Also available at http://amc.imfm.si
ISSN 1855-3966 (printed edn.), ISSN 1855-3974 (electronic edn.)
ARS MATHEMATICA CONTEMPORANEA 5 (2012) 295–306
Side lengths of pseudoconvex fullerene patches
Christina Graves , Jennifer McLoud-Mann
The University of Texas at Tyler, Department of Mathematics, Tyler, TX 75799, USA
Received 20 September 2011, accepted 11 November 2011, published online 29 March 2012
Abstract
In this paper we consider fullerene patches with nice boundaries containing between
one and five pentagonal faces. We find necessary conditions for the side lengths of such
patches, and then prove these conditions are sufficient by constructing such patches.
Keywords: Fullerenes, fullerene patches, boundary codes, pseudoconvex patches.
Math. Subj. Class.: 05C10, 05C75, 92E10
1 Preliminaries
Mathematically speaking, a fullerene is a trivalent plane graph having only hexagonal and
pentagonal faces. As a consequence of Euler’s formula, such a graph has exactly 12 pen-
tagons. A fullerene patch is formed by taking a simple closed curve on a fullerene and
deleting all vertices and edges on one side of the curve. The simple curve used to create
the patch will be called the boundary. Equivalently, we can think of a fullerene patch as
being a plane graph having only hexagonal and at most 12 pentagonal faces in addition
to one external face with all internal vertices having valence 3 and all vertices on the the
external face having valence 2 or 3. Such patches have been studied in [2], [3], and [7].
These patches are also closely related to nanocone tips (see [3] and [9]).
The boundary code of a fullerene patch is described by a sequence of 2’s and 3’s cor-
responding to the valence of the vertices on the boundary in cyclic order. Boundary codes
with the same cyclic permutation (same sequence starting at a different vertex) are equiv-
alent. In addition, boundary codes represented by a cyclic permutation and its inverse
(changing clockwise to counterclockwise orientation or vice versa) are equivalent. Bound-
ary codes have been studied in [6], [8], and [5] and there have been many algorithms that
decide whether a given boundary code is permissible (see [4] and [1]).
In this paper, we limit our study to fullerene patches with nice boundaries. Adopting
the terminology from [3], we summarize a few important definitions and include some of
our own. A pseudoconvex patch is a fullerene patch where the boundary code does not
E-mail addresses: cgraves@uttyler.edu (Christina Graves), jmcloud@uttyler.edu (Jennifer McLoud-Mann)
Copyright c© 2012 DMFA Slovenije
296 Ars Math. Contemp. 5 (2012) 295–306
contain the subsequence (3,3). In a pseudoconvex patch, a break edge is an edge on the
boundary that has two vertices of degree 2. The section of the boundary between two break
edges is a side of the patch and the length of a side is the number of vertices of degree 3
between the two break edges. A face is a (`i, `j)-corner if it is the face on both a side of
length `i and a side of length `j . In this paper, we are only studying pseudoconvex patches,
so we use the term patches if the meaning is clear. Due to Euler’s formula, the number of
sides and the number of pentagons in a pseudoconvex patch are closely related. [2]
Lemma 1.1. In a pseudoconvex patch, the number of sides s and the number of pentagons
p is related by
s = 6− p
Because the boundary codes can be quite long with pseudoconvex patches, we use a
new method to describe the boundary.
Definition 1.2. The side parameters of a pseudoconvex patch with side lengths `1, . . . , `s
in cyclic order are given by [`1, . . . , `s].
The side parameters of a psuedoconvex patch come directly from the boundary code by
counting the number of 3’s between two pairs of consecutive 2’s where three 2’s in a row
correspond to a side of length 0, and four 2’s in a row correspond to two sides of length 0.
For example, a patch with the boundary code
222(323)222(3)22(323)
has side parameters [0, 2, 0, 1, 2]. To go to the boundary code from side parameters, we fol-
low a similar process. If a patch has s sides and side parameters [`1, . . . , `s], the boundary
code is given by
2(23)`1(2)(23)`2 · · · (2)(23)`s .
