PATH COVER NUMBER, MAXIMUM NULLITY, AND ZERO

FORCING NUMBER OF ORIENTED GRAPHS AND OTHER

SIMPLE DIGRAPHS

ADAM BERLINER∗, CORA BROWN† , JOSHUA CARLSON‡ , NATHANAEL COX∗,

LESLIE HOGBEN§ , JASON HU¶, KATRINA JACOBS‖, KATHRYN MANTERNACH∗∗,

TRAVIS PETERS††, NATHAN WARNBERG‡ , AND MICHAEL YOUNG‡

April 30, 2014

Abstract. An oriented graph is a simple digraph obtained from a simple graph by choosing1

exactly one of the two arcs (u, v) or (v, u) to replace each edge {u, v}. A simple digraph describes2

the zero-nonzero pattern of off-diagonal entries of a family of (not necessarily symmetric) matrices.3

Minimum rank of a simple digraph is the minimum rank of this family of matrices; maximum nullity4

is defined analogously. The simple digraph zero forcing number and path cover number are related5

parameters. We establish bounds on the range of possible values of all these parameters for oriented6

graphs, establish connections between the values of these parameters for a simple graph G, for various7

orientations ~G and for the doubly directed digraph←→G , and establish an upper bound on the number8

of arcs in a simple digraph in terms of the zero forcing number.9

Keywords. zero forcing number, maximum nullity, minimum rank, path cover num-10

ber, simple digraph, oriented graph11

AMS subject classifications. 05C50, 05C20, 15A0312

1. Introduction. The maximum nullity and zero forcing number of simple di-13

graphs are studied in [10] and [5]. We study connections between these parameters14

and path cover number, and we study all these parameters for special types of di-15

graphs derived from graphs, including oriented graphs and doubly directed graphs.16

Section 2 considers oriented graphs. We establish a bound on the difference of values17

of the parameters path cover number, maximum nullity, and zero forcing number for18

∗Department of Mathematics, Statistics, and Computer Science, St. Olaf College, Northfield, MN55057 (berliner@stolaf.edu), (coxn@stolaf.edu). Research of N. Cox supported by NSF DMS 0750986†Department of Mathematics, Carleton College, Northfield, MN 55057 (brownc@carleton.edu).

Research supported by NSF DMS 0750986.‡Department of Mathematics, Iowa State University, Ames, IA 50011 (jmsdg7@iastate.edu),

(warnberg@iastate.edu), (myoung@iastate.edu). Research of J. Carlson supported by ISU Holl funds.Research of M. Young supported by NSF DMS 0946431.§Department of Mathematics, Iowa State University, Ames, IA 50011, (lhogben@iastate.edu) and

American Institute of Mathematics, 360 Portage Ave, Palo Alto, CA 94306 (hogben@aimath.org).¶Department of Mathematics, University of California, Berkeley, CA 94720-3840 (ja-

son.hu@berkeley.edu). Research supported by NSF DMS 0750986.‖Department of Mathematics, Pomona College, 610 North College Avenue, Claremont, CA 91711

(klj02011@mymail.pomona.edu). Research supported by NSF DMS 0750986.∗∗Department of Mathematics and Computer Science, Central College, Pella, IA 50219 (manter-

nachk1@central.edu). Research supported by ISU Holl funds.††Natural and Mathematical Science Division, Culver-Stockton College, Canton, MO 63435

(tpeters319@gmail.com).

1

two orientations of one graph, and determine the range of values of these parameters19

for orientations of paths and cycles, and some of the possible values for tournaments.20

We establish connections between these parameters for a simple graph and its dou-21

bly directed digraph in Section 3. In Section 4, we establish an upper bound on the22

number of arcs of a simple digraph in terms of the zero forcing number. We also show23

that several results for simple graphs fail for oriented graphs, including the Graph24

Complement Conjecture and Sinkovic’s theorem that maximum nullity is at most25

path cover number for outerplanar graphs.26

All graphs and digraphs are taken to be simple. We use G = (V (G), E(G)) to27

denote a graph and Γ = (V (Γ), E(Γ)) to denote a digraph, often using V and E when28

G or Γ is clear. For a digraph Γ and R ⊆ V , the induced subdigraph Γ[R] is the29

digraph with vertex set R and arc set {(v, w) ∈ E : v, w ∈ R}; an analogous definition30

is used for graphs. The subdigraph induced by the complement R is also denoted31

by Γ − R, or in the case R is a single vertex v, by Γ − v. A digraph Γ = (V,E) is32

transitive if for all u, v, w ∈ V , (u, v), (v, w) ∈ E implies (u,w) ∈ E.33

For a digraph Γ = (V,E) having v, u ∈ V and (v, u) ∈ E, u is an out-neighbor of v34

and v is an in-neighbor of u. The out-degree of v, denoted by deg+(v), is the number35

of out-neighbors of v in Γ, and analogously for in-degree, denoted by deg−(v). Define36

δ+(Γ) = min{deg+(v) : v ∈ V } and δ−(Γ) = min{deg−(v) : v ∈ V }. For a digraph Γ,37

the reversal ΓT is obtained from Γ by reversing all the arcs.38

Let G be a graph. A path in G is a subgraph P = ({v1, . . . , vk}, E(P )) where39

E(P ) = {{vi, vi+1} :≤ i ≤ k − 1}; this path is often denoted by (v1, . . . , vk) and its40

length is k−1. We say that a path in G is an induced path if it is an induced subgraph41

of G. A path cover of a graph G is a set of vertex disjoint induced paths that includes42

all vertices of G.43

Now suppose Γ is a digraph. A path in Γ is a subdigraph P = ({v1, . . . , vk}, E(P ))44

where E(P ) = {(vi, vi+1) : 1 ≤ i ≤ k − 1}; this path is often denoted by (v1, . . . , vk),45

its length is k − 1, and the arcs of E(P ) are called path arcs. If (v1, . . . , vk) is a path46

in Γ, v1 is called the initial vertex and vk is the terminal vertex. We say vertex u has47

access to v in Γ if there is a path from u to v. A path (v1, . . . , vk) in Γ is an induced48

path if E does not contain any arc of the form (vi, vj) with j > i + 1 or i > j + 1.49

We note this does not necessarily imply that the path subdigraph is induced, because50

any of the arcs in {(vi+1, vi) : 1 ≤ i ≤ k − 1} are permitted. A path (v1, . . . , vk) in Γ51

is Hessenberg if E does not contain any arc of the form (vi, vj) with j > i + 1. Any52

induced path is Hessenberg but not vice versa. A path cover for Γ is a set of vertex53

disjoint Hessenberg paths that includes all vertices of Γ [10].54

For graphs G and digraphs Γ, the path cover number P(G) or P(Γ) is the minimum55

number of paths in a path cover (induced for a graph, Hessenberg for a digraph) and56

a minimum path cover is a path cover with this minimum number of paths.57

Zero forcing was introduced in [1] for (simple) graphs. We define zero forcing58

for (simple) digraphs as in [10]. Let Γ be a digraph with each vertex colored either59

