Also available at ISSN (printed edn.), ISSN (electronic edn.) ARS MATHEMATICA CONTEMPORANEA 3 (2010)

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1 Also available at ISSN (pinted edn.), ISSN (electonic edn.) ARS MATHEMATICA CONTEMPORANEA 3 (2010) Fulleene patches I Jack E. Gave Syacuse Univesity, Depatment of Mathematics, Syacuse, NY , USA Chistina M. Gaves The Univesity of Texas at Tyle, Depatment of Mathematics, Tyle, TX 75799, USA Received 10 Decembe 2009, accepted 17 Apil 2010, published online 10 May 2010 Abstact Lage cabon molecules, discoveed at the end of the last centuy, ae called fulleenes. The most famous of these, C 60 has the stuctue of the socce ball: the seams epesent chemical bonds and the points whee thee seams come togethe epesent the cabon atoms. We use the tem fulleene to epesent all mathematically possible stuctues fo the chemical fulleenes: tivalent plane gaphs with only hexagonal and pentagonal faces. A simple consequence of Eule s fomula is that each fulleene has exactly 12 pentagonal faces; the only estiction on the numbe of hexagonal faces is that it not be 1. The chemical motivation fo this pape is to answe the question: When is it possible to alte a fulleene by changing the stuctue inside a egion of the fulleene bounded by a simple closed cicuit? Such a egion o patch is said to be ambiguous if alteations may be made to its inteio without distubing the stuctue of the fulleene outside of the egion. In this pape, we show that, elative to the minimum distance between pentagonal faces, thee ae no small ambiguous patches. Keywods: Fulleenes, fulleene patches, ambiguous patches, gaphite patches. Math. Subj. Class.: 05C10, 05C75, 92E10 1 Intoduction Let Γ = (V, E, F ) be a fulleene, a tivalent plane gaph with only hexagonal and pentagonal faces. Let B = (V (B), E(B)) be an elementay cicuit in Γ. The cicuit B patitions the sphee into two open hemisphees. A patch Π of Γ is constucted by deleting all vetices, edges and faces inteio to one of the hemisphees and eplacing them by a single face bounded by B. Ou notational convention Π = (V Π, E Π, F Π, B) should be intepeted as follows: (V Π, E Π ) is the gaph emaining afte the deletion; F Π is the collection of faces addesses: jegave@sy.edu (Jack E. Gave), cgaves@uttyle.edu (Chistina M. Gaves) Copyight c 2010 DMFA Slovenije

2 110 As Math. Contemp. 3 (2010) of Γ emaining afte the deletion. Hence, ˆΠ = (VΠ, E Π, ˆF Π ), whee ˆF Π is the collection of faces F Π along with the outside face, is a plane gaph. The cicuit B is called the bounday of the patch. Gaphite patches ae defined to be plane gaphs with all hexagonal faces save one outside face and with all vetices on the bounday of the outside face having degee 2 o 3 and with all othe (intenal) vetices having degee 3. While investigating gaphite patches, Guo, Hansen and Zheng [6] poduced two nonisomophic patches with the same bounday code, that is the same sequence of vetex degees aound the bounday. A fulleene o gaphite patch that is not uniquely detemined by its bounday is said to be ambiguous. When the bounday cuve of the Guo, Hansen and Zheng ambiguous patches is taced in the hexagonal tessellation of the plane o on a fulleene it intesects itself, that is, its bounday is not an elementay cicuit. In fact, it was shown in [5] that an ambiguous gaphite patch must tiple cove some hexagon when pojected onto the hexagonal tessellation of the plane. Ou patches have elementay cicuits fo boundaies when dawn on the fulleene. Howeve, the bounday cuve of a fulleene patch with only hexagonal faces may have self intesections when pojected onto the hexagonal tessellation of the plane. In spite of this, it seems to be the case that fulleene patches with only hexagonal faces o at most one pentagonal face cannot be ambiguous. But this has not been poved. In this pape, we show that ambiguous fulleene patches containing at most one pentagonal face, if they exist, must be elatively lage - oughly as lage o lage than the smallest patch of the fulleene containing two pentagonal faces. In the papes [1], by Binkmann, Capoossi and Hansen, [2], by Binkmann, Fiedichs and von Nathusius, and [3], by Binkmann, Gave and Justus, it is shown that the numbe of faces in an ambiguous patch containing at most one pentagonal face is uniquely detemined. The smallest example of an ambiguous fulleene patch is due to Endo and Koto [4] and is pictued in Figue 1. These two patches have the same bounding cicuit and contain two pentagons. Howeve, the elative positions of the pentagons ae diffeent in these two patches. In fact, most of the patches enclosing two o moe pentagonal faces ae ambiguous. Figue 1: The Endo-Koto ambiguous patches. Given any patch consisting of two non-adjacent pentagons joined by a simple path of hexagons, we may constuct a diffeent patch with the same bounday. This genealization of the Endo-Koto constuction is illustated in Figue 2. Let Π = (V Π, E Π, F Π, B) be a patch of a fulleene Γ. If f is a face of Π, we denote the bounday cicuit of f by B(f). The vetices of the bounday, B, of Π have degee 2 o 3 in Π and one easily sees that evey patch of size geate than 1 admits at least two bounday vetices of degee 3. The paths in the cicuit B joining two consecutive vetices of degee 3 ae called the segments of B. The poof of the following lemma is staight fowad.

