Power Domination and Zero Forcingusers.monash.edu/~gfarr/research/slides/Vasilevska... · by...

. . . . . .

Power DominationZero Forcing

Connection between PD and ZFComputing PD and ZF #s

......Power Domination and Zero Forcing

Violeta VasilevskaUtah Valley University

Violeta.Vasilevska@uvu.edu

Discrete Maths Seminar TalkMonash University, Melbourne, Australia

January 29, 2018

Violeta Vasilevska Power Domination and Zero Forcing

. . . . . .

.. AIM, ICERM, NSF, REUF Collaborators (2015)

REUF – Research Experience for Undergraduate Faculty

“a program for undergraduate faculty who are interested in mentoring undergraduate research.”

Dr. Katherine Benson (Westminister College)Dr. Daniela Ferrero (Texas State University)Dr. Mary Flagg (University of St. Thomas)Dr. Veronica Furst (Fort Lewis College)Dr. Leslie Hogben (Iowa State University)Dr. Brian Wissman (University of Hawaii at Hilo)

“Zero Forcing and Power Domination for Graph Products.”Australasian J. Combinatorics 70 (2018), 221-235

. . . . . .

.. Outline

Power Domination (PD)

Zero Forcing (ZF)

Connection between PD and ZF processes

Computing PD and ZF numbers

. . . . . .

Real-world ApplicationsModeling the ProblemExamples

P O W E R D O M I N A T I O N

. . . . . .

.. Monitoring Electrical Networks

Electric power companies need to monitor the state of theirnetworks continuously.

Solution: Place Phase Measurement Units (PMUs) atelectrical nodes, where transmission lines, loads, andgenerators are connected.A PMU placed at an electrical node measures the voltageat the node and all current phasors at the node.Problem: PMUs are costly, so it is important to minimizethe number of PMUs used.Where should those PMUs be placed to observe the entiresystem?

. . . . . .

Electric power companies need to monitor the state of theirnetworks continuously.Solution: Place Phase Measurement Units (PMUs) atelectrical nodes, where transmission lines, loads, andgenerators are connected.

A PMU placed at an electrical node measures the voltageat the node and all current phasors at the node.Problem: PMUs are costly, so it is important to minimizethe number of PMUs used.Where should those PMUs be placed to observe the entiresystem?

. . . . . .

Electric power companies need to monitor the state of theirnetworks continuously.Solution: Place Phase Measurement Units (PMUs) atelectrical nodes, where transmission lines, loads, andgenerators are connected.A PMU placed at an electrical node measures the voltageat the node and all current phasors at the node.

Problem: PMUs are costly, so it is important to minimizethe number of PMUs used.Where should those PMUs be placed to observe the entiresystem?

. . . . . .

Electric power companies need to monitor the state of theirnetworks continuously.Solution: Place Phase Measurement Units (PMUs) atelectrical nodes, where transmission lines, loads, andgenerators are connected.A PMU placed at an electrical node measures the voltageat the node and all current phasors at the node.Problem: PMUs are costly, so it is important to minimizethe number of PMUs used.

Where should those PMUs be placed to observe the entiresystem?

. . . . . .

Electric power companies need to monitor the state of theirnetworks continuously.Solution: Place Phase Measurement Units (PMUs) atelectrical nodes, where transmission lines, loads, andgenerators are connected.A PMU placed at an electrical node measures the voltageat the node and all current phasors at the node.Problem: PMUs are costly, so it is important to minimizethe number of PMUs used.Where should those PMUs be placed to observe the entiresystem?

. . . . . .

.. Modeling the Problem

An electric power network − modeled by a graphThe electrical nodes − graph vertices

Transmission lines joining − graph edgestwo electrical nodes

http://kk.org/thetechnium/Electricity Network.jpg

. . . . . .

.. The Power Domination Problem in Graphs

Find a minimum set of vertices from where the entire graph canbe observed according to certain propagation rules.

First studied by

Haynes at al. (“Domination in graphs applied to electric powernetworks” (2002))

. . . . . .

Start with a graph G whose vertices are colored eitherwhite or black.

Let S be the set of all vertices colored black. Color allneighbors of vertices in S black.

