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The Multi-level Distance Number for Symmetric

Lobster-like Trees about the Weight Center

∗

Liancui Zuo, Hongfang Guo, and Chunhong Shang

†

Abstract

The multi-level distance labeling for a network G is a functionf :V(G)→ {0,1,2,· · · }so that

|f(u)−f(v)| ≥diam(G) + 1−d(u, v)

for any u, v ∈ V(G), where diam(G) is the diameter of G and d(u, v) is the distance between u and v. The span of f is defined asmax{f(u)−f(v)|u, v∈V(G)}. The multi-level distance number of G is the minimum span of all multi-level distance labelings for G. In the present paper, a class of symmetric lobster-like trees about the weight center is studied, and its multi-level distance number is obtained.

Keywords: multi-level distance number; multi-level dis-tance labeling; symmetric lobster-like tree about the weight center; the minimum span; network

1

Introduction

In recent years, many parameters and classes of graphs are consideried. For example, in [9], different proper-ties of the intrinsic order graph were obtained, namely those dealing with its edges, chains, shadows, neighbors and degrees of its vertices, and some relevant subgraphs, as well as the natural isomorphisms between them. In [17], the n-dimensional cube-connected complete graph is studied. In [22], the linear(n−1)-arboricity ofKn(m)is

obtained. In [23, 24], the hamiltonicity, path t-coloring, and the shortest paths of Sierpin´ski-like graphs are re-searched. In [25], the vertex arboricity of integer distance graph G(Dm,k)is obtained. In [26], it is obtained that

la4(Kn,n) =d5n/8eforn≡0( mod 5).

Multi-level distance labeling (or radio labeling) is moti-vated by the channel assignment problem introduced by Hale[1]. Given a set of stations (or transmitters) in a communication network, a valid channel assignment is a function that assigns to each station with a channel (nonnegative integer) such that interference is avoided. The level of interference is related to the locations of the stations–the closer the two stations, the stronger the interference that might occur. In order to avoid inter-ference, the separation between the channels assigned a

∗_{Manuscript received Dec 23,2014. This work was supported by}

NSFC for youth with code 61103073. Email:[email protected]. Lian-cui Zuo and Chunhong Shang are with College of Mathematical Science, Tianjin Normal University, Tianjin, 300387, China; Hong-fang Guo is with Department of Mathematics, Institute of Tech-nology, East China Normal University, Shanghai,200241,China.

†_{The corresponding author:[email protected]}

pair of near-by stations must be large enough, and the amount of the required separation depends on the dis-tance between the two stations. The task is to find a valid channel assignment with the minimum span of channels used.

A graph model for this problem is to represent each sta-tion by a vertex, and connect any pair of close stasta-tions by an edge. A multi-level distance labeling (radio label-ing) of a connected graph G is a function f : V(G) → {0,1,2,· · · }, such that for any u, v∈V(G),

|f(u)−f(v)| ≥diam(G) + 1−d(u, v),

where diam(G)is the diameter (the maximum distance over all pairs of vertices) ofG. The span off is defined as max{f(u)−f(v) | u, v ∈ V(G)}. The multi-level distance number for a graph G, denoted by rn(G), is the minimum span of all multi-level distance labeling for

G. Multi-level distance labeling is a generalization of the distance-two labeling which has been studied extensively ([2]-[12]), and multi-level distance labeling can be better to reflect the nature of radio channels assignment. In [14, 15, 18, 19], it was studied that the multi-level distance labeling of paths and cycles, square of paths and square of cycles, and in [13], it was determined the radio number of the complete m-ary tree. In [16], it was studied the multi-level distance labeling of trees, and got a lower bound for trees’ radio number.

LetT be a tree rooted at a vertexr. For any two vertices

u and v, if u is on the (r, v)-path, then u is called an ancestor ofv, and v is called a descendent ofu. Define the level function on V(T) by lr(u) = d(r, u) for any u∈V(T). For anyu, v∈V(T), define

ϕ(u, v) = max{lr(t) :tis a common ancestor of u and v}.

For each vertexwin a tree T, the weight ofT rooted at

wis defined by

ωT(w) = X

u∈V(T)

lw(u).

The weight ofT is the smallest weight among all possible roots ofT

ω(T) = min

w∈V(T){ωT(w)}.

A vertex w0 is called a weight center of T ifωT(w0) =

ω(T). It can be abbreviated asl(u) =lw0(u) if there is

no confusion. It is obvious that the weight center cut the tree into a number of branches.

IAENG International Journal of Applied Mathematics, 45:1, IJAM_45_1_05

Lemma 1.1. [16] Suppose that w0 is the weight center ofT. For anyu, v∈V(T), the following two conclusions hold:

(1)d(u, v) =l(u) +l(v)−2ϕ(u, v), and

(2)ϕ(u, v) = 0if and only if uandv belong to different branches (unless one of them isw0).

Lemma 1.2. [16] Let T be ann-vertex tree with diam-eter d. Then

rn(T)≥(n−1)(d+ 1) + 1−2ω(T).

An arrangement for V(G) can be derived from the radio labeling f, denoted by V(G) = U(f) =

{u0, u1,· · · , u|V|−1}, which satisfies

0 =f(u0)< f(u1)< f(u2)<· · ·< f(u|V|−1). (1)

Iff is a radio labelling, then the span off isf(u|V|−1).

If deleting all suspension vertices and associated edges of T we get a road or an isolated vertex, then we call

T a "caterpillar". If deleting all suspension vertices and associated edges of T we get a caterpillar, then we call

T a "lobster tree". In 2009, Guo and Zuo gave the exact value of multi-level distance number for a special class of caterpillars in [20]. In 2011, Hou and Zuo [21] obtained the exact value of the multi-level distance number of a class of symmetric lobster trees about weight center. In [13], it was given the following concept.

Definition 1.3. Let f be a multi-level distance labeling for G, and the vertices of G about f have the sequence as (1). For every 0≤i≤ |V| −2, let

Jf(ui, ui+1) =f(ui+1)−f(ui)

−[diam(G) + 1−d(ui, ui+1)].

We call Jf(ui, ui+1) a k-jump from ui to ui+1 if

Jf(ui, ui+1) =k≥0 and say that f has a k-jump from

ui toui+1. Define the total number of jumps as

J(f) =

|V|−2

X

i=0

Jf(ui, ui+1).

If deleting all suspension vertices and edges associated of

T we can get a lobster tree, then we call it a "lobster-like tree". In the present paper, we mainly study the multi-level distance labeling of lobster-like trees, and obtain the multi-level distance number for a special class of lobster-like trees.

2

A lower bound of the radio number of

a class of symmetric lobster-like trees

about weight center

In the following, we denote a lobster-like tree’s all sus-pension vertices as the C layer points, the correspond-ing lobster tree’s all suspension vertices as the B layer points, and all suspension vertices of its corresponding caterpillar asAlayer points.

LetT = (t4, t5, t6,· · ·, ti,· · ·, tk−4, tk−3), and

R= (r2,0, r3,0, r4,1, r4,2,· · · , r4,2t4−4,· · ·,

ri,j,· · · , rk−3,1,· · ·, rk−3,2tk−3−4, rk−2,0, rk−1,0),

where k≥7, ti≥3, r2,0, r3,0, ri,j, rk−2,0, rk−1,0 ≥1, 1≤

j≤2ti−4, and4≤i≤k−3.

