The Complexity of Nonrepetitive Coloring

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1 The Complexity of Nonrepetitive Coloring Dániel Mrx Deprtment of Computer Science nd Informtion Theory Budpest University of Technology nd Econonomics Budpest H-1521, Hungry Mrcus Schefer Deprtment of Computer Science DePul University Chicgo, IL Astrct A coloring of grph is nonrepetitive if the grph contins no pth tht hs color pttern of the form xx (where x is sequence of colors). We show tht determining whether prticulr coloring of grph is nonrepetitive is conp-hrd, even if the numer of colors is limited to four. The prolem ecomes fixed-prmeter trctle, if we only exclude colorings xx up to fixed length k of x. 1 Squres nd Nonrepetitive Colorings In 1906 Axel Thue pulished his pper Üer unendliche Zeichenreihen which showed the remrkle result tht there is n infinite word over the lphet Σ = {0,1,2} tht does not contin squre, nmely suword of the form xx: Remrkle, ecuse over inry lphet there re only six squrefree words: 0, 1, 01, 10, 010, 101. Remrkle lso, ecuse it is rre Reserch prtilly supported y the Mgyry Zoltán Felsőokttási Közlpítvány nd the Hungrin Ntionl Reserch Fund (Grnt Numer OTKA 67651). 1 For the construction see, for exmple, Lothire s ook [11]. 1

2 instnce of pttern voidnce theorem: counter-exmple to Rmsey theory pulished when Rmsey ws three yers old. Thue s result points in two directions: the study of ptterns in words nd the study of repetition. Comintorics on words hs ecome n ctive reserch field, not lest through its importnce to computer science [11, 12, 13]. In this pper we wnt to follow the second direction studying repetition in structures more generl thn words. There re recent surveys y Grytczuk [8] nd Currie [5] on voiding repetition in vrious res of mthemtics including grph theory, geometry, nd numer theory. One nturl vrint of words is circulr words, tht is, words whose lst letter is djcent to its first letter. Currie [5] showed tht there re squrefree circulr words of every length n 18 on the lphet {0,1,2}. Currie s result cn e rephrsed s sying tht the cycle C n on n 18 vertices cn e colored using 3 colors so tht no supth of C n hs coloring of the form xx. We cll such coloring nonrepetitive. The coloring point of view ws introduced y Alon, Grytczuk, H luszczk, nd Riordn in 2002 pper [1], which lso contined the definition of the Thue chromtic numer of grph, π(g), s the smllest numer of colors needed in nonrepetitive coloring of G. In this terminology, Currie proved tht π(c n ) = 3 for n 18. Remrk One cn distinguish etween vertex nd edge Thue numers depending on whether one studies nonrepetitive vertex or edge colorings of grphs. The originl pper [1] introduced oth vrints ut emphsized the edge-coloring vrint. Susequent ppers seem to hve given more ttention to the vertex-coloring vrint. Here we use the term Thue chromtic numer to suggest vertex coloring. Mny prolems relted to the Thue chromtic numer re still open. For exmple, it is not yet known whether π(g) is ounded y some constnt for ll plnr grphs G, prticulrly intriguing prolem. Kündgen nd Pelsmjer [10] showed tht grphs of treewidth t most k hve Thue chromtic numer t most 4 k, settling the specil cse of outerplnr grphs. Moreover, π(g) 36 2, s ws shown y Alon, Grytczuk, H luszczk, nd Riordn [1]. It is lso known tht every grph hs sudivision whose Thue chromtic numer is t most 4 (shown y Grytczuk [9] for 5 nd Brát nd Wood for 4 [9, 3]). We look t the Thue chromtic numer from the point of view of computtionl complexity. Deciding whether π(g) k is n -question: is there coloring such tht no supth of the grph hs squre coloring. Deciding question of this form elongs to the complexity clss Σ p 2 = NPNP, the 2

