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On Node Ranking of Graphs

Yung-Ling Lai, Sheng-Hsiung Wang and Wei-Chun Lin

Department of Computer Science and Information Engineering

National Chiayi University, Chiayi, Taiwan

{ yllai, s0942783, s0942810}@mail.ncyu.edu.tw

Abstract

For a connected graph ( , )G V E= , a node

ranking labeling { }: 1,2,...C V k is a mappingsuch that for every path between any two vertices

u, v with ( ) ( )C u C v= , there exists at leastone vertex w on the path with

( ) ( ) ( )C w C u C v> = . The value ( )C v iscalled the rank of the vertex v and G is said to

be k-rankable if there exists a node ranking

labeling of G with maximum rank k . The

node ranking problem is the problem of findingminimum k such that G is k-rankable. There

are two versions of node ranking problem, namely

offline and online. In off-line version, the node

ranking problem deals with a graph where all

vertices and edges are given in advance. In

on-line version, the vertices are given one by one

in an arbitrary order and only the edges of the

induced subgraph of given vertices are known

when the rank of each vertex has to be chosen.

The rank can not be changed after it is assigned.

This paper solved node ranking problem of corona

graphs and mesh in off-line version, and gave

on-line node ranking algorithm for sun graphs.

1 Introduction

Let ( , )G V E= be an undirected graph where V

denotes the set of vertices and E denotes the set

of edges. G is said to be k-rankable if there is a

mapping { }: 1, 2,...C V k such that every pathbetween any two vertices u , v with

( ) ( )C u C v= , there exists at least one vertex w on the path with ( ) ( ) ( )C w C u C v> = . Themapping C is called a node ranking labeling of

G and the value ( )C v is called the rankof thevertex v . The node ranking number ( )r G isthe smallest integer k such that G is

k-rankable. A node ranking labeling is an

optimal node ranking labeling of G if its

maximum rank is ( )r G . The node rankingproblem is the problem of finding the node

ranking number ( )r G of a graph G . Wesaid the node ranking problem is solved if a

formula for the node ranking number is obtained

or an algorithm of an optimal node ranking

labeling is given. The node ranking problem is

applicable in communication network

design[3][15]; finding the minimum height of a

node separator tree of a graph [3][8][15], which

are extensively used in VLSI layout [11][14]; and

developing algorithms for planning efficient

assembly of products in manufacturing systems [4]

etc. There are two versions of node ranking

problem, off-line and on-line versions. In

off-line version, the graph with all vertices and

edges are given in advance. In on-line version,

the vertices are given one by one in an arbitrary

order together with the edges adjacent to the

vertices that are already given. The rank has to

be assigned in real time and once the rank is

assigned to a vertex, it is not changeable. Similar

to the offline version, the on-line node ranking

number is denoted as ( )*r G , which is thesmallest integer k such that G is on-line

k-rankable.

Node ranking problem has been studied since

1980s. In the offline version, it is known that for

a path nP , 1n , ( ) 2log 1r nP n = + [5].For a cycle

nC , 3n , ( )r 2log 1nC n = +

[1]. For wheel graphs 1,nW with n vertices on

the cycle, 3n , 2log)( 21 +=+ nWnr [7]; thenode ranking number of the complete bipartite

graph ,m nK is ( ), 1r m nK m = + for m n [8];and star graphs )(

1,1 =

nnKS with n vertices,

2)( =nr

S for 3n [13]. An upper bound of

the node ranking number for an arbitrary tree with

n nodes was given by [3], they proposed an

( )2logO n n time optimal node ranking algorithm,which was further improved to ( )O n by [12].There are fewer results in the on-line version. [1]

gave tight bounds for paths and cycles as

( )*2 21 2 1log logr nPn n+ < < + for 1n ;and ( )*2 2log 1 2 log 1r nn C n+ < < for

3n , respectively. The on-line version of a

complete bipartite graph with m n was proved

to be ( ) { }* ,1 min 1,2 1r m nm K n m+ + + [8]and the on-line node ranking number of star graph

is given by [13] as 3)(* =nr

S for 3n .

There are several papers worked on on-line trees,

T e 26t Wor s op on Comi natori a Mat emati cs an Computati on T eory

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the best known on-line node ranking algorithm for

trees which has ( )2O n time complexity and( )O n space complexity was given by [6]. [9]

proposed an online ranking algorithm for trees

using CREW PRAM model with ( )3 2logO n n processors. [10] provided an optimal online

ranking algorithm for stars and an

( )

3O n time

algorithm for arbitrary trees.

