Cloud Storage Data Sharing Based on Attribute Encryption in Smart Healthcare
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摘要: 在电子病历系统中，为了实现多用户环境下的数据搜索，该文提出一种属性基可搜索加密方案。该文将密文和安全索引存储在医疗云，当用户请求医疗数据时，利用属性基可搜索加密算法进行数据搜索，实现了细粒度访问控制。同时方案引入了密文验证算法，解决了半诚实且好奇的云服务器模型下搜索结果不正确的问题。利用数据去重技术实现了重复数据的消除，减少占用医疗云的存储空间。方案同时实现了访问策略的隐藏，保证了数据用户的隐私安全。安全性分析表明，所提方案能很好地保护用户的隐私以及数据的安全。性能分析表明，该方案具有较好的性能，更加适用于智慧医疗等多对多应用场景，有效实现了医生和第三方数据用户在不侵犯患者隐私的前提下共享患者电子病历。Abstract: In order to achieve multi-user data search in electronic medical record system, an attribute based searchable encryption scheme is proposed. In this scheme, ciphertext and secure indexes are stored in the medical cloud. When the users want to access the medical data, the attribute-base searchable encryption algorithm is used for data search, and the fine-grained access control is realized. At the same time, ciphertext verification algorithm is introduced into the scheme, it solves the problem of incorrect search results under the semi-honest and curious cloud server models. The scheme uses data deduplication technology to eliminate duplicate data and reduce the storage space of the medical cloud. The scheme also realizes the hiding of access policy, and the privacy security of data users is guaranteed. The security analysis shows that the proposed scheme can well protect the confidentiality of medical data and the anonymity of users. The performance analysis demonstrate that the proposed scheme has better performance; Hence, it is more suitable for many-to-many application scenarios such as smart healthcare. It effectively realizes the sharing of patient electronic medical records between doctors and third-party users without infringing on patient privacy.
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表 3 存储量比较
文献 [10] 文献 [12] 本文 KeyGen $(2s + 3)\left| {{G_1}} \right| $ $(2l + 1)\left| {{G_1}} \right| + \left| {{Z_q}} \right| $ $ (2s + 1)\left| {{G_1}} \right| + (s + 1)\left| {{Z_q}} \right|$ Encrypt $(2l + 2 + m)\left| {{G_1}} \right| $ $ (2l + 1)\left| {{G_1}} \right| + \left| {{G_2}} \right|$ $(2l + 3 + m)\left| {{G_1}} \right| + 2\left| {{Z_q}} \right| $ Trap $ (3s + 4)\left| {{G_1}} \right|$ $(3l + 3)\left| {{G_1}} \right| + \left| {{Z_q}} \right| $ $(2s + 3)\left| {{G_1}} \right| + \left| {{Z_q}} \right| $ Search $(l + 1)\left| {{G_1}} \right| + (s + 3)\left| {{G_2}} \right| $ $(l + 3)\left| {{G_2}} \right| $ $(s + 4)\left| {{G_2}} \right| $ 
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