Is the Algorithmic Kadison-Singer Problem Hard?
We study the following ๐ช๐ฒ_2(c) problem: let c โโ^+ be some constant, and v_1,โฆ, v_mโโ^d be vectors such that v_i^2โคฮฑ for any iโ[m] and โ_i=1^m โจ v_i, xโฉ^2 =1 for any xโโ^d with x=1. The ๐ช๐ฒ_2(c) problem asks to find some Sโ [m], such that it holds for all x โโ^d with x = 1 that |โ_i โ Sโจ v_i, xโฉ^2 - 1/2| โค cยทโ(ฮฑ), or report no if such S doesn't exist. Based on the work of Marcus et al. and Weaver, the ๐ช๐ฒ_2(c) problem can be seen as the algorithmic Kadison-Singer problem with parameter cโโ^+. Our first result is a randomised algorithm with one-sided error for the ๐ช๐ฒ_2(c) problem such that (1) our algorithm finds a valid set S โ [m] with probability at least 1-2/d, if such S exists, or (2) reports no with probability 1, if no valid sets exist. The algorithm has running time O(mnยทpoly(m, d))ย ย n = O(d/ฯต^2log(d) log(1/cโ(ฮฑ))), where ฯต is a parameter which controls the error of the algorithm. This presents the first algorithm for the Kadison-Singer problem whose running time is quasi-polynomial in m, although having exponential dependency on d. Moreover, it shows that the algorithmic Kadison-Singer problem is easier to solve in low dimensions. Our second result is on the computational complexity of the ๐ช๐ฒ_2(c) problem. We show that the ๐ช๐ฒ_2(1/(4โ(2))) problem is ๐ฅ๐ญ๐ฏ-hard for general values of d, and solving the ๐ช๐ฒ_2(1/(4โ(2))) problem is as hard as solving the ๐ญ๐ ๐ค3๐ฒ๐ ๐ณ problem.
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