Using symbol Pk,T ,R to represent the lobster-like tree that the diameter is k−1, the degree of the ith vertex

vi,0 on the longest path Pk isti, and the others’ degree

are

r2,0+ 1, r3,0+ 1,· · ·, ri,j+ 1,· · ·, rk−2,0+ 1, rk−1,0+ 1,

respectively, except the suspension vertices, that is, the degree ofv3,0, vk−2,0arer3,0+ 1, rk−2,0+ 1, respectively,

the degree of its descendant vertices that belong to B layers arer2,0+ 1, rk−1,0+ 1, respectively, and among A

layer points, the degree of each vi,j is ri,j+ 1, and the

degree of its descendant vertices that belong to B layers isri,j+ti−2+ 1for1≤j≤ti−2and4≤i≤k−3. Thus

the number of vertices of Pk,T ,R is

|V(Pk,T ,R)|= k−3

P

i=4

[

ti−2

P

j=1

ri,j(ri,j+ti−2+ 1) + (ti−2)] +(r2,0+ 1)r3,0+ (rk−2,0+ 1)rk−1,0+k−4.

(2)

If r2,0 =r3,0 =ri,j =rk−2,0 =rk−1,0 =r(≥1), we will

denote the lobster-like tree that the degree of all vertices are r+ 1 except for the k−6 vertices in the middle of

Pk and the suspension vertices asPk,(t4,···,ti,···,tk−3),r. If

r = 1, then it is denoted by Pk,(t4,···,ti,···,tk−3), and we

have

A={(v3,0, vi,j, vk−2,0)|1≤j≤ti−2,4≤i≤k−3},

B={(v2,0, vi,j, vk−1,0)|ti−1≤j≤2ti−4,4≤i≤k−3},

and

C={(v1,0, vi,j, vk,0)|2ti−3≤j≤3ti−6,4≤i≤k−3}.

In this paper, we mainly study the multi-level distance number of Pk,(t4,···,ti,···,tk−3). Note that

diam(Pk,(t4,···,ti,···,tk−3)) =k−1.

Define the vertices of Pk,(t4,···,ti,···,tk−3) successively v1,0, v2,0,· · · , vk,0,vi,p(1 ≤p≤ti−2,4 ≤i≤k−3) is

thepth vertex ofvi,0that belong toAlayer,vi,q(ti−1≤

q ≤2ti−4,4 ≤i ≤k−3) is the (q−ti+ 2)th vertex of vi,0 that belong to B layer, and vi,s (2ti−3 ≤ s ≤

3ti−6,4≤i≤k−3)is the(s−2ti+ 4)th vertex ofvi,0

that belong to Clayer. Please see Fig. 1.

If Pk,T ,R is symmetric about the weight center, then vk+1

2 ,0

is the weight center whenk is odd, vi,0(1≤i ≤

k−1

2 )and its descendent vertices and the associated edges

are called "upper branch" of Pk,T ,R, vi,0(k+3₂ ≤ i≤ k)

and its descendent vertices and the associated edges are called "lower branch" of Pk,T ,R, and vk+1

2 ,0 and its

de-scendent vertices and the associated edges are called "middle" of Pk,T ,R. "Middle" belong to both the up-per branch and lower branch. In this paup-per, we always

IAENG International Journal of Applied Mathematics, 45:1, IJAM_45_1_05

assume that they belong to the upper branch. To dis-tinguish them from the upper branch vertices and refer to them as "upper middle" of Pk,T ,R. When k is even, both vk

2,0 and v

k

2+1,0 are the weight centers. In this

paper, we select vk

2,0 for the weight center for even k,

vi,0 (1 ≤ i ≤ k₂) and its descendent vertices and the

associated edges are called "upper branch" of Pk,T ,R,

vi,0 (k₂ + 1 ≤ i ≤ k) and its descendent vertices and

the associated edges are called "lower branch" ofPk,T ,R,

vk

2,0and its descendent vertices and the associated edges

are called "upper middle" of Pk,T ,R, andvk

2+1,0 and its

descendent vertices and the associated edges are called "lower middle" ofPk,T ,R.

r r r r r r r r r r r r _v1,0

v2,0

v3,0

v4,0

v5,0

v6,0

v7,0

v8,0

v9,0

v10,0

v11,0

v12,0

P P P P P_P r r r r r r r r r r r r r r r r r r r r r r

v4,2

v4,1

v4,6

v4,5

v4,10

v4,9

v4,3

v4,4

v4,7

v4,8

v4,11

v4,12

P P P P P_P r r r r r r r r r r r r r r r r r r r r r r

v5,2

v5,1

v5,6

v5,5

v5,10

v5,9

v5,3

v5,4

v5,7

v5,8

v5,11

v5,12

P P P P P_P r r r r r r r r r r r r r r r r r r r r r r

v6,2

v6,1

v6,6

v6,5

v6,10

v6,9

v6,3

v6,4

v6,7

v6,8

v6,11

v6,12

P P P P P_P r r r r r r r r r r r r r r r r r r r r r r

v7,2

v7,1

v7,6

v7,5

v7,10

v7,9

v7,3

v7,4

v7,7

v7,8

v7,11

v7,12

P P P P P_P r r r r r r r r r r r r r r r r r r r r r r

v8,2

v8,1

v8,6

v8,5

v8,10

v8,9

v8,3

v8,4

v8,7

v8,8

v8,11

v8,12

P P P P P_P r r r r r r r r r r r r r r r r r r r r r r

v9,2

v9,1

v9,6

v9,5

v9,10

v9,9

v9,3

v9,4

v9,7

v9,8

v9,11

v9,12

Figure 1. A special symmetric lobster-like tree

Theorem 2.1. Let G=Pk,T ,R be a symmetric lobster-like tree about the weight center. By direct calculation, we can obtain that

|V(G)|=

                             2 k−1 2 P i=4 [

ti−2

P

j=1

ri,j(ri,j+ti−2+ 1) + (ti−1)]

+

tk+1 2

−2

P

l=1

rk+1 2 ,l

(rk+1 2 ,l+tk+1

2

−2+ 1)

+tk+1

2 + 2r2,0[r3,0+ 1] + 1f or odd k≥9,

2 k 2 P i=4 [

ti−2

P

j=1

ri,j(ri,j+ti−2+ 1) + (ti−1)] +2r2,0(r3,0+ 1) + 2 f or even k,

and

ω(G) =

                                             k−1 2 P i=4 {

ti−2

P

j=1

ri,j[(k+ 7−2i)ri,j+ti−2 +(k+ 5−2i)] + (k+ 3−2i)(ti−1)}

+

tk+1 2

−2

P

l=1

rk+1 2 ,l(3r

k+1 2 ,l+tk+1

2

−2+2)

+tk+1 2

+ [(k−1)r3,0+ (k−3)]r2,0

f or odd k≥9, k

2

P

i=4 {

ti−2

P

j=1

ri,j[(k+ 7−2i)ri,j+ti−2 +(k+ 5−2i)] + (k+ 3−2i)(ti−1)}

+[(k−1)r3,0+ (k−3)]r2,0+ 1

f or even k.