3 J. E. Gave and C. Gaves: Fulleene Patches I 111 Π Π Figue 2: Two ambiguous patches using the genealized Endo-Koto constuction. Lemma 1.1. Let Π = (V Π, E Π, F Π, B) be a patch of size at least 2. The path S = x 0, e 0, x 1,..., e k, x k+1 is a segment of B if and only if S is a connected component of B B(f) fo some face f of Π. A face f is called an i-face if B B(f) consists of a single segment and that segment has length i. By the distance between two faces in a fulleene we mean the distance between the coesponding vetices in the dual gaph. By the spead of a fulleene we mean the minimum distance between pentagonal faces taken ove all pais of pentagonal faces. By a disk of adius, we mean a patch consisting of all faces of distance o less fom a cental face (the cente of the disk) as long as its bounday is an elementay cicuit; see Figue 3. Fo any patch, its diamete is the maximum of the distances (measued in the entie fulleene) between pais of faces in the patch. Finally, the width of a patch is the diamete of the disk of least diamete containing it. Figue 3: Thee disks of adius 4. The main esult of this pape is: Theoem 1.2. Ambiguous disks contain at least two pentagonal faces. In paticula, all disks with diamete less than the spead of the fulleene ae unambiguous. As a simple coollay of the main esult we have:

4 112 As Math. Contemp. 3 (2010) Coollay 1.3. In a fulleene, any patch with width less than the spead of the fulleene is unambiguous. 2 Poof of the main esult We stat this section by poving seveal basic lemmas. All patches that we investigate will be consideed as stand-alone patches. Hence they can be extended o alteed even though the extension o alteation cannot be caied out in the fulleene fom which they came. Fo the patch Π = (V Π, E Π, F Π, B), p(π) will denote the numbe of pentagonal faces in the patch and s i, fo i = 1 to 5, will denote the numbe of its i-faces. Lemma 2.1. The numbe of pentagonal faces in a patch Π, p(π), is given by p(π) = 6 + s 1 s 3 2s 4 3s 5. Poof. Let Π = (V Π, E Π, F Π, B) be a patch in the fulleene Γ = (V, E, F ). Denote the numbe of vetices on the bounday of Π that have degee j (j = 2, 3) in Π by d j. Summing the vetex degees, we have 3 V Π d 2 = 2 E Π o 6 V Π 4 E Π = 2d 2. Summing the face degees, we have 6h+5p(Π)+ B = 2 E Π, whee h denotes the numbe of hexagonal faces of Π. Combining these two equations gives: 6 V Π 6 E Π = 2d 2 (6h + 5p(Π) + B ) = d 2 d 3 6h 5p(Π). The numbe of face of the plane gaph ˆΠ is h + p(π) + 1 and we then have 6 ˆF Π = 6h + 6p(Π) + 6. Adding this and the pevious equation gives: 6 V Π 6 E Π + 6 ˆF Π = d 2 d 3 + p(π) + 6. By Eule s fomula, the left-hand side equals 12; so, we have p(π) = 6 d 2 + d 3. Finally, one easily wites d 2 and d 3 in tems of s 1,..., s 5 : d 3 = 5 i=1 s i while d 2 = 5 i=1 (i 1)s i giving p(π) = 6 + s i (i 1)s i = 6 (i 2)s i i=1 i=1 i=1 The next lemma then follows at once. Lemma 2.2. If Π and Π ae a pai of ambiguous patches then p(π) = p(π ). By a side of a patch Π we mean a set of bounday faces between two i-faces with i 3. Note that s 3 + s 4 + s 5 is then the numbe of sides of Π. We define the length of a side in tems of the distance between the end faces in the dual. Specifically, if all of the faces of a side have bounding segments of length 2 and if thee ae l 1 of them, we say the side has length l o is an l-side. We will adopt the convention that a bounding 4-face is a side of length 0; this will eliminate the need fo special cases of the fomulas that we will develop late. (5-faces will soon be eliminated fom consideation, hence we adopt no special convention fo them.) See Figue 4. A side is an h k-side (h and k positive integes) if when otating in the clockwise diection its fist h 1 faces and last k 1 faces ae 2-faces and they ae sepaated by a single 1-face.

