Parity of an odd dominating set

For a simple graph $G$ with vertex set $V(G)=\{v_1,...,v_n\}$, we define the closed neighborhood set of a vertex $u$ as $N[u]=\{v \in V(G) \; | \; v \; \text{is adjacent to} \; u \; \text{or} \; v=u \}$ and the closed neighborhood matrix $N(G)$ as the matrix obtained by setting to $1$ all the diagonal entries of the adjacency matrix of $G$. We say a set $S$ is odd dominating if $N[u]\cap S$ is odd for all $u\in V(G)$. We prove that the parity of an odd dominating set of $G$ is equal to the parity of the rank of $G$, where the rank of $G$ is defined as the dimension of the column space of $N(G)$. Using this result we prove several corollaries in one of which we obtain a general formula for the nullity of the join of graphs.


Introduction
Let N[u] denote the closed neighborhood set of a vertex u in a simple graph G, i.e.; Then, we say a subset S of vertices is odd (even) dominating if N[u]∩S is odd (even) for all u ∈ V (G).In general, for an arbitrary subset C of vertices, we say a set S is a C-parity set if N[u] ∩ S is odd for all u ∈ C and even otherwise [2].If there is a C-parity set for a given set C, we say that C is solvable.If there exists a C-parity set for every set C of vertices in a graph G, then we say G is always solvable.
Let n be the order of G, V (G) = {v 1 , ..., v n } and W be a subset of V (G).The column vector x W = (x 1 , ..., x n ) t , which is defined as x i = 1 if v i ∈ W and x i = 0 otherwise, is called the characteristic vector of W .The closed neighbourhood matrix N = N(G) of a graph G is obtained by setting to 1 all the diagonal entries of the adjacency matrix of G.
Let us denote the vectors whose components are all 0 and all 1 by 0 and 1, respectively.Then the following are equivalent.(a1 ) S is an odd dominating set, (a2 ) S is a V (G)-parity set, (a3 ) N(G)x S = 1.Similarly, (b1 ) S is an even dominating set, (b2 ) S is a ∅-parity set, (b3 ) N(G)x S = 0, are equivalent statements.Note that every graph has an even dominating set, which is ∅.On the other hand, it is proved by Sutner that every graph has an odd dominating set as well [9] (see also [6], [7], [8]).
From the matrix equation ( 1), we see that G is always solvable if and only if ν(G) = 0.Moreover, ν(G) > 0 if and only if G has a nonempty even dominating set.
We write pr(a) to denote the parity function of a number a, i.e.; pr(a) = 0 if a is even and pr(a) = 1 if a is odd.In the case where A is a matrix, pr(A) is the parity function of the sum of its entries.For a set S, we write pr(S) to denote the parity function of the cardinality of S and say the parity of S instead of the parity of the cardinality of S. Note that pr(S) = pr(x S ).It was first noticed by Amin et al. [ [1], Lemma 3], and follows immediately from Sutner's theorem, that for a given graph, the parity of all odd dominating sets are the same.Hence, the value of pr(S), where S is an odd dominating set of a graph is independent of the particular odd dominating set S taken into account.
Our main result Theorem 2.4 states that the parity of an odd dominating set is equal to the parity of the rank of the graph.

Main result
Lemma 2.1.Let A be a n × n, symmetric, invertible matrix over the field Z 2 with diagonal entries equal to 1. Then pr(A −1 ) = pr(A) = pr(n).
Proof.In the proof, all algebraic operations are considered over the field Z 2 .First of all, note that since A is a symmetric matrix with nonzero diagonal entries, we have Similarly, On the other hand, We call a vertex a null vertex of a graph G if it belongs to an even dominating set of G. Since the set of all characteristic vectors for even dominating sets of G is a subspace of the vector space of all binary n-tuples, if v is a null vertex of G, then precisely half of the even dominating sets of G contain v. Lemma 2.2.Let G be a graph and v be a null vertex of G. Then there exists an odd dominating set of G which does not contain v.
Proof.Let R be an even dominating set containing v and S 1 be an odd dominating set of G. Assume S 1 contains v, otherwise we are done.Let S 2 be the symmetric difference of S 1 and R. Clearly S 2 is an odd dominating set which does not contain v. Proof.We prove the claim by applying induction on the nullity of the graph.Let n be the order of G.In the case where ν(G) = 0, there exists a unique odd dominating set S such that Nx S = 1.Note that N satisfies the conditions of Lemma 2.1.Hence, together with the rank nullity theorem, we have pr(S) = pr(x S ) = pr(N −1 1) = pr(N −1 ) = pr(N) = pr(n) = pr(ρ(G)).Now assume that ν(G) > 0 and the claim holds true for all graphs with nullity less than ν(G).Since ν(G) is nonzero, there exists a nonempty even dominating set.Hence, there exists a null vertex v of G.By Lemma 2.2, there is an odd dominating set S of G which does not contain v. Since S does not contain v, it is also an odd dominating set of the graph G − v.Moreover, by Lemma 2.3, nd

