On colour-blind distinguishing colour pallets in regular graphs

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1 J Comb Optim ( : DOI /s x On colour-blin istinguishing colour pallets in regular graphs Jakub Przybyło Publishe online: 25 October 2012 The Author(s This article is publishe with open access at Springerlink.com Abstract Consier a graph G (V, E an a colouring of its eges with k colours. Then every vertex v V is associate with a pallet of incient colours together with their frequencies, which sum up to the egree of v. We say that two vertices have istinct pallets if they iffer in frequency of at least one colour. This is always the case if these vertices have istinct egrees. We consier an apparently the worse case, when G is regular. Suppose further that this coloure graph is being examine by a person who cannot name any given colour, but istinguishes one from another. Coul we colour the eges of G so that a person suffering from such colour-blinness is certain that colour pallets of every two ajacent vertices are istinct? Using the Lopsie Lovász Local Lemma, we prove that it is possible using colours for every -regular graph with 960. Keywors Neighbour-istinguishing colouring Lopsie Lovász Local Lemma Colour pallet 1 Distinguishing colour pallets by colour-blin Consier a simple graph G (V, E an an ege colouring c : E {1, 2,...,k}, not necessarily proper. Such colouring is calle neighbour istinguishing (or vertex colouring, see e.g., Aario-Berry et al if for every ege uv E, the multiset of colours incient with u is istinct from the multiset of colours incient with v. In other wors, if for every vertex v we set c(v (a 1,...,a k, where a i {w : wv E, c(wv i} for i 1,...,k, then the colouring c is neighbour istinguishing if c(u c(v for each ege uv of G. Clearly, one can fin such colouring if a graph J. Przybyło (B AGH University of Science an Technology, al. A. Mickiewicza 30, Krakow, Polan przybylo@wms.mat.agh.eu.pl

2 J Comb Optim ( : contains no isolate eges, e.g., by painting each ege ifferently. Karoński et al. (2004 first prove that in fact a finite number of 183 colours are always sufficient, or even 30 if the minimum egree δ of G is at least This was then greatly improve by Aario-Berry et al. (2005, who showe that four colours are sufficient, an these can be ecrease to three if δ 1, 000. Suppose now that such coloure graph is examine by a colour-blin person, i.e., someboy who cannot name colours but istinguishes one from another. Can such iniviual istinguish neighbours then? The answer is affirmative in many cases. It is ue to the fact that given a set of coloure eges, they are able to ivie it into monochromatic subsets an count their carinalities. Given any sequence c(v (a 1,...,a k, let us re-orer it non-ecreasingly. The obtaine sequence c (v ( 1,..., k we shall call a pallet of v. Note that there is a bijection between the set of all possible pallets one may obtain for a vertex v of egree an the set of all k-partitions of the integer, i.e., the set P(, k {( 1, 2,..., k N k : k an 0 i i+1 for i 1,...,k 1}. We say that a colour-blin person can istinguish neighbours in our colouring c : E {1,...,k} if c (u c (v for every ege uv E. The smallest integer k for which such colouring exists is calle the colourblin inex of G, an is enote by al(g. This notion refers to the English chemist John Dalton, who in 1798 wrote the first paper on colour-blinness. In fact, because of Dalton s work, the conition is often calle altonism. It has to be note that this parameter is unefine for some classes of graphs, in particular we must exclue graphs with isolate eges. However, thus far all known graphs with unefine colour-blin inex have minimum egree at most three, see Kalinowski et al. for etails. It has been prove there that given a fixe R > 1, there always exists δ R such that al(g 6 for every graph with imum egree Rδ, provie that δ δ R. Unfortunately δ R tens to infinity along with R. It is thus not even known whether graphs with δ δ 0 have well efine colour-blin inex for any constant δ 0, though Kalinowski et al. conjecture that it is so (maybe even with δ 0 4. Situation with this mysterious parameter changes if we restrict ourselves to regular graphs exclusively. Using a Lovász Local Lemma, Kalinowski et al. prove that al(g 6 for every -regular graph G if its egree is greater than a huge constant, namely, if An application of the probabilistic metho in this context meets unusual obstacles. Unlike in many other similar problems, increasing the number of colours, helps only until a certain point. Then the probability of a ba event that vertices are inistinguishable for a colour-blin person (e.g., when the ege colouring is proper grows. In this paper we optimize this probabilistic approach in orer to significantly reuce the threshol for at the cost of a few more colours. We shall thus prove the following theorem. Theorem 1 For every -regular graph G of egree 960, al(g. The proof is base on the following variation of the Lovász Local Lemma, ue to Erős an Spencer (1991, sometimes referre to as the Lopsie Local Lemma. We