It was shown in [7] that the side parameters of a patch with only hexagons satisfy certain
conditions. It is not hard to follow the proof backward to show that the converse is also
true. The result is summarized below.
Lemma 1.3. The sequence of integers [`1, `2, `3, `4, `5, `6] are the side parameters of a
pseudoconvex patch consisting solely of hexagonal faces if and only the the following three
equalities are satisfied:
(1.3.1) `1 + `2 = `4 + `5
(1.3.2) `2 + `3 = `5 + `6
(1.3.3) `3 + `4 = `6 + `1
Equivalently, this lemma says that the sequence 2(23)`1 · · · 2(23)`6 is the boundary
code for a patch if and only if the three equalities above hold.
The goal of the rest of this paper is to continue in the manner of Lemma 1.3 and find
necessary and sufficient conditions on the side parameters of any pseudoconvex patch.
Note that we only need focus on patches containing zero to five pentagons since a patch
containing seven to twelve pentagons is a complement of an aforementioned patch. Many
of these proofs involve exploring the effect of adding or deleting rows of hexagons from a
given patch.
C. Graves and J. McLoud-Mann: Side lengths of pseudoconvex fullerene patches 297
2 Necessary conditions on side parameters
To find conditions on patches with one or more pentagons, we will use inductive arguments
that involve making a patch smaller. We accomplish this by deleting all faces on one side
of a patch which will be called stripping off the side. It is clear that stripping off a side
results in another pseudoconvex patch [7]. Also, adding a row of hexagons along a side
results in another pseudoconvex patch. The following lemma tells us how these processes
affect the side parameters of a patch under certain conditions.
Lemma 2.1. Let Π be a pseudoconvex patch with side parameters [`1, . . . , `s].
1. If the `1 side contains only hexagons, `2 and `s are positive, and s ≥ 3, then stripping
off the `1 side results in a new pseudoconvex patch with parameters
[`1 + 1, `2 − 1, `3, . . . , `s−1, `s − 1].
2. If `1 > 0 and s ≥ 3, then adding a row of hexagons to the `1 side results in a new
pseudoconvex patch with side parameters
[`1 − 1, `2 + 1, `3, . . . , `s−1, `s + 1].
There are many extensions of this lemma including what happens if adjacent sides have
length 0, or what happens if the side to strip has a pentagon on it. All arguments are similar,
though, and the list would be too extensive to include here. We will be using many of these
arguments throughout the rest of the paper.
Because many of the arguments throughout this paper are inductive, a special type of
pseudoconvex patch is important to understand. A linear patch is a patch with boundary
code
(23)`12r1(23)`12r2 .
In other words, a linear patch is a straight chain of adjacent hexagons capped off at each
end with either a pentagon or hexagon. Thus, the possible side parameters of a linear patch
are [`1, 0, 0, `1, 0, 0] (if both ends are hexagons), [`1, 0, 0, `1, 0] (if one end is a hexagon
and one is a pentagon), and [`1, 0, `1, 0] (if both ends are pentagons). In fact, any time a
pentagon in a patch has four edges on the boundary, the patch must be linear. Also, any
patch whose side parameters contain two consecutive zeros must be linear.
The necessary conditions on the side parameters of a pseudoconvex patch containing
exactly one pentagon are stated in the following lemma.
Lemma 2.2. Let Π be a pseudoconvex patch with side parameters [`1, `2, `3, `4, `5]. Then
the following inequalities hold:
(2.2.1) `1 + `2 ≥ `4
(2.2.2) `2 + `3 ≥ `5
(2.2.3) `3 + `4 ≥ `1
(2.2.4) `4 + `5 ≥ `2
(2.2.5) `5 + `1 ≥ `3
298 Ars Math. Contemp. 5 (2012) 295–306
Proof. Let Π be as given. Let d be the minimum distance from the pentagon to a boundary
face of Π and without loss of generality assume that the pentagon is closest to the `1 side.
We will induct on d.
For the base case where d = 0, the pentagon must be on a boundary face of the patch.
