Optimal 2-pebbling number Hung-Hsing Chiang Department of Mathematics Chung Yuan Christian University Advisor : Chin-Lin Shiue and Mu- Ming Wong
Jan 02, 2016
Optimal 2-pebbling number
Hung-Hsing ChiangDepartment of Mathematics
Chung Yuan Christian University Advisor : Chin-Lin Shiue and Mu-Ming
Wong
Definitions
A distribution δ of pebble of G is a function : V(G) →N∪ { 0 } δ ( v ) be the number of pebbles
distribution to v.
δ ( G ) be the number of pebbles that be distributed vertices of G .
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1.Pebbling move consists of removing two pebbles from one vertex and then placing one pebble at an adjacent vertex.
2.If a distribution δ of pebbles lets us move at least one pebbles to each vertex v by applying pebbling moves repeatly(if necessary), then δ is called a pebbling of G.
3. δG(v) denote the maximum number of pebbles which can be moved to v by applying pebbling moves on G.
4.A pebbling type αof G is a mapping from
V(G) onto N∪ { 0 }
5.A distribution δ is called an α-pebbling if whenever we choose a target vertex v, we can move α(v) pebbles to v by applying pebbling moves.
6.An ι-pebbling of G is an α-pebbling of G for which α(v) = ιfor each v in G, where ιis a positive integer.
7.The optimal α pebbling number of G f’α (G) is the minimum number of pebbles used in an α-pebbling of G.
Note : optimal 1 pebbling number of G is optimal pebbling number of G ,f’ (G)
Example
δ(u1)=1 δ(u2)=0 δ(u3)=3
δ(u4)=2 δ(u5)=2 δ(G)=8
δG(u1)=3 δG(u2)=3 δG(u3)=4
δG(u4)=4 δG(u5)=3
δ is a 3-pebbling of G.
u 5
1
0
32
2
u 4 u 3
u 2
u 1
G= Cn
Preliminaries
Definetion 2.1 Let x be a nonnegative integer. We define
and for each i > 0
21
xx i
i
xx 0
Cycle graph Cn
Cn =u0u1…un-1u0
where subscript is modulo n Two path P+(ui , uj)=uiui+1…uj and
P-(ui , uj)=uiui-1…u1unun-1…uj
Let δ be a distribution of Cn.. If we let δ+ be a distribution of P+(ui , uj) such that δ+(ui )=x ≦δ(ui ) and δ+(v)= δ(v) for each v≠ui in P+(ui , uj) then
m+(δ,x, ui , uj) denotes
where P+= P+(ui , uj) .
The similarly m-(δ,y, ui , uj) = where P-= P-(ui , uj) .
j jPu u
j jPu u
Fact 1
(i)We can choose some vertives w of Cn and two integers x≧2 and y ≧ 2, x+y=δ(w), such that
δCn(u) = m+(δ,x,w,u)+ m-(δ,y,w,u)+ δ(u) (ii)We can choose some vertives w of Cn with δ(w) ≧2 such that
δCn(u) = m+(δ, δ(w) ,w,u)+ δ(u) or
δCn(u) = m-(δ, δ(w) ,w,u)+ δ(u)
Fact 1
(iii)We can choose two vertives ui
and uj in V(Cn) where δ(ui)≧2 ,δ(uj) ≧ 2 and P+(ui , u) ∩ P-(uj, u)= { u } , such that δCn(u) = m+(δ, δ(ui), ui,u)+ m-(δ, δ(uj) ,uj,u)+ δ(u)
(iv)δCn(u) = δ(u)
Lemma 2.3
Let δ be a distribution of Cn such that δ(u) ≦2 for each u in Cn and let path P be a subgraph of Cn with an endvertex v .Then δP(v) ≦δ(v) +1
Corollary 2.4
Let δ be a distribution of Cn such that δ(u) ≦2 for each u in Cn and let v and w be two distinct vertices of Cn.. Then m+(δ,x,w,v)≦1 and m-(δ,y,w,v)≦1 for any two nonnegative integers x and y with x+y=δ(w)
The optimal 2-pebbling number of Cn
Proposition3.1 nCf n '
2
u20
2
0
11
1
1
1
1
11
2
u4
u3
u1
otherwise 1
or if 2
or if 0
42
31
iii
iii
ui
The prove of Lemma3.2
1.Let δ be an optimal 2-pebbling of Cn and δ(uk)=max{ δ(v) |v ∊ V(Cn) } .
2.Pk=uk-muk-m+1…uk…uk+l-1uk+l where δ(uk-m)=
δ(uk+l)=0 and δ(v) > 0 for each internal vertex of Pk.
3.Let δ’ be a distribution of Cn defined by δ’(uk)= δ(uk)-2, δ’(uk-m)= δ(uk-m)+1, δ’(uk+l)= δ(uk+l)+1 and δ’(v)= δ(v) for v∊V(Cn)∖ { uk-m , uk , uk+l} .
4.Claim 1. δ’Cn(u) ≧2 for each vertex u∊V(Pk)
Claim 2. δ’Cn(u) ≧2 for each vertex u∊ V(Cn) ∖ (Pk)
Lemma3.3
Let δ is an optimal 2-pebbling of Cn
with δ(u) ≦2 for each u in Cn . If δ(uj)=0 then exists a path P= uiui+1…ujuj+1…uk such that δ(ui)=δ(uk)=2 and δ(u)=1 for each vertex u≠ui ,uj ,uk in P.
The prove of Lemma3.3
1. There is a path P= uiui+1…ujuj+1…uk such that δ(ui)=δ(uk)=2 and δ(u) ≦1 for each vertex u≠ui ,uj ,uk in P .
2. Suppose there exists a vertex ul for i < l < j such that δ(ul) =0
3. By Fact1 and Corollary 2.4 δCn(uj) ≦1
Proposition3.4
Proof: Let δ is an optimal 2-pebbling of Cn with
δ(u) ≦2 for each u in Cn . Assume δ(Cn) =k ≦n-1. Let A and B are two subsets of V(Cn) where
A= { u ∊ V(Cn) |δ(u)=0 } and B= { u ∊ V(Cn) |δ(u)=2 } . Since δ(Cn)≦n-1 and δ(u) ≦2 for each u ∊ V(Cn), |A|≧|B|+1. But |A|≦|B| by Lemma3.3, it is a contradiction. Hence δ(Cn)≧n.
2
'nf C n