Similarly, the following conclusions can be obtained.

Corollary 2.2. Let G=Pk,(t4,···,ti,···,tk−3),r. Then

|V(G)|=

                 [2 k−1 2 P i=4

(ti−2) +tk+1 2 ](r

2_{+r+ 1)}

+k−6 f or odd k≥9,

[2 k

2

P

i=4

(ti−2) + 2](r2+r+ 1)

+k−6 f or even k,

and

ω(G) =

                             k−1 2 P i=4

{[(k+ 7−2i)r2_{+ (k+ 5−2i)r](ti−2)}

+(k+ 3−2i)(ti−1)}

+(tk+1 2

−2)(3r2_{+ 2r) +t}

k+1 2

+(k−1)r2_{+ (k−3)r f or odd k≥9,}

k

2

P

i=4

{[(k+ 7−2i)r2+ (k+ 5−2i)r](ti−2) +(k+ 3−2i)(ti−1)}+ (k−1)r2

+(k−3)r+ 1 f or even k.

Corollary 2.3. Let G=Pk,(t4,···,ti,···,tk−3). Then

|V(G)|=

             6 k−1 2 P i=4

(ti−2) + 3tk+1 2

+k−6

f or odd k≥9,

6 k

2

P

i=4

(ti−2) +k f or even k,

and

ω(G) =

             k−1 2 P i=4

[(3k+ 15−6i)(ti−2)] + 6tk+1 2

+k₄2 −49

4 f or odd k≥9,

k

2

P

i=4

[(3k+ 15−6i)(ti−2)] + k₄2 f or even k.

By Lemma 1.2 and equation (2), it is easy to verify that

rn(G)≥(3k−12)tk+1 2 +k

2_{−7k+ 1f or k= 7.(3)}

By Theorem 2.1, Corollaries 2.2-2.3, and Lemma 1.2, we have the following conclusions.

IAENG International Journal of Applied Mathematics, 45:1, IJAM_45_1_05

Theorem 2.4. Let G=Pk,T ,R be a symmetric

lobster-like tree about the weight center. Then

rn(G)≥

                    

                   

k−1 2

P

i=4 {

ti−2

P

j=1

ri,j[(4i−14)ri,j+ti−2+ (4i−10)] +(4i−6)(ti−1)}

+

tk+1 2

−2

P

l=1

rk+1 2 ,l

[(k−6)rk+1 2 ,l+tk+1

2 −2

+(k−4)] + (k−2)tk+1 2

+2r2,0(r3,0+ 3) + 1 f or odd k≥9,

k

2

P

i=4 {

ti−2

P

j=1

ri,j[(4i−14)ri,j+ti−2+ (4i−10)] +(4i−6)(ti−1)}+2r2,0(r3,0+ 3)

+k−1 f or even k.

Theorem 2.5. Let G=Pk,(t4,···,ti,···,tk−3),r. Then

rn(G)≥

                 

                

k−1 2

P

i=4

{[(4i−14)r2_{+ (4i−10)r](ti−2)}

+(4i−6)(ti−1)}

+[(k−6)r2_{+ (k−4)r](t}

k+1 2

−2) +(k−2)tk+1

2 + 2r

2_{+ 6r+ 1}

f or odd k≥9, k

2

P

i=4

{[(4i−14)r2_{+ (4i−10)r](t}

i−2)

+(4i−6)(ti−1)}

+2r2+ 6r+k−1 f or even k.

Theorem 2.6. Let G=Pk,(t4,···,ti,···,tk−3). Then

rn(G)≥

        

       

6 k−1

2

P

i=4

[(2i−5)(ti−2)] + (3k−12)tk+1 2

+1₂(k−7)2+ 1 f or odd k≥9,

6 k

2

P

i=4

[(2i−5)(ti−2)]

+k₂2 −k+ 1 f or even k.

In order to improve the lower bound of the multi-level distance number ofGfor oddk≥9, we give the following lemma.

Lemma 2.7. Suppose that G=Pk,(t4,···,ti,···,tk−3),f is

a one-to-one non-negative integer function onV(G), and the vertices inGaboutf have the sequence as (1). Then

f is a radio labeling ofGif and only if for any consecutive subset of vertices{ui, ui+1,· · · , uj},0≤i < j≤ |V| −1,

the following results hold:

(1) if ui, uj belong to different branches ofG, then

j−1

P

t=i

[Jf(ut, ut+1) + 2ϕ(ut, ut+1)]

≥2

j−1

P

t=i+1

l(ut)−k(j−i−1),

(2) Ifui, uj belong to the same branches ofG, then

(i)when any one isn’t the ancestor of another one, we

have

j−1

P

t=i

[Jf(ut, ut+1) + 2ϕ(ut, ut+1)]

≥

                                       

                                      

2

j−1

P

t=i+1

l(ut)−k(j−i−1) +2 min{l(ui)−1, l(uj)−1}

if ui ∈A or deg(ui) =ti, uj ∈A,

2

j−1

P

t=i+1

l(ut)−k(j−i−1) +2 min{l(ui)−1, l(uj)−2}

if ui ∈A or deg(ui) =ti, uj ∈B,

2

j−1

P

t=i+1

l(ut)−k(j−i−1) +2 min{l(ui)−1, l(uj)−3}

if ui ∈A or deg(ui) =ti, uj ∈C,

2

j−1

P

t=i+1

l(ut)−k(j−i−1)

+2 min{l(ui)−2, l(uj)−2} if ui, uj ∈B,

2

j−1

P

t=i+1

l(ut)−k(j−i−1)

+2 min{l(ui)−2, l(uj)−3}if ui ∈B, uj∈C,

2

j−1

P

t=i+1

l(ut)−k(j−i−1)

+2 min{l(ui)−3, l(uj)−3} if ui, uj ∈C,

whereui anduj may exchange their positions.

(ii) when one vertex is the ancestor of another one, we have

j−1

P

t=i

[Jf(ut, ut+1) + 2ϕ(ut, ut+1)]

≥2

j−1

P

t=i+1

l(ut)−k(j−i−1) + 2 min{l(ui), l(uj)},

Proof. Suppose thatf is a multi-level distance labeling ofGwithdiam(G) =k−1. For any0≤i < j≤ |V| −1, add up the following equations

Jf(ut, ut+1) + 2ϕ(ut, ut+1)

=f(ut+1)−f(ut) +l(ut+1) +l(ut)−diam(G)−1,

i≤t≤j−1,

we obtain that

j−1

P

t=i

[Jf(ut, ut+1) + 2ϕ(ut, ut+1)]

=f(uj)−f(ui)−k(j−i) +2[

j−1

P

t=i+1

l(ut)] +l(ui) +l(uj).

By the definition off, we have

f(uj)−f(ui)≥k−l(ui)−l(uj),

and then the condition (1) holds.

