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Relations between global forcing number and maximum anti-forcing number of a graph111This work is supported by NSFC (Grant No. 11871256)

Yaxian Zhang and Heping Zhang Corresponding author.
(School of Mathematics and Statistics, Lanzhou University, Lanzhou, Gansu 730000, P.R. China
E-mails: [email protected], [email protected]
)
Abstract

The global forcing number of a graph GG is the minimal cardinality of an edge subset discriminating all perfect matchings of GG, denoted by gf(G)gf(G). For any perfect matching MM of GG, the minimal cardinality of an edge subset SE(G)MS\subseteq E(G)\setminus M such that GSG-S has a unique perfect matching is called the anti-forcing number of MM, denoted by af(G,M)af(G,M). The maximum anti-forcing number of GG among all perfect matchings is denoted by Af(G)Af(G). It is known that the maximum anti-forcing number of a hexagonal system equals the famous Fries number.

We are interested in some comparisons between the global forcing number and the maximum anti-forcing number of a graph. For a bipartite graph GG, we show that gf(G)Af(G)gf(G)\geq Af(G). Next we mainly extend such result to non-bipartite graphs. Let 𝒢\mathcal{G} be the set of all graphs with a perfect matching which contain no two disjoint odd cycles such that their deletion results in a subgraph with a perfect matching. For any G𝒢G\in\mathcal{G}, we also have gf(G)Af(G)gf(G)\geq Af(G) by revealing further property of non-bipartite graphs with a unique perfect matching. As a consequence, this relation also holds for the graphs whose perfect matching polytopes consist of non-negative 11-regular vectors. In particular, for a brick GG, de Carvalho, Lucchesi and Murty [4] showed that G𝒢G\in\mathcal{G} if and only if GG is solid, and if and only if its perfect matching polytope consists of non-negative 11-regular vectors.

Finally, we obtain tight upper and lower bounds on gf(G)Af(G)gf(G)-Af(G). For a connected bipartite graph GG with 2n2n vertices, we have that 0gf(G)Af(G)12(n1)(n2)0\leq gf(G)-Af(G)\leq\frac{1}{2}(n-1)(n-2); For non-bipartite case, 12(n2n2)gf(G)Af(G)(n1)(n2)-\frac{1}{2}(n^{2}-n-2)\leq gf(G)-Af(G)\leq(n-1)(n-2).

Keywords:  Perfect matching; Perfect matching polytope; Solid brick; Maximum anti-forcing number; Global forcing number.

1 Introduction

In this paper, we only consider finite simple graphs with at least one perfect matching. Let GG be a graph with vertex set V(G)V(G) and edge set E(G)E(G). We denote the order of GG by v(G)=|V(G)|v(G)=|V(G)|, and the size by e(G)=|E(G)|e(G)=|E(G)|. An edge subset MM of GG is a perfect matching (1-factor or a Kekulé structure in chemical literature), if every vertex of GG is incident with exactly one edge in MM.

In 1987, Klein and Randić [15] introduced the innate degree of freedom of a Kekulé structure, which plays an important role in the resonance theory in chemistry. Harary et al. [13] called it as forcing number: If a subset SS of a perfect matching MM in a graph GG is contained in exactly one perfect matching of GG, then SS is a forcing set of MM. The minimum cardinality over all forcing sets of MM is called the forcing number of MM, denoted by f(G,M)f(G,M). The minimum (resp. maximum) forcing number of GG is denoted by f(G)f(G) (resp. F(G)F(G)). For a hypercube QnQ_{n}, Diwan [9] showed that f(Qn)=2n2f(Q_{n})=2^{n-2}, confirming a conjecture by Pachter and Kim [24]. For a hexagonal system, Xu et al. [32] and Zhou and Zhang [35] showed that the maximum forcing number is equal to its Clar number (resonant number). For polyomino graphs and (4,6)-fullerenes (also called Birkoff-Von Neumann graphs) the same result also holds (cf. [34, 36, 27]).

As early as 1997, Li [19] raised the concept of forcing single edge (i.e. anti-forcing edge). In 2007, Vukičević and Trinajstić [30, 31] introduced the anti-forcing number of a graph GG. In general, recently Lei et al. [18] and Klein and Rosenfeld [16] independently defined the anti-forcing number of a single perfect matching MM of a graph GG. A subset of E(G)ME(G)\setminus M is an anti-forcing set of MM if its removal results in a graph with a unique perfect matching. The minimum cardinality of an anti-forcing set of MM is called the anti-forcing number of MM in GG, denoted by af(G,M)af(G,M). The minimum (resp. maximum) anti-forcing number of GG is denoted by af(G)af(G) (resp. Af(G)Af(G)). Lei et al. [18] and Shi et al. [27] respectively showed that the maximum anti-forcing number of hexagonal systems and (4,6)-fullerenes are equal to their Fries numbers. For upper bounds on maximum anti-forcing number of graphs, see [7, 28]

The concept of “global forcing” was introduced by Vukičević [29] to distinguish all perfect matchings of a graph. An edge subset SS of GG is called a global forcing set of GG, if no two distinct perfect matchings of GG coincide on SS. The minimum cardinality of a global forcing set of GG is the global forcing number of GG, denoted by gf(G)gf(G). Since then, some scholars studied the global forcing numbers of some chemical graphs, see [10, 1, 20, 33, 26, 29].

In this paper, we readily find that gf(G)F(G)gf(G)\geq F(G) for any graph GG. A relation Af(G)F(G)Af(G)\geq F(G) has been already revealed [18]. Došlić [10] and Deng and Zhang [8] respectively proved that the global forcing number and the maximum forcing number of a graph have the cyclomatic number as a common upper bound. Motivated by the above facts, it is natural to consider relations between the global forcing number and the maximum anti-forcing number of a graph.

For (4,6)-fullerene graph GG, a plane cubic graph whose faces are hexagons and squares, Cai and Zhang [1] gave a sharp lower bound on global forcing number: gf(G)2f3gf(G)\geq\lceil\frac{2f}{3}\rceil, and Shi and Zhang [27] obtained a formula for the maximum anti-forcing number: Af(G)=v3+2Af(G)=\left\lfloor\frac{v}{3}\right\rfloor+2, where ff and vv are the numbers of faces and vertices of GG respectively. By Euler’s formula we have 2f=v+42f=v+4, which implies that 2f3=v3+2\lceil\frac{2f}{3}\rceil=\left\lfloor\frac{v}{3}\right\rfloor+2. Accordingly we have that gf(G)Af(G)gf(G)\geq Af(G) for any (4,6)-fullerene graph GG.

For a general bipartite graph GG with 2n2n vertices, luckily we can prove that gf(G)Af(G)gf(G)\geq Af(G) by finding a special edge ee of GG with gf(G)gf(Ge)+1gf(G)\geq gf(G-e)+1 and Af(Ge)+1Af(G)Af(G-e)+1\geq Af(G). Moreover, we get that gf(G)Af(G)12(n1)(n2)gf(G)-Af(G)\leq\frac{1}{2}(n-1)(n-2), and equality holds if and only if GG is isomorphic to Kn,nK_{n,n}.

The next question is whether we can extend the above relation gf(G)Af(G)gf(G)\geq Af(G) to non-bipartite graphs. In this paper we give a negative answer by constructing a matching covered graph GkG_{k} for any positive integer kk such that gf(Gk)Af(Gk)=kgf(G_{k})-Af(G_{k})=-k. However we can extend such a relation to conformal graphs which has a close connection with classical problems as perfect matching polytope and tight cut decomposition in matching theory of graphs [22, 12, 11, 21].

