Cycle Length Parities and the Chromatic Number

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1 Cycle Length Parities and the Chromatic Number Christian Löwenstein, Dieter Rautenbach and Ingo Schiermeyer 2 Institut für Mathematik, TU Ilmenau, Postfach 00565, D Ilmenau, Germany, s: christian.loewenstein@tu-ilmenau.de, dieter.rautenbach@tu-ilmenau.de 2 Institut für Diskrete Mathematik und Algebra, Technische Universität Bergakademie Freiberg, Freiberg, Germany, schierme@math.tu-freiberg.de Abstract. In 966 Erdős and Hajnal proved that the chromatic number of graphs whose odd cycles have lengths at most l is at most l +. Similarly, in 992 Gyárfás proved that the chromatic number of graphs which have at most k odd cycle lengths is at most 2k + 2 which was originally conjectured by Bollobás and Erdős. Here we consider the influence of the parities of the cycle lengths modulo some odd prime on the chromatic number of graphs. As our main result we prove the following: Let p be an odd prime, k N and I {0,,..., p } with I p. If G is a graph such that the set of cycle lengths of G contains at most k elements which are not in I modulo p, then χg) Ramsey number. + I ) k + pp )r2p, 2p) + ) + where rp, q) denotes the ordinary Keywords. Graph; colouring; chromatic number; cycle Introduction We consider finite, simple and undirected graphs G = V, E) and denote by LG) the set of cycle lengths of G. In this paper we continue the study of the influence of LG) on the chromatic number χg) of G which was essentially initiated by Erdős and Hajnal in 966. Theorem Erdős and Hajnal [2]) If G is a graph and l is the maximum odd element in LG), then χg) l +. Bollobás and Erdős [] conjectured the following strengthening which was eventually proved by Gyárfás in 992. Theorem 2 Gyárfás [3]) If G is a graph and LG) contains k odd elements, then χg) 2k + 2. In 2004 Mihok and Schiermeyer proved an analogous result for even cycle lengths. Theorem 3 Mihok and Schiermeyer [7]) If G is a graph and LG) contains k even elements, then χg) 2k + 3.

2 The extremal graphs for all three results are complete graphs of suitable order. The result of Erdős and Hajnal has recently been refined by excluding large cliques [4]. Similarly, the result of Gyárfás has been strengthened for graphs G for which LG) contains two specified odd elements [5, 6, 9]. The starting point for our new results was the following observation which we made together with Frank Göring. Observation 4 Let p be an odd prime and k N 0 = N {0}. If G is a graph such that LG) contains at most k elements which are not divisible by p, then χg) 2k + 3. Proof: We prove the result by induction on the order n of G. For n 2k + 3 the result is trivial and we may assume that n > 2k + 3. If G has a vertex u of degree at most 2k + 2, then we can extend a 2k + 3)-coloring of G u to G. Hence we may assume that the minimum degree δ of G satisfies δ 2k + 3. Let P : v v 2... v l be a longest path in G. Let x, x 2,..., x 2k+3 be the first 2k + 3 neighbours of x 0 on P ordered according to increasing distance from x 0 on P, i.e. if for 0 i < j 2k + 3 the subpath P i, j) of P between x i and x j is of length di, j), then = d0, ) < d0, 2) <... < d0, 2k + 3). Since P, i) together with the two edges x 0 x and x 0 x i forms a cycle of length d, i)+2, there are at least 2k + 3) k = k + 2 indices i with 2 i 2k + 3 and d, i) mod p. Let i 0 < i <... < i k+ be k + 2 such indices. Now di 0, i j ) + 2 = d, i j ) + 2) d, i 0 ) + 2) mod p and P i 0, i j ) together with the two edges x 0 x i0 and x 0 x ij forms a cycle of length di 0, i j )+2 for every j k +. Hence G contains k + cycles of different lengths not divisible by p which is a contradiction and the proof is complete. In the next section we prove bounds for the chromatic number of graphs whose cycle lengths are either small or have a fixed parity modulo some odd prime p Theorem 5), whose cycle lengths are either small or have a parity different from 2 modulo p Proposition 6), whose cycle lengths are either small or have a parity modulo p belonging to some set of allowed parities Corollary 8), and that have a bounded number of cycle lengths which do not have a parity modulo p belonging to some set of allowed parities Theorem 7). 2