Since

f(uj)−f(ui) =k(j−i)−2[

j−1

P

t=i+1

l(ut)]−l(ui)−l(uj)

+

j−1

P

t=i

[Jf(ut, ut+1) + 2ϕ(ut, ut+1)]

IAENG International Journal of Applied Mathematics, 45:1, IJAM_45_1_05

≥

                       

                      

k−l(ui)−l(uj) + 2 min{l(ui), l(uj)}

if ui(uj)is the ancestor of uj(ui), k−l(ui)−l(uj) + 2 min{l(ui)−1, l(uj)−1}

if ui ∈A or deg(ui) =ti, uj ∈A, k−l(ui)−l(uj) + 2 min{l(ui)−1, l(uj)−2}

if ui ∈A or deg(ui) =ti, uj ∈B, k−l(ui)−l(uj) + 2 min{l(ui)−1, l(uj)−3}

if ui ∈A or deg(ui) =ti, uj ∈C, k−l(ui)−l(uj) + 2 min{l(ui)−2, l(uj)−2}

if ui, uj ∈B, k−l(ui)−l(uj) + 2 min{l(ui)−2, l(uj)−3}

if ui∈B, uj ∈C, k−l(ui)−l(uj) + 2 min{l(ui)−3, l(uj)−3}

if ui, uj∈C,

the condition (2) holds.

Assume that f satisfies conditions (1) and (2). If ui, uj

are in different branches, then

d(ui, uj) =l(ui) +l(uj),

and

f(uj)−f(ui)≥k−l(ui)−l(uj) =k−d(ui, uj).

Ifui, uj are in a same branch, then

d(ui, uj) =

                       

                      

l(ui) +l(uj)−2 min{l(ui), l(uj)}

if ui(uj)is the ancestor of uj(ui), l(ui) +l(uj)−2 min{l(ui)−1, l(uj)−1} if ui ∈A or deg(ui) =ti, uj∈A, l(ui) +l(uj)−2 min{l(ui)−1, l(uj)−2} if ui ∈A or deg(ui) =ti, uj∈B, l(ui) +l(uj)−2 min{l(ui)−1, l(uj)−3}

if ui ∈A or deg(ui) =ti, uj∈C, l(ui) +l(uj)−2 min{l(ui)−2, l(uj)−2}

if ui, uj∈B, l(ui) +l(uj)−2 min{l(ui)−2, l(uj)−3}

if ui∈B, uj ∈C, l(ui) +l(uj)−2 min{l(ui)−3, l(uj)−3}

if ui, uj∈C,

and thus f(uj)−f(ui) ≥ k−d(ui, uj). Hence f is a multi-level distance labeling ofG.

Now we revise the lower bound of the multi-level distance number for the symmetric lobster-like tree about weight center in Theorem 2.6 for odd k≥9.

Theorem 2.8. LetG=Pk,(t4,···,ti,···,tk−3)be a

symmet-ric lobster-like tree about weight center. For odd k ≥9

andt4=tk+1

2 =tk−3, we have

rn(G)≥6 k−1

2

P

i=4

[(2i−5)(ti−2)]

+(3k−12)tk+1 2 +

1 2(k−7)

2_{+ 2.}

Proof. Letf be a radio labeling ofG, and the vertices in

G aboutf have the sequence as (1). By Definition 1.3

and Lemma 2.7, we have

f(u_|V|−1) = |V|−2

P

i=0

[f(ui+1)−f(ui)]

=k(|V| −1)− |V|−2

P

i=0

d(ui, ui+1) + |V|−2

P

i=0

Jf(ui, ui+1)

=k(|V| −1)−2ω(G) +l(u0) +l(u|V|−1) +σ(f) ≥k[6

k−1 2

P

i=4

(ti−2) + 3tk+1 2

+k−7] + 1 +σ(f)

−2{

k−1 2

P

i=4

[(3k+ 15−6i)(ti−2)] + 6tk+1 2 +

k2

4 − 49

4}

= 6 k−1

2

P

i=4

[(2i−5)(ti−2)] + (3k−12)tk+1 2

+1₂(k−7)2_{+ 1 +σ(f),}

where σ(f) =

|V|−2

P

i=0

[Jf(ui, ui+1) + 2ϕ(ui, ui+1)]. The

weights of all vertices appear twice except for l(u0)and

l(u|V|−1). Note thatl(ui)≥0for0≤i≤ |V|−1.

There-fore, only ifu0=vk+1 2 ,0

and the distance fromu|V|−1to

u0is one, that is,l(u0) = 0andl(u|V|−1) = 1, the

right-hand side of above formulae gets its minimum value.

Claim 1. Under the conditiont4 =tk+1

2 =tk−3, there

must exist a vertex

ui∈ {v1,0, v4,j, vk−3,j, vk,0|2t4−3≤j≤3t4−6}

such that ui−1, ui+1 6= vk+1 2 ,q

, where 2tk+1 2

−3 ≤ q ≤

3tk+1 2

−6.

Assume that every vertex

ui∈ {v1,0, v4,j, vk−3,j, vk,0|2t4−3≤j≤3t4−6}

forG=Pk,(t4,···,ti,···,tk−3)(k≥9)is adjacent to one

ver-tex vk+1

2 ,q for some2t

k+1

2 −3≤q≤3t

k+1

2 −6, then

tk+1 2 −

2≥ 1

2(2tk+12 −

4 + 2) =tk+1 2 −

1,

a contradiction. Hence the claim holds.

Claim 2. Suppose thatui satisfies Claim 1, then

σ(f) =

|V|−2

X

t=0

[Jf(ut, ut+1) + 2ϕ(ut, ut+1)]≥1.

Consider a consecutive subset of vertices{ui−1, ui, ui+1}

with2≤i≤ |V| −2. Then it is clear that there are two vertices belong to the same branch.

(1) (i) Ifui−1, ui+1 belong to the same branch ofGand

any one vertex isn’t the ancestor of the other one, then

σ(f) =

|V|−2

P

t=0

[Jf(ut, ut+1) + 2ϕ(ut, ut+1)]

≥

i

P

t=i−1

[Jf(ut, ut+1) + 2ϕ(ut, ut+1)]

IAENG International Journal of Applied Mathematics, 45:1, IJAM_45_1_05

≥

                   

                  

2l(ut)−k+ 2 min{l(ui)−1, l(uj)−1}, if ui∈A or deg(ui) =ti, uj∈A

2l(ut)−k+ 2 min{l(ui)−1, l(uj)−2}, if ui∈A ordeg(ui) =ti, uj ∈B,

2l(ut)−k+ 2 min{l(ui)−1, l(uj)−3}, if ui∈A ordeg(ui) =ti, uj ∈C.

2l(ut)−k+ 2 min{l(ui)−2, l(uj)−2}, if ui, uj ∈B,

2l(ut)−k+ 2 min{l(ui)−2, l(uj)−3}, if ui∈B, uj∈C,

2l(ut)−k+ 2 min{l(ui)−3, l(uj)−3}, if ui, uj ∈C,

≥2·k−1

2 −k+ 2 = 1.

Note that the upper result also holds when we exchange the positions ofui anduj.

(ii) Ifui−1, ui+1belong to the same branch ofGand one

vertex is the ancestor of another one, then

σ(f) =

|V|−2

P

t=0

[Jf(ut, ut+1) + 2ϕ(ut, ut+1)]

≥

i

P

t=i−1

[Jf(ut, ut+1) + 2ϕ(ut, ut+1)] ≥2l(ui)−k+ 2 min{l(ui−1), l(ui+1)} ≥2·k−1

2 −k+ 2 = 1.