Let 𝒢\mathcal{G} denote the set of graphs GG with a perfect matching such that GG contains no two disjoint odd cycles CC and CC^{\prime} such that GCCG-C-C^{\prime} has a perfect matching. We reveal a novel substructure of a graph with a unique perfect matching MM and the minimum degree larger than 11, which consists of two disjoint odd cycles and an odd path connecting them such that MM contains a perfect matching of it. Based on this substructure, we show that gf(G)Af(G)gf(G)\geq Af(G) for a graph G𝒢G\in\mathcal{G} by finding a 11-degree vertex from a special subgraph of GG with a unique perfect matching. As a corollary, we have that for a graph GG whose perfect matching polytope consists of non-negative 11-regular vectors, G𝒢G\in\mathcal{G} and thus gf(G)Af(G)gf(G)\geq Af(G).

In particular, for a brick GG (3-connected bicritical graph), de Carvalho, Lucchesi and Murty [4] showed that G𝒢G\in\mathcal{G} if and only if the perfect matching polytope of GG consists of non-negative 11-regular vectors, and if and only if GG is solid (without non-trivial separating cuts). For a general graph, the above three conditions are not necessarily equivalent. For more details on solid bricks, see [2, 23, 12, 3, 6, 5]. In 2013, Kawarabayashi [14] gave a simpler proof for the characterization of the graphs without two disjoint odd cycles.

Finally, for a connected graph GG, we get tight upper and lower bounds on gf(G)Af(G)gf(G)-Af(G): 12(n2n2)gf(G)Af(G)(n1)(n2)-\frac{1}{2}(n^{2}-n-2)\leq gf(G)-Af(G)\leq(n-1)(n-2), where the second equality holds if GG is isomorphic to complete graph K2nK_{2n}.

2 Definitions and some preliminary results

For a vertex vv in GG, we denote the set of all neighbors of vv by NG(v)N_{G}(v). And the cardinality of NG(v)N_{G}(v) is said to be the degree of vv, denoted by dG(v)d_{G}{(v)}. Let Δ(G)\Delta(G) and δ(G)\delta(G) denote the maximum and minimum degree of GG respectively. Lei et al. [18] showed the following relations between the forcing number and the anti-forcing number of a perfect matching of a graph.

Theorem 2.1.

[18] Let GG be a graph with a perfect matching MM. Then f(G,M)af(G,M)(Δ(G)1)f(G,M)f(G,M)\leq af(G,M)\leq(\Delta(G)-1)f(G,M), and thus F(G)Af(G)(Δ(G)1)F(G)F(G)\leq Af(G)\leq(\Delta(G)-1)F(G).

The cyclomatic number of a connected graph GG is c(G)=e(G)v(G)+1c(G)=e(G)-v(G)+1. In 2017, Deng and Zhang [8] proved that the maximum forcing number of GG is less than or equal to c(G)c(G).

Theorem 2.2.

[8] Let GG be a connected graph with a perfect matching. Then Af(G)c(G)Af(G)\leq c(G). If GG is non-bipartite, then Af(G)<c(G)Af(G)<c(G).

For two sets AA and BB, the symmetric difference of AA and BB is defined by AB=ABABA\oplus B=A\cup B-A\cap B. Let GG be a graph with at least one perfect matching MM. We say a subgraph HH of GG is nice if GV(H)G-V(H) has a perfect matching. For convenience, we denote GV(H)G-V(H) by GHG-H. An even cycle CC of GG is called an M-alternating cycle, if the edges of CC appear alternately in MM or not. An even cycle CC is a nice cycle of GG if and only if GG has two perfect matchings M1M_{1} and M2M_{2} such that M1M2=E(C)M_{1}\oplus M_{2}=E(C). In 2007, Došlić [10] gave the following result.

Theorem 2.3.

[10] Let GG be a connected graph with a perfect matching. Then gf(G)c(G)gf(G)\leq c(G), and equality holds if and only if all cycles of GG are nice.

We denote the set of all perfect matchings of a graph GG by \mathcal{M}, and Φ(G)=||\Phi(G)=|\mathcal{M}|. Došlić gave the following lower bound on the global forcing number.

Theorem 2.4.

[10] Let GG be a graph with a perfect matching. Then gf(G)log2Φ(G)gf(G)\geq\lceil\log_{2}\Phi(G)\rceil.

The following results give equivalent conditions for forcing sets, anti-forcing sets and global forcing sets of a graph.

Lemma 2.5.

[25] Let GG be a graph with a perfect matching MM. Then an edge subset SMS\subseteq M is a forcing set of MM if and only if SS contains an edge from every MM-alternating cycle.

Lemma 2.6.

[7] Let GG be a graph with a perfect matching MM. Then an edge subset SE(G)MS\subseteq E(G)\setminus M is an anti-forcing set of MM if and only if SS contains at least one edge of every MM-alternating of GG.

Lemma 2.7.

[1] Let GG be a graph with a perfect matching. Then an edge subset SS is a global forcing set of GG if and only if SS intersects each nice cycle of GG.

From these equivalent definitions, we can find some correlations.

Theorem 2.8.

Let GG be a graph with a perfect matching. Then gf(G)F(G)gf(G)\geq F(G).

Proof.

Let SS be a minimum global forcing set of GG and let MM be a perfect matching of GG with f(G,M)=F(G)f(G,M)=F(G). Then gf(G)=|S|gf(G)=|S|, and SS intersects every MM-alternating cycle by Lemma 2.7. If SMS\subseteq M, then SS is a forcing set of MM from Lemma 2.5. Thus gf(G)=|S|f(G,M)=F(G)gf(G)=|S|\geq f(G,M)=F(G). Otherwise, for each edge ee in SS not in MM there exists an edge ee^{\prime} in MM which is adjacent to ee. Replacing all edges ee in SMS\setminus M with the edges ee^{\prime} in MM and the other edges of SS remaining unchanged we get a new edge subset SS^{\prime}. Then SMS^{\prime}\subseteq M and SS^{\prime} intersect every MM-alternating cycle. So SS^{\prime} is a forcing set of MM by Lemma 2.5, and gf(G)=|S||S|f(G,M)=F(G)gf(G)=\left|S\right|\geq\left|S^{\prime}\right|\geq f(G,M)=F(G). ∎

Corollary 2.9.

Let GG be a graph with a perfect matching. Then gf(G)Af(G)(2Δ(G))F(G)gf(G)-Af(G)\geq(2-\Delta(G))F(G).

Proof.

It is immediate from Theorems 2.1 and 2.8. ∎

Let HH be a subgraph of GG. For any FE(G)F\subseteq E(G), we denote FE(H)F\cap E(H) by F|HF|_{H}.

Lemma 2.10.

[1] An edge subset SS is a global forcing set of a graph GG if and only if for each nice subgraph HH of GG, S|HS|_{H} is a global forcing set of HH.

Lemma 2.11.

[1] Let GG be a connected graph with a perfect matching. Then SE(G)S\subseteq E(G) is a minimum global forcing set of GG if and only if T=GST=G-S is a maximum (spanning) connected subgraph of GG without any nice cycle of GG.

The lemma gives a characterization for the complement of a minimum global forcing set SS of GG. The following result shows that the addition of some edges in GG to GSG-S results in a graph with a unique perfect matching.

Lemma 2.12.

Let GG be a graph with a perfect matching. Then for any minimum global forcing set SS of GG, we can find an edge subset FSF\subseteq S such that G(SF)G-(S\setminus F) has a unique perfect matching.

Proof.

Let T=GST=G-S. Then TT is a connected spanning subgraph of GG, containing no nice cycles of GG by Lemma 2.11. If TT has a perfect matching, then TT has exactly one perfect matching and we are done. Otherwise, we can pick two different perfect matchings of TT, their symmetry difference forms at least one cycle of TT, a nice cycle in GG, a contradiction.

