We discuss a branch of Ramsey theory concerning vertex Folkman numbers and how computer algorithms have been used to compute a new Folkman number. We write G ! (a1, . . . , ak)v if for every vertex k-coloring of an undirected simple graph G, a monochromatic Kai is forced in color i 2 {1, . . . , k}. The vertex Folkman number is defined as Fv(a1, . . . , ak; p) = min{|V (G)| : G ! (a1, . . . , ak)v ^ Kp 6  G}. Folkman showed in 1970 that this number exists for p \u3e max{a1, . . . , ak}. Let m = 1+Pk i=1(ai−1) and a = max{a1, . . . , ak}, then Fv(a1, . . . , ak; p) = m for p \u3e m, and Fv(a1, . . . , ak; p) = a +m for p = m. For p \u3c m the situation is more difficult and much less is known. We show here that, for a case of p = m−1, Fv(2, 2, 3; 4) = 14

Coles, Jonathan

Radziszowski, Stanislaw

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Computing the Folkman number F_v(2,2,3;4)

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Rochester Institute of TechnologyRIT Scholar WorksArticles2006Computing the Folkman number F_v(2,2,3;4)Jonathan ColesStanislaw RadziszowskiFollow this and additional works at: http://scholarworks.rit.edu/articleThis Article is brought to you for free and open access by RIT Scholar Works. It has been accepted for inclusion in Articles by an authorizedadministrator of RIT Scholar Works. For more information, please contact ritscholarworks@rit.edu.Recommended CitationJournal of Combinatorial Mathematics and Combinatorial Computing 58 (2006) 13-22Computing the Folkman NumberFv(2, 2, 3; 4)Jonathan Coles∗ and Stanis law P. RadziszowskiDepartment of Computer ScienceRochester Institute of TechnologyRochester, NY 14623{jpc1870,spr}@cs.rit.eduAbstractWe discuss a branch of Ramsey theory concerning vertex Folkmannumbers and how computer algorithms have been used to computea new Folkman number. We write G → (a1, . . . , ak)v if for everyvertex k-coloring of an undirected simple graph G, a monochromaticKai is forced in color i ∈ {1, . . . , k}. The vertex Folkman numberis defined as Fv(a1, . . . , ak; p) = min{|V (G)| : G → (a1, . . . , ak)v ∧Kp 6⊆ G}. Folkman showed in 1970 that this number exists for p >max{a1, . . . , ak}. Letm = 1+Pki=1(ai−1) and a = max{a1, . . . , ak},then Fv(a1, . . . , ak; p) = m for p > m, and Fv(a1, . . . , ak; p) = a+mfor p = m. For p < m the situation is more difficult and much less isknown. We show here that, for a case of p = m−1, Fv(2, 2, 3; 4) = 14.1 IntroductionLet G be a simple, undirected graph with vertex set V (G) and edge setE(G). The chromatic number of G will be denoted by χ(G), and theindependence number of G by α(G). For positive integers ai, we writeG → (a1, . . . , ak)v if for every vertex k-coloring of G, a monochromaticKai is forced in some color i ∈ {1, . . . , k}. LetHv(a1, . . . , ak; p) = {G : G→ (a1, . . . , ak)v ∧Kp 6⊆ G} .The graphs in the set H = Hv(a1, . . . , ak; p) are called Folkman graphs.Folkman [2] (also [6]) showed thatH is non-empty for p > max{a1, . . . , ak}.∗Currently at the University of Zu¨rich, Institute for Theoretical Physics, Winterthur-erstr. 190, CH-8057 Zu¨rich. jonathan@physik.unizh.chHv(a1, . . . , ak; p;n) will denote the set of Folkman graphs with n vertices.Folkman graphs are maximal when the addition of any other edge willcreate the forbidden Kp. Similarly, a Folkman graph is minimal when thedeletion of any edge causes the graph to lose the Folkman property. Wedefine the vertex Folkman numbers byFv(a1, . . . , ak; p) = min{|V (G)| : G ∈ Hv(a1, . . . , ak; p)} .The Ramsey number R(r, l) is the smallest number n such that all edge2-colorings of Kn contain either a monochromatic Kr in the first color or amonochromatic Kl in the second color [3]. A graph G is an (r, l)-Ramseygraph if G has no Kr and α(G) < l. The set R(r, l;n) is the set of all(r, l)-Ramsey graphs on n vertices.Let m = 1 +∑ki=1(ai − 1). If G → (a1, . . . , ak; p) then χ(G) ≥ m [13].By the pigeon-hole principle it is easy to see that Fv(a1, . . . , ak; p) = mfor p > m.  