On the diameter and zero forcing number of some graph classes in the Johnson, Grassmann and Hamming association scheme

We determine the diameter of generalized Grassmann graphs and the zero forcing number of some generalized Johnson graphs, generalized Grassmann graphs and the Hamming graphs. Our work extends several previously known results.


Introduction
Graph classes in the Johnson, Grassmann and Hamming association scheme have received a considerable amount of attention over the last decades, and several graph properties and parameters have been investigated for these families.Examples are the work of Alspach, who showed that the Johnson graphs are Hamilton connected [6], the results of Bailey and Meagher, who studied the metric dimension of Grassmann graphs [24], and the coloring results on generalized Kneser graphs by Balogh, Cherkashin and Kiselev [7].However, many parameters of these graphs are still unknown.In this paper, we focus on studying their diameter and zero forcing number.
Zero forcing is a propagation process on a graph where the vertices are initially partitioned into two sets of black and white vertices.A white vertex is colored black (forced ) if it is the unique white neighbor of a black vertex.The minimum number of initial black vertices needed to force all vertices of a graph G is called the zero forcing number.Zero forcing was formally introduced in the AIM workshop [4] as an upper bound for the minimum rank, a connection which was also established by Alon [5].
The study of this parameter is motivated by the investigation of quantum networks, influence in social networks and power dominating sets, see for example the work of Burgarth et al. [15] and Dean et al. [18].
In this paper, we establish the diameter of generalized Grassmann graphs, extending the known results for Kneser graphs [26] and generalized Johnson graphs [3] by Valencia-Pabon and Vera, and Agong et al., respectively.We also determine the zero forcing number of some generalized Johnson and generalized Grassmann graphs.As a corollary of our work, we obtain the known zero forcing results for Johnson graphs on 2-sets by Fallat et al. [20] and for Kneser graphs by Bresar, Kos, and Torres [11].Aazami posed a conjecture on the zero forcing of hypercubes [1,Conjecture 2.4.4] which was shown by Alon [5] and, independently, by several authors in a AIM workshop [4].Here we prove the exact zero forcing number of Hamming graphs, extending the mentioned known results.