So suppose that TT has no perfect matchings. But we can choose a minimum subset FSF\subseteq S such that T=T+FT^{\prime}=T+F has a perfect matching. Next, we will show that TT^{\prime} has exactly one perfect matching. If not, we can get two different perfect matchings of TT^{\prime}. Their symmetry difference contains a nice cycle CC of TT^{\prime}, which is also a nice cycle of GG. By Lemma 2.7, there exists an edge ee in SE(C)S\cap E(C). So eE(T)e\notin E(T) and eFe\in F. Let F=F{e}F^{\prime}=F-\left\{e\right\}. Then FF^{\prime} is smaller than FF but T+FT+F^{\prime} still has a perfect matching, a contradiction. ∎

If GG has a 1-degree vertex uu, then uu is called a pendant vertex and uvuv a pendant edge, where vv is the neighbor of uu. The deletion of uu and vv with their incident edges is called a leaf matching operation (simply LMLM operation). Next, we can see that LMLM operation has no effect on the minimum global forcing sets of a graph.

Lemma 2.13.

Let GG be a graph with a perfect matching. If GG^{\prime} is a graph obtained from GG by a series of LMLM operations, then the minimum global forcing sets of GG are the same as the ones of GG^{\prime}.

Proof.

If uvuv is a pendant edge of GG, let G=G{u,v}G^{\prime}=G-\left\{u,v\right\}. Since uvuv belongs to all perfect matchings of GG^{\prime}, it follows that a cycle of GG is nice in GG if and only if it is also nice in GG^{\prime}. By Lemma 2.7, the minimum global forcing sets of GG corresponds to that of GG^{\prime}. When we make repeatedly LMLM operations from GG, the same result holds always. ∎

3 Bipartite graphs

We consider some relations between the global forcing number and the maximum anti-forcing number of graphs by starting bipartite graphs in this section. The following structure of a bipartite graph with a unique perfect matching will play an important role.

Theorem 3.1.

[22] If G=(U,V)G=(U,V) is a bipartite graph with a unique perfect matching MM and 2n2n vertices, then we can label all vertices in UU and VV such that E(G){uivj|1ijn}E(G)\subseteq\left\{u_{i}v_{j}|1\leq i\leq j\leq n\right\}. Hence GG has pendant vertices v1v_{1} and unu_{n}, and e(G)12n(n+1)e(G)\leq\frac{1}{2}n(n+1), equality holds if and only if E(G)={uivj|1ijn}E(G)=\left\{u_{i}v_{j}|1\leq i\leq j\leq n\right\}.

Combining Theorem 3.1 and Lemma 2.12, we can obtain the following critical result that connects a minimum global forcing set of a bipartite graph GG with anti-forcing sets of a perfect matching of GG.

Lemma 3.2.

Let GG be a bipartite graph with at least two perfect matchings. For any perfect matching MM of GG, GG has a minimum global forcing set S0S_{0} such that S0MS_{0}\setminus M\neq\varnothing.

Proof.

Let GG^{\prime} be a graph obtained from GG by a series of LM operations so that δ(G)2\delta(G^{\prime})\geq 2. Then M:=ME(G)M^{\prime}:=M\cap E(G^{\prime}) is a perfect matching of GG^{\prime}. By Lemma 2.13, it suffices to prove that the result holds for GG^{\prime} and MM^{\prime}. Let SS be a minimum global forcing set of GG^{\prime}. Then T=GST=G^{\prime}-S is a maximum subgraph of GG without any nice cycle of GG by Lemma 2.11. We claim that TT contains a pendant vertex. From Lemma 2.12, we can find an edge subset FSF\subseteq S such that T+FT+F has a unique perfect matching. By Theorem 3.1, we know that T+FT+F has a pendant vertex uu. Hence uu is also a pendant vertex of TT. Let eE(T)e\in E(T) and eMe^{\prime}\in M^{\prime} be the two edges incident with uu in GG^{\prime} (ee and ee^{\prime} may be the same). Then T0=Te+eT_{0}=T-e+e^{\prime} also contains no nice cycles of GG^{\prime}. Since T0T_{0} has the same size as TT, by Lemma 2.11 we have that S0=E(G)E(T0)S_{0}=E(G)-E(T_{0}) is a new minimum global forcing set. Since δ(G)2\delta(G^{\prime})\geq 2, S0S_{0} has an edge incident with uu other than ee^{\prime} in GG^{\prime}, so S0MS_{0}\setminus M^{\prime}\neq\varnothing. ∎

From the lemma, we can obtain a main result in comparing the global forcing number and the maximum anti-forcing number of a bipartite graph.

Theorem 3.3.

Let GG be a bipartite graph with a perfect matching. Then gf(G)Af(G)gf(G)\geq Af(G).

Proof.

We apply induction on e(G)e(G). If e(G)=v(G)2e(G)=\frac{v(G)}{2}, then gf(G)=Af(G)=0gf(G)=Af(G)=0 and we are done. Next we consider bipartite graph GG with larger size. Suppose that for any bipartite graph with sizes smaller than GG, the result holds. If GG has a unique perfect matching, then it is trivial as the initial case. Otherwise, by Lemma 3.2 we have that for any perfect matching MM of GG with af(G,M)=Af(G)af(G,M)=Af(G), GG has a minimum global forcing set SS such that SMS\setminus M\neq\varnothing. Take an edge ee in SMS\setminus M, and let G=GeG^{\prime}=G-e.

We claim that Af(G)Af(G)1Af(G^{\prime})\geq Af(G)-1. Since MM is also a perfect matching of GG^{\prime}, Af(G)af(G,M)Af(G^{\prime})\geq af(G^{\prime},M). Let SS^{\prime} be a minimum anti-forcing set of MM in GG^{\prime}. Since any MM-alternating cycle in GG either is contained in GG^{\prime} or contains ee, S{e}S^{\prime}\cup\left\{e\right\} is an anti-forcing set of MM in GG by Lemma 2.6. So

af(G,M)=|S|=|S{e}|1af(G,M)1=Af(G)1,af(G^{\prime},M)=\left|S^{\prime}\right|=\left|S^{\prime}\cup\left\{e\right\}\right|-1\geq af(G,M)-1=Af(G)-1,

which implies that Af(G)af(G,M)Af(G)1Af(G^{\prime})\geq af(G^{\prime},M)\geq Af(G)-1 and the claim holds.

From Lemma 2.10, we know that S{e}S\setminus\{e\} is a global forcing set of GG^{\prime}, which implies that |S{e}|gf(G)|S\setminus\{e\}|\geq gf(G^{\prime}). By the induction hypothesis, we get that gf(G)Af(G)gf(G^{\prime})\geq Af(G^{\prime}). Thus

gf(G)=|S|=|S{e}|+1gf(G)+1Af(G)+1Af(G).gf(G)=|S|=|S\setminus\{e\}|+1\geq gf(G^{\prime})+1\geq Af(G^{\prime})+1\geq Af(G).

Corollary 3.4.

If GG is a connected graph with Af(G)=c(G)Af(G)=c(G), then gf(G)=c(G)gf(G)=c(G).

Proof.

By Theorem 2.2, we know that GG is a bipartite graph. So by Theorem 3.3 we have that gf(G)Af(G)=c(G)gf(G)\geq Af(G)=c(G). By Theorem 2.3, gf(G)c(G)gf(G)\leq c(G). So gf(G)=c(G)gf(G)=c(G). ∎

Next we will fix the order of a graph, but ignore its size to estimate the difference between the global forcing number and the maximum anti-forcing number of bipartite graphs.

Theorem 3.5.

If GG is a connected bipartite graph with a perfect matching and 2n62n\geq 6 vertices, then

0gf(G)Af(G)12(n1)(n2),0\leq gf(G)-Af(G)\leq\frac{1}{2}(n-1)(n-2),

and the right equality holds if and only if GG is isomorphic to Kn,nK_{n,n}.

Proof.

