# On connectivity of basis graph of splitting matroids

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## Abstract

For a set \(\mathcal {C}\) of circuits of a matroid *M*, \(G(B(M), \mathcal {C})\) is defined by the graph with one vertex for each basis of *M*, in which two basis \(B_1\) and \(B_2\) are adjacent if \(B_1\cup B_2\) contains exactly one circuit and this circuit lies in \(\mathcal {C}\). For two elements of *a* and *b* of ground set of a binary matroid *M* a splitting matroid \(M_{a,b}\) is constructed. It is specified by two collections of circuits \(\mathcal {C}_0\) and \(\mathcal {C}_1\) dependent with collections of circuits of *M*. We want to study connectivity of \(G(B(M_{a,b}),\mathcal {C}_0)\) and \(G(B(M_{a,b}),\mathcal {C}_1)\).

## Keywords

Basis graph Splitting matroid Connected matroid## 1 Introduction

We assume that reader is familiar with the basic concepts of matroid theory and graph theory. For more details one can see [6] for matroid theory and [3] for graph theory. By considering bases of a matroid *M*, the basis graph of a matroid *M*, *G*(*B*(*M*)), is a graph in which each vertex is labeled as a basis of *M* and two bases (vertices) \(B_1\) and \(B_2\) are adjacent if \(|B_1\bigtriangleup B_2|=2\). In other words, two bases are adjacent if they differ in only one element. In [4] Holzmann and Harary proved that *G*(*B*(*M*)) is hamiltonian and therefore connected.

Many different variations of the basis graph have been studied, for instance, Li et al. [5] defined the basis graph of a matroid *M* that related to a set \(\mathcal {C}\) of circuits. It is a spanning subgraph of *G*(*B*(*M*)) such that two bases \(B_1\) and \(B_2\) are adjacent if they are adjacent in *G*(*B*(*M*)) and the unique circuit of *M* contained in \(B_1\cup B_2\) is a circuit of \(\mathcal {C}\). In an special case when *M* is a binary matroid, they found a sufficient condition and a necessary condition for this graph to be connected and they proved that given an element *e* of a binary matroid *M* and by supposing \(\mathcal {C}_e\) is the family of all the circuits that contains *e*, \(G(B(M), \mathcal {C}_e)\) is connected. Figueroa et al. [1] generalized this result and they proved that it is true for every matroid of *M* with given \(\mathcal {C}_e\).

The splitting operation is defined for both graph and binary matroid. Fleischner [2] defined splitting operation for graph by the following way; let *G* be a connected graph and suppose that *v* is a vertex with degree at least 3. Let \(a=vv_1\) and \(b=vv_2\) be two edges incident at *v*, then splitting away of *a*, *b* from *v* results in a new graph \(G_{a,b}\) obtained from *G* by deleting the edges *a* and *b*, and adding a new vertex \(v'\) adjacent to \(v_1\) and \(v_2\). The transition from *G* to \(G_{a,b}\) is called the *splitting operation* on *G*. We also denote the new edges \(v'v_1\) and \(v'v_2\) in \(G_{a,b}\) by *a* and *b*, respectively.

*a*,

*b*be two elements of

*E*. Let

*M*by adding an extra row to binary matrix representation of

*M*, in which arrays respect to

*a*,

*b*are 1 and remind arrays are 0.

Shikare and Azadi [8] characterized the collection of bases of a splitting binary matroid as the following theorem.

### **Theorem 1.1**

*Let* \(M = (E, \mathcal {C})\) *be a binary matroid and* \(a,b\in E\). *Let* *B* *be the set of all bases of* *M*. *Let* \(\mathcal {B}_{a,b} = \big \{B\cup \{\alpha \} : B\in \mathcal {B}, \alpha \in E-B\ {\text {and the unique circuit contained in}}\ B\cup \{\alpha \}\ {\text {contains either}}\ a\ {\text {or}}\ b\big \}\). *Then* \(\mathcal {B}_{a,b}\) *is a set of bases of* \(M_{a,b}\).

## 2 Main results

We shall want to consider two cases of \(G(B(M_{a,b}), \mathcal {C})\) respect the two collections of circuits of \(M_{a,b}\). First, we start with a helpful lemma.

### **Lemma 2.1**

*Let* *M* *be a binary matroid and* \(T=\{a,b,c\}\) *be a triangle of* *M*. *Let* \(B_1\) *be a basis of* *M* *contains two elements of* *T*. *Let* \(B_2\subseteq E(M)\) *such that* \(|B_2|=|B_1|\) *and* \(B_2-B_1=\{e\}\) *where* \(e\in T\), *then* \(B_2\) *is a basis of* *M*.

