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Benzenoids with branching graphs being trees
Journal of
MOLECULAR STRUCTURE ELSEVIER
Journal of Molecular Structure 415 (1997) 239-247
Benzenoids with branching graphs being trees Chen Rong-Si College of Finance and Economics, Fuzhou University, Fuzhou 350002, People's Republic of China
Received 13 November 1996; revised 11 February 1997; accepted 12 February 1997
Abstract
The branching graph of a benzenoid hydrocarbon molecule is the subgraph consisting of vertices of degree 3 and the bonds among them. In this paper, a practical method is established for recognizing trees embedded in the infinite hexagonal lattice which can be branching graphs of benzenoids. Moreover, a simple criterion is given to determine when a tree embedded in the infinite hexagonal lattice is the branching graph of a catacondensed benzenoid. © 1997 Elsevier Science B.V. Keywords: Benzenoid hydrocarbons; Branching graphs; Catacondensed benzenoid; Graph theory
1. Introduction A benzenoid is a connected finite subgraph of the infinite hexagonal lattice which has no cut vertices or non-hexagonal interior faces [1]. Benzenoids are of chemical significance since a benzenoid with perfect matchings (Kekul6 structures in chemical language) is the skeleton of a benzenoid hydrocarbon molecule. The branching graph of a benzenoid G is a special subgraph. Each vertex of the benzenoid G appears in its branching graph B(G) if and only if it is of degree 3. Each edge of G appears in B(G) if and only if it connects two such branching vertices. The concept of the branching graph was introduced [2] as a practical aid for diagnosing whether a graph is or is not traceable (a graph is traceable if it has a Hamilton path, i.e. a path that visits every vertex just once). Traceability has chemical relevance in the field of structural information transmission, since a structure that is traceable is often easier to encode from a keyboard than one that is not [3], and in the field of predicting the magnetic properties of
molecules by using the concept of spanning trees for the methods available for calculating r-electron ring currents in a conjugated system depend on whether or not the system is traceable [4-8]. It was also claimed in Ref. [2] that traceability must be among the factors that determine what kinds of structure it is possible or impossible to form by intro-molecular cross-linking of a linear polymer. In a further study [9] branching graphs have provided some insight into why there are so few sextet two-factorable benzenoids among all the benzenoids that are theoretically possible. As the branching graphs appear to have some use in chemistry, it is desirable to understand and characterize branching graphs as much as possible. Many problems about branching graphs have been raised [10[. In this paper we answer the following questions. (1) Which trees embedded in the infinite hexagonal lattice are branching graphs of benzenoids? (2) When is a tree embedded in the infinite hexagonal lattice the branching graph of a catacondensed benzenoid? We present a quick and practical way to recognize trees embedded in the infinite hexagonal lattice which can
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Fig. 1. A tree T embedded in the infinite hexagonal lattice (the vertices of T are labelled by positive integers).
be branching graphs of benzenoids. Moreover, a simple criterion is given to determine when a tree embedded in the infinite hexagonal lattice is the branching graph of a catacondensed benzenoid.
2. Definitions Let T be a tree embedded in the infinite hexagonal lattice. The boundary b(T) of T is defined to be the closed walk that travels T clockwise and each edge of T is traversed twice. For the tree depicted in Fig. 1, the boundary is b(T) = (1,2) (2,3) (3,2) (2,4) (4,5) (5,6) (6,7) (7,8) (8,9) (9,8) (8,10) (10,8) (8,7) (7,6) (6,5) (5,4) (4,2) (2,1). Note that since b(T) is a closed walk, it is not important which vertex of T is the starting point of b(T). For example, we can say that the boundary b(T) of the tree T depicted in Fig. 1 is b(T) = (5,6) (6,7) (7,8) (8,9) (9,8) (8,10) (10,8) (8,7) (7,6) (6,5) (5,4) (4,2) (2,1) (1,2) (2,3) (3,2) (2,4) (4,5). It is clear that each edge e of a tree T embedded in the infinite hexagonal lattice is contained in two hexagons of the infinite hexagonal lattice, denoted by s'(e) and s"(e). A set S of hexagons of the infinite hexagonal lattice is said to be the associate hexagon set of tree T if S = {s'(e), s"(e)le ~ T}. For an illustration, see Fig. 2.
Fig. 3. An odd-even labelling with respect to pendant bond ~.
3. Odd-even labelling Let T be a tree with associate hexagon set S. Along the boundary b(T) clockwise, we label the hexagons of S as odd or even consecutively, according to the following rules: 1. Choose optionally a pendant bond (i.e. an edge with one end-vertex being of degree 1) e of T, label the two hexagons s'(e) and s"(e) as odd. 2. Suppose that hexagon s has been labelled as odd (even). Hexagon ~ appears right after s along the boundary b(T) clockwise, and has not yet been labelled. If the common edge ofs and ~ is a pendant bond of T, then ~ is labelled as odd (even); otherwise, 3 is labelled as even (odd). An illustration is given in Fig. 3, where hexagons each with an " o " are odd hexagons, while those each with an " e " are even hexagons.
4. Lemmas
Lemma 4.1. Assume that T is the branching graph of benzenoid B, S is the associate hexagon set of T. Hexagons s and 3 belong to S and appear consecutively along the boundary b(T) clockwise.If the common edge of s and ~ is a pendant bond of T, then both s and $ belong to B; otherwise, one and only one of s and 3 belongs to B.
Fig. 2. A tree T and the associate hexagon set S = {s~, s2, s3, s4, ss, $6, $7~ $8, $9}.
Fig. 4. Two consecutive hexagons s and 3 on b(T).
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Fig. 5. Illustration of the proof of Lemma 4.2.
Proof. Since s and 3 appear consecutively along b(T)
Remark 4.3. By the above lemma, if Tis the branching
clockwise, s and ~ have an common edge, say e (see Fig. 4). If e is a pendant bond of T, then both s and must belong to B since vertex v' is of degree 3 in B. If e is not a pendant bond of T, e cannot be an edge of T. Hence s and 3 cannot belong to B simultaneously. If none of them belongs to B, then vertex v is of degree 2 in B, and cannot belong to T (bear in mind that T is the branching graph of B), a contradiction. Therefore, if e is not a pendant bond of T, one and only one of s and belongs to B.
graph of benzenzoid B, then the o d d - e v e n labelling is independent of the choice of pendant bond of T. Namely, if a hexagon of S is labelled as odd (even) under the o d d - e v e n labelling with respect to pendant bond e, then it is also labelled as odd (even) under the o d d - e v e n labelling with respect to any pendant bond e' :~ e.