By Theorem 3.3, we directly get the left inequality. So we now consider the right inequality. For any perfect matching MM of GG, we can get a maximum subgraph TMT_{M} of GG such that MM is the unique perfect matching of TMT_{M}. So af(G,M)=e(G)e(TM)af(G,M)=e(G)-e(T_{M}). Let TT be a maximum subgraph of GG without nice cycles of GG. Then gf(G)=e(G)e(T)gf(G)=e(G)-e(T) and e(T)2n1e(T)\geq 2n-1 by Lemma 2.11. By Theorem 3.1, we have that e(TM)12n(n+1)e(T_{M})\leq\frac{1}{2}n(n+1). So

gf(G)Af(G)\displaystyle gf(G)-Af(G) \displaystyle\leq gf(G)af(G,M)=e(TM)e(T)\displaystyle gf(G)-af(G,M)=e(T_{M})-e(T) (3.1)
\displaystyle\leq 12n(n+1)(2n1)=12(n1)(n2).\displaystyle\frac{1}{2}n(n+1)-(2n-1)=\frac{1}{2}(n-1)(n-2).
Refer to caption
(a) TMT_{M} and a nice cycle of GG missing exactly unu_{n} and v1v_{1} for n=10n=10.
Refer to caption
(b) A nice cycle of GG missing exactly utu_{t} and vrv_{r} for n=10n=10.
Refer to caption
(c) A nice cycle of GG missing exactly unu_{n} and vrv_{r} for n=10n=10.
Fig. 1: Illustration for the proof of Theorem 3.5

If GG is isomorphic to Kn,nK_{n,n}, then by Theorem 3.1, for every perfect matching MM of GG, we have e(TM)=12n(n+1)e(T_{M})=\frac{1}{2}n(n+1). Since every cycle of GG is nice, TT is a spanning tree of GG, i.e. e(T)=2n1e(T)=2n-1. Thus gf(G)Af(G)=12(n1)(n2)gf(G)-Af(G)=\frac{1}{2}(n-1)(n-2).

Conversely, if gf(G)Af(G)=12(n1)(n2)gf(G)-Af(G)=\frac{1}{2}(n-1)(n-2), then both equalities in Ineq. (3.1) hold. so we have that e(T)e(T) =2n1=2n-1 and for every perfect matching MM of GG, e(TM)=12n(n+1)e(T_{M})=\frac{1}{2}n(n+1). Next we want to show that GG is isomorphic to Kn,nK_{n,n}, that is, dG(ut)=nd_{G}(u_{t})=n for each 1tn1\leq t\leq n. By Theorem 2.3, every cycle of GG is nice. By Theorem 3.1, we can label all vertices in GG such that M={uivi|1in}M=\left\{u_{i}v_{i}|1\leq i\leq n\right\} and E(TM)={uivj|1ijn}E(T_{M})=\left\{u_{i}v_{j}|1\leq i\leq j\leq n\right\}. Since v2u2v3u3vn1un1vnu1v2v_{2}u_{2}v_{3}u_{3}\ldots v_{n-1}u_{n-1}v_{n}u_{1}v_{2} is a nice cycle of GG missing exactly unu_{n} and v1v_{1} (see Fig. 1(a)), unv1u_{n}v_{1}\in E(G)E(G).

For any 2r<tn12\leq r<t\leq n-1, v1u1v2u2vr1ur1vr+1urvr+2ur+1vtut1vt+1ut+1vnv_{1}u_{1}v_{2}u_{2}\ldots v_{r-1}u_{r-1}v_{r+1}u_{r}v_{r+2}u_{r+1}\ldots v_{t}u_{t-1}v_{t+1}u_{t+1}\ldots v_{n} unv1u_{n}v_{1} is a nice cycle of GG missing exactly utu_{t} and vrv_{r} (see Fig. 1(b)), which implies that utvrE(G)u_{t}v_{r}\in E(G) and G{u1,v1,un,vn}G-\{u_{1},v_{1},u_{n},v_{n}\} induces a complete bipartite graph. It remains to show that the degrees of v1v_{1} and unu_{n} must be nn.

We can get a new perfect matching MM^{\prime} by replacing the edges u1v1u_{1}v_{1}, unvnu_{n}v_{n} by u1vnu_{1}v_{n} and unv1u_{n}v_{1} in MM. Theorem 3.1 implies that TMT_{M^{\prime}} contains an edge not in MM^{\prime} such that its two end vertices have degree nn. So GG contains another nn-degree vertex except for u1u_{1} and vnv_{n}, say uku_{k}. Then unu_{n} must have degree nn, since GG contains a path from v1v_{1} to vnv_{n} and missing exactly uku_{k}, unu_{n} and vrv_{r}, and a cycle missing exactly unu_{n} and vrv_{r} formed by this path adding v1ukvnv_{1}u_{k}v_{n} (see Fig. 1(c)) for any 2rn12\leq r\leq n-1. ∎

Shi et al. [28] gave an upper bound of the maximum anti-forcing number of graphs.

Theorem 3.6.

[28] Let GG be a graph with a perfect matching MM. Then af(G,M)Af(G)2e(G)v(G)4af(G,M)\leq Af(G)\leq\frac{2e(G)-v(G)}{4}.

A perfect matching MM of GG is said to be nice, if MM satisfies all equalities in Theorem 3.6. A characterization for a nice perfect matching of a graph was given as follows [28].

Theorem 3.7.

[28] For a perfect matching MM of a graph GG, MM is nice if and only if for any two edges e1=xye_{1}=xy and e2=uve_{2}=uv in MM, xuE(G)xu\in E(G) if and only if yvE(G)yv\in E(G), and xvE(G)xv\in E(G) if and only if yuE(G)yu\in E(G).

Corollary 3.8.

Let GG be a connected graph with a nice perfect matching and 2n2n vertices. Then gf(G)n1gf(G)\geq n-1.

Proof.

Let HH be a minimum connected spanning subgraph of GG with a nice perfect matching. Then by Theorem 3.7, e(H)=n+2(n1)=3n2e(H)=n+2(n-1)=3n-2, so c(H)=n1c(H)=n-1. By Theorem 3.6, Af(H)=2e(H)v(H)4=n1Af(H)=\frac{2e(H)-v(H)}{4}=n-1. So Af(H)=c(H)=n1Af(H)=c(H)=n-1. According to Corollary 3.4, we know gf(H)=n1gf(H)=n-1. By Lemma 2.10, we get gf(G)gf(H)=n1gf(G)\geq gf(H)=n-1. ∎

In the following, we will give a graph GG such that gf(G)Af(G)gf(G)-Af(G) is one less than the upper bound in Theorem 3.5, i.e. gf(G)Af(G)=12(n23n)gf(G)-Af(G)=\frac{1}{2}(n^{2}-3n). A cycle CC in a graph GG is a Hamilton cycle if V(C)=V(G)V(C)=V(G).

Proposition 3.9.

Let GG be a connected bipartite graph with 2n82n\geq 8 vertices. If GG is isomorphic to Kn,neK_{n,n}-e for any eE(Kn,n)e\in E(K_{n,n}), then gf(G)=n22n1gf(G)=n^{2}-2n-1 and gf(G)Af(G)=12(n23n)gf(G)-Af(G)=\frac{1}{2}(n^{2}-3n). Conversely, if gf(G)=n22n1gf(G)=n^{2}-2n-1, then GG is isomorphic to Kn,neK_{n,n}-e.

Proof.

If GG is isomorphic to Kn,neK_{n,n}-e for any eE(Kn,n)e\in E(K_{n,n}), let uu and vv be nonadjacent vertices in different partite sets of GG. Then GuvG-u-v is isomorphic to Kn1,n1K_{n-1,n-1}. We claim that for any cycle CC in GG, it is nice in GG if and only if it is not a Hamilton cycle of GuvG-u-v. If CC is a Hamilton cycle of GuvG-u-v, it is not nice in GG. If either uu or vv belongs to CC, then GCG-C is either a null graph or a complete bipartite graph, so CC is a nice cycle of GG. Otherwise, neither uu nor vv are in CC, and GCG-C can be obtained from a complete bipartite with order at least 44 by deleting an edge. Hence GCG-C has a perfect matching, i.e. CC is a nice cycle of GG, and the claim is verified. Let TT be a maximum subgraph of GG without any nice cycle of GG. Since two Hamilton cycles of GuvG-u-v together contain a shorter cycle, which is a nice cycle of GG, there is exactly one cycle in TT, so e(T)=2ne(T)=2n. By Lemma 2.11, gf(G)=e(G)e(T)=(n21)2n=n22n1gf(G)=e(G)-e(T)=(n^{2}-1)-2n=n^{2}-2n-1.

