An outerplanar G graph such that $G+k_1$ is not planar graph











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An outerplanar graph is an undirected graph that can be drawn in the plane without crossings in such a way that all of the vertices belong to the unbounded face of the drawing. That is, no vertex is totally surrounded by edges. Alternatively, a graph G is outerplanar if the graph formed from G by adding a new vertex, with edges connecting it to all the other vertices, is a planar graph.



The join $G=G_1+G_2$ of graphs $G_1$ and $G_2$ with disjoint point sets $V_1$ and $V_2$ and edge sets $X_1$ and $X_2$ is the graph union $G_1$ union $G_2$ together with all the edges joining $V_1$ and $V_2$.



Does it exist An outerplanar $G$ graph such that $G+ k_1$ is not planar graph? where $k_1$ is a vertex.
can you help me?










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  • 3




    What is definition of $k1$?
    – coffeemath
    Nov 22 at 8:08






  • 1




    Isn't $G+k_1$ planar when $G$ is outerplanar, precisely because of the sentence "Alternatively, a graph G is outerplanar if the graph formed from G by adding a new vertex, with edges connecting it to all the other vertices, is a planar graph."?
    – Servaes
    Nov 22 at 11:48










  • I agree with that, iff $k_1$ is really supposed to mean the regular graph on one vertex (aka graph with a single vertex and no edges)
    – Ingix
    Nov 22 at 12:10










  • A graph is outerplanar if and only if it does not contain $K_4$ or $K_{2,3}$ as minor.
    – mathnoob
    Nov 24 at 11:15















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An outerplanar graph is an undirected graph that can be drawn in the plane without crossings in such a way that all of the vertices belong to the unbounded face of the drawing. That is, no vertex is totally surrounded by edges. Alternatively, a graph G is outerplanar if the graph formed from G by adding a new vertex, with edges connecting it to all the other vertices, is a planar graph.



The join $G=G_1+G_2$ of graphs $G_1$ and $G_2$ with disjoint point sets $V_1$ and $V_2$ and edge sets $X_1$ and $X_2$ is the graph union $G_1$ union $G_2$ together with all the edges joining $V_1$ and $V_2$.



Does it exist An outerplanar $G$ graph such that $G+ k_1$ is not planar graph? where $k_1$ is a vertex.
can you help me?










share|cite|improve this question




















  • 3




    What is definition of $k1$?
    – coffeemath
    Nov 22 at 8:08






  • 1




    Isn't $G+k_1$ planar when $G$ is outerplanar, precisely because of the sentence "Alternatively, a graph G is outerplanar if the graph formed from G by adding a new vertex, with edges connecting it to all the other vertices, is a planar graph."?
    – Servaes
    Nov 22 at 11:48










  • I agree with that, iff $k_1$ is really supposed to mean the regular graph on one vertex (aka graph with a single vertex and no edges)
    – Ingix
    Nov 22 at 12:10










  • A graph is outerplanar if and only if it does not contain $K_4$ or $K_{2,3}$ as minor.
    – mathnoob
    Nov 24 at 11:15













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An outerplanar graph is an undirected graph that can be drawn in the plane without crossings in such a way that all of the vertices belong to the unbounded face of the drawing. That is, no vertex is totally surrounded by edges. Alternatively, a graph G is outerplanar if the graph formed from G by adding a new vertex, with edges connecting it to all the other vertices, is a planar graph.



The join $G=G_1+G_2$ of graphs $G_1$ and $G_2$ with disjoint point sets $V_1$ and $V_2$ and edge sets $X_1$ and $X_2$ is the graph union $G_1$ union $G_2$ together with all the edges joining $V_1$ and $V_2$.



Does it exist An outerplanar $G$ graph such that $G+ k_1$ is not planar graph? where $k_1$ is a vertex.
can you help me?










share|cite|improve this question















An outerplanar graph is an undirected graph that can be drawn in the plane without crossings in such a way that all of the vertices belong to the unbounded face of the drawing. That is, no vertex is totally surrounded by edges. Alternatively, a graph G is outerplanar if the graph formed from G by adding a new vertex, with edges connecting it to all the other vertices, is a planar graph.



The join $G=G_1+G_2$ of graphs $G_1$ and $G_2$ with disjoint point sets $V_1$ and $V_2$ and edge sets $X_1$ and $X_2$ is the graph union $G_1$ union $G_2$ together with all the edges joining $V_1$ and $V_2$.



Does it exist An outerplanar $G$ graph such that $G+ k_1$ is not planar graph? where $k_1$ is a vertex.
can you help me?







graph-theory planar-graph






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share|cite|improve this question













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edited Nov 22 at 11:45

























asked Nov 22 at 7:43









user546115

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  • 3




    What is definition of $k1$?
    – coffeemath
    Nov 22 at 8:08






  • 1




    Isn't $G+k_1$ planar when $G$ is outerplanar, precisely because of the sentence "Alternatively, a graph G is outerplanar if the graph formed from G by adding a new vertex, with edges connecting it to all the other vertices, is a planar graph."?
    – Servaes
    Nov 22 at 11:48










  • I agree with that, iff $k_1$ is really supposed to mean the regular graph on one vertex (aka graph with a single vertex and no edges)
    – Ingix
    Nov 22 at 12:10










  • A graph is outerplanar if and only if it does not contain $K_4$ or $K_{2,3}$ as minor.
    – mathnoob
    Nov 24 at 11:15














  • 3




    What is definition of $k1$?
    – coffeemath
    Nov 22 at 8:08






  • 1




    Isn't $G+k_1$ planar when $G$ is outerplanar, precisely because of the sentence "Alternatively, a graph G is outerplanar if the graph formed from G by adding a new vertex, with edges connecting it to all the other vertices, is a planar graph."?
    – Servaes
    Nov 22 at 11:48










  • I agree with that, iff $k_1$ is really supposed to mean the regular graph on one vertex (aka graph with a single vertex and no edges)
    – Ingix
    Nov 22 at 12:10










  • A graph is outerplanar if and only if it does not contain $K_4$ or $K_{2,3}$ as minor.
    – mathnoob
    Nov 24 at 11:15








3




3




What is definition of $k1$?
– coffeemath
Nov 22 at 8:08




What is definition of $k1$?
– coffeemath
Nov 22 at 8:08




1




1




Isn't $G+k_1$ planar when $G$ is outerplanar, precisely because of the sentence "Alternatively, a graph G is outerplanar if the graph formed from G by adding a new vertex, with edges connecting it to all the other vertices, is a planar graph."?
– Servaes
Nov 22 at 11:48




Isn't $G+k_1$ planar when $G$ is outerplanar, precisely because of the sentence "Alternatively, a graph G is outerplanar if the graph formed from G by adding a new vertex, with edges connecting it to all the other vertices, is a planar graph."?
– Servaes
Nov 22 at 11:48












I agree with that, iff $k_1$ is really supposed to mean the regular graph on one vertex (aka graph with a single vertex and no edges)
– Ingix
Nov 22 at 12:10




I agree with that, iff $k_1$ is really supposed to mean the regular graph on one vertex (aka graph with a single vertex and no edges)
– Ingix
Nov 22 at 12:10












A graph is outerplanar if and only if it does not contain $K_4$ or $K_{2,3}$ as minor.
– mathnoob
Nov 24 at 11:15




A graph is outerplanar if and only if it does not contain $K_4$ or $K_{2,3}$ as minor.
– mathnoob
Nov 24 at 11:15















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