3 350 J Comb Optim ( : recall its symmetric versions from Alon an Spencer (2000 (see Corollary an the comments below. Theorem 2 (Lopsie Symmetric Local Lemma Let A be a family of (typically ba events in any probability space an let D (A, E be a irecte graph with imum out-egree +. Suppose that for each A A an every C A N + (A, Pr(A C C C p, (1 where ep( (2 Then Pr( A A A>0. 2 Proof of Theorem Ranom process an epenency igraph Suppose we are given a -regular graph G (V, E with 960. For each ege e E we inepenently an ranomly choose a colour from the available set {1, 2,...,}, each with equal probability, an enote it by c(e. In other wors, the eges of G are associate with a set of inepenent ranom variables (X e e E, each taking one of the values 1, 2,..., with probability 1/. Outcomes for these etermine an ege colouring of G, each occurring with probability 1/ E within the associate prouct probability space. By a ba event A e in our ranom process of generating c we shall mean obtaining c (u c (v for some ege e uv E. If no ba event occurs, the corresponing colouring shall meet our requirements. We thus nee to show that the probability of the event e E A e is positive in our probability space. We efine a igraph D (A, E, so calle epenency igraph, in the following manner. Let A {A e : e E}. Now for every ege e uv (i.e. e {u,v}ofg,we arbitrarily choose one of its en vertices, say v. Equivalently, we choose an orientation e (u,v of every ege e E, an the obtaine orientation of G we enote by G (V, E. Then for every ege e E with orientation e (u,v,werawan arc between A e an every event A e such that e is at istance at most 2 from v in a graph G e (where an ege incient with a vertex is at istance 1 from it, i.e., e is incient with some neighbour of v ifferent from u. The set of all such arcs we enote by E. Note that then + ( 1 2 1, (3 where + is the imum out-egree of D.

4 J Comb Optim ( : Conitional probability of a ba event Consier any event A e with e (u,van some family of events C A (N + (A e {A e }, where N + (A e is the set of out-neighbours of A e in D.WeassumethatA e / C, since inequality (1 is obvious otherwise (for every p 0. Note that every event C A f C (hence also C is etermine by the values of the ranom variables X e with e at istance at most 1 from f (i.e., sharing a vertex with f. By our construction of D, neither of such e is incient with v, except possibly when f e.lete 1, e 2,...,e enote the eges incient with v, where e e, an let e +1,...,e m enote the remaining eges of G. Then the event C C C is etermine by the outcomes for (part of the ranom variables X ei with i. Denote [] {1, 2,...,} an let Z be a set of all (partial colourings of the eges e,...,e m for which C C C hols, i.e., the set of vectors c (c,...,c m [] m +1 such that (X e,...,x em c guarantees C C C. Then (if C C C, hence Pr( C C C>0: Pr(A e C Pr(A e C C C Pr( C C C C C c Z Pr(A e (X e,...,x em cpr((x e,...,x em c Pr( C C C Pr(A e (X e,...,x em c c Z Pr((X e,...,x em c c Z Pr( C C C Pr(A e (X e,...,x em c c Z Pr(A e (X e,...,x em c. c [] m +1 Since A e is etermine by the outcomes for ranom variables associate with eges at istance at most 1 from e, an the pallet of v by the outcomes for X e1,...,x e,we thus obtain: Pr(A e C C C Pr(c (v c X e c, (4 where the imum is taken over all partitions c ( 1,..., of an all c []. Note however that since a colour blin person cannot name a single colour, in particular the colour c, then the boun from (4 is equivalent to the following one: Pr(A e C C C c P(, Pr(c (v c. (5 (Note that the same upper boun as in (5 hols also for Pr(A e. Consier any fixe c ( 1,..., P(,, an enote the lengths of its consecutive imal subsequences of ientical integers by l 1, l 2,...,l q,where

