Given a graph G, a function f: V(G) -\u3e {1, 2, ..., k} is a k-ranking of G if f(u) = f(v) implies every u - v path contains a vertex w such that f(w) \u3e f(u). A k-ranking is minimal if the reduction of any label greater than 1 violates the described ranking property. The a-rank number of G, denoted u,(G) equals the largest k such that G has a minimal k-ranking. We establish new results involving minimal rankings of paths and in particular we determine u(Pn), a problem suggested by Laskar and Pillone in 2000. We show u(Pn) = [log2 (n + 1)] + [log2(n + 1 - (2^([log2n]-1))] (Refer to PDF file for exact formulas)

Kostyuk, Victor

Narayan, Darren

Shults, Victoria

English

Name not available

Rochester Institute of TechnologyRIT Scholar WorksArticles8-17-2003Minimal k-rankings and the a-rank number of apathVictor KostyukDarren NarayanVictoria ShultsFollow this and additional works at: http://scholarworks.rit.edu/articleThis Article is brought to you for free and open access by RIT Scholar Works. It has been accepted for inclusion in Articles by an authorizedadministrator of RIT Scholar Works. For more information, please contact ritscholarworks@rit.edu.Recommended CitationKostyuk, Victor; Narayan, Darren; and Shults, Victoria, "Minimal k-rankings and the a-rank number of a path" (2003). Accessed fromhttp://scholarworks.rit.edu/article/1437Minimal !-rankings and the a-rank number of a pathVictor Kostyuk¤, Darren A. Narayany, and Victoria A. Shults¤Department of Mathematics and Statistics, Rochester Institute of TechnologyAugust 17, 2003AbstractGiven a graph !" a function # : $ (!)! f1" 2" %%%" &g is a &-ranking of ! if #(') = #(() implies every'¡ ( path contains a vertex ) such that #()) * #(')% A &-ranking is minimal if the reduction of anylabel greater than 1 violates the described ranking property. The a-rank number of !" denoted +!(!)equals the largest & such that ! has a minimal &-ranking. We establish new results involving minimalrankings of paths and in particular we determine +!(,")" a problem suggested by Laskar and Pillone in2000. We show +!(,") = blog2 (-+ 1)c+jlog2³-+ 1¡³2blog2 "c¡1´´k%1 IntroductionA labeling ! : " (#) ! f1$ 2$ %%%$ &g is a &-ranking of a graph # if and only if !(') = !(() implies thatevery '¡ ( path contains a vertex ) such that !()) * !(')% A &-ranking ! is minimal if for all (! 2 " (#)$a function + satisfying +(() = !(() when ( 6= (! and +((!) , !((!), is not a ranking. That is, if any labelin a minimal ranking is replaced with a smaller label the new labeling is not a ranking. Note that for anyranking ! there exists a minimal ranking - such that -(() · !(() for every ( 2 " (#)% The rank number ofa graph denoted ."(#), is de…ned to be the smallest & such that # has a minimal &-ranking, and the aranknumber of a graph denoted /"(#) is de…ned to be the largest & such that # has a minimal &-ranking. Whenthe value of & is unimportant, we will refer to a &-ranking as simply a ranking.The rank number of a graph has been well studied, partially due to its applications to VLSI (Very LargeScale Integration) Layouts and scheduling problems for manufacturing systems [1], [5], [8]. While the ranknumber has been determined for various families of graphs, the arank number is only known for a few classesof