3 second level of the polynomil-time hierrchy (see [15] for more informtion on the polynomil-time hierrchy). We conjecture tht the Thue chromtic numer prolem is complete for tht clss. As first result towrds settling this conjecture we show in Section 2 tht determining whether given coloring of grph is nonrepetitive is conp-complete (in other words, deciding whether coloring is repetitive is NP-complete). Indeed, the prolem remins conp-complete even when restricted to four colors, s we show in Section 3. As n illustrtion of our technique, we otin new proof of the Grytczuk-Brát-Wood result tht every grph hs sudivision with Thue chromtic numer t most 4. Deciding whether given two-coloring of grph is nonrepetitive (s well s deciding whether given grph cn e nonrepetitively two-colored) is esy, since two-coloring is nonrepetitive if nd only if it is proper coloring nd the grph does not contin pth of length t lest 4. This rises the question of how hrd it is to determine whether coloring of grph with three colors is nonrepetitive. This prolem looks difficult; for exmple, y Currie s result, we cn tke word w tht is squre-free s circulr word of ny length n 18. Then pth of length 2n with coloring ww is not squre-free, ut we hve to look t lock of length n to find this out. This exmple suggests studying nonrepetitiveness with restricted locklengths. Let π k (G) e the smllest numer of colors in coloring of G which does not contin pth of length t most 2k with repetitive coloring. This is nturl prmeteriztion of the prolem, π 1 (G) equls the chromtic numer of G, nd π 2 (G) is the str-chromtic numer of G, introduced y Vince [16]. We complement the result tht deciding the nonrepetitiveness of coloring is conp-hrd y showing how to decide in time k O(k) n 5 log n whether coloring of grph on n vertices contins pth of length t most 2k with repetitive coloring. Using the terminology of prmeterized complexity [6, 7], for ounded lock-lengths, nonrepetitiveness of coloring is fixedprmeter trctle: the exponent of the polynomil running time does not depend on the prmeter k. Complementing our results, Fedor Mnin [14] hs recently nnounced tht determining whether given edge coloring of grph is nonrepetitive is conp-hrd nd tht deciding the edge Thue numer of grph is Σ p 2 - complete. He lso estlished tht the edge version of π k (G) is NP-complete for fixed k. Mnin s nd our results do not seem to imply ech other. 3

4 2 Nonrepetitiveness of Coloring A word x is squre if x = ww for some non-empty word w. A word is nonrepetitive if it does not contin squre s suword. A repetitive sequence in grph with vertex-coloring is pth in the grph whose coloring, s red long the pth, is squre. A grph coloring is nonrepetitive if it does not contin repetitive sequence. Theorem 2.1 Deciding whether coloring of grph is nonrepetitive is conp-complete. Proof We reduce from the Hmiltonin Pth prolem. Let G = (V,E) e grph with V = {v 1,...,v n }. We construct grph H nd coloring tht is nonrepetitive if nd only if G does not hve Hmiltonin pth. The grph H consists of two prts. In the first prt, for ech v i tke K 2,n nd color the two element prtition using colors nd, nd the n-element prtition using colors c i,j (for 1 j n). Next, for every i j we introduce new vertex colored d i,j nd connect it to the vertex of the K 2,n elonging to v i nd the vertex elonging to v j. Also, we connect ll the vertices colored to new vertex colored c. We construct the second prt of H s follows: for ech 1 i,j n, we tke pth P i,j on three vertices, coloring the vertices on P i,j y,c i,j,. We connect the vertex colored y c to the vertices of the pths P i,1 (1 i n). For every P i,j (1 i n, 1 j < n) nd every edge v i v i E we dd new vertex of color d i,i nd connect it to the vertex of P i,j nd the vertex of P i,j+1. Finlly, we connect ll the -vertices of P i,n to new vertex colored c (1 i n). This finishes the construction of H nd its coloring (for n exmple see Figure 1, where G is the dimond, i.e. K 4 e). We clim tht G contins Hmiltonin pth if nd only if the coloring of H we constructed is repetitive. This implies tht deciding the nonrepetitiveness of grph coloring is conpcomplete. To prove the clim, let us first ssume tht G hs Hmiltonin pth v π(1),...,v π(n). Consider the following pth through H: we strt t the K 2,n ssocited with v π(1), trversing it so we see colors,c π(1),1,. We continue vi the vertex colored d π(1),π(2) to the K 2,n ssocited with v π(2), trversing it s,c π(2),2,, etc. until we rech the vertex in the K 2,n elonging to v π(n). We then continue to the vertex colored c, nd trverse the second hlf of H s follows: P π(1),1, vertex colored d π(1),π(2), P π(2),2, vertex colored d π(2),π(3), etc. finishing with P π(n),n nd the vertex colored c. Since v π(1),...,v π(n) is Hmiltonin pth, this trversl of H is possile, nd, compring the 4