5 352 J Comb Optim ( : l i 1fori 1,...,q an l 1 + +l q. In other wors, c is of the form ( 1,..., }{{} 1, l1 +1,..., l1 +1,..., lq +1,..., lq +1. Then }{{}}{{} l 1 l 2 Pr(c (v c! l 1!...l q!, (6 where is just the number of istinct partitions of elements (eges 1... into ( (enumerate subsets S 1,...,S of carinalities 1,...,, resp., hence! ! 2!!, the factor! appears ue to the colour-blinness of a person trying to istinguish neighbours, for which every (bijective assignment of colours 1,..., to the sets S 1,...,S yiels the same pallet, while l 1!...l q! counts how many times a given colouring has been taken into account in our calculations. Let us enote by r( 1,..., (or r(c the number of repetitions in c, where we call i a repetition if i j for some j < i (hence r( 1,..., q, an note that by (6, Pr(c (v c! 1... l q 2 r(c 1. (7 In fact such estimation has (almost no influence on the result we are able to prove, but significantly simplifies calculations. By (1, (2, (3, (5 an (7, in orer to prove Theorem 1, it is sufficient to show that e 2! r(c 1 1 (8 for every c ( 1,..., P(,. We shall prove this inequality consecutively for the elements of an ascening family P 0 (, P 1 (,...P 14 (, P(, of subsets of P(,, where P r (, {c P(, : r(c r} is just the set of all -partitions of with at most r repetitions, r 0,..., Partitions without repetitions We first consier c P 0 (,, for which all i are istinct. We shall prove that a : e 2 ( 1,..., P 0 (, 1! ( for every 960. Given a not necessarily monotone sequence of non-negative integers k 1,...,k summing up to,bytightening its two elements k i, k j satisfying k j k i + 2 we shall mean substituting these with the elements k i + 1 an k j 1. Note that such operation

6 J Comb Optim ( : always increase the value of, since if without lost of generality, i k 1...k ( 1 an j, i.e, k k > k 1 + 1, then k 1 + 1k 2...k 14 k 1 k k 1 +1 > ( k 1...k k 1...k. Moreover, the minimum an imum of this sequence shall be calle its leftan right borers, resp., an every integer which oes not appear in the sequence, but is between its borers shall be calle a gap. Let c,0 ( 1,..., be an element of P 0 (, for which the value of is imal. Then this sequence has at most one gap, since otherwise we 1... coul tighten the element preceing the smallest gap an the element succeeing the largest gap creating no repetitions in the obtaine one. For every (sufficiently large there is only one such sequence, ue to the fact that its elements must sum up to. Namely, for s (mo, s {0,...,14}, wehave: a e 2! ( s + 8 s! ( s 7! ( s 6!... ( s + 8!! 1. We shall first show that the sequence (a 960 consists of ecreasing subsequences (a n+ j n 64, j 0, 1,...,14. For this purpose consier the following proportion (for s (mo : a a 2 ( 2 2 ( 2 [ 14 ] ( i0 ( i s + 8 s 7( s 6... ( 1 s + 8 ( s 14 i0 ( i ( 7 ( 6... ( The last inequality is obvious for s 0, while for the remaining s, the fact that s s s s s + 9 s or, equivalently, log 0.5 ( 7 + log 0.5 ( 6 + +log 0.5 ( + 7 s log log 0.5 ( s + 9 s + +log 0.5 ( s + 7 s + +log 0.5 ( s + 8 (10

7 354 J Comb Optim ( : is a simple consequence of the following Theorem 3 on so calle majorization inequality, applie for the function f (x log 0.5 x. Theorem 3 (Karamata s inequality, Kaelburg et al Let I be an interval of the real line an let f enote a real-value convex function efine on I. If x 1 x 2 x n an y 1 y 2 y n are numbers in I such that (x 1,...,x n majorizes (y 1,...,y n, i.e., x 1 + +x n y y n an x 1 + +x i y y i for i 1,...,n 1, then By (10 we thus have: f (x f (x n f (y f (y n. a a 2 14 i0 ( i ( 2 ( 7 ( 6...( + 7 [ 5 ( ] ( 2i + 1( 2i ( (i + 2( + (i + 2 i1 ( ( < 1, ( 11( 12 ( ( ( + ( 13( 14 ( ( 30( + 30 (11 because ( 11( 12 ( ( ( < 0 for 32, ( 13( 14 ( ( 30( < 0for 76, an (since 960 > 4 2, 2 (4i 1+2i(2i 1 < 2 (4i (16i (i (i ( 2i+1( 2i ( (i+2(+(i+2 < 1fori 1,...,5. 2 (i Computing a < , a , a , a , a , a , a , a , a , a , a , a , a , a , a , by inequality (11 we thus obtain a < for 960, i.e., (9 hols. Now we shall exten this result, an show that e 2 ( 1,..., P i (, P i 1 (,! i 1 (12 for every 960 an i 0, 1,...,14 (where P 1 (, :. This will imply (8 an finalize the proof. It is then the more sufficient to show that a,i : e 2 ( 1,..., P i (,! i 1 (13 for 960 an i 0, 1,...,14. The proof of this fact shall be inuctive with respect to i. The case of i 0 being alreay consiere, let us fix i 1, i 14, an assume that (13 hols for a,i with i < i. Letc,i ( 1,..., be an element