Lemma 4.4. Assume that T is the branching graph of benzenoid B. Then a hexagon of S belongs to B if and only if it is odd.
Lemma 4.2. Assume that T is the branching graph of benzenoid B, the o d d - e v e n labelling of the associate hexagon set S is with respect to the pendant bond e' of T. Then for each pendant bond e of T, both s'(e) and s"(e) are odd.
Proof. Let the pendant bonds of T be arranged as e' =el, e2 ..... et, such that along b(T) clockwise em appears after ei for i = 1,2 ..... p - 1; and e~ appears after e r. By the o d d - e v e n labelling rule, both s'(et) and s"(el) are odd. If s"(el)=s'(e2), then both s'(e2) and s"(e2) are odd (see Fig. 5). Now suppose that s"(el ) 4: s'(e2). If there are no hexagons of S between s"(e l) and s'(e2) along b(T) clockwise, then by Lemma 4.1 s'(e2) does not belong to B. Yet, again by Lemma 4.1, s'(e2) must belong to B since e2 is a pendant bond of T, a contradiction. Hence, there are some hexagons of S, say sl ..... s t (t >- 1), between s"(eO and s'(e2) along b(T) clockwise. We claim that t is an odd number (cf. Fig. 5). If t is an even number, by Lemma 4.1 sl, s2 ..... St_l do not belong to B, and st belongs to B. Thus s'(e2) does not belong to B, again a contradiction. Therefore, t is an odd number. By the o d d - e v e n labelling rule, s'(e2) and s"(e2) are odd hexagons. By the same argument, we can prove that if s'(ei) and s"(ei) are odd, then s'(ei+l) and (S"(ei+l) are also odd for i = 2 ..... p - 1. The lemma follows.
Proof. By Lemma 4.2 all hexagons of S corresponding to pendant bonds of T are odd, and by Lemma 4.1 they belong to B. Now we consider those hexagons of S which do not correspond to pendant bonds of T. Suppose that e and ~ are two consecutive pendant bonds of T along b(T) clockwise, and s 1, $ 2 . . . . . S t are consecutive hexagons of S between s"(e) and s'(O) along b(T) clockwise. As mentioned in the proof of Lemma 4.2, t is an odd number. By Lemma 4.1, even hexagons s 1, s3 ..... st do not belong to B; while odd hexagons s2, s4 ..... st-i belong to B. The lemma is thus proved.
Lemma 4.5. Assume that T is the branching graph of benzenoid B, e is an edge of T. Then at least one of s'(e) and s"(e) is odd under the o d d - e v e n labelling.
Proof. Since B is a benzenoid, each edge of B must belong to at least one hexagon of B. If none of s'(e) and s"(e) is odd under the o d d - e v e n labelling, then none of them belongs to B (Lemma 4.4). Hence e does not belong to any hexagon of B, a contradiction which establishes the lemma.
Lemma 4.6. Assume that T is the branching graph of benzenoid B, v 1 and v2 are two vertices of T. If v i and
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C. Rong-Si/Journal of Molecular Structure 415 (1997) 239-247 v5
',
',
'-..../'3
v2
v2
(a)
(b)
Fig. 6. Two possibilities of the relative positions of e t and e2.
( (
v2 are connected by an edge e in the infinite hexagonal lattice, then e belongs to T. Fig. 8. Illustration of the proof of Lemma 4.7. Proof. Since T is the branching graph of B, v l and v2 are two vertices of T; evidently v i and v2 are of degree 3 in B. Hence e = (vl, v2) must belong to B, and thus T. Lemma 4.7. Assume that T is the branching graph of benzenoid B, e l and e 2 are two edges of T, and s'(e l) and s'(e2) are odd under the o d d - e v e n labelling. If s'(eO A s'(e2) = e, then e belongs to T; if s'(el)=s'(e2)=s, then et and e2 are either adjacent or are connected by a path of T which is on the boundary ofs. Proof. By Lemma 4.4, both s'(eO and s'(e2) belong to B. If s ' ( e l ) N s ' ( e 2 ) = e , then e evidently connects two vertices of degree 3 of B, and hence e belongs to T. Now suppose that s'(el ) = s'(e2) = s. If e j and e 2 are not adjacent, then there are effectively two possibilities as shown in Fig. 6. In case a, by Lemma 4.6 edge (vl, v2) belongs to T. Thus el and e2 are connected by a path of T which is on the boundary of s. Now consider case b. If one of the edges (vl, v2) and (v2, v3) belongs to T, then by Lemma 4.6 the other must also belong to T. Thus e j and e2 are connected by a path of T which is on the boundary of s. Similarly, if one of the edges (v4, vs) and (vs, v6) belongs to T, then el and e2 are connected by a path of T which is on the boundary of s [see Fig. 6(b)]. Now suppose that none of the edges (v 1, v2), (v2, v3), (v4, vs) and (vs, v6)
V
~"
V
Fig. 7. Illustration of the proof of Lemma 4.7.
belong to T. Then hexagons s~, $2, s 3 and s4 do not belong to B (see Fig. 7). Namely, el and e2 are not connected by a path of T which is on the boundary of s. We claim that both el and e2 are pendant bonds of T. If e 1 is not a pendant bond of T, then e l' belongs to T. By Lemmas 4.4 and 4.5, s l' belongs to B. Edge el' cannot be a pendant bond of T since s3 does not belong to B (Lemma 4.1). If e2' belongs to T, by Lemmas 4.4 and 4.5, s2' belongs to B. If e 3' belongs to T, then by the o d d - e v e n labelling rule, s2' belongs to B. By the analogous argument, s3' belongs to B. Then the boundary of s l' belongs to T (see Fig. 7), contradicting the assumption that T is a tree. This contradiction shows that el must be a pendant bond of T. Similarly, e2 must be a pendant bond of T. Then the odd hexagons except s of the associate hexagon set S constitute a benzenoid/~. Since s~, s2, s3 and s4 do not belong to B, B=/~ U {s} must contain a " h o l e " , contradicting the assumption that B is a benzenoid (cf. Fig. 8). The lemma is thus proved.
