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Forcing polynomials of benzenoid parallelogram and its related benzenoids
Applied Mathematics and Computation 284 (2016) 209–218
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Forcing polynomials of benzenoid parallelogram and its related benzenoids✩ Shuang Zhao, Heping Zhang∗ School of Mathematics and Statistics, Lanzhou University, Lanzhou, Gansu 730000, P.R. China
a r t i c l e
i n f o
Keywords: Forcing polynomial Perfect matching Innate degree of freedom Forcing number Benzenoid
a b s t r a c t Klein and Randic´ introduced the innate degree of freedom (forcing number) of a Kekulé structure (perfect matching) M of a graph G as the smallest cardinality of subsets of M that are contained in no other Kekulé structures of G, and the innate degree of freedom of the entire G as the sum over the forcing numbers of all perfect matchings of G. We proposed the forcing polynomial of G as a counting polynomial for perfect matchings with the same forcing number. In this paper, we obtain recurrence relations of the forcing polynomial for benzenoid parallelogram and its related benzenoids. In particular, for benzenoid parallelogram, we derive explicit expressions of its forcing polynomial and innate degree of freedom by generating functions. © 2016 Elsevier Inc. All rights reserved.
1. Introduction The forcing number of a perfect matching in benzenoid systems has been introduced by Harary et al. [11] more than twenty years ago. The roots of this concept can be found in an earlier paper by Klein and Randic´ [12] using the name ‘innate degree of freedom’. The forcing concept in Kekulé structures plays an important role in the resonance theory of theoretic chemistry [14,15]. In particular, for all benzenoid systems, recently Xu et al. [22] showed that the maximum forcing number is equal to the Clar number, and Lei et al. [13] showed that the maximum anti-forcing number is equal to the Fries number. Both indices can measure the stability of benzenoid hydrocarbons. We may refer to a survey [2] on this topic. A benzenoid system (benzenoid, or hexagonal system) is a 2-connected finite plane graph such that every interior face is a regular hexagon of side length one [5]. The study of perfect matching in benzenoid systems is of chemical relevance since a benzenoid system with a perfect matching can be regarded as carbon skeleton of a benzenoid hydrocarbon molecule [5]. Also, many well-known combinatorial problems have been encountered in perfect matching count of benzenoid systems. For example, the number of k-combinations of k + m elements equals perfect matching count of benzenoid parallelogram L(k, m) [7,8], and (n + 1 )-st Catalan number equals perfect matching count of triangular benzenoid T (n, n; n − 1 ) [7,19]. Let G be a graph with a perfect matching M. A forcing set S of M is a subset of M such that S is contained in no other perfect matchings of G. The forcing number of M, denoted by f(G, M), is the smallest cardinality over all forcing sets of M. The innate degree of freedom of G, denoted by IDF(G), is the sum over the forcing numbers of all perfect matchings of G. An edge of G is called forcing if it is contained in exactly one perfect matching of G. The maximum and minimum values of
✩ ∗
Supported by NSFC (grant no. 11371180) and the Fundamental Research Funds for the Central Universities (grant no. lzujbky-2015-211). Corresponding author. Tel.: +86 18993168369; fax: +86 9318912481. E-mail addresses: [email protected] (S. Zhao), [email protected] (H. Zhang).
http://dx.doi.org/10.1016/j.amc.2016.03.008 0 096-30 03/© 2016 Elsevier Inc. All rights reserved.
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S. Zhao, H. Zhang / Applied Mathematics and Computation 284 (2016) 209–218
a
b
Fig. 1. Illustration of L(k, m) (a) and L(k, m) ai (b), for k = 4, m = 5, i = 2.
f(G, M) over all perfect matchings M of G are denoted by F(G) and f(G), respectively. For other related concepts we refer the reader to [2,5]. Zhang and Li [23,24], and Hansen and Zheng [10] independently characterized benzenoid systems with a forcing edge, which include the above mentioned benzenoids. Zhang and Zhang [25] gave a novel computation of perfect matching count for such benzenoids, and Zhang and Deng [26] showed that their forcing numbers of all perfect matchings form either the integer interval from 1 to the Clar number or with only the gap 2. In particular, for benzenoid parallelogram, some other indices and properties about forcing were also obtained, such as global forcing number [17], anti-Kekulé number [21], anti-forcing number [21], and Clar cover polynomial [9]. In [27], we proposed the forcing polynomial of a graph to study the number of perfect matchings that have the same forcing number, i.e.
F (G, x ) =
x f (G,M ) =
M∈M(G )
F (G )
w(G, i )xi ,
(1)
i= f ( G )
where M(G ) denotes the set of all perfect matchings of G and w(G, i ) denotes the number of perfect matchings of G with forcing number i. It is easy to show that
F (G, x )|x=1 = (G ), the perfect matching count of G,
d F (G, x ) dx
= IDF (G ), the innate degree of freedom of G.
(2)
(3)
x=1
For some other graph polynomials, see a new book [18] and recent papers [3,4,6]. In this paper, we will apply this polynomial to benzenoid parallelogram and its related benzenoid systems. In Section 2, we first derive recurrence relation of the forcing polynomial for benzenoid parallelogram by equivalent definition of forcing set, and then obtain its explicit expression by generating function. As corollaries, we give explicit expressions of its perfect matching count and innate degree of freedom, and asymptotic behavior of its innate degree of freedom as one variable approaches infinity. In the end of this section, we present a new combinational explanation for the coefficients of forcing polynomial for benzenoid parallelogram with (0, 1)-sequence. In Section 3, we get recurrence relation of the forcing polynomial for pentagonal benzenoid, and explicit expressions of the forcing polynomial for benzenoid parallelograms with one additional hexagon. 2. Benzenoid parallelogram First recall some basic results on forcing set of a perfect matching of a graph. Let G be a graph with a perfect matching M. A cycle of G is called M-alternating if its edges appear alternately in M and E(G)M, where E(G) denotes the edge set of G. Lemma 2.1. [1,16] Let G be a graph with a perfect matching M. A subset S ⊆ M is a forcing set of M if and only if each Malternating cycle of G contains at least one edge of S. For an edge e ∈ M, let G e be the subgraph obtained from G by first deleting e with the end vertices, and then deleting recursively pendant edges (with an end of degree one) with the ends until the remaining graph has no pendant edges or is empty (no vertices). We call the deleted edges of M in the above procedure determined by e.
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Lemma 2.2. [27] Let G be a graph with a perfect matching M. If e ∈ M is contained in a minimum forcing set of M, then
f (G, M ) = f (G e, M ∩ E (G e )) + 1. A benzenoid system in a parallelogram-like shape, called benzenoid parallelogram and denoted by L(k, m), consists of k × m hexagons, arranged in m rows, each row containing k hexagons, shifted by a half hexagon to the right from the row immediately below, for k, m ≥ 1. In Fig. 1(a), we give as an illustrative example a benzenoid parallelogram L(k, m) for k = 4 and m = 5. For convenience, we make a convention that M(k, m ) = M(L(k , m )) and F (k , m ) = F (L(k, m ), x ) (see Eq. (1)). Since L(m, k) is obtained from L(k, m) by rotating 120 degree counter-clockwise and then inverting upside down, we have F (m, k ) = F (k, m ). For simplicity of presentation, we give some notations first. From left to right, we in turn denote the hexagons in the lowermost row by Ci , and their right ‘vertical’ edges by ai , for i = 1, 2, . . . , k. From the bottom up, we in turn denote the leftmost hexagons in each row by Dj , and their top right ‘oblique’ edges by bj , for j = 1, 2, . . . , m. Denote the left ‘vertical’ edge of C1 (= D1 ) by f and its bottom left ‘oblique’ edge by e (see Fig. 1(a)). We now divide M(k, m ) into two subsets Me and M f , where Me = {M ∈ M(k, m ) | e ∈ M} and M f = {M ∈ M(k, m ) | f ∈ M}. Further, subdivide the two sets as follows: Mei = {M ∈ Me | ai ∈ M}, for i = 1, 2, . . . , k, and M j = {M ∈ M f | b j ∈ M}, f
f
for j = 1, 2, . . . , m. Then we have Me = ∪ki=1 Mei and M f = ∪m M j . By Eq. (1), we have j=1
F (k, m ) =
x f (L(k,m ),M ) +
M∈Mei ,1ik
x f (L(k,m ),M ) .