8 J Comb Optim ( : of P i (, for which the value of is imal. If c 1...,i P i 1 (,, then (13 hols by inuction hypothesis. Assume then that our c,i contains exactly i repetitions (i 1. Observe then that c,i cannot contain any gaps, since otherwise we coul tighten two of its element creating no aitional repetitions (contraicting imality of c,i. Inee, if any repetition of c,i was larger than some gap, then we coul tighten this repetition an the element of c,i preceing its smallest (left-most gap, an analogously in the opposite case (i.e., when a repetition was smaller than some gap. Finally note that to prove (13 for our fixe i, by inuction hypothesis, it is sufficient to prove for every 960 that at least one of the following two conitions hols: ( 1,..., P i (, ( 1,..., P i (, ( 1,..., ( 1,..., P i 1(, P i 2(, ( (the later for i 2. In fact this is exactly what we shall o for almost every i. We will have to by slightly more careful with the case of i 1 though. (14 ( 2.4 Partitions with one repetition Assume that i 1, hence c,i c,1 ( 1, 1 + 1,..., 1 + t, 1 + t, 1 + t + 1,..., for some t {0,...,13}. Consequently, t , an hence Note that our repetition must be closer to one of the borers of c,1, an let us enote the smaller of these two istances by b, i.e., b {0,...,6} is an integer such that {t, 13 t} {b, 13 b}. Then we may make of c,1 a partition of without repetitions by substituting its elements 1 + b an b with 1 1 an 1 +14, respectively (an orering non-ecreasingly. For the obtaine partition ( 1,..., we then have: b j b + j 1 + j j j. 1 + j Since j j0 1 + j is a ecreasing function of 1 (for 1 1 an (for 960, we thus obtain: j j ( j , but since ( 1,..., P 0(,, by(9 wehave: e 2! 1.

9 356 J Comb Optim ( : Consequently, thus (13 hols for i ! 1 a,1 e e 2! , 2.5 Partitions with at least two repetitions Assume now that i 2. Since 1 an i are the borers of c,i, then analogously as above, 1 + ( ( i + i( i 1 + ( i(14 i 2 + i(14 i 1 + (+i(14 i (for i 2, hence Consier first the case when at least one of the repetitions of c,i,say 1 + t,is at istance at most 3 from one of the borers, i.e., there exists an integer b {0,...,3} such that {t, 14 i t} {b, 14 i b}. Then we may make of c,i a partition of with at most i 1 repetitions by substituting its elements 1 + b an i b with 1 1 an 1 + i, respectively. For the obtaine partition ( 1,..., P i 1 (, we then have: b j0 b 1... j j j ( 1,..., P i 1(, 1 + i b + j 1 + j b + j 1 + j j 1 + j j j 1,... hence (13 hols by (14. Assume then that every repetition of c,i is between an i (hence 2 i 6. Choose any two repetitions 1 +t an 1 +t of c,i with 4 t t 10 i (possibly t t, an let a {t, 14 i t }, hence t t +a 8, an thus a 8. Then we may make of c,i a partition of with at most i 2 repetitions by substituting its elements 1 +t an 1 +t with 1 +t a 1 < 1 an 1 +t +a+1 > i, respectively. For the obtaine partition ( 1,..., P i 2(, we then have:

10 J Comb Optim ( : a 1 + t j j0 1 + t a + j a 1 + (8 + t a j 1... j0 1 + t a + j t j 1... j0 1 + t 8 + j j j0 4 + j ( 1,..., P i 2(, 1,... hence (13 hols by (. The proof of Theorem 1 is thus complete. Acknowlegments This research was partly supporte by the National Science Centre Grant No. DEC- 2011/01/D/ST1/044 an by the Polish Ministry of Science an Higher Eucation. I enclose special thanks to Anrzej Żak for fruitful iscussions in 306A. Open Access This article is istribute uner the terms of the Creative Commons Attribution License which permits any use, istribution, an reprouction in any meium, provie the original author(s an the source are creite. References Aario-Berry L, Alre REL, Dalal K, Ree BA (2005 Vertex colouring ege partitions. J Combin Theory Ser B 94(2: Alon N, Spencer JH (2000 The probabilistic metho, 2n en. Wiley, New York Erős P, Spencer J (1991 Lopsie Lovász Local Lemma an Latin transversals. Discret Appl Math 30:1 4 Kaelburg Z, Duki D, Luki M, Mati I (2005 Inequalities of Karamata, Schur an Muirhea, an some applications. Teach Math 8(1:31 45 Kalinowski R, Pilśniak M, Przybyło J, Woźniak M, Can colour-blin istinguish colour pallets? (submitte Karoński M, Łuczak T, Thomason A (2004 Ege weights an vertex colours. J Combin Theory Ser B 91:1 7

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