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The sum of independent discrete Gaussians is a discrete Gaussian

I am currently learning about lattice-based cryptography and, reading from "A Decade of Lattice Cryptography" by Peikert, specifically section 2.3, it emerges that if the parameter s is greater or equal than the smoothing parameter of a lattice, then the sum of independent discrete gaussians (over that lattice) is a discrete gaussian itself. I am::Listen

I am currently learning about lattice-based cryptography and, reading from "A Decade of Lattice Cryptography" by Peikert, specifically section 2.3, it emerges that if the parameter s is greater or equal than the smoothing parameter of a lattice, then the sum of independent discrete gaussians (over that lattice) is a discrete gaussian itself.

I am looking for the formal statement (and proof) of that fact without any success. Is anyone able to point me to the appropriate reference?

Toss in your job and make $300K working for a DAO? Here’s how

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