(4)
M∈M jf ,1 jm
For convenience, we define L(0, n) and L(n, 0) to be empty graphs, for n ≥ 0, this yields F (0, n ) = 1 [27]. Then we can derive the following result. Theorem 2.3. The forcing polynomial of a benzenoid parallelogram L(k, m) (k, m ≥ 1) has the following recurrence relation:
F (k, m ) =
k−1
F (i, m − 1 )x +
m −1
i=0
F (k − 1, j )x,
(5)
j=0
with initial conditions F (0, n ) = 1, for n ≥ 0. f
Proof. For k, m ≥ 1, we claim that for an arbitrary M ∈ Mei (resp. M j ), ai (resp. bj ) is contained in some minimum forcing set of M, for i = 1, 2, . . . , k (resp. j = 1, 2, . . . , m ). In fact, we can see that ai (resp. bj ) is contained in M-alternating hexagon Ci (resp. Dj ), and every M-alternating cycle that contains any edge of M ∩ E(Ci ) (resp. M ∩ E(Dj )) must contain ai (resp. bj ), for i = 1, 2, . . . , k (resp. j = 1, 2, . . . , m). By Lemma 2.1 and the definition of minimum forcing set, we can verify the claim. It is observed from Fig. 1(b) that L(k, m ) ai = L(k − i, m − 1 ), for i = 1, 2, . . . , k, where we give as an illustrative example L(k, m) ai for k = 4, m = 5, i = 2, and use double lines to mark the edges which ai determines. Similarly, we have L(k, m ) b j = L(k − 1, m − j ), for j = 1, 2, . . . , m. By Lemma 2.2, we have f (L(k, m ), M ) = f (L(k − i, m − 1 ), M ∩ E (L(k − i, m − 1 ))) + f
1, for M ∈ Mei , i = 1, 2, . . . , k, and f (L(k, m ), M ) = f (L(k − 1, m − j ), M ∩ E (L(k − 1, m − j ))) + 1, for M ∈ M j , j = 1, 2, . . . , m. These yield
x f (L(k,m ),M ) =
M∈Mei
x f (L(k−i,m−1),M∩E (L(k−i,m−1)))+1
M∈Mei
=
x f (L(k−i,m−1),M ) · x = F (k − i, m − 1 )x,
M∈M(k−i,m−1 )
for i = 1, 2, . . . , k, and
x f (L(k,m ),M ) =
M∈M jf
x f (L(k−1,m− j ),M∩E (L(k−1,m− j )))+1
M∈M jf
=
x f (L(k−1,m− j ),M ) · x = F (k − 1, m − j )x,
M∈M(k−1,m− j )
for j = 1, 2, . . . , m. Substituting the above equations into Eq. (4), we immediately obtain Eq. (5).
To obtain explicit expression of the forcing polynomial for benzenoid parallelogram, a little manipulation is needed as the following lemma. Lemma 2.4. The forcing polynomial of a benzenoid parallelogram L(k, m) (k, m ≥ 1) has the following recurrence relation:
(1 − 2x )F (k, m ) + xF (k, m − 1 ) + xF (k − 1, m ) + F (k + 1, m + 1 ) = F (k + 1, m ) + F (k, m + 1 ). Proof. We need two new polynomials:
Gk,m =
k i=0
F (i, m ), Gm,k =
m j=0
F ( j, k ),
(6)
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for k, m ≥ 0. According to F (m, k ) = F (k, m ) and Eq. (5), it follows that
Gk,m − Gk−1,m = F (k, m ) = Gm,k − Gm−1,k ,
(7)
Gk,m − Gk−1,m = F (k, m ) = xGk−1,m−1 + xGm−1,k−1 .
(8)
for k, m ≥ 1. We derive
F (k, m ) = −xGk,m−1 + xGk−1,m−1 − xGm,k−1 + xGm−1,k−1 + xGk,m−1 + xGm,k−1 = −xF (k, m − 1 ) − xF (k − 1, m ) + xGk,m−1 + xGm−1,k + xGm,k−1 + xGk−1,m − xGm−1,k − xGk−1,m = −xF (k, m − 1 ) − xF (k − 1, m ) + F (k + 1, m ) + F (k, m + 1 ) −xGm−1,k + xGm,k − xGk−1,m + xGk,m − xGm,k − xGk,m = −xF (k, m − 1 ) − xF (k − 1, m ) + F (k + 1, m ) + F (k, m + 1 ) + 2xF (k, m ) − F (k + 1, m + 1 ), for k, m ≥ 1, which implies Eq. (6).
Theorem 2.5. The forcing polynomial of a benzenoid parallelogram L(k, m) (k, m ≥ 1) can be expressed in the following explicit form:
F (k, m ) =
min{k,m}
i=1
−
j=1
min{k,m} i−1 i=2
i i+1 j
j=1
i j
j+k−i−1 k−i
j+k−i−1 k−i
m− j i x + m−i
max{k, m} min{k,m} x |m − k|
m− j−1 i x − m−i
max{k, m} − 1 min{k,m} x . |m − k|
Proof. We define the generating function of F(k, m) to be
V (s, t ) =
∞ ∞
F (k, m )sk t m .
k=0 m=0
In the following we will derive a closed expression of V(s, t). From Eq. (6), we obtain
( 1 − 2x )
∞ ∞
F (k, m )sk+1 t m+1 + x
k=1 m=1
+x =
∞ ∞
∞ ∞
F (k, m − 1 )sk+1t m+1
k=1 m=1
F (k − 1, m )sk+1t m+1 +
∞ ∞
k=1 m=1 ∞ ∞
k=1 m=1 ∞ ∞
k=1 m=1
k=1 m=1
F (k + 1, m )sk+1t m+1 +
F (k + 1, m + 1 )sk+1t m+1
F (k, m + 1 )sk+1t m+1 .
(9)
Computing the first term, we have ∞ ∞
F (k, m )sk+1 t m+1 = st
k=1 m=1
∞ ∞
F (k, m )sk t m
k=1 m=1
= st
∞ ∞
F (k, m )s t − k m
k=0 m=0
= stV (s, t ) −
∞ m=0
F (0, m )t − m
∞
F (k, 0 )s
k
k=1
st s2 t − . 1−t 1−s
For the other terms, similarly we have, ∞ ∞
F (k, m − 1 )sk+1t m+1 = st 2V (s, t ) −
st 2 , 1−t
F (k − 1, m )sk+1t m+1 = s2 tV (s, t ) −
s2 t , 1−s
k=1 m=1 ∞ ∞ k=1 m=1 ∞ ∞
F (k + 1, m + 1 )sk+1t m+1 = V (s, t ) −
k=1 m=1 ∞ ∞ k=1 m=1
1 xs(2t − t 2 ) 1 xt (2s − s2 ) − − − + 1 + 2xst, 2 1−t 1−s (1 − t ) (1 − s )2
F (k + 1, m )sk+1t m+1 = tV (s, t ) −
t s2 t 2t − t 2 − ts − xst − , 2 1−t 1 −s (1 − t )
S. Zhao, H. Zhang / Applied Mathematics and Computation 284 (2016) 209–218 ∞ ∞
213
s st 2 2s − s2 − ts − xst − . 2 1−s 1 −t (1 − s )
F (k, m + 1 )sk+1t m+1 = sV (s, t ) −
k=1 m=1
Substituting these expressions into Eq. (9), we have 1 1−t
V (s, t ) =
1 −1 1−s . xts − 1xts 1−s −t
+
1−
(10)
Next we shall expand the above function V(s, t) so that we derive the explicit expression of the forcing polynomial F(k, m). By binomial theorem, we know
V (s, t ) =
i+1 i ∞ ∞ 1 1 1 1 + xi t i si − + xi t i si , 1−s 1−t 1−s 1−t i=0
and
1
1 + 1−s 1−t
i
=
i
j 1 i − j 1 i 1−t
j 1 i − j i−1 1 i j=1
=
1−s
j
j=0
=
i−1
1−s
j
i j
j=1
∞
w=0
1−t
+ ∞
1−t
1 i
+
1−s
i=1 j=1
i=0
+
1−s
i 1
∞ w=0
1−t
xt s =
∞ i−1
i j
i=2 j=1
+
∞ ∞ i=1 u=0
Extracting the coefficient of
F (k, m ) =
s k tm
−
min{k,m} i−1
j=1
for k, m ≥ 1, where
⎧ m ⎪ ⎪ k xk , ⎪ ⎨ ξk,m = 2xk , ⎪ ⎪ ⎪ ⎩ k xm ,
∞
w=0
∞
u=0
i=0 w=0
w=0
i+w−1 w s , w
∞
i+w w i i i s xt s, w
j+w−1 w i− j+u−1 u i i i s t xt s w u u=0
i+u−1 u i i i t xt s + u
i+1 j
i j
j+k−i−1 k−i
∞ ∞
i=1 w=0
i+w−1 w i i i s x t s + 1. w
j+k−i−1 k−i
m− j i x + ξk,m m−i
m− j−1 i x − ηk,m m−i
⎧ m−1 k ⎪ if k < m, ⎪ m−k x , ⎪ ⎨ k if k = m, ηk,m = 2 x , ⎪ ⎪ ⎪ ⎩ k − 1 xm , if k > m, k−m
m
The proof is completed.
j=1
i=2
∞
from both sides of Eq. (11), we can show
min{k,m} i i=1
j+w−1 w i− j+u u i i i s t xt s w u
i=0 u=0
i i i
u=0
∞ ∞ ∞ ∞ i+u u i i i + t xt s + u
1
∞
u=0
i+1 ∞ ∞ i 1 1 i+1 + xi t i si = j 1−s 1−t i=0
1 i
j+w−1 w i− j+u−1 u i+u−1 u s t + t + w u u
for i ≥ 1. Computing every term of Eq. (11), we have
∞
(11)
i=0
if k < m, if k = m, if k > m.