This paper established offline node ranking

number for mesh and the corona of any two

graphs and provided an on-line algorithm with

tight bound of online node ranking number for sun

graphs.

2 Offline Results for Corona Graphs

Given graphs G and H with n and m vertices

respectively, the corona of G with respect to H

denoted as HG , is the graph with vertex set

{( ) ( )V G H V G n = distinct copies of ( )V H denoted as ( ) ( )1 2, ,V H V H ( )}nV H and theedge set ( )E G H {( )E G n= distinct copiesof ( )E H denoted as ( ) ( )1 2, ,E H E H

( )}nE H {( , ) : ( ), ( ),i i iu v u V G v V H }1 i n . Figure 1 shows an example of graph

3 4P P .

Figure1: An example of graph3 4P P .

Theorem 1: Let HG denote the corona of

G with respect to H. Then the node rankingnumber ( ) ( ) ( )

r r rG H G H = + .

Proof: First we show that ( )r G H

( ) ( )r rG H + . Let 1f and 2f be optimal

node ranking labeling of G and H

respectively. Let ( ) { }1 2, , , nV u u uG = and

( ) { }1 2, , , mV H v v v= , ( ) ( )V G H V G =

{ ,i jv which is the i-th copy of the j-th vertex of

H , }1 , 1i n j m . Define f as

( ) ( ) ( )1i i rf u f u H= + for ( )iu V G and

( ) ( ), 2i j jv vf f= for ( ),i j iv V H , then f is a

node ranking labeling of HG with maximum

rank ( ) ( )r rG H + , which implies

( ) ( ) ( )r r rG H G H + .

Next we show that ( ) ( ) ( )r r rG H G H + .Suppose to the contrary, that

( ) ( ) ( )r r rG H G H < + . Let g be an

optimal node ranking of G H .

Case1: 1 ,1i n j m , ( ) ( ),i ji vg u g> .Then ( ) ( )i rg u H> , ( )iu V G . Define a

labeling 'g of G as '( ) ( ) ( )i i r

g u g u H = ,

then 'g is a node ranking labeling of G with

maximum rank ( ) ( ) ( )r r rG H GH for all 1 ,1i n j m . Note

that for each exchange, since it starts from the

smallest label, the order of the ranks are not

changed in graphi

H , and for ( ) ( )i kh u h u= we

must have somej that ( ) ( ),i j kg v g u= , since g is

a node ranking labeling of graph HG , there

must be a w oni k

u u path with

( )( ) ( )i kh w h u h u> = , which implies h is a node

ranking labeling of HG

. Since themaximum label ofh is the same as the maximum

label of g , which lead us back to case 1 and

produce a contradiction.

Let ,n mS be the graph which contains a n-cycle

and each cycle vertex has m hairs. Since the

graph has the shape of sun, we named it as sun

graph. Figure 2 shows an example of graph

5,2S .

Figure 2: An example of graph5,2S .

Since the sun graph ,n mS may be viewed as

n mC K , following corollary comes directly from

Theorem 1 by knowing that 2( ) log 1r nC n = +

(from [1]) and ( ) 1r mK = .


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Corollary 1: Let ,n mS be the sun graph of

order ( 1)n m + . Then the node ranking number

, 2( ) log 2r n mS n = + .

3 Offline Results for Mesh Graphs

In this section, we proposed an off-line ranking

algorithm for mesh graphsn n

P P . The

algorithm produced a node ranking labeling of

n nP P for odd n and 15n , for even n and

18n . Figure 3 shows an example of the graph

6 6P P .

Figure 3: An example of graph6 6

P P .

Algorithm Offline_Ranking_Mesh

Input : , 15n n if n is odd and 18n if n is

even. Each vertex is denoted as,i j

v for

1 ,i j n .

Output : An off-line rank assignment ofn n

P P .

Method:

[ ] [ ]array y x holds the label of vertex ,y xv If( n is odd ) Then

Step1: If x y even+ = Then [ ] [ ] 1array y x = .Step2: For 1x = to 2n

do

Dividing the graph by x y n+ = .

For 2x n= to n doDividing the graph by 2x y n+ = + .

Step3: For 2x = to 2n do

Dividing the graph by 1x y = .

For 2 1x n= + to 1n doDividing the graph by 1y x = .

// The graph is then divided into four isomorphic

subgraphs 1 2 3 4, , , as shown in Figure 4.

Figure 4: An example of graphs 1 2 3 4, , , .

Step4: In the 1 graph, 2 1i n= ; call Alpha

( i ); use the label for corresponding vertices in 2

to4

.

// i means the number of nodes on the edge of

subgraph1

.

