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摘要:
从智能手机、智能手表等小型终端智能设备，到智能家居、智能网联车等大型应用，再到智慧生活、智慧农业等，人工智能已经逐渐步入人们的生活，改变传统的生活方式. 各种各样的智能设备会产生海量的数据，传统的云计算模式已无法适应新的环境. 边缘计算在靠近数据源的边缘侧实现对数据的处理，可以有效降低数据传输时延，减轻网络传输带宽压力，提高数据隐私安全等. 在边缘计算架构上搭建人工智能模型，进行模型的训练和推理，实现边缘的智能化，对于当前社会至关重要. 由此产生的新的跨学科领域——边缘智能（edge intelligence，EI），开始引起了广泛的关注. 全面调研了边缘智能相关研究：首先，介绍了边缘计算、人工智能的基础知识，并引出了边缘智能产生的背景、动机及挑战. 其次，分别从边缘智能所要解决的问题、边缘智能模型研究以及边缘智能算法优化3个角度对边缘智能相关技术研究展开讨论. 然后，介绍边缘智能中典型的安全问题. 最后，从智慧工业、智慧生活及智慧农业3个层面阐述其应用，并展望了边缘智能未来的发展方向和前景.
Abstract:From smart terminal devices such as smart phones and smart watches, to large-scale intelligent applications, such as smart homes, Internet of vehicles, intelligent life and intelligent agriculture. Artificial intelligence (AI) has gradually entered and changed the life of human being. In this context, various of intelligent devices have produced massive amount of data, making traditional cloud computing paradigm unable to adapt to the unprecedented challenge. Instead, edge computing which aims to process the data at the edge of the network has the great potential to reduce latency and bandwidth pressure, as well as protect data privacy and security. Building AI models upon edge computing architecture, training and inferring the model, realizing the intelligence of the edge are crucial to the current social. As a result, a new interdisciplinary field, edge intelligence (EI), has begun to attract widespread attention. We make a comprehensive study on EI. Specifically, firstly introduce the basic knowledge of edge computing and AI, which leads to the background, motivation and challenges of EI. Secondly, the research on EI related technologies is discussed from three aspects, namely, the problems, the models and the algorithm. Further, the typical security problems in EI are introduced. Next, the applications of EI are described from three aspects of intelligent industry, intelligent life and intelligent agriculture. Finally, we propose the direction and prospect of EI in the future development.
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自2005年Sahai等人[1]提出模糊身份基加密方案后，属性基密码体制成为研究的热点. 属性基密码体制使用一组关联属性代替用户的身份信息，密文或密钥与一个事先定义的访问策略或是谓词结构相关联，当用户的属性满足访问策略或谓词结构时就可以进行解密. 因此属性基密码克服了身份基密码一对一的通信限制，只要用户拥有满足访问策略或谓词结构的属性集就可以进行通信，从而实现了一对多的通信并实现了细粒度的访问控制[2-3]. 为了满足不同的应用需求，一些新型的属性基加密方案[4-11]和属性基签名方案[12-15]相继被提出.
在属性基签名方案中，由于使用一组属性集代替用户，隐藏了真实的身份信息从而获得了匿名性. 签名者根据属性授权机构颁发的属性密钥对消息进行签名，属性密钥由属性授权机构产生并秘密发送给签名者，一旦属性密钥发生泄露或密钥传输时遭受主动攻击被截获，那么获得密钥的任何人都能产生一个有效签名. 与此同时，签名数据中可能包含一些敏感信息，例如身份证号、手机号或者个人金融交易记录等. 这些敏感信息泄露可能会带来个人隐私泄露甚至是国家机密泄露的极大风险. 因此属性基签名中的密钥泄露和敏感信息泄露问题是亟待解决的关键问题.
本文的主要贡献包括3个方面：
1） 提出了前向安全的高效属性基可净化签名（efficient and forward-secure attribute-based sanitizable signature, FABSS）方案，并在标准模型下证明该方案的安全性. 方案的安全性可规约到η-DHE（η-Diffie-Hellman exponent）困难问题假设.
2） 提出的方案利用属性集合和谓词结构提供细粒度访问控制，保护签名者的隐私；利用前向安全技术解决了密钥泄露问题；利用可净化签名技术对原始数据进行脱敏，解决了敏感数据泄露问题.
3） 提出的方案具有固定签名长度，并且在验证阶段只需要常数个配对运算，使得通信开销和计算开销低，因此提出的方案具有高效性.
1. 相关工作
属性基加密方案根据访问策略的不同布置可分为密钥策略的属性基加密方案[2]和密文策略的属性基加密方案[16]. 在密钥策略的属性基加密方案中，用户访问策略与密钥关联，一组属性集与密文相关联. 当密文中的属性集满足访问策略时，用户可以正确解密该密文. 在密文策略的属性基加密方案中，事先定义的一个访问策略嵌入到密文中，密钥由用户属性集标识，只有当标识密钥的属性集满足密文中的访问策略时用户才能正确解密.