5 J. E. Gave and C. Gaves: Fulleene Patches I 113 Side of length 3 1-side 0-side 4 2-side Figue 4: Side lengths. A disk of adius that consists solely of hexagons will be called an H-disk. A disk of adius that contains exactly one pentagonal face is a P-disk. Notice that the only H- disk of adius 0 consists of a single hexagon, and the only P -disk of adius 0 is a single pentagon. Poposition 2.3. An H-disk with adius geate than o equal to 1 has six -sides. A P -disk with adius geate than o equal to 1 has eithe (i a ) five -sides, (i b ) fou -sides and one t-side with < t 2 o (ii) five -sides and one h k-side with h + k. Futhemoe, the bounday faces of an H- o P -disk of adius ae pecisely the faces of distance fom the cente face of the disk. Poof. The only H- and P -disks of adius 1 ae easily constucted and ae pictued in Figue 5. It is clea that these disks all satisfy the conclusions of the poposition. (The simplest P -patch of type (ii) is pictued in Figue 7.) Figue 5: One H-disk and two P -disks of adius one. Let Π be an H- o P -disk of adius. We fist show that all of the bounday faces ae a distance fom the cente. Suppose that f is a bounday face of Π at a distance fom

6 114 As Math. Contemp. 3 (2010) the cente whee <. Since, f is a bounday face, thee must be a face f of Γ not in Π but adjacent to f. But then the distance of f fom the cente is less than o equal to and, by the definition of a disk, f must be a face of the patch; contadiction. We poceed by induction on the adius: assume that Π is an H- o P -disk of adius > 1 and that all H-disks and P -disks of adius less than ae accuately descibed by the poposition. Delete all bounday faces of Π and any othe faces that ae a distance fom the cente face c to get Π. Since the faces of Π ae all the faces of Π at a distance of 1 o less fom c and the faces of Π include all faces of Γ at a distance of o less fom c, the faces of Π ae all faces of Γ at a distance of 1 o less fom c. Thus Π is an H- o P -disk of adius 1 and, by induction, it has the bounday stuctue descibed above. Now conside a face f of Γ that is a distance fom c. It must be a face of Π and it must shae an edge with a bounday face of Π. Hence, adding all of the faces of Γ not in Π but adjacent to a bounday face of Π, econstucts Π. In econstucting Π, we obseve that when adding only hexagons to the bounday of an l-side poduces an l + 1-side and adding only hexagons to the bounday of an h k-side poduces anothe h k-side. Thus, if Π is an H-disk of adius 1, adding only hexagons ceates anothe H-disk Π with six -sides. If Π is a P -disk, we may add only hexagons and we see immediately that: if Π is of type (i a ) with five 1-sides, Π is a P -disk of type (i a ) with five -sides; if Π is of type (i b ) with fou 1-sides and one t -side ( 1 < t < 2 2), Π is a P -disk of type (i b ) with fou -sides and one t = t + 1-side ( < t < 2 1 < 2); finally if Π is of type (ii) with five 1-sides and one h k-side (h + k 1), Π is a P -disk of type (ii) with five -sides and one h k-side (h + k 1 ). The only emaining case is that of adding a bounday of hexagons and exactly one pentagon to an H-disk of adius 1. If the pentagon is adjacent to the cental edge of a segment of length 3, the esult is a P -disk of type (i b ) with fou -sides and one 2-side. If the pentagon is not adjacent to the cental edge of a length 3 segment, then it will have just one edge on the bounday of the enlaged disk. Numbe the faces on the side of the enlaged disk that includes the pentagon f 1,..., f +1, whee f 1 and f +1 have bounding segments of length 3. If the pentagon is face f h+1, the P -disk esulting fom adding the bounding faces will be of type (ii) with five -sides and one h k-side whee k + h =. Coollay 2.4. H-disks ae unambiguous. Poof. When an H-disk is pojected onto the hexagonal tessellation of the plane its bounday cuve is an elementay