Maximum nullity and zero forcing of circulant graphs
Linh Duong, Brenda K. Kroschel, Michael Riddell, Kevin N. Vander, Meulen, Adam Van Tuyl

TL;DR
This paper investigates the zero forcing number of circulant graphs, providing bounds and characterizations that relate to the minimum rank and maximum nullity, with specific focus on bipartite, cubic, and torus product circulants.
Contribution
It offers new bounds and characterizations for the zero forcing number of various circulant graphs, enhancing understanding of their minimum rank and nullity.
Findings
Bounds on zero forcing number for bipartite circulants
Characterizations of zero forcing number for cubic circulants
Conditions for equality in zero forcing number and nullity
Abstract
It is well-known that the zero forcing number of a graph provides a lower bound on the minimum rank of a graph. In this paper we bound and characterize the zero forcing number of certain circulant graphs, including some bipartite circulants, cubic circulants, and circulants which are torus products, to obtain bounds on the minimum rank and the maximum nullity. We also evaluate when the zero forcing number will give equality.
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Maximum nullity and zero forcing of circulant graphs
Linh Duong
Department of Mathematics, University of St. Thomas, St. Paul, MN, 55105, USA
,
Brenda K. Kroschel
Department of Mathematics, University of St. Thomas, St. Paul, MN, 55105, USA
,
Michael Riddell
Department of Mathematics & Statistics
McMaster University
Hamilton, ON, L8S 4L8, Canada 386 Whitney Ave. Hamilton, ON L8S 2H4 [email protected], [email protected]
,
Kevin N. Vander Meulen
Department of Mathematics
Redeemer University College, Ancaster, ON, L9K 1J4, Canada
and
Adam Van Tuyl
Department of Mathematics & Statistics
McMaster University
Hamilton, ON, L8S 4L8, Canada
Abstract.
It is well-known that the zero forcing number of a graph provides a lower bound on the minimum rank of a graph. In this paper we bound and characterize the zero forcing number of certain circulant graphs, including some bipartite circulants, cubic circulants, and circulants which are torus products, to obtain bounds on the minimum rank and the maximum nullity. We also evaluate when the zero forcing number will give equality.
Key words and phrases:
zero forcing, minimum rank, maximum nullity, circulant graph, bipartite graph, graph product.
2010 Mathematics Subject Classification:
05C50, 05C75, 05C76, 15A03
1. Introduction
Let be a simple finite graph with vertex set and edge set . Suppose in the graph some vertices are filled and some are unfilled. The filling rule is as follows: if a vertex is filled and has exactly one unfilled neighbor, , then vertex forces to be filled, and is referred to as a forcing vertex. Given , the final filling of is the set of filled vertices obtained by initially filling the vertices of and leaving every vertex in unfilled and applying the filling rule until no more vertices can be filled. The set is called a zero forcing set if the final filling of is . The terminology of zero forcing arose in the context of forcing entries of a null vector to be zero as first described in [2]. An example of a zero forcing set is given in Figure 1.
In various applications, it is of interest to find the cardinality of a smallest zero forcing set in (which always exists since is a trivial zero forcing set). The zero forcing number of , denoted , is the minimum cardinality of a zero forcing set for a graph . Determining is NP-hard [1] in general, but has been calculated for some well-known classes of graphs (see, for example, [2, 14, 16]). Variations of zero forcing have been useful in communication complexity, quantum mechanics, graph theory, and some inverse eigenvalue problems (see [14] for references). The zero forcing number originated in [2] as a technique to find a bound on the minimum rank of a symmetric matrix associated with a graph. Let denote the set of symmetric matrices over whose graph is . In particular, if is a graph with vertices , then if is a real symmetric matrix such that for , if and only if is adjacent to in . Note that if , then there is no restriction on the diagonal entries of . Let . It was demonstrated in [2] that provides an upperbound on .
Theorem 1.1**.**
[2]** Let be a graph and let be a zero forcing set of . Then , and thus .