3 2 Results We immediately proceed to our results. Theorem 5 Let p be an odd prime, k N with k 3 and q {0,,..., p }. If G is a graph in which all cycles of length { at least kp have a length which is equivalent 0, if q = 0 to q modulo p, then χg) kp ɛ for ɛ =, if q > 0. Proof: We prove the result by induction on the order n = V. As in the previous proof, we may assume that n > kp ɛ and also that the minimum degree δ of G satisfies δ kp ɛ. Let P : v v 2... v l be a longest paths in G. If v i N G v ) and i 3, then v v 2... v i v is a cycle of length i mod p. Therefore, all neighbours v i of v with i kp satisfy i q mod p, i.e. {i v i N G v ) and kp i l} k + N 0 ) p + q. Since δ kp ɛ, the vertex v has at least one neighbour v i with i kp and t = max{i N 0 v k+i)p+q N G v )} is well-defined. Note that q = 0 implies t. If v i N G v ) for some 2 i tp + q + 2 with i 2 mod p, then v v i v i+... v k+t)p+q v is a cycle of length k + t)p + q i + 2 kp which is not equivalent to q modulo p. Hence If tp + q + 2 < kp, then {i v i N G v ) and i tp + q + 2} N 0 p + 2. δ N G v ) = N G v ) {v i i tp + q + 2} + N G v ) {v i tp + q + 3 i kp } + N G v ) {v i kp i l} t + ) + kp ) tp + q + 3) + ) + t + ) = kp q tp 2) < kp ɛ. Hence, we may assume that tp + q + 2 kp. If q 2, then δ N G v ) = N G v ) {v i i kp } + N G v ) {v i kp i tp + q + 2} + N G v ) {v i tp + q + 3 i l} k k 2k < kp ɛ. 3

4 Hence, we may assume that q = 2 which implies We assume that P is chosen such that {i v i N G v ) and i l} N 0 p + 2. j = min{i N v ip+2 N G v )} is minimum possible. Since v jp+ v jp... v v jp+2 v jp+3... v l is a longest path, this implies that Let {i v i N G v jp+ ) and i l} {jp} j + N 0 ) p + 2 s = min{i i jp + 3, v i N G v ) N G v jp+ )}. Let x {, jp + } be such that v s N G v x ) and let {y} = {, jp + } \ {x}. Let t = max{i i jp + 3, v i N G v y )}. Since {i i jp + 3, v i N G v y )} δ 2 kp 3, we have t s kp 3)p kp and v x... v y v t... v s v x is a cycle of length more than kp which is equivalent 4 modulo p. This contradiction completes the proof. Proposition 6 Let p be an odd prime and k N with k 3. If G is a graph in which all cycles of length at least kp have a length which is not equivalent to 2 modulo p, then χg) kp +. Proof: We proceed as in the proof of Theorem 5. Therefore, we may assume that the minimum degree δ of G satisfies δ kp + and consider a longest paths P : v v 2... v l in G. By the pigeon hole principle, there is some q {0,,..., p } such that the set I = {i i l, v i N G v ), i q mod p} has at least k + elements. If s = min I and t = max I, then v v s v s+... v t v is a cycle of length at least kp which is equivalent to 2 modulo p. This contradiction completes the proof. Theorem 5 and Proposition 6 are best-possible in view of complete graphs of suitable order. Theorem 7 Let p be an odd prime, k N and I {0,,..., p } with I p. If G is a graph such that LG) contains at most k elements which are not in I modulo p, then χg) + I ) k + pp )r2p, 2p) + ) + where rp, q) denotes the ordinary Ramsey number. 4

5 Proof: Let f 0 p) = pp )r2p, 2p) + ) +, f p) = pp )r2p, 2p) ) +, f 2 p) = p )r2p, 2p) ) +, f 3 p) = r2p, 2p), and f 4 p) = 2p. We prove the result by induction on the order n of G. As in the previous proofs, we may assume that n > the minimum degree δ of G satisfies δ + I + I ) k + f 0 p). ) k + f 0 p) and also that Let P : v v 2... v l be a longest path in G. To simplify our terminology, we extend the order and parity of the indices to the vertices on P. For example, for i < j l we say that v i is smaller than v j and write v i < v j. Similarly, we say that v i is equivalent to some q modulo p if i is equivalent to q modulo p). Let N denote the set of the f p) smallest neighbours of v on P. We assume that P is chosen such that max N is smallest possible. By the pigeon hole principle, there is a set N 2 N with N 2 = f 2 p) such that all elements of N 2 are equivalent modulo p. Let v i N 2. The path v i v i 2... v v i v i+... v l is a longest path in G. Hence the vertex v i has all its neighbours on P. Furthermore, the choice of P implies that v i has at most f p) neighbours which are smaller or equal to max N. Therefore, if M v i ) denotes the set of neighbours of v i note the little index shift) which are larger than max N, then M v i ) + I ) k + f 0 p) f p) = + I ) k + 2pp ). By the hypothesis, there are at most k neighbours v j M v i ) for which the cycle v i v i... v j v i has a length j i + 2 which is not in I modulo p. This implies that, if M 2 v i ) denotes the set of all v j M v i ) for which j i + 2 is in I modulo p, then M 2 v i ) + I ) k + 2pp ) k = I k + 2pp ). By the pigeon hole principle, there is a set M 3 v i ) M 2 v i ) with M 3 v i ) k + 2p such that all vertices in M 3 v i ) are equivalent modulo p. 5