2

white or blue1. The color change rule is: If u is a blue vertex of Γ, and exactly one60

out-neighbor v of u is white, then change the color of v to blue. In this situation,61

we say that u forces v and write u → v. Given a coloring of Γ, the final coloring is62

the result of applying the color change rule until no more changes are possible. A63

zero forcing set for Γ is a subset of vertices B such that, if initially the vertices of64

B are colored blue and the remaining vertices are white, the final coloring of Γ is all65

blue. The zero forcing number Z(Γ) is the minimum of |B| over all zero forcing sets66

B ⊆ V (Γ).67

For a given zero forcing set B for Γ, we create a chronological list of forces by68

constructing the final coloring, listing the forces in the order in which they were69

performed. Although for a given set of vertices B the final coloring is unique, B need70

not have a unique chronological list of forces. Suppose Γ is a digraph and F is a71

chronological list of forces for a zero forcing set B. A forcing chain is an ordered set72

of vertices (w1, w2, . . . , wk) where wj → wj+1 is a force in F for 1 ≤ j ≤ k − 1. A73

maximal forcing chain is a forcing chain that is not a proper subset of another forcing74

chain. The following results will be used.75

Lemma 1.1. [10] Suppose Γ is a digraph and F is a chronological list of forces76

of a zero forcing set B. Then, every maximal forcing chain is a Hessenberg path that77

starts with a vertex in B.78

For a fixed chronological list of forces F of a zero forcing set B of Γ, the chain79

set is the set of all maximal forcing chains. By Lemma 1.1, the chain set of F is a80

path cover, called a zero forcing path cover and the maximal forcing chains are also81

called forcing paths.82

Proposition 1.2. [10] For any digraph Γ, P(Γ) ≤ Z(Γ).83

A cycle of length k ≥ 3 in a graph G or digraph Γ is a sub(di)graph consisting of84

a path (v1, . . . , vk) and the additional edge or arc {vk, v1} or (vk, v1).85

Lemma 1.3. Suppose P = (v1, . . . , vk) is a Hessenberg path in a digraph Γ. Then86

P is an induced path or Γ[V (P )] contains a (digraph) cycle of length at least 3.87

Proof. Suppose P is not an induced path. Then Γ must contain an arc of the88

form (vi, vj) with j > i + 1 or i > j + 1. Since P is Hessenberg, Γ does not contain89

an arc of the form (vi, vj) with j > i + 1. Thus Γ must contain an arc of the form90

(vi, vj) with i > j + 1. Then (vj , vj+1, . . . , vi, vj) is a (digraph) cycle in Γ[V (P )].91

Let F be a field. For a square matrix A = [aij ] ∈ Fn×n, the digraph of A,92

denoted Γ(A) = (V,E), is the (simple) digraph described by the off-diagonal zero-93

nonzero pattern of the entries: the set of vertices is V = {1, 2, . . . , n} and the set of94

arcs is E = {(i, j) : aij 6= 0, i 6= j}. Note that the value of the diagonal entries of A95

does not affect Γ(A).96

Conversely, given any simple digraph Γ (along with an ordering of the vertices),97

1The early literature uses the color black rather than blue.

3

we may associate with Γ a family of matrices MF (Γ) = {A ∈ Fn×n : Γ(A) = Γ}.98

The minimum rank over F of a digraph Γ is mrF (Γ) = min{rankA : A ∈ MF (Γ)}99

and the maximum nullity over F of Γ is MF (Γ) = max{null A : A ∈ MF (Γ)}. It is100

immediate that that mrF (Γ) + MF (Γ) = n.101

Similarly, symmetric matrices and undirected graphs are associated. For a sym-102

metric matrix A = [aij ] ∈ Fn×n, the graph of A is the (simple) graph G(A) = (V,E)103

with V = {1, 2, . . . , n} and E = {{i, j} : i 6= j and aij 6= 0}. The family of symmetric104

matrices associated with G is SF (G) = {A ∈ Fn×n : AT = A,G(A) = G}, and105

minimum rank and maximum nullity are similarly defined for undirected graphs.106

For the much of this paper, we let F = R and we write S(G),M(Γ),M(Γ), and107

mr(Γ) rather than SR(G),MR(Γ),MR(Γ), and mrR(Γ), etc. If a graph or digraph108

parameter that depends on matrices does not change regardless of the field F , then109

we say that that parameter is field independent; in this case MF (Γ) = M(Γ) for every110

field F .111

Remark 1.4. Clearly mrF (ΓT) = mrF (Γ), and Z(ΓT) = Z(Γ) is known [5].112

Because the reversal of a Hessenberg path is a Hessenberg path, P(ΓT) = P(Γ).113

2. Oriented graphs. In this section we establish results for minimum rank,114

maximum nullity, zero forcing number, and path cover number of oriented graphs.115

Given a graph G, an orientation ~G of G is a digraph obtained by replacing each116

edge {u, v} by exactly one of the arcs (u, v) and (v, u) (so a graph G has 2|E(G)|117

orientations, some of which may be isomorphic to each other).118

2.1. Range over orientations. We consider the range of values of β(~G) over119

all possible orientations, for the parameters β = mr,M,Z,P.120

Theorem 2.1. Suppose β is a positive-integer-valued digraph parameter with the121

following properties for every oriented graph ~G:122

(1) β(~GT) = β(~G).123

(2) If (u, v) ∈ E(~G) and ~G0 is obtained from ~G by replacing (u, v) by (v, u) (i.e.,124

reversing the orientation of one arc), then |β(~G0)− β(~G)| ≤ 1.125

Then for any two orientations ~G1 and ~G2 of the same graph G,126

|β(~G2)− β(~G1)| ≤⌊E(G)

2

⌋.127

Furthermore, every integer between β(~G2) and β(~G1) is attained as β(~G) for some128

orientation ~G of G.129

Proof. Without loss of generality, β(~G2) ≥ β(~G1). Let e = |E(G)|. Because130

~G1 and ~G2 share the same underlying graph, it is possible to obtain ~G2 from ~G1 by131

reversing some of the arcs of ~G1. Let ` be the number of arcs we need to reverse to132

obtain ~G2 from ~G1. By hypothesis, reversing the direction of one arc changes the133

value of β by at most one, so β(~G2) − β(~G1) ≤ `. The number of arcs that must be134

4

reversed to obtain ~GT2 from ~G1 is e − `, so β(~GT

2 ) − β(~G1) ≤ e − `. By hypothesis,135

β(~GT2 ) = β(~G2), so β(~G2)−β(~G1) ≤ b e2c. The last statement follows from hypothesis136

(2) and the fact that we can go from ~G1 to ~G2 by reversing one arc at a time.137

Corollary 2.2. If ~G1 and ~G2 are both orientations of the graph G, then:138

|mr(~G2)−mr(~G1)| ≤ bE(G)

2c,139

|M(~G2)−M(~G1)| ≤ bE(G)

2c,140

|Z(~G2)− Z(~G1)| ≤ bE(G)

2c, and141

|P(~G2)− P(~G1)| ≤ bE(G)