Luczak et al. [7] showed that Fv(a1, . . . , ak; p) = ak + m forp = m. For p < m much less is known. Only one nontrivial value is knownfor p = m − 2: Fv(2, 2, 2, 2; 3) = 22 computed by Jensen and Royle [4] in1995. Most research has focused on the case p = m− 1. For a summary ofother results see [6].In 2000, Nenov [13] showed that 10 ≤ Fv(2, 2, 3; 4) ≤ 14; he provedthe upper bound using the 14-node graph Γ3, depicted in Figure 1. Nograph with fewer than 14 vertices is known to exist in Hv(2, 2, 3; 4). Itis shown here that there are no such graphs and thus, Fv(2, 2, 3; 4) = 14.Nenov and Nedialkov also studied several other parameter situations in[10, 11, 12, 13, 14, 15, 16, 17]. For the purposes of this paper, references toFolkman graphs will mean graphs in Hv(2, 2, 3; 4) unless otherwise stated.Proving exact values of Folkman numbers by hand is often very difficultsince deriving the lower bound requires a non-existence proof. Computerscan be of great help, but because showing non-existence often entails largesearches, the algorithms must be carefully designed to work as efficientlyas possible.To find the exact value of Fv(2, 2, 3; 4) with the aid of computers, thereare two main approaches. One approach is to check all graphs of order < 14for inclusion in Hv(2, 2, 3; 4), starting with graphs of order 10. If a graphis found to be Folkman, then the current order is the Folkman number. Ifnot, the next higher order must be checked. A second approach is to findall Folkman graphs on 14 vertices and drop a vertex from each one in allpossible ways. If one of the resulting graphs is Folkman, then 13 is the newupper bound and the process can be repeated on the Folkman graphs with13 vertices. Once no smaller Folkman graphs are obtained, the Folkmannumber has been found.Figure 1: The Nenov graph Γ3 from [13].2 Testing for the Folkman PropertyRegardless of the approach, we must have an algorithm for testing whetheror not a graph is in Hv(2, 2, 3; 4). From the set parameters, any graph thatcontains K4 can be discarded. For the remaining graphs, any colorings ofa graph that don’t have a K2 in the first or second colors must force aK3 in the third, otherwise the graph is not Folkman. Such colorings occurwhen two independent sets are colored with the first and second colors,forcing any remaining vertices not in the union of the two independent setsto contain a K3 in the third color.The algorithm for determining if G ∈ Hv(2, 2, 3; 4) is fairly straightfor-ward: Discard G if it contains a K4 or χ(G) ≤ 4. Otherwise, check thatfor each pair of independent sets A,B ⊂ V (G), the induced subgraph onV (G) \ (A ∪ B) contains a K3. A pseudo-code outline of the algorithmis presented as Algorithm 2.1, InH2234. For this to be efficient, we needLemma 2.1.Algorithm 2.1: InH2234(G)if K4 ⊆ G or χ(G) ≤ 4then return ( false )M = AllMaxCliques(G)T = {{a, b, c} : a, b, c ∈ V (G) and abc is a triangle in G}for each A,B ∈MdoC ← V (G) \ (A ∪B)if C doesn’t contain a triangle in Tthen return ( false )return ( true )Lemma 2.1 In InH2234 it is sufficient to consider only maximal inde-pendent sets when determining whether G ∈ Hv(2, 2, 3; 4).Proof. Let A,B be maximal independent sets in G and let V (G)\ (A∪B)induce a subgraph S in G. In