6 Again by the pigeon hole principle, there is a subset N 3 N 2 with N 3 = f 3 p) such that all vertices in v i N 3 M 3 v i ) are equivalent modulo p. By the Ramsey theorem, there is a subset N 4 N 3 with f 4 p) elements N 4 = { } v j < v j 2 <... < v j f 4 p) such that, if for i f 4 p), then either or First, we assume that holds cf. Figure ). M 3 v j i) = { } v j i < v j i 2 <... j < j 2 <... < j f 4p) j j 2... j f 4p). j < j 2 <... < j f 4p) j j 2 j 3 j f 4p) Figure j j 2 j 3 j f 4p) 3) For 0 r p and s k + 2p let Cr, s) denote the cycle cf. Figures 2 and Cr, s) : v v j v j v j v j +... v j 2 v j 2 v j 2... v j 3 v j 3 v j v j 4 v j 4 v j 4... v j 5 v j 5 v j v j 2r v j 2r v j 2r... v j 2r+ v j 2r+v j 2r v... v f j 4 p) + j f 4 v s p) j f 4 p) v j f p)v 4. v j f 4 p) 6

7 j j f 4p) j Figure 2 C0, s) j f 4p) j f 4p) s j j 2 j 3 j f 4p) Figure 3 C, s) j j 2 j 3 j f 4p) j f 4p) s For 0 r p 2 and s, s 2 k + 2p the lengths of the two cycles Cr, s ) and Cr +, s 2 ) differ exactly by 2 modulo p. Furthermore, for 0 r p and s < s 2 k + 2p the cycle Cr, s 2) is longer than the cycle Cr, s ). This implies the existence of ) k ) + 2p > k cycles of different lengths not in I modulo p, which is a contradiction. In the second case j j 2... j f 4p) a very similar construction also leads to a contradicton. In order to ensure the appropriate vertices to be different one can discard the smallest 2p i) elements from every set M 3 v j i) for all i 2p. This completes the proof. Corollary 8 Let p be an odd prime, k N and I {0,,..., p } with I p. If G is a graph in which all cycles of length at least kp have a length which is in I modulo p, then χg) kp + pp )r2p, 2p) + ) +. Proof: The result follows immediately from Theorem 7, because the hypothesis implies that LG) contains at most k) many cycle lengths which are not in I modulo p. ) While the complete graphs show that the term + I in Theorem 7 and the term kp in Corollary 8 are best possible, the additive term depending only on p can clearly be improved. In fact, it is conceivable that it could be replaced by a constant. Acknowledgement: We thank Frank Göring for fruitful discussions and valuable comments. 7

8 References [] P. Erdős, Some of my favourite unsolved problems, in: A. Baker, B. Bollobás, A. Hajnal Eds.), A Tribute to Paul Erdős, Cambridge University Press, Cambridge, 990, pp [2] P. Erdős and A. Hajnal, On chromatic number of graphs and setsystems, Acta Math. Acad. Sci. Hung ), [3] A. Gyárfás, Graphs with k odd cycle lengths, Discrete Math ), [4] S. Kenkre and S. Vishwanathan, A bound on the chromatic number using the longest odd cycle length, J. Graph Theory ), [5] S. Matos Camacho, Colourings of graphs with prescribed cycle lengths, Diploma Thesis, TU Bergakademie Freiberg, [6] S. Matos Camacho and I. Schiermeyer, Colourings of graphs with two consecutive odd cycle lengths, to appear in Discrete Math. [7] P. Mihok and I. Schiermeyer, Cycle lengths and chromatic number of graphs Discrete Math ), [8] B. Randerath and I. Schiermeyer, Vertex Colouring and Forbidden Subgraphs A Survey, Graphs Combin ), -40. [9] S.S. Wang, Colouring graphs with only small odd cycles, Preprint, Mills College,