Else If( n is even and 0 mod4n ) Then

Step1: If x y even+ = Then [ ] [ ] 1array y x = .Step2: For 4 3x n= + to 2 4n + do

Dividing the graph by 2 5x y n+ = + .

For 4 3x n= + to 2 1n + do


For 2x n= to 2 1n + doDividing the graph by 1x y n+ = + .

For 2x n= to 3 4 2n do

Dividing the graph by 1y x = .

For 2 3x n= to 3 4 2n do

Dividing the graph by

03 2 3x y n u+ = = .

// The graph is divided into two isomorphic

subgraphs1 2, as shown in Figure 5.

Figure 5: An example of graphs1 2, .

Step3: In the1

graph, Dividing the graph by

12 5x y n u+ = + = . (Repeat the same processes

for the corresponding vertices in2

.)// The graph is divided into two subgraphs

1 2, asshown in Figure 6.


Step4: Dividing1 and 2 by 1x y = .and

02 1y x n v = = respectively.

// The1

graph is divided into two subgraphs

1 2, and the 2 graph is divided into two

subgraphs1 2, as shown in Figure 7.


Step5: In the1

graph, 4i n= ; callAlpha ( i );In the

2 graph, ( )3 12 8j n= ; call

Alpha ( j );

In1

and2

, there are lines of vertices

denoted as 1 2, , , le e e , such that


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( )1 0 12 , 1 2 1x y u c c u u+ = + where

( )3 4 4 3 12 8l n n= . Let 2j l= .

Dividing1

and2

byj

e and1j

e+

respectively.

Let2

u x y= + for the vertices ,y xv on line

je and

3u x y= + for the vertices ,y xv on line

1je

+.

// The1

graph is divided into two subgraphs

11 12, and the 2 graph is divided into twosubgraphs

21 22, as shown in Figure 8.


Step6: In11

, Dividing the graph by

12 2 2 1y x n n v = + = .// Graph

11 is divided into two subgraphs 11 11,

as shown in Figure 9.

In12

, Dividing the graph by

( ) 22 2 4 16 1y x n n v = + + = .// Graph

12 is divided into two subgraphs

12 12, as shown in Figure 9.

In21

, ( )3 1 2 1p u u= ; ( )0 1 2q v= ;callBeta ( ,p q );

In22 , ( )0 3 2 1p u u= ; ( )0 3 2q v= ;

callBeta ( ,p q );

Figure 9: An example of graphs11 11 12 12, , , .

Step7: In11

, ( )0 1 2 2i u v= ; callAlpha ( i );In

11 , let ( )2 1 2 1p u u= ;

( )1 0 2 1q v v= ; callBeta ( ,p q );In 12 , ( )0 2 2 2i u u= ; callAlpha ( i );

In 12 , let ( )0 2 2 1p u u= ;( )2 0 2 1q v v= ; callBeta ( ,p q );

Else //n is even and 0 mod 4n .

Step1: If x y even+ = Then [ ] [ ] 1array y x = .Step2: For 4 3x n= + to 2 3n + do

Dividing the graph by 2 4x y n+ = + .

For 4 3x n= + to 2 1n + do


For 2x n= to 2 1n + do

Dividing the graph by 1x y n+ = + .

For 2x n= to 3 4n doDividing the graph by 1y x = .

For 2 2x n= to 3 4n doDividing the graph by

03 2 2x y n u+ = = .

// The graph is divided into two isomorphicsubgraphs

1 2, which is the same as in Figure 5.Step3: In the

1 graph, Dividing the graph by

( ) 12 2 2 8x y n n u+ = + = . (Repeat the sameprocesses for the corresponding vertices in

2 .)

// The graph is divided into two subgraphs1 2, as

shown in Figure 6.

Step4: Dividing 1 and 2 by 1x y = and

02y x n v = = respectively.

// The graph1

is divided into two subgraphs

1 2, and the graph 2 is divided into twosubgraphs

1 2, as shown in Figure 7.Step5: In

1 , 4i n= ; callAlpha ( i );

In2

,

( )3 4 1 2i n =

; callAlpha ( i );

In1 and 2 , there are lines of vertices

1 2, , , le e e such that 1 2x y u c+ = + , where

( )0 11 2 1c u u , ( ) ( )2 8 2 4 3l n n= + + + .Let 2j l= , dividing 1 and 2 by

and 1je + respectively.

Let 2u x y= + for the vertices ,y xv on line

je and 3u x y= + for the vertices ,y xv on line

1je

+.

// Graphs 1 and 2 are divided into subgraphs

11 12, and 21 22, respectively, as shown inFigure 8.