5 c 1,1 c 1,2 c 1,3 c 1,4 d 1,2 d1,3 d 1,4 c 2,1 c 2,2 c 2,3 c 2,4 d 2,1 d 2,4 d 2,3 c 1,1 c 2,1 d 2,1 d 1,2 d 1,3 d 1,4 c 1,2 c 2,2 d 1,2 d 2,1 d 1,3 d 1,4 c 1,3 c 2,3 d 1,2 d 2,1 d 1,3 d 1,4 c 1,4 c 2,4 c 3,1 c 3,2 c 3,3 d 3,2 c d 3,1 c 3,1 d 2,3 d 3,1 d 3,2 c 3,2 d 2,3 d 3,1 d 3,2 c 3,3 d 2,3 d 3,1 d 3,2 c 3,4 c c 3,4 d 3,4 d 3,4 d 3,4 d 3,4 c 4,1 c 4,1 d 4,1 c 4,2 d 4,1 c 4,3 d 4,1 c 4,4 c 4,2 c 4,3 d 4,3 d 4,2 d 4,1 d 4,3 d 4,3 d 4,3 c 4,4 Figure 1: The grph H corresponding to the grph ({1, 2, 3, 4}, {{1,2},{1,3},{1,4},{2, 3},{3,4}}). colors in the two hlves of H, we see tht they re the sme, nd, therefore, the coloring is repetitive. For the reverse direction, ssume tht H contins pth P such tht the colors long P re of the form ww for some word w. Let us first suppose tht w does not contin the color c. Then P is entirely contined within the first or the second hlf of H. In either cse we cn rgue tht no repetition is possile, since ll the colors except nd re unique nd vertices with colors nd re not djcent. We cn therefore ssume tht w contins c. Consequently, P must contin oth vertices z,z colored c (let z e the vertex connecting the two hlves). Without loss of generlity, we cn ssume tht P strts in the first hlf of H, nd thus there re pths Q,Q, nd Q such tht P = QzQ z Q. The first vertex of Q hs color, while ll 5

6 neighors of z hve color, which mens tht Q is empty, nd, therefore, P = QzQ z. Now ny pth from z to z must, for every j, pss through some vertex colored c i,j, 1 i n. In prticulr Q must do so nd therefore contin t lest n vertices colored c i,j for some i,j. Therefore Q must lso pss through n vertices colored c i,j. Now Q cnnot pss through two vertices colored c i,j nd c i,j since they elong to the sme K 2,n nd this would Q to hve c i,j or c i,j s n endpoint, which is not possile s Q hs endpoints colored nd. Since Q hs to pss through n vertices colored c i,j this implies tht for every i there is exctly one j such tht Q psses through vertex colored c i,j. Since we lso rgued tht for every j there is n i so tht Q nd, therey, Q pss through vertex colored c i,j, there is permuttion π such tht c π(j),j occurs on Q. By the construction of the second hlf of H, v π(1),...,v π(n) is Hmiltonin pth of G. We note tht the proof used n unounded numer of colors to chieve the coding. This cn e remedied s we will see in the next section. 3 The Cse of 4 Colors We reduce the numer of colors in the construction y replcing colors with long nonrepetitive sequences on fixed set of colors. As n illustrtion, we first prove simple grph-theoretic result. Proposition 3.1 (Grytczuk, Brát nd Wood) Every grph hs sudivision which cn e nonrepetitively colored with t most 4 colors. Remrk Grytczuk [9] proved tht every grph hs sudivision which cn e colored with t most 5 colors; Brát nd Wood improved his result to 4 colors [3]. Our construction is closer in spirit to Grytczuk s originl proof. The following lemm constructs fmily of nonrepetitive sequences with useful properties. We write x R for the reverse of the sequence x. Lemm 3.2 We cn in polynomil time construct m nonrepetitive sequences of length O(m) on colors 1, 2 nd 3 so tht (i) for ny two sequences x nd y, if we split ech sequence into two hlves of equl length, x = x 1 x 2 nd y = y 1 y 2, then x i y i nd x i y R 3 i (for i = 1,2), 6