If the pentagon is the only face in the patch, then the parameters are [0, 0, 0, 0, 0] which
clearly satisfy all the inequalities. If the pentagon has four edges on the boundary, then
the patch must be linear with parameters [0, a, 0, 0, a] and these parameters also satisfy
all required inequalities. Notice that any sequence of side parameters that contains two
consecutive zeros falls into these categories. Thus our work on the base case begins in
earnest when when `1 > 0 and there are no two consecutive 0’s.
Let t be the distance from the pentagon to the (`1, `5)-corner. If neither `2 nor `5 are
zero, then we strip off the `1 side to create a new patch containing only hexagons. This
patch has side parameters [t, `1 − t, `2 − 1, `3, `4, `5 − 1] (see Figure 1). Using the result
from Lemma 1.3, we have the following three equalities:
`1 = `3 + `4
`1 + `2 = `4 + `5 + t
`2 + `3 = `5 + t.
Using these equalities, we see that (2.2.1) is a consequence of the middle equality,
(2.2.2) is derived from the bottom equality and (2.2.3) comes from top equality. Since
t ≤ `1, inequality (2.2.4) comes from the middle equation. Finally, (2.2.5) is obtained by
noticing that `1 + `5 = `3 + `4 + `5 (from the top equation) which is clearly greater than
or equal to `3.
t `1 − t
t `1 − t
`2`5
Figure 1: Stripping off a side with a pentagon distance t from the corner and adjacent sides
having nonzero length. Red lines indicate the new boundary.
If `2 = 0 and `5 is nonzero, we similarly strip off the `1 side to create a new patch
containing only hexagons. This time, though, the patch has side parameters [t, `1 − t −
1, 0, `3 − 1, `4, `5 − 1]. We again use the equalities from Lemma 1.3 to get the same three
equations as the previous case. Finally, we consider the case where `2 = `5 = 0. If the
C. Graves and J. McLoud-Mann: Side lengths of pseudoconvex fullerene patches 299
pentagon is on the (`2, `1)-corner or (`5, `1)-corner, we refer to the linear case described
in the first paragraph of the proof. Otherwise, we again strip off the `1 side of the patch to
create a new patch with only hexagons and side parameters [t−1, `1− t−1, 0, `3−1, `4−
1, 0]. The three equalities that must be satisfied by this patch are the same as above thus
proving the base case.
Now assume that d (the distance from the pentagon to the `1 side) is greater than 0.
Since the pentagon is closest to the `1 side, it is clear that `1 is nonzero. Because the `1
side is only hexagons, stripping the `1 side off the patch results in a new patch containing
exactly one pentagon. The side parameters of the new patch are:
[`1 + 1, `2 − 1, `3, `4, `5 − 1] if `2 6= 0, `5 6= 0;
[`1, 0, `3 − 1, `4, `5 − 1] if `2 = 0, `5 6= 0;
[`1, `2 − 1, `3, `4 − 1, 0] if `2 6= 0, `5 = 0;
[`1 − 1, 0, `3 − 1, `4 − 1, 0] if `2 = `5 = 0.
In all of these new patches, the distance from the pentagon to the `1 side is d − 1, so by
induction we know that the inequalities of the lemma are satisfied. It is not hard to list
these inequalities and verify that the original patch must have satisfied the conditions as
well.
We now list necessary conditions on the side parameters of a pseudoconvex patch con-
taining two pentagons.
Lemma 2.3. Let Π be a pseudoconvex patch with side parameters [`1, `2, `3, `4]. Then
there are no consecutive zeros and the following inequalities must hold.
(2.3.1) 2`1 + `2 + `4 > `3
(2.3.2) 2`2 + `3 + `1 > `4
(2.3.3) 2`3 + `4 + `2 > `1
(2.3.4) 2`4 + `1 + `3 > `2
Proof. First, notice that two consecutive zeros implies we have a linear patch with a hexa-
gon on one end which is impossible. Let Π be a pseudoconvex patch with side paramters
[`1, `2, `3, `4] and define the length of Π to be ` = `1 + `2 + `3 + `4. The remainder of this
proof will go by induction on `. The smallest patch containing two pentagons is the patch
with two pentagons touching and no hexagons. Such a patch has side parameters [1, 0, 1, 0]
which is length 2 and satisfies the inequalities. The only other sequence that has length 2 is
[2, 0, 0, 0] which has two consecutive zeros and is therefore not permissible. Thus the base
case is proven.