Recent work [3] describes families of graphs for which equality holds in Theorem 1.1, that is, families of graphs with . If we let , then the rank theorem tells us that . Hence Theorem 1.1 demonstrates that zero forcing can provide a lower bound on the minimum rank of any symmetric matrix associated with a graph.
Before going forward, we state some known facts about zero forcing, which can be found in [2].
Lemma 1.2**.**
[2]**
- (1)
For any -regular graph , . 2. (2)
For , . 3. (3)
For , . 4. (4)
For , , where denotes the complement of the graph . 5. (5)
For disjoint graphs and , 6. (6)
For the Cartesian product of graphs and , . 7. (7)
For ,
In this paper we explore the zero forcing number for various classes of circulant graphs. Section 2 defines circulant graphs and reviews some of their properties. In addition, we extend Deaett and Meyer’s results on consecutive circulants [10]. The maximum nullity and zero forcing number of circulant graphs that are bipartite is explored in Section 3. For every bipartite circulant considered in this section, . Section 4 introduces the torus product of a graph, noting that certain circulant graphs can be viewed as a torus product. The section explores the zero forcing number and maximum nullity for several cases of torus products. The Möbius ladder is a special case of a torus product. We note that for many circulant graphs which are torus products, the numbers and are again equal, but there are still cases for which this is an open question. In Section 5, we show that for all cubic circulant graphs , , and we compute this value.
2. Properties of circulant graphs
We recall some of the properties of circulant graphs, and derive some basic results on the zero forcing number and minimal rank for this family. For standard graph theory terminology, see [21].
Given an integer and a subset , a circulant graph is a graph with vertex set and edge set and , taking subscripts modulo . Note that if , then we will abuse notation and write instead of . Furthermore, we assume that . Some examples of circulant graphs are given in Figures 2 and 3.
Since circulant graphs are vertex transitive, they are regular graphs. In particular, if , then is -regular if , and -regular otherwise. Combining this observation with Lemma 1.2(1) gives a lower bound on the zero forcing number of a circulant graph:
Theorem 2.1**.**
Suppose . If , then . If , then .
Not every circulant graph is a connected graph (see Figure 4). The connected circulant graphs were characterized by Boesch and Tindell [6]; in the statement below, we write to denote disjoint copies of the graph .
Theorem 2.2**.**
If then is connected if and only if . If , then .
Using the function , defined by , Muzychuk [19] proved the following graph isomorphism between circulant graphs.
Lemma 2.3**.**
If and , then .
As an example of Lemma 2.3, if and , then .
As noted in the introduction, the zero forcing number (and minmal rank) of a number of families of graphs are known. Some of these families (e.g., complete graphs, cycles) are special cases of circulant graphs. The next theorem summarizes some of these known results.
Theorem 2.4**.**
Let be a circulant graph.
- (1)
If and , then . 2. (2)
If , , and , then . 3. (3)
If , then .
Proof.
(1) The circulant graph , the -cycle. By Lemma 2.3, . The result then follows from Lemma 1.2(3).
(2) The graph is the complement of . If follows by Lemma 2.3, that if and , then . Now apply Lemma 1.2(4).
(3) Under this hypothesis, ; the conclusion follows from Lemma 1.2(2). ∎
Given and , the graphs are known as consecutive circulant graphs (e.g., see Figure 5).
Deaett and Meyer determined the zero forcing number and maximum nullity of consecutive circulants.
Theorem 2.5**.**
[10*, Theorems 5.4 and 5.7]**
If , then .*
By combining the above result with Lemma 2.3, one can determine and for some other families of circulant graphs:
Corollary 2.6**.**
Suppose for .
- (1)
If , then . 2. (2)
If , then .
Proof.
(1) Suppose . By Theorem 2.2, since , the graph is disconnected. In particular, with . By Theorem 2.5 , so the result follows from Lemma 1.2(5).