Remark 2.6. With the help of generating function of Gk,m , we have an alternative method to derive the expression of V(s, t). We define the generating function of Gk,m to be
W (s, t ) =
∞ ∞ k=0 m=0
Gk,m sk t m , W (t, s ) =
∞ ∞ k=0 m=0
Gm,k sk t m .
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S. Zhao, H. Zhang / Applied Mathematics and Computation 284 (2016) 209–218
Now we give two new equations. From Eqs. (7) and (8), we obtain ∞ ∞ k=1 m=1 ∞ ∞
Gk,m sk t m − Gk,m sk t m −
∞ ∞ k=1 m=1 ∞ ∞
k=1 m=1
∞ ∞
Gk−1,m sk t m =
k=1 m=1 ∞ ∞
Gk−1,m sk t m = x
k=1 m=1
∞ ∞
Gm,k sk t m −
k=1 m=1
Gk−1,m−1 sk t m + x
k=1 m=1
Gm−1,k sk t m , ∞ ∞
Gm−1,k−1 sk t m .
(12)
(13)
k=1 m=1
Computing every term, we have ∞ ∞
Gk,m sk t m =
k=1 m=1
∞ ∞ k=0 m=0
and similarly,
Gk,0 sk
k=1
s
(1 − s )2
,
1 1 − + 1, 1−s (1 − t )2
Gm,k sk t m = W (t, s ) −
k=1 m=1 ∞ ∞
∞
1 1 − + 1, 1−t (1 − s )2
Gk−1,m sk t m = sW (s, t ) −
k=1 m=1 ∞ ∞
G0,m t m −
m=0
= W (s, t ) − ∞ ∞
∞
Gk,m sk t m −
t
Gm−1,k sk t m = tW (t , s ) −
(1 − t )2
k=1 m=1 ∞ ∞
,
Gk−1,m−1 sk t m = stW (s, t ),
k=1 m=1 ∞ ∞
Gm−1,k−1 sk t m = stW (t , s ).
k=1 m=1
Substituting these expressions into Eqs. (12) and (13), we have
W (t, s ) =
1 1 + 1−s −1 1−s 1−t W (s, t ), W (s, t ) = . 1−t 1 − s − xts − 11−s xts −t
The relationship between generating functions V(s, t) and W(s, t) is
V (s, t ) =
∞ ∞
∞ ∞
Gk,m sk t m −
k=1 m=0
Gk−1,m sk t m +
k=1 m=0
∞
F (0, m )t m = (1 − s )W (s, t ),
m=0
which yields Eq. (10). The following is an immediate consequence of Theorem 2.5. Corollary 2.7. The forcing polynomial of a benzenoid parallelogram L(k, m) (1 ≤ k ≤ m) is
k i i+1 F (k, m ) = j i=1 j=1
−
k i−1 i=2 j=1
i j
j+k−i−1 k−i
m− j i x + m−i
j+k−i−1 k−i
m k x k
m− j−1 i x − m−i
m−1 k x , m−k
For 1 ≤ k ≤ m, from the above corollary we can find that the maximum forcing number F (L(k, m )) = k and the minimum forcing number f (L(k, m )) = 1, and the corresponding number of perfect matchings are [27]
k k w(L(k, m ), k ) = j j=0
w(L(k, m ), 1 ) =
m− j , m−k
2, m + 1,
if k 2, if k = 1.
For Eq. (10), differentiating both sides with respect to x and setting x to 1 gives the generating function of IDF(L(k, m)) ∞ ∞ k=0 m=0
IDF (L(k, m ))s t
k m
=
1 1−t
+
1 1−s
1−
−1 ts 1−s
−
ts 1−s ts 1−t
+
2
ts 1−t
=
ts 1−s
1−
ts 1−s
+
ts
1−t . − 1ts−t (1 − s − t )
S. Zhao, H. Zhang / Applied Mathematics and Computation 284 (2016) 209–218
215
Setting x to 1 gives the generating function of (L(k, m)) ∞ ∞
1 1−t
(L(k, m ))skt m =
1−
k=0 m=0
1 −1 1−s ts ts − 1−s 1−t
+
1 , 1−s−t
=
which implies the following corollary. Corollary 2.8. [8] The perfect matching count of a benzenoid parallelogram L(k, m) (k, m ≥ 1) is
k+m . k
(L(k, m )) =
Applying Corollary 2.7 to Eqs. (2) and (3), we obtain the following two corollaries. Corollary 2.9. The perfect matching count of a benzenoid parallelogram L(k, m) (1 ≤ k ≤ m) is
(L(k, m )) =
k i i+1 j i=1 j=1
−
k i−1
i j
i=2 j=1
j+k−i−1 k−i
j+k−i−1 k−i
m− j m−i
+
m k
m− j−1 m−i
m−1 . m−k
−
(14)
Corollary 2.10. The innate degree of freedom of a benzenoid parallelogram L(k, m) (1 ≤ k ≤ m) is
k i i+1 j
IDF (L(k, m )) =
i=1 j=1
−
k i−1
i j
i=2 j=1
j+k−i−1 k−i
j+k−i−1 k−i
m− j i+ m−i
m k k
m− j−1 i− m−i
m−1 k. m−k
(15)
Corollary 2.11. For a benzenoid parallelogram L(k, m) (1 ≤ k ≤ m), we have
lim
m→∞
IDF (L(k, m )) = k. (L(k, m ))
Proof. It is obvious that
(L(k, m )) =
m+k k
By Eq. (15), we have
IDF (L(k, m )) =
∼
mk as m → ∞. k!
k i i+1 j i=1 j=1
−
k i−1 i=2 j=1
i j
j+k−i−1 k−i
j+k−i−1 k−i
m− j i+ i− j
m k k
m− j−1 i− i− j−1
mk m−1 k∼ k as m → ∞, k−1 k!
since the second term is an infinite large quantity dominating the other terms. So the corollary holds.
From the above corollary, we know that the average forcing number per perfect matching approaches k when k is fixed and m goes to infinity. However, we have not obtained the asymptotic behavior of the innate degree of freedom of a benzenoid parallelogram when both k and m go to infinity. We conclude this section with a combinatorial explanation of the forcing polynomial. Let a(k, m; i) be the number of (0, 1)-sequences of k 1’s and m 0’s with maximum number of disjoint pairs “01” (consecutive elements in the corresponding (0, 1)-sequence) being i, for k, m ≥ 1, 1 ≤ i ≤ min {k, m}. We find that the coefficients of the forcing polynomial for benzenoid parallelogram can be expressed as such combinatorial count; Precisely, see Theorem 2.12. To obtain it we place a benzenoid parallelogram as shown in Fig. 2 such that an edge-direction is vertical and it has exactly one peak (vertex whose neighbors are below it) and one valley (vertex whose neighbors are above it). A monotone path in it is a path connecting the peak with the valley in which, when starting at the peak, one always goes downwards. This is easy to see in such monotone paths, vertical and oblique (may happen two directions as left-downward and rightdownward) edges appear alternately. A monotone path is exemplified in Fig. 2. Such monotone paths have one-one correspondence with perfect matchings. The corresponding is as follows: take all oblique edges in the monotone and vertical edges not in the monotone path and a perfect matching is obtained [8]. Theorem 2.12. w(L(k, m ), i ) = a(k, m; i ) (k, m 1, 1 i min{k, m} ).
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Fig. 2. Illustration for the proof of Theorem 2.12.
Fig. 3. Illustration of T(k, m; p) (left) and T (n, n; n − 1 ) (right), for k = 4, m = 5, p = 2, n = 5.
