Spherical Fullerenes - Part 2

Theo Schiminovich

Introduction

This article is a continuation of a previous article of mine published in the March 2021 issue of Math&CS Research, called “Bucky Balls." I use terminology in this article that I defined in that article, so I recommend you read that article first.

Narrowing Down the Spiral Methods

In Part \(1\), I tested three spiral methods, the Double Spiral Method, the Single Spiral Method, and the Modified Single Spiral Method, on each spherical fullerene. However, this is not necessary.

The Double Spiral Method and Single Spiral method are very similar, except in the Double Spiral method you have to keep track of two spirals, while in the Single Spiral method you only have to keep track of one. This means there are more potential points of failure in the Double Spiral method, as one has to ensure that both spirals continue, not just that one spiral does. Therefore, the Single Spiral method can be used in place of the Double Spiral method.

The Modified Single Spiral method will work whenever the Single Spiral method does, as it is nearly the same, but with the inclusion of a possible alternative in case things go wrong. Therefore, the Modified Single Spiral method can be used in place of the Single Spiral method.

Testing the Modified Single Spiral Method on Larger Cases

The Modified Single Spiral method worked on the three spherical fullerenes I looked at in part 1: C20 (the (1,0) fullerene), C60 (the (1,1) fullerene), and C80 (the (2,0) fullerene). However, as I only decided to test three spherical fullerenes (the \((m,n)\) fullerenes with \(m + n \le 2\)) I could not draw any conclusions.

C140 or the (2,1) fullerene

C140

The modified spiral method did not work in the (2,1) fullerene.

C180 or the (3,0) fullerene

C180 C180 spiral

The modified spiral method worked in the (3,0) fullerene.

C240 or the (2,2) fullerene

C240 C240 spiral

The modified spiral method worked in the (2,2) fullerene. In fact, the single spiral method would work in it too.

C260 or the (3,1) fullerene

C260

The modified spiral method did not work in the (3,1) fullerene.

C320 or the (4,0) fullerene

C320 C320 spiral

The modified spiral method worked in the (4,0) fullerene.

Analysis

This revealed that the modified spiral method doesn’t work in all cases. I looked into the (2,1) fullerene and the (3,1) fullerene, the two that didn’t work, to see what went wrong. It turned out that an assumption I made earlier was wrong. When I created the rings of polygons that were a distance \(d\) away from some fixed pentagon, I assumed that no three polygons in a ring would share a vertex. However, as the images below show, that is not the case for the (2,1) fullerene and the (3,1) fullerene.

C140 Rings C260 rings

I will try to prove that the modified single spiral method always works when no three polygons in a ring share a vertex (these are cases in which every polygon in a ring borders exactly two other polygons in that ring, which don’t border each other).

Showing that the Modified Single Spiral Method actually creates a Hamiltongon

First, it’s important to show that the Modified Single Spiral Method actually creates a hamiltongon. Otherwise, there would be no point in using it. I’ll show that it satisfies every property of a hamiltongon.

Connectedness

In the process of creating a hamiltongon through the modified single spiral method, a pentagon is added to the hamiltongon and then hexagons and pentagons that adjoin the edge are repeatedly added. No pentagon or hexagon will ever be added that is isolated from the rest. Therefore, in the final product, no pentagon or hexagon will be isolated from the rest.

No Looping (or connectedness of the -hamiltongon)

For a loop to appear in a hamiltongon through the single spiral method, a polygon would have to be added that would "complete the loop" by bordering two polygons.

When moving up a ring, this cannot occur because the polygon added must be clockwise of the most clockwise polygon in the +hamiltongon in the previous ring, or counterclockwise of the most counterclockwise +hamiltongon in the previous ring. This means it can only border one polygon in the +hamiltongon in the previous ring, so it can’t complete a loop. When moving along a ring, this also cannot occur, because the instructions say to move along a ring until you would break a rule, so polygons would stop being added before this could happen. Therefore, this will not happen.

The Perimeter Must Go Through Every Vertex

As was shown in the first article, an equivalent statement is to say that no vertex is entirely outside or entirely inside the +hamiltongon.

Every vertex is between two rings, or bordering two rings. Otherwise, the vertex would be completely surrounded by polygons in a ring, which is illegal in the cases for which I am proving this.

Each vertex is surrounded by three faces. This means if we call the rings on either side of a vertex ring \(A\) and ring \(B\) (where ring \(A\) is one closer to the origin pentagon than ring \(B\)), there are \(a=1\) or \(2\) polygons surrounding it in ring \(A\), and \(b=3-a\) (\(1\) or \(2\)) polygons surrounding it in ring \(B\). \(1\) or \(2\) of those polygons must be in the +hamiltongon.

Because of the way the modified single spiral method is constructed, there will be a section of ring \(A\) where all of the polygons are in the +hamiltongon, and in the rest of ring \(A\) all of the polygons are in the -hamiltongon. Similarly, there is a section of ring \(B\) where all of the polygons are in the +hamiltongon, and in the rest of ring \(B\) all of the polygons are in the -hamiltongon. The section in ring \(B\) that’s in the +hamiltongon touches one polygon in the section of ring \(A\) that’s in the +hamiltongon, and then moves opposite to the section of ring \(A\) in the -hamiltongon until it ends one polygon before the end of that section (this is because it goes as far as possible while preventing a loop).