We now assume that ` is bigger than 2. Since Π has four sides and two pentagons, either
the pentagons are on opposite corners of the patch or there is some side with no pentagons.
If the two pentagons on opposite corners, then we can remove the two pentagons resulting
in a hexagonal patch with side parameters [1, `1 − 1, `2 − 1, 1, `3 − 1, `4 − 1]. This patch
must satisfy the equalities given for hexagonal convex patches in Lemma 1.3, and we see
that `1 = `3 and `2 = `4. Thus
2`1 + `2 + `4 = 2`3 + 2`2 > `3
300 Ars Math. Contemp. 5 (2012) 295–306
since `2 and `3 cannot both be zero. The other inequalities follow similarly.
If the two pentagons are not on opposite corners, there is some side that consists solely
of hexagons. Without loss of generality, we will assume this side is the `1 side. Stripping
this side from the patch yields a new patch with parameters
[`1 + 1, `2 − 1, `3, `4 − 1] if `2 6= 0, `4 6= 0;
[`1, 0, `3 − 1, `4 − 1] if `2 = 0, `4 6= 0;
[`1, `2 − 1, `3 − 1, 0] if `2 6= 0, `4 = 0;
[`1 − 1, 0, `3 − 2, 0] if `2 = `4 = 0.
Notice the new total length of all of these patches is less than `. In the first case, the
new length is `− 1 and we use induction to see that the following inequalities hold:
2`1 + `2 + `4 > `3;
2`2 + `3 + `1 > `4;
2`3 + `4 + `2 − 3 > `1;
2`4 + `1 + `3 > `2.
Thus the original patch Π must have satisfied the inequalities stated in the lemma. The
other cases can be done similarly.
Lemma 2.4. Let Π be a pseudoconvex patch with side parameters [`1, `2, `3]. Then [`1, `2,
`3] is not of the form [0, 0, k] or [0, 1, k] for k ≥ 0.
Proof. We begin by noticing that there can be no two consecutive zeros because otherwise
the patch would be linear and could not contain three pentagons. Now, assume that Π is a
patch with side parameters [0, 1, k] for some k > 0. The face on the side of length 0 must
be a hexagon (because otherwise the patch would be linear), so we delete that face to create
a new patch with side parameters [1, 0, k − 1]. We continue deleting the the face on the
side of length 0 until we reach a patch with side parameters [0, 1, 0] which is an impossible
configuration to realize.
Lemma 2.5. If Π is a pseudoconvex patch with side parameters [`1, `2], then `1 + `2 ≥ 6
or [`1, `2] ∈ {[2, 2], [1, 4], [2, 3]}. If Π is a pseduoconvex patch with side parameter [`1],
then `1 ≥ 5.
Proof. In the case where Π has four pentagons, we know from [3] that the minimal pseu-
doconvex patches have side parameters [2, 2] and [2, 3]. Thus, we only need to check that
the patches with parameters [1, 3], [0, 4], and [0, 5] are not realizable. Consider a patch with
side parameters [1, 3]. If one of the corner faces is a pentagon, we remove it to get a patch
with three pentagons and parameters [1, 0, 2] which is not possible. If neither corner is a
pentagon, then the side of length 1 is only hexagons and we strip off this side yielding a
[2, 1] patch which is not realizable.
A patch with side parameters [0, 4] (similarly [0, 5]) must have a hexagon on the side of
length 0. Removing this face results in a patch with side parameters [1, 2] (or [1, 3]) which
is not possible. If Π has 5 pentagons, we also refer to [3] to see that the minimal such patch
has side length 5.
C. Graves and J. McLoud-Mann: Side lengths of pseudoconvex fullerene patches 301
Figure 2: Patches with boundary codes [5], [2, 2], [2, 3], and [2, 4] respectively.