为了解决数据完整性、认证性以及用户细粒度访问控制问题， Maji等人[17]在2011年首次提出属性基签名方案，并在一般群模型中证明了该方案的安全性. Okamoto等人[18]基于CDH （computational Diffie-Hellman）困难问题假设，提出了标准模型下证明安全的属性基签名方案. 标准模型下证明安全的方案通常需要大量的计算开销，其中配对运算的代价尤其高昂. 为了提高效率，Gagn等人[19]设计了具有短配对运算的高效属性基签名方案. 为了进一步提高效率，Anada等人[20]提出了无配对运算的属性基签名方案. 然而文献[19-20]中的方案仅仅考虑性能的提升而没有考虑密钥泄露问题. 在密钥颁发和存储过程中，可能会遭受主动攻击或者由于管理不当造成密钥泄露，恶意攻击者在获得密钥后就能产生任意时间片段签名. 为了解决密钥泄露问题，在2015年，Wei等人[21]提出了门限结构的前向安全属性基签名方案，并在标准模型下给出了安全性证明. 另一个解决密钥泄露的方法是密钥隔离技术，2017年，Rao[22]提出一个签名策略的属性基密钥隔离签名，将密钥分为长期密钥和短期密钥，并将长期密钥保存在一个安全的设备中，从而保证了密钥的安全. 然而在签名方案中，可能发生泄露的不仅仅有签名者密钥，同时还包括消息中的一些敏感信息，例如个人医疗记录信息、金融机构交易信息以及政府部门政务信息等. 这些信息一旦发生泄露将会给个人、金融市场或者政府部门带来极大的安全风险. 因此我们需要对数据中的敏感数据进行编辑从而隐藏真实信息，这样的方法可称之为“净化”. 在可净化签名中，净化者可以在不知道签名者密钥的前提下对数据进行编辑并重新生成一个有效签名. Ateniese等人[23]首次提出可净化签名概念，利用变色龙哈希设计了可净化签名方案并在随机预言模型下给出了安全性证明. Agrawal等人[24]提出了在标准模型下证明安全的可净化签名方案，方案的安全性规约到CDH困难问题假设. 但该方案需要大量的配对运算和指数运算，具有较高的运算开销. 为了改进效率，Pöhls等人[25]提出了高效的可净化方案. 可审计性要求签名者可以对净化者的行为进行追责，2017年，Beck等人[26]提出一个具有强审计性的可净化签名，不仅实现对净化者的追责，同时也防止签名者对净化者的恶意指责. 为了获得细粒度访问控制和以及签名者隐私，刘西蒙等人[27]给出属性基可净化签名方案的构造并在标准模型下给出了方案的安全性证明. 文献[25]利用门限结构作为访问策略，为了获得更丰富和灵活的访问结构，莫若等人[28]和Mo等人[29]先后给出了基于树形访问结构的属性基可净化签名方案和具有灵活访问结构的属性基可净化签名方案，方案支持与门、或门和门限结构. 为了同时获得访问控制和可审计性，Samelin等人[30]提出了属性基可净化签名并实现了对净化者的追责. 为了解决属性基签名中签名者滥用签名问题，李继国等人[31]提出了可追踪的属性基可净化签名方案，不仅实现了恶意用户追踪，而且还保证了敏感数据的隐私.
2. 预备知识
本节介绍FABSS方案中使用的相关密码学知识，其中包括双线性映射、拉格朗日插值、η-DHE假设.
2.1 双线性映射
令G1和G2是2个p阶乘法循环群，p是大素数.g是G1的一个生成元. 一个双线性映射e:G1×G1→G2具有3个性质：
1） 双线性. 对任意a,b∈Zp，都有e(ga,gb)=e(g,g)ab.
2） 非退化性. e(g,g)≠1.
3） 可计算性. 对所有g1,g2∈G1，存在多项式时间算法计算e(g1,g2).
2.2 拉格朗日插值
设p为素数，S⊆Zp，拉格朗日系数定义为ΔSi(x)=∏j∈S,j≠ix−ji−j，其中i∈Zp. 给定Zp上的d个点(1,q1), (2,q2), …, (d,qd)，d−1次多项式q(x)可以重构为q(x)=∑i∈Sq(i)ΔSi(x)，其中|S|=d.
2.3 η-DHE问题和困难问题假设
η-DHE问题. G1是一个p阶群，g是G1的一个生成元，随机选取a∈Zp.给定元组(g,ga,ga2,…,gaη,gaη+2,…,ga2η)，计算gaη+1.
ε-(η-DHE)困难问题假设. 若不存在多项式时间算法以不可忽略的概率ε解决G1上的η-DHE困难问题，则称ε-η-DHE困难问题假设在群G1上是成立的.
3. 形式化定义和安全模型
借鉴文献[21]中前向安全的属性基签名的形式化定义，本节给出FABSS方案的形式化定义和安全模型.
3.1 FABSS方案的形式化定义
FABSS方案包括设置、密钥生成、密钥更新、签名、净化和验证6个算法，每个算法的定义为：
1）设置. 算法输入安全参数λ、系统时间片段总数T、系统门限值d，输出公共参数params和主密钥msk.
2）密钥生成. 该算法由属性授权中心执行. 算法输入公共参数params、主密钥msk、签名者属性集wa以及初始时间片段t0，输出初始时间片段密钥SKt0.
3）密钥更新. 该算法由签名者执行. 算法输入公共参数params、当前时间片段tj的密钥SKtj以及时间片段tj′，其中tj<tj′.算法输出时间片段tj′的密钥SKtj′.
4）签名. 算法输入公共参数params、当前时间片段tj的密钥SKtj、消息M、签名者属性集wa、净化者属性集wτ以及签名策略Γd,S(⋅). 若签名者属性集wa满足签名策略Γd,S(⋅)，即|wa∩S|≥d，算法输出消息M的签名σ以及秘密值集合SI. 其中d是门限值，S是谓词结构中的属性集合.
5）净化. 该算法由净化者执行. 签名者公开声明允许净化的消息索引集合IN⊆{1,2,…,nm}，其中N≤nm.净化者获得由签名者发送的秘密值集合SI. 算法输入消息M、签名σ、签名者属性集wa、净化者属性集wτ以及秘密值集合SI. 算法输出净化消息M′和净化签名σ′.
6）验证. 算法输入公共参数params、当前时间片段tj、消息M′以及签名σ′. 若验证签名有效，输出accept；否则，输出reject.