5. A criterion for a tree to be the branching graph of a benzenoid Theorem 5.1. Assume that T is a tree embedded in the infinite hexagonal lattice; the o d d - e v e n labelling of the associate hexagon set S is respect to a pendant bond ~ of T. Then T is the branching graph of a benzenoid B if and only if the following conditions hold: 1. For any pendant bond e of T, both s'(e) and s"(e) are odd. 2. For any edge e of T, at least one of s'(e) and s"(e) is odd. 3. Assume that vl and v2 are two vertices of T. If v~
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Fig. 9. Illustration of the proof of sufficiency of Theorem 5.1.
and v2 are connected by an edge e of the infinite hexagonal lattice, then e belongs to T. 4. Assume that el and e2 are two edges of T, and both s ' ( e O and s'(e2) are odd. If s ' ( e l ) A s'(e2) =e, then e belongs to T. I f s ' ( e l ) = s ' ( e 2 ) = s , then el and e2 are either adjacent, or are connected by a path of T which is on the boundary of s. Proof. By Lemmas 4.2, 4.5, 4.6 and 4.7, conditions 1 - 4 are necessary. In the following we prove that they are also sufficient. Suppose that B is the graph consisting of all the odd hexagons of S. By condition 2, all the edges and vertices of T belong to B. Thus all the vertices of degree 3 of T are also vertices of degree 3 of B. By condition 1, all the vertices of degree 1 of T are of degree 3 in B. By the odd-even labelling rule and condition 2, all the vertices of degree 2 of T are of degree 3 in B. Hence all the vertices of T are of degree 3 in B. Now we prove that all the vertices of degree 3 of B belong to T. Assume that v is a vertex of degree 3 of B. By condition 2, B contains no edge which does not belong to any hexagon of B. Thus at least two of the three hexagons sb s2 and s3 belong to B (see Fig. 9), say s i and s2. Hence, by condition 4, e belongs to T, and therefore v belongs to T. Note that any edge
Fig. 10. T contains a cycle if for any hexagon on the boundary of the hole of B, the removal of it does not open the hole.
of B connecting two vertices of degree 3 also belongs to B. This is guaranteed by condition 3. We have already proved that T is the branching graph of B. Now we want to prove that B is a benzenoid. If B is not a benzenoid, then B contains a "hole", i.e. an internal face which is bounded by a polygon C with more than six edges. If, for any hexagon on the boundary C of the hole, the removal of it does not open the hole, then the branching graph T of B contains a cycle and is not a tree, a contradiction (see Fig. 10). Hence there is a hexagon of B on the boundary of the hole, say s, and how they are connected to s, such that the removal of s from B will open the hole. According to how many hexagons of B are connected to s, there are effectively six modes, as shown in Fig. 11. Since the branching graph of B is a tree, modes d, e and f are impossible. Now let sbs2 ..... s e be hexagons of B such that the removal of each of them will open the hole. Suppose that none of them are of mode a. We first remove all the vertices of degree 2 on the perimeter of B together with their incident edges. Evidently, the resulting
0
f Fig. 11. Six modes of hexagon s.
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( Fig. 14. Illustration of the proof of the validity of condition 1.
T. If B is catacondensed, then the following conditions are satisfied:
Fig. 12. If B contains no hexagon of mode a on the boundary of the hole, then T contains a cycle.
graph still has a hole. Then we remove all the vertices of degree 2 on the boundary C of the hole together with their incident edges. Now the resulting graph is the branching graph of B, i.e.T. Yet T contains a cycle since none of the hexagons s 1,s2..... sp are of mode a (cf. Fig. 12), contradicting the assumption that T is a tree. Hence B has a hexagon of mode a on the boundary of the hole. Now we find two edges el and e2 of T (see Fig. 13) such that s'(el ) = s'(e2) = s, but e l and e2 are neither adjacent nor connected by a path of T on the boundary of s, contradicting condition 4. This contradiction shows that B cannot contain a hole. Therefore, B is a benzenoid. The proof is completed.
6. Catacondensed benzenoids with branching graphs being trees In this section we present a criterion to determine when a tree is the branching graph of a catacondensed benzenoid. Recall that a benzenoid is said to be catacondensed if all its vertices lie on the perimeter of it.
1. v is connected to one and only one vertex of degree 1 ofT. 2. v is connected to at most one vertex of degree 3 of T. 3. If v is connected to a vertex of degree 3 of T by a chain of T with length greater than 1, then the length is odd; if v is connected to a vertex of degree 1 of T by a chain of T with length greater than 1, then the length is even.
P r o o f Suppose that v is a vertex of degree 3 of T. If v is connected to two vertices of degree 1 of T, say v' and v" (see Fig. 14), then by L e m m a 4.1, hexagons s 1, s2 and s3 belong to B. Thus v is a vertex which does not lie on the perimeter of B, contradicting the assumption that B is catacondensed. Hence v is connected to at most one vertex of degree 1 of T. Now suppose that v is connected to no vertex of degree 1 of T. Then none of v' and v" are of degree 1 of T. Since B is catacondensed, Sl, s2 and s3 cannot belong to B simultaneously. By Lemmas 4.4 and 4.5, two of sl, s2 and s3, say sl and s2, must belong to B. W e claim that s4 must belong to B. Since v" is not of degree 1 in T, at least one of edges el and e2 belongs to T. If ej belongs to T, by Lemmas 4.4 and 4.5, s4 belongs to B. If e i does not belong to T, then e2 belongs to T. Then by Lemma 4.1, s4 must belong to B. By an analogous argument, s6 must belong to B, and at least one of s5
Theorem 6.1. Assume that tree T is the branching graph of benzenoid B, and v is a vertex of degree 3 of
V0V Fig. 13. A situation violating condition 4.
Fig. 15. Illustration of the proof of the validity of condition 2.
