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On the mean number of encryptions for tree-based broadcast encryption schemes

机译:基于树的广播加密方案的平均加密数

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The challenge of stateless-receiver broadcast encryption lies in minimizing storage and the number of encryptions while maintaining system security. Tree-based key distribution schemes offer the best known trade-off between the two parameters. Examples include the complete subtree scheme [D. Wallner, et al., Internet draft, http://www.ietf.org/ID.html [10]; C.K. Wong, et al., in: Proc. SIG-COMM, 1998, pp. 68-79 [11]], the subset difference scheme [D. Naor, et al., in: CRYPTO 2001, Lecture Notes in Comput. Sci., vol. 2139, 2001, pp. 41-62 [7]], and the layered subset difference scheme [D. Halevy, A. Shamir, in: CRYPTO 2002, Lecture Notes in Comput. Sci., vol. 2442,2002, pp. 47-60 [5]]. We introduce generating functions for this family of schemes, which lead to analysis of the mean number of encryptions over all privileged sets of users. We also derive the mean number of encryptions when the number of privileged users is fixed. We expect that the techniques introduced as well as the results in this work will find applications in related areas.
机译:无状态接收器广播加密的挑战在于,在保持系统安全性的同时,将存储空间和加密次数降至最低。基于树的密钥分发方案在两个参数之间提供了最广为人知的折衷方案。示例包括完整的子树方案。 Wallner等人,Internet草案,http://www.ietf.org/ID.html [10]; C.K. Wong et al。,刊于:Proc.Natl.Acad.Sci.USA。 SIG-COMM,1998年,第68-79页[11],子集差异方案[D。 Naor等人,在:CRYPTO 2001,计算机中的讲义。科学,卷。 2139,2001,pp.41-62 [7],以及分层子集差异方案[D. Halevy,A。Shamir,在:CRYPTO 2002,计算机讲义中。科学,卷。 2442,2002,pp.47-60 [5]。我们为该系列方案引入了生成函数,从而导致对所有特权用户集的平均加密数量进行分析。当特权用户数量固定时,我们还可以得出平均加密数量。我们希望介绍的技术以及这项工作的结果将在相关领域找到应用。

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