cicuit. Hence it is unambiguous. To pove that P -disks ae unambiguous, we ae going to need to know just which bounday cuves ae possible fo cetain classes of hexagonal patches - patches with only hexagonal faces. By Lemma 2.1, when a hexagonal patch has s 1 = s 4 = s 5 = 0 it has exactly six sides. If the lengths of these sides ae a, b, c, d, e, f in cyclic ode (counteclockwise), we say the patch has side paametes, o simply paametes, (a, b, c, d, e, f). If a hexagonal patch has s 1 = s 5 = 0, s 4 1, then s 3 = 6 2s 4 and it will have 6 s 4 sides of positive length. By ou convention, each 4-face is a side of length 0. Hence such a patch can also be descibed by six side paametes (a, b, c, d, e, f) whee non-adjacent vaiables could take on the value 0; fo example, the patch with paametes (a, 0, c, d, e, f) whee a, c, d, e, f > 0 is a 5-sided hexagonal patch with the a and c sides sepaated by a 4-face.

7 J. E. Gave and C. Gaves: Fulleene Patches I 115 Lemma 2.5. Let Π be a hexagonal patch with s 1 = s 5 = 0 and side paametes (a, b, c, d, e, f). Then thee ae no consecutive sides of length 0 and the following thee equalities must hold: a + b = d + e, b + c = e + f and c + d = f + a. Poof. If we poject Π onto the hexagonal tessellation, then the bounday of the patch can be epesented in the dual (tiangula) tessellation by the cicuit joining consecutive bounding faces. In this epesentation, a side of length x is epesented as staight lines of x dual edges; these staight lines make 120 angles at 3-faces and 60 angles at 4-faces. Since s 1 = s 5 = 0 and s 3 + 2s 4 = 6 (by Lemma 2.1), we see that Π has at most thee 4-faces. If Π has thee 4-faces, then Π is a tiangula patch. It is clea that this tiangle must be an equilateal tiangle and that all sides of Π ae the same length. Thus, the side paametes of Π ae (a, 0, a, 0, a, 0) which do satisfy the above equalities. (See Figue 6.) b=0 b=0 a b=0 f=0 a e=a c=a d=0 a=c+d c c d d d f=0 e=c d 0 e=0 f=c d=a c e=0 f d a a+b b c b b b+c e+f f a+b e e e d c a b b b b+c d+e Figue 6: Hexagonal Patch Boundaies. If Π has exactly two 4-faces, called f 1 and f 2, then it must have two 3-faces, g 1 and g 2. If f 1 and f 2 ae sepaated by only 2-faces, then pojecting the patch onto the hexagonal tessellation has as its diagam a tapezoid with sides of length (a, 0, c, d, e, 0) whee a is the side between f 1 and f 2. It is appaent that sides a and d ae paallel and that b + c = c = e = e+f. We can enlage the patch by extending sides c and e to sides of length c+d. This new patch is an equilateal tiangle, and thus a+b = a = e+d and a+f = a = c+d. On the othe hand, if f 1 and f 2 ae sepaated by 2-faces and one 3-face, then pojecting Π onto the hexagonal tessellation yields a paallelogam with side paametes (a, 0, c, d, 0, f) which must satisfy a = d and c = f giving a + f = d + c, a + b = a = d = e + d and b + c = c = f = e + f. If Π has exactly one 4-face, then pojecting Π onto the hexagonal tessellation yields a patch with side paametes (a, b, c, d, 0, f) whee sides f and c ae paallel and sides a and d ae paallel. By extending sides a and c by b hexagons, we get a paallelogam which must satisfy a+b = d and b+c = f. It follows that a+b = d = e+d and b+c = f = e+f. Adding a + b = d and f = b + c gives a + b + f = b + c + d and canceling the b gives c + d = f + a. Finally, if Π has no 4-faces, then Π is a patch with side paametes (a, b, c, d, e, f)