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and s7 must belong to B. Now if s5 belongs to B, then hexagon s2 belongs to T; if sv belongs to B, then hexagon s l belongs to T. Each case contradicts the proposition that Tis a tree. Hence v must be connected to at least one vertex of degree 1 of T. Consequently, v is connected to one and only one vertex of degree 1 of T, i.e. condition 1 holds. Suppose that v is a vertex of degree 3 of T which is connected to two vertices of degree 3 of T, say v~ and v2 (see Fig. 15). By condition 1, v' is a vertex of degree 1 of T. Thus s~ and s2 belong to B (Lemma 4.1). Hence s does not belong to B since B is catacondensed. Therefore, edge (v l, v3) is not a pendant bond of T (Lemma 4.1), and Vl' must be of degree 1 in T (condition 1). By Lemma 4.1, s3 belongs to B. Similarly, S4 belongs to B. Note that v3 is not a vertex of degree 1 in T. If edge (v3, u) belongs to T, by Lemma 4.6, edge (u, v4) belongs to T, and T contains a cycle, i.e. the boundary of s, a contradiction. Thus edge (v3, u') belongs to T. By Lemma 4.1 s5 belongs to B. Similarly, s6 belongs to B. Hence s belongs to B, again a contradiction. Consequently, v cannot be connected to two vertices of degree 3 in T, i.e. condition 2 holds. Now by conditions 1 and 2 a vertex v of degree 3 of T is connected by a chain of T with length greater than 1 to at least a vertex of degree 3 or degree 1 of T. Consider the chain C = vvl v 2 . . . v t , where vt is a vertex of degree 3 or 1 of T, and v i ( i = 2 . . . . . t - 1) is a vertex of degree 2 of T. Without loss of generality, we may assume that v' is a vertex of degree 1 of T (condition 1, see Fig. 16(a)). Then s l and s2 belong to B, and s does not belong to B. If e is on the chain C, then s3 belongs to B (Lemmas 4.4 and 4.5), and e t belongs to T (Lemma 4.7), contradicting the assumption that v l is of degree 2 in T. Hence e~ is on the chain C, and s3 belongs to B (Lemma 4.1). If e 2 is on C, then e 3 cannot belong to T. Otherwise, by Lemma 4.6, e4 belongs to T, thus s2 belongs to T, a contradiction. Hence, e5 is on C. Since s2 belongs to B, s5 does not belong to B (Lemma 4.1). Consider es. By Lemmas 4.4 and 4.5, s4 belongs to B, contradicting the idea that B is catacondensed. Hence e2 cannot be on C and e 6 is on C. By repeated reasoning, it can be seen that C has the form as shown in Fig. 16(b). If the end vertex of C, i.e. v,, is a vertex of degree 3 of T, the length t must be odd. If not, t is even; then by Lemma 4.1, hexagons sl, s2 ..... sin- l and s m belong to B [cf. Fig. 16(c)]. By
245
Av, . ,sL ¥
(a)
(b)
VI
#
(c)
(d) Fig. 16. Illustration of the proof of the validity of condition 3.
condition 1, one of v' and v" is of degree 1 in T, and s' must belong to B by Lemma 4.1. Then vt does not lie on the perimeter of B, a contradiction. Now suppose the vt is a vertex of degree 1 of T, then the length t must be even. If t is odd, again by Lemma 4.1, hexagons sl, s2 ..... sn-1, sn and s" belong to B. Thus vt-I
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C. Rong-Si/Journal of Molecular Structure 415 (1997) 239-247
A.
,/~, -.
e ~
(a)
V wt
) Fig. 17. Illustration of the proof of Theorem 6.2. does not lie on the perimeter of B, again a contradiction. The proof is completed. T h e o r e m 6.2. Assume that tree T is the branching graph of benzenoid B. B is catacondensed if the following conditions are satisfied by T:
1. For any vertex v of degree 3 of T, v is connected to one and only one vertex of degree 1 of T. 2. For any vertex v of degree 3 of T, v is connected to at most one vertex of degree 3 of T. 3. If C is a chain connecting two vertices of degree 3 of T, then the length of C is odd; if C is a chain connecting one vertex of degree 3 and one vertex of degree 1 of T, then the length of C is even.
contradicting Lemmas 4.4 and 4.5. Hence e 2 belongs to C. Edge e3' does not belong to C. Otherwise, by Lemma 4.6, e4' belongs to C, and thus hexagons sl belongs to T, contradicting the possibility that T is a tree. Therefore, e3 belongs to C. By repetition of the above discussion, all the edges e t, e2 ..... et are seen to belong to C [see Fig. 17(a)]. If ~ is connected by C to a vertex of degree 1 of T, then by condition 3, the length of C is even. Now by Lemma 4.1, sl, s2 ..... s,,, sn+l
Proof. Suppose that T is the branching graph of ben-
zenoid B which satisfies conditions 1-3, but B is not catacondensed. Then there is a vertex of degree 3 of B, say ~, which is not on the perimeter of B, and is also a vertex of degree 3 of T (see Fig. 17). By conditions 1 and 2, one of the three vertices adjacent to ~, say v', is of degree 2 in T, namely v' is on a chain of T connecting ~ and another vertex of degree 3 of T, or a vertex of degree 1 of T. We distinguish two cases. Case 1. Edge e I belongs to C. If e2' belongs to C, then by Lemma 4.1 neither s~' nor s2 belongs to B,
Fig. 18. Illustration of Case 2.
C. Rong-Si/Journal of Molecular Structure 415 (1997) 239-247
belong to B. Note that et is a pendant bond; s' must belong to B. Hence ~ belongs to T (Lemma 4.7). Thus v" is of degree 3 in T, contradicting any suggestion that v" is on C and is of degree 2. Therefore, ~ can only be connected by the chain C to a vertex of degree 3 of T, say vl" [see Fig. 17(b)]. Then vl" is also a vertex of degree 3 of T, and does not lie on the perimeter of B. Repeat the above discussion for v l". v ~" is connected to, or is connected by a chain to, another vertex of degree 3 of T. Since B is finite, we will eventually find a vertex v" of degree 3 of T which is connected to a vertex of degree 1 of T by a chain with even length, and deduce a contradiction as before. Case 2. Edge el' belongs to C [see Fig. 17(a)]. We can discuss this in a similar way as in case l, and deduce a contradiction (cf. Fig. 18). Consequently, B cannot have a vertex of degree 3 which does not lie on the perimeter of B, and is thus catacondensed. The proof is completed.
247
References [1] I. Gutman, S.J. Cyvin, Introduction to the Theory of Benezenoid Hydrocarbons, Springer, Berlin, 1989. [2] (a) E.C. Kirby, Journal of Mathematical Chemistry 4 (1990) 31; (b) E.C. Kirby, Journal of Mathematical Chemistry 8 (1991) 77. [3] E.C. Kirby, Journal of Mathematical Chemistry l (1987) 175. [4] R.B. Mallion, Proceedings of the Royal Society A341 (1975) 429. [5] A. Baba Ahmed, J. Gayoso, International Journal of Quantum Chemistry 23 (1983) 71. [6] B. O'Leary, R.B. Mallion, Journal of Mathematical Chemistry 3 (1989) 323. [7] N. Mizoguchi, Bulletin of the Chemical Society of Japan 60 (1987) 2005. [8] C.W. Halgh, R.B. Mallion, Croatia Chimica Acta 62 (1989) 1. [9] E.C. Kirby, Journal of the Chemistry Society, Faraday Transactions 86 (1990) 447. [ l 0] E.C. Kirby, I. Gutman, Journal of Mathematical Chemistry 13 (1993) 359.