FABSS系统框架如图1所示. 签名者将属性集wa以及初始时间片段t0发送给属性授权中心，属性授权中心为签名者生成时间片段t0的密钥SKt0. 签名者用私钥SKt0对消息M进行签名获得σ，并生成秘密值集合SI，将(M,σ,SI)通过安全信道发送给净化者. 净化者对允许净化范围内的消息进行修改，重新生成关于净化后消息M′的签名σ′. 净化者将(M′,σ′)发送给验证者，验证者通过验证算法判断签名是否有效. 此后，签名者通过密钥更新算法生成时间片段t1的密钥SKt1，并重复上述过程.
3.2 安全模型
借鉴文献[21]的思想，给出FABSS方案的前向安全性和不变性安全模型.
3.2.1 前向安全性
FABSS方案满足传统ABS方案不可伪造性的同时达到了前向安全性. FABSS方案的前向安全性可以通过挑战者B和敌手A之间的游戏来刻画.
基于文献[21]给出的安全模型，定义FABSS的前向安全性游戏.
1）初始化. A将需要挑战的签名策略Γd∗,S∗(⋅)和时间片段tj∗发送给B.
2）设置. B运行设置算法，生成公共参数params和主密钥msk，设置初始时间片段t0. 挑战者B将公共参数params发送给A，主密钥msk保密.
3）密钥生成询问. A自适应选择属性wa和时间片段tj，将wa和tj交给B. 通过密钥生成算法，B生成对应的密钥SKtj并发送给A.
4）密钥更新询问. A随机选择一个新时间片段tj并要求B执行密钥更新算法，此时当前时间片段tj被更新为后一时间片段tj′，B将更新后的密钥SKtj′发送给A.
5）签名询问. A自适应地选择签名者属性集wa，净化者属性集wτ，消息M和签名策略Γd,S(⋅)并发送给B，B通过签名算法产生当前时间片段tj的签名σ，并发送给A.
6）伪造. A生成关于消息M∗={m∗1,m∗2,…,m∗nm}，签名策略Γd∗,S∗(⋅)在时间片段tj∗的签名σ∗. 若满足条件①~③，则称A赢得前向安全性游戏.
① σ∗是一个有效签名；
② A没有对(wa,tj)进行密钥生成询问，其中属性集wa满足签名策略Γd∗,S∗(⋅)并且tj≤tj∗；
③ A没有在时间片段tj∗对消息M∗={m∗1,m∗2,…,m∗nm}进行签名询问.
定义1. 对于任意概率多项式时间t的敌手，如果赢得上述游戏的概率ε是可忽略的，那么就称FABSS方案满足前向安全性.
3.2.2 不变性
FABSS方案的不变性要求净化者只能对允许净化范围内的消息进行修改，无法对净化范围之外的消息进行任何操作. 不变性可以通过敌手A和挑战者B之间的游戏来刻画.
1）初始化. A将挑战索引集合I∗N和签名策略Γd∗,S∗(⋅)发送给B，其中I∗N表示净化者可以执行净化操作的消息索引集合.
2）设置. B执行设置算法产生公开参数params和主密钥msk，将公开参数params发送给A，主密钥msk保密.
3）询问. A 自适应地进行多项式次密钥生成询问，密钥更新询问和签名询问. 其中A可以进行qs次签名询问，在第j次签名询问中，A询问关于消息Mj={mj,1,mj,2,…,mj,nm}的签名σj. B将签名σj和秘密值集合SI发送给A.
4）伪造. A输出关于消息M∗={m∗1,m∗2,…,m∗nm}，时间片段tj∗和签名策略Γd∗,S∗(⋅)的签名σ∗，若满足条件①~③，则称A赢得不变性游戏.
① σ∗是一个有效签名；
② A没有对(wa,tj)进行密钥生成询问，其中属性集合wa满足签名策略Γd∗,S∗(⋅)并且tj≤tj∗；
③ 对于任何j∈{1,2,…,qs}，存在i∉I∗N使得mj,i≠m∗i.
定义2. 如果任意概率多项式时间t的敌手进行至多qk次密钥询问和至多qs次签名询问，最终赢得不变性游戏的概率ε是可忽略的，则FABSS方案具有ε-不变性.
4. 方案构造
根据文献[32]给出的二叉树结构，利用该结构分配时间片段. 在二叉树结构中，如图2所示，将完整时间片段T分解为t0,t1,…,tT−1时间片段. 每个时间片段对应一个层数为l的满二叉树的叶子节点. 其中根节点用一个空串γ标记，k(1≤k≤l)层上的每个节点v用一个二进制比特串bv∈{0,1}k表示，bv与节点v到根节点的路径相关，其中0表示左子节点，1表示右子节点. 对每个二进制串b∈{0,1}k，都对应二叉树第k层上的一个节点，将这个节点记为vb，并令bv[i]表示bv中的第i位. 例如初始时间片段t0对应节点vt0，bvt0=0l；t1时间片段的节点为vt1，bvt1=0l−11. 用Pathv表示节点v到根节点路径上包含的所有节点的集合，R(v)表示v的右子节点. 对于时间片段tj及其对应的节点vtj，定义集合Vtj={R(v)|v∈Pathvtj,R(v)∉Pathvtj}∪{vtj}. 如图2所示，Pathvt0={γ,v0,v00,vt0}，Vt0={v1,v01,vt1,vt0}. 基于上述构造，可得引理1.
引理1. 存在时间tj和tj′，若tj′>tj，对于每个节点v′∈Vtj′，存在一个节点v∈Vtj，有bv′=bv||b∗. 其中，b∗∈{0,1}k，k=|bv′|−|bv|.