8 116 As Math. Contemp. 3 (2010) whee sides a and d ae paallel, sides b and e ae paallel, and sides f and c ae paallel. By extending sides a and c by d hexagons, and extending sides f and d by e hexagons, we ceate a paallelogam (see Figue 6). This paallelogam gives a+b = d+e, b+c = e+f and a + b + e + f = b + c + d + e diectly and a + f = c + d by canceling the b and e in the last equation. If a P -disk Π wee ambiguous, thee exists a patch Π with the same bounday. So we intoduce a class of patches that will include all of these possible altenatives. By an -patch, we mean a patch of one of the following two types. An -patch of the fist type is a patch with s 1 = s 4 = s 5 = 0, s 3 = 5 and with 4 sides of length and one side of length t (t ). That is, an -patch of the fist type has side paametes (,,,, t), whee t. An -patch of the second type is a patch with s 4 = s 5 = 0, s 1 = 1 and s 3 = 6, with five -sides and one h k-side. That is, an -patch of the second type has side paametes (,,,,, h k), whee h and k ae positive integes. It is an immediate coollay of Lemma 2.1 that an -patch contains exactly one pentagon. One easily checks that thee ae only two 1-patches: the two P -disks of adius 1 with paametes (1, 1, 1, 1, 1) and (1, 1, 1, 1, 2). These ae both -patches of the fist type and ae pictued in Figue 5. The simplest -patch of the second type has paametes (2, 2, 2, 2, 2, 1 1) and is pictued in Figue 7; note that it is a P -disk of adius 2. Figue 7: The -patch of the second type with paametes (2, 2, 2, 2, 2, 1 1). The following lemma gives anothe popety of -patches namely that we can delete all bounding faces of an -patch ( > 1) and still get a legitimate patch. It will always be possible to emove all of the faces on a side unless one of those faces actually meets the bounday in two distinct segments its emoval would then leave a disconnected patch. It will always be possible to emove all of the faces on the bounday unless two faces on diffeent sides shae an edge. See Figue 8. Lemma 2.6. Let Π = (V Π, E Π, F Π, B) be an -patch > 1. i. Let f be a face of Π. Then, B(f) B consists of at most one path. ii. Let f and g be faces of Π that ae not consecutive as one cyclically taveses the bounday. Then, B(f) B(g) =.

9 J. E. Gave and C. Gaves: Fulleene Patches I 117 Figue 8: Impossible -patch boundaies. Poof. (i) Assume thee exists a face f in Π consisting of 2 disjoint segments, a cut-face. Clealy both segments of f have length 2 o less. If both segments have length 2, then both faces adjacent to f ae also cut-faces. Extending to the left, the sting of cut-faces must teminate. Teminating with a single face is impossible since this is an -patch and such a face would be a 4-face o 5-face. Hence the leftmost cut-face must meet two faces on its left and must have at least one bounding segment of length 1. If f meets two faces on its ight, then it would have two bounding segments of length 1. If f meets only one face on its ight then the ight-most cut face in the sting of cut-faces to the ight must also have at least one bounding segment of length 1. Hence this patch must have at least two bounding segments of length 1 contadicting the assumption that it is an -patch. (ii) Conside any -patch Π and suppose that it has a pentagonal 3-face f on its bounday. If the patch is of type (,,,, t), eplacing f by a hexagon will esult in a hexagonal patch with bounday lengths (, 0,,,, t), (,, 0,,, t) o (,,,, 0, t); all of which violate the conditions of Lemma 2.5. If the patch is of type (,,,,, h k), eplacing f by a hexagon will esult in a hexagonal patch with bounday lengths (, 0,,,,, h k), (,, 0,,,, h k) o (,,,,, 0, h k). Now we can eliminate the segment of length 1 in the h k-side by attaching an h k paallelogam of hexagons (see the left-hand diagam in Figue 9, also see Figue 10). The esulting six-sided hexagonal patches will then have bounday lengths (h +, 0,,,, + k), (h +,,, 0,, + k) o (h +,,,,, k); all of which violate the conditions of Lemma 2.5. Hence Π has no pentagonal 3-face f on its bounday. Let f and g be adjacent, non-consecutative bounday faces - a cut-pai. Since they ae non-consecutive, the face labeled y in the ight-hand diagam of Figue 9 must belong to the patch. If neithe of the faces labeled x and z belong to the patch, then y would be a hexagonal 