?






      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.
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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






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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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      share|cite|improve this question







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      asked 1 hour 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.






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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
            35 mins ago






          • 1




            $begingroup$
            Bravo! (+1).... the key seems to be reversing chirality.
            $endgroup$
            – David G. Stork
            23 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
            2 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
            16 mins ago










          • $begingroup$
            You may well be right. Could be that there's a Z/2Z obstruction hiding in here somewhere.
            $endgroup$
            – John Hughes
            8 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
            35 mins ago






          • 1




            $begingroup$
            Bravo! (+1).... the key seems to be reversing chirality.
            $endgroup$
            – David G. Stork
            23 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
            2 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
            35 mins ago






          • 1




            $begingroup$
            Bravo! (+1).... the key seems to be reversing chirality.
            $endgroup$
            – David G. Stork
            23 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
            2 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 35 mins ago

























          answered 41 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
            35 mins ago






          • 1




            $begingroup$
            Bravo! (+1).... the key seems to be reversing chirality.
            $endgroup$
            – David G. Stork
            23 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
            2 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
            35 mins ago






          • 1




            $begingroup$
            Bravo! (+1).... the key seems to be reversing chirality.
            $endgroup$
            – David G. Stork
            23 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
            2 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
          35 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
          35 mins ago




          1




          1




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





          $begingroup$
          Bravo! (+1).... the key seems to be reversing chirality.
          $endgroup$
          – David G. Stork
          23 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
          2 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
          2 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 21 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












          • 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
            16 mins ago










          • $begingroup$
            You may well be right. Could be that there's a Z/2Z obstruction hiding in here somewhere.
            $endgroup$
            – John Hughes
            8 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
            16 mins ago










          • $begingroup$
            You may well be right. Could be that there's a Z/2Z obstruction hiding in here somewhere.
            $endgroup$
            – John Hughes
            8 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 44 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
            16 mins ago










          • $begingroup$
            You may well be right. Could be that there's a Z/2Z obstruction hiding in here somewhere.
            $endgroup$
            – John Hughes
            8 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
            16 mins ago










          • $begingroup$
            You may well be right. Could be that there's a Z/2Z obstruction hiding in here somewhere.
            $endgroup$
            – John Hughes
            8 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
          16 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
          16 mins ago












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




          $begingroup$
          You may well be right. Could be that there's a Z/2Z obstruction hiding in here somewhere.
          $endgroup$
          – John Hughes
          8 mins 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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          मुम्बईबौद्ध तीर्थ स्थलभीमराव आंबेडकर को समर्पित स्मारक मुंबईदादरभारतीय संविधानडॉ भीमराव आंबेडकरबौद्ध धर्मदिल्लीमहापरिनिर्वाणमुंबईभदन्त आनन्द कौशल्यायनअरबी ...
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