order for G ∈ Hv(2, 2, 3; 4), S must containa K3. Now let A′ ⊂ A and B′ ⊂ B. Since K3 is a subgraph of S, it followsthat the induced subgraph on V (G) \ (A′ ∪B′) also contains a K3. Thus,in Algorithm 2.1 it is sufficient to consider only the maximal independentsets A and B. For graphs of order 14, the number of maximal independent sets isusually very small (around 40), so the complexity of the algorithm is nota great obstacle. The algorithm AllMaxCliques [5] is used to find thesesets. Although AllMaxCliques returns maximal cliques, if the input isG, the result will be maximal independent sets in G.For efficiency, it is also necessary to have a precomputed table of tri-angles in G. This table is used to check if an induced subgraph containsa triangle. The table is not very large, with an upper bound of 364 ele-ments for K14. The algorithm to build the table is a simple triple-nestedsearch over all vertices. With at most 14 vertices, the O(n3) complexity isinsignificant.3 Computing Fv(2, 2, 3; 4)For a single graph, InH2234 is virtually instantaneous, but the number ofgraphs that need to be examined explodes as the order increases. For exam-ple, there are 12,346 graphs on 8 vertices, which takes about 0.97 secondsto analyze. However, to analyze all 165,091,172,592 on 12 vertices wouldtake about 96 days on a 1 GHz Pentium III CPU. Although distributing thecomputation over many computers is possible with graphs on 12 vertices,it would still not be a practical solution for graphs on 13 vertices.There is an alternative, however, that avoids examining all graphs onless than 14 vertices. If it were possible to generate all the Folkman graphson 14 vertices, the existence of Folkman graphs on 13 vertices could bedecided by dropping a vertex in all possible ways from each graph andtesting for the Folkman property.We write G − v to denote a graph with deleted vertex v and edgesincident to v. Let S = Hv(2, 2, 3; 4; 14) and D = {G−v : G ∈ S}, then thereis a Folkman graph on 13 vertices if and only if D ∩Hv(2, 2, 3; 4; 13) 6= ∅.The difficulty lies in generating S, but two observations help: (1) TheRamsey number R(3, 4) = R(4, 3) = 9 guarantees that all graphs of order≥9 have either a K4 or a K3, and (2) the Folkman graphs in Hv(2, 2, 3; 4; 14),by definition, do not have a K4, so they must have a K3 due to (1).To find all maximal Folkman graphs we extend those graphs on 11 ver-tices without K4 and χ ≥ 4 to graphs on 14 vertices. The restriction on χis possible because all the Folkman graphs have χ ≥ 5. Each graph is ex-tended by connecting three independent vertices to all triangle-free subsetsin all possible ways. Subsets with triangles are avoided so that a K4 is notformed. Furthermore, it is enough to consider extending only those graphswhere the addition of any edge forms a triangle. The result will includeall maximal Folkman graphs but Hv(2, 2, 3; 4) can easily be recovered fromthis subset. Figure 2 illustrates the extension. The algorithm is calledExtend:1. For each graph G of order 11 (a) that is K4-free, (b) that has chro-matic number at least 4, and (c) where the addition of any edge formsa triangle, do steps 2, 3, 4, and 5 below. The graphs are generatedusing geng from the nauty software package [8] and then filtered for(a), (b), and (c) using simple, custom algorithms. [easy]2. Extend G to 14 vertices by efficiently adding K3 and incident edges.Each new vertex is incident with a maximal triangle-free subset toavoid creating a K4. This is done in all possible ways with obvi-ous isomorphs skipped (e.g., permutations of the new vertices). Theoutput will contain all maximal