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Study on Power Domination of Graphs 圖圖圖圖圖圖圖圖圖圖圖 圖圖圖 圖圖圖 圖圖圖圖 圖圖圖 Student Chien-Cheng Chuang Advisor Gerard Jennhwa Chang Department of Mathematics, National Taiwan University June 2008 1

Study on Power Domination of Graphs 圖上電力支配問題的研究 研究生:莊建成 指導教授:張鎮華 Student : Chien-Cheng Chuang Advisor : Gerard Jennhwa Chang

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Study on Power Domination of Graphs圖上電力支配問題的研究

研究生:莊建成指導教授:張鎮華

Student: Chien-Cheng ChuangAdvisor: Gerard Jennhwa Chang

Department of Mathematics,National Taiwan University

June 2008

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Outline

• Introduction• Previous work and results• Results in this article– Cartesian product of two cycles– Co-graphs– Trees

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Introduction

• Electric power companies need to monitor the state of their power system.

• Let G = (V, E) represents a power system– A vertex: an electric node (a substation bus)– An edge: a transmission line joining two nodes

• One method of monitoring the system is to place phase measurement units (PMUs) in the power system.

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• Each PMU is placed on one vertex, and the observation rules of an PMU are as follows:– (1) The vertex where a PMU is placed and its

incident edges are observed.– (2) The vertex that is incident to an observed edge is

observed.– (3) The edge joining two observed vertices is

observed.– (4) If a vertex is incident to k>1 edges and k-1 edges

are observed, then the remaining edge is observed.

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• Example:

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• The system is observed if all vertices and edges are observed by a set of PMUs.

• G=(V,E), S is a power dominating set (PDS) if all vertices and edges are observed by S.

• The minimum cardinality of a power dominating set of G is called power domination number, denoted by Gp

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• Simpler version for power domination: all vertices and edges are observed if and only if all vertices are observed.– (1) The vertex where a PMU is placed is observed.– (2) All neighbors of the vertex where the PMU is

placed are observed.– (3) If a vertex has k>1 neighbors, and k-1

neighbors are observed, then the remaining neighbor is observed.

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• Example:

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Previous work and results (1)

1. (Haynes , 2002) for any of order

at least 3,

2. (Xu-Kang-Shan-Zhao, 2006) for any

of order at least 3,

3. (Chien, 2004; Brueni-Hea

tree

connected block

grap

3

3th

h

,

p

p

et. al T n

G n

nT

nG

2005; Zhao-Kang-Chang, 2006)

for of order at least 3,

4. (Zhao-Kang-Chang, 2006) for a

any connected graph

connected, 3-regular,

claw-free gr

ny

oaph ,

3

f 4

p

p

G n

G n

nG

nG

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• Solve power domination problem by algorithm– NP-complete:

• Bipartite graphs• Chordal graphs• Split graphs• Circle graphs• Planar graphs

– Polynomial-time:• Trees• Block graphs• Interval graphs• Graphs of bounded treewidth ( Partial k-tree )

Previous work and results (2)

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Results in this article

• Determine the power domination numbers of Cartesian product of two cycles

• Find a minimum PDS for co-graphs• Find a minimum PDS for trees

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Cartesian product of two cycles

• Some results about Cartesian product of two graphs:

1. ,

1 for all ,

2. ,

1 for 1 or , 2,3 ;

2 for other cases

3. ,

2

1

3

p

p

p

m

n m

p

n

n m

G n m

G n n m

G

G K P

G K C m

G K K m n

G n

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• The power domination number on grid graphs: (Dorfling-Henning, 2006)

Theorem:

If 1 , then

1 if 4 mod 8 ,

4

otherwise.4

n m

p

G P P m n

nn

Gn

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• Applying the method for grid graphs, we have the following theorem:

Theorem:

If 3 , then

+1 if 2 mod 4 ,2

otherwise.2

n m

p

G C C m n

nn

Gn

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Co-graphs

• Disjoint union ( sum ) of two graphs• Join of two graphs• Definition of co-graphs– (1)

– (2)4 is a co-graph iff is -freeG G P

1, disjoint union, join, K

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1 2

p p1

If + + + is the disjoint union

of graphs, then .

n

n

ii

G G G G

n G G

1G 2GnG

Proposition 1

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1 2

p

1 2 p

p 1

1 1 1

1. If is the join of 2 graphs,

then 2.

2. with 1 if and only if

1 1

2 , 1p

G G G

G

G G G G

G

G C u C

Proposition 2

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Parse tree: – the construction process of a given co-graph

a co-graph G parse tree of GT G

1v

2v

4v

5v1v 2v

3v 4v

5v2w

3w1w

4w

1

1

0

0

3v

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1. for 1 to do

1; ( ) 0; ;

end for

2. for 1 to 1 do

Let and be the 2 children of in ;

if 0

then { if ( ) ( ) 1, then ( ) 1, else (

ip i i v i

i G

i

i i

i n

v Type v S v

i n

x y w T

L w

Type x Type y Type w Type w

) 0;

;

;

if 2 and ( ) 0 ,

then { if ( ) 0, then , else } }

if 1

then { ( ) 1;

i

i i

p i p p

w x y

p i i

w y w x

i

i

w x y

S S S

w Type w

Type x T S T S

L w

Type w

{ if 1, then 1;

elseif 1, then 1;

elseif 2 and ( ) 0 , then 1;

elseif 2 and ( ) 0 , then 1

i

i

i

p p i w x

p p i w y

p p i w x

p p i

x w S S

y w S S

x Type x w S T

y Type y w

11

;

else 2; a vertex from and a vertex from }}

end for

3. ;

4. output and

i

i

n

w y

p i w x y

p p n w

p

S T

w S S S

G w S S

G S

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Trees

• Haynes, Hedetniemi, Hedetniemi, and Henning (2002) gave an algorithm for the power domination problem on trees.

• Chien (2004) gave another algorithm for trees.

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(0,B) (0,B)

(0,B)(0,B)(0,B)(0,B)(0,B)

(0,B) (0,B) (0,B)

(0,B)

(0,B)

(1,B)(2,B)

(0,F) (1,B)(2,B) (1,B) (1,B)

(0,F)(1,F)(2,F)

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1. for 1 to 1 do

; parent of ;

if 2, then { ; }

if 1 and , then 1

if , = 0, , then

end for

2.

if 2 or , then

3.

i i

v u

v v u u

v v u

n

v v

i n

v v u v

a D D v b F

a b B a a

a b F b F

v v

a b B D D v

output D

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The EndThanks for your listening