Finding $K_{3,3}$-subdivision when adding edge to maximal planar graph.












5












$begingroup$


I'm working on the following problem.




Show that adding a new edge to a maximal planar graph
of order at least 6 always produces subdivisions of both $K_5$ of $K_{3,3}$.




I had no trouble getting the subdivision of $K_5$ (I had to modify my attempt a bit based in the sketch of a solution I found here, since there were some cases I hadn't covered).



But I have trouble getting the subdivision of $K_{3,3}$.



So, with the notation from the link, I am here:



Let $G$ be maximal planar, $v_1,v_2$ two vertices; we want to build some edge between them and see this ends up with a graph having a subdivision of $K_{3,3}$.



We also have three $u-v$ vertex disjoint paths, say, $P_1,P_2,P_3$, each of them only having one neighbor of $v$, say, $u_1,u_2,u_3$. We also have that the neighbors of $v$ form a cycle since $G$ is a triangulation of the plane.
Also as they say, we have a new vertex $w$, which we may suppose is (given, if necessary, a rotation of the sphere which interchanges some faces) inside one of the regions bounded by the cycles determined by $P_i$ and $P_j$, say, $P_1$ and $P_2$. In this case, by Menger's theorem we have three paths to $u_1,u_2$ and $v_2$.
Anyways, we have the following sketch of the situation:
enter image description here
The curly 'edges' mean they are paths, the red dashed edge means it's the edge we want to add. In the solution they say that with this we get a subdivision of $K_{3,3}$ with partition given by $A={u,w,u_3}$ and $B={u_1,u_2,v}$. In the picture suppose the $u_1-u_2,u_2-u_3,u_3-u_1-$ paths are $zeta_1,zeta_2,zeta_3$. If the three paths from $w$ don't meet $zeta_2$ or $zeta_3$ we don't have much trouble and that's indeed a partition of those six vert which, with the edges and paths we already have and $v_1v_2$ make a subdivision of $K_{3,3}$. But if the paths from $w$ somehow meet $zeta_2$ or $zeta_3$ we may need to change the vertices which make the subdivision.



So, are those both cases possible? If they are, is there some way to handle them?



Edit: There is also this question but they only talk about how some solution (a bit different from mine) is wrong.






graph-theory planar-graph







share|cite|improve this question











$endgroup$

















    5












    $begingroup$


    I'm working on the following problem.




    Show that adding a new edge to a maximal planar graph
    of order at least 6 always produces subdivisions of both $K_5$ of $K_{3,3}$.




    I had no trouble getting the subdivision of $K_5$ (I had to modify my attempt a bit based in the sketch of a solution I found here, since there were some cases I hadn't covered).



    But I have trouble getting the subdivision of $K_{3,3}$.



    So, with the notation from the link, I am here:



    Let $G$ be maximal planar, $v_1,v_2$ two vertices; we want to build some edge between them and see this ends up with a graph having a subdivision of $K_{3,3}$.



    We also have three $u-v$ vertex disjoint paths, say, $P_1,P_2,P_3$, each of them only having one neighbor of $v$, say, $u_1,u_2,u_3$. We also have that the neighbors of $v$ form a cycle since $G$ is a triangulation of the plane.
    Also as they say, we have a new vertex $w$, which we may suppose is (given, if necessary, a rotation of the sphere which interchanges some faces) inside one of the regions bounded by the cycles determined by $P_i$ and $P_j$, say, $P_1$ and $P_2$. In this case, by Menger's theorem we have three paths to $u_1,u_2$ and $v_2$.
    Anyways, we have the following sketch of the situation:
    enter image description here
    The curly 'edges' mean they are paths, the red dashed edge means it's the edge we want to add. In the solution they say that with this we get a subdivision of $K_{3,3}$ with partition given by $A={u,w,u_3}$ and $B={u_1,u_2,v}$. In the picture suppose the $u_1-u_2,u_2-u_3,u_3-u_1-$ paths are $zeta_1,zeta_2,zeta_3$. If the three paths from $w$ don't meet $zeta_2$ or $zeta_3$ we don't have much trouble and that's indeed a partition of those six vert which, with the edges and paths we already have and $v_1v_2$ make a subdivision of $K_{3,3}$. But if the paths from $w$ somehow meet $zeta_2$ or $zeta_3$ we may need to change the vertices which make the subdivision.