Step6: Dividing the graph 11 by

( ) 12 2 2 8 2y x n n v = + + =

.// Graph 11 is divided into two subgraphs 11 11, as shown in Figure 9.

Dividing graph 12 by

( ) 22 2 2 16y x n n v = + + = .// Graph 12 is divided into two subgraphs 12 12, as shown in Figure 9.

In graph 21 , let ( )3 1 2 1p u u= ;( )0 1 2q v= ; callBeta ( ,p q );

In graph 22 , let ( )0 3 2 1p u u= ;( )0 3 2q v= ; callBeta ( ,p q );

Step7: In graph 11 , let ( )0 1 2 2i u v= ; callAlpha ( i );

In graph 11 , let ( )2 1 2 1p u u= ;

( )1 0 2 1q v v= ; callBeta ( ,p q );In 12 , let ( )0 2 2 2i u u= ; callAlpha ( i );In graph 12 , let ( )0 2 2 1p u u= ;

( )2 0 2 1q v v= ; callBeta ( ,p q );End of Algorithm Offline_Ranking_Mesh_Graph

Procedure Alpha ( int i )

If ( )3i Then {3p i= ; 1q i p= ;


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// p means the number of nodes of the

dividing line P.

1 1p p p= + ;

Dividing the graph by1: 2 2 1I y x p c = + .

Dividing the graph by1: 2 2 1P x y p c+ = + + .

// The graph1 is divided into three

subgraphs 11, 12, 13 as shown in Figure 10.

Figure 10: Subgraphs 11, 12, 13 .

In graph11

, callAlpha (p );

If ( ) ( )( )( 11r r pP p + and( ) ( )( )11 2 1r r pP and( )( )

)11r pP Then {

In the 12 graph, dividing thegraph by 0 : 4a x y n+ = .

2q q= ;

}

Else If ( ) ( )( )( 11r r pP p + and( ) ( )( )11 2 1r r pP > and ( )( ))11r pP

Then {

Dividing graph 12 by

0 : 6a x y n+ = .

3q q= ;

}

Else {

Dividing graph12 by

0 : 8a x y n+ = .4q q= ;

}

// The graph 12 is divided into twosubgraphs 1 2, as shown in Figure 11.


In 1 , callBeta ( ,p q );1i i p= ; c + + ;

In 2 , callBeta ( ,p q );In

13 , callAlpha ( i );

Return ( )1r ; // ranking of 1 }

Else If ( )2i = Then Rank the vertices individuallywith 2, 3, 4.

Else Rank the vertex with 2. // 1i =

End of Procedure Alpha

Procedure Beta ( int p , int q )

If ( )3q Then {In

1 , there are dividing lines : 1 2, , , qa a a

as shown in Figure 12.

Figure 12: An example of dividing lines

1 2, , , qa a a .

If q is odd Then {

Dividing the graph by2q

a

.

// The dividing line divides the graph into

two the same graphs.

2q q= ;

}If q is even Then {

Dividing the graph by q 2 +1a .

2q q= ;

}

}

Else If ( )2q = Then call Gamma( p );Else Rank by optimal ranking as path // 1q =

callBeta( ,p q );

End of Procedure Beta

Procedure Gamma ( int p )

If ( )3p Then {

// graph 1 is divided into two subgraphs: 1 2, by1a .

In1

, there are dividing lines: 1 2, , , pb b b .

Figure 13 shows1 2, and 1 2, , , pb b b .

Figure 13: Graphs1 2, and

1 2, , ,

p

b b b .

If p is odd Then {

Dividing the graph byp 2

b

.

// The graph is divided into two identical

subgraphs.

2p p= ;

}

If p is even Then {


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Dividing the graph by 2 1pb + .

2p p= ;

}

}

Else If ( )2p = Then Rank the nodes using 2 to 5.Else Rank the nodes individually with 2 and 3.

call Gamma (p );

End of Procedure Gamma

Theorem 2: The labeling given by Algorithm

Offline_Ranking_Mesh is a node ranking labeling of

n nP P for odd n and 15n , for even n and

18n .

Proof: In the algorithm, we always assign greater

rank for the dividing vertices, and on every path

between two nodes of the same rank there is at least

one dividing line, so the labels given by the

algorithm must follow the node ranking condition for

n nP P .

