SN Applied Sciences

, 1:444

# On connectivity of basis graph of splitting matroids

• M. Pourbaba
• H. Azanchiler
Research Article
Part of the following topical collections:
1. 3. Engineering (general)

## 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.

The notion of the splitting operation extends to binary matroid in the following way [7]. Let $$M = (E, \mathcal {C})$$ be a binary matroid and a, b be two elements of E. Let
\begin{aligned} \mathcal {C}_0=&\{C\in \mathcal {C} : a,b\in C\ {\text {or}}\ a,b\not \in C\}\\ \mathcal {C}_1=&\{C_1\cup C_2: C_1, C_2\in \mathcal {C}, \ C_1\cap C_2=\emptyset , \ \\&a\in C_1, \ b\in C_2\ {\text {and}}\ C_1\cup C_2\\&{\text {contains no member of}} \ \mathcal {C}_0\}. \end{aligned}
Let $$\mathcal {C}'=\mathcal {C}_0\cup \mathcal {C}_1$$, then $$M_{a,b}=(E,\mathcal {C}')$$ is a binary matroid. As the collection of cycles of splitting graph is the same with the collection of circuits of splitting binary matroid defined above, we used the same notation. In [7], the authors showed that $$M_{a,b}$$ obtains from 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

For the graph 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

Consider the graph 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

Let $$C_1$$ be the only circuit of $$\mathcal {C}_1$$. Since $$\mathcal {C}_0$$ is non-empty, there is a basis of $$M_{a,b}$$ called $$B_1$$ such that
\begin{aligned} |C_1-B_1|\ge 2. \end{aligned}
(1)
Let $$B_2$$ be a basis of $$M_{a,b}$$ in which contains $$C_1-e$$, where $$e\in C_1$$. We claim that vertices $$B_1$$ and $$B_2$$ of $$G(B(M_{a,b}),\mathcal {C}_1)$$ are not adjacent. Assume the contrary, let $$B_1\cup B_2$$ contains a circuit of $$\mathcal {C}_1$$, that is $$C_1$$. As $$B_2=(B_1-f)\cup g$$, where $$f\in B_1$$ and $$g\in B_2$$, by (1), $$|C_1-(B_1\cup B_2)|\ge 1$$. Hence $$C_1$$ is not contained in $$B_1\cup B_2$$, a contradiction. In fact $$B_1$$ has no adjacent vertex, then $$G(B(M_{a,b}),\mathcal {C}_1)$$ is disconnected. $$\square$$

## Notes

### Acknowledgements

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

### Conflict of interest

The authors declare that they have no conflict of interest.

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