2c.142

Furthermore, every integer between β(~G2) and β(~G1) is attained as β(~G) for some143

orientation ~G of G when β is any of the parameters mr,M,Z,P.144

Proof. The first hypothesis of Theorem 2.1, β(~GT) = β(~G), is established for these145

parameters in Remark 1.4. To show these parameters satisfy the second hypothesis146

of Theorem 2.1, suppose arc (u, v) of ~G is reversed to obtain ~G0 from ~G. In each case,147

the process is reversible, so it suffices to prove β(~G0) ≤ β(~G) + 1.148

For minimum rank, suppose Γ(A) = ~G and rankA = mr(~G). Define B by buu =149

bvv = buv = bvu = −auv and bij = 0 for all other entries of B. Then Γ(A+ B) = ~G0150

and rank(A + B) ≤ rankA + 1. Thus mr(~G0) ≤ mr(~G) + 1. The statement for151

maximum nullity is equivalent.152

For zero forcing number, choose a minimum zero forcing set B and chronological153

list of forces F of ~G. If the force u→ v is in F , then B ∪ {v} is a zero forcing set for154

~G0. If u 6→ v and for some w, v → w is in F , then B ∪ {u} is a zero forcing set for155

~G0. If v does not perform a force and u→ v is not in F , then B is a zero forcing set156

for ~G0. Thus, Z(~G0) ≤ Z(~G) + 1.157

For path cover number, suppose P = {P (1), P (2), . . . , P (k)} is a path cover of ~G158

and |P| = P(~G). If (u, v) is not an arc in one of the paths in P, then P is a path159

cover for ~G0 and P(~G0) ≤ P(~G). So suppose (u, v) is an arc in some path P (`). Then160

we construct a path cover for ~G0 by replacing P (`) by the two paths resulting from161

deleting the arc (u, v). Thus, P(~G0) ≤ P(~G) + 1.162

2.2. Hierarchal orientation. We establish a method for finding an orientation163

~G of a graph G for which P(~G) = P(G). Let P = {P (1), P (2), . . . , P (k)} be any164

path cover of a graph G. A rooted path cover of G, R = {R(1), R(2), . . . , R(k)}, is165

obtained from P by choosing one endpoint as the root of P (i) for each i = 1, . . . , k.166

R is a minimum rooted path cover if |R| = P(G). In the case P is a zero forcing167

path cover of a zero forcing set B, then the root of P (i) is automatically chosen to168

be the unique element of B that is a vertex of P (i). A rooted path cover obtained169

5

from P naturally orders V (P (i)), starting with the root, and we denote this order by170

R(i) = (r(i)1 , r

(i)2 , . . . , r

(i)si ) where si − 1 is the length of P (i). Observe that if a rooted171

path cover is formed from a zero forcing path cover of a zero forcing set, the ordering172

within each rooted path coincides with the forcing order in that path.173

Definition 2.3. Given a rooted path cover R of a graph G, the hierarchal174

orientation_GR of G resulting from R is defined by orienting G as follows:175

(1) Orient R(i) as r(i)1 → r

(i)2 → · · · → r

(i)si ; that is, replace the edge {r(i)j , r

(i)j+1}176

by the arc (r(i)j , r

(i)j+1) for j = 1, . . . , si − 1.177

(2) For any edge between R(i) and R(j) with i < j, orient as i → j; that is, if178

i < j, replace the edge {r(i)`i , r(j)`j} by the arc

(r(i)`i, r

(j)`j

).179

Since by definition, the paths in a path cover of a graph are induced, all the edges of180

G have been oriented by these two rules.181

Observation 2.4. For any rooted path cover R of G, R is a path cover of_GR182

(with each path originating at its root), so P(_GR) ≤ |R|.183

Proposition 2.5. An oriented graph ~G is the hierarchal orientation_GR of G184

for some rooted path cover R of G (not necessarily minimum) if and only if ~G does185

not contain a digraph cycle.186

Proof. Suppose R = {R(1), . . . , R(k)} is rooted path cover of G. Since each path187

R(i) is induced, in order for_GR to have a digraph cycle, V (

_GR) would have to include188

vertices from at least two paths R(i) and R(j), with i < j. But by definition of_GR,189

there are no arcs from vertices in R(j) to vertices in R(i).190

Suppose ~G does not contain a digraph cycle. Then we may order the vertices191

{v1, . . . , vn}, vj does not have access to vi whenever j > i. Then if V (R(i)) = {vi},192

R = {R(1), . . . , R(n)} is a rooted path cover and ~G =_GR.193

Theorem 2.6. Suppose R = {R(1), . . . , R(k)} is a rooted path cover of G and194

_GR is the hierarchal orientation of G resulting from R. Then any path cover for

_GR195

is a path cover for G. If R is a minimum rooted path cover, then P(G) = P(_GR).196

Proof. Let P be a Hessenberg path in_GR. By Proposition 2.5,

_GR does not197

contain a digraph cycle, so by Lemma 1.3, P is an induced path. Thus, any path198

cover for_GR is a path cover for G, and this implies P(G) ≤ P(

_GR). IfR is a minimum199

rooted path cover of G, then P(_GR) ≤ |R| = P(G) ≤ P(

_GR), so P(G) = P(

_GR).200

Example 2.7. Not every orientation ~G having P(~G) = P(G) is a hierarchal201

orientation. The oriented graph ~G shown in Figure 2.1 has P(~G) = 2 = P(G), but ~G202

is not a hierarchal orientation because ~G contains a digraph cycle.203

6

321 4

Fig. 2.1. An oriented graph ~G that is not a hierarchal orientation but has P( ~G) = P(G).

Although for any graph G we can find an orientation so that P(~G) = P(G), this204

is not always the case for zero forcing number or maximum nullity.205

Example 2.8. Consider K4, the complete graph on four vertices. It is well206

known that M(K4) = Z(K4) = 3, whereas we show that for any orientation ~K4 of207

K4, 2 ≥ Z( ~K4) ≥ M( ~K4). If ~K4 contains a directed 3-cycle, then any one vertex on208

the 3-cycle and the remaining vertex form a zero forcing set. If ~K4 has no directed209

3-cycle, then we may order the vertices {u1, u2, u3, u4} where uj does not have access210

to ui whenever j > i. Then, {u1, u3} is a zero forcing set.211

Observation 2.9. If R = {R(1), . . . , R(k)} is a rooted path cover for G, then the212

set of roots {r(1)1 , . . . , r(k)1 } is a zero forcing set of the digraph

_GR, as zero forcing can213

be done in path order along R(k), followed by R(k−1), etc.214

Theorem 2.10. Suppose G is a graph and R is a minimum rooted path cover of215

G. Then Z(_GR) = P(

_GR) = P(G).216

Proof. From Theorem 2.6, Proposition 1.2, Observation 2.9, and the hypotheses,217

P(G) = P(_GR) ≤ Z(

_GR) ≤ |R| = P(G).218

Whenever P(G) = Z(G), we can use a minimum rooted path cover to find an219

orientation of G realizing Z(G) as its zero forcing number.220

Corollary 2.11. Suppose G is a graph such that P(G) = Z(G) and R is a221

minimum rooted path cover of G. Then Z(_GR) = Z(G).222

Because P(T ) = Z(T ) for every (simple undirected) tree T [1], we have the fol-223

lowing corollary.224

Corollary 2.12. If T is a tree, then there exists an orientation ~T of T such225

that Z(~T ) = Z(T ).226

If we allow a path cover that is not a minimum path cover, it is not difficult to find227

a graph and rooted path cover R with P(_GR) < Z(

_GR) (in fact, P(

_GR) < M(

_GR)).228

Example 2.13. Let G be the double triangle graph shown in Figure 2.2(a)229

and consider the rooted path cover of G defined by R = {R(1), R(2), R(3)} where230

V (R(1)) = {1}, V (R(2)) = {3}, and V (R(3)) = {2, 4} with 2 as the root of R(3). The231

hierarchal orientation_GR is shown in Figure 2.2(b).232

7

1

2

3

4

1

2

3

4

(a) (b)

Fig. 2.2. An hierarchal orientation_GR having P(

_GR) < M(

_GR) = Z(

_GR).