Let
. By the induction hypothesis pr(S) = pr(ρ(G − v)).On the other hand, using the rank nullity theorem we obtain ρ We complete the proof by noting that all odd dominating sets in G have the same parity.

Some corollaries
Corollary 3.1.Let G be an always solvable graph of order n.Then the odd dominating set of G has odd (even) cardinality if n is odd (even).
Note that if every vertex of a graph G has even degree, then V (G) itself is an odd dominating set.This, together with Theorem 2.4, gives the following.Corollary 3.2.If every vertex of a graph G has even degree, then ν(G) is even.
Corollary 3.3.If the number of even degree vertices of a tree T is at most one, then every odd dominating set of T has odd cardinality.
Proof.Let n be the order of T .By [ [3], Theorem 3] if every vertex of T has odd degree, then ν(T ) = 1.By the handshaking lemma, n must be even, hence ρ(T ) is odd.By [ [3], Theorem 4], if exactly one vertex of T has even degree, then ν(T ) = 0. Since n must be odd, ρ(T ) is also odd.Hence in either case, every odd dominating set has odd cardinality by Theorem 2.4.Corollary 3.4.Every odd dominating set of a graph G has an odd (even) number of vertices of odd degree if and only if ν(G) is odd (even).In particular, the odd dominating set of an always solvable graph has an even number of odd degree vertices.

Proof. Observe that for any subsets
Then we have x A c = x A + 1.Now let S be an odd dominating set of G and D be the set of vertices with odd degree.Observe that ).On the other hand, pr(S c ) = pr(ν(G)) by Theorem 2.4.Hence, the result follows.
We define the join G 1 ⊕ ... ⊕ G m of m pairwise disjoint graphs G 1 , ..., G m as follows.We take the vertex set as Together with Theorem 2.4, the above proposition implies the following. (2) Equivalently, Equivalence of ( 2) and ( 3) follows from the rank nullity theorem.Expressing the nullity/rank of G 1 ⊕ G 2 as a single formula involving nullities/ranks of G 1 and G 2 as above enables us to extend this result and to write a formula for the nullity/rank of the join of arbitrary number of graphs as follows.
Proposition 3.6.Let {G 1 , ..., G m } be a collection of pairwise disjoint graphs.Let j be the number of graphs in {G 1 , ..., G m } with odd rank.Then Equivalently, Proof.We prove (5), then (4) follows from the rank nullity theorem.If j = 0, then all graphs have even rank and the result follows applying (3) successively.Now let j = 0. Without loss of generality, we can assume that the first j graphs have odd rank.Then, by (3 , which is odd, and so on, yielding ρ(G , which is odd.Since the rank of the joins of the m − j even ones is the sum of the ranks (which is even), the join of all m of them is the sum of the ranks minus (j − 1).

Declaration of Competing Interests
The author declares that he has no known competing financial interests or personal relationships that could have appeared to influence the work reported in this paper.
G−v denote the graph obtained by removing a vertex v and all its incident edges from a graph G.The number nd(v) := ν(G−v)−ν(G) is called the null difference number.It turns out that nd(v) can be either −1, 0, or 1.Moreover, Ballard et al. proved the following lemma in [ [5], Proposition 2.4.].Lemma 2.3 ( [5]).Let v be a vertex of a graph G. Then v is a null vertex if and only if nd(v) = −1.Now we are ready to state our main result.Theorem 2.4.Let G be a graph and S be an odd dominating set of G. Then pr(S) = pr(ρ(G)).Equivalently, pr(V (G)\S) = pr(ν(G)).