graphs, such as stars and split graphs. An important property of the arank number is that it implies anecessary condition for a given ranking to be minimal. That is, if a ranking contains a label greater than/"(#) it cannot be a minimal ranking.The problem of determining the arank number of a path was suggested by Laskar and Pillone [7]. InTheorem 13 we provide a complete solution to this problem. In addition, we provide a general resultinvolving necessary conditions for a ranking of a path to be minimal. In Theorem 7 we prove that morethan half of the vertices in a minimal ranking of 0# must be labeled 1 or 2.¤Research partially supported by JetBlue Airways, Kay & Tony Carlisi, and Timothy GilbertyPartially supported by a 2002 RIT COS Dean’s Summer Research Fellowship Grant12 BackgroundWe will use 0# to denote the Hamiltonian path (1$ (2$ %%%$ (# and h!((1)$ !((2)$ %%%$ !((#)i to explicitlydescribe the labels in a ranking ! . For a given ranking let 1! represent the independent set of all verticeslabeled 2. Given a graph # and a set 1 µ " (#) the reduction of # is a graph #¤such that " (#¤) = " (#)¡1and for vertices ' and (, ('$ () 2 3(#¤) if and only if there exists a ' ¡ ( path in #% Note that if # is apath, #¤ is also a path. An example of a reduction is given in Figure 1.Figure 1: A reduction with # = 07 and 1 = 11.For a ranking ! of a graph #, !¤j$¤ will represent the ranking of #¤ where !¤j$¤(() = !(() ¡ 1 for all( 2 " (#) with !(() * 1% For any other unde…ned notation, see the graph theory text by D. B. West [9].We continue with a series of lemmas involving the frequency and locations of small labels that mustappear in a minimal ranking. We restate the following two lemmas from [2].Lemma 1 Let # be a graph and ! be a minimal ranking of #% If 4 2 " (#) and !(4) = 2$ then there existsa vertex ' adjacent to 4 such that !(') = 1%Lemma 2 If 4 is a pendant vertex of a graph # and 5 is adjacent to 4$ then in any minimal ranking ! of#$ either !(4) = 1 or !(5) = 1%In the context of paths, this last lemma states that for any minimal ranking one of the …rst two vertices(or last two) must be labeled 1. If 6 ¸ 4$ we can use operation of reduction to show that one of the …rstfour (or last four) vertices must be labelled 2% This is presented in our next lemma.Lemma 3 Let ! be a minimum ranking of a path 0# = (1$ (2$ %%%$ (# with 6 ¸ 4. Then !((!) = 2 for some1 · 2 · 4. Furthermore if !((!) 6= 2 for 1 · 2 · 3, then !((1) = !((3) = 1%Proof. Assume the smallest 2 such that !((!) = 2 is greater than 4% Then at least two of the …rst fourvertices in the path are labeled with integers greater than 2% It follows that in !j%¤! an end vertex and itsneighbor will both have labels greater than 1$ contradicting Lemma 2. For the second part, assume !((!) 6= 2for 1 · 2 · 3 and !((4) = 2% Suppose that either !((1) 6= 1 or !((3) 6= 1. Then two of the vertices (1$ (2and (3 will have labels greater than 2% Then again, the pedant vertex and its neighbor will be mapped to avalue greater then 1 by !j%¤! , contradicting Lemma 2.We next give a bound on the maximum size of a subpath with end vertices labeled ) and all internalvertices labelled 7 6= )%2Lemma 4 If ! is a minimal ranking of 0# then any subpath of order 2&+1 has a vertex ( such that !