(2) (i) If ui−1, ui belong to the same branch ofG and

one vertex is the ancestor of the other one, then

σ(f) =

|V|−2

P

t=0

[Jf(ut, ut+1) + 2ϕ(ut, ut+1)] ≥Jf(ui−1, ui) + 2ϕ(ui−1, ui) ≥2 min{l(ui−1), l(ui)} ≥2.

(ii) Ifui−1, ui belong to the same branch ofG and any

one vertex isn’t the ancestor of the another one, then

σ(f) =

|V|−2

P

t=0

[Jf(ut, ut+1) + 2ϕ(ut, ut+1)] ≥Jf(ui−1, ui) + 2ϕ(ui−1, ui)

≥

                   

                  

2 min{l(ui)−1, l(uj)−1},

if ui∈A or deg(ui) =ti, uj∈A

2 min{l(ui)−1, l(uj)−2},

if ui∈A ordeg(ui) =ti, uj ∈B,

2 min{l(ui)−1, l(uj)−3},

if ui∈A ordeg(ui) =ti, uj ∈C,

2 min{l(ui)−2, l(uj)−2},

if ui, uj∈B,

2 min{l(ui)−2, l(uj)−3},

if ui ∈B, uj∈C,

2 min{l(ui)−3, l(uj)−3},

if ui, uj∈C.

≥2,

where the positions ofuiand uj may exchange.

(3) For the same reason as (2), ifui, ui+1 belong to the

same branch, then we have

σ(f) =

|V|−2

P

t=0

[Jf(ut, ut+1) + 2ϕ(ut, ut+1)] ≥Jf(ui, ui+1) + 2ϕ(ui, ui+1)≥2.

Thus Claim 2 holds.

By arbitrary off, we have

rn(G)≥6 k−1

2

P

i=4

[(2i−5)(ti−2)] +(3k−12)tk+1

2 + 1 2(k−7)

2_{+ 2.}

3

The radio number of a class of

sym-metric lobster-like trees about weight

center

The radio number ofG=Pk,(t4,···,ti,···,tk−3)is given

be-low.

Theorem 3.1. Let G=Pk,(t4,···,ti,···,tk−3) which

satis-fies

(i) all ti have the same parity,

(ii) t4 = tk+1 2

= tk−3 and tk+1 2 −j

≥ t4+j for 1 ≤ j ≤

k−7 4

, whenk is odd,

(iii)tk

2 ≥ 1

2(t4−2)+2andtk₂−j≥t4+jfor1≤j≤k−8₄

if both k and ti are even, and tk

2 ≥ 1

2(t4−1) + 2 and

tk

2−j ≥t4+j for1≤j≤

k−8 4

ifkis even andti is odd.

Then

rn(G) =

          

         

(3k−12)tk+1 2 +k

2_{−7k+ 1 f or k= 7,}

6 k−1

2

P

i=4

[(2i−5)(ti−2)] + (3k−12)tk+1 2

+1₂(k−7)2_{+ 2 f or odd k≥9,}

6 k

2

P

i=4

[(2i−5)(ti−2)] + k₂2 −k+ 1

f or even k.

Proof. By Theorems 2.6, 2.8, and equation (3), it is only need to prove the opposite inequality. Rearrange the sequence of vertices of G as V(G) = U(f) =

{u0, u1,· · ·, u|V|−1}. In the following, we will use an

algorithm to construct a multi-level distance labeling

f. The symbol → (l) indicates that the jump between the two successive vertices ui and ui+1 is l, that is,

Jf(ui, ui+1) = l, and the symbol → indicates that the

jump between the two successive verticesui andui+1 is

0, that is,Jf(ui, ui+1) = 0. Letf(u0) = 0, and

f(ui+1) =f(ui)+diam(G)+1−d(ui, ui+1)+Jf(ui, ui+1)

for0≤i≤ |V| −2.

Case 1. Ifk is even,ti is odd, tk

2 ≥ 1

2(t4−1) + 2 and

tk

2−j ≥ t4+j for 1 ≤ j ≤

k−8 4

, then the algorithm is defined as following:

