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arXiv 2610.20173cs.CR
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基于不可提取随机性的秘密共享

Secret Sharing Despite Non-Extractable Randomness

National University of Singapore(新加坡国立大学)

Divesh Aggarwal

AI总结:

该研究针对两方秘密共享的公开问题给出否定答案,构造了无需可提取随机性的完美秘密共享方案,证明共享足够长的秘密不依赖可提取随机性。

摘要:

Bosley与Dodis(TCC 2007)证明,一类足以对长度大于密钥长度对数的消息进行信息论加密的非完美随机性源,必须允许对近似均匀比特进行确定性提取,并询问两方秘密共享是否也满足该结论。Aggarwal、Chung、Obremski与Ribeiro(TCC 2022)对额外具备抗泄漏性或非延展性的方案给出了肯定回答,但原问题仍未解决。我们给出否定答案:对任意b≥2,在{0,1}^{b+2}上构造一类平坦源,每个源的最小熵至少为b,同时给出一个完全正确、完全私密的两方秘密共享方案,可利用该类中的任意源将b比特秘密共享为两个(b+1)比特份额;且证明从{0,1}^{b+2}到{0,1}的任意确定性映射,在该类的某个源上与均匀分布的统计距离至少为1/6。这意味着共享足够长的秘密不需要可提取随机性,证明为简短的奇偶性论证。该构造思路可扩展至其他阈值:对n方中的两方共享和r方中的r方共享,我们给出完美方案,其份额为(b+1)比特,且对无法以低于1/(4n-2)和1/(2(2^r-1))的误差提取单个比特的平坦源类具备私密性。

英文摘要:

Bosley and Dodis (TCC 2007) proved that a class of imperfect randomness sources that suffices for information-theoretic encryption of messages longer than the logarithm of the key length must admit deterministic extraction of nearly uniform bits, and asked whether the same is true of two-out-of-two secret sharing. Aggarwal, Chung, Obremski, and Ribeiro (TCC 2022) gave a positive answer for schemes that are in addition leakage-resilient or non-malleable; the original question remained open. We answer it in the negative. For every $b\ge2$ we construct a class of flat sources on $\{0,1\}^{b+2}$, each of min-entropy at least $b$, together with a perfectly correct and perfectly private two-out-of-two scheme that shares a $b$-bit secret into two $(b+1)$-bit shares using any source in the class, and we show that every deterministic map from $\{0,1\}^{b+2}$ to $\{0,1\}$ is at statistical distance at least $1/6$ from uniform on some source in the class. This implies that sharing a long enough secret does not require extractable randomness. The proof is a short parity argument. The idea of the construction extends to other thresholds: for two-out-of-$n$ and $r$-out-of-$r$ sharing we give perfect schemes with $(b+1)$-bit shares, private for classes of flat sources from which no single bit can be extracted with error below $1/(4n-2)$ and $1/(2(2^r-1))$, respectively.

机构、总结、中文摘要均由AI分析而来,可能存在错误,请以论文正文为准。

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