(() = 8%Proof. The proof is by induction on 8% The case where 8 = 1 was shown in [7]. The inductive stepfollows using reduction.It is not di¢cult to show that if 0 0 is an induced subpath of a path 0 , then /"(0 0) · /"(0 )% We restatea lemma from [4] which shows that this monotonicity property holds in general.Lemma 5 Let 9 be an induced subgraph of graph #. Then /"(9) · /"(#)%Proof. An alternate proof is found in [4]. Let ! be a minimal &-ranking of 9%We construct a labeling of+ where +(() = !(() for all ( 2 9 and labeling all other vertices arbitrarily &+1$ &+2$ %%%$ &+j" (#)j¡j" (9)j %To see that + is a ranking note that if two vertices in # have identical labels then both vertices must be in9, and use the fact that ! is a ranking. Although + may not be a minimal ranking, no label of a vertex in9 may can be replaced with a smaller label since ! is a minimal ranking. Replacing labels in " (#)¡" (9)with smaller lables, if needed, will result in a minimal ranking of # that uses at least & labels.We conclude this section by restating a lemma from [2] that will play a central role later in our proof ofTheorem 7.Lemma 6 Let # be a graph and let ! be a minimal /"-ranking of #% If 11 = f4 : !(4) = 1g then /"¡#¤'1¢=/" (#)¡ 1%3 Minimal !-rankings of pathsIn our last section we noted many necessary conditions for a given ranking of a path to be minimal inlemmas , 2, 3, 4, and 6 All of these lemmas involve the proximity of vertices labeled 1 or 2 in a minimalranking. This leads to our main result, which states that in any minimal ranking of a path, more than halfof the vertices must be labeled 1 or 2.Theorem 7 If ! is a minimal ranking of 0# then j11 [ 12j * #2 %Proof. Let " (0#) = (1$ (2$ %%%$ (#. The vertices in 12 partition 0# into parts :1$ :2$ %%%$ :( where each4 2 12 is the last vertex in some part :!$ 1 · 2 · ; ¡ 1 and :( consists of the remaining vertices. Weillustrate this in Figure 2.Figure 2. Partitioning of 012.3We note that by Lemma 3, j" (:1)j · 4 and by Lemma 4 j" (:!)j · 8 for all 2 = 2$ 3$ %%%$;% Ourstrategy will be as follows: we will prove that j:1 \ (11 [ 12)j * j) (*1)j2 and j:! \ (11 [ 12)j ¸ j) (*")j2 for all2 = 2$ 3$ %%%$;% Combining these inequalities will yield j" (0#) \ (11 [ 12)j = j11 [ 12j * #2 %First we establish the inequality j:1 \ (11 [ 12)j * j) (*1)j2 . By Lemma 3 the …rst 2 must appearsomewhere among the …rst four vertices. We consider four cases and show the inequality holds in each one.² (!((1) = 2) Then :1 = (1 and j" (:1) \ (11 [ 12)j * j) (*1)j2 %² (!((2) = 2) By Lemma 2 !((1) = 1 and j" (:1) \ (11 [ 12)j * 1 = j) (*1)j2 %² (!((3) = 2) By Lemma 2, either !((1) = 1 or !((2) = 1% Hence j" (:1) \ (11 [ 12)j * j) (*1)j2 %² (!((4) = 2) By Lemma 3, !((1) = 1 and !((3) = 1% Hence j" (:1) \ (11 [ 12)j * j) (*1)j2 %We use a similar argument for :( to show j" (:() \ (11 [ 12)j ¸ j) (*# )j2 % Next we show j" (:!) \ (11 [ 12)j¸ j) (*")j2 for all 2 = 2$ 3$ %%%$; ¡ 1% Consider :! for some 2$ 2 · 2 · ;% Let (!+1$ (!+2$ %%%$ (!$j) (*")j be thevertices of :! keeping the same ordering as in 0#. The inequality is clear when j" (:!)j = 2% By Lemma4, j" (:!)j · 8% We consider cases for the various possible lengths of :!% For completeness we include thedetails.