Prove or disprove that If $G$ is a graph with one or more triangles and contains no subdivision of$ K_5$ as a subgraph, then $G$ is planar.Planar and non-planar graphs, and Kuratowski's TheoremExtending Kuratowski's planarity theorem on finite graphs to countable infinite graphs.Prove that the tesseract graph is non-planarDoes adding more edge to the Triakis Tetrahedral Graph make it non-planar because it contains the a subgaph homeomorphic to $K_3,3$?Prove that if a graph contains a $K_5$ minor, then it's not planarProve that $G$ contains a subdivision of $K_5$ or $K_3,3$ if and only if G contains a $K_5$ or $K_3,3$ minorIf $G$ is a graph of order $n ge 5$ and size $mge 3n - 5$ then $G$ contains a subgraph with minimum degree 4.Disprove: “If a graph $G$ does not have a $K_3,3$ or a $K_5$ as an induced subgraph, then it is planar”Graph Theory - Show that every graph with at most three cycles is planarShow that there exists a $5$-regular planar graph and a $5$-regular nonplanar graph.
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Prove or disprove that If $G$ is a graph with one or more triangles and contains no subdivision of$ K_5$ as a subgraph, then $G$ is planar.
Planar and non-planar graphs, and Kuratowski's TheoremExtending Kuratowski's planarity theorem on finite graphs to countable infinite graphs.Prove that the tesseract graph is non-planarDoes adding more edge to the Triakis Tetrahedral Graph make it non-planar because it contains the a subgaph homeomorphic to $K_3,3$?Prove that if a graph contains a $K_5$ minor, then it's not planarProve that $G$ contains a subdivision of $K_5$ or $K_3,3$ if and only if G contains a $K_5$ or $K_3,3$ minorIf $G$ is a graph of order $n ge 5$ and size $mge 3n - 5$ then $G$ contains a subgraph with minimum degree 4.Disprove: “If a graph $G$ does not have a $K_3,3$ or a $K_5$ as an induced subgraph, then it is planar”Graph Theory - Show that every graph with at most three cycles is planarShow that there exists a $5$-regular planar graph and a $5$-regular nonplanar graph.
$begingroup$
I'm working in the following graph theory excercise.
Prove or disprove that If $G$ is a graph with one or more triangles and contains no subdivision of$ K_5$ as a subgraph, then $G$ is planar.
I'm thinking about the graph $K_6$ and how is non-planar by Kuratowski's theorem, so the answer would be that the statement is false. But I'm not sure about what does "no subdivision of$ K_5$ as a subgraph" means, any hint or help will be really appreciated.
graph-theory
asked 2 days ago
mrazmraz
44319
$endgroup$
add a comment |
$begingroup$
I'm working in the following graph theory excercise.
Prove or disprove that If $G$ is a graph with one or more triangles and contains no subdivision of$ K_5$ as a subgraph, then $G$ is planar.
I'm thinking about the graph $K_6$ and how is non-planar by Kuratowski's theorem, so the answer would be that the statement is false. But I'm not sure about what does "no subdivision of$ K_5$ as a subgraph" means, any hint or help will be really appreciated.
graph-theory
asked 2 days ago
mrazmraz
44319
$endgroup$
add a comment |
$begingroup$
I'm working in the following graph theory excercise.
Prove or disprove that If $G$ is a graph with one or more triangles and contains no subdivision of$ K_5$ as a subgraph, then $G$ is planar.
I'm thinking about the graph $K_6$ and how is non-planar by Kuratowski's theorem, so the answer would be that the statement is false. But I'm not sure about what does "no subdivision of$ K_5$ as a subgraph" means, any hint or help will be really appreciated.
graph-theory
asked 2 days ago
mrazmraz
44319
$endgroup$
I'm working in the following graph theory excercise.
Prove or disprove that If $G$ is a graph with one or more triangles and contains no subdivision of$ K_5$ as a subgraph, then $G$ is planar.
I'm thinking about the graph $K_6$ and how is non-planar by Kuratowski's theorem, so the answer would be that the statement is false. But I'm not sure about what does "no subdivision of$ K_5$ as a subgraph" means, any hint or help will be really appreciated.
graph-theory
graph-theory
asked 2 days ago
mrazmraz
44319
asked 2 days ago
mrazmraz
44319
asked 2 days ago
mrazmraz
44319
asked 2 days ago
mrazmraz
44319
asked 2 days ago
mrazmraz
44319
44319
add a comment |
add a comment |
1 Answer
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$begingroup$
Subdivisions are defined here; essentially you can subdivide a graph by adding extra vertices along edges (as you choose). This adds a bunch of extra vertices with degree $2$.
When the question says the graph "has no subdivision of $K_5$", it means that no subgraph of the graph is of this form. As a non-example, $K_6$ indeed has a subdivision of $K_5$, as if we remove $3$ edges coming from a single vertex (so that it now has degree $2$), then the resulting graph is a subdivision of $K_5$.
To give you a hint, if this were true, then we could take any non-planar graph without a subdivision of $K_5$, add in three extra vertices connected in a triangle but disconnected from the rest of the graph, and suddenly it would be planar. That is, every graph with a subdivision of $K_5$ would have to be non-planar. Compare this with Wagner's Theorem (often mistakenly attributed to Kuratowski), to find a non-planar graph without a $K_5$ subdivision, and use it as above to form a counterexample.
answered 2 days ago
Theo BenditTheo Bendit
19.4k12353
$endgroup$
$begingroup$
@MishaLavrov Thanks.