Proof. We proceed by establishing a one-to-one correspondence between all perfect matchings of L(k, m) and all (0, 1)sequences of k 1’s and m 0’s, for k, m ≥ 1, then by proving that the forcing number of each perfect matching equals the maximum number of disjoint pairs “01” of its corresponding (0, 1)-sequence. Besides giving the entire monotone path to obtain a perfect matching in L(k, m), we could only give the direction of oblique edges in the monotone path when starting at the peak. Treat the left-downward edge as 1, and right-downward as 0, we could get a (0, 1)-sequence of k 1’s and m 0’s corresponding to a perfect matching. For example, the perfect matching in Fig. 2 is corresponding to sequence 1011010 0 0. It is known that the forcing number of perfect matching M equals the maximum number of disjoint M-alternating hexagons of L(k, m) [26], since L(k, m) has a forcing edge. And it is obvious that all the M-alternating hexagons of L(k, m) intersect the corresponding monotone path at two oblique edges since all the oblique edges in M appear only in the monotone path. Consecutive oblique edges of different direction in the monotone path together with some vertical edge not in the monotone path constitute the matching in M-alternating hexagon. These imply a one-to-one corresponding between M-alternating hexagons and pairs “01” of the (0, 1)-sequence corresponding to M. For example, the maximum set of disjoint M-alternating hexagons of the perfect matching in Fig. 2, illustrated using bold lines in the first graph, is corresponding to three pairs “10” (1st and 2nd entries), “10” (4th and 5th entries) and “10” (6th and 7th entries) of 1011010 0 0. Then the claim holds.
3. Related benzenoids A benzenoid system in a pentagon-like shape, called pentagonal benzenoid and denoted by T(k, m; p), consists of two parts: one is m − p − 1 rows with k hexagons in each row, shifted by a half hexagon to the right from the row immediately below; another is p + 1 rows with the number of hexagons in a row decreasing by one from k in the lowermost row to k − p in the uppermost row, each row shifted by one and a half hexagon to the right from the row immediately below; the two parts adhere at the uppermost row of the (m − p − 1 )’s and the lowermost row of ( p + 1 )’s, the upper shifted by a half hexagon to the right from the lower, for 1 ≤ p < k, m. In particular, a pentagonal benzenoid in a triangle-like shape with n rows, is called triangular benzenoid and denoted by T (n, n; n − 1 ), for n ≥ 2. In Fig. 3, we give as illustrative examples a pentagonal benzenoid T(k, m; p) and a triangular benzenoid T (n, n; n − 1 ), for k = 4, m = 5, p = 2 and n = 5.
S. Zhao, H. Zhang / Applied Mathematics and Computation 284 (2016) 209–218
217
Fig. 4. Illustration of L (k, m), L (k, m) and L (k, m) (from left to right), for k = 4 and m = 5.
By similar arguments used for benzenoid parallelogram, we can derive recurrence relation of the forcing polynomial for pentagonal benzenoid. For convenience, we make a convention that F (T (k, m; p), x ) = F (k, m; p) and define T(k, m; 0) to be L(k, m). Theorem 3.1. The forcing polynomial of a pentagonal parallelogram T(k, m; p) (1 ≤ p < k, m) has the following recurrence relation: k−1
F (k, m; p) =
F (i, m − 1; i − k + p)x +
k−p−1
F (i, m − 1 )x +
i=0
i=k−p
m −1
F (k − 1, j; p − 1 )x,
(16)
j= p
with initial conditions F (l, n; 0 ) = F (l, n ), for l, n ≥ 1. As some special cases, we have
F (k, m; 1 ) = F (k − 1, m − 1 )x +
k−2
F (i, m − 1 )x +
i=0
F (n, n; n − 1 ) =
n−1
m −1
F (k − 1, j )x = F (k, m ) − x,
j=1
F (i, n − 1; i − 1 )x + x + F (n − 1, n − 1; n − 2 )x,
(17)
i=1
for k, m, n ≥ 2. For convenience, we define T(n , n; n) to be T (n − 1, n; n − 1 ), for n > n ≥ 1. For Eq. (16), setting x to 1 gives the perfect matching count of a pentagonal benzenoid
(k, m; p) =
k−1
(i, m − 1; i − k + p) +
k−p−1
(i, m − 1 ) +
i=0
i=k−p
m −1
(k − 1, j; p − 1 ),
j= p
simplifying it gives
(T (k, m; p)) = (T (k − 1, m; p − 1 )) + (T (k, m − 1; p)), a tedious calculation gives [19],
(T (k, m; p)) =
m+k k
−
m+k , p−1
for 1 ≤ p < k, m. For Eq. (17), setting x to 1 gives the perfect matching count of a triangular benzenoid
(n, n; n − 1 ) =
n−1
(i, n − 1; i − 1 ) + 1 + (n − 1, n − 1; n − 2 ),
i=1
simplifying it gives
(T (n, n; n − 1 )) = (T (n − 1, n; n − 2 )) + (T (n − 1, n − 1; n − 2 )), a tedious calculation gives [19],
(T (n, n; n − 1 )) = for n ≥ 2.
1 2n + 2 , n+2 n+1
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We now introduce another related benzenoids. A single hexagon can be added to a benzenoid parallelogram in two ways: either to its one edge or to its two adjacent edges. However, in the latter case the obtained benzenoids possess no perfect matchings, since their order are odd. In the former case, three classes of benzenoids can be generated depending on which edge in L(k, m) the hexagon is attached, for k, m ≥ 2. These three classes of benzenoids, denoted by L (k, m), L (k, m), L (k, m), are depicted in Fig. 4, respectively. By similar arguments used for benzenoid parallelogram, we can derive explicit expressions of the forcing polynomial for L (k, m), L (k, m) and L (k, m). Theorem 3.2. The forcing polynomials of benzenoid parallelograms with one additional hexagon L (k, m), L (k, m) and L (k, m) (k, m ≥ 2) are
F (L (k, m ), x ) = 2F (k, m − 1 )x +
k−1
F (i, m − 1 )x,
i=0
F (L (k, m ), x ) = 2F (k − 1, m )x +
m −1
F (k − 1, j )x,
j=0
F (L (k, m ), x ) = 2F (k, m )x − 2x2 + x. For the above three equations, setting x to 1 gives the perfect matching counts of L (k, m), L (k, m) and L (k, m) [20]
(L (k, m )) = 2
m+k−1 k
+
m+k−1 m
+
(L (k, m )) = 2
(L (k, m )) = 2
m+k k
m+k−1 , m m+k−1 , k
− 1.
References [1] P. Adams, M. Mahdian, E.S. Mahmoodian, On the forced matching numbers of bipartite graphs, Discr. Math. 281 (2004) 1–12. [2] Z. Che, Z. Chen, Forcing on perfect matchings – a survey, MATCH Commun. Math. Comput. Chem. 66 (2011) 93–136. [3] Q. Chou, H. Liang, F. Bai, Computing the permanental polynomial of the high level fullerene C70 with high precision, MATCH Commun. Math. Comput. Chem. 73 (2015) 327–336. [4] C.P. Chou, H.A. Witek, Closed-form formulas for the Zhang–Zhang polynomials of benzenoid structures: chevrons and generalized chevrons, MATCH Commun. Math. Comput. Chem. 72 (2014) 105–124. [5] S.J. Cyvin, I. Gutman, Kekulé Structures in Benzenoid Hydrocarbons, Springer, Berlin, 1988. [6] M. Dehmer, F. Emmert-Streib, Y. Shi, M. Stefu, S. Tripathi, Discrimination power of polynomial-based descriptors for graphs by using functional matrices, PLoS ONE 10 (10) (2015) e0139265. [7] T. Došlic´ , Perfect matchings in lattice animals and lattice paths with constraints, Croat. Chem. Acta 78 (2005) 251–259. [8] M. Gordon, W.H.T. Davison, Theory of resonance topology of fully aromatic hydrocarbons. I, J. Chem. Phys. 20 (1952) 428–435. ´ [9] I. Gutman, B. Borovicanin , Zhang–Zhang polynomial of multiple linear hexagonal chains, Z. Naturforsch. 61a (2006) 73–77. [10] P. Hansen, M. Zheng, Bonds fixed by fixing bonds, J. Chem. Inf. Comput. Sci. 34 (1994) 297–304. [11] F. Harary, D.J. Klein, T.P. Živkovic´ , Graphical properties of polyhexes: perfect matching vector and forcing, J. Math. Chem. 6 (1991) 295–306. [12] D.J. Klein, M. Randic´ , Innate degree of freedom of a graph, J. Comput. Chem. 8 (1987) 516–521. [13] H. Lei, Y. Yeh, H. Zhang, Anti-forcing numbers of perfect matchings of graphs, Discr. Appl. Math. 202 (2016) 95–105. [14] M. Randic´ , Aromaticity of polycyclic conjugated hydrocarbons, Chem. Rev. 103 (2003) 3449–3605. [15] M. Randic´ , D.J. Klein, Mathematical and Computational Concepts in Chemistry, John Wiley & Sons, New York, 1985, pp. 274–282. [16] M.E. Riddle, The minimum forcing number for the torus and hypercube, Discr. Math. 245 (2002) 283–292. [17] J. Sedlar, The global forcing number of the parallelogram polyhex, Discr. Appl. Math. 160 (2012) 2306–2313. [18] Y. Shi, M. Dehmer, I. Gutman, X. Li, Graph Polynomials, CRC Press, 2016. [19] R. Tošic´ , S.J. Cyvin, Enumeration of Kekulé structures in benzenoid hydrocarbons: “flounders”, J. Math. Chem. 3 (1989) 393–401. [20] D. Vukicˇ evic´ , I. Lukovits, N. Trinajstic´ , Counting Kekulé structures of benzenoid parallelograms containing one additional benzene ring, Croat. Chem. Acta 79 (2006) 509–512. [21] D. Vukicˇ evic´ , N. Trinajstic´ , On the anti-forcing number of benzenoids, J. Math. Chem. 42 (2007) 575–583. [22] L. Xu, H. Bian, F. Zhang, Maximum forcing number of hexagonal systems, MATCH Commun. Math. Comput. Chem. 70 (2013) 493–500. [23] F. Zhang, X. Li, Forcing bonds of a benzenoid system, Acta Math. Appl. Sin. 12 (1996) 209–215. [24] F. Zhang, X. Li, Hexagonal systems with forcing edges, Discr. Math. 140 (1995) 253–263. [25] F. Zhang, H. Zhang, A new enumeration method for kekulé structures of hexagonal systems with forcing edges, J. Mol. Struct. (Theochem) 331 (1995) 255–260. [26] H. Zhang, K. Deng, Spectrum of matching forcing numbers of a hexagonal system with a forcing edge, MATCH Commun. Math. Comput. Chem. 73 (2015) 457–471. [27] H. Zhang, S. Zhao, R. Lin, The forcing polynomial of catacondensed hexagonal systems, MATCH Commun. Math. Comput. Chem. 73 (2015) 473–490.