Apply the following color-change rule as many times aspossible.

Color-change Rule:If there is a black vertex that has exactly one white neighbor- color that neighbor black.

. . . . . .

The set S is called a power dominating set of a graph Gif at the end of applying the propagation rule all vertices inG are colored black.

A minimum power dominating set is a power dominatingset with minimum number of vertices.

Power domination number for G, denoted γP(G), is thenumber of vertices in a minimum power domination set.

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E X A M P L E S

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.. Path P4

γP(P4) = 1

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.. Path P4

γP(P4) = 1

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.. Path P4

γP(P4) = 1

. . . . . .

.. Path P4

γP(P4) = 1

. . . . . .

.. Circle C6

γP(C6) = 1

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.. Circle C6

γP(C6) = 1

. . . . . .

.. Circle C6

γP(C6) = 1

. . . . . .

.. Circle C6

γP(C6) = 1

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.. Grid

γP(G) = 2

. . . . . .

.. Grid

γP(G) = 2

. . . . . .

.. Grid

γP(G) = 2

. . . . . .

.. Grid

γP(G) = 2

. . . . . .

.. Cylinder

γP(G) = 3

. . . . . .

.. Cylinder

γP(G) = 3

. . . . . .

.. Cylinder

γP(G) = 3

. . . . . .

.. Cylinder

γP(G) = 3

. . . . . .

Examples

Z E R O F O R C I N G

. . . . . .

Examples

A zero forcing set for a graph G is a subset of vertices B suchthat if initially the vertices in B are colored black and thereminding vertices are colored white, repeated application ofthe color change rule can color all vertices of G black.

A minimum zero forcing set is a zero forcing set withminimum number of vertices.

Zero forcing number for G, denoted Z (G), is the number ofvertices in a minimum zero forcing set.

. . . . . .

Examples

The zero forcing number was introduced

by mathematicians Hogben et al. (“Zero forcing sets andthe minimum rank of graphs,” (2008))

and independently by mathematical physicists studyingcontrol of quantum systems

and later by computer scientists studying graph searchalgorithms.

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Examples

E X A M P L E S

. . . . . .

Examples

.. Path P4

Z (P4) = 1

. . . . . .

Examples

.. Path P4

Z (P4) = 1

. . . . . .

Examples

.. Path P4

Z (P4) = 1

. . . . . .

Examples

.. Path P4

Z (P4) = 1

. . . . . .

Examples

.. Circle C6

Z (C6) = 2

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Examples

.. Circle C6

Z (C6) = 2

. . . . . .

Examples

.. Circle C6

Z (C6) = 2

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ObservationDetermining PD #Research ProblemRelation between PD and ZF #s

C O N N E C T I O N B E T W E E N

PD A N D ZF N U M B E R S

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.. Observation

The power domination process on a graph G can be describedas

choosing a set S ⊆ V (G) andapplying the zero forcing process to the closedneighborhood N[S] of S.

The set S is a power dominating set of G if and only if N[S] is azero forcing set for G.

. . . . . .

.. Observation

The power domination process on a graph G can be describedas

choosing a set S ⊆ V (G) andapplying the zero forcing process to the closedneighborhood N[S] of S.

The set S is a power dominating set of G if and only if N[S] is azero forcing set for G.

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.. Determining PD #

The power domination number of several families of graphs hasbeen determined using the following two-step process:

find an upper bound:The upper bound is usually obtained by providing a patternto construct a set, together with a proof that constructed setis a power dominating set;

find a lower bound:The lower bound is usually found by exploiting structuralproperties of the particular family of graphs, and it usuallyconsists of a very technical and lengthy process.

. . . . . .

.. Research Problem

Finding good general lower bounds for the power dominationnumber.

An effort in that direction is the work by Stephen et al. (“Powerdomination in certain chemical structures,” (2015))

. . . . . .

.Theorem (1)..

......

Let G be a graph that has an edge.Then ⌈

∆(G)

⌉≤ γP(G),

where ∆(G) = max{deg v : v ∈ V (G)} is the maximum degreeof G.

This bound is tight.

. . . . . .