(2) By Lemma 2.3, . Now apply Theorem 2.5. ∎
3. Families of bipartite circulants
In this section, we determine the zero forcing number and minimal rank of some families of bipartite circulant graphs using the work of Meyer [18]. Note that Meyer investigates the family of bipartite graphs whose biadjacency matrix is a circulant matrix. These graphs are sometimes called generalized bipartite circulants, although in [18], for expediency, they are simply called bipartite circulants. This usage is different than our usage of the term bipartite circulant graph. More precisely, a bipartite circulant (as used in this paper) is a circulant graph which is bipartite. In particular, the family of bipartite circulants is a subclass of the generalized bipartite circulants, the family of graphs studied in [18]. The reader should be aware of the two usages when consulting [18].
Our starting point is the following characterization of bipartite graphs due to Heuberger.
Theorem 3.1**.**
[15*, Theorem 1]**
Let be a connected circulant. Then is bipartite if and only if is even and are odd.*
Note that partitioning the vertices of a bipartite circulant into parts based on the parity of their index will provide a bipartition of the vertex set. Two bipartite circulant graphs are given in Figure 6.
Following Meyers [18], we can represent a bipartite circulant graph using its biadjacency matrix. Recall that if is a biparitite graph with bipartition and and , then we can represent by the matrix where is if there is an edge between vertex and vertex , and 0 otherwise. The matrix is the biadjacency matrix of .
The next lemma describes how to represent the biadjacency matrix of a bipartite circulant graph. Below, denotes the permutation matrix corresponding to the cycle . Note that and if .
Lemma 3.2**.**
Let be a bipartite circulant graph.
- (1)
If , then the biadjacency matrix of is
[TABLE] 2. (2)
If , then the biadjacency matrix of is
[TABLE]
Proof.
This result is implicit in the proof of [18, Theorem 2.2]. In particular, it is shown that if , then contributes the matrices and to the biadjacency matrix of (there is a typo in [18] where the author has an instead of an ). Note that . The result now follows by noting that if , then each gives two distinct matrices, but when , the two matrices and are the same matrix. ∎
Remark 3.3**.**
Note that in Lemma 3.2, the exponents of the the matrices satisfy
[TABLE]
We recall two further results from Meyer’s paper [18]; we have specialized his results to bipartite circulant graphs of the form .
Lemma 3.4**.**
[18*, Theorem 2.4]**
Suppose that is a connected bipartite circulant graph with biadjacency matrix . Then for each unit and element , the graph is isomorphic to the graph with biadjacency matrix where the exponents are computed modulo .*
Theorem 3.5**.**
[18*, Corollary 3.5]**
Suppose that is a connected bipartite circulant graph with biadjacency matrix . Then .*
We now come to the main results of this section.
Theorem 3.6**.**
Fix with and .
- (1)
If is odd and , then . 2. (2)
If is even and , then .
Proof.
(1) Set . Then by Lemma 3.2 (2), the biadjacency matrix of has the form
[TABLE]
Let . Then by Lemma 3.4, is isomorphic to the graph with biadjacency matrix
[TABLE]
Since , by Theorem 3.5 we get .
The proof of (2) is similar. Let . By Lemma 3.2, the biadjacency matrix of has the form
[TABLE]
Using Theorem 3.4, this graph is isomorphic to the graph with biadjacency matrix
[TABLE]
So, Theorem 3.5 gives us the conclusion . ∎
For our last result, we require the following result about complete bipartite graphs.
Theorem 3.7**.**
Fix with , , and .
- (1)
If and , then . 2. (2)
If and , then .
Proof.
The proof of both statements are similar to the proof of Theorem 3.6.
(1) By Lemma 3.2, the biadjacency matrix of has the form
[TABLE]
Adding to each exponent, by Lemma 3.4, the graph is isomorphic to the graph with the biadjacency matrix
[TABLE]
Employing Theorem 3.5 gives us .
(2) If , then the biadjacency matrix is , and consequently, by Theorem 3.5. Alternatively, one notes that , and so the result follows, for example, by [4, Observation 3]. ∎
4. Circulants which are torus products
In this section we extend the definition of the Möbius ladder to a type of torus product. The zero forcing number for the Möbius ladder was calculated in [2] to be four (e.g., see Lemma 1.2(7)). We compute the zero forcing number for our torus products, and as a corollary, we are able to compute the zero forcing number for a new family of circulant graphs. We also give evidence for a conjecture on the minimal rank of this family.
