Long Monochromatic Cycles in Edge-Colored Complete Graphs

Transcription

1 Long Monochromatic Cycles in Edge-Colored Complete Graphs Linda Lesniak Drew Uniersity and Western Michigan Uniersity May 2012 L. Lesniak (Drew, WMU) Long Monochromatic Cycles May / 20

2 The circumference c(g) of a graph G is the length of a longest cycle in G. Theorem 1 (Faudree, Lesniak & Schiermeyer 2009) Let G be a graph of order n 6. Then and this bound is sharp. max{c(g), c(g)} 2n 3 Comment: The proof of Theorem 1 depended heaily on the ramsey number for two een cycles. L. Lesniak (Drew, WMU) Long Monochromatic Cycles May / 20

3 Notation (Fujita 2011): Let f (r, n) be the maximum integer k such that eery r-edge-coloring of K n contains a monochromatic cycle of length at least k. (For i {1, 2}, we regard K i as a cycle of length i.) Thus, Faudree et al. proed f (2, n) = 2n 3 for n 6. L. Lesniak (Drew, WMU) Long Monochromatic Cycles May / 20

4 Comments: 1 To show that f (r, n) k we must show that eery r-edge-coloring of K n contains a monochromatic cycle of length at least k. 2 To show that f (r, n) k we must exhibit one r-edge-coloring of K n in which the longest monochromatic cycle has length (exactly) k. L. Lesniak (Drew, WMU) Long Monochromatic Cycles May / 20

5 Using affine planes of index r 1 and order (r 1) 2 Gyárfas exhibited r-edge-colorings of K n in which the longest monochromatic cycle had length n r 1. In fact, f (r, n) n r 1 for infinitely many r and, for each such r, infinitely many n. L. Lesniak (Drew, WMU) Long Monochromatic Cycles May / 20

6 Using affine planes of index r 1 and order (r 1) 2 Gyárfas exhibited r-edge-colorings of K n in which the longest monochromatic cycle had length n r 1. In fact, f (r, n) n r 1 for infinitely many r and, for each such r, infinitely many n. Conjecture 2 (Faudree, Lesniak & Schiermeyer 2009) For r 3, f (r, n). n r 1 L. Lesniak (Drew, WMU) Long Monochromatic Cycles May / 20

7 Howeer, Conjecture 2 is not true as stated. For example, K 2r can be factored into r hamiltonian paths. Giing each path a different color shows that f (r, 2r) 2, whereas the conjecture would gie f (r, 2r) = 3. 2r r 1 L. Lesniak (Drew, WMU) Long Monochromatic Cycles May / 20

8 Theorem 3 (Fujita 2011) For 1 r n, f (r, n) n r. Proof. Arbitrarily color the edges of K n with r colors. Let G i be the maximal monochromatic spanning subgraph of G with color i for 1 i r. Then some G i0 contains at least ( n 2) /r edges. So, Thus G i0 ( n ( E(G i0 ) 2) n = r r ) ( n 1 2 ) > ( n r contains a cycle of length at least n r. ) 1 (n 1). 2 (Erdös & Gallai 1959) L. Lesniak (Drew, WMU) Long Monochromatic Cycles May / 20

9 Fujita determined f (r, n) for n 2r + 1 and proposed: Problem 4 (Fujita 2011) For r 2 and n 2r + 2 determine f (r, n). Rest of the talk: 1 Determine f (r, 2r + 2). 2 Determine f (r, sr + c) for r sufficiently large with respect to s and c. 3 Open questions. L. Lesniak (Drew, WMU) Long Monochromatic Cycles May / 20

10 Theorem 5 (Ray-Chaudhury & Wilson 1971) For any t 1, the edge set of K 6t+3 can be partitioned into 3t + 1 parts, where each part forms a graph isomorphic to 2t + 1 disjoint triangles. Theorem 6 (Fujita, Lesniak & Toth 2012+) For r 3, f (r, 2r + 2) = 3. For r = 1, 2, f (r, 2r + 2) = 4. Outline of proof for r 4. By Theorem 3, f (r, 2r + 2) 2r+2 r = 3. We show f (r, 2r + 2) 3 by exhibiting an r-edge-coloring of K 2r+2 in which the longest monochromatic cycle is a triangle. L. Lesniak (Drew, WMU) Long Monochromatic Cycles May / 20

11 Case 1: r=3k+1. Then n = 2r + 2 = 6k + 4. Begin with a coloring of the edges of K 6k+3 on the ertices 1, 2,..., 6k+3 with colors c 1, c 2,..., c 3k+1 according to Theorem 5. c 1 : K 6k+3 : c 2 : c 3k+1 : 1 4 6k k+2 6k+3 }{{} 2k+1 L. Lesniak (Drew, WMU) Long Monochromatic Cycles May / 20

12 Case 1 (continued). c 3k+1 : k+2 6k+1 6k+3 6k+4 L. Lesniak (Drew, WMU) Long Monochromatic Cycles May / 20

13 Case 1 (continued). c 3k+1 : k+2 6k+1 6k+3 1 6k+4 c 1 : 2 c 2k+1 : 6k+1 etc. 6k+4 2k + 1 3k 6k+2 6k+4 L. Lesniak (Drew, WMU) Long Monochromatic Cycles May / 20