    So, are those both cases possible? If they are, is there some way to handle them?



    Edit: There is also this question but they only talk about how some solution (a bit different from mine) is wrong.






    graph-theory planar-graph







    share|cite|improve this question











    $endgroup$















      5












      5








      5


      1



      $begingroup$


      I'm working on the following problem.




      Show that adding a new edge to a maximal planar graph
      of order at least 6 always produces subdivisions of both $K_5$ of $K_{3,3}$.




      I had no trouble getting the subdivision of $K_5$ (I had to modify my attempt a bit based in the sketch of a solution I found here, since there were some cases I hadn't covered).



      But I have trouble getting the subdivision of $K_{3,3}$.



      So, with the notation from the link, I am here:



      Let $G$ be maximal planar, $v_1,v_2$ two vertices; we want to build some edge between them and see this ends up with a graph having a subdivision of $K_{3,3}$.



      We also have three $u-v$ vertex disjoint paths, say, $P_1,P_2,P_3$, each of them only having one neighbor of $v$, say, $u_1,u_2,u_3$. We also have that the neighbors of $v$ form a cycle since $G$ is a triangulation of the plane.
      Also as they say, we have a new vertex $w$, which we may suppose is (given, if necessary, a rotation of the sphere which interchanges some faces) inside one of the regions bounded by the cycles determined by $P_i$ and $P_j$, say, $P_1$ and $P_2$. In this case, by Menger's theorem we have three paths to $u_1,u_2$ and $v_2$.
      Anyways, we have the following sketch of the situation:
      enter image description here
      The curly 'edges' mean they are paths, the red dashed edge means it's the edge we want to add. In the solution they say that with this we get a subdivision of $K_{3,3}$ with partition given by $A={u,w,u_3}$ and $B={u_1,u_2,v}$. In the picture suppose the $u_1-u_2,u_2-u_3,u_3-u_1-$ paths are $zeta_1,zeta_2,zeta_3$. If the three paths from $w$ don't meet $zeta_2$ or $zeta_3$ we don't have much trouble and that's indeed a partition of those six vert which, with the edges and paths we already have and $v_1v_2$ make a subdivision of $K_{3,3}$. But if the paths from $w$ somehow meet $zeta_2$ or $zeta_3$ we may need to change the vertices which make the subdivision.



      So, are those both cases possible? If they are, is there some way to handle them?



      Edit: There is also this question but they only talk about how some solution (a bit different from mine) is wrong.






      graph-theory planar-graph







      share|cite|improve this question











      $endgroup$




      I'm working on the following problem.




      Show that adding a new edge to a maximal planar graph
      of order at least 6 always produces subdivisions of both $K_5$ of $K_{3,3}$.




      I had no trouble getting the subdivision of $K_5$ (I had to modify my attempt a bit based in the sketch of a solution I found here, since there were some cases I hadn't covered).



      But I have trouble getting the subdivision of $K_{3,3}$.



      So, with the notation from the link, I am here:



      Let $G$ be maximal planar, $v_1,v_2$ two vertices; we want to build some edge between them and see this ends up with a graph having a subdivision of $K_{3,3}$.



      We also have three $u-v$ vertex disjoint paths, say, $P_1,P_2,P_3$, each of them only having one neighbor of $v$, say, $u_1,u_2,u_3$. We also have that the neighbors of $v$ form a cycle since $G$ is a triangulation of the plane.
      Also as they say, we have a new vertex $w$, which we may suppose is (given, if necessary, a rotation of the sphere which interchanges some faces) inside one of the regions bounded by the cycles determined by $P_i$ and $P_j$, say, $P_1$ and $P_2$. In this case, by Menger's theorem we have three paths to $u_1,u_2$ and $v_2$.
      Anyways, we have the following sketch of the situation:
      enter image description here
      The curly 'edges' mean they are paths, the red dashed edge means it's the edge we want to add. In the solution they say that with this we get a subdivision of $K_{3,3}$ with partition given by $A={u,w,u_3}$ and $B={u_1,u_2,v}$. In the picture suppose the $u_1-u_2,u_2-u_3,u_3-u_1-$ paths are $zeta_1,zeta_2,zeta_3$. If the three paths from $w$ don't meet $zeta_2$ or $zeta_3$ we don't have much trouble and that's indeed a partition of those six vert which, with the edges and paths we already have and $v_1v_2$ make a subdivision of $K_{3,3}$. But if the paths from $w$ somehow meet $zeta_2$ or $zeta_3$ we may need to change the vertices which make the subdivision.