Contents lists available at ScienceDirect
Applied Mathematics and Computation journal homepage: www.elsevier.com/locate/amc
Forcing polynomials of benzenoid parallelogram and its related benzenoids✩ Shuang Zhao, Heping Zhang∗ School of Mathematics and Statistics, Lanzhou University, Lanzhou, Gansu 730000, P.R. China
a r t i c l e
i n f o
Keywords: Forcing polynomial Perfect matching Innate degree of freedom Forcing number Benzenoid
a b s t r a c t Klein and Randic´ introduced the innate degree of freedom (forcing number) of a Kekulé structure (perfect matching) M of a graph G as the smallest cardinality of subsets of M that are contained in no other Kekulé structures of G, and the innate degree of freedom of the entire G as the sum over the forcing numbers of all perfect matchings of G. We proposed the forcing polynomial of G as a counting polynomial for perfect matchings with the same forcing number. In this paper, we obtain recurrence relations of the forcing polynomial for benzenoid parallelogram and its related benzenoids. In particular, for benzenoid parallelogram, we derive explicit expressions of its forcing polynomial and innate degree of freedom by generating functions. © 2016 Elsevier Inc. All rights reserved.
1. Introduction The forcing number of a perfect matching in benzenoid systems has been introduced by Harary et al. [11] more than twenty years ago. The roots of this concept can be found in an earlier paper by Klein and Randic´ [12] using the name ‘innate degree of freedom’. The forcing concept in Kekulé structures plays an important role in the resonance theory of theoretic chemistry [14,15]. In particular, for all benzenoid systems, recently Xu et al. [22] showed that the maximum forcing number is equal to the Clar number, and Lei et al. [13] showed that the maximum anti-forcing number is equal to the Fries number. Both indices can measure the stability of benzenoid hydrocarbons. We may refer to a survey [2] on this topic. A benzenoid system (benzenoid, or hexagonal system) is a 2-connected finite plane graph such that every interior face is a regular hexagon of side length one [5]. The study of perfect matching in benzenoid systems is of chemical relevance since a benzenoid system with a perfect matching can be regarded as carbon skeleton of a benzenoid hydrocarbon molecule [5]. Also, many well-known combinatorial problems have been encountered in perfect matching count of benzenoid systems. For example, the number of k-combinations of k + m elements equals perfect matching count of benzenoid parallelogram L(k, m) [7,8], and (n + 1 )-st Catalan number equals perfect matching count of triangular benzenoid T (n, n; n − 1 ) [7,19]. Let G be a graph with a perfect matching M. A forcing set S of M is a subset of M such that S is contained in no other perfect matchings of G. The forcing number of M, denoted by f(G, M), is the smallest cardinality over all forcing sets of M. The innate degree of freedom of G, denoted by IDF(G), is the sum over the forcing numbers of all perfect matchings of G. An edge of G is called forcing if it is contained in exactly one perfect matching of G. The maximum and minimum values of
✩ ∗
Supported by NSFC (grant no. 11371180) and the Fundamental Research Funds for the Central Universities (grant no. lzujbky-2015-211). Corresponding author. Tel.: +86 18993168369; fax: +86 9318912481. E-mail addresses: [email protected] (S. Zhao), [email protected] (H. Zhang).
http://dx.doi.org/10.1016/j.amc.2016.03.008 0 096-30 03/© 2016 Elsevier Inc. All rights reserved.
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S. Zhao, H. Zhang / Applied Mathematics and Computation 284 (2016) 209–218
a
b
Fig. 1. Illustration of L(k, m) (a) and L(k, m) ai (b), for k = 4, m = 5, i = 2.
f(G, M) over all perfect matchings M of G are denoted by F(G) and f(G), respectively. For other related concepts we refer the reader to [2,5]. Zhang and Li [23,24], and Hansen and Zheng [10] independently characterized benzenoid systems with a forcing edge, which include the above mentioned benzenoids. Zhang and Zhang [25] gave a novel computation of perfect matching count for such benzenoids, and Zhang and Deng [26] showed that their forcing numbers of all perfect matchings form either the integer interval from 1 to the Clar number or with only the gap 2. In particular, for benzenoid parallelogram, some other indices and properties about forcing were also obtained, such as global forcing number [17], anti-Kekulé number [21], anti-forcing number [21], and Clar cover polynomial [9]. In [27], we proposed the forcing polynomial of a graph to study the number of perfect matchings that have the same forcing number, i.e.
F (G, x ) =
x f (G,M ) =
M∈M(G )
F (G )
w(G, i )xi ,
(1)
i= f ( G )
where M(G ) denotes the set of all perfect matchings of G and w(G, i ) denotes the number of perfect matchings of G with forcing number i. It is easy to show that
F (G, x )|x=1 = (G ), the perfect matching count of G,
d F (G, x ) dx
= IDF (G ), the innate degree of freedom of G.
(2)
(3)
x=1
For some other graph polynomials, see a new book [18] and recent papers [3,4,6]. In this paper, we will apply this polynomial to benzenoid parallelogram and its related benzenoid systems. In Section 2, we first derive recurrence relation of the forcing polynomial for benzenoid parallelogram by equivalent definition of forcing set, and then obtain its explicit expression by generating function. As corollaries, we give explicit expressions of its perfect matching count and innate degree of freedom, and asymptotic behavior of its innate degree of freedom as one variable approaches infinity. In the end of this section, we present a new combinational explanation for the coefficients of forcing polynomial for benzenoid parallelogram with (0, 1)-sequence. In Section 3, we get recurrence relation of the forcing polynomial for pentagonal benzenoid, and explicit expressions of the forcing polynomial for benzenoid parallelograms with one additional hexagon. 2. Benzenoid parallelogram First recall some basic results on forcing set of a perfect matching of a graph. Let G be a graph with a perfect matching M. A cycle of G is called M-alternating if its edges appear alternately in M and E(G)M, where E(G) denotes the edge set of G. Lemma 2.1. [1,16] Let G be a graph with a perfect matching M. A subset S ⊆ M is a forcing set of M if and only if each Malternating cycle of G contains at least one edge of S. For an edge e ∈ M, let G e be the subgraph obtained from G by first deleting e with the end vertices, and then deleting recursively pendant edges (with an end of degree one) with the ends until the remaining graph has no pendant edges or is empty (no vertices). We call the deleted edges of M in the above procedure determined by e.