Then P(_GR) = 2 because paths (1, 2) and (3, 4) cover all vertices, and the vertices233

1 and 3 must each be initial vertices of any path they are in. Let A =

0 1 0 1

0 1 0 1

0 1 0 1

0 0 0 0

.234

Then Γ(A) =_GR and nullityA = 3. The set {1, 2, 3} is a zero forcing set for

_GR so235

3 ≤ M(_GR) ≤ Z(

_GR) ≤ 3.236

Every graph G we have examined has M(_GR) = P(

_GR) for minimum rooted path237

covers R, but these examples all involve a small number of vertices.238

Question 2.14. Does M(_GR) = P(

_GR) if R is a minimum rooted path cover of239

G?240

2.3. Tournaments. A tournament is an orientation of of the complete graph241

Kn. In this section we consider the possible values of path cover number, maximum242

nullity, and zero forcing number for tournaments.243

Example 2.15. We create an orientation of Kn by labeling the vertices {1, . . . , n}244

and by orienting the edges {u, v} as (u, v) if and only if v < u− 1 or v = u+ 1. The245

resulting orientation is called the Hessenberg tournament of order n, denoted ~K(H)n .246

This is the Hessenberg path on n vertices containing all possible arcs except those247

of the form (u+ 1, u) for 1 ≤ u ≤ n − 1. Since the zero forcing number of any248

Hessenberg path is one, P( ~K(H)n ) = M( ~K

(H)n ) = Z( ~K

(H)n ) = 1. Observe that ~K

(H)n is249

self-complementary.250

Example 2.16. Label the vertices of Kn by {1, . . . , n} and orient the edges {u, v}251

as (u, v) if and only if u < v. The resulting orientation is the transitive tournament,252

denoted ~K(T )n . We show that P( ~K

(T )n ) = Z( ~K

(T )n ) = M( ~K

(T )n ) =

⌈n2

⌉for any n. Let A253

be the adjacency matrix of ~K(T )n , and let D = diag(0, 1, 0, 1, . . . ). Then Γ(A+D) =254

~K(T )n and nullity(A + D) =

⌈n2

⌉because A + D has

⌊n2

⌋duplicate rows and if n is255

odd, an additional row of zeros. The set of odd numbered vertices, B = {1, 3, . . . , },256

is a zero forcing set. Thus,⌈n2

⌉≤ M( ~Kn) ≤ Z( ~Kn) ≤

⌈n2

⌉. Furthermore, from the257

8

definition of ~K(T )n , no more than 2 vertices can be on the same Hessenberg path.258

Proposition 2.17. For any tournament ~Kn, 1 ≤ P( ~Kn) ≤⌈n2

⌉, and for every259

integer k with 1 ≤ k ≤⌈n2

⌉, there is an orientation ~Kn having P( ~Kn) = k.260

For every integer k with 1 ≤ k ≤⌈n2

⌉, there is an orientation ~Kn having Z( ~Kn) = k.261

Proof. For both P and Z, ~K(H)n (Example 2.15) realizes the lower bound and ~K

(T )n262

(Example 2.16) realizes the upper bound. For the upper bound on attainable path263

cover numbers, partition the vertices of ~Kn into⌈n2

⌉sets of size two or one. Each264

pair of vertices and the arc between them forms a path. The assertion that all values265

for P and Z between 1 and⌈n2

⌉are possible follows from Corollary 2.2.266

For n ≤ 7, the transitive tournament ~K(T )n achieves the highest zero forcing267

number; i.e., for all orientations ~Kn, Z( ~Kn) ≤⌈n2

⌉(this has been verified using the268

program [12] written in Sage). But for n = 8, there exists a tournament having max-269

imum nullity greater than that of the transitive tournament, as in the next example.270

Example 2.18.

1

2 3

4

5

67

8

Fig. 2.3. A tournament ~K8 having M( ~K8) = Z( ~K8) = 5 > d 82e.

271

Let ~K8 be the tournament shown in Figure 2.3. Observe that {1, 2, 3, 4, 8} is a272

zero forcing set for ~K8 so Z( ~K8) ≤ 5. The matrix273

A =

0 1 2 1 2 3 1 2

0 0 1 1 1 1 0 0

0 0 1 1 1 1 0 0

0 0 0 0 1 1 1 1

0 0 0 0 1 1 1 1

0 0 0 0 0 0 0 0

0 1 1 0 0 1 0 1

0 1 1 0 0 1 0 1

274

has rankA = 3 and Γ(A) = ~K8, so 5 ≤ M( ~K8). Thus, Z( ~K8) = M( ~K8) = 5 > 4 =275

d 82e. We also show that P( ~K8) = 3: Since {(2, 4, 8), (3, 5, 7), (1, 6)} is a path cover,276

9

P( ~K8) ≤ 3. There are no induced paths of length greater than two in ~K8, so by277

Lemma 1.3 any path of length three or more must have a cycle. Thus vertices 1 and278

6 must be in paths of length at most two. If they are in separate paths in a path279

cover P, then |P| ≥ 3. So assume (1, 6) is a path in P. Since ~K8 − {1, 6} is not a280

(Hessenberg) path, |P| ≥ 3.281

There are also examples of tournaments ~Kn for which M( ~Kn) < Z( ~Kn).282

Proposition 2.19. The tournament ~K7 shown in Figure 2.4 has P( ~K7) = 2,283

M( ~K7) = 3, and Z( ~K7) = 4.

1 2

3

4

56

7

Fig. 2.4. A tournament ~K7 having M( ~K7) = 3 < 4 = Z( ~K7).

284

Proof. Because {(4, 6, 1, 3), (2, 5, 7)} is a path cover for ~K7, and ~K7 is not a285

Hessenberg path, P( ~K7) = 2.286

Next we show M( ~K7) ≤ 3. Suppose Γ(A) = ~K7. The nonzero pattern of A is287

? ∗ ∗ ∗ ∗ 0 0

0 ? ∗ ∗ ∗ 0 0

0 0 ? ∗ ∗ ∗ ∗0 0 0 ? ∗ ∗ ∗0 0 0 0 ? ∗ ∗∗ ∗ 0 0 0 ? ∗∗ ∗ 0 0 0 0 ?

