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removed the detail about cyclically-4-edge connected graphs, as I found an example
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Zach Hunter
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Second edit: On further thought, an example is the (unique) cubic polyhedral graph with 6 squares and 3 hexagons. (constructed by having 2 clumps of 3 pairwise incident squares divided by a ring of hexagons) Presumably there is also an example which is cyclically-4-edge connected, else Barnette's Conjecture is proved, due to an equivalence by Kelmans which can be found in page 9 of Thoughts on Barnette's Conjecture. Finding such an example is of interest.

Second edit: On further thought, an example is the (unique) cubic polyhedral graph with 6 squares and 3 hexagons. (constructed by having 2 clumps of 3 pairwise incident squares divided by a ring of hexagons) Presumably there is also an example which is cyclically-4-edge connected, else Barnette's Conjecture is proved, due to an equivalence by Kelmans which can be found in page 9 of Thoughts on Barnette's Conjecture. Finding such an example is of interest.

Second edit: On further thought, an example is the (unique) cubic polyhedral graph with 6 squares and 3 hexagons. (constructed by having 2 clumps of 3 pairwise incident squares divided by a ring of hexagons)

mentioned Kelman's equivalence, and asked for an example in this case.
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Zach Hunter
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Context: Barnette's Conjecture is that every bipartite cubic polyhedral graph is Hamiltonian. I have been interested by this problem for a long time, and I recently came up with a result. From my perspective, it doesn't seem too impressive, the argument is rather elementary, but I can't find another paper which states what I found. The closest I could find was Jan Florek's [On Barnette's Conjecture][1]On Barnette's Conjecture, which is about a subcase which I was previously aware of.

Second edit: On further thought, an example is the (unique) cubic polyhedral graph with 6 squares and 3 hexagons. (constructed by having 2 clumps of 3 pairwise incident squares divided by a ring of hexagons) [1]: Presumably there is also an example which is cyclically-4-edge connected, else Barnette's Conjecture is proved, due to an equivalence by Kelmans which can be found in page 9 of https://www.sciencedirect.com/science/article/pii/S0012365X10000403Thoughts on Barnette's Conjecture. Finding such an example is of interest.

Context: Barnette's Conjecture is that every bipartite cubic polyhedral graph is Hamiltonian. I have been interested by this problem for a long time, and I recently came up with a result. From my perspective, it doesn't seem too impressive, the argument is rather elementary, but I can't find another paper which states what I found. The closest I could find was Jan Florek's [On Barnette's Conjecture][1], which is about a subcase which I was previously aware of.

Second edit: On further thought, an example is the (unique) cubic polyhedral graph with 6 squares and 3 hexagons. (constructed by having 2 clumps of 3 pairwise incident squares divided by a ring of hexagons) [1]: https://www.sciencedirect.com/science/article/pii/S0012365X10000403

Context: Barnette's Conjecture is that every bipartite cubic polyhedral graph is Hamiltonian. I have been interested by this problem for a long time, and I recently came up with a result. From my perspective, it doesn't seem too impressive, the argument is rather elementary, but I can't find another paper which states what I found. The closest I could find was Jan Florek's On Barnette's Conjecture, which is about a subcase which I was previously aware of.

Second edit: On further thought, an example is the (unique) cubic polyhedral graph with 6 squares and 3 hexagons. (constructed by having 2 clumps of 3 pairwise incident squares divided by a ring of hexagons) Presumably there is also an example which is cyclically-4-edge connected, else Barnette's Conjecture is proved, due to an equivalence by Kelmans which can be found in page 9 of Thoughts on Barnette's Conjecture. Finding such an example is of interest.

provided counterexample
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Zach Hunter
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Context: Barnette's Conjecture is that every bipartite cubic polyhedral graph is Hamiltonian. I have been interested by this problem for a long time, and I recently came up with a result. From my perspective, it doesn't seem too impressive, the argument is rather elementary, but I can't find another paper which states what I found. The closest I could find was Jan Florek's On Barnette's Conjecture[On Barnette's Conjecture][1], which is about a subcase which I was previously aware of.

While I am aware that there are non 2-connected bipartite planar graphs, such as the tetrahedron with all edges subdivided, I cannot immediately come up with any non-constructible graphs that are also cubic and polyhedral. Presumably, there are such graphs, but if not, then Barnette's conjecture is solved!

SoSecond edit: On further thought, if one could provide any examples of non-constructiblean example is the (unique) cubic polyhedral bipartite graphs, that would be of interestgraph with 6 squares and 3 hexagons. (constructed by having 2 clumps of 3 pairwise incident squares divided by a ring of hexagons) [1]: https://www.sciencedirect.com/science/article/pii/S0012365X10000403

Context: Barnette's Conjecture is that every bipartite cubic polyhedral graph is Hamiltonian. I have been interested by this problem for a long time, and I recently came up with a result. From my perspective, it doesn't seem too impressive, the argument is rather elementary, but I can't find another paper which states what I found. The closest I could find was Jan Florek's On Barnette's Conjecture, which is about a subcase which I was previously aware of.

While I am aware that there are non 2-connected bipartite planar graphs, such as the tetrahedron with all edges subdivided, I cannot immediately come up with any non-constructible graphs that are also cubic and polyhedral. Presumably, there are such graphs, but if not, then Barnette's conjecture is solved!

So, if one could provide any examples of non-constructible cubic polyhedral bipartite graphs, that would be of interest.

Context: Barnette's Conjecture is that every bipartite cubic polyhedral graph is Hamiltonian. I have been interested by this problem for a long time, and I recently came up with a result. From my perspective, it doesn't seem too impressive, the argument is rather elementary, but I can't find another paper which states what I found. The closest I could find was Jan Florek's [On Barnette's Conjecture][1], which is about a subcase which I was previously aware of.

While I am aware that there are non 2-connected bipartite planar graphs, such as the tetrahedron with all edges subdivided, I cannot immediately come up with any non-constructible graphs that are also cubic and polyhedral.

Second edit: On further thought, an example is the (unique) cubic polyhedral graph with 6 squares and 3 hexagons. (constructed by having 2 clumps of 3 pairwise incident squares divided by a ring of hexagons) [1]: https://www.sciencedirect.com/science/article/pii/S0012365X10000403

Added second question
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Zach Hunter
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