1 / 15

k t ( G ) = # of cliques of order t in a graph G

( ). n. t. e t = lim e t ( n ). n. ¥. Erdös’ conjecture on multiciplities of complete subgraphs revisited (lower upper bound for cliques of size 5 and 6) F. Franek, McMaster University, Hamilton, Ontario, Canada. k t ( G ) = # of cliques of order t in a graph G

ross-woodard
Download Presentation

k t ( G ) = # of cliques of order t in a graph G

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  1. ( ) n t et=limet(n) n ¥ Erdös’ conjecture on multiciplities of complete subgraphs revisited(lower upper bound for cliques of size 5 and 6) F. Franek, McMaster University, Hamilton, Ontario, Canada kt(G)= # of cliques of order t in a graph G kt(n)=min{kt(G)+kt(Ĝ):|G|=n} Ĝdenotes the complement of G kt(n) et(n) = CS-2001 (Sedmihorky) Slide 1

  2. t et = 21- ( ) 2 et minimum proportion of monochromatic Kt’s in a edge-colouring of Kn with two colours. An old (1962) conjecture of Erdös related to Ramsey theory states that It follows from Goodman’s (1959) work that the conjecture is true for t = 3. Erdös & Moon (1964) showed that the conjecture modified for bipartite cliques is true. Sidorenko (1980’s) showed that the conjecture modified for cycles is true, while not true for certain incomplete subgraphs. CS-2001 (Sedmihorky) Slide 2

  3. The conjecture is true for random and “pseudo-random” graphs (Graham & Spencer 1971, Frankl, Rödl, & Wilson 1988, Thomason 1985). Franek & Rödl 1993 showed that the conjecture holds not only for “pseudo-random” graphs, but for graphs obtained from “pseudo-random” graphs by small “perturbations”. Thomason showed the original conjecture false for t ³4in 1989. He obtained following upper bounds: e4< 0.976 x 2-5 e5< 0.906 x 2-9 CS-2001 (Sedmihorky) Slide 3

  4. t et< 0.936 x 21-() for t > 5 2 As for the lower bound Giraud 1979 showed that e4 > 1 46 Franek & Rödl 1993 used simple Cayley graphs together with a computer search to disapprove the conjecture for t = 4 with virtually the same upper bound as Thomason. Searches for higher values for t were not computationally feasible at that time. In this paper we will present the same technique for t = 5 and t = 6 obtaining better upper bounds that Thomason: CS-2001 (Sedmihorky) Slide 4

  5. e5< 0.886 x 2-9 < 0.906 x 2-9 e6< 0.745 x 2-14 < 0.936 x 2-14 The method: For a fixed graph G = (V, S)Gn = (Vn,Sn) is defined in the following way - every vertex v from V is “blown up” to av, a set of size n. Every two elements of av are joined by an edge; xÎ av and yÎ aw are joined by an edge iff v and w are joined by an edge in G. CS-2001 (Sedmihorky) Slide 5

  6. ( ) ( ) ( ) ( ) ( ) ( ) n n n 3 n n 2 2 2 2 3 4 2 nk2(G) + k1(G) k2(G) + G Gn Now we can calculate the number of k-cliques in Gn from number of m-cliques in G. k4(Gn) = n4k4(G) + n2k3(G) + k4(Ĝn) = n4k4(Ĝ) CS-2001 (Sedmihorky) Slide 6

  7. 2 2 3 k2(G) + k2(G) + nk3(G) + ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) n n n n n n n n n n n n n n n n n n 2 3 5 5 2 4 2 3 2 2 3 4 3 3 2 4 2 2 2 3 4 4 3 2 5 n3k4(G) + n4k5(G) + n2k3(G) + nk2(G) + n3k4(G) + k1(G) nk2(G) + n2k3(G) + 3 2 nk3(G) + k2(G) + k5(Gn) = n5k5(G) + k5(Ĝn) = n5k5(Ĝ) k6(Gn) = n6k6(G) + CS-2001 (Sedmihorky) Slide 7

  8. k3(G) + n2k4(G) + ( ) ( ) ( ) ( ) ( ) ( ) ( ) n n n 4 n n n 2 2 2 2 2 6 2 lim k1(G) k4(Gn)+k4(Ĝn) ( ) |Gn| 4 n ¥ k6(Ĝn) = n6k6(Ĝ) This leads immediately to the following limit formulas = 24(k4(G)+k4(Ĝ))+36k3(G)+14k2(G)+k1(G) overk1(G)4 CS-2001 (Sedmihorky) Slide 8