The section of ring \(A\) in the +hamiltongon and the section of ring \(B\) in the +hamiltongon are only connected by one pair of polygons. This means that the form a chain of polygons, where no two polygons not adjacent in the chain border each other. This means there is no vertex completely surrounded by the +hamiltongon, as that would require two polygons not adjacent in the chain in the +hamiltongon to border each other.

Since the section of ring \(B\) in the +hamiltongon moves opposite to the section of ring \(A\) in the -hamiltongon until one before the end, the section of ring \(A\) in the -hamiltongon and the section of ring \(B\) in the -hamiltongon are also only connected by one pair of polygons. This means they also form a chain of polygons in which no two polygons not adjacent in the chain border each other, which means no vertex can be completely surrounded by the -hamiltongon (or completely outside of the +hamiltongon).

Conclusion

As the shape created by the modified single spiral method satisfies all the properties of a hamiltongon, it is a hamiltongon.

Proof

To show that a modified single spiral hamiltongon can always be created when no vertex is contained entirely within a ring, I’ll show that no step in the construction of the modified single spiral method fails, going through one step at a time.

  1. Start by choosing a pentagon on the spherical fullerene. Call it \(P\). Group the polygons in the figure into rings based on their distance from \(P\).

    • The first step cannot fail, as a pentagon can always be chosen.

  2. Add \(P\) to the +hamiltongon.

    • The second step cannot fail, as a pentagon can always be added to the +hamiltongon.

  3. Add one of the polygons in ring \(1\) to the +hamiltongon.

    • The third step cannot fail, as an adjoining polygon can always be added to the hamiltongon

  4. There will be one or more polygons adjoining the polygon that was just added in ring \(2\). Add the one that is clockwise of the rest.

    • The fourth step cannot fail. This polygon borders one pentagon in ring \(0\) and two polygons in ring \(1\). Therefore, it borders at least two polygons in ring \(2\). This means there will be a polygon clockwise of the others, which can be added.

  5. Continue adding polygons in a clockwise direction in ring \(2\) until adding one would break a rule.

    • The fifth step cannot fail, as no vertex will be enclosed by it since no three polygons in a ring share a vertex, and the +hamiltongon won’t enclose some isolated piece of the -hamiltongon, because the addition of polygons stops when a rule would be broken.

  6. Add the polygon of the adjoining ones in the ring above that is clockwise of the rest and keep adding polygons clockwise of it until you would break a rule OR if that is not possible, add the polygon of the adjoining ones in the ring above that is counterclockwise of the rest and keep adding polygons counterclockwise of it until you would break a rule.

    • The sixth step is a bit more complicated. This is because of the part where you must add an adjoining polygon in the ring above which is clockwise or counterclockwise of the rest (depending on direction). If there is only one such adjoining polygon, it will share a vertex with two of the polygons that have already been added to the +hamiltongon in the row below, therefore breaking a rule. This could be avoided if you can always add from either the polygon in the ring below that is clockwise of the rest, or the polygon in the ring below that is counterclockwise of the rest. The section of a ring that’s in the +hamiltongon is always exactly \(\frac{0}{5}\), \(\frac{1}{5}\), \(\frac{2}{5}\), \(\frac{3}{5}\), or \(\frac{4}{5}\) of the polygons in that ring. This can be shown by induction. A spherical fullerene is symmetric across a \(72 \degree\) rotation, which means the polygon at the clockwise end of the section of a ring \(A\) (which I’ll call \(R\)) in the +hamiltongon is equivalent to the polygon that is one counterclockwise of the polygon at the counterclockwise end of the section in ring \(A\) in the +hamiltongon. This means for both directions to fail, the polygon at the counterclockwise end (which I’ll call \(P\)) and the polygon counterclockwise of that (which I’ll call \(Q\)) must only border one polygon in the ring above, which I’ll call ring \(B\). This polygon must then border four polygons in ring \(A\): the two polygons that only border one polygon in ring \(B\), as well as the polygons on either side of them. As it must border two polygons in ring \(B\), it borders no polygons in ring \(C\). This means the two polygons it borders in ring \(B\) are adjacent to each other. However, that means those three polygons surround a vertex, so a vertex is entirely contained within ring \(B\). This is a contradiction. Therefore, \(P\) and \(Q\) cannot each border only one polygon in ring \(B\), which means polygons \(P\) and \(R\) cannot each border only one polygon in ring \(B\) (as \(Q\) and \(R\) are equivalent). This means step \(6\) is always possible.

  7. Repeat step \(6\) until you are forced to stop. If this works, you should be forced to stop on the diametrically opposed pentagon to \(P\) or one of the five polygons adjoining it, and you should have created two hamiltongons on the spherical fullerene.

    • If step \(6\) works in general, it should work repeatedly.

Works Cited

  1. Dodecahedrane. (2004, September 16). Retrieved April 11, 2021, from https://pubchem.ncbi.nlm.nih.gov/compound/Dodecahedrane

  2. Buckminsterfullerene. (n.d.). Retrieved February 09, 2021, from https://pubchem.ncbi.nlm.nih.gov/compound/Buckminsterfullerene#section=Structures

  3. Walton, D. R., & Kroto, H. W. (n.d.). Fullerene. Retrieved February 09, 2021, from https://www.britannica.com/science/fullerene