Preliminaries
Let G = (V, E) be a graph with vertex set V and edge set E. Throughout this paper, we consider simple graphs, i.e. undirected, loopless graphs without multiple edges.Adjacency of vertices v and w will be denoted by v ∼ w and the open and closed neighborhood of a vertex v by G(v) and G[v] respectively.The induced subgraph G[S] on a subset S ⊆ V is the graph with vertices S and edges {e ∈ E | e ⊆ S}.The Cartesian product of two graphs G and H, denoted by G H, is the graph on The distance d(v, w) between two vertices v and w is the length of a shortest walk from v to w.The diameter diam(G) is the largest possible distance between two vertices of G.The girth g (G) is the length of a shortest cycle.
Let n, k be positive integers with n ≥ k and let S ⊆ {0, 1, . . ., k−1}.The generalized Johnson graph J S (n, k) has as vertices the k-subsets of {1, 2, . . ., n} =: [n].Two vertices are adjacent if their intersection size is in S. Well-known subfamilies include the Kneser graph K(n, k) = J {0} (n, k) and the Johnson graph J(n, k) = J {k−1} (n, k).Note that if n ≤ 2k, the Kneser graph is either empty or a disjoint set of edges, hence we assume that n ≥ 2k + 1 for these graphs.
The generalized Johnson graphs have the following q-analogue.Let F q be the finite field of order q, where q is a prime power.A subspace of F q of dimension k is called a k-subspace for short.If we let S ⊆ {0, 1, . . ., k − 1}, the generalized Grassmann graph J q,S (n, k) has as vertices the k-subspaces of F n q and vertices are adjacent if the dimension of their intersection is in S. Particular examples include the q-Kneser graph K q (n, k) = J q,{0} (n, k) and the Grassmann graph J q (n, k) = J q,{k−1} (n, k).The number of vertices of J q,S (n, k) is the Gaussian binomial coefficient n k q , which equals the number of k-subspaces of an n-dimensional vector space over F q .
In Johnson graphs, vertices are associated with sets and adjacency is determined by the cardinality of their intersection.If we replace sets by ordered tuples, we obtain the well-known Hamming graphs H(n, q), where n denotes the tuple length and the entries range from 0 to q − 1. Tuples are connected by an edge whenever they coincide in n − 1 coordinates, i.e., when their Hamming distance is 1.Alternatively, one may view H(n, q) as the Cartesian product of n copies of K q .
Zero forcing is a graph coloring problem in which an initial subset of vertices is colored black, while the others are colored white.A black vertex v colors a white neighbor w black if G(v) \ w is entirely black, i.e. if w is the unique white neighbor of v.A black vertex which initiates a color change is called a pivot and we refer to the initial set of black vertices as the leader set.If all white vertices can be colored black under the given coloring rule, we call the leader set a zero forcing set.The zero forcing number of a graph G = (V, E), denoted by Z(G), is the cardinality of a smallest subset of V which is a zero forcing set.
By modifying the color changing rule or adding restrictions to the zero forcing set, several interesting variations on zero forcing can be obtained.Below, we list the ones that are relevant for the results in this article.A more complete overview of these parameters and their mutual relations can be found in [8,23].
The connected zero forcing number Z c (G) is the cardinality of the smallest zero forcing set S such that G[S] is connected [14].The total zero forcing number Z t (G) is the cardinality of the smallest zero forcing set S such that G[S] has no isolated vertices [17].Both Z c (G) and Z t (G) upper bound the zero forcing number.A single vertex can only be a zero forcing set if G is a path, hence if For a graph G, let S F (G) be the set of symmetric matrices M over the field F with m ij = 0 whenever vertices i = j are not adjacent.The maximum nullity of G over F, denoted by M F , is the largest multiplicity of eigenvalue zero for any matrix in S(G) F .It was shown in [4] and [5] that the maximum nullity of a graph over any field lower bounds the zero forcing number.Lemma 1 ([4], Proposition 2.4 and [5], Theorem 2.1).For any graph G and field F, Besides maximum nullity, zero forcing is closely related to other graph parameters, such as metric dimension [19], power domination [18] and Grundy domination [12] In other words, a vertex w is footprinted by v i if i is the smallest integer such that v i dominates w.Note that every vertex has a unique footprinter.The maximum length of a Grundy dominating sequence, denoted by γ gr (G), is referred to as the Grundy domination number of G. Grundy domination is known to give a lower bound on the zero forcing number, a fact that we will use in some of our proofs.
For an overview of Grundy domination variants and their relation to zero forcing, we refer to [12].

Diameter and girth of generalized Grassmann graphs
We extend the results of [3] on the girth and diameter of Johnson graphs to generalized Grassmann graphs.Our main tool will be the following (folklore) result.

Lemma 3 ([9]
).Let n ≥ k + m.Given at most q n−k−m+1 k-spaces in F n q , we can always find an m-space that intersects them trivially.
In the statement below, a "nontrivial subset" denotes a subset that is neither empty nor equal to the whole set.
Theorem 4. Let n ≥ 2k and suppose that S is a nontrivial subset of {0, 1, . . ., k − 1} with least element s.The generalized Grassmann graph J q,S (n, k) has diameter Proof.Let v and w be two arbitrary vertices with intersection dimension t.We will show that If s = 0, then we always have t ≥ s and the diameter is to construct a k-space that intersects both v and w in an s-space.Choose an s-space π through v ∩ w in v and an s-space τ through v ∩ w in w.These two span a (2s − t)-space, which we can extend to a k-space that intersects v and w in just this space, by applying Lemma 3 to its residue.Therefore, Case 1(b): k + t < 2s.We first show that d(v, w) ≤ d ′ by constructing a walk of length d ′ from v to w. Choose a basis {x 1 , x 2 , . . ., x t } of v ∩ w and expand it to a basis {x 1 , x 2 , . . ., x t , y 1 , y 2 , . . ., y k−t } of v and to a basis {x 1 , x 2 , . . ., x t , z 1 , z 2 , . . ., z k−t } of w.Define The intersection of u d ′ −2 and w is at least 2s − k, so by applying the first case, we can extend the walk to w by only two more steps, which results in a walk of length d ′ between v and w.
. . .In order to show the converse inequality ) is an integer.We do this by induction on where we used that u ∼ w in the second step, and applied the Grassmann identity in the third step.We conclude that d(v, w) ≥ k−t k−s .
Case 2: t ≥ s.If t ≥ s, we can choose an s-space π in the intersection of v and w and construct a k-space u that intersects v and w in π using Lemma 3, like we did in Case 1.This construction provides us with a walk of length 2, so the distance d(v, w) is at most 2.
Allowing n to be smaller than 2k leads to the following, more general statement.Note that the condition n ≥ 2k − max(S) below is just there to ensure connectivity.
Case 2: n < 2k.By definition, J q,S (n, k) ∼ = J q,{s+n−2k|s∈S} (n, n − k).Applying the first case on the latter graph (since We end this section with a short proof of the girth of generalized Grassmann graphs.Proposition 6.Every generalized Grassmann graph J q,S (n, k) with S = ∅ has girth 3.
Proof.Let J q,S (n, k) be a nontrivial Grassmann graph and let s ∈ S. Recall that we may assume that n ≥ 2k without loss of generality.Choose two k-spaces v and w that intersect in an s-space π.Since 2 ≤ q ≤ q n−2k+1 ≤ q (n−s)−2(k−s)+1 , we can apply Lemma 3 to the residual projective space of π to find a third k-space u that intersects v and w in π.Then u, v and w are mutually adjacent, i.e. there is a triangle in the graph.We conclude that the girth must be 3.