Recall that the Cartesian product of the graphs and with and is the graph with vertex set with two vertices , adjacent if either and is adjacent to in , or and is adjacent to in . We position the vertices of in a grid such that the -th column contains the vertices , for , and the -th row contains the vertices , for . Then essentially consists of copies of as columns and copies of as rows. The product can be pictured as a lattice on a torus (see [17]). For , define the torus product graph to consist of copies, , of with having vertices with edges between copies as follows: for and , is adjacent to and, with subscript addition modulo , is adjacent to . Then the Möbius ladder is simply the torus product (see for example Figures 7 and 8). Note that the torus product is referred to as a twisted torus in [17].
The following proof takes advantage of the fact that the torus product is locally similar to the Cartesian product . Let for . It was shown in [2] that . Below we give an alternate argument that ; the same argument applies to the torus product .
Theorem 4.1**.**
Let or . If , then . If and , then . If and , then .
Proof.
Let or .
First consider the case . Assume the vertices are in a grid as described before the theorem. The argument uses the fact that locally, about a column of vertices, the graphs of and both have the subgraph structure . (In fact there is an automorphism of that takes column to for any .)
Observe that if vertices of two adjacent copies of are filled, this set is a zero forcing set of . Thus, .
Let be a minimal zero forcing set for . Pick a forcing vertex . All but one neighbour of must be in . Since is –regular, . Once a force is made from , the filled vertices are all those vertices in some copy of (a column), plus two additional vertices in some copy of (a row). These filled vertices are in three consecutive columns, say and , the middle column being completely filled.
In order for a vertex outside of these three columns to force some other vertex, it must be in a column that already has filled vertices. If these vertices were originally in , then . Thus, before a vertex outside of the three columns can do any forcing, some vertex in one of the three columns must first force a vertex outside these columns.
Without loss of generality, let be the first vertex in column used to force a vertex outside the three columns. Then the remaining vertices in are already filled. Suppose vertices of column are in . Then the remaining vertices in column must have been forced from vertices in column . For any vertex in column to force a vertex in column , there must already be a vertex in column , in the same row, that is filled. This implies that there must be at least vertices in column that are in . Hence . Therefore .
Now consider the case . Let be a forcing set. As noted above, for a vertex to force another vertex, there must be at least other vertices of in the same column as . As labelled above, must start off with either (a) vertices in or (b) vertices in . Note that in case (a), there must be at least one row with vertices of in both and . In either case, after a force is made, the set of all filled vertices must then contain all the vertices of and two vertices, one in and one in , adjacent to a common vertex in . If the zero forcing set contains vertices of and vertices of , then the vertices of could force at most vertices of and vertices of . Thus, after forcing, at most vertices of (and their corresponding vertices in are filled). If , then no further forcing can occur. Thus . In fact, if is the number of rows that have a vertex of in both and , then . In case (a), and in case (b) . It follows that To construct a minimal zero forcing set, let consist of the first vertices of (with as column 1), the first vertices of (with as column 2), and the first vertex of (column 3). Then we claim that is a forcing set with exactly vertices. In particular can force and then can force . From here the remaining vertices in can be forced by vertices in . Thus, .
Finally, for the case and , is a Möbius ladder and so by [2], . ∎
Remark 4.2**.**
As seen above, the proof takes advantage of the shared local structure of the Cartesian and torus products. Note the only difference between the two graphs is the particular permutation of adjacencies between the first column and the last column As such, the same result holds true for a much larger class of graphs, if all of the permutations of the adjacencies between the first and last column are considered, not just the two specified by and .
Theorem 4.1 can now be applied to the study of circulant graphs. In the next theorem since if then the graph considered is a consecutive circulant which is already discussed in Theorem 2.5.
Theorem 4.3**.**
Fix , and let with . If , then . If , then . If , then .