MOLECULAR STRUCTURE ELSEVIER
Journal of Molecular Structure 415 (1997) 239-247
Benzenoids with branching graphs being trees Chen Rong-Si College of Finance and Economics, Fuzhou University, Fuzhou 350002, People's Republic of China
Received 13 November 1996; revised 11 February 1997; accepted 12 February 1997
Abstract
The branching graph of a benzenoid hydrocarbon molecule is the subgraph consisting of vertices of degree 3 and the bonds among them. In this paper, a practical method is established for recognizing trees embedded in the infinite hexagonal lattice which can be branching graphs of benzenoids. Moreover, a simple criterion is given to determine when a tree embedded in the infinite hexagonal lattice is the branching graph of a catacondensed benzenoid. © 1997 Elsevier Science B.V. Keywords: Benzenoid hydrocarbons; Branching graphs; Catacondensed benzenoid; Graph theory
1. Introduction A benzenoid is a connected finite subgraph of the infinite hexagonal lattice which has no cut vertices or non-hexagonal interior faces [1]. Benzenoids are of chemical significance since a benzenoid with perfect matchings (Kekul6 structures in chemical language) is the skeleton of a benzenoid hydrocarbon molecule. The branching graph of a benzenoid G is a special subgraph. Each vertex of the benzenoid G appears in its branching graph B(G) if and only if it is of degree 3. Each edge of G appears in B(G) if and only if it connects two such branching vertices. The concept of the branching graph was introduced [2] as a practical aid for diagnosing whether a graph is or is not traceable (a graph is traceable if it has a Hamilton path, i.e. a path that visits every vertex just once). Traceability has chemical relevance in the field of structural information transmission, since a structure that is traceable is often easier to encode from a keyboard than one that is not [3], and in the field of predicting the magnetic properties of
molecules by using the concept of spanning trees for the methods available for calculating r-electron ring currents in a conjugated system depend on whether or not the system is traceable [4-8]. It was also claimed in Ref. [2] that traceability must be among the factors that determine what kinds of structure it is possible or impossible to form by intro-molecular cross-linking of a linear polymer. In a further study [9] branching graphs have provided some insight into why there are so few sextet two-factorable benzenoids among all the benzenoids that are theoretically possible. As the branching graphs appear to have some use in chemistry, it is desirable to understand and characterize branching graphs as much as possible. Many problems about branching graphs have been raised [10[. In this paper we answer the following questions. (1) Which trees embedded in the infinite hexagonal lattice are branching graphs of benzenoids? (2) When is a tree embedded in the infinite hexagonal lattice the branching graph of a catacondensed benzenoid? We present a quick and practical way to recognize trees embedded in the infinite hexagonal lattice which can
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Fig. 1. A tree T embedded in the infinite hexagonal lattice (the vertices of T are labelled by positive integers).
be branching graphs of benzenoids. Moreover, a simple criterion is given to determine when a tree embedded in the infinite hexagonal lattice is the branching graph of a catacondensed benzenoid.
2. Definitions Let T be a tree embedded in the infinite hexagonal lattice. The boundary b(T) of T is defined to be the closed walk that travels T clockwise and each edge of T is traversed twice. For the tree depicted in Fig. 1, the boundary is b(T) = (1,2) (2,3) (3,2) (2,4) (4,5) (5,6) (6,7) (7,8) (8,9) (9,8) (8,10) (10,8) (8,7) (7,6) (6,5) (5,4) (4,2) (2,1). Note that since b(T) is a closed walk, it is not important which vertex of T is the starting point of b(T). For example, we can say that the boundary b(T) of the tree T depicted in Fig. 1 is b(T) = (5,6) (6,7) (7,8) (8,9) (9,8) (8,10) (10,8) (8,7) (7,6) (6,5) (5,4) (4,2) (2,1) (1,2) (2,3) (3,2) (2,4) (4,5). It is clear that each edge e of a tree T embedded in the infinite hexagonal lattice is contained in two hexagons of the infinite hexagonal lattice, denoted by s'(e) and s"(e). A set S of hexagons of the infinite hexagonal lattice is said to be the associate hexagon set of tree T if S = {s'(e), s"(e)le ~ T}. For an illustration, see Fig. 2.
Fig. 3. An odd-even labelling with respect to pendant bond ~.
3. Odd-even labelling Let T be a tree with associate hexagon set S. Along the boundary b(T) clockwise, we label the hexagons of S as odd or even consecutively, according to the following rules: 1. Choose optionally a pendant bond (i.e. an edge with one end-vertex being of degree 1) e of T, label the two hexagons s'(e) and s"(e) as odd. 2. Suppose that hexagon s has been labelled as odd (even). Hexagon ~ appears right after s along the boundary b(T) clockwise, and has not yet been labelled. If the common edge ofs and ~ is a pendant bond of T, then ~ is labelled as odd (even); otherwise, 3 is labelled as even (odd). An illustration is given in Fig. 3, where hexagons each with an " o " are odd hexagons, while those each with an " e " are even hexagons.
4. Lemmas
Lemma 4.1. Assume that T is the branching graph of benzenoid B, S is the associate hexagon set of T. Hexagons s and 3 belong to S and appear consecutively along the boundary b(T) clockwise.If the common edge of s and ~ is a pendant bond of T, then both s and $ belong to B; otherwise, one and only one of s and 3 belongs to B.
Fig. 2. A tree T and the associate hexagon set S = {s~, s2, s3, s4, ss, $6, $7~ $8, $9}.
Fig. 4. Two consecutive hexagons s and 3 on b(T).
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241
Fig. 5. Illustration of the proof of Lemma 4.2.
Proof. Since s and 3 appear consecutively along b(T)
Remark 4.3. By the above lemma, if Tis the branching
clockwise, s and ~ have an common edge, say e (see Fig. 4). If e is a pendant bond of T, then both s and must belong to B since vertex v' is of degree 3 in B. If e is not a pendant bond of T, e cannot be an edge of T. Hence s and 3 cannot belong to B simultaneously. If none of them belongs to B, then vertex v is of degree 2 in B, and cannot belong to T (bear in mind that T is the branching graph of B), a contradiction. Therefore, if e is not a pendant bond of T, one and only one of s and belongs to B.
graph of benzenzoid B, then the o d d - e v e n labelling is independent of the choice of pendant bond of T. Namely, if a hexagon of S is labelled as odd (even) under the o d d - e v e n labelling with respect to pendant bond e, then it is also labelled as odd (even) under the o d d - e v e n labelling with respect to any pendant bond e' :~ e.
