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Given a cubic bipartite graph $G$, living happily on an orientable surface with Euler characteristic $\chi$. Euler's formula then reads: $$ F+V=E+\chi , $$ where $F=\sum f_k$, the number of $k$-gons. Now, let's restrict $G$ to $f_6$ hexagons and $f_8$ octagons only. Since its $3$-regularity it follows that: $$ 0\cdot f_6 -2 \cdot f_8=6\chi $$ all along similar lines as given here...

How does the simplest non-trivial ($\chi=-2 \rightarrow f_8=6$) example look like?

Feel free to add as countably many hexagons, in case, you need them...

draks ...
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  • Looking at my construction again, it appears to have an octagon and four squares instead of three octagons. Sorry! – Misha Lavrov Dec 23 '18 at 23:58
  • @MishaLavrov hmm but this doesn't work out on the double torus according to $2f_4-2f_8=-6$...strange... – draks ... Dec 24 '18 at 10:15
  • I don't understand "a bipartite graph" on a surface. Can you give an example? – san Dec 29 '18 at 17:16
  • If you don't insist on bipartite, whatever it means, you can cover the double torus with exactly 6 octagons (and any number of hexagons) such that edges and vertices form a 3-regular graph. – san Dec 30 '18 at 05:44
  • This is forced by Euler's formula. On the other hand one can construct the graph with exactly 6 octagons and no hexagons. I' m not completely sure about the "any number of hexagons" part. Maybe there is a restriction on the number of hexagons. But there must always be exactly 6 octagons. – san Dec 30 '18 at 06:10
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    Bipartite means the vertices are 2-colorable. In other words, that every face has even degree. – Zach Hunter Dec 30 '18 at 07:21
  • @MishaLavrov can you verify my version of Euler's formula? Thanks... – draks ... Dec 30 '18 at 21:03
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    Related: http://www.weddslist.com/groups/genus/2/8-3/8-3.php – Josh B. Dec 30 '18 at 21:26
  • @JoshB. Thanks for your link. It induced a follow-up question... – draks ... Jan 10 '19 at 11:12

2 Answers2

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Take two hexagonal lattices of torii. In each lattice, choose a hexagon, $h$, and then "glue" these two faces together into a new cycle $h*$.

Every vertex in $h*$ now has a degree of 4. To fix this, in $h*$, remove three edges of the same color. We get 3 new faces, $f'$, with $2\cdot(6-1)=10$ degrees.

No other faces or vertices were modified, so we arrive at a cubic graph with $\chi=-2$ composed only of hexagons and three 10-gons. By removing all the edges in $h*$, and then contracting all the vertices in $h*$, we get 6 octagons and hexagons everywhere else.

For a visual example, this is how you could create the 6 octagons part:

Start with two of these:

enter image description here

Then draw green edges connecting the highlighted (black) vertices in the previous picture, like so:

enter image description here

These faces with green edges all have degree 8. Elsewhere, each torus is unchanged, and should still have hexagons everywhere. Subdividing each green edge and adding there more edges can create 6 new hexagons and 3 10-gons.

Regarding your modified Euler's formula, it seems the $1+\chi$ part is wrong. According to this, $\chi = V-E-F$. So using all your other work, $\sum (6-k)f_k = 6\chi$. This is consistent with what is known about polyhedra on the sphere.

Zach Hunter
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  • Hmm since Henning didn't complain about my formula in the linked post I assume(d) it to be right... – draks ... Dec 30 '18 at 11:45
  • Can you draw your construction? – draks ... Dec 30 '18 at 11:45
  • I've added a visual aid, forgive me, but my artistic abilities are limited. – Zach Hunter Dec 30 '18 at 16:25
  • thanks for your contribution – draks ... Dec 31 '18 at 12:44
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    Using your method one can obtain lattices with 6 octagons and $k=4,7,10,11,13,14,15,16,17,...$ hexagons. For $1,2,3,5,6,8,9$ or $12$ hexagons, I'm not sure if there are lattices. – san Jan 01 '19 at 21:26
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    @san in $\chi = 2$ you can have 6 squares and any number of hexagons except 1. – Zach Hunter Jan 01 '19 at 22:37
  • I suspect that in the present case you can have any number of hexagons, but your construction provides only the numbers above. There should be other constructions providing the rest of the numbers. – san Jan 01 '19 at 22:50
  • Stretching your method a little bit, one can get 6, 9 and 12 hexagons(together with 6 octagons). Do you know if the stuff for $\chi=2$ is published somewhere? Or at least written up? – san Jan 04 '19 at 15:20
  • It was mentioned by Grünbaum, I believe within his translation/summary of “Morphology Der Polymer” by Eherhard. It’s been a year at this point, I’ll try to find it. – Zach Hunter Jan 04 '19 at 15:25
  • Correction: I believe it was in one of his many papers on Eberhard-like theorems, either way, in the paper, I recall it taking no more than two sentences to mention. I think you can simply get the result that there does not exist a polyhedron with 1 hexagon by simply starting with one hexagon and maxing it cubic by adding squares – Zach Hunter Jan 04 '19 at 15:36
  • And how do you get for example 4 hexagons(and 6 squares)? – san Jan 04 '19 at 17:51
  • Turn two opposite faces of a cube into a column of three squares. Make these two columns be pointed orthogonal directions. – Zach Hunter Jan 04 '19 at 18:20
  • Nice. Is there a general method for doing this or is it just trial and error? – san Jan 04 '19 at 18:48
  • I do not know if the initial result was constructive or not. – Zach Hunter Jan 04 '19 at 19:34
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enter image description hereIf you cover the double torus with only hexagons and octagons such that edges and vertices form a 3-regular graph, then you need exactly 6 octagons.

$$ f_6+f_8+(6f_6+8f_8)/3=(6f_6+8f_8)/2 + \chi $$ Since $\chi=-2$, we have $f_8=6$. From Euler's formula there is no restriction for the number of hexagons, but maybe there are geometric restrictions. The picture shows a covering with 6 octagons and no hexagon.

san
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  • +1 very illustrative. Thanks. Concerning the octagons at the edge of the double torus (at the outside of the handles) it looks like that they split in two non connected squares, although the "face" is bounded by 8 vertices. Interesting... how does that relate to the linked stuff in Josh's comment? – draks ... Dec 31 '18 at 12:48
  • I don't see the squares. The regular maps in the link in Josh comment require the faces to be embedded,i.e. no self-intersection, so my covering is not a regular map. – san Dec 31 '18 at 14:16
  • Squares are the cycles of length 4 at the cutted handles. Thanks pointing out the difference to Josh... – draks ... Jan 01 '19 at 13:13
  • Maybe you mean the 3 cycles? – san Jan 02 '19 at 00:43
  • Ah I thought that squares. But then I don't get how to form the fifth and sixth octagon... – draks ... Jan 02 '19 at 09:32
  • The most left and the most right octagons have half circles that match the half circles of the 3 cycles – san Jan 03 '19 at 02:10