### *Proof*

Let \(B_1=\{a,b,1,2,\ldots ,n\}\) be a basis of *M* and let \(B_2=\{a,t,1,2,\ldots ,n\}\) be the assumed subset of the lemma. It is clear that \(X=\{a,1,2,\ldots ,n\}\) is independent, we prove \(X\cup \{t\}=B_2\) is too. Assume the contrary and let \(X\cup \{t\}\) contains a circuit *C*. Clearly \(t\in C\). If \(a\in C\), as *M* is binary and \(\{a,b,t\}\) is a triangle, then \(a+b=t\). So \(C'=b\cup C-\{t,a\}\) is a circuit and \(C'\subseteq B_1\), a contradiction. If \(a\not \in C\), then \(C''=\{a,b\}\cup C-\{t\}\) is a circuit and \(C''\subseteq B_1\), a contradiction. Thus \(X\cup \{t\}\) is independent and therefore \(B_2\) is a basis of *M*. \(\square\)

The following theorem is our main result.

### **Theorem 2.2**

*Let* *M* *be a binary matroid and* \(T=\{a,b\}\), *where* \(a,b\in E(M)\). *Let* \(\mathcal {C}_0\) *be the collection of circuits in* *M* *in which meet* *T* *at even elements, then a sufficient condition for* \(G(B(M_{a,b}),\mathcal {C}_0)\) *to be connected is that* *T* *lies in a triangle in* *M*.

### *Proof*

If \(\mathcal {C}_1\) at the collection of circuits of splitting matroids is empty set, then \(M_{a,b}=M\) and hence \(\mathcal {C}_0=\mathcal {C}(M)\). Therefore \(G(B(M_{a,b}),\mathcal {C}_0)=G(B(M))\) and *G*(*B*(*M*)) always is connected. Thus, suppose \(\mathcal {C}_1\) is non-empty set. Let \(\{a,b,t\}\) be the assumed triangle. We consider four cases about vertices of \(G(B(M_{a,b}),\mathcal {C}_0)\). In these cases \(B_i\) are bases of \(M_{a,b}\) as characterized in Theorem 1.1 and \(x,y\in E(M_{a,b})-T\).

**Case 1** \(B_1=\{a,1,2,\ldots ,x\}\) and \(B_2=\{a,1,2,\ldots ,y\}\).

Since \(M_{a,b}\) does not have a circuit with just an element of *T*, then \(B_1\cup B_2\) contains a circuit *C* that avoids *a*. Hence \(C\cap T=\emptyset\), so \(C\in \mathcal {C}_0\). We conclude \(B_1\) and \(B_2\) are adjacent in \(G(B(M_{a,b}),\mathcal {C}_0)\).

**Case 2** \(B_1=\{a,1,2,\ldots ,x\}\) and \(B_2=\{a,1,2,\ldots ,b\}\).

Suppose \(B_1\) and \(B_2\) are not adjacent. Then \(t\not \in B_1\cup B_2\); otherwise the triangle \(\{a,b,t\}\) contained in \(B_1\cup B_2\), contradicting the fact that \(B_1\) and \(B_2\) are not adjacent. By Lemma 2.1 \(B_3=\{a,1,2,\ldots ,t\}\) is a basis of \(M_{a,b}\). Now \(B_2\cup B_3\) contains a unique circuit in which *t* and *b* belong to it and since *a* belongs to that union, the circuit is the triangle. Thus, it is on \(\mathcal {C}_0\). Then \(B_2\) and \(B_3\) are adjacent. By the first case \(B_1\) and \(B_3\) are adjacent. Then there is a path from \(B_1\) to \(B_2\).

**Case 3** \(B_1=\{a,b,1,2,\ldots ,x\}\) and \(B_2=\{a,b,1,2,\ldots ,y\}\).

If a circuit *X* in \(B_1\cup B_2\) is a member of \(\mathcal {C}_0\), the result is trivial and this two vertices are adjacent in \(G(B(M_{a,b}),\mathcal {C}_0)\). By the Lemma 2.1, \(B_3=\{a,t,1,2,\ldots ,x\}\) and \(B_4=\{a,t,1,2,\ldots ,y\}\) are two bases of \(M_{a,b}\). By the second case \(B_1=\{a,b,1,2,\ldots ,x\}\) and \(B_3=\{a,t,1,2,\ldots ,x\}\) are adjacent and similarly two bases of \(B_2=\{a,b,1,2,\ldots ,y\}\) and \(B_4=\{a,t,1,2,\ldots ,y\}\) are adjacent by the second case. In the other hand \(B_3\) and \(B_4\) are adjacent by the first case. Therefore there is a path between \(B_1\) and \(B_2\).