3 Sufficient conditions on side parameters
The goal of this section is prove that all necessary conditions stated in the previous section
are actually sufficient conditions. This direction is much more difficult because we will be
working with a sequence of numbers that satisfy some conditions which we must then use
to construct a patch. We begin constructing a patch that has five pentagons.
Lemma 3.1. There exists a pseudoconvex patch with side parameters [`1] for all `1 ≥ 5.
Proof. If `1 = 5, then the desired patch can be seen in Figure 2. For `1 = 5+k with k ≥ 1,
we add k rings of hexagons to the patch in Figure 2 to yield a patch with side parameters
[`1].
The next step is to construct pseudoconvex patches with four pentagons. Given a patch
with side parameters [`1, `2] with `1 > 0, we see that adding a row of hexagons to the `1
side creates a new patch with side parameters [`1 − 1, `2 + 2]. This technique will be used
in the next proof.
Lemma 3.2. There exists a pseudoconvex patch with side parameters [`1, `2] when `1 +
`2 ≥ 6 or when [`1, `2] ∈ {[2, 2], [1, 4], [2, 3]}.
Proof. Three small cases ([2, 2], [2, 3], and [2, 4]) are shown in Figure 2. To draw the rest,
we first assume that `1 ≤ `2. Let d = (`2 − `1) mod 3. Write `2 − `1 = 3k + d. Starting
with [2, 2+d], add a row of hexagons to the side of length 2 followed by a row of hexagons
to the side of length 2 + d. Continue adding these two rows a total of of `1 + k − 2 times.
This yields the patch [`1 + k, `1 + k + d]. Now add k rows of hexagons to the `1 + k + d
side to get a patch with side parameters [`1, `1 + 3k + d] = [`1, `2].
Before creating patches with three pentagons based on their side parameters, it is useful
to remember how adding rows of hexagons changes the side parameters of a patch with
three pentagons. Consider a patch with side parameters [`1, `2, `3]. Notice that if we want
to add a row of hexagons to the `1 side of this patch, we need `1 to be greater than 0, and
the new patch has side parameters [`1 − 1, `2 + 1, `3 + 1]. If we want to add a row of
hexagons to two sides at once, this is also possible. In this case, we already know from
Lemma 2.4 that no two of the sides can be 0. If only one of the sides is 0, we can still add
to both sides at the same time by essentially adding to the nonzero side first. Thus, adding
a row of hexagons to both the `1 side and the `2 side of a patch with parameters [`1, `2, `3]
results in a new patch with parameters [`1, `2, `3 + 2]. This last operation will be especially
important because it allows us to keep two sides of a patch the same size while increasing
the length of the third side.
302 Ars Math. Contemp. 5 (2012) 295–306
Figure 3: Patches with three pentagons and boundary codes [1, 1, 1] and [1, 1, 2].
Lemma 3.3. There exists a pseudoconvex patch with side parameters [`1, `2, `3] if [`1, `2,
`3] is not of the form [0, 0, k] or [0, 1, k] for k ≥ 0.
Proof. We begin by showing the patches with parameters [1, 1, 1] and [1, 1, 2] which are
depicted in Figure 3. From these two figures we can make a few other essential patches. A
[2, 2, 1] patch can be constructed by adding a row of hexagons to a side of length 1 in the
[1, 1, 2] patch. A patch with side parameters [2, 2, 2] can be constructed from the [1, 1, 1]
patch by adding rows of hexagons to all three sides. A [0, 2, 2] patch can be constructed by
adding a row of hexagons on one side of the [1, 1, 1] patch. A patch with side parameters
[0, 2, 3] can be constructed by adding a row of hexagons to a side of length 1 in the [1, 1, 2]
patch. A [0, 3, 3] patch can be constructed by adding two rows of hexagons to the side of
length 2 on the [1, 1, 2] patch. For the general construction, assume [`1, `2, `3] is not of the
form [0, 0, k] or [0, 1, k] for k ≥ 0 and that `1 is the smallest side parameter.