From Theorem 3.5, we know that Af(G)gf(G)[12(n23n+2)1]=12(n2)(n+1)Af(G)\geq gf(G)-\left[\frac{1}{2}(n^{2}-3n+2)-1\right]=\frac{1}{2}(n-2)(n+1). By Theorem 3.6, we get that Af(G)2e(G)v(G)4=n2n12=12(n2)(n+1)Af(G)\leq\lfloor\frac{2e(G)-v(G)}{4}\rfloor=\lfloor\frac{n^{2}-n-1}{2}\rfloor=\frac{1}{2}(n-2)(n+1). So Af(G)=12(n2)(n+1)Af(G)=\frac{1}{2}(n-2)(n+1) and gf(G)Af(G)=12(n23n)gf(G)-Af(G)=\frac{1}{2}(n^{2}-3n).

Conversely, if gf(G)=n22n1gf(G)=n^{2}-2n-1, then e(G)gf(G)+v(G)1=(n22n1)+2n1=n22e(G)\geq gf(G)+v(G)-1=(n^{2}-2n-1)+2n-1=n^{2}-2 by Theorem 2.3. Because GG is bipartite, e(G)n2e(G)\leq n^{2}. If e(G)=n2e(G)=n^{2}, then GG is isomorphic to Kn,nK_{n,n}. By Theorem 3.5, gf(Kn,n)=n22n+1gf(K_{n,n})=n^{2}-2n+1, a contradiction. If e(G)=n22e(G)=n^{2}-2, then GKn,n{e,e}G\cong K_{n,n}-\left\{e,e^{\prime}\right\} for distinct e,eE(Kn,n)e,e^{\prime}\in E(K_{n,n}). Since gf(G)=c(G)gf(G)=c(G), by Theorem 2.3, all cycles in GG are nice. If ee and ee^{\prime} are adjacent, then GV(e)G-V(e) has a Hamilton cycle, not a nice cycle of GG, a contradiction. So ee and ee^{\prime} are not adjacent.

Let A={u1,u2,,un}A=\left\{u_{1},u_{2},\ldots,u_{n}\right\} and B={v1,v2,,vn}B=\left\{v_{1},v_{2},\ldots,v_{n}\right\} be a bipartite sets of V(G)V(G). We can assume e=u1v1e=u_{1}v_{1} and e=u2v2e^{\prime}=u_{2}v_{2}. Let P1=u4v2u3v3u2v4P_{1}=u_{4}v_{2}u_{3}v_{3}u_{2}v_{4} and P2=u4v5u5v6un1vnunv4P_{2}=u_{4}v_{5}u_{5}v_{6}\ldots u_{n-1}v_{n}u_{n}v_{4} (see Fig. 2). Obviously, P1P2P_{1}\cup P_{2} is a cycle of GG missing exactly u1u_{1} and v1v_{1}, which is not a nice cycle of GG, a contradiction. Thus, e(G)=n21e(G)=n^{2}-1. ∎

Refer to caption
Fig. 2: A cycle in GG missing ee and ee^{\prime} for n=10n=10

4 Generalizations of Theorem 3.3

A connected graph GG is called matching covered, if every edge of GG belongs to a perfect matching of GG. For a graph GG, E(G)\mathbb{R}^{E(G)} denotes the set of all vectors whose entries are indexed by the edges of GG. For any perfect matching MM of GG, the incidence vector of MM is denoted by qMq^{M}. The perfect matching polytope of GG is the convex hull of {qM:M}\left\{q^{M}:M\in\mathcal{M}\right\}, denoted by PM(G)PM(G). In a landmark paper, Edmonds [11] gave a linear inequality description of PM(G)PM(G). For any 𝒙E(G)\bm{x}\in\mathbb{R}^{E(G)} and FE(G)F\subseteq E(G), let 𝒙(F)=eF𝒙(e)\bm{x}(F)=\sum\limits_{e\in F}\bm{x}(e). For a nonempty subset UV(G)U\subset V(G), G(U)\partial_{G}(U) represents for a cut, the set of edges which have exactly one end vertex in UU. If UU is a singleton, then G(U)\partial_{G}(U) is a trivial cut.

Theorem 4.1.

[11] A vector 𝒙\bm{x} in E(G)\mathbb{R}^{E(G)} belongs to PM(G)PM(G) if and only if it satisfies the following system of linear inequalities:

𝒙\displaystyle\bm{x} \displaystyle\geq 0(nonnegativeity),\displaystyle 0\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (non-negativeity),
𝒙(G(v))\displaystyle\bm{x}(\partial_{G}(v)) =\displaystyle= 1forallvV(G)(degreeconstraints),\displaystyle 1\ \ \ for\ all\ v\in V(G)\ \ \ \ \ \ \ \ (degree\ constraints),
𝒙(G(S))\displaystyle\bm{x}(\partial_{G}(S)) \displaystyle\geq 1foralloddSV(G)(oddsetconstraints).\displaystyle 1\ \ \ for\ all\ odd\ S\subset V(G)\ \ (odd\ set\ constraints).

A vector 𝒙\bm{x} in E(G)\mathbb{R}^{E(G)} is 1-regular if 𝒙(G(v))=1\bm{x}(\partial_{G}(v))=1, for all vV(G)v\in V(G). For a cut C:=G(X)C:=\partial_{G}{(X)} of GG, we denote the graph obtained from GG by shrinking the shore X¯=V(G)X\overline{X}=V(G)-X to a vertex by G{X}G\left\{X\right\}. We refer to G{X}G\left\{X\right\} and G{X¯}G\left\{\overline{X}\right\} as the CC-contractions of GG. A cut CC of a matching covered graph GG is a separating cut if both of the CC-contractions of GG are also matching covered, and is tight if |CM|=1|C\cap M|=1 for all MM\in\mathcal{M}. Every tight cut is separating, but the converse is not true. A matching covered graph is solid if all separating cuts of it are tight. A brick is a non-bipartite matching covered graph satisfying that all tight cuts of it are trivial.

Theorem 4.2.

[4] For a brick GG, the following three statements are equivalent.
(i) GG is solid,
(ii) G𝒢G\in\mathcal{G}, and
(iii) PM(G)PM(G) consists of non-negative 11-regular vectors.

de Carvalho et al. [4] pointed out that in 2003 Reed and Wakabayashi had already showed that (i) and (ii) are equivalent in Theorem 4.2, and gave a new proof to it and proved the equivalence of (i) and (iii).

In 1959, Kotzig [17] showed the following property of the graphs with a unique perfect matching.

Theorem 4.3.

[17] If GG is a connected graph with a unique perfect matching, then GG has a cut-edge belonging to the perfect matching.

For the graphs GG with a unique perfect matching, we come to a deeper structure property: GG contains either a pendant vertex as Theorem 3.1 in the bipartite case or an odd dumbbell (see Fig. 3): two disjoint odd cycles connected by one path of odd length.

Refer to caption
Fig. 3: An odd dumbbell: The bold (red) edges form the unique perfect matching.
Theorem 4.4.

Let a graph GG have a unique perfect matching MM without pendant vertices. Then GG contains an odd dumbbell as a subgraph so that MM contains the perfect matching of the subgraph.

Proof.

By Theorem 4.3, GG has a cut edge uvMuv\in M. Since δ(G)2\delta(G)\geq 2, the components GuG_{u} and GvG_{v} of GuvG-uv containing uu and vv respectively, must be non-trivial. Consider the longest MM-alternating path in each component, starting with uu and vv, respectively. This must end with an edge in MM, and the last vertex must be adjacent to some vertex in the path. This must form an odd cycle and gives the required subgraph. ∎

Now we can extend the relation gf(G)Af(G)gf(G)\geq Af(G) to the class of graphs 𝒢\cal G. First we extend Lemma 3.2 as follows.