1）设置. 选取安全参数λ，生成p阶双线性群G1和G2，其中p是大素数；e:G1×G1→G2是双线性映射. 令T=2l为总时间片段，U={1,2,…,n+d}表示属性域，其中n为常数. Ω={ω1,ω2,…,ωd−1}为缺省属性集，ωi∈Zp. 设S⊆Zp，且i∈S，定义拉格朗日系数ΔSi(x)=∏j∈S,j≠ix−ji−j. 随机选取α∈Z∗p，计算Z=e(g,g)α，其中g是G1的生成元. 随机选取群元素fa,fτ和群元素集合H={h1,h2,…,hl}，W={w1,w2,…,wnm}，F={f1,f2,…,fη}，其中nm是消息长度，η=n+d−1. 则params= {G1,G2,e,g,h0,w0,fa,fτ,H,W,F,T,U,Ω,Z}是公共参数，主密钥为α.
2）密钥生成. 算法输入签名者属性集wa⊆U，主密钥α，公共参数params和初始时间片段t0. 首先选择一个d−1次多项式q(x)，满足q(0)=α. 随机选取ri∈Zp，其中i∈wa；随机选取ri,v∈Zp，其中v∈Vt0. 计算μi=gri；φi={fri1,fri2,…,frii−1,frii+1,..,friη}；ski,v=(gq(i)(fafi)ri⋅(h0|bv|∏k=1hbv[k]k)ri,v,gri,v,hri,v|bv|+1,…,hri,vl).因此t0时间片段的密钥SKt0={μi,φi,{ski,v|v∈Vt0}}，其中i∈wa.
3）密钥更新. 算法输入当前时间片段tj的密钥SKtj，后续时间片段tj′和公共参数params. 将当前时间片段密钥SKtj表示成：
ski,v={ai,0,ai,1,ai,|bv|+1,…,ai,l}，SKtj={μi,φi,{ski,v|v∈Vtj}}，因为tj′>tj，由文献[32]可得，对每个节点v′∈Vtj′，一定存在节点v∈Vtj，有b∗满足bv′=bv||b∗. 随机选取r′i∈Zp，其中i∈wa；随机选取ri,v′∈Zp，其中v′∈Vtj′. 计算μ′i=μi×gr′i；φ′i={fri1fr′i1,fr′i2,…,frii−1fr′ii−1,frii+1fr′ii+1,…,friηfr′iη}；ski,v′ = {ai,0(fafi)r′i(h0|bv′|∏k=1hbv′[k]k)ri,v′|bv′|∏k=|bv|+1abv′[k]i,k,ai,1gr′i,ai,|bv′|+1hri,v′|bv′|+1,…,ai,lhri,v′l} = {a′i,0,a′i,1,a′i,2,a′i,|bv′|+1,…,a′i,l}. 时间片段tj′的密钥为SKtj′={μ′i,φ′i,{ski,v′|v′∈Vtj′}}，删除当前时间片段tj的密钥SKtj.
4）签名. 算法输入消息M={m1,m2,…,mnm}，签名策略Γd,S(⋅)，签名者属性集wa，净化者属性集^wτ，密钥SKtj，要求属性集wa满足Γd,S(wa)=1，即|wa∩S|≥d. 因此存在属性w′a⊆wa∩S，其中|w′a|=d. 选取缺省属性集Ω′⊂Ω，满足w′a∩Ω′=∅. 令^wa=w′a∪Ω′，当i∈^wa，计算ˉai,0=ai,0∏j∈^wa,j≠ifrij=(h0l∏k=1hbvtj[k]k)ri,vtjgq(i)(fa∏j∈^wafj)ri；a0=∏i∈^wa(ˉai,0)Δ^wai(0)=(h0l∏k=1hbvtj[k]k)r(fa∏j∈^wafj)r′⋅gα；a1=∏i∈^wa(ai,1)Δ^wai(0)=gr；μ′=∏i∈^wa(μi)Δ^wai(0)=gr′. 此时有r=∑i∈^waΔ^wai(0)⋅ri,vtj；r′=∑i∈^waΔ^wai(0)⋅ri. 随机选取ra,s,z,rτ∈Zp，计算σ0=a0(fa∏j∈^wafj)ra(h0l∏k=1hbvtj[k]k)s(w0nm∏j=1wmij)z⋅(fτ∏j∈^wτfj)rτ；σ1=a1gs；σ2=μ′⋅gra；σ3=grτ；σ4=gz. 因此在当前时间片段tj产生的签名为σ={σ0,σ1,σ2,σ3,σ4}. 签名者计算秘密值SIi=wzi，其中i∈IN. 用SI表示秘密值集合，即SI={SI1,SI2,…,SI|IN|}，IN={1,2,…,N}表示签名者允许净化的消息索引集合，其中N≤nm.
5）净化. 净化者获得签名σ和秘密值集合SI，首先通过验证算法判断签名是否有效，若是有效签名，定义此次需要净化的消息索引集I⊆IN. 令I1={i∈I:mi=0,m′i=1}；I2={i∈I:mi=1,m′i=0}. 净化者随机选取r′a,s′,z′,r′τ∈Zp，计算σ′0=σ0(fa∏j∈^wafj)r′a(h0l∏k=1hbvtj[k]k)s′⋅ (fτ∏j∈^wτfj)r′τ∏i∈I1SIi∏i∈I2SIi(nm∏j=1wm′ij⋅w0)z′；σ′1=σ1gs′；σ′2=σ2gr′a；σ′3=σ3gr′τ；σ′4=σ4gz′. 净化后的签名为σ′={σ′0,σ′1,σ′2,σ′3,σ′4}.