If `1 is nonzero, consider the patch [S1, S2, S3] where Si = (1 + `i) mod 2 + 1 that
was constructed above. For i = 1, 2, and 3 add b `i−12 c rows of hexagons to the two sides
Sj where j 6= i thus keeping the lengths of two sides fixed while increasing the length of
the Si side by 2b `i−12 c. Then the ith side of the new patch has length
(1 + `i) mod 2 + 1 + 2
⌊
`i − 1
2
⌋
= `i.
If `1 = 0, consider the patch [0, S2, S3] where Si = `2 mod 2 + 2 that was constructed
in the beginning of proof. For i = 2 and i = 3, add b `i−22 c rows of hexagons to the two
sides Sj where j 6= i (making sure to add to the nonzero side first). Adding these rows to
two sides keeps the length of those two sides fixed while the ith side of the new patch has
length
`i mod 2 + 2 + 2
⌊
`i − 2
2
⌋
= `i.
It still remains to be shown that the necessary conditions on the patches with one or
two pentagons are also sufficient conditions. Before creating patches with two pentagons,
we examine how adding rows of hexagons changes the side parameters of such a patch.
Consider a patch with side parameters [`1, `2, `3, `4]. Notice that the conditions in Lemma
2.3 imply that there are not two consecutive 0’s. If we want to add one row of hexagons to
three sides of this patch at once, we can think of this as adding a row to a nonzero side first
and then the other two sides. If the three sides we add a row to are the `1, `2, and `3 side,
then our new patch has side parameters [`1, `2 + 1, `3, `4 + 2].
C. Graves and J. McLoud-Mann: Side lengths of pseudoconvex fullerene patches 303
Lemma 3.4. Let [`1, `2, `3, `4] be a sequence of numbers satisfying 2`i+`i+1+`i+3 > `i+2
where i is cyclic in 1 to 4. Then there exists a pseudoconvex patch with side parameters
[`1, `2, `3, `4].
Proof. Because the side parameters are cyclic and invertible, we can rearrange the side
parameters so that `1 ≥ `3, `2 ≥ `4, and 2`3 − `1 ≤ 2`4 − `2. We say the parameters are
of Type k if `1 + `3 − `2 − `4 = k mod 3. We also introduce the notation
ri =
2`i−2 + `i−1 + `i+1 − `i
3
for i from 1 to 4 where indices are cyclic in 1 through 4. The hypotheses of the lemma
tell us that all ri’s are positive and are only integers for Type 0 conditions. Also, `2 ≥ `4
implies that r2 ≤ r4; `1 ≥ `3 implies that r1 ≤ r3; and 2`3 − `1 ≤ 2`4 − `2 implies that
r3 ≥ r4. There are two essential patches that we will build our new patch from. These
patches have side parameters [1, 1, 1, 1] and [1, 0, 1, 0] and are shown in Figure 4. The
overall construction will be to add either bri − 1c or bric rows of hexagons to a jth side of
an essential patch depending on these conditions. We must take some care to make sure we
are never adding rows of hexagons to a side of length 0.
Start by drawing the patch [1, 1, 1, 1] for Type 0 conditions, [0, 1, 0, 1] for Type 1 con-
ditions, and [1, 0, 1, 0] for Type 2 conditions. Label the sides of the patch S1, S2, S3, and
S4 in cyclic order.
We begin modifying these essential patches by first adding rows of hexagons to sides
S2, S3, and S4. Under Type 0 conditions we add br2 − 1c rows whereas we add br2c for
Type 1 or 2 conditions. This new patch has side parameters
[2r2 − 1, 1, r2, 1] for Type 0 conditions,
[2r2 −
2
3
, 1, r2 −
1
3
, 1] for Type 1 conditions,
[2r2 −
1
3
, 0, r2 +
1
3
, 0] for Type 2 conditions.
