Forcing parameters for graphs
David Amos
November 5, 2014
1
Basic definitions
Here we collect a few necessary graph theory definitions. To begin, a graph
is a pair G = (V, E) of sets. The set V is the vertex set of G. We always
assume that V is finite, and usually take V to be the set {1, 2, . . . , n}. The
set E is the edge set of G, and is a subset of {{u, v} : u, v ∈ V }. In particular,
we only consider undirected graphs without loops or multiple edges between
vertices.
If {u, v} ∈ E, we say that u and v are neighbors, or that u is adjacent to
v. The degree of a vertex v ∈ V , denoted deg(v), is the number of neighbors
of v. A graph is said the be d-regular if every vertex has degree d.
Let G = (V, E) be a graph. If X ⊆ V and Y ⊆ E for which every edge in
Y is between vertices of X, then the graph H = (X, Y ) is called a subgraph
of G. If Y contains every edge between vertices in X, we say that H is an
induced subgraph of G. We also that that H is the subgraph induced by X.
A path in a graph G from a vertex u to a vertex w is a sequence of vertices
u = v1 , v2 , . . . , vk = w such that vi is adjacent to vi+1 for all 1 ≤ i ≤ k − 1.
A graph G is connected if there exists a path from each vertex of G to any
other vertex of G.
2
Zero forcing
Zero forcing was introduced independently by D. Burgarth and V. Giovanneti
in 2007 (see [2]) and the AIM Minimum Rank - Special Graphs Work Group
(the “AIM Group” for short) in 2008 (see [3]). Burgarth and Giovanneti
introduced zero forcing in the study of control of quantum systems; the AIM
1
group introduced zero forcing in order to bound the minimum rank of a graph
(the minimum rank problem is discussed in section 3).
Definition 1 (AIM Group, [3]). Let G be a simple, finite and undirected
graph. Color each vertex of G either black or white.
(a) Color change rule: If u is a black vertex of G and u has exactly one
white neighbor v, then u “forces” v to become black.
(b) Derived coloring: The derived coloring of G is the coloring obtained by
applying the color change rule as many times as possible.
(c) Zero forcing set: A nonempty set Z of vertices is called a zero forcing
set if, when all of the vertices in Z are colored black and all other
vertices of G are colored white, the derived coloring of G has no white
vertices.
(d) The zero forcing number of G, denoted Z(G) is the minimum |Z| over
all zero forcing sets Z.
Example 1. Consider the following graph G and coloring of G:
The set of black vertices is easily seen to be a zero forcing set: 4 forces 3, 3
(or 5) forces 6, and 6 (or 2) forces 1. Moreover, no set of one or two vertices
is a zero forcing set, so Z(G) = 3.
Example 2. The zero forcing number of a path is 1 (a single degree one
vertex is a zero forcing set).
Example 3. The zero forcing number of a cycle C is 2 (clearly no single
vertex forces, so Z(C) ≥ 2; any set containing a pair of adjacent vertices is
a zero forcing set, so Z(C) = 2).
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Remark. For any graph G with n vertices and at least one edge, Z(G) ≤ n−1
since the set containing every vertex of G except for one of the vertices
incident to an edge is a zero forcing set.
Proposition 1. For any graph G with minimum degree δ, Z(G) ≥ δ.
Proof. Let Z be a smallest zero forcing set for G. If Z contains every vertex
of G, we are done, since n ≥ δ + 1. If Z does not contain every vertex of
G, then there exists a u in Z that forces one of its neighbors v. This means
that all but one of u’s neighbors is in Z so that |Z| ≥ deg(u). Since Z is a
smallest zero forcing set, Z(G) ≥ δ.
Example 4. Let Kn be the complete graph on n-vertices. Then Z(Kn ) =
n − 1. To see this, note that by Proposition 1, Z(Kn ) ≥ n − 1, since the
minimum degree of Kn is n − 1. On the other hand, any set of n − 1 vertices
is a zero forcing set for Kn , and thus Z(Kn ) = n − 1.