.. Sketch of Proof:

Choose a minimum PD set {u1,u2, . . . , ut}. Hence t = γP(G).Then

∑ti=1 deg ui ≤ t∆(G).

If G has no isolated vertices: Dean et al. (“On the powerdominating sets of hypercubes,” (2011)):

Z (G) ≤t∑

deg ui .

If G has isolated vertices, they contribute one to each ZFnumber and PD number.

Bound is tight:

Z (Kn) = ∆(Kn) = n − 1 and γP(Kn) = 1. �

. . . . . .

.. Sketch of Proof:

Choose a minimum PD set {u1,u2, . . . , ut}. Hence t = γP(G).Then

∑ti=1 deg ui ≤ t∆(G).

If G has no isolated vertices: Dean et al. (“On the powerdominating sets of hypercubes,” (2011)):

Z (G) ≤t∑

deg ui .

If G has isolated vertices, they contribute one to each ZFnumber and PD number.

Bound is tight:

Z (Kn) = ∆(Kn) = n − 1 and γP(Kn) = 1. �

. . . . . .

Computing PD #s for Tensor ProductsComputation of ZF #s for Lexicographic Products

C O M P U T I N G

PD N U M B E R S

T E N S O R P R O D U C T S

. . . . . .

.. Tensor Products

G = (V (G),E(G)) and H = (V (H),E(H))

be disjoint graphs.

The tensor product (also called the direct product) of G and His denoted by G × H:

vertex set: V (G)× V (H)

edge set:a vertex (g,h) is adjacent to a vertex (g′,h′) in G × H

if{g,g′} ∈ E(G) and {h,h′} ∈ E(H).

. . . . . .

.. Computing PD # for Tensor Products

Dorbec et al. (“Power domination in product graphs,” (2008))(the power domination problem for the tensor product of twopaths)

Question:

What is the power domination number for a tensor product of apath and a complete graph and of a cycle and a completegraph?

. . . . . .

.. Computing PD # for Tensor Products

Dorbec et al. (“Power domination in product graphs,” (2008))(the power domination problem for the tensor product of twopaths)

Question:

What is the power domination number for a tensor product of apath and a complete graph and of a cycle and a completegraph?

. . . . . .

.Theorem..

......

Let t ≥ 3 and G = Pt or G = Ct .Suppose t is odd and n ≥ t , or suppose t is even and either

...1 G = Pt and n ≥ t2 + 2, or

...2 G = Ct and n ≥ t2 .

γP(G × Kn) =

⌉if t ̸≡ 2 mod 4,

t2 or t

2 + 1 if t ≡ 2 mod 4.

. . . . . .

.. Sketch of Proof:

Upper bound on γP(G × Kn):.Theorem..

......

Let n ≥ 3. If G = Pt with t ≥ 2 or G = Ct with t ≥ 3, then

γP(G × Kn) ≤

⌉if t ̸≡ 2 mod 4,

t2 + 1 if t ≡ 2 mod 4.

A lower bound on γP(G × Kn):

Use known ZF #s...

. . . . . .

.. Sketch of Proof:

......

γP(G × Kn) ≤

⌉if t ̸≡ 2 mod 4,

t2 + 1 if t ≡ 2 mod 4.

Use known ZF #s...

. . . . . .

.. Sketch of Proof:

......

γP(G × Kn) ≤

⌉if t ̸≡ 2 mod 4,

t2 + 1 if t ≡ 2 mod 4.

Use known ZF #s...

. . . . . .

.. Sketch of Proof - continue.Theorem..

......

...1 (Fernandes, da Fonseca)If t ≥ 1 is odd and n ≥ 2, then Z (Pt × Kn) = (n − 2)t + 2.

...2 (V. et al.)If t ≥ 2 is even and n ≥ 3, then Z (Pt × Kn) = (n − 2)t .

.Theorem..

......

If n, t ≥ 3, then

Z (Ct × Kn) =

(n − 2)t + 2 if t is odd,

(n − 2)t + 4 if t is even.

. . . . . .

.. Sketch of Proof - continue

Observation:

deg(g,h) = degG(g)degH(h) for (g,h) ∈ E(G × H).