Next we add r4 − r2 rows of hexagons to the S3 and S4 sides of the patch creating a
patch with side parameters
[r4 + r2 − 1, r4 − r2 + 1, r2, 1] for Type 0 conditions,
[r4 + r2 −
2
3
, r4 − r2 + 1, r2 −
1
3
, 1] for Type 1 conditions,
[r4 + r2 −
1
3
, r4 − r2, r2 +
1
3
, 0] for Type 2 conditions.
The next step is to add br3c−br4c rows of hexagons to the S3 side of the patch yielding
a patch with side parameters
[r4 + r2 − 1, r3 − r2 + 1, r2 − r3 + r4, r3 − r4 + 1] for Type 0 conditions,
[r4 + r2 −
2
3
, r3 − r2 +
2
3
, r2 − r3 + r4, r3 − r4 +
2
3
] for Type 1 conditions,
[r4 + r2 −
1
3
, r3 − r2 +
1
3
, r2 − r3 + r4, r3 − r4 +
1
3
] for Type 2 conditions.
304 Ars Math. Contemp. 5 (2012) 295–306
Figure 4: Patches with two pentagons and boundary sequences [1, 0, 1, 0] and [1, 1, 1, 1].
Note that the inequalities on ri’s allow us to decrease the third side parameter as needed.
Finally, we add rows of hexagons to S1. We add br1−1c rows under Type 0 conditions
and br1c rows under Type 1 or 2 conditions. Note that the inequalities on ri’s allow us
to decrease the first side parameter as needed. Under any conditions the ith side becomes
ri−1 + ri+1 − ri = `i, thus our new patch has side parameters [`1, `2, `3, `4] as desired.
We now consider pseudoconvex patches with one pentagon. This construction of such
patches will be similar to the previous proof. Consider a patch that has [`1, `2, `3, `4, `5] as
its side parameters with either `1 or `2 being nonzero. Then we can add a row of hexagons
to these two adjacent sides by adding to the nonzero side first creating a new patch with side
parameters [`1, `2, `3 + 1, `4, `5 + 1]. If both `1 and `2 are zero, then we have a linear patch
with parameters [0, 0, `3, 0, `3]. In order to add a row of hexagons to the `1 and `2 side, we
simply add a hexagon to the end of the linear patch creating a new patch with parameters
[0, 0, `3 + 1, 0, `3 + 1]. This idea of adding rows of hexagons to two sides simultaneously
plays a large role in the following proof.
Lemma 3.5. Let `1, `2, `3, `4, and `5 be given nonnegative integers satisfying the condi-
tions that `i + `i+1 − `i+3 ≥ 0 for i from 1 to 5 where subscripts are cyclic in 1 to 5. Then
there is a pseudoconvex patch with side parameters [`1, `2, `3, `4, `5].
Proof. Without loss of generality, we assume that `1 is the largest of the side parameters
and that `2 ≥ `5. We begin by drawing a single pentagon which is a patch with side
parameters [0, 0, 0, 0, 0]. Label the sides of this patch S1, S2, S3, S4, and S5 in cyclic order.
The overall construction will be to add `i+2 + `i+3 − `i rows of hexagons to side i of the
single pentagon patch.
We begin by adding `3 + `4 − `1 rows of hexagons to the S1 and S5 sides of the patch
creating a patch with side parameters
[0, `3 + `4 − `1, 0, `3 + `4 − `1, 0].
We then add (`2 − `5) + (`1 − `4) rows of hexagons to sides S4 and S5 yielding a patch
with side parameters
[`2 − `5 + `1 − `4, `3 + `4 − `1, `2 − `5 + `1 − `4, `3 + `4 − `1, 0].
The next step is to add `5 rows of hexagons to the S3 and S4 sides which creates a patch
with side parameters
[`2 − `5 + `1 − `4, `3 + `4 + `5 − `1, `2 − `5 + `1 − `4, `3 + `4 − `1, `5].
C. Graves and J. McLoud-Mann: Side lengths of pseudoconvex fullerene patches 305
There are now two cases to consider. The first case is if `1 + `2 ≤ `3 + `4 + `5. In this case
we add `1 − `3 rows of hexagons to sides S2 and S3. This patch has side parameters
[2`1 + `2 − `3 − `4 − `5, `3 + `4 + `5 − `1, `1 + `2 − `4 − `5, `4, `5].
