Zero-knowledge Location Proof Based on Blockchain
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摘要: 地理位置虚拟软件泛滥、民用卫星定位信号易模拟或篡改，致使地理位置可信认证难以实现。针对已有位置证明方案采用中心化架构存在单点失效和易引起集中攻击等安全风险，该文引入去中心化范式思路，利用区块链具有的去中心化、不可篡改、可追溯等特点，并结合零知识证明协议，提出了基于区块链的零知识位置证明方法，实现了以去中心化、保护隐私、高度准确、审查抵制的地理位置认证服务，从而确保用户所提供位置的准确性。该方法不仅能消除中心化位置证明的弊端，确保位置数据的机密性，而且被证明位置数据一旦上链后不可篡改，实现了不可抵赖性。测试分析结果表明：完整的证明流程(包含证明生成验证和上链全过程)实际测试每次平均用时约5 s，其中证明生成和验证的总耗时是50.5～55.5 ms。因此，算法具有较好的性能开销，可满足实际应用需求。Abstract: Due to the proliferation of geographic location virtual software and the easy simulation or tampering of civil satellite positioning signals, it is difficult to realize the trusted authentication of geographic location. In view of the security risk of single-point failure in the existing location certification scheme using centralized architecture, a zero-knowledge location certification method based on blockchain is proposed, combining with zero knowledge certification protocol, to achieve a decentralized, privacy protected, highly accurate, review offset geographic location certification service, so as to ensure the accuracy of the location provided by users. This method not only ensures the confidentiality of the location data, but also proves that the location data can not tamper once it is linked. The results of the test analysis show that the average performance of the whole proving process is about 5 s/time, and the total time of proof generation and verification is 50.5～55.5 ms. Therefore, the algorithm has better performance overhead, which can meet the actual application requirements.
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Key words:
- Location proof /
- Blockchain /
- Zero-knowledge proof /
- Smart contract
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表 1 零知识证明生成算法1
输入：证明者和见证者的经度、纬度和海拔、与以见证者位置为圆心的半径$R$， 输出：零知识证明结果${\rm{pf}}$； (1) 挑选随机大数$a,{b_1},{b_2},{b_3},{b_4},{b_5},{e_1},{e_2},{e_3},{e_4},{f_1},{f_2},{f_3},{f_4},k,{l_1},{l_2},{l_3},{l_4},n,{q_1},{q_2},s$； (2) ${\alpha _1},{\alpha _2},{\alpha _3} \leftarrow $经纬度海拔之差，$D \leftarrow {\rm{getDis}}({\alpha _1},{\alpha _2},{\alpha _3})$； (3) if ${R^2} \ge {D^{\rm{2} } }$ do //判断是否在圆内 (4) 挑选两个大素数，相乘得到N，并舍弃掉两个素数；
(5) $\displaystyle\sum\nolimits_{i = 1}^4 {c_i^2} \leftarrow {R^2} - {D^2}$；${d_1} \leftarrow \displaystyle\sum\nolimits_{i = 1}^3 {e_i^2} + \displaystyle\sum\nolimits_{i = 1}^4 {f_i^2} (od N)$；(6) ${d_2} \leftarrow \displaystyle\sum\nolimits_{i = 1}^4 { {c_i}{f_i} } + \displaystyle\prod\nolimits_{i = 1}^3 { {e_i}{\alpha _i} } (od N)$；$m \leftarrow b_4^n \displaystyle\prod\nolimits_{i = 1}^4 {l_i^{ {f_i} } } (od N)$； (7) $g \leftarrow \displaystyle\prod\nolimits_{i = 1}^4 {b_i^{ {e_i} } } (od N)$；$h \leftarrow b_4^k \displaystyle\prod\nolimits_{i = 1}^4 {l_i^{ {c_i} } } (od N)$； (8) $p \leftarrow b_4^{ - {d_1}}b_5^{{q_1}}(od N)$；$r \leftarrow b_4^{ - 2{d_2}}b_5^{{q_2}}(od N)$； (9) ${x_i} \leftarrow s \cdot {\alpha _i} + {e_i}(od N)(i = 1,2,3)$，${x_4} \leftarrow s \cdot a + {e_4}(od N)$； (10) ${\beta _i} \leftarrow s{c_i} + {f_i}(od N)(i = 1,2,3,4)$；$A \leftarrow \displaystyle\prod\nolimits_{i = 1}^3 {b_i^{ {\alpha _i} } } b_4^a(od N)$； (11) $\gamma \leftarrow sk + n(od N),\lambda \leftarrow s{q_1} + {q_2}(od N)$； (12) ${\rm{pf}} \leftarrow \{ N,A,s,{b_i},{x_i},g,R,{\beta _i},\lambda ,p,r,\gamma ,h,{l_i},m\} $； (13) else (14) ${\rm{pf}} \leftarrow {\rm{\{ \} }}$. 
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表 2 零知识证明验证算法2
输入：零知识证明${\rm{pf}}$， 输出：验证结果$R$； (1) ${v_1} \leftarrow {A^{ - s} }\displaystyle\prod\nolimits_{i = 1}^4 {b_i^{ {x_i} } } (od N)$； (2) if ${v_1}! = g$ do (3) $R \leftarrow F$；//返回验证失败
(4) ${v_2} \leftarrow {s^2}{R^2} - \displaystyle\sum\nolimits_{i = 3}^3 {x_i^2} - \displaystyle\sum\nolimits_{i = 1}^4 {\beta _i^2}$;
${v_3} \leftarrow b_4^{{v_2}}b_5^\lambda (od N)$；(5) if ${v_3}! = {\rm{p}}{{\rm{r}}^s}(od N)$ do
(6) $R \leftarrow F$；(7) ${v_4} \leftarrow b_4^\gamma {h^{ - s} }\displaystyle\prod\nolimits_{i = 1}^4 {l_i^{ {\beta _i} } } (od N)$； (8) if ${v_4}! = m$ do (9) $R \leftarrow F$； (10) $R \leftarrow T$；//返回验证通过。
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