A sufficient condition for planar graph











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Let $G$ a graph with $v$ vertices and $e$ edges.
Right, I know that if $e>3v-6$ then $G$ is not planar.
Do you know any theorem like "If $e<f(v)$ then $G$ is planar"?






topological-graph-theory







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    The problem I see is $K_{3,3}$. Such a function $f$ can be at most $9$. So the only theorem of this kind would be "If $e<9$ then $G$ is planar"
    – Federico
    Nov 22 at 17:47















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Let $G$ a graph with $v$ vertices and $e$ edges.
Right, I know that if $e>3v-6$ then $G$ is not planar.
Do you know any theorem like "If $e<f(v)$ then $G$ is planar"?






topological-graph-theory







share|cite|improve this question




















  • 1




    The problem I see is $K_{3,3}$. Such a function $f$ can be at most $9$. So the only theorem of this kind would be "If $e<9$ then $G$ is planar"
    – Federico
    Nov 22 at 17:47













up vote
0
down vote

favorite









up vote
0
down vote

favorite











Let $G$ a graph with $v$ vertices and $e$ edges.
Right, I know that if $e>3v-6$ then $G$ is not planar.
Do you know any theorem like "If $e<f(v)$ then $G$ is planar"?






topological-graph-theory







share|cite|improve this question















Let $G$ a graph with $v$ vertices and $e$ edges.
Right, I know that if $e>3v-6$ then $G$ is not planar.
Do you know any theorem like "If $e<f(v)$ then $G$ is planar"?






topological-graph-theory




topological-graph-theory






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edited Nov 22 at 17:43









Federico

4,228512




4,228512










asked Nov 22 at 17:38









Michele Impedovo

1




1








  • 1




    The problem I see is $K_{3,3}$. Such a function $f$ can be at most $9$. So the only theorem of this kind would be "If $e<9$ then $G$ is planar"
    – Federico
    Nov 22 at 17:47














  • 1




    The problem I see is $K_{3,3}$. Such a function $f$ can be at most $9$. So the only theorem of this kind would be "If $e<9$ then $G$ is planar"
    – Federico
    Nov 22 at 17:47








1




1




The problem I see is $K_{3,3}$. Such a function $f$ can be at most $9$. So the only theorem of this kind would be "If $e<9$ then $G$ is planar"
– Federico
Nov 22 at 17:47




The problem I see is $K_{3,3}$. Such a function $f$ can be at most $9$. So the only theorem of this kind would be "If $e<9$ then $G$ is planar"
– Federico
Nov 22 at 17:47










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I want to report an error (I think) in the book
"Graph Theory with applications", by C. Vasudev, New Age International Publishers
Chapter 2, Problem 2.15. Find a graph G with degree sequence [4,4,3,3,3,3] such that G is not planar.



Solution. It is not possible to draw such a non planar graph because $$2e=4+4+3+3+3+3=20$$$$e=10$$ and the inequality $$e leq 3v-6$$ does not hold.



It is false. The graph G:={{1, 3}, {1, 4}, {1, 5}, {1, 6}, {2, 4}, {2, 5}, {2, 6}, {3, 4}, {3, 5}, {3, 6}} has 6 vertices, 10 edges, the degree sequence [4,4,3,3,3,3] and it is not planar. The author uses the wrong theorem "If $$e leq 3v-6$$ then G is planar."






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    I want to report an error (I think) in the book
    "Graph Theory with applications", by C. Vasudev, New Age International Publishers
    Chapter 2, Problem 2.15. Find a graph G with degree sequence [4,4,3,3,3,3] such that G is not planar.



    Solution. It is not possible to draw such a non planar graph because $$2e=4+4+3+3+3+3=20$$$$e=10$$ and the inequality $$e leq 3v-6$$ does not hold.



    It is false. The graph G:={{1, 3}, {1, 4}, {1, 5}, {1, 6}, {2, 4}, {2, 5}, {2, 6}, {3, 4}, {3, 5}, {3, 6}} has 6 vertices, 10 edges, the degree sequence [4,4,3,3,3,3] and it is not planar. The author uses the wrong theorem "If $$e leq 3v-6$$ then G is planar."






    share|cite|improve this answer

























      up vote
      0
      down vote













      I want to report an error (I think) in the book
      "Graph Theory with applications", by C. Vasudev, New Age International Publishers
      Chapter 2, Problem 2.15. Find a graph G with degree sequence [4,4,3,3,3,3] such that G is not planar.



      Solution. It is not possible to draw such a non planar graph because $$2e=4+4+3+3+3+3=20$$$$e=10$$ and the inequality $$e leq 3v-6$$ does not hold.



      It is false. The graph G:={{1, 3}, {1, 4}, {1, 5}, {1, 6}, {2, 4}, {2, 5}, {2, 6}, {3, 4}, {3, 5}, {3, 6}} has 6 vertices, 10 edges, the degree sequence [4,4,3,3,3,3] and it is not planar. The author uses the wrong theorem "If $$e leq 3v-6$$ then G is planar."






      share|cite|improve this answer























        up vote
        0
        down vote










        up vote
        0
        down vote









        I want to report an error (I think) in the book
        "Graph Theory with applications", by C. Vasudev, New Age International Publishers
        Chapter 2, Problem 2.15. Find a graph G with degree sequence [4,4,3,3,3,3] such that G is not planar.



        Solution. It is not possible to draw such a non planar graph because $$2e=4+4+3+3+3+3=20$$$$e=10$$ and the inequality $$e leq 3v-6$$ does not hold.



        It is false. The graph G:={{1, 3}, {1, 4}, {1, 5}, {1, 6}, {2, 4}, {2, 5}, {2, 6}, {3, 4}, {3, 5}, {3, 6}} has 6 vertices, 10 edges, the degree sequence [4,4,3,3,3,3] and it is not planar. The author uses the wrong theorem "If $$e leq 3v-6$$ then G is planar."






        share|cite|improve this answer












        I want to report an error (I think) in the book
        "Graph Theory with applications", by C. Vasudev, New Age International Publishers
        Chapter 2, Problem 2.15. Find a graph G with degree sequence [4,4,3,3,3,3] such that G is not planar.



        Solution. It is not possible to draw such a non planar graph because $$2e=4+4+3+3+3+3=20$$$$e=10$$ and the inequality $$e leq 3v-6$$ does not hold.



        It is false. The graph G:={{1, 3}, {1, 4}, {1, 5}, {1, 6}, {2, 4}, {2, 5}, {2, 6}, {3, 4}, {3, 5}, {3, 6}} has 6 vertices, 10 edges, the degree sequence [4,4,3,3,3,3] and it is not planar. The author uses the wrong theorem "If $$e leq 3v-6$$ then G is planar."







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Dec 5 at 17:10









        Michele Impedovo

        1




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