Zero forcing number
Zero forcing is known to be an NP-complete problem.The proof of this result is often wrongly attributed to Aazami [1,2] (see, e.g.[13,14,23,25]), who showed NPcompleteness for weighted zero forcing.However, this is incorrect, since the unweighted version does not follow from the weighted 0, 1 version that Aazami established in [1,Theorem 2.3.1].NP-completeness of the unweighted variant is proved in [27], where it is shown that fast-mixed searching is NP-complete, and it is known that this problem is equivalent to zero forcing [21].Despite its NP-completeness, the exact zero forcing number is known for certain graph classes [4,10,14].In this section, we determine the zero forcing number of several families of generalized Johnson graphs, generalized Grassmann graphs and Hamming graphs.

Generalized Johnson graphs
First, we investigate the zero forcing number of generalized Johnson graphs, generalizing known results on Johnson and Kneser graphs from [20] and [11].A natural way to generalize the Johnson graphs, is to extend their connection set {k − 1} to S = {s, s + 1, . . ., k − 1}.We determine the zero forcing number of these graphs by studying the more general case where min(S) = s.Note that if n < 2k − s, every two k-sets intersect in strictly more than s elements, so not all options in S can actually occur.We therefore assume that n ≥ 2k − s.To prove a lower bound on the zero forcing number, we will make use of the following variant of Bollobás's Theorem.
∈ v} and B := V \W.Note that B is well-defined, since n ≥ 2(k − s).This set has size n k − n−2(k−s) s and we will show that it is a zero forcing set.
Consider a white vertex Every other white vertex has at most s − 1 elements in common with w, because their intersection must be a subset of {v 1 , v 2 , . . ., v s }.Therefore, v is the unique white neighbor of w and is forced black.This holds for any v ∈ W, so B is a zero forcing set.
It is also a connected zero forcing set, since we can construct a black path between any two black vertices as follows.Let v, v ′ ∈ B. For any choice of S, vertices are adjacent if they differ in k − s elements.Using this property, we can replace one element a ∈ v at the time with a new element b to eventually arrive at v ′ : choose a (2k − s)-set X containing v ∪ {a, b} and a subset Y ⊂ v such that |Y | = k − s, a ∈ Y .Then there exists a path v → X \ Y → (v \ a) ∪ b, where at each step, we replace exactly k − s elements.Moreover, by choosing Y appropriately we can ensure that all path vertices are black.Therefore, the leader set is connected.Now assume that S = {s, s + 1, . . ., k − 1}.To prove that the upper bound is tight, we use the Grundy domination number of J S (n, k).Consider a Grundy dominating sequence v 1 , v 2 , . . ., v m of maximum length and pair each v i with a vertex w i that is footprinted by it.Define the sets Moreover, if j > i, we know that |v i ∩ w j | ≤ s − 1, otherwise w j is either v i or one of its neighbors, so it would have been dominated by v i before v j .Hence |X i ∩ Y j | ≥ k − s + 1 for i < j.Lemma 7 implies that γ gr (J S (n, k)) where the first inequality follows from Lemma 2.
If s = k − 1, we obtain the zero forcing number of the classical Johnson graphs.
The Kneser graphs can be generalized in a similar way by extending the intersection set to {0, 1, . . ., s}.This generalization also appears in [7], where the coloring number of generalized Kneser graphs is studied, and [16], which shows that some of these graphs are not determined by their spectrum.As usual for Kneser graphs, we require n ≥ 2k+1.