Folkman graphs in addition to otherFolkman and non-Folkman graphs. [hard]3. Eliminate isomorphs using nauty software tools. [easy]4. Collect graphs for which the addition of any edge forms a K4. [easy]5. Filter for Folkman property. This is a more expensive test so it shouldbe performed after step 4. [easy]Independent SetSetTriangle−Free{GMaximalFigure 2: Illustration of the extension process.In addition to the custom built algorithms InH2234 and Extend, thesoftware packages nauty [8] and Condor [1] were crucial. nauty includeshighly optimized and efficient tools for handling graphs. Developed byBrendan McKay from the Australian National University, nauty containsprograms to quickly generate all non-isomorphic graphs of a given orderas well as identify and eliminate isomorphic graphs. nauty has been usedin numerous research projects for many years. Condor is a distributedprocessing package created at the Computer Science Department at theUniversity of Wisconsin.The advantage of the procedure Extend is that finding all graphs on 11vertices without K4 is feasible: There are 138,892,304 such graphs. The ex-tension process is computationally challenging, yet easy to parallelize; thework was divided over the machines in the RIT Computer Science Depart-ment labs using Condor. Various types of machines were used: Sun Blade150, Sun Blade 1500, and Sun Fire 880. With the distributed processingonly moderate computational effort was required.To find all Folkman graphs on 14 vertices, the maximal Folkman graphswere reduced using the algorithm ReduceSize. This algorithm inputs agraph and removes edges in all possible ways, outputting only those that areFolkman. ReduceSize was applied to all of the maximal Folkman graphs.The resulting set of non-isomorphic graphs combined with the maximalgraphs is the complete set of Folkman graphs on 14 vertices. Isomorphswere eliminated using nauty. The set included the Nenov graph Γ3. Theprocessing time for this stage was small.To verify the correctness of the results, the approach just described wasslightly modified: All the Folkman graphs on 14 vertices were generated byextending graphs on 10 vertices instead of 11. The process is as follows:Extend graphs on 10 vertices by adding an independent set of 4 vertices andconnecting them to maximal triangle-free subsets of the graph in all possibleways. However, by using an independent set of order 4, the extensions avoidgraphs which have no independent set of order 4. This is not a problem,as the missing graphs are those Folkman graphs which are in the Ramseygraph set R(4, 4; 14), computed in [9].Since the extension from 10 to 14 vertices is guaranteed to generateall maximal Folkman graphs with α(G) ≥ 4, the other maximal Folkmangraphs were extracted from R(4, 4; 14). This set was then reduced, usingReduceSize as before. The final set of non-isomorphic graphs was exactlythe same as previously found after extending from 11 vertices and reducing.To ensure that no graph on 13 vertices is in Hv(2, 2, 3; 4), the last algo-rithm ReduceOrder drops a vertex in all possible ways from each of theFolkman graphs on 14 vertices and the algorithm InH2234 then checks forthe Folkman property. No Folkman graphs were found on 13 vertices, thusproving that Fv(2, 2, 3; 4) = 14.4 Results|E(G)| maxdeg(G) mindeg(G) α(G) |Aut(G)|# # # # #42 1 7 527 4 451 3 1507 1 1136743 6 8 11080 5 5759 4 10557 2 80244 51 9 393 6 5996 5 160 4 4445 453 10 227 7 21 6 2 7 146 2279 7 1 8 1047 4555 14 248 3628 16 149 113850 11451 2Table 1: Properties of graphs in Hv(2, 2, 3; 4; 14).G |E(G)| maxdeg(G) mindeg(G) α(G) |Aut(G)|Γ3 42 8 4 7 14F16 45 