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Lemma 2.2. [27] Let G be a graph with a perfect matching M. If e ∈ M is contained in a minimum forcing set of M, then
f (G, M ) = f (G e, M ∩ E (G e )) + 1. A benzenoid system in a parallelogram-like shape, called benzenoid parallelogram and denoted by L(k, m), consists of k × m hexagons, arranged in m rows, each row containing k hexagons, shifted by a half hexagon to the right from the row immediately below, for k, m ≥ 1. In Fig. 1(a), we give as an illustrative example a benzenoid parallelogram L(k, m) for k = 4 and m = 5. For convenience, we make a convention that M(k, m ) = M(L(k , m )) and F (k , m ) = F (L(k, m ), x ) (see Eq. (1)). Since L(m, k) is obtained from L(k, m) by rotating 120 degree counter-clockwise and then inverting upside down, we have F (m, k ) = F (k, m ). For simplicity of presentation, we give some notations first. From left to right, we in turn denote the hexagons in the lowermost row by Ci , and their right ‘vertical’ edges by ai , for i = 1, 2, . . . , k. From the bottom up, we in turn denote the leftmost hexagons in each row by Dj , and their top right ‘oblique’ edges by bj , for j = 1, 2, . . . , m. Denote the left ‘vertical’ edge of C1 (= D1 ) by f and its bottom left ‘oblique’ edge by e (see Fig. 1(a)). We now divide M(k, m ) into two subsets Me and M f , where Me = {M ∈ M(k, m ) | e ∈ M} and M f = {M ∈ M(k, m ) | f ∈ M}. Further, subdivide the two sets as follows: Mei = {M ∈ Me | ai ∈ M}, for i = 1, 2, . . . , k, and M j = {M ∈ M f | b j ∈ M}, f
f
for j = 1, 2, . . . , m. Then we have Me = ∪ki=1 Mei and M f = ∪m M j . By Eq. (1), we have j=1
F (k, m ) =
x f (L(k,m ),M ) +
M∈Mei ,1ik
x f (L(k,m ),M ) .
(4)
M∈M jf ,1 jm
For convenience, we define L(0, n) and L(n, 0) to be empty graphs, for n ≥ 0, this yields F (0, n ) = 1 [27]. Then we can derive the following result. Theorem 2.3. The forcing polynomial of a benzenoid parallelogram L(k, m) (k, m ≥ 1) has the following recurrence relation:
F (k, m ) =
k−1
F (i, m − 1 )x +
m −1
i=0
F (k − 1, j )x,
(5)
j=0
with initial conditions F (0, n ) = 1, for n ≥ 0. f
Proof. For k, m ≥ 1, we claim that for an arbitrary M ∈ Mei (resp. M j ), ai (resp. bj ) is contained in some minimum forcing set of M, for i = 1, 2, . . . , k (resp. j = 1, 2, . . . , m ). In fact, we can see that ai (resp. bj ) is contained in M-alternating hexagon Ci (resp. Dj ), and every M-alternating cycle that contains any edge of M ∩ E(Ci ) (resp. M ∩ E(Dj )) must contain ai (resp. bj ), for i = 1, 2, . . . , k (resp. j = 1, 2, . . . , m). By Lemma 2.1 and the definition of minimum forcing set, we can verify the claim. It is observed from Fig. 1(b) that L(k, m ) ai = L(k − i, m − 1 ), for i = 1, 2, . . . , k, where we give as an illustrative example L(k, m) ai for k = 4, m = 5, i = 2, and use double lines to mark the edges which ai determines. Similarly, we have L(k, m ) b j = L(k − 1, m − j ), for j = 1, 2, . . . , m. By Lemma 2.2, we have f (L(k, m ), M ) = f (L(k − i, m − 1 ), M ∩ E (L(k − i, m − 1 ))) + f
1, for M ∈ Mei , i = 1, 2, . . . , k, and f (L(k, m ), M ) = f (L(k − 1, m − j ), M ∩ E (L(k − 1, m − j ))) + 1, for M ∈ M j , j = 1, 2, . . . , m. These yield
x f (L(k,m ),M ) =
M∈Mei
x f (L(k−i,m−1),M∩E (L(k−i,m−1)))+1
M∈Mei
=
x f (L(k−i,m−1),M ) · x = F (k − i, m − 1 )x,
M∈M(k−i,m−1 )
for i = 1, 2, . . . , k, and
x f (L(k,m ),M ) =
M∈M jf
x f (L(k−1,m− j ),M∩E (L(k−1,m− j )))+1
M∈M jf
=
x f (L(k−1,m− j ),M ) · x = F (k − 1, m − j )x,
M∈M(k−1,m− j )
for j = 1, 2, . . . , m. Substituting the above equations into Eq. (4), we immediately obtain Eq. (5).
To obtain explicit expression of the forcing polynomial for benzenoid parallelogram, a little manipulation is needed as the following lemma. Lemma 2.4. The forcing polynomial of a benzenoid parallelogram L(k, m) (k, m ≥ 1) has the following recurrence relation:
(1 − 2x )F (k, m ) + xF (k, m − 1 ) + xF (k − 1, m ) + F (k + 1, m + 1 ) = F (k + 1, m ) + F (k, m + 1 ). Proof. We need two new polynomials:
Gk,m =
k i=0
F (i, m ), Gm,k =
m j=0
F ( j, k ),
(6)
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for k, m ≥ 0. According to F (m, k ) = F (k, m ) and Eq. (5), it follows that
Gk,m − Gk−1,m = F (k, m ) = Gm,k − Gm−1,k ,
(7)
Gk,m − Gk−1,m = F (k, m ) = xGk−1,m−1 + xGm−1,k−1 .
(8)
for k, m ≥ 1. We derive
F (k, m ) = −xGk,m−1 + xGk−1,m−1 − xGm,k−1 + xGm−1,k−1 + xGk,m−1 + xGm,k−1 = −xF (k, m − 1 ) − xF (k − 1, m ) + xGk,m−1 + xGm−1,k + xGm,k−1 + xGk−1,m − xGm−1,k − xGk−1,m = −xF (k, m − 1 ) − xF (k − 1, m ) + F (k + 1, m ) + F (k, m + 1 ) −xGm−1,k + xGm,k − xGk−1,m + xGk,m − xGm,k − xGk,m = −xF (k, m − 1 ) − xF (k − 1, m ) + F (k + 1, m ) + F (k, m + 1 ) + 2xF (k, m ) − F (k + 1, m + 1 ), for k, m ≥ 1, which implies Eq. (6).
Theorem 2.5. The forcing polynomial of a benzenoid parallelogram L(k, m) (k, m ≥ 1) can be expressed in the following explicit form:
F (k, m ) =
min{k,m}
i=1
−
j=1
min{k,m} i−1 i=2
i i+1 j
j=1
i j
j+k−i−1 k−i
j+k−i−1 k−i
m− j i x + m−i
max{k, m} min{k,m} x |m − k|
m− j−1 i x − m−i
max{k, m} − 1 min{k,m} x . |m − k|
Proof. We define the generating function of F(k, m) to be
V (s, t ) =
∞ ∞
F (k, m )sk t m .
k=0 m=0
In the following we will derive a closed expression of V(s, t). From Eq. (6), we obtain
( 1 − 2x )
∞ ∞
F (k, m )sk+1 t m+1 + x
k=1 m=1
+x =
∞ ∞
∞ ∞
F (k, m − 1 )sk+1t m+1
k=1 m=1
F (k − 1, m )sk+1t m+1 +
∞ ∞
k=1 m=1 ∞ ∞
k=1 m=1 ∞ ∞
k=1 m=1
k=1 m=1
F (k + 1, m )sk+1t m+1 +
F (k + 1, m + 1 )sk+1t m+1
F (k, m + 1 )sk+1t m+1 .