Example 5. The zero forcing number of the Peterson graph (shown below)
is 5. The five vertices on the outer cycle constitute a zero forcing set, and
thus the zero forcing number is at most 5. It follows from Proposition 1 that
the zero forcing number is at least 3. That the zero forcing is not 3 or 4 is
left as an exercise.
3
Zero forcing and the minimum rank problem
Definition 2. Let A = (aij ) be a symmetric matrix over a field F .
(a) The graph of A, denoted G(A), is the graph with vertex set {1, 2, . . . , n}
and edge set {{i, j} : aij 6= 0, 1 ≤ i < j ≤ n}. Note that we do not
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consider the diagonal of A when determining G(A). Moreover, for any
graph G and field F , there exists a symmetric matrix A whose graph
is G (the adjacency matrix, for example).
(b) The minimum rank of a graph G over a field F , denoted mrF (G), is
the minimum rank over all matrices A for which G(A) ' G.
(c) The maximum nullity of a graph G over a field F , denoted M F (G), is
the maximum nullity over all matrices A for which G(A) ' G.
Remark. For any field F and graph G with n vertices, the rank-nullity relation
implies that mrF (G) + M F (G) = n. If F = R, we will omit the F in our
notation for minimum rank and maximum nullity. Finally, observe that for
any field F , mrF (G) ≤ n − 1.
The minimum rank problems asks, given a graph G and field F , what is
mrF (G).
Example 6. Let F = R and let P
mr(P ) ≤ 3. If we label the vertices of
of P is

0 1
1 0

0 1
0 0
be the path on 4 vertices. Clearly
P sequentially, the adjacency matrix

0 0
1 0

0 1
1 0
which has rank 3. This minimizes the rank among all matrices whose graph is
P , so mr(P ) = 3. In general, if Pn is the path on n vertices, mr(Pn ) = n − 1.
Example 7. Let F = R and let Kn be the complete graph on n vertices.
Clearly mr(Kn ) ≥ 1. Kn is the graph of the n × n matrix of all 1s, which has
rank 1, and thus mr(Kn ) = 1. Conversely, if G is a connected graph with n
vertices and mr(G) = 1, then G ' Kn .
We now explore the relationship between MF (G) and Z(G).
Definition 3. Let F be a field. The support of a vector x = (x1 , x2 , . . . , xn )
with components in F , is the set
supp(x) = {i : xi 6= 0}.
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Proposition 2 (AIM Group, [3]). If F is a field, A an n × n matrix with
entries in F and null(A) > k, then there is a nonzero vector x ∈ ker(A)
vanishing in any k specified positions.
The proof is omitted. The interested reader may try to prove the proposition
themselves, or find the proof in [3].
Proposition 3 (AIM Group, [3]). Let F be a field, Z a zero forcing set for
a graph G = (V, E), and let A be a symmetric matrix with entries in F such
that G(A) ' G. If x ∈ ker(A) and supp(x) ∩ Z = ∅, then x = 0.
Proof. If Z = V , we are done. If Z 6= V , then since Z is a zero forcing set for
G, we must be able to perform a color change. So there is a black vertex u
(note that xu is required to be 0) with exactly one neighbor v colored white
(so there is no restriction on xv , yet). Consider the u-th component of the
vector Ax. By the definition of multiplication of a vector by a matrix, this
is given by
X
auw xw .
(Ax)u =
w∈V
Writing u ∼ w to indicate that u is adjacent to w, we may expand the right
hand side of the above equation as
X
X
X
auw xw = auu xu + auv xv
auw xw +
auw xw = auu xu +
w∈V
w6∼u
w∼u
since auw = 0 for every w 6∼ u, xw = 0 if w ∈ Z and every neighbor of u is
in Z except for v. Then since xu = 0 and x ∈ ker(A), we must have xv = 0.
Each color change requires another component of x to be 0, and since Z is a
zero forcing set, this implies that x = 0.
Remark. The previous proposition explains the name zero forcing.