      So, are those both cases possible? If they are, is there some way to handle them?



      Edit: There is also this question but they only talk about how some solution (a bit different from mine) is wrong.






      graph-theory planar-graph




      graph-theory planar-graph






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Jan 10 at 2:49









      Henning Makholm

      239k17304541




      239k17304541










      asked Jan 5 at 2:51









      David MolanoDavid Molano

      1,376720




      1,376720






















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          $begingroup$

          I know it has a bit of irony that I found a way to solve it just after putting the bounty. I had proven before that every $3-$connected nonplanar graph contains a $K_{3,3}-$minor, but hadn't really realized that it also has a $K_{3,3}-$subdivision (In my proof there was a part when I couldn't find such subdivision but I still was able to get a minor so since that's what I wanted to find I didn't worry about that; now, after verifying that part, I see that such subdivision can be easily found given what I already had).



          So, I can use the proof of the lemma 3.4 here.



          In this case I can find a subdivision of $K_{3,3}$ in any $3-$connected nonplanar graph, in particular one built by adding an edge to a maximal planar graph.






          share|cite|improve this answer











          $endgroup$













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            $begingroup$

            I know it has a bit of irony that I found a way to solve it just after putting the bounty. I had proven before that every $3-$connected nonplanar graph contains a $K_{3,3}-$minor, but hadn't really realized that it also has a $K_{3,3}-$subdivision (In my proof there was a part when I couldn't find such subdivision but I still was able to get a minor so since that's what I wanted to find I didn't worry about that; now, after verifying that part, I see that such subdivision can be easily found given what I already had).



            So, I can use the proof of the lemma 3.4 here.



            In this case I can find a subdivision of $K_{3,3}$ in any $3-$connected nonplanar graph, in particular one built by adding an edge to a maximal planar graph.






            share|cite|improve this answer











            $endgroup$


















              2












              $begingroup$

              I know it has a bit of irony that I found a way to solve it just after putting the bounty. I had proven before that every $3-$connected nonplanar graph contains a $K_{3,3}-$minor, but hadn't really realized that it also has a $K_{3,3}-$subdivision (In my proof there was a part when I couldn't find such subdivision but I still was able to get a minor so since that's what I wanted to find I didn't worry about that; now, after verifying that part, I see that such subdivision can be easily found given what I already had).



              So, I can use the proof of the lemma 3.4 here.



              In this case I can find a subdivision of $K_{3,3}$ in any $3-$connected nonplanar graph, in particular one built by adding an edge to a maximal planar graph.






              share|cite|improve this answer











              $endgroup$
















                2












                2








                2





                $begingroup$

                I know it has a bit of irony that I found a way to solve it just after putting the bounty. I had proven before that every $3-$connected nonplanar graph contains a $K_{3,3}-$minor, but hadn't really realized that it also has a $K_{3,3}-$subdivision (In my proof there was a part when I couldn't find such subdivision but I still was able to get a minor so since that's what I wanted to find I didn't worry about that; now, after verifying that part, I see that such subdivision can be easily found given what I already had).



                So, I can use the proof of the lemma 3.4 here.



                In this case I can find a subdivision of $K_{3,3}$ in any $3-$connected nonplanar graph, in particular one built by adding an edge to a maximal planar graph.






                share|cite|improve this answer











                $endgroup$



                I know it has a bit of irony that I found a way to solve it just after putting the bounty. I had proven before that every $3-$connected nonplanar graph contains a $K_{3,3}-$minor, but hadn't really realized that it also has a $K_{3,3}-$subdivision (In my proof there was a part when I couldn't find such subdivision but I still was able to get a minor so since that's what I wanted to find I didn't worry about that; now, after verifying that part, I see that such subdivision can be easily found given what I already had).



                So, I can use the proof of the lemma 3.4 here.



                In this case I can find a subdivision of $K_{3,3}$ in any $3-$connected nonplanar graph, in particular one built by adding an edge to a maximal planar graph.







                share|cite|improve this answer














                share|cite|improve this answer



                share|cite|improve this answer








                edited Jan 10 at 4:58

























                answered Jan 7 at 4:05









                David MolanoDavid Molano

                1,376720




                1,376720






























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