  9. lim lim k5(Gn)+k5(Ĝn) k6(Gn)+k6(Ĝn) ( ) ( ) |Gn| |Gn| 6 5 n n ¥ ¥ = 120(k5(G)+k5(Ĝ))+240k4(G)+150k3(G)+30k2(G)+k1(G) overk1(G)5 = 720(k6(G)+k6(Ĝ))+1800k5(G)+1560k4(G)+540k3(G)+62k2(G)+k1(G) overk1(G)6 CS-2001 (Sedmihorky) Slide 9

  10. Consider a Cayley graph GX,E : vertices are subsets of a set X,EÍ {1,2,...,|X|}, a family of sizes.a, bÍ X are joined by an edge iff |a Db| Î E a Db denotes the symmetric difference. How to count cliques in GX,E : <a0,a1,...,am-1> is an X,E,m-sequence iff each ai is a subset of X, |ai| Î X, and |aiDaj| Î E whenever i¹ jcm(X,E) denotes the set of all X,E,m-sequences. CS-2001 (Sedmihorky) Slide 10

  11. k4((GX,E)n)+k4((ĜX,E)n) k5((GX,E)n)+k5((ĜX,E)n) = = ( ) ( ) |(GX,E)n| |(GX,E)n| 4 5 2 |X| cm(X,E) km+1(GX,E) = (m+1)! lim lim n n ¥ ¥ Observation: Thus c3(X,E)+c3(X,Ê)+6c2(X,E)+7c1(X,E)+1 over 23|X| CS-2001 (Sedmihorky) Slide 11

  12. k6((GX,E)n)+k6((ĜX,E)n) = ( ) |(GX,E)n| 6 lim n ¥ c4(X,E)+c4(X,Ê)+10c3(X,E)+25c2(X,E)+15c1(X,E)+1 over 24|X| c5(X,E)+c5(X,Ê)+15c4(X,E)+65c3(X,E)+90c2(X,E)+31c1(X,E)+1over 25|X| Sequences can be calculated by generating them and tallying them along the process. Since done by a computer program, must be careful not to miss any. CS-2001 (Sedmihorky) Slide 12

  13. Illustration of the method a0 a1 y01 y0 y1 y012 y02 y12 y2 a2 <a0,a1,a2> is an X,E,3-sequence iff |y0|+ |y01|+ |y02|+ |y012| ÎE|y1|+ |y01|+ |y12|+ |y012| ÎE|y2|+ |y02|+ |y12|+ |y012| ÎE|y0|+ |y01|+ |y2|+ |y12| ÎE|y0|+ |y02|+ |y1|+ |y01| ÎE|y1|+ |y01|+ |y2|+ |y02| ÎE CS-2001 (Sedmihorky) Slide 13

  14. |X|-m0m1 |X|-m0-m1 m2 |X|-m0-m1-m2 m01 ì üî þ ì üî þ ì üî þ |X|-m0-m1 -m2 -m01 m02 |X|-m0-m1 -m2 -m01 -m02 m12 ì üî þ ì üî þ |X|-m0-m1 -m2 -m01 -m02 -m12 m012 ì üî þ |X|m0 ì üî þ Generate all sequences <m0,m1,m2,m01,m02,m12,m012> so that m0 + m01 + m02 + m012ÎEm1 + m01 + m12 + m012ÎEm2 + m02 + m12 + m012ÎEm0 + m01 + m2 + m12ÎEm0 + m02 + m1 + m01ÎEm1 + m01 + m2 + m02ÎE for each sequence calculate product CS-2001 (Sedmihorky) Slide 14

  15. For X = {1,2,3,4,5,6,7,8,9,10}E = {1,3,4,7,8,10}, Ê = {2,5,6,9} we calculated c5(E) = 13677741000 c5(Ê) = 25382760480 c4(E) = 742203000c4(Ê) = 1009617840 c3(E) = 14734170c3(Ê) = 17273850 c2(E) = 125730 c1(E) = 506 Thus e4£0.97650119 x 2-5 e5£ 0.88583369880 x 2-9 e6£ 0.744513803200 x 2-14 CS-2001 (Sedmihorky) Slide 15

More Related