**Case 4** \(B_1=\{1,2,\ldots ,a\}\) and \(B_2=\{1,2,\ldots ,b\}\).

Suppose \(B_1\) and \(B_2\) are not adjacent in \(G(B(M_{a,b}),\mathcal {C}_0)\) . The uniqe circuit *X* in \(B_1\cup B_2\) is disjoint union of two circuits \(C_1\) and \(C_2\) of *M* such that each of them meets *T* precisely in one element. In fact \(C_1=\{a,a_1,a_2,\ldots ,a_m\}\subseteq B_1\) and \(C_2=\{b,b_1,b_2,\ldots ,b_n\}\subseteq B_2\). It is clear that \(t\not \in B_1\cup B_2\). We construct \(B_3\) by the following way; we delete a member of \(C_1\) like \(a_1\) and add *t* in it. Without loss of generality, we can assume that \(a_1=1\). Hence \(B_3=\{t,2,\ldots ,a\}\). Since \(a+b=t\) and \(B_3-\{a,t\}\subseteq B_2\) it is clear that \(B_3\) is a basis of \(M_{a,b}\). Now by the second case \(B_1\) and \(B_3\) are connected by a path. Suppose \(B_4=\{b,2,\ldots ,a\}\) is constructed by deleting *t* and adding *b*. By the Lemma 2.1, \(B_4\) is a basis of \(M_{a,b}\). By the second case \(B_4\) and \(B_3\) are connected by a path too. Now if we apply the same procedure for the basis \(B_2\) and without loss of generality by considering \(b_1=2\), we get bases \(B_5=\{1,t,\ldots ,b\}\) and \(B_6=\{1,a,\ldots ,b\}\) that by the second case there is a path between \(B_2\) to \(B_5\) and a path between \(B_5\) to \(B_6\). Now notice that \(B_4\bigtriangleup B_6=\{1,2\}\) and \(B_4\) and \(B_6\) are connected by a path by the second case. Hence there is a path between \(B_1\) and \(B_2\).

Now suppose that \(B_1\) and \(B_2\) are two arbitrary bases of \(M_{a,b}\). As \(G(B(M_{a,b}))\) is connected, there is a path from \(B_1\) to \(B_2\) in \(G(B(M_{a,b}))\). Then \(B_1\) has an adjacent vertex in \(G(B(M_{a,b}))\). This two vertices are connected with a path by using four cases mentioned above. Thus we conclude \(B_1\) and \(B_2\) in \(G(B(M_{a,b}),\mathcal {C}_0)\) are connected by a path, then the graph \(G(B(M_{a,b}),\mathcal {C})\) is connected. \(\square\)

Evidently the sufficient condition in the theorem is not necessary. Consider the following example.

### *Example 2.3*

*G*in the following Figure (a) \(G_{a,b}\) is shown in Figure (b). There is no triangle in

*G*but one can show \(G(B(M(G)_{a,b}),\mathcal {C}_0)\) is connected.

### **Corollary 2.4**

*Let* *M* *be a matroid and* \(\mathcal {C}_0\) *be specified collection of circuits in the last Theorem*. *Let* *S* *be a triangle of* *M*. *Suppose* \(T=\{a,b\}\) *and* *T*, *S* *are disjoint. If* \(M_{x,y}=M_{a,b}\), *where* \(x,y\in S\), *then* \(G(B(M),\mathcal {C}_0)\) *is connected.*

### **Theorem 2.5**

*Let* *a* *and* *b* *be two elements of a binary matroid* *M*. *If* \(\mathcal {C}_0\) *is empty, then* \(G(B(M_{a,b}),\mathcal {C}_1)\) *is connected.*

### *Proof*

Suppose \(\mathcal {C}_0\) is empty. If \(\mathcal {C}_1\) is empty, then \(M_{a,b}\) is a free matroid and clearly \(G(B(M_{a,b}),\mathcal {C}_1)\) is connected. Thus, we can assume that \(\mathcal {C}_1\) is non-empty. Now suppose *C* and \(C'\) are two circuit of \(M_{a,b}\). As \(\mathcal {C}_0\) is empty, so *a* and *b* belong to both of *C* and \(C'\). Since \(M_{a,b}\) is binary then \(C\bigtriangleup C'\) contains a circuit and none of *a* or *b* belongs to this circuit, hence \(\mathcal {C}_0\) is non-empty, a contradiction. Thus \(M_{a,b}\) just has one circuit. Therefore \(G(B(M_{a,b}),\mathcal {C}_1)\) is a complete graph and hence it is connected.\(\square\)

Notice that the converse of the last theorem is not true generally, for instance consider following example.

### *Example 2.6*

*G*and its splitting graph \(G_{a,b}\) in following figure. One can easily check that there is a path between every two vertices of \(B(M(G)_{a,b},\mathcal {C}_1)\). Hence \(B(M(G)_{a,b},\mathcal {C}_1)\) is connected while \(\mathcal {C}_0\) is non-empty.

The converse of Theorem 2.5 can be true with a special condition.

### **Theorem 2.7**

*Let* *a* *and* *b* *be two elements of a binary matroid* *M*. *Let* \(\mathcal {C}_0\) *be non-empty and* \(\mathcal {C}_1\) *has only one circuit. Then* \(G(B(M_{a,b}),\mathcal {C}_1)\) *is disconnected.*

### *Proof*

## Notes

### Acknowledgements

The authors gratefully acknowledge the Faculty of Science of Urmia University for helpfull support given.

### Compliance with ethical standards

### Conflict of interest

The authors declare that they have no conflict of interest.

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