288

where ∗ denotes a nonzero entry and ? may have any real value. Rows 1, 5, and 7 are289

necessarily linearly independent, by considering columns 2, 3, and 6.290

For A to achieve nullity 4, rankA = 3 and thus all the remaining rows must be in291

the span of rows 1, 5, and 7. We show this is impossible, implying that mr( ~K7) ≥ 4292

and M( ~K7) ≤ 3. Once that is done, construct a matrix A with Γ(A) = ~K7 and293

rankA = 4 by setting all nonzero off-diagonal entries to 1 and setting the diagonal294

entries as aii = 0 for i odd and aii = 1 for i even, so M( ~K7) = 3.295

If a11 = 0, then row 3 cannot be expressed as a linear combination of rows 1,296

5, and 7: By considering column 1, the coefficient of row 7 must be zero, which297

10

implies that then the coefficient of row 1 must be zero by considering column 2, but298

by considering column 4, row 3 is not a multiple of row 5. Thus a11 6= 0.299

If a77 6= 0, then row 2 cannot be expressed as a linear combination of rows 1, 5,300

and 7: By considering column 6 the coefficient of row 5 must be zero, which implies301

that the coefficient of row 7 must be zero by considering column 7, but by considering302

column 1, row 2 is not a multiple of row 1 (because a11 6= 0). Thus a77 = 0.303

If a55 = 0, then row 4 cannot be expressed as a linear combination of rows 1, 5,304

and 7: By considering column 3 the coefficient of row 1 must be zero, which implies305

that the coefficient of row 7 must be zero by considering column 1. But row 4 is not306

a multiple of row 5 (because a55 = 0). Thus a55 6= 0.307

Now row 6 cannot be expressed as a linear combination of rows 1, 5, and 7: By308

considering column 3 the coefficient of row 1 must be zero. By considering column309

5 the coefficient of row 5 must be zero. Now by considering column 7, row 6 is not310

a scalar multiple of row 7. Therefore row 6 is not a linear combination of rows 1, 5,311

and 7, rankA ≥ 4, and mr( ~K7) ≥ 4.312

Finally we show that Z( ~K7) = 4. Observe that any zero forcing set must contain a313

vertex from the set {1, 2}: If 1 and 2 are initially colored white, the only vertices that314

can force them are 6 and 7, but 1 and 2 are both out-neighbors of 6 and of 7. Observe315

that any zero forcing set must contain a vertex from the set {6, 7}: If 6 and 7 are316

initially colored white, the only vertices that can force them are 3, 4, and 5, but 6 and317

7 are both out-neighbors of 3, of 4, and of 5. Observe that any zero forcing set must318

contain a vertex from the set {3, 4}: If 3 and 4 are initially colored white, the only319

vertices that can force them are 1 and 2, but 3 and 4 are both out-neighbors of 1 and320

of 2. Observe that any zero forcing set must contain a vertex from the set {4, 5}: If 4321

and 5 are initially colored white, the only vertices that can force them are 1, 2, and 3,322

but 4 and 5 are both out-neighbors of 1, of 2, and of 3. Hence, a zero forcing set must323

contain at least four vertices, unless vertex 4 is the only vertex from {3, 4} and {4, 5}324

selected. However, by inspection the sets {1, 4, 6}, {1, 4, 7}, {2, 4, 6}, {2, 4, 7} are not325

zero forcing sets. The set {1, 2, 4, 6} is a zero forcing set for ~K7, and so Z( ~K7) = 4.326

2.4. Orientations of paths. In this section we consider the possible values of327

path cover number, maximum nullity, and zero forcing number for orientations of328

paths.329

Example 2.20. Starting with the path Pn, label the vertices in path order by330

{1, . . . , n} and orient the edge {i, i + 1} as arc (i, i + 1) for i = 1, . . . , n − 1. The331

resulting orientation is the path orientation of Pn, denoted ~P(H)n . Then P(~P

(H)n ) =332

M(~P(H)n ) = Z(~P

(H)n ) = 1.333

Example 2.21. Starting with the path Pn, label the vertices in path order by334

{1, . . . , n} and orient the edges as follows: Orient {1, 2} as (1, 2). For i = 1, . . . ,⌊n2

⌋−335

1, orient {2i+ 1, 2i} and {2i+ 1, 2i+ 2} as (2i+ 1, 2i) and (2i+ 1, 2i+ 2). If n is odd,336

orient {n− 1, n} as (n, n− 1). The resulting orientation is the alternating orientation337

11

of Pn, denoted ~P(A)n . Note that all odd-numbered vertices have in-degree zero. So338

P(~P(A)n ) = M(~P

(A)n ) = Z(~P

(A)n ) =

⌈n2

⌉because the odd vertices form a minimum zero339

forcing set, there is no directed path of length greater than one, and the adjacency340

matrix A of ~P(A)n has rank

⌊n2

⌋.341

Proposition 2.22. For any oriented path ~Pn, 1 ≤ P(~Pn) ≤ dn2 e, 1 ≤ M(~Pn) ≤342

dn2 e, and 1 ≤ Z(~Pn) ≤ dn2 e. For every integer k with 1 ≤ k ≤⌈n2

⌉, there are (possibly343

three different) orientations ~Pn having P(~Pn) = k, M(~Pn) = k, and Z(~Pn) = k.344

Proof. The proof of the second statement follows from Examples 2.20 and 2.21345

and Corollary 2.2. To complete the proof we show that Z(~Pn) ≤⌈n2

⌉for every346

orientation ~Pn. Apply Corollary 2.2 to the given orientation ~Pn and ~P(H)n . Then347

Z(~Pn)−Z(~P(H)n ) ≤

⌊n−12

⌋, so Z(~Pn) ≤

⌊n−12

⌋+ 1. If n is odd,

⌊n−12

⌋+ 1 = n−1

2 + 1 =348 ⌈n2

⌉. If n is even,

⌊n−12

⌋+ 1 = (n2 − 1) + 1 =

⌈n2

⌉. Therefore Z(~Pn) ≤

⌈n2

⌉.349

2.5. Orientations of cycles. In this section we consider the possible values350

of path cover number, maximum nullity, and zero forcing number for orientations of351

cycles of length at least 4 (since a cycle of length 3 is a complete graph).352

Example 2.23. Starting with the cycle Cn, label the vertices in cycle order by353

{1, . . . , n} and orient the edge {i, i+ 1} as arc (i, i+ 1) for i = 1, . . . , n (where n+ 1354

is interpreted as 1). The resulting orientation is the cycle orientation of Cn, denoted355

~C(H)n . Then P(~C

(H)n ) = M(~C

(H)n ) = Z(~C

(H)n ) = 1.356

Example 2.24. Starting with Cn, label the vertices in cycle order by {1, . . . , n}357

and orient the edges as follows: Orient {1, 2} and {1, n} as (1, 2) and (1, n). For358

i = 1, . . . ,⌊n2

⌋− 1, orient {2i + 1, 2i} and {2i + 1, 2i + 2} as arcs (2i + 1, 2i) and359

(2i + 1, 2i + 2). If n is odd, orient the edge {n − 1, n} as (n, n − 1). The resulting360

orientation is the alternating orientation of Cn, denoted ~C(A)n . If n is odd there is361

one path of length 2, so P(~C(A)n ) =

⌊n2

⌋. Let S be the set of odd numbered vertices362

(with the exception of vertex n if n is odd), so every vertex in S has in-degree zero363

and |S| =⌊n2

⌋. Clearly S ⊆ B for any zero forcing set, and every vertex in S has364

two out-neighbors not in S, so every zero forcing set must have cardinality at least365 ⌊n2