Proof.
Let with . Then can be obtained by combining the edges of and . One can then observe that for , is the torus product (see for example, Figures 8 and 9). The zero forcing number can then be obtained from Theorem 4.1.
If , then is a –regular graph and by Lemma 1.2(1), . Taking any vertex and all but one of its neighbours provides a zero forcing set of size . In particular, will force its only unfilled neighbour. The remaining unfilled vertices can be forced consecutively from the neighbours of with subscripts that have the same parity as that of . ∎
We expect that for all the graphs in Theorem 4.3. Using special matrices, Theorem 4.5 and Theorem 4.7 demonstrate that for these graphs when and .
An circulant Hankel matrix is a matrix for which each row is shifted one position to the left from the row above it with a wrap around to the end of the row. In particular, if the first row of is , then the th row of is . For example,
[TABLE]
is a circulant Hankel. Note that the reverse diagonals of a Hankel matrix are constant and consequently the matrix is symmetric. Let
[TABLE]
Note that if is a circulant Hankel matrix, then and is itself a circulant Hankel matrix.
Lemma 4.4**.**
For , there exists an orthogonal circulant Hankel matrix such that both and have no zero entries.
Proof.
Let be a circulant Hankel matrix with first row and . We claim that has orthogonal rows. Since is circulant, it is enough to show that the first row of is orthogonal to every other row of . If is row of , , then and
[TABLE]
Thus with . Therefore is an orthogonal circulant Hankel matrix with no zero entries. The fact that has no zero entries follows from the fact that for where denotes row of . ∎
Theorem 4.5**.**
Given , .
Proof.
Let be an orthogonal circulant Hankel matrix as in Lemma 4.4. Let
[TABLE]
with . Note that is a circulant Hankel matrix and since . Also, since and are symmetric. Let
[TABLE]
Then, using the fact that and ,
[TABLE]
Since is invertible, it follows that nulllity. Note that is a symmetric matrix with graph , since , , and are symmetric matrices with no zero entries. Therefore, and thus by Theorem 4.1, . ∎
Since is the circulant with , we have the following:
Corollary 4.6**.**
If with , then .
Theorem 4.7**.**
Given , .
Proof.
Let be an orthogonal circulant Hankel matrix as in Lemma 4.4. Let
[TABLE]
The graph of is . If
[TABLE]
then
[TABLE]
noting that Since that first and last rows of are zero, and is invertible, it follows that and by Theorem 4.1, . ∎
Corollary 4.8**.**
If then .
Remark 4.9**.**
Based upon Corollaries 4.6 and 4.8, we wonder if implies in general.
There is another circulant graph that is isomorphic to a torus product. In particular, the graph . (Note that for , is not isomorphic to .)
The following theorem mimics a result [5] on the zero forcing number of the Cartesian product , which is not surprising since locally, the graph is the same as . In particular, it was shown in [5] that, with , then if is odd, and otherwise .
We do not know if in general for , but the zero forcing number is bounded above in same was as the Cartesian product, except when and is even. In this case, . The argument is similar to that in [11, Theorem 2.18].
Theorem 4.10**.**
Suppose and . Then
Proof.
Let . First note that if two consecutive columns of vertices of are in a set , then is a zero forcing set of . In particular, each of these two columns can force all the vertices on a neighbouring column. Likewise, if two consecutive rows of are in some set , then is a zero forcing set. In particular, suppose two consecutive rows of vertices of are in . By symmetry, one can assume the first two rows of are in . In this case, one can force the whole third row, left to right: in particular, can force as ranges from 1 to . (Note that cannot force until has been filled.) Consequently, each of the subsequent rows can also be forced. Therefore,
Now suppose . Let . Suppose consists of the vertices of column , and vertices of column , namely . Then column can force vertices in column , in rows 2 through . Now the two columns with vertices can each force vertices in the columns and respectively, namely in rows 3 though . This can be repeated, filling two less vertices in each column until the end columns are reached. If is odd, then the forcing above will result in the two vertices in each of columns 1 and being filled. In particular, rows and will be completely filled, and so the resulting set will force the remaining rows to be filled, as noted at the beginning of the proof. If is even, then column 1 will have the three vertices and filled but column will only have vertex filled. However, can force . At this point, rows and are completely filled, and so the remainder of the vertices can be forced. Therefore, . ∎
Corollary 4.11**.**
If , then , and if , then for each
Corollary 4.11 deals with the case that (An earlier version of the case is found in [12].) For the case with , see Theorem 5.2. Determining for the graphs in Corollary 4.11 does not seem to be straightforward. For the case with , the following theorem demonstrates that .
