Ciphertext Sorting Search Scheme Based on B+ Tree Index Structure on Blockchain
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摘要: 为了克服云存储不可信及云存储中密文检索效率低的问题，该文提出区块链上基于B+树的密文排序可搜索加密方案。该方案结合区块链技术解决了在互不了解的多方建立可靠信任的问题；使用向量空间模型降低了文本的复杂性实现了高效的文本检索系统；采用B+树的索引结构提高了区块链上密文交易的检索速度；利用加权统计(TF-IDF)算法实现了多关键词查询结果的排序。在随机预言机模型下，证明该方案是适应性不可区分安全的，通过效率对比分析，表明该方案在区块链上实现了高效的密文检索。Abstract: In order to overcome the problem that cloud storage is not trusted and the low efficiency of ciphertext retrieval in cloud storage, a searchable ciphertext sorting encryption scheme based on B+ tree on the block chain is proposed. Combined with the blockchain technology, the problem of establishing reliable trust in multiple parties that do not understand each other is solved. A vector space model is used to reduce the complexity of the text and an efficient text retrieval system is implemented. The index structure of the B+ tree is used to improve the retrieval of ciphertext transactions on the blockchain. The ranking of multi-keyword query results is realized by the Term Frequency–Inverse Document Frequency (TF-IDF) algorithm. Under the random oracle model, it is proved that the scheme is adaptive and indistinguishable. Through the comparative analysis of efficiency, it is shown that the scheme achieves efficient ciphertext retrieval on the blockchain.
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Key words:
- Cloud storage /
- Blockchain /
- B+ tree /
- Sorting search
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表 1 B+树算法
算法1 BuildIndexTree(${{I} }$) if($u$是叶子节点)then 将密文交易标识符$\rm{TXid} $放到叶子节点，并计算叶子节点的${\text{D}}$的
向量return else 根据新节点的位置，向下查找该节点插入的子节点$\rm{ul} $ if(节点$\rm{ul} $的数值为最大)then 对节点进行分割，重新确定向下插入的子节点$\rm{ul} $ end if 
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表 2 效率对比分析
方案 $\left| {{\rm{Trapdoor}} } \right|$ $\left| {{\rm{SearchComplexity}} } \right|$ 公平性 搜索结果是否排序 文献[6] $O\left( {{m^2}} \right)$ $O\left( {{\varphi _m}\log n} \right)$ 否 是 文献[14] $O\left( 1 \right)$ $O\left( {\left| {D\left( w \right)} \right|} \right)$ 是 否 文献[16] $O\left( 1 \right)$ $O\left( {\left| {D\left( w \right)} \right|} \right)$ 否 否 本文方案 $O\left( {{m^2}} \right)$ ${O}\left( { {\varphi _m}{ {\log }_{\left\lceil {\tfrac{m}{2} } \right\rceil } }\tfrac{ {n + 1} }{2} } \right)$ 是 是
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