SN Applied Sciences

, 1:444 | Cite as

On connectivity of basis graph of splitting matroids

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


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 \(G(B(M_{a,b}),\mathcal {C}_1)\).


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.


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.


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.


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.


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
$$\begin{aligned} |C_1-B_1|\ge 2. \end{aligned}$$
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. \(\square\)



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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Authors and Affiliations

  1. 1.Department of Mathematics, Faculty of ScienceUrmia UniversityUrmiaIran

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