On Node Ranking of Graphs under Strong Orientation

The 24th Workshop on Combinatorial Mathematics and Computation TheoryOn Node Ranking of Graphs under Strong OrientationYung-Ling Lai and Yi-Ming ChenDepartment of Computer Science and Information EngineeringNational Chiayi University, Taiwan{yllai, s0950299}@mail.ncyu.edu.twAbstractFor a graph G ( V, E), a mappingC: V 1,2, k is a node ranking labeling ifevery path between any two vertices u and v, withCu ( ) Cv ( ), there is a node w on the path withCw ( ) Cu ( ) Cv ( ). The minimum possible k suchthat the node ranking labeling of G exists is the noderanking number of G. The node ranking problem isthe problem of finding the node ranking number ofgraphs. The directed node ranking problem may bedefined accordingly. The upper/lower node rankingnumber of a graph G is the maximum/minimumpossible directed node ranking number among allstrong orientations of G. This paper established theupper and lower node ranking number of wheelgraphs.1 IntroductionA graph G ( V, E)is k-rankable if there is amapping C: V 1,2, ksuch that for every u-vpath of G with Cu ( ) Cv ( ), there is a vertex w lieson the path such that Cw ( ) Cu ( ) Cv ( ) . Themapping C is called a node ranking labeling of G andthe value Cv () is called the rank or color of thevertex v. The node ranking number r( G)of agraph G is the smallest integer k such that G isk-rankable. A node ranking labeling is called anoptimal node ranking labeling of G if its maximumrank is r( G). The node ranking problem (alsocalled ordered coloring problem [7]) for a graph G isthe problem of finding the node ranking numberr( G)of G. For an arbitrary graph G, thedecision version of the node ranking problem isNP-complete [1]. In fact, this problem remainsNP-Complete even with restriction of the graph to beco-bipartite graphs [12]. Since the node rankingnumber of a graph G is equivalent to the minimumheight of the separation tree on G [3, 4], finding thenode ranking number is important when applying oncommunication network design [3, 6, 11, 14],computing Cholesky factorizations of matrices inparallel [1, 5, 10], and VLSI layout problem [9, 15]etc.Node ranking problem has been studied since1980s. It is known that r( Pn) logn 1forn 1 [7] and r( Cn) log( n1) 2forn 3 [2]. An upper bound of the node rankingnumber of an arbitrary tree with n nodes was givenby [6], they proposed an On ( log n ) time optimalnode ranking labeling algorithm of trees, which wasfurther improved to On ( ) by [13].We may define the node ranking problem ondigraph accordingly. A digraph D ( V, E)isk-rankable if there is a mapping C : V 1,2, ksuch that for every u-v directed path of D withC( u) C( v). There is a vertex w on the path withC( w) C( u) C( v). The mapping C is calleda (directed) node ranking labeling of D and the valueC () v is called the rank or color of the vertex v.The (directed) ranking number r( D)of a graph Dis the smallest integer k such that D is k-rankable.Since not every pair of vertices have a directed pathin digraph D, we only consider strong digraphs.Define the upper node ranking number among allstrong orientation of a graph G asRANK( G) max r( D) : D is a strong orientationof G . Similarly, the lower node ranking numberamong all strong orientation of G is defined asrank( G) min r( D) : Dis a strong orientation of G.This paper discuss the upper and lower node rankingnumber of wheel graphs. Let W1,ndenote a wheelgraph with a vertex in the center and n vertices forma cycle of the wheel. Figure 1 shows an example ofan optimal node ranking labeling of W1,8and showsdifferent directed node ranking labelings of a strongorientation of W1,8.-422-

The 24th Workshop on Combinatorial Mathematics and Computation TheoryW1,8W1,81321W 1,8132 11221535122112411241Figure 1: Node ranking labelings on undirected and strong orientations of W1,8.2 Main resultsLemma 1: [2] The node ranking number of a cycleCnis r( Cn) lgn 1.Since the only strong orientation of a cycle iseither a clockwise or counterclockwise directed cycle,by lemma 1, theorem 2 comes trivial.Lemma 2: Let G Cnbe a cycle on n vertices.Then the upper and lower node ranking numberRANK( G) rank( G) lgn 1.Given H be a subgraph of G, since every u-v path inH must be a u-v path in G, but not the reverse, wehave following proposition:Proposition 1: Let H be a subgraph of G. Then ( H ) ( G).rrSimilar to the undirected graph, digraphs have thesame proposition which is stated as proposition 2.Proposition 2: Let H be a subgraph of D. Then( H ) ( D).rrLemma 3: The node ranking number of a wheelW1,nis r( W1,n) lgn 2.ProofLet vcdenote the center vertex of W1,nandV { v , v ,..., v } denote the vertices on the cycle ofn1 2nthe wheel. To see that r( W1,n) lgn 2,suppose to the contrary, r( W1,n) lgn 1. LetC: V( W1,n) 1,2, , lgn1be an optimal noderanking labeling of W1,n. Since v cis adjacent tovifor all v i V n, we have Cv (c) Cv (i),vi Vn.Since every pair of vertices v , v withijCv (i) Cv (j), vi-vc-v jis a path in W 1,n, withoutloss of generality, we may have Cv (c) lgn1which implies Cv (i) lgn for all vi Vn.Consider the subgraph Cn W1,n vc, letC : V( Cn) 1,2, , lgnbe a labeling definedas C( vi) C( vi)for all v i V n. Then C is anode ranking labeling of C nwith maximum ranklg n , which implies r( Cn) lgn contradictsto Lemma 1. Hence r( W1,n) lgn 2. Nowconsider a labeling f : V( W1,n) 1,2, lgn2 which labels Vn { v1, v2,..., vn} by an optimalnode ranking labeling of C and labels the vertexvcas lg n 2 . Then f is a node ranking labelingof W1,nwith maximum rank lg n 2 whichimplies r( W1,n) lgn 2. Hence we haver( W1,n) lgn 2. Since each directed path P in a strong orientation Dof graph G is a path in G, if f is an optimal noderanking of D then f must be a node ranking labelingof G. Hence we have proposition 3 as follows.Proposition 3: Let G be a graph and D be anorientation of G. Then ( D) ( G).Since the adjacent vertices may not have the samerank, lemma 4 is trivial.Lemma 4: The node ranking number of completegraph is r( Kn) n . For any tournament D withn vertices, r( D) n .Lemma 5: [8] Le G be a graphC: V( G) {1,2, k}be an optimal node rankinglabeling of G . Then i,1i k , there is a vertexv V( G)such that Cv () i.rnr-423-