Proposition 4 (AIM Group, [3]). Let G = (V, E) be a graph and let Z ⊆ V
be a zero forcing set for G. Then M F (G) ≤ |Z|, and thus M F (G) ≤ Z(G).
Proof. Suppose M F (G) > |Z| and let A be a symmetric matrix with entries
in F such that G(A) ' G with null(A) > |Z|. By Proposition 2, there is a
nonzero vector x ∈ ker(A) that vanishes on all vertices in Z. But x = 0 by
Proposition 3, a contradiction.
Since mrF (G) + M F (G) = n, Proposition 4 implies that
mrF (G) = n − M F (G) ≥ n − Z(G).
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4
k-forcing
The k-forcing number is a generalization of the zero forcing number, and was
introduced by myself, Yair Caro, Randy Davila and Ryan Pepper in [1].
Definition 4 (DA, Y. Caro, R. Davila and R. Pepper, [1]). Let G be a
simple, finite and undirected graph and k a positive integer. Color each
vertex of G either black or white.
(a) k-color change rule: If u is a black vertex of G and u has at most k
white neighbors, then u “k-forces” each of these neighbors to become
black.
(b) Derived coloring: The derived coloring of G is the coloring obtained by
applying the k-color change rule as many times as possible.
(c) k-forcing set: A nonempty set F of vertices is called a k-forcing set if,
when all of the vertices in F are colored black and all other vertices of
G are colored white, the derived coloring of G has no white vertices.
(d) The k-forcing number of G, denoted Fk (G) is the minimum |F | over
all k-forcing sets F .
Remark. Any 1-forcing set is a zero forcing set, and any zero forcing set is a
1-forcing set. Thus F1 (G) = Z(G) for any graph G.
Example 8. Let k = 2 and consider the following graph G and coloring of
G:
The set of black vertices is easily seen to be a 2-forcing set, since 4 2-forces 3
and 5, and then 2 can 2-force 6 and 1. Thus F2 (G) ≤ 2. However, F2 (G) = 1,
since the set containing only the vertex 4 (or only the vertex 1) is a 2-forcing
set.
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Example 9. The k-forcing number of the path Pn is 1 for all k, since for
any k the set containing one of the degree one vertices of Pn is a k-forcing
set.
Remark. Any k-forcing set is a (k + 1)-forcing set, and thus Fk ≥ Fk+1 for
all positive integers k. Also, if G is connected and k is at least the maximum
degree of G, then Fk (G) = 1. For this reason, we often restrict k to be at
most the maximum degree.
Example 10. For any positive integer k, if Kn is the complete graph on n
vertices, then Fk (Kn ) = max{n − k, 1}. To see this, note that if k ≥ n − 1,
then Fk (G) = 1. Otherwise, if k ≤ n − 2, color any set of n − k vertices black
and the remaining vertices white. Each black vertex has exactly k white
neighbors, and thus k-forces all of the white vertices. Thus Fk (Kn ) ≤ n − k.
On the other hand, if at most n − k − 1 vertices are colored black, then each
black vertex has at least k + 1 white neighbors, so no vertex can k-force.
The following proposition gives a generalization of the lower bound for zero
forcing in Proposition 1.
Proposition 5 ([1]). Let G = (V, E) be a graph with minimum degree δ and
let k be a positive integer. Then Fk (G) ≥ δ − k + 1.
Proof. Let F be a smallest k-forcing set for G and color the vertices of F black
(and all other vertices white). If F = V , then |F | = |V |, so the inequality
is satisfied trivially. If F 6= V , then let v ∈ F − V be a white vertex that is
forced by u ∈ F . Since u forces, u must have at most deg(u) − k ≥ δ − k
black neighbors. Hence, counting u, there are at least δ − k + 1 vertices in
F.
Proposition 6 ([1]). If G is connected with maximum degree ∆ ≥ 1, then
F∆−1 (G) ≤ 2
with equality holding if and only if G is ∆-regular.