6） 验证. 为了验证签名是否有效，需要计算等式 Z=e(σ′0,g)e(σ′1,h0l∏k=1hbvtj[k]k)e(σ′2,fτ∏j∈^wτfj)⋅1e(σ′3,fτ∏j∈^wτfj)e(σ′4,w0nm∏j=1wm′ij)是否成立. 若等式成立，则签名有效；否则拒绝该签名. 验证算法不仅可用于验证净化消息签名对，同时也可以验证非净化的消息签名是否有效.
5. 安全性分析
本节将分别给出FABSS方案的安全性分析.
5.1 正确性
验证方程既能验证原始签名σ，同时也能验证净化签名σ′. 首先给出对原始签名σ的验证过程，在5.2节中给出净化签名的净化性分析. 给定签名σ={σ0,σ1,σ2,σ3,σ4}，通过证明等式（1）成立，表明FABSS方案满足正确性要求. 下面分别验证方程中的每一部分.
(σ0,g)=e(g,(fa∏j∈^wafj)ra+r′(h0l∏k=1hbvtj[k]k)s+r⋅(fτ∏j∈^wτfj)rτ(w0nm∏j=1wmij)zgα)=e(gα,g)e((w0nm∏j=1wmij)z,g)e(g,(fτ∏j∈^wτfj)rτ⋅(fa∏j∈^wafj)ra+r′)e(g,(h0l∏k=1hbvtj[k]k)s+r)；e(σ1,h0l∏k=1hbvtj[k]k)=e(h0l∏k=1hbvtj[k]k,gr+s)=e((h0l∏k=1hbvtj[k]k)r+s,g)；e(σ2,fa∏j∈^wafj)=e(gra+r′,fa∏j∈^wafj)=e(g,(fa∏j∈^wafj)ra+r′)；e(σ3,fτ∏j∈^wτfj)=e(grb,fτ∏j∈^wτfj)=e(g,(fτ∏j∈^wτfj)rτ)；e(σ4,w0nm∏j=1wmij)=e(w0nm∏j=1wmij,gz)=e(g,(w0nm∏j=1wmij)z)； 因此有
e(σ0,g)e(σ1,h0l∏k=1hbvtj[k]k)e(σ2,fa∏j∈^wafj)e(σ3,fτ∏j∈^wτfj)⋅1e(σ4,w0nm∏j=1wmij)=e(g,g)α=Z. (1) 综上所述，方案满足正确性.
5.2 净化性
净化者操作后的净化签名为σ′={σ′0,σ′1,σ′2,σ′3,σ′4}.当i∈I1时，m′i−mi=1，σ′记为1；当i∈I2时，m′i−mi=−1，σ′记为0. σ′0=σ0(fa∏j∈^wafj)r′a(h0l∏k=1hbvtj[k]k)s′⋅(fτ∏j∈^wτfj)r′τ∏i∈I1SIi∏i∈I2SIi(w0nm∏j=1wm′ij)z′=(fτ∏j∈^wτfj)rτ+r′τ⋅gα⋅ (fa∏j∈^wafj)ra+r′+r′a(h0l∏k=1hbvtj[k]k)s+r+s′\cdot {\left({w}_{0}\displaystyle\prod\limits _{j=1}^{{n}_{m}}{w}_{j}^{{m}_{i}' }\right)}^{z+{z}' }；{\sigma }_{1}' = {\sigma }_{1}{g}^{{s}' }={g}^{r+s+{s}' }， {\sigma }_{2}' ={\sigma }_{2} {g}^{{r}_{\mathrm{a}}' }={g}^{{r}_{\mathrm{a}}+{r}' +{r}_{\mathrm{a}}' } ，{\sigma }_{3}' ={\sigma }_{3}{g}^{{r}_{\mathrm{\tau }}' }= {g}^{{r}_{\mathrm{\tau }}+{r}_{\mathrm{\tau }}' }， {\sigma }_{4}' ={\sigma }_{4}{g}^{{z}' }={g}^{z+{z}' } . 综上所述，净化后的签名{\sigma }' = \{{\sigma }_{0}' ,{\sigma }_{1}' ,{\sigma }_{2}' ,{\sigma }_{3}' ,{\sigma }_{4}' \}与原始签名\sigma =\{{\sigma }_{0},{\sigma }_{1},{\sigma }_{2}, {\sigma }_{3}, {\sigma }_{4}\}有相同的分布. 因此签名 {\sigma }' 和 \sigma 都能通过验证方程.
5.3 前向安全性
定理1. 在{\varepsilon }' \text{-}(\eta \text{-}\mathrm{D}\mathrm{H}\mathrm{E})困难问题假设下，提出的FABSS方案具有\left(\varepsilon ,{q}_{\mathrm{s}}\right) \text{-}\mathrm{前}\mathrm{向}\mathrm{安}\mathrm{全}\mathrm{性}. 其中{\varepsilon }' \ge \dfrac{\varepsilon }{4T\times {q}_{\mathrm{s}}\times ({n}_{m}+1)}， T 是时间片段总数， {n}_{m} 是消息的长度， {q}_{\mathrm{s}} 是敌手A进行签名询问的次数.
证明. 通过敌手A和挑战者B之间的交互游戏证明定理1.
1）初始化. 给B一个 \eta \text{-}\mathrm{D}\mathrm{H}\mathrm{E} 困难问题的随机实例 \{g,{g}_{1}={g}^{a},{g}_{2}={g}^{{a}^{2}},… ,{g}_{\eta }={g}^{{a}^{\eta }},{g}_{\eta +2}={g}^{{a}^{\eta +2}},… ,{g}_{2\eta }={g}^{{a}^{2\eta }}\} ，其中 g 是素数阶群 {G}_{1} 的生成元， a\in {\mathbb{Z}}_{p} . A选择挑战签名谓词{\varGamma }_{{d}^{*},{S}^{*}}(\cdot )和时间片段 {t}_{{j}^{*}} 并发送给B，其中 0\le {t}_{{j}^{*}}\le T={2}^{l}-1 . 同时定义属性域 U=\{\mathrm{1,2},… ,n+d\} ，其中 n 是常数. 选择缺省属性集\varOmega =\left\{{\omega }_{1},{\omega }_{2},… ,{\omega }_{d-1}\right\}. 在以下交互中，B尝试计算得到 {g}_{\eta +1}={g}^{{a}^{\eta +1}} .
