ISSN 2820-6657 (electronic edn.)
DISCRETE MATHEMATICAL CHEMISTRY 1 (2025) #P1.02
https://doi.org/10.26493/2820-6657.4.a42
(Also available at http://dmc-journal.eu)
Continuous forcing spectra of perfect matchings
of convex hexagonal systems*
Bo Zhang, Yaxian Zhang , Heping Zhang †
School of Mathematics and Statistics, Lanzhou University,
Lanzhou, Gansu 730000, P.R. China
Received 7 August 2022, accepted 3 May 2023, published online 18 September 2025
Abstract
A convex hexagonal system is a hexagonal system whose inner dual graph has a convex
polygonal boundary. A forcing set S for a perfect matching M of a graph G is a subset
of M that is contained in no other perfect matchings of G. The smallest cardinality of a
forcing set of M is called the forcing number of M , denoted by f(G,M). The forcing
spectrum of G is defined as: Spec(G) = {f(G,M)|M is a perfect matching of G}. In this
paper, we show that for any convex hexagonal system O(m, k, n) with a perfect matching,
its forcing spectrum is continuous (or an integer interval).
Keywords: Convex hexagonal system, perfect matching, forcing number, forcing spectrum.
Math. Subj. Class. (2020): 05C70, 05C92, 92E10
1 Introduction
A hexagonal system (or benzenoid system) H is a plane 2-connected bipartite graph whose
interior faces are regular hexagons with length 1. A perfect matching (or Kekulé structure
in chemical literature) M of a graph G is a set of disjoint edges covering all vertices of G.
A forcing set of a perfect matching M of a graph G is a subset S of M that is contained
in no other perfect matchings of G. The smallest cardinality of a forcing set of M is called
the forcing number of M , denoted by f(G,M). The minimum (resp. maximum) forcing
number of G is the minimum (resp. maximum) value of forcing numbers of all perfect
matchings of G, denoted by f(G) (resp. F (G)).
*This work is supported by NSFC (Grant No. 12271229). The authors are very grateful to the referees for
their valuable suggestions and comments in improving the presentation of this manuscript.
†Corresponding author.
E-mail addresses: 1879190934@qq.com (Bo Zhang), zhangyax@lzu.edu.cn (Yaxian Zhang),
zhanghp@lzu.edu.cn (Heping Zhang)
cb This work is licensed under https://creativecommons.org/licenses/by/4.0/
2 Discrete Math. Chem. 1 (2025) #P1.02
The forcing number of a perfect matching of a graph was introduced by Harary et
al. [7], and originally appeared in earlier papers of Randić and Klein [8, 11] under the
name “innate degree of freedom” of Kekulé structures, which plays an important role in
the resonance theory in chemistry. Xu et al. [14] proved that the maximum forcing number
of a hexagonal system is equal to its Clar number, which can measure the stability of
benzenoid hydrocarbons within Clar’s aromatic sextet theory [5]. For polyomino graphs
and (4,6)-fullerene graphs, the same results still hold [13, 23] in an extensive sense.
A spanning subgraph C of a hexagonal system H is called a Clar cover if each compo-
nent of C is either a hexagon or an edge. The set of hexagons of a Clar cover C of H is a
resonant set or sextet pattern of H . Let h(C) denote the number of hexagons of C. A Clar
cover of H with the maximum number of hexagons is called a Clar structure, and the Clar
number Cl(H) of H is the number of hexagons in a Clar structure.
Afshani et al. [2] defined the forcing spectrum of a graph G as:
Spec(G) = {f(G,M)|M is a perfect matching of G}
and showed that any finite set of positive integers can be the forcing spectrum of a planar
bipartite graph. So it is an interesting problem to determine whether a graph G has the
continuous forcing spectrum, i.e. Spec(G) forms an integer interval from the minimum to
maximum forcing numbers of G. Lam and Pachter [9] determined the forcing spectra of
polyomino graphs of stop signs that are continuous. For any simply-connected polyomino
graphs Zhang and Jiang [19] showed their forcing spectra are always continuous by using
Z-transformation graph. Che and Chen [4] conjectured that the forcing spectrum of any
fullerene graph is continuous. A hexagonal system is called forced if it has a forcing edge
(an edge in exactly one perfect matching). Zhang and Li [17] determined the forced hexag-
onal systems. Zhang and Deng [18] further got the forcing spectra of forced hexagonal
systems, which are either continuous or with only gap 2. For a recent survey on matching
forcing and related topics, see [20].
The inner dual graph T (H) of a hexagonal system H is a plane graph such that the
vertices are the centers of all the hexagons of H , and two vertices are joined by a (straight)
edge providing the corresponding hexagons have a common edge in H . The boundary of
T (H) means the boundary of the exterior face of T (H). A hexagonal system H is convex
if the boundary of T (H) bounds a convex polygon or a straight line segment. Cyvin and
Gutman [6] showed that a convex hexagonal system has a perfect matching if and only if
the periphery of T (H) is either a path, a parallelogram or a big hexagon with each angle
equal to 120◦ and each pair of opposite sides having the same length. Therefore any convex
hexagonal system with a perfect matching can be uniquely determined by an integer triplet
(m, k, n) and denoted by O(m, k, n) with conventions m ≤ k ≤ n; For example, see
Figure 1.
For a convex hexagonal system O(m, k, n), Cyvin and Gutman [6] and Bodroža et
al. [3] gave and proved a formula for the number of Kekulé structures respectively. Zhang
[17] got an expression of its Clar number, i.e. maximum forcing number. Zhang and
Zhang [21] proved that its minimum forcing number is equal to m by using monotonic path
systems. Further they [22] proved that a regular convex hexagonal system O(m,m,m) has
the continuous forcing spectrum.