14 Case 2: r=3k+2. Then n = 2r + 2 = 6k + 6. Begin with a coloring of the edges of K 6k+3 on the ertices 1, 2,..., 6k+3 with colors c 1, c 2,..., c 3k+1 according to Theorem 5. c 1 : K 6k+3 : c 2 : c 3k+1 : 1 4 6k k+2 6k+3 }{{} 2k+1 L. Lesniak (Drew, WMU) Long Monochromatic Cycles May / 20

15 Case 2: r=3k+2. Then n = 2r + 2 = 6k + 6. Begin with a coloring of the edges of K 6k+3 on the ertices 1, 2,..., 6k+3 with colors c 1, c 2,..., c 3k+1 according to Theorem 5. c 1 : K 6k+3 : c 2 : c 3k+1 : 1 4 6k k+2 6k+3 }{{} 2k+1 6k+4 6k+5 6k+6 One unused color c 3k+2 L. Lesniak (Drew, WMU) Long Monochromatic Cycles May / 20

16 Case 3: r=3k. Then n = 2r + 2 = 6k + 2. Begin with a coloring of the edges of K 6k+3 on the ertices 1, 2,..., 6k+3 with colors c 1, c 2,..., c 3k+1 according to Theorem 5. c 1 : K 6k+3 : c 2 : c 3k+1 : 1 4 6k k+2 6k+3 } {{ } 2k+1 In this case we used one extra color and hae one extra ertex. We recolor to remoe color c 3k+1 so that the longest monochromatic cycle is a triangle. Thus f (r, 2r + 3) 3. L. Lesniak (Drew, WMU) Long Monochromatic Cycles May / 20

17 Case 3: r=3k. Then n = 2r + 2 = 6k + 2. Begin with a coloring of the edges of K 6k+3 on the ertices 1, 2,..., 6k+3 with colors c 1, c 2,..., c 3k+1 according to Theorem 5. c 1 : K 6k+3 : c 2 : c 3k+1 : 1 4 6k k+2 6k+3 } {{ } 2k+1 In this case we used one extra color and hae one extra ertex. We recolor to remoe color c 3k+1 so that the longest monochromatic cycle is a triangle. Thus f (r, 2r + 2) f (r, 2r + 3) 3. L. Lesniak (Drew, WMU) Long Monochromatic Cycles May / 20

18 We e determined f (r, n) for n 2r + 2. Problem 7 Determine f (r, sr + c) for integers s, c 2. Comment: f (r, sr + c) sr+c r s + 1 by Theorem 3. We ll show that f (r, sr + c) = s + 1 for r sufficiently large with respect to s and c. L. Lesniak (Drew, WMU) Long Monochromatic Cycles May / 20

19 Theorem 8 (Chang 2000) Let q 3. Then for t sufficiently large, the edge set of K q(q 1)t+q can be partitioned into qt + 1 parts, where each part is isomorphic to (q 1)t + 1 disjoint copies of K q. Comment: Chang s result was stated in terms of resolable balanced incomplete block designs. Thank you, Wal Wallis. Comment: q = 3 in Theorem 8 is our preious Theorem 5. L. Lesniak (Drew, WMU) Long Monochromatic Cycles May / 20

20 Theorem 9 (Fujita, Lesniak & Toth 2012+) For any pair of integers s, c 2 there is an R such that f (r, sr + c) = s + 1 for all r R. Comment: For s = c = 2, R = 3 (Theorem 8). Outline of Proof. 1 We show f (r, sr + c) s + 1 with an r-edge-coloring of K sr+c in which the longest monochromatic cycle has length s Since f (r, n) is monotone increasing in n, we may assume that sr + c = (s + 1)st + (s + 1) for some t. L. Lesniak (Drew, WMU) Long Monochromatic Cycles May / 20

21 Outline of Proof (continued). 3 Color the edges of K (s+1)st+(s+1) = K sr+c with (s + 1)t + 1 = r + c 1 s colors using Chang s Theorem for q = s + 1. c 1 : c (s+1)t+1 : K S+1 K S+1 K S+1 K S+1 K S+1 K S+1 } {{ } st+1 (s + 1)t + 1 = r + c 1 s L. Lesniak (Drew, WMU) Long Monochromatic Cycles May / 20

22 Outline of Proof (continued). 4 For r sufficiently large with respect to s and c we can recolor to reduce the number of colors by c 1 s without creating monochromatic cycles of length greater than s + 1. Comment: To remoe one color class, we need at most ( s log s(s+1)+1 s(s+1) s + 1 r + c ) s + 1 other classes. Thus we can aoid c 1 s classes if ( ) c 1 log s(s+1)+1 2 s(s+1) color class with the remaining r c ) r, s + 1 ( s s + 1 r + which is true for sufficiently large r compared with s and c. L. Lesniak (Drew, WMU) Long Monochromatic Cycles May / 20

23 Some open questions: 1 Determine f (r, n) for n large with respect to r. Comment: We know f (r, n) n r for all 1 r n. Comment: We know f (r, n) n r 1 for infinitely many r and, for each such r, infinitely many n. L. Lesniak (Drew, WMU) Long Monochromatic Cycles May / 20

24 Some open questions (continued): 2 The proof that f (2, n) = 2n 3 for n 6 depended heaily on the ramsey number for two een cycles. Can the ramsey number for three een cycles be used to determine f (3, n) for n sufficiently large? Theorem 10 (Beneides & Skokan 2009) There exists an n 1 such that for eery een n n 1, r(c n, C n, C n ) = 2n. L. Lesniak (Drew, WMU) Long Monochromatic Cycles May / 20