2）设置. B通过如下方式生成公共参数 params 和主密钥 msk . B随机选取 {\alpha }' ,{\delta }_{\mathrm{a}},{\delta }_{\mathrm{\tau }},{\delta }_{1},…,{\delta }_{\eta }\in {\mathbb{Z}}_{p} ，计算 {f}_{i}={g}^{{\delta }_{i}}{g}_{\eta -i+1} ，其中 1\le i\le \eta ；选择缺省属性子集{\varOmega }^{*{'}}\subset \varOmega，计算{f}_{\mathrm{a}}={g}^{{\delta }_{\mathrm{a}}}\displaystyle\prod\limits _{i\in {S}^{*}\cup {\varOmega }^{*{'}}}{f}_{i}^{-1}；令 {w}_{\mathrm{\tau }}\subseteq U ，计算 {f}_{\mathrm{\tau }}={g}^{{\delta }_{\mathrm{\tau }}}\displaystyle\prod\limits _{i\in {w}_{\mathrm{\tau }}}{f}_{i}^{-1} ；随机选取 {\theta }_{0},{\theta }_{1},… ,{\theta }_{l}\in {\mathbb{Z}}_{p} ，计算 {h}_{k}={g}^{{\theta }_{k}}{g}_{\eta -k+1}^{-1} ， {h}_{0}={g}^{{\theta }_{0}}\displaystyle\prod\limits _{k=1}^{l}{g}_{\eta -k+1}^{{b}_{{v}_{{t}_{{j}^{*}}}}\left[k\right]} ，其中 1\le k\le l ；随机选取 \zeta \in \{\mathrm{0,1},… ,{n}_{m}\} 以及2个随机数集合 X=\{{x}_{0},{x}_{1},…,{x}_{{n}_{m}}\} 和 Y=\{{y}_{0},{y}_{1},…,{y}_{{n}_{m}}\} ，其中 {x}_{i}\in {\mathbb{Z}}_{2{q}_{\mathrm{s}}-1} ， {y}_{i}\in {\mathbb{Z}}_{p} ；计算 {w}_{i}={{g}_{1}}^{{x}_{i}}{g}^{{y}_{i}} ， {w}_{0}={{g}_{1}}^{{x}_{0}-2\zeta {q}_{\mathrm{s}}}{g}^{{y}_{0}} ，其中 1\le i\le {n}_{m} ；计算Z=e\left({g}_{1},{g}_{\eta }\right) e{\left(g,g\right)}^{{\alpha }' }=e{\left(g,g\right)}^{{\alpha }' +{a}^{\eta +1}}. 最后B设置params= \{{G}_{1},{G}_{2},e,g,{f}_{\mathrm{a}},{f}_{\mathrm{\tau }},{h}_{0},{w}_{0},U,\varOmega ,T,H,W, Z\}为公共参数，主密钥 msk 为 \alpha ={\alpha }' +{a}^{\eta +1} . 定义2个函数，J\left(M\right)={x}_{0}+\displaystyle\sum _{j=1}^{{n}_{m}}{x}_{j}{m}_{j}-2\zeta {q}_{\mathrm{s}}；K\left(M\right)={y}_{0}+\displaystyle\sum _{j=1}^{{n}_{m}}{y}_{j}{m}_{j}. 此时{w}_{0}\displaystyle\prod\limits _{j=1}^{{n}_{m}}{w}_{j}^{{m}_{i}}={g}_{1}^{J\left(M\right)}{g}^{K\left(M\right)}.
3）密钥生成询问. A最多进行 {q}_{\mathrm{k}} 次密钥生成询问. A询问属性集 {w}_{\mathrm{a}} 在时间片段 {t}_{j} 的密钥 {S K}_{{t}_{j}} ，此时必须满足 |{w}_{\mathrm{a}}\cap {S}^{*}| < {d}^{*} 或者 {|w}_{\mathrm{a}}\cap {S}^{*}|\ge {d}^{*} ， {t}_{j} > {t}_{{j}^{*}} . 下面分别讨论这2种情况.
①当 |{w}_{\mathrm{a}}\cap {S}^{*}| < {d}^{*} 时，B定义3个属性集合 \varGamma , {\varGamma }' , S ，使\varGamma =\left({w}_{\mathrm{a}}\cap {S}^{*}\right)\cup {\varOmega }^{*{'}}， \varGamma \subseteq {\varGamma }' \subseteq S ，其中 \left|{\varGamma }' \right|={d}^{*}-1 . 令 S={\varGamma }' \cup \left\{0\right\} . 同时随机选取一个 {d}^{*}-1 次多项式 q\left(x\right) ，满足 q\left(0\right)=\alpha ={\alpha }' +{a}^{\eta +1} .
