6 research outputs found

    Discrepancy of Products of Hypergraphs

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    For a hypergraph H=(V,E)\mathcal{H} = (V,\mathcal{E}), its dd―fold symmetric product is ΔdH=(Vd,{Ed∣E∈E})\Delta^d \mathcal{H} = (V^d,\{ E^d | E \in \mathcal{E} \}). We give several upper and lower bounds for the cc-color discrepancy of such products. In particular, we show that the bound disc(ΔdH,2)≤disc(H,2)\textrm{disc}(\Delta^d \mathcal{H},2) \leq \textrm{disc}(\mathcal{H},2) proven for all dd in [B. Doerr, A. Srivastav, and P. Wehr, Discrepancy of Cartesian products of arithmetic progressions, Electron. J. Combin. 11(2004), Research Paper 5, 16 pp.] cannot be extended to more than c=2c = 2 colors. In fact, for any cc and dd such that cc does not divide d!d!, there are hypergraphs having arbitrary large discrepancy and disc(ΔdH,c)=Ωd(disc(H,c)d)\textrm{disc}(\Delta^d \mathcal{H},c) = \Omega_d(\textrm{disc}(\mathcal{H},c)^d). Apart from constant factors (depending on cc and dd), in these cases the symmetric product behaves no better than the general direct product Hd\mathcal{H}^d, which satisfies disc(Hd,c)=Oc,d(disc(H,c)d)\textrm{disc}(\mathcal{H}^d,c) = O_{c,d}(\textrm{disc}(\mathcal{H},c)^d)

    Eight Biennial Report : April 2005 – March 2007

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    Discrepancy of products of hypergraphs

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    For a hypergraph H = (V, E), its d–fold symmetric product is ∆ d H = (V d, {E d |E ∈ E}). We give several upper and lower bounds for the c-color discrepancy of such products. In particular, we show that the bound disc( ∆ d H, 2) ≤ disc(H, 2) proven for all d in [B. Doerr, A. Srivastav, and P. Wehr, Discrepancy of Cartesian products of arithmetic progressions, Electron. J. Combin. 11(2004), Research Paper 5, 16 pp.] cannot be extended to more than c = 2 colors. In fact, for any c and d such that c does not divide d!, there are hypergraphs having arbitrary large discrepancy and disc( ∆ d H, c) = Ωd(disc(H, c) d). Apart from constant factors (depending on c and d), in these cases the symmetric product behaves no better than the general direct product H d, which satisfies disc(H d, c) = Oc,d(disc(H, c) d)
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