Theorem 4.12**.**
If , then .
Proof.
Let . By Corollary 4.11, . By Theorem 1.1, it is enough to show that Let
[TABLE]
Then, and rank( and, therefore, . ∎
For the family of circulant graphs of Corollary 4.11, we can also find a lower bound using the following result based on the girth of a graph. The girth of a graph is size of the smallest cycle in .
Theorem 4.13**.**
[7]** Let a graph with girth and minimal degree . Then
[TABLE]
The above theorem was first conjectured in [8], and proved in some special cases. The proof of Theorem 4.13 was completed in [7] (and see the references within this paper). This result is now applied to .
Theorem 4.14**.**
Let with If , then
Proof.
Let . The graph contains a four cycle: Since, , in order for to have a cycle of length three, then . In particular, would contain the -cycle . The theorem then follows from Theorem 4.13 since is 4-regular. ∎
In the case the zero forcing number equals the lower bound given in Theorem 4.14.
Theorem 4.15**.**
If , then
Proof.
When , the result follows from Theorem 4.12. When , the upper bound of Corollary 4.11 and the lower bound of Theorem 4.14 agree. ∎
In this section we considered graphs with In the next section we consider the case .
5. The cubic circulant graphs
A circulant graph is cubic if it is three regular. If is a cubic circulant graph, then for some with . The cubic circulant graphs were characterized in [9].
Theorem 5.1**.**
Let with , and let .
- (1)
If is even, then . 2. (2)
If is odd, then .
Theorem 5.1 demonstrates that the zero forcing number of a cubic circulant is going to be a multiple of the zero forcing number of a circulant of the form or . The zero forcing number is calculated for these classes in Theorems 5.2 and 5.3 respectively.
Theorem 5.2**.**
Suppose , and . Then
Proof.
Let . If , then and, hence, by Lemma 1.2(2). Suppose . Then is isomorphic to the Möbius ladder , and hence as noted in [2]. ∎
Theorem 5.3**.**
Suppose and .
- (1)
If is odd, then . 2. (2)
If is even, then
Proof.
Let . Suppose for some positive integer . Then , and so . Thus, by Theorem 5.2.
Suppose is odd. Note that is a subgraph of consisting of two disjoint cycles of length . Observe that the cycle containing vertex consists of the vertices with even subscripts. The other cycle will consist of the vertices with odd subscripts. For any vertex , its neighbours will be . Thus, besides the edges of the two aforementioned cycles, the graph also contains the perfect matching consisting of edges of the form . It follows that . The result follows from Lemma 1.2(7). ∎
Theorem 5.4**.**
Let with , and let .
- (1)
If is even, then 2. (2)
If is odd, then
Proof.
The theorem follows from Theorem 5.1 and Theorems 5.2 and 5.3, along with Lemma 1.2(5). ∎
Note that when and are odd, then the cubic circulant graphs in Theorem 5.4 are further examples of bipartite circulants discussed in Section 3.
6. Concluding comment
For every circulant graph for which we have calculated and , these two numbers have been equal. Equality also holds for the extreme cases; when or . We wonder if equality holds for every circulant graph in general.
Acknowledgments To draw the circulant graphs in this paper, we used the LaTeX script of Eastman [13]. Some of the results in this paper first appeared the MSc thesis of Riddell [20]. This research was supported in part by grants held by the last two authors, NSERC RGPIN-2016-03867 and NSERC RGPIN-2019-05412, respectively.
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