B随机选取 {r}_{i},{\rho }_{i}\in {\mathbb{Z}}_{p} ，令 q\left(i\right)={\rho }_{i} ，其中 i\in {\varGamma }' . 随机选取 {r}_{i,v}\in {\mathbb{Z}}_{p} ，其中 v\in {V}_{{t}_{j}} . 计算密钥{S K}_{{t}_{j}} = \{{\mu }_{i},{\varphi }_{i},\{{sk}_{i,v}|v\in {V}_{{t}_{j}}\}\}，其中 {\mu }_{i}={g}^{{r}_{i}} ，{\varphi }_{i}= \{{f}_{1}^{{r}_{i}},{f}_{2}^{{r}_{i}},… ,{f}_{i-1}^{{r}_{i}},{f}_{i+1}^{{r}_{i}},…,{f}_{\eta }^{{r}_{i}}\} ，{sk}_{i,v}= \left\{{g}^{{\rho }_{i}} ({f}_{a}{f}_{i})^{{r}_{i}}\left({h}_{0} \displaystyle\prod\limits _{k=1}^{\left|{b}_{v}\right|}{h}_{k}^{{b}_{v}\left[k\right]}\right)^{{r}_{i,v}},{g}^{{r}_{i,v}},{h}_{\left|{b}_{v}\right|+1}^{{r}_{i,v}},… ,{h}_{l}^{{r}_{i,v}}\right\} = \{{a}_{i,0}, {a}_{i,1},… , {a}_{i,\left|{b}_{v}\right|+1},… ,{a}_{i,l}\} . B随机选取 {r}_{i}' \in {\mathbb{Z}}_{p} ，其中i\in \left({w}_{\mathrm{a}}\cap \varOmega \right)/{\varGamma }'. 令 {r}_{i}={r}_{i}' -{\varDelta }_{0}^{S}\left(i\right){a}^{i} . 由拉格朗日插值可得q\left(i\right)=\displaystyle\sum _{j\in S}q\left(j\right){\varDelta }_{j}^{S}\left(i\right). 随机选取 {r}_{i,v}\in {\mathbb{Z}}_{p} ，其中 v\in {V}_{{t}_{j}} . 计算密钥 {S K}_{{t}_{j}}=\left\{{\mu }_{i},{\varphi }_{i},\left\{{sk}_{i,v}|v\in {V}_{{t}_{j}}\right\}\right\} ，其中 {\mu }_{i}={g}^{{r}_{i}}={g}^{{r}_{i}' }{g}^{-{\varDelta }_{0}^{S}\left(i\right)} ；{\varphi }_{i}=\left\{{f}_{1}^{{r}_{i}},\;{f}_{2}^{{r}_{i}},\;… ,\;{f}_{i-1}^{{r}_{i}},\;{f}_{i+1}^{{r}_{i}},\;…,\;{f}_{\eta }^{{r}_{i}}\right\}=\{{f}_{1}^{{r}_{i}' }{\left({g}^{{\delta }_{1}}{g}_{\eta }\right)}^{-{\varDelta }_{0}^{S}\left(i\right){a}^{i}},\;…, {f}_{i-1}^{{r}_{i}' }{\left({g}^{{\delta }_{\eta }}{g}_{1}\right)}^{-{\varDelta }_{0}^{S}\left(i\right){a}^{i}}({g}_{\eta -i+2} {{g}^{{\delta }_{i-1}})}^{-{\varDelta }_{0}^{S}\left(i\right){a}^{i}}{f}_{i+1}^{{r}_{i}' }{\left({g}^{{\delta }_{i+1}}{g}_{\eta -i}\right)}^{-{\varDelta }_{0}^{S}\left(i\right){a}^{i}} , {f}_{\eta }^{{r}_{i}' }\}= \{{f}_{1}^{{r}_{i}' } {\left({g}^{{\delta }_{1}}{g}_{\eta +i}\right)}^{-{\varDelta }_{0}^{S}\left(i\right)},\;…,\;{f}_{i-1}^{{r}_{i}' }{\left({g}^{{\delta }_{i-1}}{g}_{\eta +2}\right)}^{-{\varDelta }_{0}^{S}\left(i\right)}, {f}_{i+1}^{{r}_{i}' }({g}_{\eta } {{g}^{{\delta }_{i+1}})}^{-{\varDelta }_{0}^{S}\left(i\right)}, {f}_{\eta }^{{r}_{i}' }{\left({g}^{{\delta }_{\eta }}{g}_{i+1}\right)}^{-{\varDelta }_{0}^{S}\left(i\right)}\} ；{sk}_{i,v}=\big\{{g}^{q\left(i\right)}\left({f}_{\mathrm{a}} {f}_{i}\right)^{{r}_{i}}{\left({h}_{0}\displaystyle\prod\limits _{k=1}^{\left|{b}_{v}\right|}{h}_{k}^{{b}_{v}\left[k\right]}\right)}^{{r}_{i,v}}, {g}^{{r}_{i,v}}, {h}_{\left|{b}_{v}\right|+1}^{{r}_{i,v}},\;… ,\;{h}_{l}^{{r}_{i,v}}\}\;=\;\{{a}_{i,0},\; {a}_{i,1},\;{a}_{i,\;\left|{b}_{v}\right|+1},\;… ,\;{a}_{i,l}\} . 此时，计算可得 {a}_{i,0}\;\;=\;\;{g}^{q\left(i\right)} {\left({f}_{0}{f}_{i}\right)}^{{r}_{i}}{\left({h}_{0}\displaystyle\prod\limits _{k=1}^{\left|{b}_{v}\right|}{h}_{k}^{{b}_{v}\left[k\right]}\right)}^{{r}_{i,v}} = {\left({f}_{0}{f}_{i}\right)}^{{r}_{i}' -{\varDelta }_{0}^{S}\left(i\right){a}^{i}} \cdot {g}^{{\sum }_{j\in {\varGamma }' }q\left(j\right){\varDelta }_{j}^{S}\left(i\right)+q\left(0\right){\varDelta }_{0}^{S}\left(i\right)} ； {\left({h}_{0}\displaystyle\prod\limits _{k=1}^{\left|{b}_{v}\right|}{h}_{k}^{{b}_{v}\left[k\right]}\right)}^{{r}_{i,v}}={g}^{{\sum }_{j\in {\varGamma }' }{\rho }_{i}{\varDelta }_{j}^{S}\left(i\right)}{g}^{{\alpha }' {\varDelta }_{0}^{S}\left(i\right)}\cdot {g}_{\eta +1}^{{\varDelta }_{0}^{S}\left(i\right)}{\left({f}_{\mathrm{a}}{f}_{i}\right)}^{{r}_{i}' }{g}_{i}^{-{\delta }_{j}{\varDelta }_{0}^{S}\left(i\right)}\left(\displaystyle\prod\limits _{k=1}^{\left|{b}_{v}\right|}{h}_{k}^{{b}_{v}\left[k\right]} {h}_{0}\right)^{{r}_{i,v}}{\left({g}_{i}^{{\delta }_{\mathrm{a}}}\displaystyle\prod\limits _{j\in {S}^{*}\cup {\varOmega }^{*{'}}}{g}_{j}^{{\delta }_{j}}{g}_{\eta -j+i+1}\right)}^{{\varDelta }_{0}^{S}\left(i\right)}. 综上所述，模拟的密钥与原始方案生成的密钥具有相同的分布，因此对敌手而言模拟的密钥与原始密钥不可区分.
