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Improved Classical and Quantum Algorithms for the Shortest Vector Problem via Bounded Distance Decoding

2020/02/19 by Aggarwal, Divesh, Chen, Yanlin, Kumar, Rajendra +1 · 3 citations
#Cryptography and Security (cs.CR) #Data Structures and Algorithms (cs.DS) #FOS: Computer and information sciences

paper · doi:10.48550/arxiv.2002.07955

Abstract

The most important computational problem on lattices is the Shortest Vector Problem (SVP). In this paper, we present new algorithms that improve the state-of-the-art for provable classical/quantum algorithms for SVP. We present the following results. \bullet A new algorithm for SVP that provides a smooth tradeoff between time complexity and memory requirement. For any positive integer 4≤ q≤ √(n), our algorithm takes q13n+o(n) time and requires poly(n)⋅ q16n/q2 memory. This tradeoff which ranges from enumeration (q=√(n)) to sieving (q constant), is a consequence of a new time-memory tradeoff for Discrete Gaussian sampling above the smoothing parameter. \bullet A quantum algorithm for SVP that runs in time 20.950n+o(n) and requires 20.5n+o(n) classical memory and poly(n) qubits. In Quantum Random Access Memory (QRAM) model this algorithm takes only 20.835n+o(n) time and requires a QRAM of size 20.293n+o(n), poly(n) qubits and 20.5n classical space. This improves over the previously fastest classical (which is also the fastest quantum) algorithm due to [ADRS15] that has a time and space complexity 2n+o(n). \bullet A classical algorithm for SVP that runs in time 21.669n+o(n) time and 20.5n+o(n) space. This improves over an algorithm of [CCL18] that has the same space complexity. The time complexity of our classical and quantum algorithms are obtained using a known upper bound on a quantity related to the lattice kissing number which is 20.402n. We conjecture that for most lattices this quantity is a 2o(n). Assuming that this is the case, our classical algorithm runs in time 21.292n+o(n), our quantum algorithm runs in time 20.750n+o(n) and our quantum algorithm in QRAM model runs in time 20.667n+o(n).

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