Theorem 3: The maximum rank given by Algorithm

Offline_Ranking_Mesh for odd n and 15n is

( ) ( ) ( )2 1 6 1 6 8 1 12n n n n n+ + + + + + 10 x + ; and for even n, 0 mod4n and 20n

is ( ) ( ){4 2 8 1 max 3 4 2 8 ,n n n n n+ + + + }2 4 8 5n n+ ; and for even n 0 mod4n and

18n is ( )4 2 8 1n n n+ +

( ){ ( )max 3 4 2 8 ,2 4 3 12n n n n+ + + + ( ) ( ) }3 12 8 3 24 9n n x+ + + + , where

0 if 0, 2

2 if 4

4 if 6,8

5 if 10

y

yx

y

y

=

==

= =

for ( )( )3 12 3 12y n n= + .

Theorem 3 gave us an upper bound for the node

ranking number of a meshn n

P P for 17n .

For 16n , Table 1 shows the node ranking number

ofn n

P P which is obtained by a brute force

computer testing, and is denoted by the formula that

represents the relation between n and ( )r n nP P .

Table 1: ( )r n nP P for 16n .n ( )r n nP P

1 7n 2 1n

8 10n 2n

11 2 1n +

12 2n

13 2 2n +

14 2 1n +

15 2 3n +

16 2 2n +

4 Online Results

In this section, we proposed an on-line ranking

algorithm for sun graphs,n m n mS C K= , and

analyze both time and space complexities of our

algorithm. Notice that there are ( 1)n m +

vertices in ,n mS . Let 1v to ( 1)n mv + represent

the order of the coming vertices.

Algorithm Online_Ranking_Sun_Graph

We use structure array Sun[] to record the

information of each node, and use value_record[]

as the counter of the index values appearance.

Input:

n: the number of vertices on the cycle of,n mS .

m: the number of vertices on the hairs of ,n mS .

iv for 1 ( 1)i n m + the vertices together with

the adjacent verticesj

v for 1 j i < .

Output:

An online node ranking labeling of ,n mS .

Method:For 1i = to ( )1m n+ do {

Ifi

v has no adjacent vertices, ( ) 2i

rank v =

Else ifi

v has only one neighborj

v for

j i< andj

v has two neighbors with

degree >1 then ( ) 1i

rank v =

Else {

1. Set the occurrence record to zero except rank 1(which is set to 2)

2. Applying online tree algorithm given in [5] toloop through each adjacent path, go as long as

possible, to find the max value j that has

occurred at least twice

3.

Find minimum available rank k that is greaterthanj

4. Set ( )i

rank v k =

}

}

End of Algorithm Online_Ranking_Sun_Graph

Theorem 4: Algorithm Online_Ranking_Sun has

( )( )3O mn time complexity and ( )O mn spacecomplexity.

Proof:

First we analyze the time complexity of the

algorithm. Since a Sun graph with ( )1m n+ verticeshas exactly ( )1m n+ edges, each time when insert anew node iv , the looping for finding max j usedonline tree algorithm which takes 2( )O i times , so

after ( )1m n+ vertices inserted, the whole processwill be at most ( )( )

22 21 2 1 nm+ + + +

( )( )3O mn= times. For the space complexity, weuse structure array with ( )1m n+ elements to storethe node plus one array to record the rank occurrence,

the total space complexity then is

( ) ( ) ( )O mn O mn O mn+ = .


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Theorem 5: For a sun graph,n mS with 7n is

( )*2 , 2log 2 log 6r n mn S n m+ + .Proof:

Since the online node ranking number is at least as

much as the offline node ranking number, we know

that ( )*2 ,log 2 r n mn S+ . Let ,n mS be thegraph which contains a n-cycle for 1

iv i n and

each cycle vertex has m hairs denoted as, for 1 ,1i ju i n j m . If we use adversary

analysis, for the best we can do, saving rank 1 for the

vertices which are known as hair, assume that the

vertices of,n mS are coming as follows:

1 2 3 4, , ,v v v v coming in first, so they get labels

2,3,2,4.

For 3 3i n , all ,i ju for 1 j m comes before

2iv

+.

Then comes2, for 1n ju j m and

, for 1,2, 1 and 1i ju i n j m= .

At last,n

v then, for 1n ju j m .

In this coming order, beforen

v comes in, no vertex

can be recognized as end vertex, hence online tree

algorithm worked for this case which is the worst

case and our algorithm guarantee the maximum rank

is no more than 2log 6n m+ , hence we have

( )* , 2log 6r n mS n m + , which complete the proof.

5 Conclusion

There are several papers discussed the node

ranking problem. In this paper, we presented

off-line node ranking number for the corona graphs

and gave labeling algorithm for mesh. We also

gave on-line algorithm for sun graphs. Both timeand space complexities of our algorithms are

analyzed.

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