7 (ii) ll sequences egin 31 nd end 13, nd (iii) ll sequences hve the sme length. To see tht the lemm is true, tke nonrepetitive sequence x of length 1764m+13 nd permute the colors so it strts with 31. We clim tht every suword of 14 letters hs to contin the sequences 13 nd 31, clim we will verify lter. So if we let x i e the suword of x tht strts with the i-th 31 in x, nd ends with the first 13 t lest 1176m 1 positions lter, we know tht 1176m x i 1176m+13 nd we hve fulfilled condition (ii). In this fshion we cn pick 42m sequences x i from x (1 i 42m), since x 42m ends no lter thn position 42m m + 13 = 1764m Note tht ny two of these sequences y nd z overlp in t lest 588m + 13 positions in x, ecuse x 42m must contin position 14 (42m 1) + 1 = 588m 13 nd x 1 ends no erlier thn position 1176m, so there is string of length 588m+13 common to ll x i. Since y nd z hve length t most 1176m+13 the overlp of length t lest 588m + 13 etween them forces their first hlves, s well s their second hlves to overlp. Therefore, the first hlves of y nd z must differ from ech other, s must the second hlves (otherwise, x would contin squre). Among the 42m sequences, we cn pick 3m sequences of the sme length, fulfilling condition (iii). While it is possile tht for two of these sequences y nd z, the first hlf of y equls the reverse of the second hlf of z, it is not possile tht the first hlf of y equls the reverse of the second hlf of two other sequences z nd z, since in tht cse the second hlves of z nd z would e identicl, which we excluded. Similrly, the second hlf of y cn e equl to the reverse of t most one other sequence. Hence we cn pick m = 3m/3 sequences fulfilling condition (i). We re left with the proof of the clim tht ny nonrepetitive sequence of length 14 contins the susequence 13, nd, consequently, every other two-digit susequence. So let x e nonrepetitive 14-digit string over the lphet {1,2,3}. A 1 must occur within the first four digits of x. If tht 1 is followed y 3 we re done, so we know tht there is sequence 12 strting within the first four positions of x. Suppose tht sequence continued with 1, i.e. we see 121. Then the next digit cnnot e 2 gin, so we hve 1213, nd, therefore, 13 within the first seven digits of x. In other words, we know tht there is sequence 123 strting within the first four positions of x. There re two cses: suppose the next digit fter 123 is 1, i.e. we hve 1231, the next digit hs to e 2 (otherwise we hve 13), followed y 1 (since the word is nonrepetitive): The next digit cnnot e 2, since the word is nonrepetitive, so it hs to e 3 nd we re done, since we hve found 13 within the first nine positions of x. In the second cse, 7