u0=vk

2,0→vk,0→v1,0→vk2+1,2tk

2+1 −3

→v4,2t4−3→vk−3,2tk−3−3→vk 2,2tk

2 −3→

vk−3,2tk−3−2→v4,2t4−2→vk₂+1,2tk

2+1 −2

→ · · · →vk

2+1,

t₄

2+2tk

2+1 −9

2

→v4,3t4−6→

IAENG International Journal of Applied Mathematics, 45:1, IJAM_45_1_05

vk−3,3tk−3−6→vk

2,

t₄

2+2tk

2 −9

2

→vk−1,0→

v2,0→vk

2+1,

t₄

2+2tk

2+1 −7

2 →vk

2,

t₄

2+2tk

2 −7

2 → · · · →vk

2+1,3tk

2+1

−6→vk

2,3tk

2

−6→vk−2,0 →vk

2−2,0→v

k

2+2,2tk

2+2

−3→v5,2t5−3→ · · · →vk

2+2,3t5−6→v5,3t5−6→vk−3,0→v

k

2−3,0 → · · · →v_b3k−4

4 c,2t_b3k−4 4 c

−3

→v_bk+8 4 c,2t_bk+8

4 c

−3→ · · · →vb3k−4 4 c,3t_bk+8

4 c −6

→v_bk+8 4 c,3t_bk+8

4 c

−6→vb3k+2 4 c,0

→v_bk+2 4 c,0 →v_b3k

4c,2tb3k

4c

−3→vbk+12

4 c,2tbk+12 4 c

−3→ · · · →v_b3k

4c,3t_b3k

4c

−6→vbk+12 4 c,3t_b3k

4c −6

→v_b3k−2 4 c,0→

v_bk−2

4 c,0→ · · · →

vk−4,2tk−4−3→vk₂−1,2tk

2−1

−3→ · · ·

→vk−4,3tk−4−6→vk₂−1,3tk−4−6→vk₂+3,0→

v3,0→vk

2+2,3t5−5→vbk+124 c,3t_b3k

4c

−5→ · · · →v_b3k−4

4 c,3t_b3k−4 4 c

−6→vk

2−1,3tk

2−1 −6→

vk

2+1,tk

2+1

−1→v4,t4−1→vk−3,tk−3−1→vk₂,tk

2 −1

→vk−3,tk−3→v4,t4→vk₂+1,tk

2+1

→ · · · →

vk

2+1,

t4 2+tk

2+1 −5

2

→v4,2t4−4→vk−3,2tk−3−4→ vk

2,

t4 2+tk

2 −5

2 →vk

2+1,

t4 2+tk

2+1 −3

2 →

vk

2,

t4 2+tk

2 −3

2

→ · · · →vk

2+1,2tk

2+1

−4→vk

2,2tk

2 −4

→vk

2+2,tk

2+2

−1→v5,t5−1→ · · · →vk

2+2,2t5−4 →v5,2t5−4→ · · · →v_b3k−4

4 c,t_b3k−4 4 c

−1

→v_bk+8 4 c,t_bk+8

4 c

−1→ · · · →vb3k−4 4 c,2t_bk+8

4 c −4

→v_bk+8 4 c,2t_bk+8

4 c

−4→vb3k

4c,t_b3k

4c −1

→v_bk+12 4 c,t_b3k

4c

−1→ · · · →vb3k

4c,2t_b3k

4c −4

→v_bk+12 4 c,2t_b3k

4c

−4→ · · · →vk−4,tk−4−1→

vk

2−1,tk

2−1

−1→ · · · →vk−4,2tk−4−4→vk₂−1,2tk−4−4

→vk

2+2,2t5−3→vbk+124 c,2t_b3k

4c

−3→ · · · →v_b3k−4

4 c,2t_b3k−4 4 c

−4→vk2−1,2tk

2−1

−4→vk

2+1,1 →v4,1→vk−3,1→vk

2,1→vk−3,2 →v4,2→vk

2+1,2→v4,3→vk−3,3→ · · · →vk

2+1,

t4−1 2

→v4,t4−2→vk−3,tk−3−2→ vk

2,

t4−1 2

→vk

2+1,

t4 +1 2

→vk

2,

t4 +1 2

→vk

2+1,

t4 +3 2

→

vk

2,

t4 +3 2

→ · · · →vk

2+1,tk

2+1

−2→vk

2,tk

2 −2

→vk

2+2,1→v5,1→v

k

2+2,2→v5,2→ · · · →vk

2+2,t5−2→v5,t5−2→ · · · →vb 3k−4

4 c,1 →v_bk+8

4 c,1

→ · · · →v_b3k−4 4 c,t_bk+8

4 c −2→

v_bk+8 4 c,t_bk+8

4 c

−2→vb3k

4c,1

→v_bk+12 4 c,1

→ · · · →

v_b3k

4c,t_b3k

4c

−2→vbk+12 4 c,t_b3k

4c

−2→ · · · →vk−4,1 →vk

2−1,1→ · · · →vk−4,tk−4−2 →vk

2−1,tk−4−2→vk2+2,t5−1→vbk+124 c,t_b3k

4c −1

→ · · · →v_b3k−4 4 c,t_b3k−4

4 c

−2→vk

2−1,tk

2−1 −2

→vk

2+2,0→v

k

2−1,0→v

k

2+1,0.

By the definition off, we know thatσ(f) = 0. Hence

f(u|V|−1) = 6

k

2

X

i=4

(2i−5)(ti−2) +k

2

2 −k+ 1.

In the following we show thatf is a multi-level distance labelling.

Because2l(ut)≤kfor0≤t≤ |V| −1, we have

2

|V|−2

X

t=1

l(ut)−k(|V| −1)≤0,

but

|V|−2

X

t=0

[Jf(ut, ut+1) + 2ϕ(ut, ut+1)]≥0,

so (1) of Lemma 2.1 holds.

In the algorithm above, by definition of f, there must be a consecutive subset of vertices {ut−1, ut, ut+1} for

2 ≤t ≤ |V| −2 so that ut−1, ut+1 belong to the same

branch. If one of ut−1, ut+1 is the others’ ancestor, by

the construction of the algorithm, we always have

2l(ut)−k+ 2 min{l(ut−1), l(ut+1)} ≤0

holds. Otherwise, we have

2l(ut)−k+ 2 min{l(ut−1)−1, l(ut+1)−1} ≤0

or

2l(ut)−k+ 2 min{l(ut−1)−1, l(ut+1)−2} ≤0

or

2l(ut)−k+ 2 min{l(ut−1)−1, l(ut+1)−3} ≤0

or

2l(ut)−k+ 2 min{l(ut−1)−2, l(ut+1)−2} ≤0

or

2l(ut)−k+ 2 min{l(ut−1)−2, l(ut+1)−3} ≤0

or

2l(ut)−k+ 2 min{l(ut−1)−3, l(ut+1)−3} ≤0

holds, and

|V|−2

X

t=0

[Jf(ut, ut+1) + 2ϕ(ut, ut+1)]≥0.

Therefore, (2) of Lemma 3 holds.

Sof is a multi-level distance labeling ofG, then

rn(G)≤f(u|V|−1) = 6

k

2

P

i=4

[(2i−5)(ti−2)] +k₂2 −k+ 1.

IAENG International Journal of Applied Mathematics, 45:1, IJAM_45_1_05

Case 2. If both k and ti are even, tk

2 ≥ 1

2(t4−2) + 2

and tk

2−j ≥t4+j for1 ≤j ≤

k−8 4

, then the algorithm is defined as following:

