Cover PebblingCover PebblingCycles and Graham’s ConjectureCycles and Graham’s Conjecture

Victor M. Moreno

California State University Channel Islands

Advisor: Dr. Cynthia Wyels

Sponsored by the Mathematical Association of America’s REU in

Mathematics at California Lutheran University; funded by NSA and NSF.

DefinitionsDefinitions

DistanceDistance DiameterDiameter Pebbling MovePebbling Move Cover Pebbling Cover Pebbling

NumberNumber Support (Support (GG)) Simple Simple

ConfigurationConfiguration

DistanceDistance

Distance,Distance, dist( dist(u,vu,v) is the length of a ) is the length of a shortest path in shortest path in GG between between uu and and vv..

u v

DiameterDiameter

DiameterDiameter, , dd((GG)) is the longest distance in a is the longest distance in a graph graph GG

u v

Pebbling MovePebbling Move

Pebbling MovePebbling Move is defined as removing is defined as removing two pebbles from a vertex and two pebbles from a vertex and subsequently placing one pebble on an subsequently placing one pebble on an adjacent vertex.adjacent vertex.

42 1

1

Cover Pebbling NumberCover Pebbling Number

Cover Pebbling Number Cover Pebbling Number of a graph of a graph GG, , , , is the minimum number of pebbles is the minimum number of pebbles needed to place a pebble on every vertex needed to place a pebble on every vertex of of GG simultaneouslysimultaneously regardless of initial regardless of initial configuration.configuration.

G

93 31 111

9 2114

15 7 3 1111

Support (Support (GG))

TheThe Support Support of a configuration is the of a configuration is the subset of vertices of the graph that have at subset of vertices of the graph that have at least one pebble.least one pebble.

u v

3 2

Simple ConfigurationSimple Configuration

Simple Configuration: Simple Configuration: we say we have a we say we have a simple configuration when the support simple configuration when the support subset consists of one vertex.subset consists of one vertex.

u

15

Cover Pebbling for PathsCover Pebbling for Paths

)(Pn

v w

),(2)( wvdist

iv

)(...)()()()( 321 nn vvvvP

1

0

2)(n

i

inP

12)(: nnPTheorem

Cover Pebbling for Complete Cover Pebbling for Complete Graphs Graphs

12)(: nKTheorem n

)( nK

1, wvdist

exother vertany

rtexsupport ve

w

v

Where Where is the number of pebbles in a is the number of pebbles in a SSimple Configuration.imple Configuration.

Simple Configuration ConjectureSimple Configuration Conjecture

Conjecture 1 There exists a Conjecture 1 There exists a Simple Simple configurationconfiguration for which . for which .)(GCS

SC

Which configurations are the largest?Which configurations are the largest?

Simple configurations are largest for both Simple configurations are largest for both Paths and Complete graphs.Paths and Complete graphs.

Cover Pebbling for CyclesCover Pebbling for Cycles

Can we generalize for all ?Can we generalize for all ?

)( nC

nC

What is its Cover Pebbling What is its Cover Pebbling Number?Number?

Is there an easier way Is there an easier way to find it?to find it?

Cover Pebbling for Cycles Cover Pebbling for Cycles . .

)( nC

Cover Pebbling for CyclesCover Pebbling for Cycles . .

Two cases: odd and even.Two cases: odd and even.

)( nC

12)( nnP

Cover Pebbling for Cycles Cover Pebbling for Cycles Case one : n is oddCase one : n is odd

1)(2)( dn PC

)( nC

12)( ddP

32)()

2

1(

n

nC

2

1n

d

1)12(2)( dnC

Cover Pebbling for Cycles Cover Pebbling for Cycles Case two : n is evenCase two : n is even

)( nC

dP 1dP

Cover Pebbling for Cycles Cover Pebbling for Cycles Case two : n is evenCase two : n is even

)( nC

1)()()( 1 ddn PPC

)12(3)( 2 n

nC

1)12()12()( 1 ddnC

2

nd

Cover Pebbling for Cycles Cover Pebbling for Cycles )( nC

even is ),12(3

odd is ,32)(

2

2

1

n

nC

n

n

n

Graham’s ConjectureGraham’s Conjecture

Graham’s Conjecture (Cover): For any two Graham’s Conjecture (Cover): For any two graphs graphs G G and and HH, ,

)()()( HGHG )()()( HfGfHGf

)()()( HfGfHGf

Graham’s Conjecture: For any two graphs Graham’s Conjecture: For any two graphs G G and and HH,,

Graham’s Conjecture (Proof)Graham’s Conjecture (Proof):)( Definition HG

productcartesian theisset vertex

egraph whos theisproduct cross thegraphs,

twoare , and , IfHHGGEVHEVG

')',(

)',(':))',').(,((

bygiven are edges whoseand

1

yyxxor

yyandxxyxyx

EE

EG

H

HG

VVVVV HGHGHG,y(x,y):x

G H

12 5

3 4

1

2 3

1

23

1 2

3

1

23

1

2

31

2

3

HG

Graham’s Conjecture (Proof) Graham’s Conjecture (Proof) continued… continued…

. toisomorphic is

form theofsubgraph each Where

H

Hvi

ii

n

HHv

Hv

Hv

Hv

HG

2

1

)(

Graham’s Conjecture (Proof) Graham’s Conjecture (Proof) continued…continued…

. ofsupport theis where Hvi

iHHvi )()(

Vv

vvdist

i

iG

Gv),(2)(

for which of vertex thebe Let

Graham’s Conjecture (Proof)Graham’s Conjecture (Proof)

HGHG

),(2)( ji vvdistiv

iHvHv ii )()()(

Vv

vvdist

i

ijHHG ),(2)()(

HGHG :Theorem

Open QuestionOpen Question

Conjecture 1 There exists a Conjecture 1 There exists a Simple Simple configurationconfiguration for which . for which .)(GCS

Where Where is the number of pebbles in a is the number of pebbles in a SSimple Configuration?.imple Configuration?.

SC

ReferencesReferences

[1] Wyels, Cynthia; “Optimal Pebbling of [1] Wyels, Cynthia; “Optimal Pebbling of Paths and Cycles” May 30, 2003, pg 6.Paths and Cycles” May 30, 2003, pg 6.

[2] Sjöstrand, Jonas; “The Cover Pebbling [2] Sjöstrand, Jonas; “The Cover Pebbling Theorem, arXiv: math.CO/0410129 v1; Theorem, arXiv: math.CO/0410129 v1; October 6.October 6.