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 $4\leq q\leq \sqrt{n}$, our algorithm takes $q^{13n+o(n)}$ time and requires $poly(n)\cdot q^{16n/q^2}$ memory. This tradeoff which ranges from enumeration ($q=\sqrt{n}$) 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 $2^{0.950n+o(n)}$ and requires $2^{0.5n+o(n)}$ classical memory and poly(n) qubits. In Quantum Random Access Memory (QRAM) model this algorithm takes only $2^{0.835n+o(n)}$ time and requires a QRAM of size $2^{0.293n+o(n)}$, poly(n) qubits and $2^{0.5n}$ 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 $2^{n+o(n)}$. $\bullet$ A classical algorithm for SVP that runs in time $2^{1.669n+o(n)}$ time and $2^{0.5n+o(n)}$ 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 $2^{0.402n}$. We conjecture that for most lattices this quantity is a $2^{o(n)}$. Assuming that this is the case, our classical algorithm runs in time $2^{1.292n+o(n)}$, our quantum algorithm runs in time $2^{0.750n+o(n)}$ and our quantum algorithm in QRAM model runs in time $2^{0.667n+o(n)}$.
翻译:在 lattices 上最重要的计算问题是 最短的 Vctor 复杂度 。 在本文中, 我们推出新的算法来改善 SVP 可变古典/quantum 算法的状态 。 我们展示了以下结果 。 $\ bulllet$ 为 SVP 提供了时间复杂性和记忆要求之间的平衡。 对于任何正整 4\leq q\leq\ srt{n} 问题来说, 我们的算法花费了 q ⁇ 13n+o(n) 美元时间, 并且需要 $(n) diople comlient comlium QQ} qdo qdent liental 16n/q%2} 存储。 这个算法从计数( qósqualent{rt{n} 到 $ququaltime $.