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Licheng Zhang
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A $1$-planar graph is a graph which has a drawing on the plane such that each edge has at most one crossing.

I used nauty to generate all 3-regular graphs up to order 12, and checked each one of them individually. They all turned out to be 1-planar graphs.

My question is whether it is possible to construct a 3-regular non-1-planar graph.

I believe that it definitely exists.

I also conducted experiments on graphs with relatively large crossing numbers, but unfortunately, they were all 1-planar graphs. For example: The smallest 6-crossing cubic graph is the Desargues graph, with 20 vertices.

Desargues graph![![Desargues graph

Or see   this link for the hexagonal graph.

enter image description here

This problem is related to the question Why is the crossing number of Tutte 12-cage 170?, because we know that the crossing number of an $n$-order 1-planar graph is less than or equal to $n-2$ (see [1]).

  • [1] Czap J, Hudák D. On drawings and decompositions of 1-planar graphs[J]. the electronic journal of combinatorics, 2013: P54-P54.

If the crossing number of Tutte 12-cage (126 vertices)is greater than or equal to $125$, then it is the 3-regular non- 1-planar graph we are looking for. Unfortunately, it seems that the crossing number of Tutte 12-cage is unknown. It would still be helpful if we knew a good lower bound of Tutte 12-cage.


Edit 1: As reminded by Joseph O'Rourke, Coxeter graph is a candidate. The following embedding is from Wikipedia and it almost has a 1-planar embedding, except for one edge that will cross twice.

enter image description here

Edit 2: Thanks to Sam Hopkins for the reminder, the above sentence in Edit 1 should be corrected to: 3 of the edges will cross twice (not just one edge).

A $1$-planar graph is a graph which has a drawing on the plane such that each edge has at most one crossing.

I used nauty to generate all 3-regular graphs up to order 12, and checked each one of them individually. They all turned out to be 1-planar graphs.

My question is whether it is possible to construct a 3-regular non-1-planar graph.

I believe that it definitely exists.

I also conducted experiments on graphs with relatively large crossing numbers, but unfortunately, they were all 1-planar graphs. For example: The smallest 6-crossing cubic graph is the Desargues graph, with 20 vertices.

Desargues graph

Or see this link .

enter image description here

This problem is related to the question Why is the crossing number of Tutte 12-cage 170?, because we know that the crossing number of an $n$-order 1-planar graph is less than or equal to $n-2$ (see [1]).

  • [1] Czap J, Hudák D. On drawings and decompositions of 1-planar graphs[J]. the electronic journal of combinatorics, 2013: P54-P54.

If the crossing number of Tutte 12-cage (126 vertices)is greater than or equal to $125$, then it is the 3-regular non- 1-planar graph we are looking for. Unfortunately, it seems that the crossing number of Tutte 12-cage is unknown. It would still be helpful if we knew a good lower bound of Tutte 12-cage.


Edit 1: As reminded by Joseph O'Rourke, Coxeter graph is a candidate. The following embedding is from Wikipedia and it almost has a 1-planar embedding, except for one edge that will cross twice.

enter image description here

Edit 2: Thanks to Sam Hopkins for the reminder, the above sentence in Edit 1 should be corrected to: 3 of the edges will cross twice (not just one edge).

A $1$-planar graph is a graph which has a drawing on the plane such that each edge has at most one crossing.

I used nauty to generate all 3-regular graphs up to order 12, and checked each one of them individually. They all turned out to be 1-planar graphs.

My question is whether it is possible to construct a 3-regular non-1-planar graph.

I believe that it definitely exists.

I also conducted experiments on graphs with relatively large crossing numbers, but unfortunately, they were all 1-planar graphs. For example: The smallest 6-crossing cubic graph is the Desargues graph, with 20 vertices.

![![Desargues graph

Or see   this link for the hexagonal graph.

enter image description here

This problem is related to the question Why is the crossing number of Tutte 12-cage 170?, because we know that the crossing number of an $n$-order 1-planar graph is less than or equal to $n-2$ (see [1]).

  • [1] Czap J, Hudák D. On drawings and decompositions of 1-planar graphs[J]. the electronic journal of combinatorics, 2013: P54-P54.

If the crossing number of Tutte 12-cage (126 vertices)is greater than or equal to $125$, then it is the 3-regular non- 1-planar graph we are looking for. Unfortunately, it seems that the crossing number of Tutte 12-cage is unknown. It would still be helpful if we knew a good lower bound of Tutte 12-cage.