Proof. Let F = {u, v} be any set of two adjacent vertices. Color each vertex
in F black and all remaining vertices white. Each vertex in F has at most
∆−1 white neighbors, and thus (∆−1)-forces. Each vertex that was (∆−1)forced to become black by either u or v has at moast ∆ − 1 white neighbors,
and consequently (∆ − 1)-forces. Again, each newly colored black vertex has
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at most ∆ − 1 white neighbors, and this will be true at each step of applying
the (∆−1)-color change rule. Thus, since G is connected, the derived coloring
colors each vertex of G black. Hence F is a (∆ − 1)-forcing set, which means
that Fk (G) ≤ 2.
To see that equality holds if and only if G is ∆-regular, first observe
that if only a single vertex u of G is colored black, and all remaining vertice
are colored white, then u does not (∆ − 1)-force since u has at ∆ white
neighbors. Hence F∆−1 (G) ≥ 2, so that F∆−1 (G) = 2. The converse is
proven by contrapositive. If G is not ∆-regular, then there exists a vertex
u of G with deg(u) < ∆. Color u black and all other vertices of G white.
Then u has at most ∆ − 1 white neighbors and thus (∆ − 1)-forces. Each
vertex of that was (∆ − 1)-forced by u to become black has at most ∆ − 1
white neighbors, and this will be true at each step that the (∆ − 1)-color
change rule is applied. Thus the derived coloring colors each vertex of G
black, which means that F∆−1 (G) = 1.
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Upper bounds for the k-forcing number
In this section we surevey, mostly without proof, some upper bounds on the
k-forcing number of a graph. The missing proofs may be found in [1]. The
first result gives a general upper bound in terms of easily computed graph
invariants.
Theorem 7. Let k be a positive integer and let G = (V, E) be a graph on
n ≥ 2 vertices with maximum degree ∆ ≥ k and minimum degree δ ≥ 1.
Then
(∆ − k + 1)n
Fk (G) ≤
.
∆ − k + 1 + min{δ, k}
When the minimum degree of G is at most k, Theorem 7 has the following
immediate corollary:
Corollary 8. Let k be a positive integer and let G = (V, E) be a graph with
maximum degree ∆ ≥ k and minimum degree 1 ≤ δ ≤ k. Then
Fk (G) ≤
(∆ − k + 1)n
.
∆+1
The upper bound in Corollary 8 is sharp for the the complete graph Kn when
k ≤ n − 1. First, recall from Example 10 that Fk (Kn ) = n − k whenever
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k ≤ n − 1. The maximum degree and minimum degree of Kn are both n − 1,
so the bound in Corollary 8 gives
Fk (Kn ) =
(n − k)n
((n − 1) − k + 1)n
=
= n − k.
(n − 1) + 1
n
In the special case that k = 1, both Theorem 7 and Corollary 8 have the
following immediate corollary:
Corollary 9. Let G = (V, E) be a graph with maximum degree ∆ ≥ 1 and
minimum degree δ ≥ 1. Then
F1 (G) = Z(G) ≤
∆
n.
∆+1
Remark. The upper bound in Corollary 9 is also sharp for Kn . The appeal of
Corollary 9 is that it bounds ratio of the zero forcing number to the number
of vertices of a graph G. For example, if G is a graph with n vertices and
≤ 34 .
maximum degree ∆ = 3, then Corollary 9 implies that Z(G)
n
If we impose the mild restriction that a graph G is connected, we can
improve the bound in Theorem 7 (the improvement is found in Theorem 12
below). First, we need a few concepts. Let k be a positive integer. We
say that a graph G = (V, E) is k-connected if the subgraph induced by the
complement of any set of at most k − 1 vertices is connected.
Next we introduce the notion of a connected k-dominating set. Let G =
(V, E) be a connected graph and let D ⊂ V be any subset of vertices that
induces a connected subgraph. We say that D is a connected k-dominating
set for G if every vertex in V − D has at least k neighbors in D. The
following theorem shows that there is a relationship between k-forcing sets
and connected k-dominating sets.
Theorem 10. Let k be a positive integer and let G = (V, E) be a k-connected
graph with n > k vertices. If F is a smallest k-forcing set such that the subgraph induced by V −F is connected, then V −F is a connected k-dominating
set.