Lemma 4.4. Assume that T is the branching graph of benzenoid B. Then a hexagon of S belongs to B if and only if it is odd.
Lemma 4.2. Assume that T is the branching graph of benzenoid B, the o d d - e v e n labelling of the associate hexagon set S is with respect to the pendant bond e' of T. Then for each pendant bond e of T, both s'(e) and s"(e) are odd.
Proof. Let the pendant bonds of T be arranged as e' =el, e2 ..... et, such that along b(T) clockwise em appears after ei for i = 1,2 ..... p - 1; and e~ appears after e r. By the o d d - e v e n labelling rule, both s'(et) and s"(el) are odd. If s"(el)=s'(e2), then both s'(e2) and s"(e2) are odd (see Fig. 5). Now suppose that s"(el ) 4: s'(e2). If there are no hexagons of S between s"(e l) and s'(e2) along b(T) clockwise, then by Lemma 4.1 s'(e2) does not belong to B. Yet, again by Lemma 4.1, s'(e2) must belong to B since e2 is a pendant bond of T, a contradiction. Hence, there are some hexagons of S, say sl ..... s t (t >- 1), between s"(eO and s'(e2) along b(T) clockwise. We claim that t is an odd number (cf. Fig. 5). If t is an even number, by Lemma 4.1 sl, s2 ..... St_l do not belong to B, and st belongs to B. Thus s'(e2) does not belong to B, again a contradiction. Therefore, t is an odd number. By the o d d - e v e n labelling rule, s'(e2) and s"(e2) are odd hexagons. By the same argument, we can prove that if s'(ei) and s"(ei) are odd, then s'(ei+l) and (S"(ei+l) are also odd for i = 2 ..... p - 1. The lemma follows.
Proof. By Lemma 4.2 all hexagons of S corresponding to pendant bonds of T are odd, and by Lemma 4.1 they belong to B. Now we consider those hexagons of S which do not correspond to pendant bonds of T. Suppose that e and ~ are two consecutive pendant bonds of T along b(T) clockwise, and s 1, $ 2 . . . . . S t are consecutive hexagons of S between s"(e) and s'(O) along b(T) clockwise. As mentioned in the proof of Lemma 4.2, t is an odd number. By Lemma 4.1, even hexagons s 1, s3 ..... st do not belong to B; while odd hexagons s2, s4 ..... st-i belong to B. The lemma is thus proved.
Lemma 4.5. Assume that T is the branching graph of benzenoid B, e is an edge of T. Then at least one of s'(e) and s"(e) is odd under the o d d - e v e n labelling.
Proof. Since B is a benzenoid, each edge of B must belong to at least one hexagon of B. If none of s'(e) and s"(e) is odd under the o d d - e v e n labelling, then none of them belongs to B (Lemma 4.4). Hence e does not belong to any hexagon of B, a contradiction which establishes the lemma.
Lemma 4.6. Assume that T is the branching graph of benzenoid B, v 1 and v2 are two vertices of T. If v i and
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C. Rong-Si/Journal of Molecular Structure 415 (1997) 239-247 v5
',
',
'-..../'3
v2
v2
(a)
(b)
Fig. 6. Two possibilities of the relative positions of e t and e2.
( (
v2 are connected by an edge e in the infinite hexagonal lattice, then e belongs to T. Fig. 8. Illustration of the proof of Lemma 4.7. Proof. Since T is the branching graph of B, v l and v2 are two vertices of T; evidently v i and v2 are of degree 3 in B. Hence e = (vl, v2) must belong to B, and thus T. Lemma 4.7. Assume that T is the branching graph of benzenoid B, e l and e 2 are two edges of T, and s'(e l) and s'(e2) are odd under the o d d - e v e n labelling. If s'(eO A s'(e2) = e, then e belongs to T; if s'(el)=s'(e2)=s, then et and e2 are either adjacent or are connected by a path of T which is on the boundary ofs. Proof. By Lemma 4.4, both s'(eO and s'(e2) belong to B. If s ' ( e l ) N s ' ( e 2 ) = e , then e evidently connects two vertices of degree 3 of B, and hence e belongs to T. Now suppose that s'(el ) = s'(e2) = s. If e j and e 2 are not adjacent, then there are effectively two possibilities as shown in Fig. 6. In case a, by Lemma 4.6 edge (vl, v2) belongs to T. Thus el and e2 are connected by a path of T which is on the boundary of s. Now consider case b. If one of the edges (vl, v2) and (v2, v3) belongs to T, then by Lemma 4.6 the other must also belong to T. Thus e j and e2 are connected by a path of T which is on the boundary of s. Similarly, if one of the edges (v4, vs) and (vs, v6) belongs to T, then el and e2 are connected by a path of T which is on the boundary of s [see Fig. 6(b)]. Now suppose that none of the edges (v 1, v2), (v2, v3), (v4, vs) and (vs, v6)
V
~"
V
Fig. 7. Illustration of the proof of Lemma 4.7.
belong to T. Then hexagons s~, $2, s 3 and s4 do not belong to B (see Fig. 7). Namely, el and e2 are not connected by a path of T which is on the boundary of s. We claim that both el and e2 are pendant bonds of T. If e 1 is not a pendant bond of T, then e l' belongs to T. By Lemmas 4.4 and 4.5, s l' belongs to B. Edge el' cannot be a pendant bond of T since s3 does not belong to B (Lemma 4.1). If e2' belongs to T, by Lemmas 4.4 and 4.5, s2' belongs to B. If e 3' belongs to T, then by the o d d - e v e n labelling rule, s2' belongs to B. By the analogous argument, s3' belongs to B. Then the boundary of s l' belongs to T (see Fig. 7), contradicting the assumption that T is a tree. This contradiction shows that el must be a pendant bond of T. Similarly, e2 must be a pendant bond of T. Then the odd hexagons except s of the associate hexagon set S constitute a benzenoid/~. Since s~, s2, s3 and s4 do not belong to B, B=/~ U {s} must contain a " h o l e " , contradicting the assumption that B is a benzenoid (cf. Fig. 8). The lemma is thus proved.
