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
edited Nov 22 at 17:43
Federico
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asked Nov 22 at 17:38
Michele Impedovo
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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
edited Nov 22 at 17:43
Federico
4,228512
asked Nov 22 at 17:38
Michele Impedovo
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
add a comment |
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
edited Nov 22 at 17:43
Federico
4,228512
asked Nov 22 at 17:38
Michele Impedovo
1
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
edited Nov 22 at 17:43
Federico
4,228512
asked Nov 22 at 17:38
Michele Impedovo
1
edited Nov 22 at 17:43
Federico
4,228512
asked Nov 22 at 17:38
Michele Impedovo
1
edited Nov 22 at 17:43
Federico
4,228512
edited Nov 22 at 17:43
Federico
4,228512
edited Nov 22 at 17:43
Federico
4,228512
4,228512
asked Nov 22 at 17:38
Michele Impedovo
1
asked Nov 22 at 17:38
Michele Impedovo
1
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
add a comment |
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
add a comment |
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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."
answered Dec 5 at 17:10
Michele Impedovo
1
add a comment |
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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."
answered Dec 5 at 17:10
Michele Impedovo
1
add a comment |
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."
answered Dec 5 at 17:10
Michele Impedovo
1
add a comment |
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."
answered Dec 5 at 17:10
Michele Impedovo
1
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."
answered Dec 5 at 17:10
Michele Impedovo
1
answered Dec 5 at 17:10
Michele Impedovo
1
answered Dec 5 at 17:10
Michele Impedovo
1
answered Dec 5 at 17:10
Michele Impedovo
1
1
add a comment |
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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