In this paper we use a simpler method to show that the forcing spectrum of a general
convex hexagonal system O(m, k, n) is also continuous (see Theorem 3.6). To this end we
give some preliminaries in Section 2 and a proof to our main result in Section 3.
B. Zhang et al.: Continuous forcing spectra of perfect matchings of convex hexagonal systems 3
2 Preliminaries
Suppose that hexagonal systems considered are drawn in the plane such that some edges
are vertical. Let H be a hexagonal system with a specific hexagon s0 at the center, O. We
establish a 3-coordinate system O − ABC on H such that O is the origin and the three
axes are perpendicular to three disjoint edges of s0 respectively (see Figures 1 and 2). The
coordinate system O − ABC divides the plane into three areas AOB, BOC and COA.
For a point W in the plane, we define its coordinates with respect to O − ABC. If W
lies in the area AOB (for the other cases we can do similarly), draw two straight lines
through W such that one is parallel to axis OB and intersects axis OA at the point WA,
and the other is parallel to axis OA and intersects axis OB at the point WB . Suppose that
the distance between two centers of any two adjacent hexagons of H is 1. The lengths of
OWA and OWB are defined as the coordinates of W on axes OA and OB respectively, and
the coordinate of W on axis OC is defined as zero (see Figure 2). If the lengths of OWA
and OWB are denoted by x and y respectively, then the coordinate of W with respect to
O −ABC is written as (x, y, 0).
n
k
center subgraph
O(m,m,m)
center subgraph
O(m0,k0,n0)
O
m
C
B
A
Figure 1. O(5, 7, 8) with 3-divisible M
and center subgraphs.
wA
wB
w
O
C
B
A
t = 6
t = 6
t = 6
t = 6
t = 6
t = 6
t = 6
t = 6
t = 6
t = 6
t  
t = 6
t = 6
t = 6
t = 6
t = 6 t = 6
t = 6
t = 6
t = 6
t = 6
t = 6
t = 6
t = 6
t = 6
t = 6
t = 6
t = 6
t = 6
t = 6
t = 6
t = 6
t = 6
t = 6
t = 6
t = 6
t = 6
t = 6
t = 6
t = 6
t = 6t = 6
t = 6
t = 6
t = 6t = 6
t = 6
t = 6
t = 6
t = 6
t = 6
t = 6
t = 6
t = 6
t = 6
t = 6
t = 6
t = 6
t = 6
t = 6
t = 6
t = 6
t = 6
t = 6
t = 6
t = 6
t = 6
t = 6
t = 6
t  
t  
t = 6
t = 6
t = 6t = 6
t = 6
t = 6t = 6
t = 6
t = 6t = 6
t = 6
Figure 2. The coordinate (4, 2, 0) of a point
W w.r.t. O −ABC.
We mark a hexagon in H with si,j,k, where (i, j, k) is the coordinate of the center of
the hexagon. We make a convention: we use double bonds, bold double bonds and grey
hexagons to represent a perfect matching, a forcing set and the hexagons of a Clar cover,
respectively.
A perfect matching M of a hexagonal system H is called 3-divisible if no edge of M
intersect an axis and the edges of M lying in the same area are mutually parallel to each
other (see Figure 1).
The Z-transformation graph or resonance graph Z(H) of a hexagonal system H is the
graph whose vertices represent the perfect matchings of H and two perfect matchings M1
and M2 of H are joined by an edge if and only if they differ in exactly one hexagon of H .
4 Discrete Math. Chem. 1 (2025) #P1.02
Lemma 2.1 ( [16], Lemma 1). Let H be a hexagonal system. Then Z(H) has a 1-degree
vertex if and only if there exists a 3-coordinate system O − ABC w.r.t. a hexagon s0 such
that a perfect matching M of H is 3-divisible w.r.t. O −ABC.
Zhang and Li [16] proved that Z(O(m, k, n)) has exactly two vertices of degree one.
For convenience, from now on we always place an O(m, k, n) on the plane and establish
a 3-coordinate system O − ABC w.r.t. s0 satisfying the condition of Lemma 2.1 and axes
OB, OA and OC toward right, upward and downward (see Figure 1).
Let M be a 3-divisible perfect matching of O(m, k, n) w.r.t. O − ABC. A subgraph
O(m0, k0, n0) of O(m, k, n) is called a center subgraph (1 ≤ m0 ≤ m,m0 ≤ k0 ≤
k, k0 ≤ n0 ≤ n) if M ∩ E(O(m0, k0, n0)) is also a 3-divisible perfect matching of
O(m0, k0, n0) w.r.t. O −ABC (see Figure 1).
Let G be a graph with a perfect matching. A subgraph G′ of a graph G is called a nice
subgraph of G if G− V (G′) has a perfect matching.
Proposition 2.2. Any center subgraph O(m0, k0, n0) of O(m, k, n) is a nice subgraph of
O(m, k, n).
For a bipartite graph G with a perfect matching M , an edge subset S ⊂ M forces an
edge uv ∈ M \ S if all neighbors of v except u are in V (S), where V (S) denotes the set
of end-vertices of edges in S. If there exists a sequence of edges {ei}ki=1 and a sequence
of edge subsets {Si}ki=0 such that S0 = S, Si = Si−1 ∪ {ei} and Si−1 forces ei for
1 ≤ i ≤ k, then we say S forces Sk. We define G ⊖ S as G − V (Sk) whenever S forces
Sk and G− V (Sk) has no 1-degree vertex.
Lemma 2.3 ([1, 12]). Let M be a perfect matching of a bipartite graph G. Then S is a
forcing set of M if and only if S forces M.
Proposition 2.4. Let M be a perfect matching of a bipartite graph G. For any S ⊆ M , if
S
′
is a forcing set of E(G ⊖ S) ∩M in the subgraph G ⊖ S, then S ∪ S′ is a forcing set
of M .