Lemma 4.5.

Let G𝒢G\in\mathcal{G} have at least two perfect matchings. Then for any perfect matching MM of GG, there exists a minimum global forcing set S0S_{0} such that S0MS_{0}\setminus M\neq\varnothing.

Proof.

We use almost the same proof as Lemma 3.2, but only replace Theorem 3.1 with Theorem 4.4. For the present graph GG, T+FT+F has a unique perfect matching. Although T+FT+F is not necessarily bipartite, T+FT+F is in 𝒢\cal G. By Theorem 4.4, T+FT+F has a pendant vertex. The others remain unchanged. ∎

Then by using the above lemma we obtain the following generalization as the proof of Theorem 3.3 .

Theorem 4.6.

If G𝒢G\in\mathcal{G}, then gf(G)Af(G)gf(G)\geq Af(G).

Combining Theorems 4.2 and 4.6, we directly get the following result.

Corollary 4.7.

If GG is a solid brick, then gf(G)Af(G)gf(G)\geq Af(G).

For general non-bipartite graphs, the three conditions in Theorem 4.2 are not equivalent. However we have the following implication.

Lemma 4.8.

If PM(G)PM(G) consists of non-negative 11-regular vectors, then G𝒢G\in\mathcal{G}.

Proof.

Suppose to the contrary that GG has disjoint odd cycles CC and CC^{\prime} such that GCCG-C-C^{\prime} has a perfect matching MM. Let 𝒙\bm{x} be a vector in E(G)\mathbb{R}^{E(G)} such that 𝒙(e)=1\bm{x}(e)=1, if eMe\in M; 𝒙(e)=12\bm{x}(e)=\frac{1}{2}, if eCCe\in C\cup C^{\prime}; 𝒙(e)=0\bm{x}(e)=0, otherwise. Note that 𝒙\bm{x} is a non-negative and 11-regular vector. So 𝒙PM(G)\bm{x}\in PM(G). But for odd cycle CC, 𝒙(G(C))=0\bm{x}(\partial_{G}(C))=0, contradicting the odd set constraint in Theorem 4.1. ∎

By Theorem 4.6 and Lemma 4.8, we can get the following corollary.

Corollary 4.9.

If PM(G)PM(G) consists of non-negative 11-regular vectors, then gf(G)Af(G)gf(G)\geq Af(G).

An example in Fig. 4 shows there exists a matching covered graph GG not in 𝒢\mathcal{G} with gf(G)Af(G)gf(G)\geq Af(G).

Refer to caption
Fig. 4: A graph G𝒢G\notin\mathcal{G}, and gf(G)=Af(G)=3gf(G)=Af(G)=3.

5 The difference gf(G)Af(G)gf(G)-Af(G)

In this section, we will give sharp lower and upper bounds on the difference gf(G)Af(G)gf(G)-Af(G) on general connected graphs GG.

Theorem 5.1.

[22] If GG has a unique perfect matching and 2n2n vertices, then e(G)n2e(G)\leq n^{2}.

Theorem 5.2.

Let GG be a connected graph with a perfect matching and 2n42n\geq 4 vertices. Then

12(n2n2)gf(G)Af(G)(n1)(n2),-\frac{1}{2}(n^{2}-n-2)\leq gf(G)-Af(G)\leq(n-1)(n-2),

and the left equality holds if and only if n=2n=2.

Proof.

(11) We first prove the left inequality. If gf(G)=0gf(G)=0, then GG has a unique perfect matching, so Af(G)=0Af(G)=0. If gf(G)=1gf(G)=1, then GG has at least two perfect matchings. By Theorem 2.4, we have that Φ(G)2gf(G)=2\Phi(G)\leq 2^{gf(G)}=2. Hence, GG has exactly two perfect matchings, denoted by M1M_{1} and M2M_{2}. Obviously, Af(G)=af(G,M1)=af(G,M2)=1Af(G)=af(G,M_{1})=af(G,M_{2})=1. So for the case of gf(G)1gf(G)\leq 1, we have that 12(n2n2)0=gf(G)Af(G)-\frac{1}{2}(n^{2}-n-2)\leq 0=gf(G)-Af(G).

Next we consider gf(G)2gf(G)\geq 2. Let TT be a maximum subgraph of GG without any nice cycle of GG. By Lemma 2.12, we can find an edge subset FE(G)E(T)F\subseteq E(G)\setminus E(T) such that T+FT+F has a unique perfect matching. From Theorem 5.1, e(T)e(T+F)n2e(T)\leq e(T+F)\leq n^{2}. By Lemma 2.11 and Theorem 3.6, we have

gf(G)Af(G)\displaystyle gf(G)-Af(G) \displaystyle\geq gf(G)2e(G)v(G)4=gf(G)gf(G)+e(T)2+n2\displaystyle gf(G)-\frac{2e(G)-v(G)}{4}=gf(G)-\frac{gf(G)+e(T)}{2}+\frac{n}{2} (5.1)
=\displaystyle= 12gf(G)12e(T)+n212(n2n2).\displaystyle\frac{1}{2}gf(G)-\frac{1}{2}e(T)+\frac{n}{2}\geq-\frac{1}{2}(n^{2}-n-2).

Now we verify the equivalence condition for the left equality. If n=2n=2, for all cases of GG, we can verify that gf(G)Af(G)=0gf(G)-Af(G)=0. Conversely, if gf(G)Af(G)=12(n2n2)gf(G)-Af(G)=-\frac{1}{2}(n^{2}-n-2), then for gf(G)1gf(G)\leq 1, we know gf(G)Af(G)=0gf(G)-Af(G)=0, which implies that n=2n=2. For gf(G)2gf(G)\geq 2, all the equalities in Ineq. (5.1) hold, so Af(G)=2e(G)v(G)4Af(G)=\frac{2e(G)-v(G)}{4} and gf(G)=2gf(G)=2. By Corollary 3.8, ngf(G)+1=3n\leq gf(G)+1=3. Next we will prove that if n=3n=3, then gf(G)2gf(G)\neq 2. For n=3n=3, we have gf(G)Af(G)=2gf(G)-Af(G)=-2, so Af(G)=4Af(G)=4, e(G)=11e(G)=11 and c(G)=6c(G)=6. By Theorem 3.7, GG has two cases G1G_{1} and G2G_{2} as shown in Fig. 55 in the sense of isomorphism, where {u1u2,u3u4,u5u6}\{u_{1}u_{2},u_{3}u_{4},u_{5}u_{6}\} is a nice perfect matching of GG.

Refer to caption
(a) Graph G1G_{1}
Refer to caption
(b) Graph G2G_{2}
Fig. 5: Two cases of GG with a nice perfect matching, n=3n=3 and e(G)=11e(G)=11.

We first consider the graph G1G_{1}. Let S1S_{1} be a minimum global forcing set of G1G_{1}. Then |S1E(G)|2|S_{1}\cap E(G^{\prime})|\geq 2, for the subgraph GG^{\prime} of G1G_{1} induced by {u1,u2,u3,u4}\left\{u_{1},u_{2},u_{3},u_{4}\right\}, which is isomorphic to K4K_{4}. Similarly, the subgraph G′′G^{\prime\prime} of G1G_{1} induced by {u3,u4,u5,u6}\left\{u_{3},u_{4},u_{5},u_{6}\right\} satisfies that |S1E(G′′)|2|S_{1}\cap E(G^{\prime\prime})|\geq 2. Thus we have gf(G1)=|S1||S1E(G)|+|S1E(G′′)|13gf(G_{1})=|S_{1}|\geq|S_{1}\cap E(G^{\prime})|+|S_{1}\cap E(G^{\prime\prime})|-1\geq 3.