Let γk,c (G) denote the cardinality of a smallest connected k-dominating set
for G. We call γk,c (G) the connected k-domination number of G. Theorem
10 has the following corollary:
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Corollary 11. Let k be a positive integer and let G = (V, E) be a k-connected
graph with n > k vertices. Then
Fk (G) ≤ n − γk,c (G).
Proof. Any set S of n − 1 vertices of G is a k-forcing set for G, and the
subgraph induced by V −S is connected trivially. Thus there exists a smallest
k-forcing set F for G for which V − F induces a connected subgraph. By
Theorem 10, V −F is a connected k-dominating set for G. Since |F | ≥ Fk (G),
it follows that γk,c (G) ≤ |V − F | = |V | − |F | ≤ n − Fk (G). Rearranging gives
the desired upper bound.
The next theorem gives an upper bound for the k-forcing number of a kconnected graph in terms of the number of vertices, the maximum degree,
and k. While the proof is omitted, we remark that the proof uses the result
of Theorem 10.
Theorem 12. Let k be a positive integer and let G = (V, E) be a k-connected
graph with n > k vertices. If G has maximum degree ∆ ≥ 2, then
Fk (G) ≤
(∆ − 2)n + 2
.
∆+k−2
Remark. The upper bound in Theorem 12 is an improvement over the upper
bound in Theorem 7 whenever k = 1 or 2, and G is a k-connected with
n > k vertices and minimum degree δ ≥ k. This can be seen by comparing
the upper bounds in Theorems 7 and 12 when k = 1 and k = 2.
When k = 1, Theorem 12 has the following immediate corollary that is
reminiscent of Corollary 9:
Corollary 13. Let G be a connected graph with maximum degree ∆ ≥ 2.
Then
(∆ − 2)n + 2
.
F1 (G) = Z(G) ≤
∆−1
Remark. Corollary 13 is easily seen to be sharp for cycles, complete graphs,
and complete bipartite graphs with equal parts (a bipartite graph is any
graph whose vertices can be partition into two parts A and B such that any
edge has one end vertex in A and the other in B; a complete bipartite graph
contains every possible edge between the parts A and B). No other families
of graphs are known to satisfy equality in the upper bound. In fact, it is
conjectured that these families characterize the case of equality.
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6
Open problems
Here we list some open problems concering zero forcing and k-forcing. It
would be interesting if there is any connection between k-forcing and the
minimum rank problem. The first problem considers one way in which such
a relationship could be made.
Problem 1. Let k be a positive integer, F a field and G = (V, E) a graph.
Is there a constant ck (G) such that M F (G) ≤ ck (G)Fk (G) ≤ Z(G)?
Another possible application of zero forcing and k-forcing is in the study
of Cayley graphs of finite groups.
Problem 2. Let G be a finite group. For every positive integer k, we could
define Fk (G) to be the k-forcing number of the Cayley graph of G. Is Fk (G)
related to any properties of the group?
The definition of zero forcing given by the AIM Group was motivated by
a problem in linear algebra. The next problem is a natural question in this
setting.
Problem 3. Let k be a positive integer, G a graph and A the adjacency
matrix of G. Is Fk (G) related to the spectrum of A? Are there any bounds
for Fk in terms of the spectrum of A?
Finally, one would expect the k-forcing number of an edge dense graph
to be larger than the k-forcing number of a graph with the same number of
vertices but few edges. How well does Fk (G) measure the edge density of G?
References
[1] David Amos, Yair Caro, Randy Davila, and Ryan Pepper, Upper bounds
on the k-forcing number of a graph, Discrete Applied Mathematics (2014),
Accepted, in press.
[2] Daniel Burgarth and Vittorio Giovannetti, Full control by locally induced
relaxation, Phys. Rev. Lett. 99 (2007), 100501.
[3] AIM Minimum Rank Special Graphs Work Group, Zero forcing sets and
the minimum rank of graphs, Linear Algebra and its Applications 428
(2008), no. 7, 1628 – 1648.
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