² 6 · j" (:!)j · 8% If j:! \ 11j , j" (:!)j ¡ 4 then :! contains at least four vertices with labelshigher than 2. Then !¤j%¤! contains labels for four consective vertices that are all greater than 1%By Lemma 4 !¤j%¤! can not be a minimal ranking, a contradiction. Hence j" (:!) \ 11j ¸ j" (:!)j ¡ 4and j" (:!) \ (11 [ 12)j ¸ j" (:!)j ¡ 3 ¸ j) (*")j2 %² j" (:!)j = 5% Bby Lemma 4 j" (:!) \ 11j ¸ 1 and the vertex labeled 1 can not be the …rst or fourthvertex of :!% Assume, without loss of generality, the second vertex is labeled 1% We use <$ =$ and >to denote the …rst, third and fourth vertices of :! respectively. If !(>) * !(=), then !(=) can be setto 2 and ! still is a ranking; thus !(>) , !(=), which implies !(>) can only equal 1 if the ranking ! isminimal. Hencej" (:!) \ (11 [ 12)j ¸ 3 ¸ j) (*")j2 %² j" (:!)j = 3 or 4% By Lemma 4, j" (:!) \ 11j ¸ 1) j" (:!) \ (11 [ 12)j ¸ 2 ¸ j) (*")j2 %In our next section we use this result to completely determine the arank number of a path.44 The "-rank number of a pathThe <-rank number of a path denoted /"(0#) has been determined for small values of 6 [2]. Thesevalues are given in Table 1.6 1 2 3 4 5 6 7 8 9 10 11/"(0#) 1 2 3 4 4 4 5 5 5 5 6Table 1: <-rank numbers for small pathsA recursive construction was given in [7] for creating a minimal (28¡ 1)-ranking of path with 2& ¡ 1vertices and a minimal (28¡ 2)-ranking of path with 2& ¡ 2&¡2 ¡ 1 vertices. The same construction wasused for both familes of paths and it was conjectured that the rankings produced by this construction were/"-rankings.The case 8 = 1 is trivial and when 8 = 3, a minimal 3-ranking of a 03 can be constructed simply bylabeling the vertices h3$ 1$ 2i % Starting with a &-ranking of a path on ) vertices, …rst delete the two endvertices. We next join two copies of the resulting path with a 03 with labels, h& ¡ 1$ &$ & ¡ 1i % Finally addone vertex to each end of the path and label one of these vertices & + 1 and the other & + 2% An exampleshowing the construction of a minimal 6-ranking of 011 is shown in Figure 2.Figure 3. Construction of a minimal 6-ranking from a minimal 4-ranking.A direct application of Lemma 6 can be used to show that the rankings produced by the constructionare in fact /"-rankings. We prove this in the following two lemmas.5Lemma 8 /"(02$¡1) = 28¡ 1 for all integers 8 ¸ 2%Proof. We proceed by induction on 8% As seen in Table 1, /"(022¡1) = 2(2)¡ 1 = 3%Assume the equality holds for 8% Given a path on 2&+1¡1 vertices, using the construction from Laskarand Pillone we can produce a (28 + 1)-ranking. Hence /"(02$+1¡1) ¸ 28 + 1% To show the reverseinequality, we assume that /"(02$+1¡1) ¸ 28+2% Then there exists a minimal 28+2-ranking for 02$+1¡1,in which case reducing 02$+1¡1 twice produces a path 0 with a (28)-ranking. By Theorem 7, 0 must haveless than 2& ¡ 1 vertices. Then Lemma 5 implies /"(02$¡1) ¸ 28 which contradicts our assumption.Lemma 9 /"(02$¡2$¡2¡1) = 28¡ 2 for all integers 8 ¸ 2%Proof. We proceed by induction on 8% As seen in Table 1, /" (024¡22¡1) = /" (011) = 6 = 2(4)¡ 2%Assume the equality holds for 8% Given a path on 2&+1¡2&¡1¡1 vertices, we can construct a 28-ranking.Hence /"(02$+1¡2$¡1¡1) ¸ 28% To show the reverse inequality, we assume that /"(02$+1¡2$¡1¡1) ¸ 28+1%Then there exists a minimal 28+ 1-ranking for 02$+1¡2$¡1¡1. Reducing 