The maximum forcing number of polyomino^{1}^{1}1Supported by NSFC
(Grant No.11301217,11171134), the Fundamental
Research Funds for the Central Universities (Grant No.2010121007),
the Natural Science Foundation of Fujian Province (Grant
No.2013J01014) and New Century Excellent
Talents in Fujian Province University(grant JA14168).

Corresponding author: xuliqiong@jmu.edu.cn

[6truemm] Yuqing Lin, Liqiong Xu and Fuji Zhang

School of Electrical Engineering and Computer Science, University of Newcastle,

NSW2308, Australia

School of Mathematical Sciences, Jimei University,

Jimei, Fujian 361005, P. R. China

School of Mathematical Sciences, Xiamen University,

Xiamen, Fujian 361005, P. R. China

Abtract The forcing number of a perfect
matching of a graph is the cardinality of the smallest
subset of that is contained in no other perfect matchings of .
For a planar embedding of a 2-connected bipartite planar graph which has a perfect matching,
the concept
of Clar number of hexagonal system had been extended by Abeledo and Atkinson as follows:
a spanning subgraph of is called a Clar cover of if each of its components is either an even face or an edge, the maximum number of even faces in Clar covers of is
called Clar number of , and the Clar cover with the maximum number of even faces is called the maximum Clar cover.
It was proved that if is a hexagonal system with a perfect matching and is a
set of hexagons in a maximum Clar cover of , then has a unique
1-factor. Using this result, Xu et. at. proved that the maximum forcing number of the elementary hexagonal system are equal to their Clar numbers, and then the maximum forcing
number of the elementary hexagonal system can be computed in polynomial time. In this paper, we show that an elementary
polyomino has a unique perfect matching when removing the set of tetragons from its maximum Clar cover. Thus the maximum forcing number of elementary polyomino equals to its Clar number and can be computed in polynomial time. Also, we have extended our result to the non-elementary polyomino and hexagonal system.

Keywords Forcing Number; Clar Number; Polyomino

## 1 Introduction

Let be a graph that has a perfect matching. A forcing set for a perfect matching of is a subset of , such that is contained in no other perfect matchings of . The cardinality of a smallest forcing set of is called the forcing number of , and is denoted by . The minimum and maximum of over all perfect matchings of is denoted by and , respectively. Given a matching in a graph , an -alternating path (cycle) is a path (cycle) in whose edges are alternately in and outside of . Let be an edge of . If is contained in all perfect matchings of , or is not contained in any perfect matchings of , then is called fixed bond. If is contained in a perfect matchings of , then is called double bond.

A polyomino (respectively, hexagonal system) is a finite connected plane graph with no cut vertex and every interior region is surrounded by a regular square (respectively, hexagon). A connected bipartite graph is called elementary (or normal) if its every edge is contained in some perfect matching. Let be a plane bipartite graph, a face of is called resonant if its boundary is an alternating cycle with respect to a perfect matching of .

The Clar number is originally defined for hexagonal systems by Clar[1]. The Clar cover polynomial had been introduced in [2,3]. The further progress of Clar’s theory can be found in two surveys [4,5]. Later, Abeledo and Atkinson[6] generalized the concept of Clar number for bipartite and 2-connected plane graphs. For a planar embedding of a 2-connected bipartite planar , a Clar cover of is a spanning subgraph with its components is either an even face or an edge, the maximum number of even faces in Clar covers of is called Clar number of , and denoted by . We call a Clar cover with the maximum number of even face the maximum Clar cover.

The idea of forcing number was inspired by practical chemistry problems. This concept was first proposed by Harary et. al. in [7]. The same idea appeared in earlier papers by Randić and Klein [8,9] in terms of ¡°innate degree of freedom¡± of a Kekulé structure. The forcing numbers of square grids, stop signs, torus, hypercube, and some special fullerene graphs have been considered in [10-16]. Adams et al.[11] proved it is NP-complete to find the smallest forcing set of a bipartite graph with maximum degree 3. Later, Afshani etal.[10] proved it is NP-complete to find the smallest forcing number of a bipartite graph with maximum degree 4, and proposed the following question:

###### Open Problem.

Given a graph , what is the computational complexity of finding the maximum forcing number of .

Recently, Xu,Bian and Zhang[17] showed that when is an elementary hexagonal system, the maximum forcing number of can be computed in polynomial time. In this paper, we first show that an elementary polyomino has a unique perfect matching when removing all tetragons of a maximum Clar cover of the polyomino. Base on this result, we know that the maximum forcing number of an elementary polyomino equals to its Clar number, thus the maximum forcing number of the elementary polyomino can be computed in polynomial time. Finally, we also show that for both elementary and non-elementary hexagonal system and polyomino, their maximum forcing number can be computed in polynomial time, which generalize the result in [17] to non-elementary hexagonal system.