Edit 1: As reminded by Joseph O'Rourke, Coxeter graph is a candidate. The following embedding is from Wikipedia and it almost has a 1-planar embedding, except for one edge that will cross twice.

enter image description here

Edit 2: Thanks to Sam Hopkins for the reminder, the above sentence in Edit 1 should be corrected to: 3 of the edges will cross twice (not just one edge).

added 194 characters in body
Source Link
Licheng Zhang
  • 1.9k
  • 6
  • 17

A $1$-planar graph is a graph which has a drawing on the plane such that each edge has at most one crossing.

I used nauty to generate all 3-regular graphs up to order 12, and checked each one of them individually. They all turned out to be 1-planar graphs.

My question is whether it is possible to construct a 3-regular non-1-planar graph.

I believe that it definitely exists.

I also conducted experiments on graphs with relatively large crossing numbers, but unfortunately, they were all 1-planar graphs. For example: The smallest 6-crossing cubic graph is the Desargues graph, with 20 vertices.

Desargues graph

Or see this link .

enter image description here

This problem is related to the question Why is the crossing number of Tutte 12-cage 170?, because we know that the crossing number of an $n$-order 1-planar graph is less than or equal to $n-2$ (see [1]).

  • [1] Czap J, Hudák D. On drawings and decompositions of 1-planar graphs[J]. the electronic journal of combinatorics, 2013: P54-P54.

If the crossing number of Tutte 12-cage (126 vertices)is greater than or equal to $125$, then it is the 3-regular non- 1-planar graph we are looking for. Unfortunately, it seems that the crossing number of Tutte 12-cage is unknown. It would still be helpful if we knew a good lower bound of Tutte 12-cage.


Edit 1: As reminded by Joseph O'Rourke, Coxeter graph is a candidate. The following embedding is from Wikipedia and it almost has a 1-planar embedding, except for one edge that will cross twice.

enter image description here

Edit 2: Thanks to Sam Hopkins for the reminder, the above sentence in Edit 1 should be corrected to: 3 of the edges will cross twice (not just one edge).

A $1$-planar graph is a graph which has a drawing on the plane such that each edge has at most one crossing.

I used nauty to generate all 3-regular graphs up to order 12, and checked each one of them individually. They all turned out to be 1-planar graphs.

My question is whether it is possible to construct a 3-regular non-1-planar graph.

I believe that it definitely exists.

I also conducted experiments on graphs with relatively large crossing numbers, but unfortunately, they were all 1-planar graphs. For example: The smallest 6-crossing cubic graph is the Desargues graph, with 20 vertices.

Desargues graph

This problem is related to the question Why is the crossing number of Tutte 12-cage 170?, because we know that the crossing number of an $n$-order 1-planar graph is less than or equal to $n-2$ (see [1]).

  • [1] Czap J, Hudák D. On drawings and decompositions of 1-planar graphs[J]. the electronic journal of combinatorics, 2013: P54-P54.

If the crossing number of Tutte 12-cage (126 vertices)is greater than or equal to $125$, then it is the 3-regular non- 1-planar graph we are looking for. Unfortunately, it seems that the crossing number of Tutte 12-cage is unknown. It would still be helpful if we knew a good lower bound of Tutte 12-cage.


Edit 1: As reminded by Joseph O'Rourke, Coxeter graph is a candidate. The following embedding is from Wikipedia and it almost has a 1-planar embedding, except for one edge that will cross twice.

enter image description here

Edit 2: Thanks to Sam Hopkins for the reminder, the above sentence in Edit 1 should be corrected to: 3 of the edges will cross twice (not just one edge).

A $1$-planar graph is a graph which has a drawing on the plane such that each edge has at most one crossing.

I used nauty to generate all 3-regular graphs up to order 12, and checked each one of them individually. They all turned out to be 1-planar graphs.

My question is whether it is possible to construct a 3-regular non-1-planar graph.

I believe that it definitely exists.

I also conducted experiments on graphs with relatively large crossing numbers, but unfortunately, they were all 1-planar graphs. For example: The smallest 6-crossing cubic graph is the Desargues graph, with 20 vertices.

Desargues graph

Or see this link .

enter image description here

This problem is related to the question Why is the crossing number of Tutte 12-cage 170?, because we know that the crossing number of an $n$-order 1-planar graph is less than or equal to $n-2$ (see [1]).

  • [1] Czap J, Hudák D. On drawings and decompositions of 1-planar graphs[J]. the electronic journal of combinatorics, 2013: P54-P54.

If the crossing number of Tutte 12-cage (126 vertices)is greater than or equal to $125$, then it is the 3-regular non- 1-planar graph we are looking for. Unfortunately, it seems that the crossing number of Tutte 12-cage is unknown. It would still be helpful if we knew a good lower bound of Tutte 12-cage.