$endgroup$
– Theo Bendit
2 days ago
add a comment |
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$begingroup$
Subdivisions are defined here; essentially you can subdivide a graph by adding extra vertices along edges (as you choose). This adds a bunch of extra vertices with degree $2$.
When the question says the graph "has no subdivision of $K_5$", it means that no subgraph of the graph is of this form. As a non-example, $K_6$ indeed has a subdivision of $K_5$, as if we remove $3$ edges coming from a single vertex (so that it now has degree $2$), then the resulting graph is a subdivision of $K_5$.
To give you a hint, if this were true, then we could take any non-planar graph without a subdivision of $K_5$, add in three extra vertices connected in a triangle but disconnected from the rest of the graph, and suddenly it would be planar. That is, every graph with a subdivision of $K_5$ would have to be non-planar. Compare this with Wagner's Theorem (often mistakenly attributed to Kuratowski), to find a non-planar graph without a $K_5$ subdivision, and use it as above to form a counterexample.
answered 2 days ago
Theo BenditTheo Bendit
19.4k12353
$endgroup$
$begingroup$
@MishaLavrov Thanks.
$endgroup$
– Theo Bendit
2 days ago
add a comment |
$begingroup$
Subdivisions are defined here; essentially you can subdivide a graph by adding extra vertices along edges (as you choose). This adds a bunch of extra vertices with degree $2$.
When the question says the graph "has no subdivision of $K_5$", it means that no subgraph of the graph is of this form. As a non-example, $K_6$ indeed has a subdivision of $K_5$, as if we remove $3$ edges coming from a single vertex (so that it now has degree $2$), then the resulting graph is a subdivision of $K_5$.
To give you a hint, if this were true, then we could take any non-planar graph without a subdivision of $K_5$, add in three extra vertices connected in a triangle but disconnected from the rest of the graph, and suddenly it would be planar. That is, every graph with a subdivision of $K_5$ would have to be non-planar. Compare this with Wagner's Theorem (often mistakenly attributed to Kuratowski), to find a non-planar graph without a $K_5$ subdivision, and use it as above to form a counterexample.
answered 2 days ago
Theo BenditTheo Bendit
19.4k12353
$endgroup$
$begingroup$
@MishaLavrov Thanks.
$endgroup$
– Theo Bendit
2 days ago
add a comment |
$begingroup$
Subdivisions are defined here; essentially you can subdivide a graph by adding extra vertices along edges (as you choose). This adds a bunch of extra vertices with degree $2$.
When the question says the graph "has no subdivision of $K_5$", it means that no subgraph of the graph is of this form. As a non-example, $K_6$ indeed has a subdivision of $K_5$, as if we remove $3$ edges coming from a single vertex (so that it now has degree $2$), then the resulting graph is a subdivision of $K_5$.
To give you a hint, if this were true, then we could take any non-planar graph without a subdivision of $K_5$, add in three extra vertices connected in a triangle but disconnected from the rest of the graph, and suddenly it would be planar. That is, every graph with a subdivision of $K_5$ would have to be non-planar. Compare this with Wagner's Theorem (often mistakenly attributed to Kuratowski), to find a non-planar graph without a $K_5$ subdivision, and use it as above to form a counterexample.
answered 2 days ago
Theo BenditTheo Bendit
19.4k12353
$endgroup$
Subdivisions are defined here; essentially you can subdivide a graph by adding extra vertices along edges (as you choose). This adds a bunch of extra vertices with degree $2$.
When the question says the graph "has no subdivision of $K_5$", it means that no subgraph of the graph is of this form. As a non-example, $K_6$ indeed has a subdivision of $K_5$, as if we remove $3$ edges coming from a single vertex (so that it now has degree $2$), then the resulting graph is a subdivision of $K_5$.
To give you a hint, if this were true, then we could take any non-planar graph without a subdivision of $K_5$, add in three extra vertices connected in a triangle but disconnected from the rest of the graph, and suddenly it would be planar. That is, every graph with a subdivision of $K_5$ would have to be non-planar. Compare this with Wagner's Theorem (often mistakenly attributed to Kuratowski), to find a non-planar graph without a $K_5$ subdivision, and use it as above to form a counterexample.
answered 2 days ago
Theo BenditTheo Bendit
19.4k12353
edited 2 days ago
answered 2 days ago
Theo BenditTheo Bendit
19.4k12353
answered 2 days ago
Theo BenditTheo Bendit
19.4k12353
answered 2 days ago
Theo BenditTheo Bendit
19.4k12353
19.4k12353
$begingroup$
@MishaLavrov Thanks.
$endgroup$
– Theo Bendit
2 days ago
add a comment |
$begingroup$
@MishaLavrov Thanks.
$endgroup$
– Theo Bendit
2 days ago
$begingroup$
@MishaLavrov Thanks.
$endgroup$
– Theo Bendit
2 days ago
$begingroup$
@MishaLavrov Thanks.
$endgroup$
– Theo Bendit
2 days ago
add a comment |
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