Proof. Since S
′
is a forcing set of (G⊖S)∩M , that is, S ∪S′ forces M . By Lemma 2.3,
we have that S ∪ S′ is a forcing set of M .
Let A and B be finite sets. Then we define the symmetric difference A△B = (A−B)∪
(B−A) = A∪B−A∩B. Let M be a perfect matching of a graph G. A cycle of G is M -
alternating, if its edges appear alternately in M or not. If C is an M -alternating cycle of G,
then M△C := M△E(C) is also a perfect matching of G. If C = {C1, C2, ..., Cm} is a set
of disjoint M -alternating cycles of G, then we define M△C = M△C1△C2△· · ·△Cm
for convenience, which is also a perfect matching of G.
Lemma 2.5 ([10]). Let M be a perfect matching in a planar bipartite graph G. Then
f(G,M) = c(M), where c(M) is the maximum number of disjoint M-alternating cycles
of G.
For a hexagon s in a hexagonal system H , the edges of s are denoted clockwise by
e1, e2, e3, f1, f2 and f3 as shown in Figure 3 (sometimes we add superscript as esi , f
s
i , i =
1, 2, 3). The hexagon s with perfect matchings {e1, f2, e3} and {f1, e2, f3} are improper
and proper respectively (see Figure 3). A perfect matching M of H can be partitioned into
three subsets, E1(M), E2(M) and E3(M), such that all edges in Ei(M) are parallel to ei
(or fi), i = 1, 2, 3.
B. Zhang et al.: Continuous forcing spectra of perfect matchings of convex hexagonal systems 5
e1
e2
e3f1
f2
f3
(a)
e1
e2
e3f1
f2
f3
(b) improper
e1
e2
e3f1
f2
f3
(c) proper
Figure 3. A hexagon s with marked edges and two states.
Lemma 2.6. Let M be a perfect matching of a hexagonal system H . If S ⊂ M can force
an arbitrary edge subset Ei(M), i = 1, 2, 3, then S is a forcing set of M .
Proof. Without loss of generality, assume S can force E1(M). Then H ⊖ S ⊆ H −
V (E1(M)). Naturally, M ′ =: E2(M)∪E3(M) is a perfect matching of H −V (E1(M)).
Since any M -alternating cycle in H contains an edge in each Ei(M) for i = 1, 2, 3, there
is no M ′-alternating cycles in H − V (E1(M)). So H − V (E1(M)) has a unique perfect
matching M ′. So ∅ is a forcing set of M ′ in H − V (E1(M)). By Lemma 2.3, S can force
M .
Let T ′(p) represent a prolate triangle polyhex with side length p, whose boundary faces
consist of two linear hexagonal chains with length p and one zigzag hexagonal chain with
length 2p− 1 (see Figure 4). Let T r(p) represent a hexagonal system obtained from T ′(p)
by adding r linear hexagonal chains with length p along one side (see Figure 5). Obvi-
ously, both T ′(p) and T r(p) are forced hexagonal systems. Zhang et al. [15] introduced an
invariant triple (x, y, z) with x ≤ y ≤ z for any perfect matching of a hexagonal system,
where x, y and z are the numbers of double bonds of M in the three different directions
and showed that x is an upper bound for the Clar number. Zhang [17] showed that the Clar
numbers of T ′(p) and T r(p) are both equal to x = p. From [18, Theorem 4.5], we get
the forcing spectra of T ′(p) and T r(p) are both continuous. For convenience, for integers
q ≤ p we can use [q, p] the integer interval consisting of all integers x with q ≤ x ≤ p.
Corollary 2.7 ([18]). Spec(T ′(p)) = [1, p].
A
B
B
O
O
A
(a) M1
A
B
B
O
O
A
(b) M2
O
A O
A
B
B
(c) M3
O
A O
A
B
B
(d) M4
Figure 4. A perfect matching sequence M1M2M3M4 of T
′
(4) and minimum forcing sets
for each Mi.
6 Discrete Math. Chem. 1 (2025) #P1.02
Now we construct a perfect matching sequence {Mi}pi=1 of T ′(p) such that
f(T ′(p),Mi) = i. Let M1 be the 3-divisible perfect matching of T ′(p) w.r.t. O − ABC.
Let Hi = {sa,b,0|a + b = i − 1}, Mi = M1△H1△· · ·△Hi−1 and Si = {es2|s ∈ Hi} =
E2(Mi) for i = 1, 2, . . . , p (see Figure 4). Then Si can force Mi by Lemma 2.6 and Hi
consists of i disjoint Mi-alternating cycles, so f(T ′(p),Mi) = i by Lemma 2.5.
Corollary 2.8. Spec(T r(p)) = [1, p].
A
B
O
B
O
A
(a) M1
T'(3)
A
B
O
B
O
A
(b) M2
A
A
B
O
O
B
(c) M3
A
A
B
O
O
B
(d) M4
Figure 5. A perfect matching sequence M1M2M3M4 of T 2(4) and minimum forcing sets
for each Mi.
Proof. For T r(p), we can construct a perfect matching sequence {Mi}pi=1 of T r(p) such
that f(T r(p),Mi) = i. Let M1 be the 3-divisible perfect matching of T r(p) w.r.t. O −
ABC. If i ≥ 2, let Mi contain f
sp−1,r,0
2 . As shown in Figure 5, T
r(p) ⊖ fsp−1,r,02 is
isomorphic to T ′(p− 1). Let {M ′i}
p−1
i=2 be a perfect matching sequence of T
′(p− 1) such
that f(T ′(p− 1),M ′i) = i− 1. Let Mi be a perfect matching of T r(p) containing M ′i and
f
sp−1,r,0
2 . We can easily get f(T
r(p),Mi) = i.
