1,680 research outputs found

    Induced Ramsey-type results and binary predicates for point sets

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    Let kk and pp be positive integers and let QQ be a finite point set in general position in the plane. We say that QQ is (k,p)(k,p)-Ramsey if there is a finite point set PP such that for every kk-coloring cc of (Pp)\binom{P}{p} there is a subset QQ' of PP such that QQ' and QQ have the same order type and (Qp)\binom{Q'}{p} is monochromatic in cc. Ne\v{s}et\v{r}il and Valtr proved that for every kNk \in \mathbb{N}, all point sets are (k,1)(k,1)-Ramsey. They also proved that for every k2k \ge 2 and p2p \ge 2, there are point sets that are not (k,p)(k,p)-Ramsey. As our main result, we introduce a new family of (k,2)(k,2)-Ramsey point sets, extending a result of Ne\v{s}et\v{r}il and Valtr. We then use this new result to show that for every kk there is a point set PP such that no function Γ\Gamma that maps ordered pairs of distinct points from PP to a set of size kk can satisfy the following "local consistency" property: if Γ\Gamma attains the same values on two ordered triples of points from PP, then these triples have the same orientation. Intuitively, this implies that there cannot be such a function that is defined locally and determines the orientation of point triples.Comment: 22 pages, 3 figures, final version, minor correction

    On path-quasar Ramsey numbers

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    Let G1G_1 and G2G_2 be two given graphs. The Ramsey number R(G1,G2)R(G_1,G_2) is the least integer rr such that for every graph GG on rr vertices, either GG contains a G1G_1 or G\overline{G} contains a G2G_2. Parsons gave a recursive formula to determine the values of R(Pn,K1,m)R(P_n,K_{1,m}), where PnP_n is a path on nn vertices and K1,mK_{1,m} is a star on m+1m+1 vertices. In this note, we first give an explicit formula for the path-star Ramsey numbers. Secondly, we study the Ramsey numbers R(Pn,K1Fm)R(P_n,K_1\vee F_m), where FmF_m is a linear forest on mm vertices. We determine the exact values of R(Pn,K1Fm)R(P_n,K_1\vee F_m) for the cases mnm\leq n and m2nm\geq 2n, and for the case that FmF_m has no odd component. Moreover, we give a lower bound and an upper bound for the case n+1m2n1n+1\leq m\leq 2n-1 and FmF_m has at least one odd component.Comment: 7 page
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