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基于區(qū)塊鏈的零知識(shí)位置證明方法研究

余榮威 周博孝 王麗娜 朱欣焰 謝輝華 謝紅軍

余榮威, 周博孝, 王麗娜, 朱欣焰, 謝輝華, 謝紅軍. 基于區(qū)塊鏈的零知識(shí)位置證明方法研究[J]. 電子與信息學(xué)報(bào), 2020, 42(9): 2142-2149. doi: 10.11999/JEIT191054
引用本文: 余榮威, 周博孝, 王麗娜, 朱欣焰, 謝輝華, 謝紅軍. 基于區(qū)塊鏈的零知識(shí)位置證明方法研究[J]. 電子與信息學(xué)報(bào), 2020, 42(9): 2142-2149. doi: 10.11999/JEIT191054
Rongwei YU, Boxiao ZHOU, Lina WANG, Xinyan ZHU, Huihua XIE, Hongjun XIE. Zero-knowledge Location Proof Based on Blockchain[J]. Journal of Electronics & Information Technology, 2020, 42(9): 2142-2149. doi: 10.11999/JEIT191054
Citation: Rongwei YU, Boxiao ZHOU, Lina WANG, Xinyan ZHU, Huihua XIE, Hongjun XIE. Zero-knowledge Location Proof Based on Blockchain[J]. Journal of Electronics & Information Technology, 2020, 42(9): 2142-2149. doi: 10.11999/JEIT191054

基于區(qū)塊鏈的零知識(shí)位置證明方法研究

doi: 10.11999/JEIT191054
基金項(xiàng)目: 國(guó)家自然科學(xué)基金(U1836112, 61876134)
詳細(xì)信息
    作者簡(jiǎn)介:

    余榮威:男,1981年生,副教授,研究方向?yàn)榭尚庞?jì)算、區(qū)塊鏈安全

    周博孝:男,1996年生,碩士生,研究方向?yàn)橄到y(tǒng)安全

    王麗娜:女,1964年生,教授,研究方向?yàn)榭尚庞?jì)算、信息隱藏

    朱欣焰:男,1963年生,教授,研究方向?yàn)镚IS、大數(shù)據(jù)安全

    謝輝華:男,1996年生,碩士生,研究方向?yàn)橄到y(tǒng)安全

    謝紅軍:男,1985年生,碩士生,研究方向?yàn)閰^(qū)塊鏈

    通訊作者:

    周博孝 boxiao@whu.edu.cn

  • 中圖分類號(hào): TN918; TP309

Zero-knowledge Location Proof Based on Blockchain

Funds: The National Natural Science Foundation of China (U1836112, 61876134)
  • 摘要: 地理位置虛擬軟件泛濫、民用衛(wèi)星定位信號(hào)易模擬或篡改,致使地理位置可信認(rèn)證難以實(shí)現(xiàn)。針對(duì)已有位置證明方案采用中心化架構(gòu)存在單點(diǎn)失效和易引起集中攻擊等安全風(fēng)險(xiǎn),該文引入去中心化范式思路,利用區(qū)塊鏈具有的去中心化、不可篡改、可追溯等特點(diǎn),并結(jié)合零知識(shí)證明協(xié)議,提出了基于區(qū)塊鏈的零知識(shí)位置證明方法,實(shí)現(xiàn)了以去中心化、保護(hù)隱私、高度準(zhǔn)確、審查抵制的地理位置認(rèn)證服務(wù),從而確保用戶所提供位置的準(zhǔn)確性。該方法不僅能消除中心化位置證明的弊端,確保位置數(shù)據(jù)的機(jī)密性,而且被證明位置數(shù)據(jù)一旦上鏈后不可篡改,實(shí)現(xiàn)了不可抵賴性。測(cè)試分析結(jié)果表明:完整的證明流程(包含證明生成驗(yàn)證和上鏈全過(guò)程)實(shí)際測(cè)試每次平均用時(shí)約5 s,其中證明生成和驗(yàn)證的總耗時(shí)是50.5~55.5 ms。因此,算法具有較好的性能開銷,可滿足實(shí)際應(yīng)用需求。
  • 圖  1  系統(tǒng)結(jié)構(gòu)圖

    圖  2  位置證明時(shí)序圖

    圖  3  零知識(shí)證明算法代碼執(zhí)行時(shí)間

    圖  4  不同系統(tǒng)中不同合約的執(zhí)行時(shí)間比較

    圖  5  各個(gè)流程的時(shí)間花銷

    表  1  零知識(shí)證明生成算法1

     輸入:證明者和見證者的經(jīng)度、緯度和海拔、與以見證者位置為圓心的半徑$R$,
     輸出:零知識(shí)證明結(jié)果${\rm{pf}}$;
     (1)  挑選隨機(jī)大數(shù)$a,{b_1},{b_2},{b_3},{b_4},{b_5},{e_1},{e_2},{e_3},{e_4},{f_1},{f_2},{f_3},{f_4},k,{l_1},{l_2},{l_3},{l_4},n,{q_1},{q_2},s$;
     (2)  ${\alpha _1},{\alpha _2},{\alpha _3} \leftarrow $經(jīng)緯度海拔之差,$D \leftarrow {\rm{getDis}}({\alpha _1},{\alpha _2},{\alpha _3})$;
     (3)  if ${R^2} \ge {D^{\rm{2} } }$ do //判斷是否在圓內(nèi)
     (4)  挑選兩個(gè)大素?cái)?shù),相乘得到N,并舍棄掉兩個(gè)素?cái)?shù);
     (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)  ${\rm{pf}} \leftarrow {\rm{\{ \} }}$.
    下載: 導(dǎo)出CSV

    表  2  零知識(shí)證明驗(yàn)證算法2

     輸入:零知識(shí)證明${\rm{pf}}$,
     輸出:驗(yàn)證結(jié)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)  $R \leftarrow F$;//返回驗(yàn)證失敗
     (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)  $R \leftarrow T$;//返回驗(yàn)證通過(guò)。
    下載: 導(dǎo)出CSV
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  • 收稿日期:  2019-12-30
  • 修回日期:  2020-08-10
  • 網(wǎng)絡(luò)出版日期:  2020-08-19
  • 刊出日期:  2020-09-27

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