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FOCS
2008
IEEE

Spherical Cubes and Rounding in High Dimensions

14 years 6 months ago
Spherical Cubes and Rounding in High Dimensions
What is the least surface area of a shape that tiles Rd under translations by Zd ? Any such shape must have volume 1 and hence surface area at least that of the volume-1 ball, namely Ω( √ d). Our main result is a construction with surface area O( √ d), matching the lower bound up to a constant factor of 2 p 2π/e ≈ 3. The best previous tile known was only slightly better than the cube, having surface area on the order of d. We generalize this to give a construction that tiles Rd by translations of any full rank discrete lattice Λ with surface area 2π ‚ ‚V −1 ‚ ‚ fb , where V is the matrix of basis vectors of Λ, and · fb denotes the Frobenius norm. We show that our bounds are optimal within constant factors for rectangular lattices. Our proof is via a random tessellation process, following recent ideas of Raz [11] in the discrete setting. Our construction gives an almost optimal noise-resistant rounding scheme to round points in Rd to rectangular lattice points.
Guy Kindler, Ryan O'Donnell, Anup Rao, Avi Wigders
Added 29 May 2010
Updated 29 May 2010
Type Conference
Year 2008
Where FOCS
Authors Guy Kindler, Ryan O'Donnell, Anup Rao, Avi Wigderson
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