8 we hve To void repetition, this sequence needs to continue If the next digit is 3, we re done, so we cn ssume we see , which cnnot e followed y 1 (repetition), so we hve , which cnnot e followed y 2 (repetition), giving us followed y 2 (otherwise we hve 13), followed y 1 (repetition), yielding Finlly, this string cnnot e followed y 2, so we see , which mens 13 within x. Proof of Proposition 3.1 It is enough to prove the theorem for the cse G = K n. Let (x i ) m i=1 e fmily of m = ( n 2) nonrepetitive sequences s descried in Lemm 3.2. Replce the i-th edge of G with pth of length x i +7 nd color it 210x i 012. Also, give ech vertex of G color 0. We clim tht this coloring of sudivision G of G is nonrepetitive. Suppose, to the contrry, tht G contins pth P with coloring of the form ww. P hs to contin the color 0, since otherwise ww would e suword of some x i which is not possile (s the x i s re nonrepetitive). There re two types of vertices colored 0: the vertices of G, ll of whose neighors re colored 2, nd the vertices introduced in the sudivision, ll of whose neighors re colored 1 nd 3. Hence, for repetition, P must contin two vertices colored 0 of the sme type, nd tht is only possile if P contins whole pth Q etween two vertices of G. It is not possile tht the coloring of Q is suword of w, since the colorings of the pths (nd their reverses) re unique. Hence, Q must contin the order etween the two hlves of P. In other words, ww hs to contin the following string: 0210v0120, where v = x i for some i (if v = x R i we reverse P), nd the oundry of ww occurs within v. Assuming tht the oundry occurs in the second hlf of v (the other cse eing similr), the first hlf of 0210v0120 must occur second time long P; ut then the first hlf of v must occur s the prefix or the reverse of the suffix of some other x j. This possiility, however, is precluded y choosing sequences x i fulfilling Lemm 3.2 (i). Corollry 3.3 Deciding whether coloring of grph is nonrepetitive is conp-complete even for colorings with t most 4 colors. Proof We will show how to replce the colors in the grph H constructed in the proof of Theorem 2.1 with just 4 colors. Using Lemm 3.2 we otin sequences x i, one for ech of the colors c, c k,j, nd d k,j. If vertex hs color c k,j or d k,j, nd it hs een ssigned sequence x i, replce the vertex 8

9 with pth of length x i + 7 nd color it 210x i 012. For the two vertices colored c, we proceed similrly, ut in this cse the vertex is replced with pth colored 130x i 031; cll the two pths replcing the c vertices C nd C (where C is the pth connecting the two hlves of G). Finlly, recolor vertices with colors or to hve color 0. This construction uses colors 0,1,2,3 only. We clim tht the coloring of the resulting grph will e nonrepetitive if nd only if the originl grph G did not hve Hmiltonin pth. The proof of one direction remins unchnged: Hmiltonin pth in G still corresponds to repetitive coloring, since we just replced colors y color sequences. Suppose then tht G contins pth P colored ww. As we rgued erlier, P hs to contin the color 0, since otherwise ww would e suword of some x i, which is nonrepetitive. We hve four types of vertices colored 0: those with neighors 1,3, those with neighors 1,2, those with two neighors colored 2 nd those with two neighors colored 3. Let us look t the lst type first. Suppose P does not contin the sequence 303 (which occurs exctly four times: twice on ech of the pths replcing c). In tht cse P cnnot trverse C (or C ), nd is therefore cught within one of the two hlves of G. We clim tht this is impossile. First of ll, oserve tht P does hve to contin t lest one vertex from C or C, since otherwise we rgue s in the proof of Theorem 2.1 tht the two hlves of the grph otined y removing C nd C do not contin squre. (Tht prt of the proof of Theorem 2.1 did not use the fct tht nd re different colors.) Suppose next tht P contins exctly one vertex from C nd/or exctly one vertex from C. Such vertex must e one of the end-vertices of C or C colored 1. Then P must contin one of the sequences 201 or 102; y chnging direction of P if necessry, we cn ensure tht P contins the sequence 201. We distinguish two cses y whether P lies in the left or right hlf of G. If P lies in the left hlf of G, we use the fct tht ll occurrences of 201 in tht hlf shre the sme vertex colored 1 (the end-vertex of C which elongs to P), so 201 cn occur t most once long P. Hence, the middle of P hs to occur either t 2 01 or In the first cse the next color fter 01 long P must e 0: 2 010, so P itself must strt with 010, however, 010 does not occur nywhere else in the left hlf of G without using the endpoint of C tht is lredy in use, so this is impossile. In the second cse, the next two colors of P fter 201 re 02: , tht is P hs to strt with the sequence 9