Proof.Define T := {n − 2k + 2s + 2, n − 2k + 2s + 3, . . ., n} and W := W 1 ∪ W 2 , where distinct vertices, so B has the desired cardinality.We will show that it is a connected zero forcing set by first forcing the vertices in W 1 and then those in W 2 .Let v ∈ W 1 and let w be the vertex so w has no white neighbors besides v.If w is black, it will therefore force v.This is the case if (T \ [2k − s]) ⊂ v, because then we cannot have w ∈ W 2 .If this should hold for any choice of v, we need |T \ [2k − s]| = n − 2k + s > k − s, which is satisfied by assumption, as n > 3k − 2s.Therefore, W 1 can be forced entirely.
Next, consider a vertex v ∈ W 2 and let w This means that v is the unique white neighbor of w, so it will be colored black.
The above construction also holds for the connected zero forcing number.The proof is analogous to that of Theorem 8.If S = {0, 1, . . ., s}, it follows from Theorem 10 that the upper bound is tight for the (connected) zero forcing number of J S (n, k).
The construction in Theorem 11 no longer works if n ≤ 3k − 2s, as some pivots of W 1 will be contained in the white set W 2 , which is forced last.However, for the extremal case n = 3k − 2s, we can construct a different zero forcing set of the same cardinality.
Proof.Consider the sets T , W and B from Theorem 11.The zero forcing process used in the construction in Theorem 11 is no longer applicable when n = 3k − 2s because the white vertex v However, we want to force W 1 before W 2 .Therefore, the zero forcing process will no longer color the entire graph.We propose the following change. Let to the leader set B, and instead color v ′ = [s] ∪ {s + 2} ∪ X ∪ {t} white.We will show that this gives a zero forcing set.
Using the same pivots as in the proof of Theorem 11, all vertices of W 1 can be colored black, except those containing X ∪ {t}.We will force those later and focus on W 2 first.
Consider the pivots from Theorem 11 corresponding to W 2 .A vertex cannot be forced if its pivot is adjacent to v ′ or a white vertex of W 1 .It meets at least one of these conditions whenever it contains X.Let v ′′ be such a vertex.Its pivot contains y, as v ′′ = v.Replace y by t, then the resulting vertex has no other white neighbors in W 2 and is not adjacent to v ′ or any white vertex from W 1 (both contain t).Moreover, it is black, so it can force v ′′ .Hence all of W 2 can be colored black.Now consider the vertex [s] ∪ (T \ ({i} ∪ [2k − s])) ∪ {y}.This vertex from W 1 has been forced already and is not adjacent to any white vertices in W 1 , which must contain at least one element from (T \ ({i} ∪ [2k − s])) ∪ {y}.It can therefore force v ′ .The remainder of W 1 can then be forced by their usual pivots.
Once again, tightness of this construction for S = {0, 1, . . ., s} follows from Theorem 10.The connectivity argument is analogous to the one in the proof of Theorem 8.

Generalized Grassmann graphs
Lemma 7 has the following analogue for subspaces over a field.Lemma 13 ([22]).Let U 1 , U 2 , . . ., U m be r-dimensional subspaces and let W 1 , W 2 , . . ., W m be s-dimensional subspaces of a linear space over a field F. Let t ≥ 0 such that Then m ≤ r+s−2t r−t .
Using the above lemma, we obtain a lower bound on Z(J q,{0,1,...,s} (n, k)), similar to the generalized Johnson case.Theorem 14.Let n ≥ 2k + 1 and S = {0, 1, . . ., s} for some s ∈ {0, 1, . . ., k − 2}.Then Proof.Consider a Grundy dominating sequence v 1 , v 2 , . . ., v s of maximum length in J q,S (n, k) and pair each v i with a vertex w i that is footprinted by it.Then dim(v i ∩ w i ) ≤ s and if j > i, dim(v i ∩ w j ) ≤ s + 1. Lemma 13 implies that γ gr (J q,S (n, k)) ≤ 2k−2s k−s , hence For a given vector space F n q , let a 1 , a 2 , . . ., a n be an orthogonal basis.Note that the proofs of Theorem 11 and 12 are still valid for K q (n, k) if we replace each subset of [n] with the corresponding set of basis vectors.Therefore, we have the following analogous result for K q (n, k).