Hence, ∆(G × H) = ∆(G)∆(H).

∆(G × Kn) = ∆(G)∆(Kn) = 2(n − 1)

. . . . . .

Observation:

deg(g,h) = degG(g)degH(h) for (g,h) ∈ E(G × H).

Hence, ∆(G × H) = ∆(G)∆(H).

∆(G × Kn) = ∆(G)∆(Kn) = 2(n − 1)

. . . . . .

Consider two cases depending on the parity of t and useTheorem 1.

t = 2k + 1 for some positive integer k

γP(G × Kn) ≥⌈(n−2)(2k+1)+2

2(n−1)

⌈k +

n − 2k2(n − 1)

⌉≥ k + 1 if n − 2k > 0.

Hence,⌈ t

⌉≤ γP(G × Kn) if t is odd and n ≥ t .

. . . . . .

t = 2k for some positive integer k .Take c = 0 for G = Pt and c = 2 for G = Ct .

γP(G × Kn) ≥⌈(n−2)(2k)+2c

2(n−1)

⌈k − k − c

n − 1

⌉= k if n − 1 > k − c.

Hence,t2 ≤ γP(G × Kn) if G = Pt and

n ≥ t2 + 2, or if G = Ct and n ≥ t

2 . �

. . . . . .

C O M P U T I N G

ZF N U M B E R S

G R A P H P R O D U C T S

. . . . . .

.. Lexicographic Product

G = (V (G),E(G)) and H = (V (H),E(H))

be disjoint graphs.

The lexicographic product of G and H is denoted by G ∗ H:

vertex set: V (G)× V (H)

edge set:two vertices (g,h) and (g′,h′) are adjacent in G ∗ H ifeither

{g, g′} ∈ E(G), org = g′ and {h,h′} ∈ E(H).

. . . . . .

.. Domination Number

A vertex v in a graph G is said to dominate itself and all of itsneighbors in G.

A set of vertices S is a dominating set of G if every vertex of Gis dominated by a vertex in S.

The minimum cardinality of a dominating set is the dominationnumber of G (denoted by γ(G)).

. . . . . .

.Theorem..

......

Let G and H be regular graphs with degree dG and dH ,respectively.

If γP(H) = 1 and γ(G) = 1, then

Z (G ∗ H) = dG|V (H)|+ dH .

.Corollary..

......

For n ≥ 2 and m ≥ 3,

Z (Kn ∗ Cm) = (n − 1)m + 2.

. . . . . .

.Theorem..

......

Let G and H be regular graphs with degree dG and dH ,respectively.

If γP(H) = 1 and γ(G) = 1, then

Z (G ∗ H) = dG|V (H)|+ dH .

.Corollary..

......

For n ≥ 2 and m ≥ 3,

Z (Kn ∗ Cm) = (n − 1)m + 2.

. . . . . .

.. Sketch of Proof

Dorbec et al., (“Power Domination in Product Graphs,”(2008))

γP(G ∗ H) = γ(G) if γP(H) = 1.

For lexicographic product

degG∗H(g,h) = (degG g)|V (H)|+ degH h

for any vertex (g,h) ∈ V (G ∗ H), hence

∆(G ∗ H) = ∆(G)|V (H)|+∆(H).

. . . . . .

By Theorem 1: Z (G ∗ H) ≤ γP(G ∗ H)∆(G ∗ H).Hence

Z (G ∗ H) ≤ γ(G)(∆(G)|V (H)|+∆(H)

)if γP(H) = 1.

Since γ(G) = 1, G is dG-regular, and H is dH -regular:

Z (G ∗ H) ≤ dG|V (H)|+ dH .

G ∗ H is (dG|V (H)|+ dH)-regular, hence

dG|V (H)|+ dH = δ(G ∗ H) ≤ Z (G ∗ H),

where δ(G) = min{deg v :v ∈ V} is the minimum degree ofG. �

. . . . . .

T HANK YOU !

Violeta.Vasilevska@uvu.edu

Power Domination and Zero Forcingusers.monash.edu/~gfarr/research/slides/Vasilevska... · by...

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