u0=vk

2,0→vk,0→v1,0→v

k

2+1,2tk

2+1 −3

→v4,2t4−3→vk−3,2tk−3−3→vk₂,2tk

2 −3

→vk−3,2tk−3−2→v4,2t4−2→vk₂+1,2tk

2+1 −2 → · · · →vk−3,3tk−3−6→v4,3t4−6→vk−1,0

→v2,0→vk

2+1,

t4 2+2tk

2+1

−4→vk

2,

t4 2+2tk

2 −4 →vk

2+1,

t4 2+2tk

2+1

−3→vk

2,

t4 2+2tk

2

−3→ · · · →

vk

2+1,3tk

2+1

−6→vk

2,3tk

2

−6→vk−2,0 →vk

2−2,0→v

k

2+2,2tk

2+2

−3→v5,2t5−3→ · · · →

vk

2+2,3t5−6→v5,3t5−6→vk−3,0 →vk

2−3,0→ · · · →vb3k4−4c,2t_b3k−4 4 c

−3→

v_bk+8 4 c,2t_bk+8

4 c

−3→ · · · →vb3k−4 4 c,3t_bk+8

4 c −6→

v_bk+8 4 c,3t_bk+8

4 c

−6→vb3k+2 4 c,0 →v_bk+2

4 c,0

→v_b3k

4c,2t_b3k

4c

−3→vbk+12

4 c,2t_bk+12 4 c

−3

→ · · · →v_b3k

4c,3t_b3k

4c

−6→vbk+12 4 c,3t_b3k

4c −6

→v_b3k−2

4 c,0→vb

k−2

4 c,0→ · · · →vk−4,2tk−4−3→

vk

2−1,2tk

2−1

−3→ · · · →vk−4,3tk−4−6→vk₂−1,3tk−4−6

→vk

2+3,0→v3,0→v

k

2+2,3t5−6→vbk+124 c,3t_b3k

4c −6

→ · · · →v_b3k−4

4 c,3t_b3k−4 4 c

−6→vk

2−1,3tk

2−1 −6→

vk

2+1,tk

2+1

−1→v4,t4−1→vk−3,tk−3−1→vk 2,tk

2 −1→

vk−3,tk−3 →v4,t4 →vk₂+1,tk

2+1

→ · · · →vk−3,2tk−3−4

→v4,2t4−4→vk

2+1,

t₄

2+tk

2+1

−2→vk

2,

t₄

2+tk

2 −2

→ · · · →vk

2+1,2tk

2+1

−4→vk

2,2tk

2

−4→vk

2+2,tk

2+2 −1

→v5,t5−1→ · · · →vk

2+2,2t5−4→v5,2t5−4 → · · · →v_b3k−4

4 c,t_b3k−4 4 c

−1→vbk+8 4 c,t_bk+8

4 c −1

→ · · · →v_b3k−4 4 c,2t_bk+8

4 c

−4→vbk+8 4 c,2t_bk+8

4 c −4

→v_b3k

4c,t_b3k

4c

−1→vbk+12 4 c,t_bk+12

4 c

−1→ · · · →

v_b3k

4c,2t_b3k

4c

−4→vbk+12 4 c,2t_b3k

4c

−4→ · · · →

vk−4,tk−4−1→vk₂−1,tk

2−1

−1→ · · · →vk−4,2tk−4−4 →vk

2−1,2tk−4−4→vk2+2,2t5−3→vbk+124 c,2t_b3k

4c −3

→ · · · →v_b3k−4

4 c,2t_b3k−4 4 c

−4→vk

2−1,2tk

2−1 −4→

vk

2+1,1→v4,1→vk−3,1→v

k

2,1→vk−3,2→v4,2 →vk

2+1,2→v4,3→vk−3,3→ · · · →vk2,

t4 2−1 →vk−3,tk−3−2→v4,t4−2→vk

2+1,

t₄

2 →vk

2,

t₄

2 →

vk

2+1,

t4 2+1

→ · · · →vk

2,tk

2

−2→vk

2+2,1 →v5,1→ · · · →vk

2+2,t5−2→v5,t5−2→ · · · →

v_b3k−4 4 c,1

→v_bk+8 4 c,1

→ · · · →v_b3k−4 4 c,t_bk+8

4 c −2

→v_bk+8 4 c,t_bk+8

4 c

−2→vb3k

4c,1

→v_bk+12 4 c,1

→ · · · →v_b3k

4c,t_b3k

4c

−2→vbk+12 4 c,t_b3k

4c

−2→ · · · →vk−4,1→vk

2−1,1→ · · · →vk−4,tk−4−2→

vk

2−1,tk−4−2→vk₂+2,t5−1→vbk+12₄ c,t b3k

4c −1

→ · · · →v_b3k−4 4 c,tb3k−4

4 c −2→

vk

2−1,tk

2−1

−2→vk

2+2,0→vk2−1,0→vk2+1,0.

Similar as Case 1, we can obtain thatf is a multi-level distance labelling of G, and then

rn(G)≤f(u|V|−1) = 6

k

2

X

i=4

[(2i−5)(ti−2)] +k

2

2 −k+ 1.

Case 3. If bothkandtiare odd, thent4=tk+1 2 =tk−3

and tk+1

2 −j ≥t4+j for1≤j ≤

k−7 4

. The algorithm is defined as following:

u0=vk+1

2 ,0→vk,0→v1,0→v

k+1 2 ,2tk+1

2

−3→v4,2t4−3 →vk−3,2tk−3−3→vk+1₂ ,2tk+1

2

−2→vk−3,2tk−3−2

→v4,2t4−2→vk+1 2 ,2tk+1

2

−1→ · · · →vk+1 2 ,3tk+1

2 −6

→v4,3t4−6→vk−3,3tk−3−6→(1)vk−1₂ ,2tk−1 2

−3

→vk−1,0→v2,0→vk+3 2 ,2tk+3

2

−3→v5,2t5−3→

vk+3 2 ,2tk+3

2

−2→v5,2t5−2→ · · · →vk+3

2 ,3t5−6→

v5,3t5−6→vk+7 2 ,0

→v3,0→vk+5 2 ,2tk+5

2

−3→v6,2t6−3 → · · · →vk+5

2 ,3t6−6→v6,3t6−6→v

k+9

2 ,0→v4,0 → · · · →v_b3k−5

4 c,2t_b3k−5 4 c

−3→vbk+9 4 c,2t_bk+9

4 c −3→ · · · →v_b3k−5

4 c,3t_bk+9 4 c

−6→vbk+9 4 c,3t_bk+9

4 c −6→

v_b3k+3 4 c,0

→v_bk+5 4 c,0

→v_b3k−1

4 c,2t_b3k−1 4 c

−3→

v_bk+13

4 c,2t_bk+13 4 c

−3→ · · · →vb3k−1

4 c,3t_b3k−1 4 c

−6→

v_bk+13

4 c,3t_b3k−1 4 c

−6→vb3k+7 4 c,0

→v_bk+9 4 c,0

→ · · · →

vk−4,2tk−4−3→vk−1 2 ,2tk−1

2

−2→ · · · →vk−1

2 ,3tk−4−6

→vk−4,3tk−4−6→vk−5 2

,0→vk−2,0→vk−1

2 ,3tk−4−5 →vk+3

2 ,3t5−5

→v_bk+13

4 c,3t_b3k−1 4 c

−5→ · · · →

vk−1 2 ,3tk−1

2

−6→vb3k−5

4 c,3t_b3k−5 4 c

−6→vk+1 2 ,tk+1

2 −1

→vk−3,tk−3−1→v4,t4−1→vk+1₂ ,tk+1 2

→v4,t4 →

vk−3,tk−3→ · · · →vk+1 2 ,2tk+1

2

−4→vk−3,2tk−3−4

→v4,2t4−4→vk+3₂ ,tk+3 2

−1→v5,t5−1→ · · · →

vk+3 2 ,2t5−4

→v5,2t5−4→ · · · →v_b3k−5 4 c,t_b3k−5

4 c −1

→v_bk+9 4 c,t_bk+9

4 c

−1→ · · · →vb3k−5 4 c,2t_bk+9

4 c −4

→v_bk+9 4 c,2t_bk+9

4 c

−4→vb3k−1 4 c,t_b3k−1

4 c −1

→v_bk+13 4 c,t_bk+13

4 c

−1→ · · · →vb3k−1

4 c,2t_b3k−1 4 c

−4

→v_bk+13

4 c,2t_b3k−1 4 c

−4→ · · · →vk−4,tk−4−1

→vk−1 2 ,tk−1

2

−1→ · · · →vk−4,2tk−4−4

→vk−1

2 ,2tk−4−4→v

k+3

2 ,2t5−3→vb

k+13

4 c,2t_b3k−1 4 c

−3

→ · · · →vk−1 2 ,2tk−1

2

−4→vb3k−5

4 c,2t_b3k−5 4 c

−4→

vk+1 2 ,1

→vk−3,1→v4,1→vk+1 2 ,2

→v4,2→vk−3,2 → · · · →vk+1

2 ,tk+1 2

−2→v4,t4−2→vk−3,tk−3−2

→v5,1→vk+3

2 ,1→ · · · →v5,t5−2→v

k+3 2 ,t5−2

IAENG International Journal of Applied Mathematics, 45:1, IJAM_45_1_05

→ · · · →v_bk+9 4 c,1

→v_b3k−5 4 c,1

→ · · · →

v_bk+9 4 c,t_bk+9

4 c

−2→vb3k−5 4 c,t_bk+9

4 c

−2→vbk+13 4 c,1 →v_b3k−1

4 c,1

→ · · · →v_bk+13 4 c,t_b3k−1

4 c −2→

v_b3k−1 4 c,t_b3k−1

4 c

−2→ · · · →vk−1

2 ,1→vk−4,1 → · · · →vk−1

2 ,tk−4−2→vk−4,tk−4−2→

v_bk+13 4 c,t_b3k−1

4 c

−1→vk+3

2 ,t5−1→ · · · →

vk−1 2 ,tk−1

2

−4→v_b3k−5 4 c,t_b3k−5

4 c

−2→vk−3 2 ,0 →vk+5

2 ,0

→vk−1 2 ,0

→vk+3 2 ,0

.