4-face, a pentagonal 3-face o a cut face. None of these is possible in an -patch. Hence we may assume that z belongs to the patch. We may also assume without loss of geneality that f and g ae the leftmost cut-pai and so x is also in the patch. By a simila agument v and one of u and w on the ight must belong to the patch - say w. Hence g is a 1-face. If u is a face of the patch, then f is also a 1-face. If u is not a face of the patch, then v and w fom a cut pai and the ight-most cut pai in this chain of cut-pais must also include a 1-face. Hence this patch must have at least two 1-faces contadicting the assumption that it is an -patch. The coe of a patch Π is the plane subgaph obtained by emoving all the bounday faces of Π. By the constuction of P -disks in Lemma 2.3, it is clea that a P -disk of adius has eithe an H-disk o a P -disk of adius 1 as its coe. By Lemma 2.6, all -patches have a legitimate patch as thei coe. We use this fact in caying out the next step to

10 118 As Math. Contemp. 3 (2010) h k k h x y f u v z g w Figue 9: Diagams fo the poof of Lemma 5 eventually show that the only -patches ae the P -disks. The next step is to show that the paametes of any -patches actually match the paametes of some P -disk. Lemma 2.7. Let Π be an -patch of the fist type with paametes (,,,, t) Then t 2 and i. if t = 2, Π has a pentagon as the cental face of the t-side; ii. if t < 2, Π has only hexagons on the bounday. Poof. Let Π have paametes (,,,, t). In the poof of the pevious lemma, we showed that a pentagon in the bounday of an -patch cannot be a 3-face. Suppose that the ith face (0 < i < ) in the fist -side is a pentagon. Replace this face by a hexagon. This new hexagonal patch has paametes (i, i,,,, t); which is impossible since i + ( i) +. We get the same kind of contadiction if we assume that any face on any of the -sides ae pentagons. Now assume that the ith face (0 < i < t) in the t-side is a pentagon. Again, we eplace the pentagon by a hexagon to get a hexagonal patch with paametes (,,,, i, t i). We must have + = + i and + = i + (t i). This is possible only if i = and t = 2. Thus, the only way a pentagon can be on the bounday is if t = 2 and the cental face on the t-side is a pentagon. To pove that t < 2 when Π has only hexagonal bounday faces, we ll go by induction on. This inequality holds fo = 1 since the only 1-patch with only hexagonal bounday faces has paametes (1, 1, 1, 1, 1). Let Π be an -patch with paametes (,,,, t) whee > 1 and with only hexagons on its bounday. We can then emove the bounday and conside its coe. If a l-side has only hexagons, emoving them poduces an (l 1)-side. Thus, the coe must be an ( 1)-patch with side paametes ( 1, 1, 1, 1, t 1). By induction, we have that (t 1) 2( 1) o t < 2. Lemma 2.8. Let Π be an -patch of the second type with paametes (,,,,, h k). Then h + k and i. if h + k =, then Π has a pentagon as the face with bounding segment of length 1; ii. if h + k <, then Π has only hexagons on the bounday. Poof. Let Π have paametes (,,,,, h k). Suppose that the ith face in the h k-side, i < h, is a pentagon. We eplace the pentagon by a hexagon to get a hexagonal patch with

11 J. E. Gave and C. Gaves: Fulleene Patches I 119 paametes (,,,,, i, (h i) k). If we attach the hexagons of a (h i) k paallelogam (see Figue 10), we get a 6-sided hexagonal patch with paametes (,,,, i+k, +h i). By Lemma 2.5, we have (i + k) + ( + h i) = 2; fom which we conclude that we must have h + k =. Consideing anothe adjacent pai, we have ( + h i) + = 2; fom which we conclude that i = h. Thus the only case in which one of the faces on the h k-side could be a pentagon is the case whee h + k = and the pentagonal face is the face with bounday segment of length 1. i i+k i h-i k h-i k +h-i Figue 10: Diagams fo the poof of Lemma 2.8 Next suppose that the ith face in fist -side is a pentagon. Replace