5. A criterion for a tree to be the branching graph of a benzenoid Theorem 5.1. Assume that T is a tree embedded in the infinite hexagonal lattice; the o d d - e v e n labelling of the associate hexagon set S is respect to a pendant bond ~ of T. Then T is the branching graph of a benzenoid B if and only if the following conditions hold: 1. For any pendant bond e of T, both s'(e) and s"(e) are odd. 2. For any edge e of T, at least one of s'(e) and s"(e) is odd. 3. Assume that vl and v2 are two vertices of T. If v~
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243
Fig. 9. Illustration of the proof of sufficiency of Theorem 5.1.
and v2 are connected by an edge e of the infinite hexagonal lattice, then e belongs to T. 4. Assume that el and e2 are two edges of T, and both s ' ( e O and s'(e2) are odd. If s ' ( e l ) A s'(e2) =e, then e belongs to T. I f s ' ( e l ) = s ' ( e 2 ) = s , then el and e2 are either adjacent, or are connected by a path of T which is on the boundary of s. Proof. By Lemmas 4.2, 4.5, 4.6 and 4.7, conditions 1 - 4 are necessary. In the following we prove that they are also sufficient. Suppose that B is the graph consisting of all the odd hexagons of S. By condition 2, all the edges and vertices of T belong to B. Thus all the vertices of degree 3 of T are also vertices of degree 3 of B. By condition 1, all the vertices of degree 1 of T are of degree 3 in B. By the odd-even labelling rule and condition 2, all the vertices of degree 2 of T are of degree 3 in B. Hence all the vertices of T are of degree 3 in B. Now we prove that all the vertices of degree 3 of B belong to T. Assume that v is a vertex of degree 3 of B. By condition 2, B contains no edge which does not belong to any hexagon of B. Thus at least two of the three hexagons sb s2 and s3 belong to B (see Fig. 9), say s i and s2. Hence, by condition 4, e belongs to T, and therefore v belongs to T. Note that any edge
Fig. 10. T contains a cycle if for any hexagon on the boundary of the hole of B, the removal of it does not open the hole.
of B connecting two vertices of degree 3 also belongs to B. This is guaranteed by condition 3. We have already proved that T is the branching graph of B. Now we want to prove that B is a benzenoid. If B is not a benzenoid, then B contains a "hole", i.e. an internal face which is bounded by a polygon C with more than six edges. If, for any hexagon on the boundary C of the hole, the removal of it does not open the hole, then the branching graph T of B contains a cycle and is not a tree, a contradiction (see Fig. 10). Hence there is a hexagon of B on the boundary of the hole, say s, and how they are connected to s, such that the removal of s from B will open the hole. According to how many hexagons of B are connected to s, there are effectively six modes, as shown in Fig. 11. Since the branching graph of B is a tree, modes d, e and f are impossible. Now let sbs2 ..... s e be hexagons of B such that the removal of each of them will open the hole. Suppose that none of them are of mode a. We first remove all the vertices of degree 2 on the perimeter of B together with their incident edges. Evidently, the resulting
0
f Fig. 11. Six modes of hexagon s.
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( Fig. 14. Illustration of the proof of the validity of condition 1.
T. If B is catacondensed, then the following conditions are satisfied:
Fig. 12. If B contains no hexagon of mode a on the boundary of the hole, then T contains a cycle.
graph still has a hole. Then we remove all the vertices of degree 2 on the boundary C of the hole together with their incident edges. Now the resulting graph is the branching graph of B, i.e.T. Yet T contains a cycle since none of the hexagons s 1,s2..... sp are of mode a (cf. Fig. 12), contradicting the assumption that T is a tree. Hence B has a hexagon of mode a on the boundary of the hole. Now we find two edges el and e2 of T (see Fig. 13) such that s'(el ) = s'(e2) = s, but e l and e2 are neither adjacent nor connected by a path of T on the boundary of s, contradicting condition 4. This contradiction shows that B cannot contain a hole. Therefore, B is a benzenoid. The proof is completed.
6. Catacondensed benzenoids with branching graphs being trees In this section we present a criterion to determine when a tree is the branching graph of a catacondensed benzenoid. Recall that a benzenoid is said to be catacondensed if all its vertices lie on the perimeter of it.
1. v is connected to one and only one vertex of degree 1 ofT. 2. v is connected to at most one vertex of degree 3 of T. 3. If v is connected to a vertex of degree 3 of T by a chain of T with length greater than 1, then the length is odd; if v is connected to a vertex of degree 1 of T by a chain of T with length greater than 1, then the length is even.
P r o o f Suppose that v is a vertex of degree 3 of T. If v is connected to two vertices of degree 1 of T, say v' and v" (see Fig. 14), then by L e m m a 4.1, hexagons s 1, s2 and s3 belong to B. Thus v is a vertex which does not lie on the perimeter of B, contradicting the assumption that B is catacondensed. Hence v is connected to at most one vertex of degree 1 of T. Now suppose that v is connected to no vertex of degree 1 of T. Then none of v' and v" are of degree 1 of T. Since B is catacondensed, Sl, s2 and s3 cannot belong to B simultaneously. By Lemmas 4.4 and 4.5, two of sl, s2 and s3, say sl and s2, must belong to B. W e claim that s4 must belong to B. Since v" is not of degree 1 in T, at least one of edges el and e2 belongs to T. If ej belongs to T, by Lemmas 4.4 and 4.5, s4 belongs to B. If e i does not belong to T, then e2 belongs to T. Then by Lemma 4.1, s4 must belong to B. By an analogous argument, s6 must belong to B, and at least one of s5
Theorem 6.1. Assume that tree T is the branching graph of benzenoid B, and v is a vertex of degree 3 of
V0V Fig. 13. A situation violating condition 4.
Fig. 15. Illustration of the proof of the validity of condition 2.