Similar as Case 1, we can show that f is a multi-level distance labelling of G.

By definition off, ifk= 7, then there is no vertex sat-isfying Claim 1, soσ(f) = 0, thus rn(G)≤f(u|V|−1) =

(3k−12)tk+1 2 +k

2_{−7k+ 1.}

Ifk≥9, then there is only one vertex

ut∈ {v1,0, v4,j, vk−3,j, vk,0|2t4−3≤j≤3t4−6}

such that ut−1, ut+1 belong to the same branch,

l(ut−1) =l(v4,3t4−6) =

k−1

2 , l(ut+1) =l(vk−1₂ ,2tk−1 2

−3) =

4 andut+1 is not the ancestor ofut−1, thus

ϕ(ut−1, ut+1) = min{l(ut−1)−3, l(ut+1)−3}= 1.

By Theorem 2.8 and the above algorithm, we have

σ(f) = 1, hence

rn(G)≤6 k−1

2

P

i=4

[(2i−5)(ti−2)] + (3k−12)tk+1 2

+1₂(k−7)2_{+ 2.}

Case 4. Ifkis odd andtiis even, thent4=tk+1 2 =tk−3

and tk+1

2 −j ≥t4+j for1≤j ≤

k−7 4

. The algorithm is defined as following:

u0=vk+1

2 ,0→vk,0→v1,0→v

k+1 2 ,2tk+1

2

−3→v4,2t4−3 →vk−3,2tk−3−3→vk+1

2 ,2tk+1 2

−2→vk−3,2tk−3−2

→v4,2t4−2→vk+1 2 ,2tk+1

2

−1→ · · · →vk+1 2 ,3tk+1

2 −6

→vk−3,3tk−3−6→v4,3t4−6→(1)vk+3₂ ,2tk+3 2

−3→

v2,0→vk−1,0→vk−1 2 ,2tk−1

2

−3→vk−4,2tk−4−3

→vk−1 2 ,2tk−1

2

−2→vk−4,2tk−4−2→ · · · →vk−1₂ ,3tk−4−6

→vk−4,3tk−4−6→vk−5

2 ,0

→vk−2,0→vk−3 2 ,2tk−3

2 −3→

vk−5,2tk−5−3→ · · · →vk−3₂ ,3tk−5−6→vk−5,3tk−5−6→ vk−7

2 ,0→vk−3,0→ · · · →vb

k+13

4 c,2t_bk+13 4 c

−3

→v_b3k−1

4 c,2t_b3k−1 4 c

−3→ · · · →vbk+13

4 c,3t_b3k−1 4 c

−6

→v_b3k−1

4 c,3t_b3k−1 4 c

−6→vbk+5 4 c,0

→v_b3k+7 4 c,0

→

v_bk+9 4 c,2t_bk+9

4 c

−3→vb3k−5

4 c,2t_b3k−5 4 c

−3→ · · · →

v_bk+9 4 c,3t_bk+9

4 c

−6→vb3k−5 4 c,3t_bk+9

4 c

−6→vbk+1 4 c,0

→

v_b3k+3 4 c,0

→ · · · →v6,2t6−3→vk+5 2 ,2tk+5

2

−3→ · · · →

v6,3t6−6→vk+5

2 ,3t6−6→v4,0→vk+92 ,0

→v5,2t5−3→

vk+3 2 ,2tk+3

2

−2→v5,2t5−2→vk+3 2 ,2tk+3

2

−1→ · · · →v5,3t5−6→vk+3

2 ,3t5−5→

v4,0→vk+9 2 ,0→

v_bk+13

4 c,3t_b3k−1 4 c

−6→vk+3

2 ,3t5−4→ · · · →

v_b3k−5

4 c,3t_b3k−5 4 c

−6→vk−1₂ ,3tk−1 2

−6→vk+1 2 ,tk+1

2 −1

→v4,t4−1→vk−3,tk−3−1→vk+1₂ ,tk+1 2

→vk−3,tk−3 →v4,t4→ · · · →vk+1

2 ,2tk+1 2

−4→vk−3,2tk−3−4

→v4,2t4−4→vk+3 2 ,tk+3

2

−1→v5,t5−1→ · · · →

vk+3

2 ,2t5−4→v5,2t5−4→ · · · →

v_b3k−5 4 c,t_b3k−5

4 c

−1→vbk+9 4 c,t_bk+9

4 c

−1→ · · · →

v_b3k−5 4 c,2t_bk+9

4 c

−4→vbk+9 4 c,2t_bk+9

4 c −4→

v_b3k−1 4 c,t_b3k−1

4 c

−1→vbk+13 4 c,t_bk+13

4 c

−1→ · · · →

v_b3k−1

4 c,2t_b3k−1 4 c

−4→vbk+13

4 c,2t_b3k−1 4 c

−4→ · · · →

vk−4,tk−4−1→vk−1

2 ,tk−1 2

−1→ · · · →vk−4,2tk−4−4→ vk−1

2 ,2tk−4−4→v

k+3

2 ,2t5−3→vb

k+13

4 c,2t_b3k−1 4 c

−3→ · · · →v_b3k−5

4 c,2t_b3k−5 4 c

−4→vk−1 2 ,2tk−1

2

−4→vk+1 2 ,1 →v4,1→vk−3,1→vk+1

2 ,2→vk−3,2→v4,2→ · · · →vk+1

2 ,tk+1 2

−2→vk−3,tk−3−2→v4,t4−2→vk+3

2 ,1 →

v5,1→ · · · →vk+3

2 ,t5−2→v5,t5−2→ · · · →vb 3k−5

4 c,1 →v_bk+9

4 c,1

→ · · · →v_b3k−5 4 c,t_bk+9

4 c −2→

v_bk+9 4 c,t_bk+9

4 c

−2→vb3k−1 4 c,1

→v_bk+13 4 c,1

→ · · · →

v_b3k−1 4 c,tb3k−1

4 c

−2→vbk+13 4 c,tb3k−1

4 c

−2→ · · · →

vk−4,1→vk−1

2 ,1→ · · · →vk−4,tk−4−2→v

k−1 2 ,tk−4−2

→vk+3

2 ,t5−1→

v_bk+13 4 c,t_b3k−1

4 c −1→ · · · →v_b3k−5

4 c,t_b3k−5 4 c

−2→vk−1₂ ,tk−1 2

−2→ · · · →

vk+5 2 ,0

→vk−3 2 ,0

→vk+3 2 ,0

→vk−1 2 ,0

.

Similar as Case 3, we can show that f is a multi-level distance labelling of G.

Ifk= 7, then

rn(G)≤(3k−12)tk+1 2 +k

2_{−7k+ 1.}

Ifk≥9,then

rn(G)≤6 k−1

2

P

i=4

[(2i−5)(ti−2)] + (3k−12)tk+1 2

+1 2(k−7)

2_{+ 2.}

Applying Theorems 2.4-2.6, and 2.8, we have shown that the theorem holds.

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IAENG International Journal of Applied Mathematics, 45:1, IJAM_45_1_05