(9)
Computing the first term, we have ∞ ∞
F (k, m )sk+1 t m+1 = st
k=1 m=1
∞ ∞
F (k, m )sk t m
k=1 m=1
= st
∞ ∞
F (k, m )s t − k m
k=0 m=0
= stV (s, t ) −
∞ m=0
F (0, m )t − m
∞
F (k, 0 )s
k
k=1
st s2 t − . 1−t 1−s
For the other terms, similarly we have, ∞ ∞
F (k, m − 1 )sk+1t m+1 = st 2V (s, t ) −
st 2 , 1−t
F (k − 1, m )sk+1t m+1 = s2 tV (s, t ) −
s2 t , 1−s
k=1 m=1 ∞ ∞ k=1 m=1 ∞ ∞
F (k + 1, m + 1 )sk+1t m+1 = V (s, t ) −
k=1 m=1 ∞ ∞ k=1 m=1
1 xs(2t − t 2 ) 1 xt (2s − s2 ) − − − + 1 + 2xst, 2 1−t 1−s (1 − t ) (1 − s )2
F (k + 1, m )sk+1t m+1 = tV (s, t ) −
t s2 t 2t − t 2 − ts − xst − , 2 1−t 1 −s (1 − t )
S. Zhao, H. Zhang / Applied Mathematics and Computation 284 (2016) 209–218 ∞ ∞
213
s st 2 2s − s2 − ts − xst − . 2 1−s 1 −t (1 − s )
F (k, m + 1 )sk+1t m+1 = sV (s, t ) −
k=1 m=1
Substituting these expressions into Eq. (9), we have 1 1−t
V (s, t ) =
1 −1 1−s . xts − 1xts 1−s −t
+
1−
(10)
Next we shall expand the above function V(s, t) so that we derive the explicit expression of the forcing polynomial F(k, m). By binomial theorem, we know
V (s, t ) =
i+1 i ∞ ∞ 1 1 1 1 + xi t i si − + xi t i si , 1−s 1−t 1−s 1−t i=0
and
1
1 + 1−s 1−t
i
=
i
j 1 i − j 1 i 1−t
j 1 i − j i−1 1 i j=1
=
1−s
j
j=0
=
i−1
1−s
j
i j
j=1
∞
w=0
1−t
+ ∞
1−t
1 i
+
1−s
i=1 j=1
i=0
+
1−s
i 1
∞ w=0
1−t
xt s =
∞ i−1
i j
i=2 j=1
+
∞ ∞ i=1 u=0
Extracting the coefficient of
F (k, m ) =
s k tm
−
min{k,m} i−1
j=1
for k, m ≥ 1, where
⎧ m ⎪ ⎪ k xk , ⎪ ⎨ ξk,m = 2xk , ⎪ ⎪ ⎪ ⎩ k xm ,
∞
w=0
∞
u=0
i=0 w=0
w=0
i+w−1 w s , w
∞
i+w w i i i s xt s, w
j+w−1 w i− j+u−1 u i i i s t xt s w u u=0
i+u−1 u i i i t xt s + u
i+1 j
i j
j+k−i−1 k−i
∞ ∞
i=1 w=0
i+w−1 w i i i s x t s + 1. w
j+k−i−1 k−i
m− j i x + ξk,m m−i
m− j−1 i x − ηk,m m−i
⎧ m−1 k ⎪ if k < m, ⎪ m−k x , ⎪ ⎨ k if k = m, ηk,m = 2 x , ⎪ ⎪ ⎪ ⎩ k − 1 xm , if k > m, k−m
m
The proof is completed.
j=1
i=2
∞
from both sides of Eq. (11), we can show
min{k,m} i i=1
j+w−1 w i− j+u u i i i s t xt s w u
i=0 u=0
i i i
u=0
∞ ∞ ∞ ∞ i+u u i i i + t xt s + u
1
∞
u=0
i+1 ∞ ∞ i 1 1 i+1 + xi t i si = j 1−s 1−t i=0
1 i
j+w−1 w i− j+u−1 u i+u−1 u s t + t + w u u
for i ≥ 1. Computing every term of Eq. (11), we have
∞
(11)
i=0
if k < m, if k = m, if k > m.
Remark 2.6. With the help of generating function of Gk,m , we have an alternative method to derive the expression of V(s, t). We define the generating function of Gk,m to be
W (s, t ) =
∞ ∞ k=0 m=0
Gk,m sk t m , W (t, s ) =
∞ ∞ k=0 m=0
Gm,k sk t m .
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S. Zhao, H. Zhang / Applied Mathematics and Computation 284 (2016) 209–218
Now we give two new equations. From Eqs. (7) and (8), we obtain ∞ ∞ k=1 m=1 ∞ ∞
Gk,m sk t m − Gk,m sk t m −
∞ ∞ k=1 m=1 ∞ ∞
k=1 m=1
∞ ∞
Gk−1,m sk t m =
k=1 m=1 ∞ ∞
Gk−1,m sk t m = x
k=1 m=1
∞ ∞
Gm,k sk t m −
k=1 m=1
Gk−1,m−1 sk t m + x
k=1 m=1
Gm−1,k sk t m , ∞ ∞
Gm−1,k−1 sk t m .
(12)
(13)
k=1 m=1
Computing every term, we have ∞ ∞
Gk,m sk t m =
k=1 m=1
∞ ∞ k=0 m=0
and similarly,
Gk,0 sk
k=1
s
(1 − s )2
,
1 1 − + 1, 1−s (1 − t )2
Gm,k sk t m = W (t, s ) −
k=1 m=1 ∞ ∞
∞
1 1 − + 1, 1−t (1 − s )2
Gk−1,m sk t m = sW (s, t ) −
k=1 m=1 ∞ ∞
G0,m t m −
m=0
= W (s, t ) − ∞ ∞
∞
Gk,m sk t m −
t
Gm−1,k sk t m = tW (t , s ) −
(1 − t )2
k=1 m=1 ∞ ∞
,
Gk−1,m−1 sk t m = stW (s, t ),
k=1 m=1 ∞ ∞
Gm−1,k−1 sk t m = stW (t , s ).
k=1 m=1
Substituting these expressions into Eqs. (12) and (13), we have
W (t, s ) =
1 1 + 1−s −1 1−s 1−t W (s, t ), W (s, t ) = . 1−t 1 − s − xts − 11−s xts −t
The relationship between generating functions V(s, t) and W(s, t) is
V (s, t ) =
∞ ∞
∞ ∞
Gk,m sk t m −
k=1 m=0
Gk−1,m sk t m +
k=1 m=0
∞
F (0, m )t m = (1 − s )W (s, t ),
m=0
which yields Eq. (10). The following is an immediate consequence of Theorem 2.5. Corollary 2.7. The forcing polynomial of a benzenoid parallelogram L(k, m) (1 ≤ k ≤ m) is
k i i+1 F (k, m ) = j i=1 j=1
−
k i−1 i=2 j=1
i j
j+k−i−1 k−i
m− j i x + m−i
j+k−i−1 k−i
m k x k
m− j−1 i x − m−i
m−1 k x , m−k
For 1 ≤ k ≤ m, from the above corollary we can find that the maximum forcing number F (L(k, m )) = k and the minimum forcing number f (L(k, m )) = 1, and the corresponding number of perfect matchings are [27]
k k w(L(k, m ), k ) = j j=0
w(L(k, m ), 1 ) =
m− j , m−k
2, m + 1,
if k 2, if k = 1.
For Eq. (10), differentiating both sides with respect to x and setting x to 1 gives the generating function of IDF(L(k, m)) ∞ ∞ k=0 m=0
IDF (L(k, m ))s t
k m
=
1 1−t
+
1 1−s
1−
−1 ts 1−s
−
ts 1−s ts 1−t
+
2
ts 1−t
=
ts 1−s
1−
ts 1−s
+
ts
1−t . − 1ts−t (1 − s − t )
S. Zhao, H. Zhang / Applied Mathematics and Computation 284 (2016) 209–218
215
Setting x to 1 gives the generating function of (L(k, m)) ∞ ∞
1 1−t
(L(k, m ))skt m =
1−
k=0 m=0
1 −1 1−s ts ts − 1−s 1−t
+
1 , 1−s−t
=
which implies the following corollary. Corollary 2.8. [8] The perfect matching count of a benzenoid parallelogram L(k, m) (k, m ≥ 1) is
k+m . k
(L(k, m )) =
Applying Corollary 2.7 to Eqs. (2) and (3), we obtain the following two corollaries. Corollary 2.9. The perfect matching count of a benzenoid parallelogram L(k, m) (1 ≤ k ≤ m) is
(L(k, m )) =
k i i+1 j i=1 j=1
−
k i−1
i j
i=2 j=1
j+k−i−1 k−i
j+k−i−1 k−i
m− j m−i
+
m k
m− j−1 m−i
m−1 . m−k
−
(14)
Corollary 2.10. The innate degree of freedom of a benzenoid parallelogram L(k, m) (1 ≤ k ≤ m) is
k i i+1 j
IDF (L(k, m )) =
i=1 j=1
−
k i−1
i j
i=2 j=1
j+k−i−1 k−i
j+k−i−1 k−i
m− j i+ m−i
m k k
m− j−1 i− m−i
m−1 k. m−k
(15)
Corollary 2.11. For a benzenoid parallelogram L(k, m) (1 ≤ k ≤ m), we have
lim
m→∞
IDF (L(k, m )) = k. (L(k, m ))
Proof. It is obvious that
(L(k, m )) =
m+k k
By Eq. (15), we have
IDF (L(k, m )) =
∼
mk as m → ∞. k!
k i i+1 j i=1 j=1
−
k i−1 i=2 j=1
i j
j+k−i−1 k−i
j+k−i−1 k−i
m− j i+ i− j
m k k
m− j−1 i− i− j−1
mk m−1 k∼ k as m → ∞, k−1 k!
since the second term is an infinite large quantity dominating the other terms. So the corollary holds.