3 Forcing spectra of convex hexagonal systems
Reference [17] found the expression of the Clar number of O(m, k, n), which equals the
maximum forcing number, and gave a construction for a Clar structure of O(m, k, n),
which plays an important role in our later proof.
Theorem 3.1 ([17]). For any convex hexagonal system O(m, k, n) (m ≤ k ≤ n) we have
Cl(O(m, k, n)) =

mk − 14{(m+ k − n)
2 − 1}, if m+ k − n ≥ 2 is odd,
mk − 14 (m+ k − n)
2, if m+ k − n ≥ 2 is even,
mk, if m+ k − n ≤ 1.
Now we recall such construction for a Clar cover C(r) of O(m, k, n) (simply, H) with
parameter r (1 ≤ r ≤ m) as follows (for details, see Proof (1) of Theorem 5 in [17]). Let
n0 =min{n,m+k−r}. If n0 = n ≤ m+k−r, there exist six prolate triangle polyhexes on
the corners of H respectively as T1 = T
′
(r), T2 = T
′
(k−r+1), T3 = T
′
(m−r+1), T4 =
T
′
(n−k+ r), T5 = T
′
(n−m+ r), T6 = T
′
(m+k−n− r+1), and their maximum res-
onant sets are sequentially labeled (s1, s2, . . . , sr), (sr, sr+1, . . . , sk), (h1, . . . , hm−r+1),
(sk, . . . , sn+r−1), (hm−r+1, . . . , hn), (hn, . . . , hk+m−r), where s1 = h1 and sn+r−1 =
hk+m−r, and s1, sr, sk, hm−r+1, sn+r−1, hn lie on the periphery of H (see Figure 6).
B. Zhang et al.: Continuous forcing spectra of perfect matchings of convex hexagonal systems 7
Figure 6. Clar cover C(r) of O(m, k, n)
with r ≤ m + k − n (for this example,
m = 5, k = 7, n = 8, r = 2).
T4
T5
T2
T3
T1
T6
s...
s...
n-n0
n-n0
n0-m+r
n0-k+r
k-r+1
r
m
s1=h1
sn+r-2
hn0
h...
h...
h...
hi
h2
s...
si+r-1
sk
sr+1
sr
s2
s1
C
B
A
O
Figure 7. Clar cover C(r) of O(m, k, n)
with r > m+k−n(for this example, m =
5, k = 7, n = 8, r = 5).
After deleting the vertices of T1, . . . T6 together with their incident edges, we get an all-
benzenoid system (i.e. its Clar structure only contains hexagons). In this case, Clar cover
C(r) of H has been determined. If n > n0 = m + k − r, the Clar cover of the center
subgraph O(m, k, n0) only depends on r (1 ≤ r ≤ m) and has been already constructed
as above and denoted by C ′(r), where T6 = T
′
(1), and O(m, k, n) may be reproduced
by adding n − n0 copies of a hexagonal chain of hexagons intersecting axes OA and OB
to O(m, k, n0) in turn as shown in Figure 7. The (n − n0)(k + m) independent edges
produced and C ′(r) form a Clar cover C(r) of H . There must be a Clar structure of H
such as C(r) [17].
Let Mr be the set of all perfect matchings obtained from the Clar cover C(r) by placing
each hexagon of C(r) in proper or improper states, where the double bonds of C(r) remain
unchanged. Partition the hexagons of Clar cover C(r) of H into m disjoint classes, where
the i-th class, denoted by Fi, consists of hexagons lying on the i-th vertical line from left
to right; for example, the first class and last class are the maximum resonant sets of T5 and
T2 respectively; See Figures 6 and 7.
Take a center subgraph O(m0, k0, n0) of H with m0 ≤ k0 ≤ n0. For 1 ≤ r ≤ m0, let
Cm0,k0,n0(r) denote the Clar cover of H that contains Clar cover C(r) of O(m0, k0, n0) so
that the part of Cm0,k0,n0(r) in H−O(m0, k0, n0) is a 3-divisible matching w.r.t. O−ABC
(see Figures 8 and 9). Let Mm0,k0,n0r be the set of all perfect matchings of H corresponding
to Cm0,k0,n0(r). In particular, we write simply Cm0,m0,m0(r) as Cm0(r) and Mm0,m0,m0r
as Mm0r .
Theorem 3.2. For any convex hexagonal system O(m, k, n) (m ≤ k ≤ n, simply H) we
have the following results.
(1) For each Clar cover C(r) of H , there exists a perfect matching Mr ∈ Mr such that
f(H,Mr) = h(C(r)) for 1 ≤ r ≤ m.
(2) For each Clar cover Cm,k0,k0(m) of H , there exists a perfect matching
M ∈ Mm,k0,k0m of H such that f(H,M) = h(Cm,k0,k0(m)) for m ≤ k0 ≤ k.
8 Discrete Math. Chem. 1 (2025) #P1.02
(3) For each Clar cover Cm0(m0) of H , there exists a perfect matching M ∈ Mm0m0 of
H such that f(H,M) = h(Cm0(m0)) +m−m0 for 1 ≤ m0 ≤ m.
Proof. (1) If r ≥ m+ k− n, let Mr ∈ Mr be a perfect matching of H such that es1 ∈ Mr
for any hexagon s in C(r) and Sr := {es1|s ∈ C(r)}(see Figure 7). By Lemma 2.5 we have
f(H,Mr) ≥ h(C(r)) = |Sr|. As shown in Figure 7, we observe that E1(Mr) = Sr ∪
(E(T1)∩E1(Mr)) and Sr ∩E(T1) can force E(T1)∩E1(Mr). So Sr can force E1(Mr).
By Lemma 2.6 we have that Sr is a forcing set of Mr. So f(H,Mr) = |Sr| = h(C(r)).