Next we claim that gf(G2)=4gf(G_{2})=4. Since {u1u2,u2u4,u3u4,u1u3}\{u_{1}u_{2},u_{2}u_{4},u_{3}u_{4},u_{1}u_{3}\} is a global forcing set of G2G_{2}, gf(G2)4gf(G_{2})\leq 4. Suppose to the contrary that gf(G2)3gf(G_{2})\leq 3. Let S2S_{2} be a minimum global forcing set of G2G_{2}. Then T1=G2S2T_{1}=G_{2}-S_{2} contains at least three cycles. Note that every even cycle in G2G_{2} is nice, with length either 44 or 66. And any three triangles in G2G_{2} must produce a 44-cycle. So T1T_{1} has a pentagon CC. Let CC^{\prime} be another cycle in T1T_{1} except for CC. Then |V(C)V(C)|2|V(C)\cap V(C^{\prime})|\geq 2. So we can take two consecutive common vertices of CC and CC^{\prime} such that there exists three internally disjoint paths between them, which must contain an even cycle, a contradiction. Thus gf(G2)=4gf(G_{2})=4.

(22) We now prove the right inequality. For any perfect matching MM of GG, let TMT_{M} be a maximum subgraph of GG with a unique perfect matching MM. Then af(G,M)=e(G)e(TM)af(G,M)=e(G)-e(T_{M}). Corollary 5.1 implies that 2n1e(TM)n22n-1\leq e(T_{M})\leq n^{2}. Let TT be a maximum spanning subgraph of GG without nice cycles of GG. Then e(T)2n1e(T)\geq 2n-1. If e(TM)n2n+1e(T_{M})\leq n^{2}-n+1, then gf(G)Af(G)gf(G)af(G,M)=e(TM)e(T)(n2n+1)(2n1)=n23n+2gf(G)-Af(G)\leq gf(G)-af(G,M)=e(T_{M})-e(T)\leq(n^{2}-n+1)-(2n-1)=n^{2}-3n+2, and the required result holds.

Next we consider the case e(TM)n2n+2e(T_{M})\geq n^{2}-n+2. Let e(TM)=n2ke(T_{M})=n^{2}-k, where 0kn20\leq k\leq n-2. By the definition of TMT_{M}, the subgraph of TMT_{M} induced by {ui,vi,uj,vj}\{u_{i},v_{i},u_{j},v_{j}\} can contain at most 4 edges. Consider the subgraphs induced by {ui,vi,ui+1,vi+1}\{u_{i},v_{i},u_{i+1},v_{i+1}\} for 1in11\leq i\leq n-1. If k+1k+1 of these subgraphs contain only 3 edges, the number of edges in TMT_{M} is at most n2(k+1)n^{2}-(k+1), a contradiction. Therefore n1kn-1-k of these must have 4 edges and give n1kn-1-k triangles whose edges can not form additional cycles. So we have e(T)2n1+n1k=3n2ke(T)\geq 2n-1+n-1-k=3n-2-k. It follows that gf(G)Af(G)e(TM)e(T)(n2k)(3n2k)=n23n+2gf(G)-Af(G)\leq e(T_{M})-e(T)\leq(n^{2}-k)-(3n-2-k)=n^{2}-3n+2 and we are done. ∎

The following discussions will show that the upper bound on gf(G)Af(G)gf(G)-Af(G) in Theorem 5.2 can be achieved.

Proposition 5.3.

gf(K2n)=2(n1)2gf(K_{2n})=2(n-1)^{2} and gf(K2n)Af(K2n)=(n1)(n2)gf(K_{2n})-Af(K_{2n})=(n-1)(n-2).

Proof.

Let TT be a maximum spanning subgraph of K2nK_{2n} which contains no any nice cycle of K2nK_{2n}. By Lemma 2.11, gf(K2n)=e(K2n)e(T)gf(K_{2n})=e(K_{2n})-e(T). Obviously, every even cycle in K2nK_{2n} is nice. So TT has no even cycles. We claim that e(T)3n2e(T)\leq 3n-2. Let C1,C2,,Cc(T)C_{1},C_{2},\ldots,C_{c(T)} be different cycles in TT. Then they are pairwise edge-disjoint. Otherwise, there are distinct CiC_{i} and CjC_{j} sharing an edge. So we can take a path PP on CiC_{i}, which has only two end vertices and no edges in CjC_{j}. Then CjPC_{j}\cup P consists of three internally disjoint paths between two vertices, which must contain an even cycle, a contradiction. For any 1ic(T)1\leq i\leq c(T), we have e(Ci)3e(C_{i})\geq 3. So

e(T)=c(T)+2n1i=1c(T)|E(Ci)|3c(T),e(T)=c(T)+2n-1\geq\sum\limits_{i=1}^{c(T)}\left|E(C_{i})\right|\geq 3c(T),

which implies that c(T)n1c(T)\leq n-1 and e(T)3n2e(T)\leq 3n-2. So the claim holds.

Refer to caption
Fig. 6: A spanning subgraph without any nice cycles of K2nK_{2n} for n=10n=10.

On the other hand, we can find n1n-1 edge-disjoint triangles in K2nK_{2n} as shown in Fig. 6, so e(T)3n2e(T)\geq 3n-2. Moreover, e(T)=3n2e(T)=3n-2 and gf(K2n)=2(n1)2gf(K_{2n})=2(n-1)^{2}. It is known [28] that Af(K2n)=n2nAf(K_{2n})=n^{2}-n, so gf(K2n)Af(K2n)=(n1)(n2)gf(K_{2n})-Af(K_{2n})=(n-1)(n-2). ∎

Finally, for any positive integer kk we will construct a matching covered graph GkG_{k} such that gf(Gk)Af(Gk)=kgf(G_{k})-Af(G_{k})=-k. Let TiT_{i} be a copy of the triangular prism, where 1ik1\leq i\leq k. Let u1(i)u2(i)u3(i)u1(i)u_{1}^{(i)}u_{2}^{(i)}u_{3}^{(i)}u_{1}^{(i)} and u4(i)u5(i)u6(i)u4(i)u_{4}^{(i)}u_{5}^{(i)}u_{6}^{(i)}u_{4}^{(i)} be two triangles in TiT_{i}, and u1(i)u4(i)u_{1}^{(i)}u_{4}^{(i)}, u2(i)u5(i)u_{2}^{(i)}u_{5}^{(i)} and u3(i)u6(i)u_{3}^{(i)}u_{6}^{(i)} are three remaining edges in TiT_{i}. We join consecutively T1,T2,,TkT_{1},T_{2},\ldots,T_{k} with edges u1(i)u1(i+1)u_{1}^{(i)}u_{1}^{(i+1)} and u4(i)u4(i+1)u_{4}^{(i)}u_{4}^{(i+1)}, for each 1ik11\leq i\leq k-1, resulting in a graph GkG_{k} (see Fig. 7(a)). Dr. Kai Deng gave gf(G1)Af(G1)=1gf(G_{1})-Af(G_{1})=-1. In general we have the following result.

Refer to caption
(a) Graph GkG_{k} for k=3k=3
Refer to caption
(b) Graph GkFG_{k}-F for k=3k=3
Fig. 7: Illustration for the proof of Proposition 5.4
Proposition 5.4.

For any integer k1k\geq 1, GkG_{k} is a matching covered graph, gf(Gk)=3k1gf(G_{k})=3k-1 and Af(Gk)=4k1Af(G_{k})=4k-1. Thus, gf(Gk)Af(Gk)=kgf(G_{k})-Af(G_{k})=-k.

Proof.