7 5 4 16Table 2: Properties of Γ3 and F16.1061133 41250211798Figure 3: The Folkman graph F16 with |Aut(G)| = 16.Table 1 lists various properties of all Folkman graphs in Hv(2, 2, 3; 4; 14).There are 12,227 such graphs in total; interestingly, all of them have chro-matic number equal to 5. Among them, there are 591 maximal graphs, 1213minimal graphs, and 8 bicritical graphs (those which are both maximal andminimal).The order of the automorphism group of the Nenov graph Γ3, picturedin Figure 1, is equal to 14. The graph F16, presented in Figure 3, hasthe largest automorphism group among all 12,227 graphs in Hv(2, 2, 3; 4),with |Aut(F16)| = 16. Let g1 = (0 1), g2 = (3 4)(6 7)(8 9)(10 11)(12 13),and g3 = (2 3)(4 5)(7 8)(11 12) be the permutations of the set {0, . . . , 13}.Then the full automorphism group of F16 is generated by g1, g2, and g3.Table 2 lists the specific properties of these two graphs in relation tothe properties shown in Table 1. Notice that Hv(3, 3; 4) ⊂ Hv(2, 2, 3; 4). Itwas shown in [12, 18] that Fv(3, 3; 4) = 14. The graph Γ3 is in Hv(3, 3; 4),but F16 is not.References[1] Condor. High throughput distributed computing software. Universityof Wisconsin-Madison, (1999). http://www.cs.wisc.edu/condor/.[2] J. Folkman. Graphs with monochromatic complete subgraphs in everyedge coloring. SIAM J. Appl. Math., 18(1970), 19–24.[3] R. L. Graham, B. L. Rothschild, and J. H. Spencer. Ramsey theory,(1990), ix+174. John Wiley & Sons Inc., New York.[4] T. Jensen and G. F. Royle. Small graphs with chromatic number 5: acomputer search. J. Graph Theory, 19(1)(1995), 107–116.[5] D. L. Kreher and D. R. Stinson. Combinatorial Algorithms: Genera-tion, Enumeration, and Search, (1999). CRC Press.[6] T.  Luczak, A. Rucin´ski, and S. Urban´ski. On minimal Folkman graphs.Discrete Math., 236(1-3)(2001), 245–262.[7] T.  Luczak and S. Urban´ski. A note on restricted vertex Ramsey num-bers. Period. Math. Hungar., 33(2)(1996), 101–103.[8] B. D. McKay. nauty user’s guide (version 1.5). Technical Report TR-CS-90-02. Australian National University, Department of ComputerScience. Software available at http://cs.anu.edu.au/∼bdm/nauty/.[9] B. D. McKay and S. P. Radziszowski. A new upper bound for theRamsey number R(5, 5). Australas. J. Combin., 5(1992), 13–20.[10] E. Nedialkov and N. D. Nenov. Computation of the vertex Folkmannumber F (4, 4; 6). Proceedings of the Third Euro Workshop on OptimalCodes and related topics, Sunny Beach, Bulgaria, (2001), 123–128.[11] E. Nedialkov and N. D. Nenov. Computation of the vertex Folk-man numbers F (2, 2, 2, 4; 6) and F (2, 3, 4; 6). Electron. J. Combin.,9(1)(2002). http://www.combinatorics.org/.[12] N. D. Nenov. An example of a 15-vertex (3, 3)-Ramsey graph withclique number 4. C. R. Acad. Bulgare Sci., 34(11)(1981), 1487–1489.[13] N. D. Nenov. On a class of vertex Folkman graphs. Annuaire Univ.Sofia Fac. Math. Inform., 94(2000), 15–25.[14] N. D. Nenov. On the 3-colouring vertex Folkman number F (2, 2, 4).Serdica Math. J., 27(2)(2001), 131–136.[15] N. D. Nenov. On the vertex Folkman number F (3, 4). C. R. Acad.Bulgare Sci., 54(2)(2001), 21–24.[16] N. D. Nenov. Lower bound for a number of vertices of some vertexFolkman graphs. C. R. Acad. Bulgare Sci., 55(4)(2002), 33–36.[17] N. D. Nenov. On the triangle vertex Folkman numbers. DiscreteMath., 271(1-3)(2003), 327–334.[18] K. Piwakowski, S. P. Radziszowski, and S. Urban´ski. Computationof the Folkman number Fe(3, 3; 5). J. Graph Theory, 32(1)(1999),41–49.

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Computing the Folkman number F_v(2,2,3;4)

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