If r < m+k−n, let Mr ∈ Mr be a perfect matching of H such that es1 ∈ Mr for each
hexagon s ∈ Fi for i > m− r and fs1 ∈ Mr for each hexagon s ∈ Fi for 1 ≤ i ≤ m− r
(see Figure 6). Let
Sr := {es1|s ∈ Fi, i > m− r} ∪ {fs1 |s ∈ Fi, 1 ≤ i ≤ m− r}.
Similarly, it suffices to prove that Sr can force E1(Mr). As shown in Figure 6, we observe
that E1(Mr) = Sr ∪ (E(T1) ∩E1(Mr)) ∪ (E(T6) ∩E1(Mr)) and Sr ∩E(T1) can force
E(T1) ∩ E1(Mr). Since m− r + 1 ≥ m+ k − n− r + 1, we can show that Sr ∩ E(T6)
can force E(T6) ∩ E1(Mr). So Sr can force E1(Mr).
k-k0
center subgraph
O(m,k0,k0)
m
k0
n-k0 O
s...
s...
n-k0
m
hk0
h...
h2
hi
s...
si+m-1
sk0
sm
s2
s1
C
B
A
t = 6
t = 6
t = 6
t = 6
t = 6
t = 6
t = 6
t = 6
t = 6
t = 6
t = 6
t = 6
t = 6
t = 6
t = 6
t = 6
t = 6
t = 6
t = 6
t = 6
t = 6
t = 6
t = 6
t = 6
t = 6
t = 6
t = 6
t = 6
t = 6
t = 6
t = 6
t = 6
t = 6
t = 6t = 6
t = 6
t = 6
t = 6
t = 6
t = 6
t = 6t = 6
t = 6
t = 6
t = 6t = 6
t = 6
t = 6
t = 6
t = 6
t = 6
t = 6
t = 6
t = 6
t = 6
t = 6
t = 6
t = 6
t = 6
t = 6
t = 6
t = 6
t = 6
t = 6
t = 6
t = 6
t = 6
t = 6
t = 6t  
t  
t = 6
t = 6
t = 6t = 6
t = 6
t = 6t = 6
t = 6
t = 6t = 6
t = 6
Figure 8. Clar cover Cm,k0,k0(m) of H
with perfect matching M .
Tk-m0(m0-1)
n-m0
center subgraph
O(m0,m0,m0)
k-m0
O
s...
m
hm0
hi
si+m0-1
sm0
s1
C
B
A
Figure 9. Clar cover Cm0(m0) of H
with perfect matching M .
(2) Let M ∈ Mm,k0,k0m be a perfect matching of H such that es1 ∈ M for each hexagon
s in Cm,k0,k0(m) and Sk0 := {es1|s ∈ Cm,k0,k0(m)}(see Figure 8). Similarly, it suffices to
show that Sk0 can force E1(M). As shown in Figure 8, we observe that E1(M) = Sk0 ∪
{esa,b,01 : a+ b ≥ k0} and Sk0 ∩{e
sa,b,0
1 : a+ b = k0−1} can force {e
sa,b,0
1 : a+ b ≥ k0}.
So Sk0 can force E1(M).
(3) Let M ∈ Mm0m0 be a perfect matching of H such that e
s
1 ∈ M for each hexagon
s in Cm0(m0) (see Figure 9). Let Lm0 := {e
s0,m0−1+i,0
1 |1 ≤ i ≤ m − m0} and Sm0 :=
{es1|s ∈ Cm0(m)} ∪ Lm0 . As shown in Figure 9, we observe that the subgraph H ⊖ Lm0
is just O(m0, k, n). Since O(m0,m0,m0) is the center subgraph of O(m0, k, n), from (2)
we know that Sm0 \ Lm0 can force M ∩ E(O(m0, k, n)). By Proposition 2.4 we have
that Sm0 is a forcing set of M . We can also get that there are |Sm0 | − |Lm0 | disjoint
B. Zhang et al.: Continuous forcing spectra of perfect matchings of convex hexagonal systems 9
M -alternating hexagons in the center subgraph O(m0, k, n) (denoted by F ′m0 ) which also
belong to the center subgraph O(m0,m0,m0). As shown in Figure 11, the periphery of
the center subgraph O(m0 + i,m0 + i,m0 + i) of H is an M -alternating cycle for all
i = 1, . . . ,m − m0 and they do not intersect each other. Obviously these cycles do not
intersect to each hexagon in F ′m0 . So f(H,M
m0
m0 ) = |Sm0 | = h(Cm0(m0))+m−m0.
References [21] and [14] have proved f(O(m, k, n)) = m and F (O(m, k, n)) =
Cl(O(m, k, n)) respectively. In the following, we divide the proof of our main theorem
(see Theorem 3.6) into two parts: one is [m, 12m(m + 1)] ⊂ Spec(H) (see Lemma 3.3);
the other is [ 12m(m + 1), Cl(H)] ⊆ Spec(H) (see Corollary 3.5, which can be directly
deduced from Lemma 3.4).
Lemma 3.3. The integer interval [m, 12m(m+ 1)] is a subset of Spec(H).
Proof. By Theorem 3.2(3), there is Mm0m0 ∈ M
m0
m0 such that f(H,M
m0
m0 ) = h(Cm0(m0))+
m −m0 for 1 ≤ m0 ≤ m. So f(H,M11 ) = h(C1(1)) +m − 1 = m and f(H,Mmm ) =
h(Cm(m)) =
1
2m(m + 1). It suffices to prove that [f(H,M
m0−1
m0−1 ), f(H,M
m0
m0 )] ⊆
Spec(H) for 2 ≤ m0 ≤ m. Obviously f(H,Mm0m0 )− f(H,M
m0−1
m0−1 ) = |C0| = m0 − 1.