Take a perfect matching MM of GkG_{k} as M={u1(i)u4(i),u2(i)u5(i),u3(i)u6(i)|1ik}M=\left\{u_{1}^{(i)}u_{4}^{(i)},u_{2}^{(i)}u_{5}^{(i)},u_{3}^{(i)}u_{6}^{(i)}|1\leq i\leq k\right\}. We can see that each edge of GkG_{k} belongs to an MM-alternating cycle in GkG_{k}, so GkG_{k} is matching covered. From Theorem 3.7, MM is a nice perfect matching of GkG_{k}, so Af(Gk)=2e(Gk)v(Gk)4=2(11k2)6k4=4k1Af(G_{k})=\frac{2e(G_{k})-v(G_{k})}{4}=\frac{2(11k-2)-6k}{4}=4k-1.

Now we consider the global forcing number of GkG_{k}. Set

F={u1(i)u4(i),u2(i)u5(i)|1ik}{u1(i)u1(i+1)|1ik1}.F=\left\{u_{1}^{(i)}u_{4}^{(i)},u_{2}^{(i)}u_{5}^{(i)}|1\leq i\leq k\right\}\cup\left\{u_{1}^{(i)}u_{1}^{(i+1)}|1\leq i\leq k-1\right\}.

Because all cycles of GkFG_{k}-F are triangles (see Fig. 7(b)), it contains no nice cycles of GkG_{k}, so gf(Gk))|F|=3k1gf(G_{k}))\leq\left|F\right|=3k-1.

Next, we prove that gf(Gk)3k1gf(G_{k})\geq 3k-1 by induction on kk. If k=1k=1, then GkG_{k} is a triangular prism. We know gf(Gk)2gf(G_{k})\geq 2. Otherwise, GkG_{k} has a global forcing set of a single edge, which belongs to at most two of three nice cycles u1(1)u2(1)u5(1)u4(1)u1(1)u_{1}^{(1)}u_{2}^{(1)}u_{5}^{(1)}u_{4}^{(1)}u_{1}^{(1)}, u2(1)u3(1)u6(1)u5(1)u2(1)u_{2}^{(1)}u_{3}^{(1)}u_{6}^{(1)}u_{5}^{(1)}u_{2}^{(1)} and u1(1)u3(1)u6(1)u4(1)u1(1)u_{1}^{(1)}u_{3}^{(1)}u_{6}^{(1)}u_{4}^{(1)}u_{1}^{(1)}, a contradiction. We now consider GkG_{k} with k2k\geq 2. Suppose that for such graphs with less than kk triangular prisms, the assertion holds. Let

F1\displaystyle F_{1} =E(T1){u1(1)u1(2),u4(1)u4(2)},\displaystyle=E(T_{1})\cup\left\{u_{1}^{(1)}u_{1}^{(2)},u_{4}^{(1)}u_{4}^{(2)}\right\},
Fk\displaystyle F_{k} =E(Tk){u1(k1)u1(k),u4(k1)u4(k)},\displaystyle=E(T_{k})\cup\left\{u_{1}^{(k-1)}u_{1}^{(k)},u_{4}^{(k-1)}u_{4}^{(k)}\right\},
Fi\displaystyle F_{i} =E(Ti){u1(i1)u1(i),u1(i)u1(i+1),u4(i1)u4(i),u4(i)u4(i+1)},where 2ik1.\displaystyle=E(T_{i})\cup\left\{u_{1}^{(i-1)}u_{1}^{(i)},u_{1}^{(i)}u_{1}^{(i+1)},u_{4}^{(i-1)}u_{4}^{(i)},u_{4}^{(i)}u_{4}^{(i+1)}\right\},\ {\rm where}\ 2\leq i\leq k-1.

Then E(Gk)=i=1kFiE(G_{k})=\bigcup\limits_{i=1}^{k}F_{i}. Let SS be a minimum global forcing set of GkG_{k}. We claim |SF1|3\left|S\cap F_{1}\right|\geq 3 or |SFk|3\left|S\cap F_{k}\right|\geq 3 or |SFi|4\left|S\cap F_{i}\right|\geq 4 for some 2ik12\leq i\leq k-1. Otherwise, |SF1|2\left|S\cap F_{1}\right|\leq 2, |SFk|2\left|S\cap F_{k}\right|\leq 2 and for all 2ik12\leq i\leq k-1, |SFi|3\left|S\cap F_{i}\right|\leq 3. In this case, we can construct a nice cycle of GkG_{k} not passing through any edge of SS. Let G=GkTkG^{\prime}=G_{k}-T_{k}. By induction hypothesis, gf(G)3k4gf(G^{\prime})\geq 3k-4. Since GG^{\prime} is a nice subgraph of GkG_{k}, by Lemma 2.10, S|GS|_{G^{\prime}} is a global forcing set of GG^{\prime}. We have

3k4\displaystyle 3k-4 \displaystyle\leq gf(G)|SE(G)|\displaystyle gf(G^{\prime})\leq\left|S\cap E(G^{\prime})\right|
\displaystyle\leq |i=1k1(SFi)|i=1k1|SFi|2+3(k2)=3k4.\displaystyle|\bigcup\limits_{i=1}^{k-1}(S\cap F_{i})|\leq\sum\limits_{i=1}^{k-1}|S\cap F_{i}|\leq 2+3(k-2)=3k-4.

So |SF1|=2\left|S\cap F_{1}\right|=2, |SFi|=3\left|S\cap F_{i}\right|=3 and (SFi1)(SFi)=(S\cap F_{i-1})\cap(S\cap F_{i})=\varnothing for all 2ik12\leq i\leq k-1. For GkT1G_{k}-T_{1}, in an analogous argument, we get |SFk|=2\left|S\cap F_{k}\right|=2 and (SFk1)(SFk)=(S\cap F_{k-1})\cap(S\cap F_{k})=\varnothing. Thus Si=1kE(Ti)S\subseteq\bigcup\limits_{i=1}^{k}E(T_{i}). For i=1i=1 and kk, since |SE(Ti)|=2\left|S\cap E(T_{i})\right|=2, we can take one path PiP_{i} in TiT_{i} from three edge-disjoint paths between u1(i)u_{1}^{(i)} and u4(i)u_{4}^{(i)} such that SE(Pi)=S\cap E(P_{i})=\varnothing. We can check that such two paths P1P_{1} and PkP_{k} with two paths P=u1(1)u1(2)u1(k)P=u_{1}^{(1)}u_{1}^{(2)}\ldots u_{1}^{(k)} and P=u4(1)u4(2)u4(k)P^{\prime}=u_{4}^{(1)}u_{4}^{(2)}\ldots u_{4}^{(k)} form a nice cycle of GkG_{k}, a contradiction. So the claim is verified.

If |SFk|3\left|S\cap F_{k}\right|\geq 3 (similarly for the case |SF1|3|S\cap F_{1}|\geq 3), as before by the induction hypothesis we have

|S|=|SFk|+|SE(G)|3+(3k4)=3k1.\left|S\right|=\left|S\cap F_{k}\right|+\left|S\cap E(G^{\prime})\right|\geq 3+(3k-4)=3k-1.

So suppose that there is an integer 2ik12\leq i\leq k-1 such that |SFi|4\left|S\cap F_{i}\right|\geq 4. Then GkTiG_{k}-T_{i} has two components, denoted by G1G_{1}^{\prime} and G2G_{2}^{\prime}. Using the induction hypothesis to G1G^{\prime}_{1} and G2G^{\prime}_{2}, gf(G1)+gf(G2)3(k1)2gf(G^{\prime}_{1})+gf(G^{\prime}_{2})\geq 3(k-1)-2. Further,

|S|=|SFi|+|SE(G1)|+|SE(G2)|4+gf(G1)+gf(G2)4+3(k1)2=3k1.\left|S\right|=\left|S\cap F_{i}\right|+\left|S\cap E(G^{\prime}_{1})\right|+\left|S\cap E(G^{\prime}_{2})\right|\geq 4+gf(G^{\prime}_{1})+gf(G^{\prime}_{2})\geq 4+3(k-1)-2=3k-1.

In conclusion, we get gf(Gk)=3k1gf(G_{k})=3k-1. Thus gf(Gk)Af(Gk)=kgf(G_{k})-Af(G_{k})=-k. ∎

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