⌋+1. Since S∪{2} is a zero forcing set, Z(~C

(A)n ) =

⌊n2

⌋+1. We construct a matrix366

A ∈ M(~C(A)n ) of nullity

⌊n2

⌋+ 1, showing that M(~C

(A)n ) =

⌊n2

⌋+ 1. Any matrix in367

M(~C(A)n ) has two nonzero off-diagonal entries in every odd row (except n if n is odd)368

and no nonzero off-diagonal entries in every even row. Define a matrix A = [aij ] with369

Γ(A) = ~C(A)n by setting aii = 0 except ann = −1 if n is odd, and in each odd row the370

first nonzero entry is 1 and the second is −1. Then nullA =⌊n2

⌋+ 1.371

Proposition 2.25. Let ~Cn be any orientation of Cn (n ≥ 4). Then 1 ≤ P(~Cn) ≤372 ⌊n2

⌋and for every integer k with 1 ≤ k ≤

⌊n2

⌋, there is an orientation ~Cn having373

P(~Cn) = k. For any orientation of a cycle ~Cn, 1 ≤ M(~Cn) ≤ Z(~Cn) ≤⌊n2

⌋+ 1 and374

for every integer k with 1 ≤ k ≤⌊n2

⌋+1, there are (possibly two different) orientations375

12

~Cn having M(~Cn) = k and Z(~Cn) = k.376

Proof. The proof of the second part of each statement follows from Examples 2.20377

and 2.21 and Corollary 2.2. To complete the proof we show that P(~Cn) ≤⌊n2

⌋and378

Z(~Cn) ≤⌊n2

⌋+ 1 for all orientations ~Cn, by exhibiting a path cover and zero forcing379

set of cardinality not exceeding this bound.380

For path cover number: If n is even, choose two adjacent vertices, cover them381

with one path, and delete them, leaving a path on n − 2 vertices which is an even382

number. By Proposition 2.22, there is a path cover of these n− 2 vertices with n2 − 1383

paths, so there is a path cover of ~Cn having n2 =

⌊n2

⌋paths. If n is odd, then for any384

orientation ~Cn, there is a path on 3 vertices. Cover these vertices with that path and385

delete them, leaving a path on n − 3 vertices (again an even number), which can be386

covered by n−32 paths, and there is a path cover of ~Cn having n−3

2 + 1 =⌊n2

⌋paths.387

For zero forcing number: Delete any one vertex v, leaving a path on n−1 vertices,388

which has a zero forcing set B with |B| =⌈n−12

⌉by Proposition 2.22. Then the set389

B′ := B ∪ {v} is a zero forcing set for ~Cn and |B′| =⌈n−12

⌉+ 1. If n is even,390 ⌈

n−12

⌉+ 1 = n

2 + 1 =⌊n2

⌋+ 1. If n is odd,

⌈n−12

⌉+ 1 = n−1

2 + 1 =⌊n2

⌋+ 1.391

3. Doubly directed graphs. Given a graph G, the doubly directed graph←→G392

of G is the digraph obtained by replacing each edge {u, v} by both of the arcs (u, v)393

and (v, u). In this section we establish results for minimum rank, maximum nullity,394

zero forcing number, and path cover number of doubly directed graphs.395

Proposition 3.1. P(G) = P(←→G ) for any graph G.396

Proof. P(G) is the minimum number of induced paths of G and P(←→G ) is the397

minimum number of Hessenberg paths in←→G . It is enough to show that all Hessenberg398

paths in←→G are induced. Suppose P is a Hessenberg path in

←→G that is not induced.399

Then there exists some arc (vi, vj) ∈ E(←→G ), where i > j+1. But because the digraph400

is doubly directed, (vj , vi) ∈ E(←→G ), which contradicts the definition of a Hessenberg401

path. Therefore, all Hessenberg paths must be induced. Thus, P(G) = P(←→G ).402

Proposition 3.2. Z(G) = Z(←→G ) for any graph G.403

Proof. The color change rule for graphs is that a blue vertex v can force a white404

vertex w if w is the only white neighbor of v. The color change rule for digraphs is405

that a blue vertex V can force a white vertex w if w is the only white out-neighbor406

of v. If G is a graph then for any vertex v ∈ V (G), w is a neighbor of v in G if and407

only if w is an out-neighbor of v in←→G . This means that v forces w in G if and only408

if v forces w in←→G . Thus B is a zero forcing set in G if and only if B is a zero forcing409

set in←→G and Z(G) = Z(

←→G ).410

Observation 3.3. For any graph G, M(G) ≤ M(←→G ), since S(G) ⊆M(

←→G ).411

Corollary 3.4. If G is a graph such that M(G) = Z(G), then M(G) = M(←→G ).412

13

Proof. Z(G) = Z(←→G ) ≥ M(

←→G ) ≥ M(G) = Z(G).413

It was established in [1] that for every tree T , P(T ) = M(T ) = Z(T ), giving the414

following corollary.415

Corollary 3.5. If T is a tree, then P(←→T ) = M(

←→T ) = Z(

←→T ).416

As the following example shows, it is possible to have M(←→G ) > M(G).417

Example 3.6. The complete tripartite graph on three sets of three vertices K3,3,3418

has V1 = {1, 2, 3}, V2 = {4, 5, 6}, V3 = {7, 8.9}, V (K3,3,3) = V1∪V2∪V3 and E(K3,3,3)419

equal to the set of all edges with one vertex in Vi and the other in Vj (i 6= j). It is420

well-known that mr(K3,3,3) = 3. Let J3 be the 3× 3 matrix with all entries equal to421

1, 03 be the 3 × 3 matrix with all entries equal to 0, and let A =

03 J3 −J3J3 03 J3J3 J3 03

.422

Then Γ(A) =←−−→K3,3,3 and rankA = 2. Thus, M(

←−−→K3,3,3) = 7 > 6 = M(K3,3,3).423

The pentasun H5 graph shown in Figure 3.1 has M(H5) = 2 < 3 = P(H5) [4],424

establishing the noncomparability of M and P (because there are many examples of425

graphs G with P(G) < M(G)). The same is true for the doubly directed pentasun.426

v

w

Fig. 3.1. The pentasun H5.