## 2 A maximum Clar cover of polyomino

First we give some theorems in the following.

###### Theorem 1 ([17]).

Let be a plane elementary bipartite graph. Then .

###### Theorem 2 ([18]).

Let be a hexagonal system with a perfect matching and be a set of hexagons in a Clar cover of with maximum cardinality of hexagons. Then has a unique 1-factor.

In [19], Hansen and Zheng formulate the computation of Clar number as an integer program problem. Later, Abeledo and Atkinson[20] proved that:

###### Theorem 3 ([20]).

Let be a 2-connected bipartite planar graph. Then the Clar number of can be computed in polynomial time using linear programming methods.

Zhang et. al. showed that

###### Theorem 4 ([21]).

Let be a plane bipartite graph with more than two vertices. Then each face of is resonant if and only if is elementary.

Using Theorems 1,2,3 and 4, Xu,Bian and Zhang [17] proved that the maximum forcing number of an elementary hexagonal system equals to its Clar number, and then the maximum forcing number of the elementary hexagonal system can be computed in polynomial time, and conjectured that

###### Conjecture.

[17] Let be an elementary polyomino, then the maximum forcing number of can be computed in polynomial time.

We now prove that the conjecture is true.

###### Theorem 5.

Let be an elementary polyomino with perfect matchings. Let be a maximum Clar cover of with tetragons, and let be the set of tetragons in . Then has a unique 1-factor.

Proof: Suppose that has more than one perfect matching. Joining any two of the perfect matchings of will give us a set of alternating cycles in the graph . Choose the two of the perfect matchings of such that one of the alternating cycles is smallest possible. Let denote the subgraph of such that the outer boundary of is the alternating cycle . Clearly, there is no alternating cycles for any perfect matchings in since is the smallest.

Now let’s look at the graph and we label the graph in the following way:

The top left vertex is labelled , and the vertices in the same row will be labelled as . The second row vertices is labelled as and could be 0 or a negative value if the vertex is on the left of the vertex . The face with vertex on its left top is labelled as , see Fig.1 for an example. Let denote the set of tetragons of in . By the choose of perfect matchings, is a nice cycle of (i.e. has a perfect matching) and has a unique perfect matching which is denoted by .

Observation 1 In , if there are two parallel edges, i,e, two opposite edges of a face, either and or and , then is not a maximum Clar cover.

Clearly, then the tetragon surrounded by these two parallel edges can be added to and then is not a maximum Clar cover.

Along the same line of reasoning, we have

Observation 2 If there are two parallel edges on the boundary of , i,e, edges such as and or and on , then is not a maximum Clar cover.

Suppose that the face is surrounded by these two parallel edges, clearly that if is a pending face, then the corresponding tetragon can be included in the , a contradiction. In other cases, removing the tetragon that corresponding to face , the has not been disturbed. Furthermore, the cycle has been disconnected into two odd length paths with a unique perfect matching. Thus, we could include the tetragon corresponding to face to increase , a contradiction.

Observation 3 In , If there is an edge parallel to the boundary, then is not a maximum Clar cover.

Now we shall prove that either there exist a pair of parallel edges in or we could replace some tetragons of by a larger set, which lead to a contradiction.

Look at the consecutive faces in the first row of . Because of Observation 2, we know that . Now we look at the faces . Let’s first see that these faces are all belong to . Suppose that one of the faces , where , is not part of the , then has two parallel edges on the boundary of . Based on Observation 2, we know it is not possible (See Fig.2 for an example).

Now, we know that are faces of . First, it is clear that none of the edges , .. are in the because of Observation 3. This implies that the edges , . are either in the perfect matching or in . And furthermore, due to Observation 2, there are no parallel edges in the .

Suppose that is odd, then we know that faces should be in , otherwise, remove those faces in of form where and then take with the remaining tetragons of to get a Clar cover of with larger number of tetragons, a contradiction.

Now we have , we shall replace with the faces which give us a Clar cover with one more tetragons than that in (See Fig.3 for an example), a contradiction.

Now we look at the even case, where the previous suggested approach doesn’t work since it will not give us a new Clar with larger number of tetragons. However, we could assume that the edge is in and faces are in , otherwise, we could make the rearrangement for that configuration to happen. If the edge is on the boundary of , then by the Observation 3, we know that is not a maximum Clar cover of . We then could assume that there are other faces on the left of

Next, we know that the edge is either in or or belong to the boundary of . First we see it is not possible for to be on the boundary, since it implies that the edges are on the cycle , based on the Observation 2, it is a contradiction. Thus we know that must belong to or .