The 24th Workshop on Combinatorial Mathematics and Computation TheoryTheorem 1 Let G W1,nbe a wheel graph.Then RANK( G) lgn 2.Proof Similar to Lemma 3, let v cdenote thecenter vertex of G and Vn { v1, v2,..., vn} denote thesuburb vertices which form a cycle on the wheel.By lemma 3 and proposition 3, we haveRANK( G) lgn 2. Consider an orientation Dof G which makes Vn { v1, v2,..., vn}to be adirected cycle and let v chave both out-neighborsand in-neighbors, then D is a strong orientation of G.Consider the same labeling f as in lemma 3 on D, it isa node ranking labeling of D with maximum ranklg n 2 which implies r( D) lgn 2. Bylemma 2 and lemma 5, since Vn[ D ] is a directedcycle which has at least lg n 1ranks and v cisadjacent with v ifor all v i V n, for any directednode ranking labeling C of D, we must haveC( v ) C( v ), v V , which implies C mustc i i nhave at least lg n 2 ranks. Then we haveRANK( G) r( D) lgn 2. Hence,RANK( G) lgn 2. Theorem 2 Let G W1,nbe a wheel graph.Then3 if n is even,rank( G) 4 if n is odd.Proof: For even n, consider an orientation D1ofG such that viis adjacent to vi1,vi 1where i isodd, 1 i n. vjis adjacent to v cfor all evenj, 2 j n, and vcis adjacent to vifor all odd i,1 i n. Then D1is a strong orientation of G.Consider a labeling C which ranks Cv (c) 3 ,Cv (i) 1 for all vi Vn, i is odd; andCv (j) 2 for all v j V n, j is even. Then C isa directed node ranking labeling of D1withmaximum rank 3, which impliesrank( G) r( D1) 3. Since G contains K3as asubgraph, by lemma 4 and proposition 2, wehave rank( G) r( K3) 3 which give usrank( G) 3 for even n.For odd n, consider an orientation D 2of G suchthat viis adjacent to v i1and vi 1where i isodd and 1 i n. vjis adjacent to v cfor alleven j, 2 j n , and v cis adjacent to v ifor allodd i, 1 i n, and the edge vv1 nmay be orientedeither way. Then D2is a strong orientation of G.Consider a labeling C which ranks Cv (c) 4Cv (n) 3 , Cv (i) 1 for all v i V n, i n andi is odd; and Cv (j) 2 for all v j V n, j iseven. Then C is a node ranking labeling of D 2with maximum rank 4, which impliesrank( G) r( D2) 4. Since V nis odd, and theadjacent vertices can not have the same rank, thereare at least 3 ranks have to be used in V n. SinceVccan not have the same rank with any vertex inVn, a directed node ranking labeling of anyorientation of G, must use at least 4 ranks. Then wehave rank( G) 4, hence rank( G) 4. 3 ConclusionThe node ranking problem of graphs may beextended to digraphs. This paper defined thedirected node ranking number and established bothupper and lower node ranking number of wheelgraphs among all strong orientations.References[1] H.L. Bodlaender, J.R. Gilbert, H.Hafsteninsson and T. Kloks, “Approximatingtreewidth, pathwidth, frontsize and shortestelimination tree,” Journal of Algorithms 18, pp.238-255, 1995.[2] E. Bruoth and M. Horák, “On-line rankingnumber for cycles and paths,” DiscussionesMathematicae, Graph Theory 19, pp. 175-197,1999.[3] P. de la Torre, R. Greenlaw, A.A. Schäffer,“Optimal ranking of trees in polynomial time,”Proceedings of the 4th Annual ACM-SIAMSymposium on Discrete Algorithms, pp.138-144, 1993.[4] J.S. Deogun, T. Kloks, D. Kratsch, and H.Muller, “On vertex ranking for permutationand other graphs,” Proc. of the 11th AnnualSymposium on Theoretical Aspects ofComputer Science, P. Enjalbert, E.W. Mayr,K.W. Wagner, Lecture Notes in ComputerScience 775, Springer-Verlag, Berlin, pp.747-758, 1994.[5] I.S. Duff and J.K. Reid, “The multifrontalsolution of indefinite sparse symmetric linearequations,” ACM Transactions onMathematical Software 9 (1983), pp. 302-325.[6] A.V. Iyer, H.D. Ratliff and G. Vijayan,“Optimal node ranking of trees,” Inform.Process. Lett. 28 225~229, 1988.-424-

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