Let C be the set of all hexagons in Cm0(m0). Let C0 := {sa,b,0|a+ b = m0 − 1, b ̸=
m0 − 1}, Lm0 := {e
s0,m0−1+i,0
1 |1 ≤ i ≤ m−m0} and Sm0 := {es1|s ∈ C − C0} ∪ Lm0 .
Suppose T k−m0(m0 − 1) consists of hexagons {sa,b,0|b ≤ m0 − 2, a+ b ≥ m0 − 1},
as shown in Figure 9. We observe that H ⊖ Sm0 = T k−m0(m0 − 1). By Corollary 2.8,
we can get a perfect matching sequence {M ′j}
m0−1
j=1 of T
k−m0(m0 − 1) and the minimum
forcing set S
′
j for each M
′
j such that f(T
k−m0(m0 − 1),M ′j) = |S′j | = j. Let Mm0j be a
perfect matching of H obtained from Mm0m0 by replacing the edges of T
k−m0(m0 − 1) in
Mm0m0 with M
′
j for all j = 1, 2, . . . ,m0 − 1. By Proposition 2.4, we know that Sm0 ∪ S
′
j is
a forcing set of Mm0j .
Tm-m0(m0-1)
e1s0,m0-1,0
O
m
s0,m0-1,0
C
B
A
Figure 10. For j = 1, |Sm0 |+ |S
′
1| disjoint
Mm01-alternating cycles.
Tm-m0(m0-1)
e1s0,m0-1,0
O
m
s0,m0-1,0
C
B
A
Figure 11. For j ≥ 2, |Sm0 |+ |S
′
1| disjoint
Mm0j-alternating cycles.
Next we shall prove that there exist |Sm0 | + |S
′
j | disjoint Mm0j-alternating cycles in
H . If it holds for O(m,m,m), then it also holds for O(m, k, n). From the structure of
Mm0j as above, we can find |Sm0 | + |S
′
j | − |Lm0 | disjoint Mm0j-alternating hexagons in
10 Discrete Math. Chem. 1 (2025) #P1.02
H denoted by F ′j . Further it needs to find m − m0 disjoint Mm0j-alternating cycles not
intersecting F ′j . If j ≥ 2, as shown in Figure 11, we observe that the periphery of the center
subgraph O(m0+i,m0+i,m0+i) of H (denoted by Cm0+i) is an M
m0
m0 -alternating cycle
for i = 1, . . . ,m − m0. Because F ′j belongs to the center subgraph O(m0,m0,m0), F ′j
and Cm0+i do not intersect. So there exist |Sm0 | + |S
′
j | disjoint Mm0j-alternating cycles
for all j = 2, . . . ,m0 − 1. If j = 1, there also exist |Lm0 | disjoint Mm01-alternating
cycles (see Figure 10), which do not intersect F ′1. So there also exist |Sm0 |+ |S
′
1| disjoint
Mm01-alternating cycles.
To sum up, for 1 ≤ m0 ≤ m, 1 ≤ j ≤ m0 − 1 we have
f(H,Mm0j) = |Sm0 |+ |S
′
j | = f(H,Mm0m0 )− (m0 − 1) + j = f(H,M
m0−1
m0−1 ) + j.
So we get [f(H,Mm0−1m0−1 ), f(H,M
m0
m0 )] ⊆Spec(H).
H1
F11
hm=s2m-1
h...
si
O
s...
s2m-i
m
h2
hi sm
s2
s1
C
B
A
Figure 12. Clar cover Cm(m) of H .
F1(k-t+1)
H2
t
m-t+1
A
B
C
m
O
Figure 13. Perfect matching M1(k−t+1).
Among all Clar covers as C(r) (1 ≤ r ≤ m) of H , we choose r as t such that C(t) is
a Clar structure of H , i.e. h(C(t)) = Cl(H) is maximum.
From [17] we get that if m − n + k ≥ 2, then t = ⌈m−n+k+12 ⌉ or ⌊
m−n+k+1
2 ⌋; if
m− n+ k ≤ 1 then t = 1. For example, for O(5, 7, 8), t = 2 or 3; See Figure 6.
For the Clar cover Cm(m) of H , recall the partition of the hexagons of Cm(m) of H
into m disjoint classes, where the i-th class, denoted by Fi0 (1 ≤ i ≤ m), consists of
hexagons lying on the i-th column left to right (see Figure 12).
Let M10 ∈ Mmm be a perfect matching of H such that es1 ∈ M for each hexagon s in
Cm(m) (each s is improper; see Figure 12). Next we define recursively a series of perfect
matchings M1j of H from M10, j = 1, 2, . . . , k − t + 1. Let M11 = M10△F10. Then
each hexagon of F10 becomes proper. This results in a series of new improper hexagons
which are adjacent to hexagons of F10 via edges e1 or e3. Let F11 be the set of such
new improper M11-alternating hexagons; In other words, F11 can be obtained from F10 by
moving one column to the left (see Figure 12). Repeating the above procedure, let M1j =
M1(j−1)△F1(j−1) for j = 1, 2, . . . , k − t + 1 and F1j be the set of all improper M1j-
alternating hexagons minus all M1(j−1)-alternating hexagons; For example, see Figure 13
for M1(k−t+1).
B. Zhang et al.: Continuous forcing spectra of perfect matchings of convex hexagonal systems 11
For i = 2, . . . ,m, j ≥ 1, let Mij = Mi0△Fi0△· · ·△Fi(j−1), where
Mi0 = M(i−1)(k−t+1) or M(i−1)(k−t) according to i ≤ m − t + 1 or not, and Fij is
the set of improper Mij-alternating hexagons minus all Mi(j−1)-alternating hexagons. It is
not difficult to find that Mm(k−t) ∈ Mt is a perfect matching of H corresponding to Clar
structure C(t), and f(H,Mm(k−t)) = Cl(H). In this procedure if i < m − t + 1, then
0 ≤ j ≤ k − t+ 1 and we shift Fi0 to the left k − t+ 1 times, else 0 ≤ j ≤ k − t and we
shift it k − t times (see Figures 6, 12 and 13).
