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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    $begingroup$


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


      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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      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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          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
            50 mins ago






          • 1




            $begingroup$
            Bravo! (+1).... the key seems to be reversing chirality.
            $endgroup$
            – David G. Stork
            39 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
            17 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
            12 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
            32 mins ago










          • $begingroup$
            You may well be right. Could be that there's a Z/2Z obstruction hiding in here somewhere.
            $endgroup$
            – John Hughes
            23 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
            50 mins ago






          • 1




            $begingroup$
            Bravo! (+1).... the key seems to be reversing chirality.
            $endgroup$
            – David G. Stork
            39 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
            17 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
            12 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
            50 mins ago






          • 1




            $begingroup$
            Bravo! (+1).... the key seems to be reversing chirality.
            $endgroup$
            – David G. Stork
            39 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
            17 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
            12 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 50 mins ago

























          answered 56 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
            50 mins ago






          • 1




            $begingroup$
            Bravo! (+1).... the key seems to be reversing chirality.
            $endgroup$
            – David G. Stork
            39 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
            17 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
            12 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
            50 mins ago






          • 1




            $begingroup$
            Bravo! (+1).... the key seems to be reversing chirality.
            $endgroup$
            – David G. Stork
            39 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
            17 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
            12 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
          50 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
          50 mins ago




          1




          1




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






          $begingroup$
          Bravo! (+1).... the key seems to be reversing chirality.
          $endgroup$
          – David G. Stork
          39 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
          17 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
          17 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
          12 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
          12 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 36 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
            32 mins ago










          • $begingroup$
            You may well be right. Could be that there's a Z/2Z obstruction hiding in here somewhere.
            $endgroup$
            – John Hughes
            23 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
            32 mins ago










          • $begingroup$
            You may well be right. Could be that there's a Z/2Z obstruction hiding in here somewhere.
            $endgroup$
            – John Hughes
            23 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 59 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
            32 mins ago










          • $begingroup$
            You may well be right. Could be that there's a Z/2Z obstruction hiding in here somewhere.
            $endgroup$
            – John Hughes
            23 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
            32 mins ago










          • $begingroup$
            You may well be right. Could be that there's a Z/2Z obstruction hiding in here somewhere.
            $endgroup$
            – John Hughes
            23 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
          32 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
          32 mins ago












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




          $begingroup$
          You may well be right. Could be that there's a Z/2Z obstruction hiding in here somewhere.
          $endgroup$
          – John Hughes
          23 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.
















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          Residenzschloss Arolsen

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          Residenzschloss Arolsen Residenzschloss Arolsen Entstehungszeit: 1710–1728 Erhaltungszustand: Vollständig Ständische Stellung: Grafen Ort: Bad Arolsen Geographische Lage 51° 22′ 51″  N , 9° 1′ 19″  O 51.380833333333 9.0219444444444 293 Koordinaten: 51° 22′ 51″  N , 9° 1′ 19″  O Höhe: 293  m ü. NHN Das Residenzschloss Arolsen ist ein barockes Schloss in Bad Arolsen im Kreis Waldeck-Frankenberg in Nordhessen (Deutschland). Das Schloss wurde als dreiflügelige Anlage gebaut, an die sich ein englischer Garten anschließt. Zentrales landschaftsarchitektonisches Gestaltungselement ist ein ausgedehntes Rondell. Inhaltsverzeichnis 1 Geschichte 1.1 Vorgängerbau 1.2 Bau und Geschichte des neuen Schlosses 1.3 Bibliothek 1.4 Finanzielle und staatsrechtliche Konsequenzen 1.5 Heutige Nutzung 2 Innenausstattung 3 Umgebung 3.1 Park 3.2 Wirtschaftsgebäude 4 Literatur 5 Weblinks 6...
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