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and s7 must belong to B. Now if s5 belongs to B, then hexagon s2 belongs to T; if sv belongs to B, then hexagon s l belongs to T. Each case contradicts the proposition that Tis a tree. Hence v must be connected to at least one vertex of degree 1 of T. Consequently, v is connected to one and only one vertex of degree 1 of T, i.e. condition 1 holds. Suppose that v is a vertex of degree 3 of T which is connected to two vertices of degree 3 of T, say v~ and v2 (see Fig. 15). By condition 1, v' is a vertex of degree 1 of T. Thus s~ and s2 belong to B (Lemma 4.1). Hence s does not belong to B since B is catacondensed. Therefore, edge (v l, v3) is not a pendant bond of T (Lemma 4.1), and Vl' must be of degree 1 in T (condition 1). By Lemma 4.1, s3 belongs to B. Similarly, S4 belongs to B. Note that v3 is not a vertex of degree 1 in T. If edge (v3, u) belongs to T, by Lemma 4.6, edge (u, v4) belongs to T, and T contains a cycle, i.e. the boundary of s, a contradiction. Thus edge (v3, u') belongs to T. By Lemma 4.1 s5 belongs to B. Similarly, s6 belongs to B. Hence s belongs to B, again a contradiction. Consequently, v cannot be connected to two vertices of degree 3 in T, i.e. condition 2 holds. Now by conditions 1 and 2 a vertex v of degree 3 of T is connected by a chain of T with length greater than 1 to at least a vertex of degree 3 or degree 1 of T. Consider the chain C = vvl v 2 . . . v t , where vt is a vertex of degree 3 or 1 of T, and v i ( i = 2 . . . . . t - 1) is a vertex of degree 2 of T. Without loss of generality, we may assume that v' is a vertex of degree 1 of T (condition 1, see Fig. 16(a)). Then s l and s2 belong to B, and s does not belong to B. If e is on the chain C, then s3 belongs to B (Lemmas 4.4 and 4.5), and e t belongs to T (Lemma 4.7), contradicting the assumption that v l is of degree 2 in T. Hence e~ is on the chain C, and s3 belongs to B (Lemma 4.1). If e 2 is on C, then e 3 cannot belong to T. Otherwise, by Lemma 4.6, e4 belongs to T, thus s2 belongs to T, a contradiction. Hence, e5 is on C. Since s2 belongs to B, s5 does not belong to B (Lemma 4.1). Consider es. By Lemmas 4.4 and 4.5, s4 belongs to B, contradicting the idea that B is catacondensed. Hence e2 cannot be on C and e 6 is on C. By repeated reasoning, it can be seen that C has the form as shown in Fig. 16(b). If the end vertex of C, i.e. v,, is a vertex of degree 3 of T, the length t must be odd. If not, t is even; then by Lemma 4.1, hexagons sl, s2 ..... sin- l and s m belong to B [cf. Fig. 16(c)]. By
245
Av, . ,sL ¥
(a)
(b)
VI
#
(c)
(d) Fig. 16. Illustration of the proof of the validity of condition 3.
condition 1, one of v' and v" is of degree 1 in T, and s' must belong to B by Lemma 4.1. Then vt does not lie on the perimeter of B, a contradiction. Now suppose the vt is a vertex of degree 1 of T, then the length t must be even. If t is odd, again by Lemma 4.1, hexagons sl, s2 ..... sn-1, sn and s" belong to B. Thus vt-I
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A.
,/~, -.
e ~
(a)
V wt
) Fig. 17. Illustration of the proof of Theorem 6.2. does not lie on the perimeter of B, again a contradiction. The proof is completed. T h e o r e m 6.2. Assume that tree T is the branching graph of benzenoid B. B is catacondensed if the following conditions are satisfied by T:
1. For any vertex v of degree 3 of T, v is connected to one and only one vertex of degree 1 of T. 2. For any vertex v of degree 3 of T, v is connected to at most one vertex of degree 3 of T. 3. If C is a chain connecting two vertices of degree 3 of T, then the length of C is odd; if C is a chain connecting one vertex of degree 3 and one vertex of degree 1 of T, then the length of C is even.
contradicting Lemmas 4.4 and 4.5. Hence e 2 belongs to C. Edge e3' does not belong to C. Otherwise, by Lemma 4.6, e4' belongs to C, and thus hexagons sl belongs to T, contradicting the possibility that T is a tree. Therefore, e3 belongs to C. By repetition of the above discussion, all the edges e t, e2 ..... et are seen to belong to C [see Fig. 17(a)]. If ~ is connected by C to a vertex of degree 1 of T, then by condition 3, the length of C is even. Now by Lemma 4.1, sl, s2 ..... s,,, sn+l
Proof. Suppose that T is the branching graph of ben-
zenoid B which satisfies conditions 1-3, but B is not catacondensed. Then there is a vertex of degree 3 of B, say ~, which is not on the perimeter of B, and is also a vertex of degree 3 of T (see Fig. 17). By conditions 1 and 2, one of the three vertices adjacent to ~, say v', is of degree 2 in T, namely v' is on a chain of T connecting ~ and another vertex of degree 3 of T, or a vertex of degree 1 of T. We distinguish two cases. Case 1. Edge e I belongs to C. If e2' belongs to C, then by Lemma 4.1 neither s~' nor s2 belongs to B,
Fig. 18. Illustration of Case 2.
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belong to B. Note that et is a pendant bond; s' must belong to B. Hence ~ belongs to T (Lemma 4.7). Thus v" is of degree 3 in T, contradicting any suggestion that v" is on C and is of degree 2. Therefore, ~ can only be connected by the chain C to a vertex of degree 3 of T, say vl" [see Fig. 17(b)]. Then vl" is also a vertex of degree 3 of T, and does not lie on the perimeter of B. Repeat the above discussion for v l". v ~" is connected to, or is connected by a chain to, another vertex of degree 3 of T. Since B is finite, we will eventually find a vertex v" of degree 3 of T which is connected to a vertex of degree 1 of T by a chain with even length, and deduce a contradiction as before. Case 2. Edge el' belongs to C [see Fig. 17(a)]. We can discuss this in a similar way as in case l, and deduce a contradiction (cf. Fig. 18). Consequently, B cannot have a vertex of degree 3 which does not lie on the perimeter of B, and is thus catacondensed. The proof is completed.
247
References [1] I. Gutman, S.J. Cyvin, Introduction to the Theory of Benezenoid Hydrocarbons, Springer, Berlin, 1989. [2] (a) E.C. Kirby, Journal of Mathematical Chemistry 4 (1990) 31; (b) E.C. Kirby, Journal of Mathematical Chemistry 8 (1991) 77. [3] E.C. Kirby, Journal of Mathematical Chemistry l (1987) 175. [4] R.B. Mallion, Proceedings of the Royal Society A341 (1975) 429. [5] A. Baba Ahmed, J. Gayoso, International Journal of Quantum Chemistry 23 (1983) 71. [6] B. O'Leary, R.B. Mallion, Journal of Mathematical Chemistry 3 (1989) 323. [7] N. Mizoguchi, Bulletin of the Chemical Society of Japan 60 (1987) 2005. [8] C.W. Halgh, R.B. Mallion, Croatia Chimica Acta 62 (1989) 1. [9] E.C. Kirby, Journal of the Chemistry Society, Faraday Transactions 86 (1990) 447. [ l 0] E.C. Kirby, I. Gutman, Journal of Mathematical Chemistry 13 (1993) 359.