10 102. Agin, the only other occurrences of 102 in the left hlf of G use the endpoints of C which is lredy in use in P, so this is not possile either. If P lies in the right hlf, the rgument is similr: the occurrence of 201 cn e either s 201 or 2 01 or 20 1 with respect to the midpoint of P. In the first cse (i.e. the middle of P does not occur within 201), the 201 must e mtched s whole which is only possile y including vertices from C. Hence the pth P needs to contin vertices from oth C nd C, implying tht ww contins string of the form 210x i 012. As we rgued in Proposition 3.1 this is impossile y the construction of the x i. Consequently, P must contin t lest two vertices from either C or C ; since we ssumed tht it does not contin 303, P must egin in C with 310 or 0310 (the lst 0 corresponding to n vertex) nd/or end in C with 013 or 0130 (the initil 0 corresponding to vertex). However ny two occurrences of ny of these sequences in the sme hlf shre vertex, so this is not possile. We conclude tht P must contin the sequence 303. This sequence occurs exctly four times, twice in C nd C. The two occurrences in the sme pth C or C cnnot mtch with ech other, since one egins 3031x i, nd the other (nd the x i s do not contin zeroes). Hence 303 from C must mtch with 303 from C. But then either ll of C or ll of C, nd therefore oth must elong to P. From this point on, we cn rgue s in the originl proof. 4 Bounded-Length Sequences Checking whether coloring of grph is nonrepetitive for lock-lengths up to some fixed vlue k cn e done in polynomil time: we hve to check ll the O(n 2k ) pths of length t most 2k. Here we present n lgorithm tht is significntly more efficient thn rute force: we show tht the prolem is fixed-prmeter trctle, i.e., it cn e solved in time O(f(k)n c ). This mens tht the exponent of n does not increse s k increses. Theorem 4.1 Given vertex-colored grph G(V, E), it cn e checked in time k O(k) V 5 log V whether G hs repetitive sequence of length 2k. Proof The lgorithm is sed on color-coding, introduced y Alon et l. [2]. Assign rndom lel from {1,...,2k} to ech vertex of G independently with uniform distriution. Assume tht we hve polynomil-time lgorithm for checking whether there is repetitive sequence v 1,..., v 2k where vertex v i hs lel i (elow we will present such n lgorithm). If the grph 10

11 hs repetitive sequence, then the sequence receives the lels 1,..., 2k with proility 1/(2k) 2k, hence the lgorithm finds such repetitive sequence with proility 1/(2k) 2k. If the grph hs no repetitive sequence, then of course no such sequence is found y the lgorithm. Therefore, the lgorithm produces correct nswer with proility t lest 1/(2k) 2k, which cn e incresed to constnt y repeting the lgorithm (2k) 2k times. Rndomized lgorithms sed on color-coding cn e derndomized using stndrd techniques, see [2] nd [6, Section 8.3]. We still need to show how to check whether there is repetitive sequence v 1,..., v 2k where vertex v i hs lel i. For given leling λ : V {1,..., 2k} of the vertices, we proceed s follows. For given vertex x, the lgorithm elow checks whether there is repetitive sequence v 1,..., v 2k where λ(v i ) = i nd v k = x. Therefore, the lgorithm hs to e repeted for every possile choice of x, i.e., V times. We uild directed grph D(U,A) where the U is suset of V V. For v,v V, the pir (v,v ) is vertex of D only if v nd v hve the sme color in G, λ(v ) = λ(v) + k, if λ(v) = k, then v = x, nd if λ(v ) = k + 1, then v is neighor of x in G. There is n rc from (v,v ) to (u,u ) in D if nd only if u is neighor of v, u is neighor of v, nd λ(u) = λ(v) + 1. Note tht, y the properties of the vertices in D, the lst requirement lso implies λ(u ) = λ(v ) + 1. It is esy to see tht D is cyclic, hence the length of the longest directed pth cn e determined in time O( A ) using stndrd techniques. We clim tht D hs directed pth on k vertices if nd only if G hs repetitive sequence on 2k vertices. Indeed, if (v 1,v 1 ), (v 2,v 2 ),..., (v k,v k ) is directed pth in D, then v 1,..., v k, v 1,..., v k is pth in G. Notice tht the i-th vertex of the pth in G hs lel i, thus vertex cnnot pper twice in the sequence. Furthermore, v i nd v i hve the sme color in G, hence the pth is repetitive. The converse sttement is lso esy to see: if v 1,..., v 2k 11