Edit 1: As reminded by Joseph O'Rourke, Coxeter graph is a candidate. The following embedding is from Wikipedia and it almost has a 1-planar embedding, except for one edge that will cross twice.

enter image description here

Edit 2: Thanks to Sam Hopkins for the reminder, the above sentence in Edit 1 should be corrected to: 3 of the edges will cross twice (not just one edge).

added 169 characters in body
Source Link
Licheng Zhang
  • 1.9k
  • 6
  • 17

A $1$-planar graph is a graph which has a drawing on the plane such that each edge has at most one crossing.

I used nauty to generate all 3-regular graphs up to order 12, and checked each one of them individually. They all turned out to be 1-planar graphs.

My question is whether it is possible to construct a 3-regular non-1-planar graph.

I believe that it definitely exists.

I also conducted experiments on graphs with relatively large crossing numbers, but unfortunately, they were all 1-planar graphs. For example: The smallest 6-crossing cubic graph is the Desargues graph, with 20 vertices.

Desargues graph

This problem is related to the question Why is the crossing number of Tutte 12-cage 170?, because we know that the crossing number of an $n$-order 1-planar graph is less than or equal to $n-2$ (see [1]).

  • [1] Czap J, Hudák D. On drawings and decompositions of 1-planar graphs[J]. the electronic journal of combinatorics, 2013: P54-P54.

If the crossing number of Tutte 12-cage (126 vertices)is greater than or equal to $125$, then it is the 3-regular non- 1-planar graph we are looking for. Unfortunately, it seems that the crossing number of Tutte 12-cage is unknown. It would still be helpful if we knew a good lower bound of Tutte 12-cage.


Edit 1: As reminded by Joseph O'Rourke, Coxeter graph is a candidate. The following embedding is from Wikipedia and it almost has a 1-planar embedding, except for one edge that will cross twice.

enter image description here

Edit 2: Thanks to Sam Hopkins for the reminder, the above sentence in Edit 1 should be corrected to: 3 of the edges will cross twice (not just one edge).

A $1$-planar graph is a graph which has a drawing on the plane such that each edge has at most one crossing.

I used nauty to generate all 3-regular graphs up to order 12, and checked each one of them individually. They all turned out to be 1-planar graphs.

My question is whether it is possible to construct a 3-regular non-1-planar graph.

I believe that it definitely exists.

I also conducted experiments on graphs with relatively large crossing numbers, but unfortunately, they were all 1-planar graphs. For example: The smallest 6-crossing cubic graph is the Desargues graph, with 20 vertices.

Desargues graph

This problem is related to the question Why is the crossing number of Tutte 12-cage 170?, because we know that the crossing number of an $n$-order 1-planar graph is less than or equal to $n-2$ (see [1]).

  • [1] Czap J, Hudák D. On drawings and decompositions of 1-planar graphs[J]. the electronic journal of combinatorics, 2013: P54-P54.

If the crossing number of Tutte 12-cage (126 vertices)is greater than or equal to $125$, then it is the 3-regular non- 1-planar graph we are looking for. Unfortunately, it seems that the crossing number of Tutte 12-cage is unknown. It would still be helpful if we knew a good lower bound of Tutte 12-cage.


As reminded by Joseph O'Rourke, Coxeter graph is a candidate. The following embedding is from Wikipedia and it almost has a 1-planar embedding, except for one edge that will cross twice.

enter image description here

A $1$-planar graph is a graph which has a drawing on the plane such that each edge has at most one crossing.

I used nauty to generate all 3-regular graphs up to order 12, and checked each one of them individually. They all turned out to be 1-planar graphs.

My question is whether it is possible to construct a 3-regular non-1-planar graph.

I believe that it definitely exists.

I also conducted experiments on graphs with relatively large crossing numbers, but unfortunately, they were all 1-planar graphs. For example: The smallest 6-crossing cubic graph is the Desargues graph, with 20 vertices.

Desargues graph

This problem is related to the question Why is the crossing number of Tutte 12-cage 170?, because we know that the crossing number of an $n$-order 1-planar graph is less than or equal to $n-2$ (see [1]).

  • [1] Czap J, Hudák D. On drawings and decompositions of 1-planar graphs[J]. the electronic journal of combinatorics, 2013: P54-P54.

If the crossing number of Tutte 12-cage (126 vertices)is greater than or equal to $125$, then it is the 3-regular non- 1-planar graph we are looking for. Unfortunately, it seems that the crossing number of Tutte 12-cage is unknown. It would still be helpful if we knew a good lower bound of Tutte 12-cage.


Edit 1: As reminded by Joseph O'Rourke, Coxeter graph is a candidate. The following embedding is from Wikipedia and it almost has a 1-planar embedding, except for one edge that will cross twice.

enter image description here

Edit 2: Thanks to Sam Hopkins for the reminder, the above sentence in Edit 1 should be corrected to: 3 of the edges will cross twice (not just one edge).

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