Example 3.7. Theorem 2.8 of [5] describes the cut-vertex reduction method for427

calculating M for directed graphs with a cut-vertex. We compute M(←→H5) = 2 by428

applying the cut-vertex reduction method to vertex v, using the notation found in429

[5]. Because M(H5−w) = Z(H5−w) = 2 and M(H5−{v, w}) = Z(H5−{v, w}) = 2,430

M(←→H5−w) = Z(

←→H5−w) = 2 and M(

←→H5−{v, w}) = Z(

←→H5−{v, w}) = 2, so rv(

←→H5−w) =431

1. Clearly mr(←→H5[{v, w}]) = 1 and mr(

←→H5[{v, w}] − v) = 0, so rv(

←→H5[{v, w}]) = 1.432

The type of the cut-vertex v of a digraph Γ, typev(Γ), is a subset of {C,R}, where433

C ∈ typev(Γ) if there exists a matrix A′ ∈ M(Γ − v) with rankA′ = mr(Γ − v) and434

a vector z in rangeA′ that has the in-pattern of v, and similarly for rows. Thus,435

typev(←→H5[{v, w}]) = ∅, so by [5, Theorem 2.8], rv(

←→H5) = 2. So mr(

←→H5) = mr(

←→H5 −436

{v, w})+mr(←→H5[{w}])+2 = 6+0+2 = 8 and M(

←→H5) = 2. Since P(

←→H5) = P(H5) = 3,437

P(←→H5) > M(

←→H5). It is easy to find an example of a digraph Γ with P(Γ) < M(Γ), e.g.438

Example 2.13, so M and P are noncomparable.439

14

Proposition 3.8. Suppose that both G and←→G have field independent minimum440

rank. Then mr(G) = mr(←→G ) and M(G) = M(

←→G ).441

Proof. Since both G and←→G have field independent minimum rank mr(G) =442

mrZ2(G) and mrZ2(←→G ) = mr(

←→G ). Furthermore, SZ2(G) =MZ2(

←→G ), so443

mr(G) = mrZ2(G) = mrZ2(←→G ) = mr(

←→G ).444

445

The converse of Proposition 3.8 is not true, however.446

Example 3.9. Let G be the full house graph, shown in Figure 3.2. It is well447

known that mrZ2(G) = 3, yet mr(G) = 2 = mr(←→G ).448

1

23

4 5

Fig. 3.2. The full house graph

4. Digraphs in general. In this section, we present some minimum rank, maxi-449

mum nullity, and zero forcing results for digraphs in general, where any pair of vertices450

may or may not have an arc in either direction. We begin with two (undirected) graph451

properties that do not extend to digraphs.452

Sinkovic [11] has shown that for any outerplanar graph G, M(G) ≤ P(G). This is453

not true for digraphs, because it was was shown in Example 2.13 that the outerplanar454

digraph_GR has M(

_GR) = Z(

_GR) = 3 > 2 = P(

_GR), and

_GR is outerplanar (although455

Figure 2.2 is not drawn that way).456

The complement of a graph G = (V,E) (or digraph Γ = (V,E)) is the graph457

G = (V,E) (or digraph Γ = (V,E)), where E consists of all two element sets of458

vertices (or all ordered pairs of distinct vertices) that are not in E. The Graph459

Complement Conjecture (GCC) is equivalent to the statement that for any graph G,460

M(G) + M(G) ≥ |G| − 2. This statement is generalized in [3]: For a graph parameter461

β related to maximum nullity, the Graph Compliment Conjecture for β, GCCβ , is462

β(G) +β(G) ≥ |G| − 2. With this notation, GCC can be denoted GCCM. The Graph463

Compliment Conjecture for zero forcing number, Z(G) + Z(G) ≥ |G| − 2, denoted464

GCCZ, is actually the Graph Complement Theorem for zero forcing [7]. However,465

as the following example shows, GCCZ does not hold for digraphs, and since for any466

15

digraph M(Γ) ≤ Z(Γ), GCCM does not hold for digraphs. A tournament provides a467

counterexample.468

Example 4.1. For the Hessenberg tournament of order n, denoted ~K(H)n ,469

Z( ~K(H)n ) = 1 because ~Hn is a Hessenberg path. Because ~K

(H)n is self-complementary,470

Z( ~K(H)n ) + Z( ~K

(H)n ) = 2,471

but for n ≥ 5, n− 2 = | ~K(H)n | − 2 ≥ 3.472

Some properties of minimum rank for graphs do remain true for digraphs. For a473

graph G, it is well-known that if Kr is a subgraph of G then M(G) ≥ r − 1 (see, for474

example, [2] and the references therein). An analogous result holds true for digraphs,475

although the proof is different than those usually given for graphs.476

Theorem 4.2. Suppose←→Kr is a subgraph of a digraph Γ. Then M(Γ) ≥ r − 1.477

Proof. First, we order the vertices of Γ so that the subdigraph induced on the478

vertices 1, 2, . . . , r is←→Kr. We will construct L ∈ M(Γ) with rankL ≤ n − r + 1,479

where L is partitioned as L =

[A B

C D

]and A ∈ M(Kr). We may choose D ∈480

M(Γ[{r + 1, . . . , n}]) so that rankD = n − r. We now choose C to be any matrix481

with the correct zero-nonzero pattern. Denote the ith column of C by ci and the jth482

column of D by dj . Since D has full rank, there exist coefficients di,1, . . . , di,n−r such483

that ci = di,1d1 + · · ·+ di,n−rdn−r for 1 ≤ i ≤ r.484

Now, we choose B to be any matrix with the correct zero-nonzero pattern and485

denote the jth column of B by bj . Then define E to be the r × r matrix whose ith486

column is equal to di,1b1 + · · · + di,n−rbn−r. Therefore, the matrix L′ =

[E B

C D

]487

has rankL′ = n− r. Let p be a real number greater than the absolute value of every488

entry of E. Define A := E+ pJr, where Jr is the r× r matrix all of whose entries are489

1, so A ∈M(Kr), L =

[A B

C D

]∈M(Γ) and rankL ≤ n− r + 1.490

In [6], the maximum number of edges in a graph G of order n with a prescribed491

zero forcing number k is shown to be kn −(k+12

). We similarly seek to bound the492

number of arcs that a digraph Γ of order n may possess given Z(Γ) = k.493

Theorem 4.3. Suppose Γ is a digraph of order n with Z(Γ) = k. Then,494

|E(Γ)| ≤(n

2

)−(k

2

)+ k(n− 1). (4.1)495

496

Proof. We prove that (4.1) holds for a digraph Γ of order n ≥ k + 1 whenever497

Z(Γ) ≤ k (since for a graph G of order n, Z(G) ≤ n − 1). The proof is by induction498

on n, for a fixed positive integer k. The base case is n = k+ 1, or k = n− 1, and the499

16

inequality (4.1) reduces to500

|E(Γ)| ≤(n

2

)−(n− 1

2

)+(n−1)2 =

n(n− 1)

2− (n− 1)(n− 2)

2+(n−1)2 = n(n−1).501

Any digraph Γ of order n has at most 2(n2

)= n2 − n arcs and thus the claim holds.502

Now assume (4.1) holds true for any digraph of order n−1 that has a zero forcing503

set of cardinality k. Let Γ be a digraph of order n with Z(Γ) ≤ k. Let B be a504

zero forcing set of Γ where |B| = k. Suppose F is a chronological list of forces for505

B and that the first force of F occurs on the arc (v, w). Then, (B \ {v}) ∪ {w} is506

a zero forcing set of cardinality k for Γ − v and the induction hypothesis applies to507