From the above corollary, we know that the average forcing number per perfect matching approaches k when k is fixed and m goes to infinity. However, we have not obtained the asymptotic behavior of the innate degree of freedom of a benzenoid parallelogram when both k and m go to infinity. We conclude this section with a combinatorial explanation of the forcing polynomial. Let a(k, m; i) be the number of (0, 1)-sequences of k 1’s and m 0’s with maximum number of disjoint pairs “01” (consecutive elements in the corresponding (0, 1)-sequence) being i, for k, m ≥ 1, 1 ≤ i ≤ min {k, m}. We find that the coefficients of the forcing polynomial for benzenoid parallelogram can be expressed as such combinatorial count; Precisely, see Theorem 2.12. To obtain it we place a benzenoid parallelogram as shown in Fig. 2 such that an edge-direction is vertical and it has exactly one peak (vertex whose neighbors are below it) and one valley (vertex whose neighbors are above it). A monotone path in it is a path connecting the peak with the valley in which, when starting at the peak, one always goes downwards. This is easy to see in such monotone paths, vertical and oblique (may happen two directions as left-downward and rightdownward) edges appear alternately. A monotone path is exemplified in Fig. 2. Such monotone paths have one-one correspondence with perfect matchings. The corresponding is as follows: take all oblique edges in the monotone and vertical edges not in the monotone path and a perfect matching is obtained [8]. Theorem 2.12. w(L(k, m ), i ) = a(k, m; i ) (k, m 1, 1 i min{k, m} ).
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S. Zhao, H. Zhang / Applied Mathematics and Computation 284 (2016) 209–218
Fig. 2. Illustration for the proof of Theorem 2.12.
Fig. 3. Illustration of T(k, m; p) (left) and T (n, n; n − 1 ) (right), for k = 4, m = 5, p = 2, n = 5.
Proof. We proceed by establishing a one-to-one correspondence between all perfect matchings of L(k, m) and all (0, 1)sequences of k 1’s and m 0’s, for k, m ≥ 1, then by proving that the forcing number of each perfect matching equals the maximum number of disjoint pairs “01” of its corresponding (0, 1)-sequence. Besides giving the entire monotone path to obtain a perfect matching in L(k, m), we could only give the direction of oblique edges in the monotone path when starting at the peak. Treat the left-downward edge as 1, and right-downward as 0, we could get a (0, 1)-sequence of k 1’s and m 0’s corresponding to a perfect matching. For example, the perfect matching in Fig. 2 is corresponding to sequence 1011010 0 0. It is known that the forcing number of perfect matching M equals the maximum number of disjoint M-alternating hexagons of L(k, m) [26], since L(k, m) has a forcing edge. And it is obvious that all the M-alternating hexagons of L(k, m) intersect the corresponding monotone path at two oblique edges since all the oblique edges in M appear only in the monotone path. Consecutive oblique edges of different direction in the monotone path together with some vertical edge not in the monotone path constitute the matching in M-alternating hexagon. These imply a one-to-one corresponding between M-alternating hexagons and pairs “01” of the (0, 1)-sequence corresponding to M. For example, the maximum set of disjoint M-alternating hexagons of the perfect matching in Fig. 2, illustrated using bold lines in the first graph, is corresponding to three pairs “10” (1st and 2nd entries), “10” (4th and 5th entries) and “10” (6th and 7th entries) of 1011010 0 0. Then the claim holds.
3. Related benzenoids A benzenoid system in a pentagon-like shape, called pentagonal benzenoid and denoted by T(k, m; p), consists of two parts: one is m − p − 1 rows with k hexagons in each row, shifted by a half hexagon to the right from the row immediately below; another is p + 1 rows with the number of hexagons in a row decreasing by one from k in the lowermost row to k − p in the uppermost row, each row shifted by one and a half hexagon to the right from the row immediately below; the two parts adhere at the uppermost row of the (m − p − 1 )’s and the lowermost row of ( p + 1 )’s, the upper shifted by a half hexagon to the right from the lower, for 1 ≤ p < k, m. In particular, a pentagonal benzenoid in a triangle-like shape with n rows, is called triangular benzenoid and denoted by T (n, n; n − 1 ), for n ≥ 2. In Fig. 3, we give as illustrative examples a pentagonal benzenoid T(k, m; p) and a triangular benzenoid T (n, n; n − 1 ), for k = 4, m = 5, p = 2 and n = 5.
S. Zhao, H. Zhang / Applied Mathematics and Computation 284 (2016) 209–218
217
Fig. 4. Illustration of L (k, m), L (k, m) and L (k, m) (from left to right), for k = 4 and m = 5.
By similar arguments used for benzenoid parallelogram, we can derive recurrence relation of the forcing polynomial for pentagonal benzenoid. For convenience, we make a convention that F (T (k, m; p), x ) = F (k, m; p) and define T(k, m; 0) to be L(k, m). Theorem 3.1. The forcing polynomial of a pentagonal parallelogram T(k, m; p) (1 ≤ p < k, m) has the following recurrence relation: k−1
F (k, m; p) =
F (i, m − 1; i − k + p)x +
k−p−1
F (i, m − 1 )x +
i=0
i=k−p
m −1
F (k − 1, j; p − 1 )x,
(16)
j= p
with initial conditions F (l, n; 0 ) = F (l, n ), for l, n ≥ 1. As some special cases, we have
F (k, m; 1 ) = F (k − 1, m − 1 )x +
k−2
F (i, m − 1 )x +
i=0
F (n, n; n − 1 ) =
n−1
m −1
F (k − 1, j )x = F (k, m ) − x,
j=1
F (i, n − 1; i − 1 )x + x + F (n − 1, n − 1; n − 2 )x,
(17)
i=1
for k, m, n ≥ 2. For convenience, we define T(n , n; n) to be T (n − 1, n; n − 1 ), for n > n ≥ 1. For Eq. (16), setting x to 1 gives the perfect matching count of a pentagonal benzenoid
(k, m; p) =
k−1
(i, m − 1; i − k + p) +
k−p−1
(i, m − 1 ) +
i=0
i=k−p
m −1
(k − 1, j; p − 1 ),
j= p
simplifying it gives
(T (k, m; p)) = (T (k − 1, m; p − 1 )) + (T (k, m − 1; p)), a tedious calculation gives [19],
(T (k, m; p)) =
m+k k
−
m+k , p−1
for 1 ≤ p < k, m. For Eq. (17), setting x to 1 gives the perfect matching count of a triangular benzenoid
(n, n; n − 1 ) =
n−1
(i, n − 1; i − 1 ) + 1 + (n − 1, n − 1; n − 2 ),
i=1
simplifying it gives
(T (n, n; n − 1 )) = (T (n − 1, n; n − 2 )) + (T (n − 1, n − 1; n − 2 )), a tedious calculation gives [19],
(T (n, n; n − 1 )) = for n ≥ 2.
1 2n + 2 , n+2 n+1
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S. Zhao, H. Zhang / Applied Mathematics and Computation 284 (2016) 209–218
We now introduce another related benzenoids. A single hexagon can be added to a benzenoid parallelogram in two ways: either to its one edge or to its two adjacent edges. However, in the latter case the obtained benzenoids possess no perfect matchings, since their order are odd. In the former case, three classes of benzenoids can be generated depending on which edge in L(k, m) the hexagon is attached, for k, m ≥ 2. These three classes of benzenoids, denoted by L (k, m), L (k, m), L (k, m), are depicted in Fig. 4, respectively. By similar arguments used for benzenoid parallelogram, we can derive explicit expressions of the forcing polynomial for L (k, m), L (k, m) and L (k, m). Theorem 3.2. The forcing polynomials of benzenoid parallelograms with one additional hexagon L (k, m), L (k, m) and L (k, m) (k, m ≥ 2) are
F (L (k, m ), x ) = 2F (k, m − 1 )x +
k−1
F (i, m − 1 )x,
i=0
F (L (k, m ), x ) = 2F (k − 1, m )x +
m −1
F (k − 1, j )x,
j=0
F (L (k, m ), x ) = 2F (k, m )x − 2x2 + x. For the above three equations, setting x to 1 gives the perfect matching counts of L (k, m), L (k, m) and L (k, m) [20]
(L (k, m )) = 2
m+k−1 k
+
m+k−1 m
+
(L (k, m )) = 2
(L (k, m )) = 2
m+k k
m+k−1 , m m+k−1 , k
− 1.
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