Is the Algorithmic Kadison-Singer Problem Hard?

05/04/2022
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by   Ben Jourdan, et al.
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We study the following ๐–ช๐–ฒ_2(c) problem: let c โˆˆโ„^+ be some constant, and v_1,โ€ฆ, v_mโˆˆโ„^d be vectors such that v_i^2โ‰คฮฑ for any iโˆˆ[m] and โˆ‘_i=1^m โŸจ v_i, xโŸฉ^2 =1 for any xโˆˆโ„^d with x=1. The ๐–ช๐–ฒ_2(c) problem asks to find some SโŠ‚ [m], such that it holds for all x โˆˆโ„^d with x = 1 that |โˆ‘_i โˆˆ SโŸจ v_i, xโŸฉ^2 - 1/2| โ‰ค cยทโˆš(ฮฑ), or report no if such S doesn't exist. Based on the work of Marcus et al. and Weaver, the ๐–ช๐–ฒ_2(c) problem can be seen as the algorithmic Kadison-Singer problem with parameter cโˆˆโ„^+. Our first result is a randomised algorithm with one-sided error for the ๐–ช๐–ฒ_2(c) problem such that (1) our algorithm finds a valid set S โŠ‚ [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/cโˆš(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 ๐–ช๐–ฒ_2(c) problem. We show that the ๐–ช๐–ฒ_2(1/(4โˆš(2))) problem is ๐–ฅ๐–ญ๐–ฏ-hard for general values of d, and solving the ๐–ช๐–ฒ_2(1/(4โˆš(2))) problem is as hard as solving the ๐–ญ๐– ๐–ค3๐–ฒ๐– ๐–ณ problem.

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