From M10 to Mm(k−t), we have already got a perfect matching sequence M10M11 · · ·
M1(k−t+1)(= M20)M21 · · ·Mi0Mi1 · · ·Mm0 · · ·Mm(k−t). Next we will prove that the
difference of forcing numbers of two consecutive perfect matchings in this sequence is no
more than 1.
Lemma 3.4. For 1 ≤ i ≤ m − t and 0 ≤ j ≤ k − t, and m − t + 1 ≤ i ≤ m and
0 ≤ j ≤ k − t− 1, we have
|f(H,Mi(j+1))− f(H,Mij)| ≤ 1. (3.1)
Proof. We recursively construct a minimum forcing set Sij of Mij so that |Sij | equals the
maximum number of Mij-alternating hexagons for 1 ≤ i ≤ m− t and 0 ≤ j ≤ k− t+ 1,
or m− t+ 1 ≤ i ≤ m and 0 ≤ j ≤ k − t.
For i = 1, 0 ≤ j ≤ k − t+ 1. Let S′1 := {es1|s ∈ Fi0, 2 ≤ i ≤ m}. Then S
′
1 can force
{esa,b,01 |a+ b ≥ m− 1, b ≥ 1} (see Figure 12) and H1 := H ⊖ S
′
1 is shown in Figure 14.
If n−m+ 1 < k − t+ 1 (in this case T6 in C(t) has length m+ k − n− t+ 1 ≥ 2 and
F1(k−t) lies on the area AOC; see Figures 13 and 14), let
S
′
1j =
{
{es1|s ∈ F1j}, if 0 ≤ j ≤ n−m,
{fs1 |s ∈ F1(j−1)}, if n−m+ 1 ≤ j ≤ k − t+ 1.
A
C
O
(a) j = 0
O
C
A
(b) j = 1
A
C
O
(c) j = k −m
A
C
O
(d) j = n−m
A
C
O
(e) j = k− t+ 1
Figure 14. Illustration for M1j ∩ E(H1) and S′1j with j = 0, 1, k −m,n−m, k − t+ 1.
Since S
′
1j can force E(H1)∩E1(M1j) (see Figure 14), S1j := S
′
1∪S
′
1j can force E1(M1j)
by Proposition 2.4. Further we get that S1j is a forcing set of M1j by Proposition 2.6.
12 Discrete Math. Chem. 1 (2025) #P1.02
On the other hand, there exist |S1j | disjoint M1j-alternating hexagons of H , so S1j is a
minimum forcing set of M1j and
f(H,M1j) = |S1j | =
{
|F1j |+ |S′1|, if 0 ≤ j ≤ n−m,
|F1(j−1)|+ |S′1|, if n−m+ 1 ≤ j ≤ k − t+ 1.
(3.2)
Combining Equation (3.2) we can get
f(H,M1(j+1))−f(H,M1j) =

|F1(j+1)| − |F1j | = 1, if 0 ≤ j ≤ k −m− 1,
|F1(j+1)| − |F1j | = 0, if k −m ≤ j
≤ n−m− 1(k < n),
|F1j | − |F1j | = 0, if j = n−m,
|F1j | − |F1(j−1)| = −1, if n−m+ 1 ≤ j ≤ k − t.
Otherwise, n−m+ 1 ≥ k − t+ 1 (see Figsures 15 and 16), let
S
′
1j = {es1|s ∈ F1j , 0 ≤ j ≤ k − t+ 1}.
Then S1j = S
′
1∪S
′
1j can force E1(M1j). Similarly we have that S1j is a minimum forcing
set of M1j . So
f(H,M1(j+1))− f(H,M1j) =
{
|F1(j+1)| − |F1j | = 1, if 0 ≤ j ≤ k −m− 1,
|F1(j+1)| − |F1j | = 0, if k −m ≤ j ≤ k − t.
To sum up, the lemma is true for i = 1.
A
BO
C
Figure 15. Perfect matching M10.
F1( k-t+1 )
O
A
B
C
Figure 16. Perfect matching Mm(k−t) and
F1(k−t+1) with n−m+ i ≥ k − t+ 1.
For 2 ≤ i ≤ m, suppose that a minimum forcing set S(i−1)j of M(i−1)j has been
already given. Next we construct a minimum forcing set Sij of Mij from Si0, which is
identical to S(i−1)(k−t+1) or S(i−1)(k−t) according to i < m− t+ 1 or not.
B. Zhang et al.: Continuous forcing spectra of perfect matchings of convex hexagonal systems 13
For i < m − t + 1, we have 0 ≤ j ≤ k − t + 1. Let Sij = S
′
i ∪ S
′
ij , where
S
′
i = S(i−1)(k−t+1) − {es1|s ∈ Fi0} and S
′
ij will be given next. If n−m+ i < k − t+ 1,
let
S
′
ij =
{
{es1|s ∈ Fij}, if 0 ≤ j ≤ n−m+ i− 1,
{fs1 |s ∈ Fi(j−1)}, if n−m+ i ≤ j ≤ k − t+ 1
else
S
′
ij = {es1|s ∈ Fij , 0 ≤ j ≤ k − t+ 1}.
For such perfect matching Mij of H , let Hi := H ⊖ S
′
i ; For example, H2 is shown in
Figures 13 and 18 (it consists of hexagons on the closed broken segment and its interior).