12 is repetitive sequence such tht λ(v i ) = i nd v k = x, then the vertices (v 1,v k+1 ), (v 2,v k+2 ),..., (v k,v 2k ) exist in D nd they form directed pth. The directed grph D contins t most V 2 vertices nd hence t most V 4 edges. Finding the longest pth in the cyclic grph D cn e done in liner time. The lgorithm hs to e repeted for every possile vertex x, thus the running time is V 5 for given leling. The derndomiztion dds fctor O(log V ) to the running time. The cse k = 2 is of specil interest. Grphs tht do not hve repetitive sequences of length t most 4 re often clled str-free or pthic. For strfree coloring, the complexity of the coloring prolem is settled: Proposition 4.2 (Colemn, Moré [4]) Deciding whether grph hs str-free coloring with three colors is NP-complete, even if the grph is iprtite. The proof is quite simple: replce ech edge of grph G with three pths of length 2. Then the originl grph is 3-colorle, if nd only if the resulting (iprtite) grph hs str-free 3-coloring. The result ws proved y Colemn nd Moré in the context of computing sprse Hessin mtrices. Acknowledgments We would like to thnk Michel Pelsmjer for crefully reviewing nd discussing the proofs nd mking severl helpful suggestions. References [1] Nog Alon, Jros lw Grytczuk, Mriusz H luszczk, nd Oliver Riordn. Nonrepetitive colorings of grphs. Rndom Structures Algorithms, 21(3-4): , Rndom structures nd lgorithms (Poznn, 2001). [2] Nog Alon, Rphel Yuster, nd Uri Zwick. Color-coding. J. Assoc. Comput. Mch., 42(4): , [3] János Brát nd Dvid R. Wood. Notes on nonrepetitive grph colouring, mth/ (ccessed Ferury 14th, 2008). 12

13 [4] Thoms F. Colemn nd Jorge J. Moré. Estimtion of sprse Hessin mtrices nd grph coloring prolems. Mth. Progrmming, 28(3): , [5] Jmes D. Currie. Pttern voidnce: themes nd vritions. Theoret. Comput. Sci., 339(1):7 18, [6] Rod G. Downey nd Michel R. Fellows. Prmeterized complexity. Monogrphs in Computer Science. Springer-Verlg, New York, [7] Jörg Flum nd Mrtin Grohe. Prmeterized Complexity Theory. Springer-Verlg, Berlin, [8] Jros lw Grytczuk. Thue type prolems for grphs, points, nd numers. Discrete Mth. (to pper). [9] Jros lw Grytczuk. Nonrepetitive grph coloring. In Grph theory in Pris, Trends Mth., pges Birkhäuser, Bsel, [10] Andre Kündgen nd Michel J. Pelsmjer. Nonrepetitive colorings of grphs of ounded treewidth. Discrete Mth., ville online 29 Septemer ( B6V00-4PSJT4D-2/2/40732ef45641d8565e3f9f ). [11] M. Lothire. Comintorics on words. Cmridge Mthemticl Lirry. Cmridge University Press, Cmridge, [12] M. Lothire. Algeric comintorics on words, volume 90 of Encyclopedi of Mthemtics nd its Applictions. Cmridge University Press, Cmridge, [13] M. Lothire. Applied comintorics on words, volume 105 of Encyclopedi of Mthemtics nd its Applictions. Cmridge University Press, Cmridge, [14] Fedor Mnin. The complexity of nonrepetitive edge coloring of grphs. (ccessed Ferury 15th, 2008), [15] Mrcus Schefer nd Chris Umns. Completeness in the polynomiltime hierrchy: Prt I: A compendium. SIGACTN: SIGACT News (ACM Specil Interest Group on Automt nd Computility Theory), 33,

14 [16] Andrew Vince. Str chromtic numer. J. Grph Theory, 12(4): ,