it in the patch by a hexagon. This esults in a hexagonal patch with paametes (i, i,,,,, h k). Filling in to eliminate the length 1 segment then yields a 6-sided hexagonal patch with paametes (i + h, i,,,, + k); which is impossible since ( i) + = + ( + k) > 2. We get the same kind of contadiction if we assume that any face on any of the -sides ae pentagons. Thus, if Π has a pentagon on the bounday, h + k = and the pentagon is the face with bounding segment of length 1. To pove that h + k < when Π has only hexagonal bounday faces, we poceed by induction on. One easily checks that thee ae no 1-patches of this second type and the only 2-patch of the second type is pictued in Figue 7 and has paametes (2, 2, 2, 2, 2, 1 1). Hence, the lemma holds fo all -patches with 2. Let Π be an -patch with paametes (,,,,, h k) fo > 2 and with only hexagons on its bounday. We can emove the bounday and conside its coe. If an -side has only hexagons, emoving them poduces an ( 1)-side; if an h k-side has only hexagons, emoving them poduces anothe h k-side. Thus, the coe must be an ( 1)-patch with sides ( 1, 1, 1, 1, 1, h k). By induction, we see that h + k 1 o h + k <. This final poposition concludes ou poof of the main esult. Poposition 2.9. P -disks ae unambiguous. Poof. This poof will poceed by induction on. It is easy to check that the only two P -disks of adius 1 ae unambiguous. Hence we assume that Π is a P -disk of adius > 1 and that P -disks of adius 1 ae unambiguous. Let Π be anothe patch with the same bounday as Π. By Poposition 2.3, Π must be an -patch with the same paametes as Π. We will show that Π is in fact the same patch as Π. Fo the fist case, we assume that Π has paametes (,,,, t) with t < 2. By Lemma 2.7, we see that Π and Π have only hexagons on thei boundaies. Stipping off

12 120 As Math. Contemp. 3 (2010) the hexagonal bounday faces of both Π and Π yields in each case a coe that is an ( 1)- patch with paametes ( 1, 1, 1, 1, t 1). By Poposition 2.3, thee is a P -disk with these paametes and by the induction hypothesis it must be the only ( 1)-patch with these paametes. Hence, Π and Π have the same P -disk as coe. Thee is only one way to add hexagons aound the bounday of a P -disk, so Π must be the same patch as Π. Now, assume that Π has paametes (,,,, 2). By Lemma 2.7, both Π and Π have a pentagon as the cental face of the 2-side. Thus the coe of both Π and Π is the H-disk of adius 1. Replacing the bounday faces, we see that Π = Π. If Π has paametes (,,,,, h k) with h+k <, both Π and Π have only hexagons on the bounday. Thus, the coe of Π is the P -disk with adius 1 and paametes ( 1, 1, 1, 1, 1, h k). Since the coe of Π is an -patch with the same paametes, it is by induction the same P -disk as the coe of Π. Again, eplacing the bounday faces, we see that Π = Π. Finally, assume Π has paametes (,,,,, h k) with h + k =. By Lemma 2.8, Π and Π must have a pentagon on the bounday and that pentagon is the bounday face with bounding segment of length 1. Thus, the coe of both Π and Π is the H-disk of adius 1. As above Π and Π have the same coe and the same bounday faces and ae theefoe equal. Refeences [1] G. Binkmann, G. Capoossi and P. Hansen, A suvey and new esults on compute enumeation of polyhex and fusene hydocabons, J. Chem. Inf. Comput. Sci. 43 (2003), [2] G. Binkmann, O. Delgado Fiedichs and U. von Nathusius, Numbes of faces and bounday encodings of patches, Gaphs and discovey, 27 38, DIMACS Se. Discete Math. Theoet. Comput. Sci., 69, AMS, Povidence, RI, [3] G. Binkmann, J. E. Gave and C. Justus, Numbes of Faces in Disodeed Patches, J. Math. Chem. 45 (2009), [4] M. Endo and H. W. Koto, Fomation of cabon nanofibes, J. Phys. Chem. 96 (1992), [5] J. E. Gave, The (m,k)-patch bounday code poblem, MATCH 48 (2003), [6] X. Guo, P. Hansen and M. Zheng, Bounday Uniqueness of Fusenes, Tech. Repot G-99-37, GERAD, 1999.