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On Lattices, Learning with Errors, Random Linear Codes, and Cryptography

机译:关于格,错误学习,随机线性代码和密码术

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Our main result is a reduction from worst-case lattice problems such as GapSVP and SIVP to a certain learning problem. This learning problem is a natural extension of the "learning from parity with error" problem to higher moduli. It can also be viewed as the problem of decoding from a random linear code. This, we believe, gives a strong indication that these problems are hard. Our reduction, however, is quantum. Hence, an efficient solution to the learning problem implies a quantum algorithm for GapSVP and SIVP. A main open question is whether this reduction can be made classical (i.e., nonquantum).rnWe also present a (classical) public-key cryptosystem whose security is based on the hardness of the learning problem. By the main result, its security is also based on the worst-case quantum hardness of GapSVP and SIVP. The new cryptosystem is much more efficient than previous lattice-based cryptosystems: the pu b lic key is of size. O(n~2) and encrypting a message increases its size by a factor of O(n) (in previous cryptosystems these values are O(n~4) and O(n~2), respectively). In fact, under the assumption that all parties share a random bit string of length O(n~2), the size of the public key can be reduced to O(n).
机译:我们的主要结果是从最坏情况的晶格问题(例如GapSVP和SIVP)减少到某个学习问题。这个学习问题是“从奇偶校验学习”问题到更高模数的自然扩展。它也可以看作是从随机线性码解码的问题。我们认为,这充分表明这些问题很难解决。但是,我们的减少是量子的。因此,对学习问题的有效解决方案意味着针对GapSVP和SIVP的量子算法。一个主要的开放性问题是这种减少是否可以设为经典(即非量子)。我们还提出了一种(经典)公钥密码系统,其安全性基于学习问题的难度。主要结果是,其安全性还基于GapSVP和SIVP的最坏情况下的量子硬度。新的密码系统比以前的基于格的密码系统更有效:公钥具有大小。 O(n〜2)和加密消息会使其大小增加O(n)(在以前的密码系统中,这些值分别为O(n〜4)和O(n〜2))。实际上,在所有各方共享长度为O(n〜2)的随机位串的假设下,公共密钥的大小可以减小为O(n)。

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