To this new patch, we add `3 + `4 + `5 − `1 − `2 rows of hexagons to side S2 creating a
patch with side parameters
[`1, `2, `3, `4, `5].
For the case where `1 + `2 > `3 + `4 + `5, we start with the patch with side parameters
[`2 − `5 + `1 − `4, `3 + `4 + `5 − `1, `2 − `5 + `1 − `4, `3 + `4 − `1, `5]
that we created above. At this point we add `4 + `5 − `2 rows of hexagons to the S2 and
S3 sides of the patch creating the patch with side parameters
[`1, `3 + `4 + `5 − `1, `2 − `5 + `1 − `4, `3 + 2`4 + `5 − `1 − `2, `5].
Finally, adding `1 + `2 − `3 − `4 − `5 rows of hexagons to the S3 side creates the desired
patch.
All of our results are summarized in the following theorem.
Theorem 3.6. The sequence of nonnegative integers [`1, . . . , `s] are the side parameters
of a pseudoconvex patch consisting of 6− s pentagons if and only if one of the following is
true:
1. s = 1 and `1 ≥ 5.
2. s = 2 and [`1, `2] ∈ {[2, 2], [1, 4], [2, 3]} or `1 + `2 ≥ 6
3. s = 3 and the side parameters are not of the form [0, 1, k] or [0, 0, k] for k ≥ 0.
4. s = 4 and there are no two consecutive zeros and the following four inequalities are
satisfied:
(a) 2`1 + `2 + `4 > `3
(b) 2`2 + `3 + `1 > `4
(c) 2`3 + `4 + `2 > `1
(d) 2`4 + `1 + `3 > `2
5. s = 5 and the following five inequalities are satisfied:
(a) `1 + `2 ≥ `4
(b) `2 + `3 ≥ `5
(c) `3 + `4 ≥ `1
(d) `4 + `5 ≥ `2
(e) `5 + `1 ≥ `3
6. s = 6 and the following three equalities are satisfied:
306 Ars Math. Contemp. 5 (2012) 295–306
(a) `1 + `2 = `4 + `5
(b) `2 + `3 = `5 + `6
(c) `3 + `4 = `6 + `1
As a corollary, we state this theorem in terms of boundary codes.
Corollary 3.7. The boundary code 2(23)`12(23)`2 · · · 2(23)`s is the boundary code for
a fullerene patch if and only if one of the conditions (1)-(6) in the previous theorem is
satisfied.
References
[1] P. Bonsma and F. Breuer, Finding fullerene patches in polynomial time, in ISAAC 2009, volume
5878 of LNCS (2009), 750–759.
[2] J. Bornhoft, G. Brinkmann and J. Greinus, Pentagon-hexagon-patches with short boundaries,
Eur. J. Combin. 24 (2003), 517–529.
[3] G. Brinkmann and N. Van Cleemput, Classification and Generation of Nanocones, Discrete Appl.
Math. 159 (2011), 1528–1539.
[4] M. Deza, P. W. Fowler and V. Grishukhin, Allowed boundary sequences for fused polycyclic
patches and related algorithmic problems, J. Chem. Inf. Comput. Sci. 41 (2001), 300–308.
[5] J. E. Graver, The (m, k)-patch boundary code problem, MATCH Commun. Math. Comput.
Chem. 48 (2003), 189–196.
[6] J. E. Graver and G. Cargo, When does a curve bound a distorted disk?, SIAM J. Discrete Math.
25 (2011), 280–305.
[7] J. E. Graver and C. Graves, Fullerene Patches I, Ars Math. Contemp. 3 (2010), 109–120.
[8] X. Guo, P. Hansen and M. Zheng, Boundary Uniqueness of Fusenes, Discrete Appl. Math. 118
(2002), 209–222.
[9] D. J. Klein and A. T. Balaban, The Eight Classes of Positive-Curvature Graphic Nanocones, J.
Chem. Inf. Model. 46 (2006), 307–320.