Hamming graphs
In this section we show the exact zero forcing number of Hamming graphs, extending the results on H(n, 2) and H(2, q) in [4] and [5], respectively.
We will use the following elementary property, which can be shown using Newton's binomial theorem.
For n = 1, the set C 1,q ∪ {(0)} is a zero forcing set for H(1, q) of size q − 1, and white vertex (q − 1) can be forced with pivot (0) / ∈ C 1,q .Suppose that we have a zero forcing set of H(n − 1, q) of size z n−1,q in which no vertex of C n−1,q acts as a pivot, but all are in the leader set.As H(n, q) = H(n − 1, q) K q , the Hamming graph H(n, q) can be constructed by taking q copies of H(n − 1, q) and connecting the corresponding vertices in each copy with an edge.The tuple corresponding to each vertex is then extended with a new entry denoting the copy it is in.In the first q − 1 copies, select the same zero forcing set of H(n − 1, q) respecting the above conditions.In the last copy, choose the same set, excluding the core.The first q − 1 copies can be forced black by applying the zero forcing process of H(n − 1, q) to each.Note that, by the induction hypothesis, the pivots that we use here are black in every copy, so they still have only one white neighbor.Now we can force the core of the last copy, using the corresponding vertices in the first copy as pivots.Finally, the zero forcing process of H(n − 1, q) can be repeated for the last copy.Note that the core of H(n, q) is included in the leader set, while none of its vertices were used as a pivot.
The constructed zero forcing set has size and therefore completes the induction.We conclude that Z(H(n, q)) ≤ z n,q .To show that this bound is tight, we now prove that Z(H(n, q)) ≥ z n,q .Define the matrix B n recursively as where I and J are the q × q unit matrix and the q × q all-one matrix respectively, and ⊗ denotes the tensor product.Then B n is in S(H(n, q)).We show that its nullity as a matrix over F 2 is at least z n,q by constructing a set of z n,q independent eigenvectors with eigenvalue zero.The result then follows from Lemma 1.Let e i be the ith standard basis vector of F q 2 and let 1 denote the all-one vector of F q 2 .This vector is an eigenvector of J with eigenvalue one if q is odd and eigenvalue zero if q is even.Let {x 1 , x 2 , . . ., x q−1 } be a set of independent eigenvectors of J with eigenvalue zero.
Case 1: q is odd.Consider all elements of {1, x 1 , x 2 , . . ., x q−1 } ⊗n that contain an even number of 1's.In other words, all tensors of the form with k even, and permutations thereof.These are all elements of the kernel of B n and, by Lemma 16, there are z n,q such vectors.Moreover, they are all linearly independent, because x 1 , x 2 , . . ., x q−1 and 1 are linearly independent.
Case 2: q is even.We construct z n,q eigenvectors with eigenvalue zero by induction on n.If n = 1, then we can choose the z 1,q = q − 1 eigenvectors x 1 , x 2 , . . ., x q−1 of J.For the induction step, suppose that X is a set of z n−1,q linearly independent vectors that nullify B n−1 .
Consider also the q n−1 vectors of the form where we used that B 2 n−1 = 0 since J 2 = qJ = 0.Moreover, the vectors x i ⊗ v and 1 ⊗ w + e 1 ⊗ (B n−1 w) are all linearly independent because the x i , 1 and e 1 are linearly independent, and all v are linearly independent by the induction hypothesis.In total, we obtain (q − 2)z n−1,q + q n−1 = 1 2 (q n + (q − 2) n ) = z n,q linearly independent eigenvectors with eigenvalue zero.