Assume that the left most vertex on the second row of is , i.e. it doesn’t exist a vertex where . We could assume that are all belong to , otherwise, we re-label the graph, take the left most as .

Now we also know that all faces are belong to , otherwise, based on the Observation 2, we could show that one of the faces , where , could be included in . See Fig.4 for detail.

As in previous cases, we know that the edges , .. are either in the perfect matching or in . If is odd, then faces must be in .

If is odd, we could then replace these faces by , and clearly we have a larger Clar cover of , see Fig.5 for an example. Clearly the left over graph has a perfect matching, i.e. there are odd length of path been removed from the boundary and no internal matchings been disturbed.

If is even, then we could assume that the edge is in the perfect matching , if is on the boundary of , then based on Observation 2, we know is not maximum. See Fig 6 for an example, now the left over case is that there are more faces on the left of . Suppose that the left most faces in the third row is .

In this case, we know that the edge is either in the or in , and alone the same line of reasoning as for the second row of , we could show that either we could get a larger Clar cover of or is in the perfect matching . The same argument terminates until the row containing the left most bottom block is encountered. i.e. the row containing the face and there is no faces with where . See Fig.7 for detail. In this case, we find a larger Clar cover which has more tetragons than , a contradiction. Consequently has a unique 1-factor.

Now Let be a perfect matching of such that . According to Theorem 2 in [20], there exist disjoint -alternating cycles, say as . Let denote the subgraph of such that the outer boundary of is the -alternating cycle . From the proof of Theorems 9 in [20], it can be seen that, if the above theorem holds then . Combines with Theorem 1, the following conclusions hold.

###### Theorem 6.

Let be an elementary polyomino. Then .

The following result follows immediately from Theorem 3 and Theorem 6.

###### Theorem 7.

Let be an elementary polyomino. Then the maximum forcing number of can be computed in polynomial time.

## 3 The complexity of computing the maximum forcing number of a non-elementary polyomino and hexagonal system

In this section we turn to the case of non- elementary polyomino and hexagonal system.

In order to find the Clar number and maximum forcing number of a non-elementary polyomino and hexagonal system, we consider the decomposition of a non-elementary polyomino and hexagonal system into a number of elementary components.

In [22], F.Zhang et al. developed an algorithm to decompose a hexagonal system into a number of regions consisting of fixed bonds and a number of elementary components. Next, H.Zhang and F.Zhang[23] also provided an algorithm to decompose a polyomino into a number of regions consisting of fixed bonds and a number of elementary components. In [24], F.Zhang and H.Zhang generalized the above results. Let be a bipartite graph whose vertices can be divided into two disjoint sets and (that is, and are each independent sets) such that every edge connects a vertex in to one in . Colored each vertex in red and each vertex in blue. Let be a prefect matching of a bipartite graph . Orient all the edges belonging to toward the red vertices and orient the other edges of toward the blue vertices, the resulting digraph is called the orientation of and denoted by . Let be an edge of . They showed that there is an algorithm of complexity to determine all elementary components and the fixed bonds of a bipartite graph . The result is in the following:

###### Theorem 8 ([24]).

Let be a bipartite graph with a perfect matching . A subgraph of is an elementary of iff the corresponding orientation of is a strongly connected component of . And there is an algorithm of complexity to decompose into a number of regions consisting of fixed bonds and a number of elementary components.

Since any non-elementary polyomino (respectively, hexagonal system) with perfect matching can be composed into a number of elementary components and fixed bonds, then the maximum forcing number of the original non-elementary polyomino (respectively, hexagonal system) with perfect matching is equal to the sum of the maximum forcing number of those elementary components, and the Clar number of the original non-elementary polyomino(hexagonal systems) with perfect matchings is equal to the sum of the Clar number of those elementary components. Thus we have the following results.

###### Theorem 9.

Let be a polyomino or a hexagonal system with perfect matchings. Then .

Since the complexity of decomposition a non-elementary polyomino (respectively, hexagonal system) with perfect matchings into a number of elementary components and fixed bonds is , the number of elementary components of a non-elementary polyomino(respectively, hexagonal system) are no more than and by Theorem 8, the maximum forcing number of every elementary components can be computed in polynomial time, then the maximum forcing number of a non-elementary polyomino or hexagonal system with perfect matching can be computed at plus polynomial time, which are also polynomial time. Thus, we obtain the following results.

###### Theorem 10.

Let be a polyomino(hexagonal systems) with perfect matchings. Then the maximum forcing number of can be computed in polynomial time.

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