02$+1¡2$¡1¡1 twice produces apath 0 with a (28¡ 1)-ranking. By Theorem 7, 0 must have less than or equal to 2&¡ 2&¡2¡ 1 vertices.Application of Lemma 5, yields /"(02$¡2$¡2¡1) ¸ 28¡ 1, which contradicts our assumption.Lemma 10 /"(02$¡2$¡2¡2) = 28¡ 3 for all integers 8 ¸ 2%Proof. We proceed by induction on 8% As seen in Table 1, /" (024¡22¡2) = /" (010) = 5 = 2(4) ¡3% Assume the equality holds for 8% Given a path on 2&+1 ¡ 2&¡1 ¡ 2 vertices, we can construct a(2(8+ 1)¡ 3)-ranking. Hence /"(02$+1¡2$¡1¡2) ¸ 28 ¡ 1% To show the reverse inequality, we assumethat /"(02$+1¡2$¡1¡1) ¸ 28% Then there exists a minimal 28-ranking for 02$+1¡2$¡1¡2. Reducing02$+1¡2$¡1¡2 twice produces a path 0 with a (28¡ 2)-ranking. By Theorem 7, 0 must have less than orequal to 2&¡ 2&¡2 ¡ 2 vertices. Then by Lemma 5 we have /"(02$¡2$¡2¡2) ¸ 28¡ 2, a contradiction.Lemma 11 /"(02$¡2) = 28¡ 2 for all integers 8 ¸ 2%Proof. We proceed by induction on 8% As seen in Table 1, /"(022¡2) = 2(2)¡ 2 = 2%Assume the equality holds for 8% Given a path on 2&+1¡2 vertices, using the construction from Laskarand Pillone we can produce a 28-ranking. Hence /"(02$+1¡2) ¸ 28% To show the reverse inequality, weassume that /"(02$+1¡2) ¸ 28+ 1% Then there exists a minimal (28+ 1)-ranking for 02$+1¡2, in whichcase reducing 02$+1¡2 twice produces a path 0 with a minimal (28)-ranking. By Theorem 7, 0 must haveless than or equal to 2& ¡ 2 vertices. Application of Lemma 5 /"(02$¡2) ¸ 28 , a contradiction.As mentioned Laskar and Pillone established an upperbound for the arank number of a path. In our nexttheorem we combine the above four lemmas with Lemma 5 to show that their upper bounds from [7] are infact tight.6Theorem 12 (<?<6& number of 0#)(i) /"(0,) = 28¡ 2 for all integers @, 2& ¡ 2&¡2 ¡ 1 · @ · 2& ¡ 2.(ii) /"(0-) = 28¡ 1 for all integers A, 2& ¡ 1 · A · 2&+1 ¡ 2&¡1 ¡ 2.Following algebraic manipulation, the above theorem can be restated as follows to give an explicit formulafor the arank number of a path.Theorem 13 Let 0# denote on a path on 6 vertices. Then /"(0#) = blog2 (6+ 1)c+¥log2¡6+ 1¡ ¡2blog2 #c¡1¢¢¦ %References[1] P. de la Torre, R. Greenlaw, and T. Przytycka, Optimal Tree Ranking is in NC. Parallel ProcessingLetters, 2(1):31–41, 1992.[2] J. Ghoshal, R. Laskar, and D. Pillone, Further Results on Minimal Rankings, Ars. Combin. 52 (1999),181-198.[3] J. Ghoshal, R. Laskar, and D. Pillone. Minimal Rankings, Networks, Vol. 28, (1996), 45-53.[4] R. Jamison, Coloring Parameters Associated with Rankings of Graphs, submitted.[5] C. E. Leiserson, Area e¢cient graph layouts for VLSI, Proc. 21st Ann. IEEE Symposium, FOCS (1980)270-281.[6] R. Laskar and D. Pillone, Extremal Results in Rankings, Congr. Numer. 149 (2001), 33-54.[7] R. Laskar and D. Pillone, Theoretical and Complexity Results for Minimal Rankings, J. Combin. Inf..Sci. 25, Nos. 1-4, 17-33.[8] A. Sen, H. Deng, and S. Guha, On a graph partition problem with application to VLSI Layouts. InfoProcess. Lett. 43 (1992) 87-94.[9] D. B. West, Introduction to Graph Theory, 2nd ed. Prentice-Hall, Englewood Cli¤s, NJ, 2001.7

Minimal k-rankings and the a-rank number of a path

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Minimal k-rankings and the a-rank number of a path

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