For n − m + i < k − t + 1, if 0 ≤ j ≤ k − m + i − 2, as shown in Figure 17 and
Figure 18 (a)-(b), we observe that E(Hi) ∩ E1(Mij) − S
′
ij appears on a linear hexagonal
chain {sa,i−1,0|m−i+j+1 ≤ a ≤ k−1} along the shifts of sm−i,i−1,0 , and e
sm−i+j,i−1,0
1
can force E(Hi)∩E1(Mij)−S
′
ij . So S
′
ij can force E(Hi)∩E1(Mij); If k−m+ i−1 ≤
j ≤ n−m+ i, as shown in Figure 18 (b) and (c) we observe that S′ij = E(Hi)∩E1(Mij),
and S
′
ij can force E(Hi) ∩ E1(Mij); if n −m + i + 1 ≤ j ≤ k − t + 1, then as shown
in Figures 17 and 18(d), we observe that E(Hi) ∩ E1(Mij) − S
′
ij appears on a linear
hexagonal chain
{sa,0,n−i|0≤a ≤ j−1−(n−m+i+i−1)}∪{s0,b,n−i|i−1−(j−1−(n−m+i)) ≤b ≤ i−1}
along the shifts of hexagon s0,i−1,m−i, and f
sj−(n−m+2i−1),0,n−i
1 or f
s0,i−1−j+(n−m+i)),n−i
1
can force E(Hi)∩E1(Mij)−S
′
ij . So S
′
ij can force E(Hi)∩E1(Mij). For n−m+i+1 ≥
k − t + 1, similarly S′ij can force E(Hi) ∩ E1(Mij). By Proposition 2.4, we know that
Sij is a forcing set of Mij . On the other hand there also exist |Sij | disjoint Mij-alternating
hexagons of H . Then Sij is the minimum forcing set of Mij i.e.
f(H,Mij) = |Sij | =
{
|Fij |+ |S′i|, if 0 ≤ j ≤ n−m+ i− 1,
|Fi(j−1)|+ |S′i|, if n−m+ i ≤ j ≤ k − t+ 1.
(3.3)
So, if n−m+ i < k − t+ 1, combining Equation (3.3) we get
f(H,Mi(j+1))−f(H,Mij) =

|Fi(j+1)| − |Fij | = 1, if 0 ≤ j ≤ k −m+ i− 2,
|Fi(j+1)| − |Fij | = 0, if k −m+ i− 1 ≤ j
≤ n−m+ i− 2(k < n),
|Fij | − |Fij | = 0, if j = n−m+ i− 1,
|Fij | − |Fi(j−1)| = −1, if n−m+ i ≤ j ≤ k − t;
else
f(H,Mi(j+1))−f(H,Mij) =
{
|Fi(j+1)| − |Fij | = 1, if 0 ≤ j ≤ k −m+ i− 2,
|Fi(j+1)| − |Fij | = 0, if k −m+ i− 1 ≤ j ≤ k − t.
The remaining case is i ≥ m − t + 1. Then 0 ≤ j ≤ k − t. Let S′i = S(i−1)(k−t) −
{es1|s ∈ Fi0} and S
′
ij = {es1|s ∈ Fij |0 ≤ j ≤ k − t}. Similarly Sij is a minimum forcing
set of Mij . So f(H,Mi(j+1)) − f(H,Mij) = 1 for all j = 0, 1, 2, . . . , (k − t − 1). The
proof is complete.
14 Discrete Math. Chem. 1 (2025) #P1.02
t
0≤j≤ k-m+i-2
n−m+i+1≤j≤k−t+1
hm
h...
si
O
s...
s2m-i
h2
hi sm
s2
s1
C
B
A
Figure 17. j shifts of hexagons sm−i,i−1,0 and s0,i−1,m−i for i = 1, 2, . . . ,m.
E(H2)∩E1(M20)-S'20
O
C
AA
C
O
(a) j = 0
E(H2)∩E1(M20)-S'20
O
C
AA
C
O
(b) j = k −
m+ 1
E(H2)∩E1(M2(k-t+1))-S'2(k-t+1)
O
C
A
A
C
O
(c) j = n −
m+ 2
E(H2)∩E1(M2(k-t+1))-S'2(k-t+1)
O
C
A
A
C
O
(d) j = k− t+1
Figure 18. Illustration for M2j∩E(H2) and S′2j with j = 0, , k−m+1, n−m+2, k−t+1.
B. Zhang et al.: Continuous forcing spectra of perfect matchings of convex hexagonal systems 15
By Theorem 3.2, we get f(H,M10) = 12m(m+1) and f(H,Mm(k−t)) = Cl(H). As
an immediate consequence of Lemma 3.4, we obtain the following result.
Corollary 3.5. The integer interval [ 12m(m+1), Cl(O(m, k, n))] is a subset of Spec(H).
By Lemma 3.3 and Corollary 3.5 we get the following main result.
Theorem 3.6. For any convex hexagonal systems O(m, k, n), then Spec(O(m, k, n)) =
[m,Cl(O(m, k, n))].
Remark 3.1. The method in obtaining Lemma 3.4 and Corollary 3.5 can be used to discuss
the forcing spectra of other types of hexagonal systems. There is an alternative approach to
obtain Corollary 3.5 as follows. Similar to the proof of Lemma 3.3, from the Clar covers
C(r) of O(m, k, n) for r ∈ [t,m], where C(t) is a Clar structure, we can construct a series
of perfect matchings to show that [ 12m(2k+1−m), Cl(O(m, k, n))] ⊂ Spec(O(m, k, n)),
and from the Clar covers Cm,k0,k0(m) of O(m, k, n) for k0 ∈ [m, k], we can construct a se-
ries of perfect matchings to show that [ 12m(m+1),
1
2m(2k+1−m)] ⊂ Spec(O(m, k, n)).
ORCID iDs
Yaxian Zhang https://orcid.org/0000-0003-0979-1438
Heping Zhang https://orcid.org/0000-0001-5385-6687
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