E(Γ−v). In order to determine an upper bound on |E(Γ)| we determine the maximum508

number of arcs incident with v in Γ. Since v forces w first in F , (v, x) ∈ E(Γ) implies509

x ∈ (B \{v})∪{w}. Furthermore, E(Γ) contains at most n−1 arcs of the form (x, v),510

one for each vertex x 6= v. Therefore,511

|E(Γ)| ≤ |E(Γ− v)|+ k + (n− 1) ≤(n− 1

2

)−(k

2

)+ k(n− 2) + k + (n− 1)512

=

(n

2

)−(k

2

)+ k(n− 1).513

514

In the paper [6], the edge bound is used to show that the zero forcing number515

must be at least half the average degree. However, a Hessenberg tournament (cf.516

Example 2.15) has half of all possible arcs and Z( ~Hn) = 1, so the analogous result is517

not true for digraphs, and any correct result of this type for digraphs is not likely to518

be useful.519

For a digraph Γ where Z(Γ) = k, Theorem 4.3 gives an upper bound for the520

number of arcs Γ may possess. However, the proof also suggests that equality is521

achievable in (4.1) when n > k. The following provides a construction of a class of522

digraphs for which (4.1) is sharp.523

Theorem 4.4. Let k be a fixed positive integer. Then for each integer n > k and524

each partition π = (n1, . . . , nk) of n, there exists a digraph Γn,k,π of order n for which525

Z(Γn,k,π) = k, the forcing chains of Γn,k,π have lengths n1, n2, . . . , nk respectively,526

and527

|E(Γn,k,π)| =(n

2

)−(k

2

)+ k(n− 1).528

529

Proof. For 1 ≤ i ≤ k, let Γi be a full Hessenberg path on ni vertices. Among all530

the Γi, there are a total ofk∑i=1

[(ni − 1) +

(ni

2

)]arcs. Within each Γi we denote the531

initial vertex of the Hessenberg path by bi and the terminal vertex of the Hessenberg532

17

path by ti. We define B = {bi : 1 ≤ i ≤ k} and T = {ti : 1 ≤ i ≤ k}, and note that533

B and T will intersect if ni = 1 for some i. To create Γn,k,π we start withk⋃i=1

Γi and534

add arcs between the Γi in the following manner:535

(1) For 1 ≤ j < i ≤ k, we add all arcs from vertices in Γi to vertices in Γj . This536

adds a total of∑i<j

ninj arcs.537

(2) Add all arcs from vertex ti to vertices in other Γj . For each i, this adds538 ∑j 6=i nj = n−ni arcs. Over all i, this adds a total of kn−

∑ki=1 ni = (k−1)n539

arcs.540

Some arcs have been double-counted, which must be reflected in the overall total.541

In particular, arcs from ti to all vertices in Γj (for j < i) have been double-counted.542

For an arc from ti to a vertex v of Γj where v 6= tj , we replace the double-counted543

arc by an arc from v to bi. Therefore, we need only remove from the total count the544

number of arcs from tj to ti for 1 ≤ i < j ≤ k. There are a total of(k2

)such arcs.545

Thus, we have546

|E| =k∑i=1

[(ni − 1) +

(ni2

)]+

∑1≤i<j≤k

ninj + (k − 1)n−(k

2

)547

= n− k +

k∑i=1

(ni2

)+

∑1≤i<j≤k

ninj

+ kn− n−(k

2

)548

=

(n

2

)−(k

2

)+ k(n− 1).549

By Theorem 4.3, we know that Z(Γn,k,π) ≥ k. We claim that B is a zero forcing550

set for Γn,k,π and that a chronological list of forces exists for which the forcing chains551

have lengths n1, . . . , nk respectively. Assume that each vertex of B is blue. Γ1 is a552

Hessenberg path and the only arcs coming from vertices of Γ1 point to vertices in B553

with the exception of arcs coming from t1. Thus, forcing may occur along Γ1, where554

t1 is the last vertex forced. We then proceed to Γ2 and so on through all Γi. When555

we get to Γi, the only arcs coming from vertices of Γi point to vertices in B or to556

the already blue vertices of Γh where h < i, with the exception of arcs coming from557

ti (which is not used to perform a force). So, forcing may occur along Γi until all558

vertices are blue. Therefore B is a zero forcing set for Γn,k,π and Z(Γn,k,π) = k.559

Although the digraph Z(Γn,k,π) achieve equality in the bound (4.3), these are not560

the only digraphs that do so.561

Example 4.5. Let Γ be the digraph of order n = 8 in Figure 4.1. Note that562

{1, 5} is a zero forcing set of Γ. Since deg+(v) ≥ 2 for all vertices v, k = Z(Γ) = 2.563

Also note that Γ has 41 arcs, the same number of arcs as each digraph Γ8,2,π.564

In all of the digraphs Γn,k,π constructed in Theorem 4.4, the forcing process may565

be completed one forcing chain at a time, and we show that this is not true for Γ.566

18

1 2 4

5 6 7 8

3

Fig. 4.1. A digraph with the maximum number of arcs for its zero forcing number.

Since deg+(v) ≥ 3 for all vertices v other than vertex 1, vertex 1 must be contained in567

any minimum zero forcing set of Γ along with one of the two out-neighbors of vertex568

1. Therefore, the only minimum zero forcing sets are B1 = {1, 5} and B2 = {1, 2}. If569

we color the vertices of B1 blue, then the first three forces must occur along the arcs570

(1, 2), (2, 3), and (5, 6), in that order. If we color the vertices of B2 blue, then the first571

three forces must occur along the arcs (1, 5), (2, 3), and (5, 6), in that order. In either572

case, neither of the two forcing chains is completely blue before the forcing process573

must begin on the other. Therefore, Γ is not equal to any of the Γn,2,π constructed574

in Theorem 4.4.575

Although M(Γ) does not necessarily equal Z(Γ) for all digraphs Γ, we get equality576

for all Γn,k,π constructed in the proof of Theorem 4.4.577

Proposition 4.6. If k and n are positive integers where k < n and Γn,k,π is one578

of the digraphs constructed in the proof of Theorem 4.4, then M(Γn,k,π) = Z(Γn,k,π).579

Proof. We adopt the notation and definitions used in the proof of Theorem 4.4.580

By construction, each of the k vertices in T has an arc to every other vertex of Γn,k,π.581

So for all vertices v 6∈ T , deg−(v) ≥ k. Now we consider ti ∈ T . If ti 6= bi, then582

there is an arc to ti from another vertex of Γi. There are also arcs to ti from all583

other vertices of T and therefore deg−(ti) ≥ k. We now consider the case where584

ti = bi. By construction, the subgraph induced on B is←→Kk and thus there is an585

arc to ti from each of the other k − 1 vertices of B. Furthermore, since n > k586

there is a vertex tj ∈ T for which tj 6= bj . By construction, there is also an arc587

from tj to ti and therefore deg−(ti) ≥ k. This implies that δ−(Γn,k,π) ≥ k. Since588

M(Γn,k,π) ≥ max{δ−(Γn,k,π), δ+(Γn,k,π)} [5], k ≤ M(Γn,k,π) ≤ Z(Γn,k,π) ≤ k, and so589

M(Γn,k,π) = Z(Γn,k,π).590

For k = n−1, Γn,k,π is the digraph←→K n, so for n ≥ 4, P(Γn,k,π) =

⌈n2

⌉< n−1 =591

Z(Γn,k,π).592

19

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