② 当 {w}_{\mathrm{a}}\cap {S}^{*}\ge {d}^{*} ， {t}_{j} > {t}_{{j}^{*}} 时，根据时间二进制树的定义可得，对节点 v\in {V}_{{t}_{j}} ，存在索引 \beta 使得 {b}_{v}\left[\beta \right]\ne {b}_{{v}_{{t}_{{j}^{*}}}}\left[\beta \right] . 为简化分析，令 \beta 为满足条件的最小索引值. B定义3个属性集合 \varGamma ， {\varGamma }' ， S ，使得\varGamma =\left({w}_{\mathrm{a}}\cap {S}^{*}\right)\cup {\varOmega }^{*{'}}， \varGamma \subseteq {\varGamma }' \subseteq S ，其中 \left|{\varGamma }' \right|={d}^{*}-1 . 令 S={\varGamma }' \cup \left\{0\right\} . 随机选取 {d}^{*}-1 次多项式 q\left(x\right) ，满足 q\left(0\right)=\alpha ={\alpha }' +{a}^{\eta +1} .
B随机选取 {r}_{i} ， {\rho }_{i}\in {\mathbb{Z}}_{p} ，令 q\left(i\right)={\rho }_{i} ，其中i\in {\varGamma }'.随机选取 {r}_{i,v}\in {\mathbb{Z}}_{p} ，其中 v\in {V}_{{t}_{j}} . 计算密钥{S K}_{{t}_{j}}=\{{\mu }_{i},{\varphi }_{i}, \{{sk}_{i,v}|v\in {V}_{{t}_{j}}\}\}，其中 {\mu }_{i}={g}^{{r}_{i}} ， {\varphi }_{i}= \{{f}_{1}^{{r}_{i}},{f}_{2}^{{r}_{i}},… ,{f}_{i-1}^{{r}_{i}},{f}_{i+1}^{{r}_{i}},…,{f}_{\eta }^{{r}_{i}}\} ，{sk}_{i,v}= \Bigg\{ ({g}^{{\rho }_{i}}{({f}_{\mathrm{a}}{f}_{i}))}^{{r}_{i}}\left({h}_{0} \displaystyle\prod\limits _{k=1}^{\left|{b}_{v}\right|}{h}_{k}^{{b}_{v}\left[k\right]}\right)^{{r}_{i,v}} , {g}^{{r}_{i,v}},{h}_{\left|{b}_{v}\right|+1}^{{r}_{i,v}}, … , {h}_{l}^{{r}_{i,v}})\Bigg\}= ({a}_{i,0},{a}_{i,1} , … , {a}_{i,\left|{b}_{v}\right|+1},… ,{a}_{i,l}) . B随机选择 {r}_{i}\in {\mathbb{Z}}_{p} ，其中i\in \left({w}_{\mathrm{a}}\cap \varOmega \right)/{\varGamma }'. 计算 {\mu }_{i}={g}^{{r}_{i}} ， {\varphi }_{i}=\{{f}_{1}^{{r}_{i}},{f}_{2}^{{r}_{i}},… ,{f}_{i-1}^{{r}_{i}},{f}_{i+1}^{{r}_{i}},…,{f}_{\eta }^{{r}_{i}}\} ；随机选取 {r}_{i,v}' \in {\mathbb{Z}}_{p} ，其中 v\in {V}_{{t}_{j}} . 令 {r}_{i,v}={a}^{\beta }{\varDelta }_{0}^{S}\left(i\right)/{b}_{v}\left[\beta \right]-{b}_{{v}_{{t}_{{j}^{*}}}}\left[\beta \right]+{r}_{i,v}' .
此时{sk}_{i,v}' =\{{a}_{i,0},\;{a}_{i,1},\;{a}_{i,\left|{b}_{v}\right|+1},\;… ,\;{a}_{i,l})= \Bigg\{ {g}^{q(i)}\left({f}_{\mathrm{a}}{f}_{i}\right)^{{r}_{i}} \cdot \left(\displaystyle\prod\limits _{k=1}^{|{b}_{v}|}{h}_{k}^{{b}_{v}[k]}{h}_{0}\right)^{{r}_{i,v}},{g}^{{r}_{i,v}},{h}_{\left|{b}_{v}\right|+1}^{{r}_{i,v}},… ,{h}_{l}^{{r}_{i,v}}\Bigg\}，其中q\left(i\right)=\displaystyle\sum _{j\in {\varGamma }' }q\left(j\right) \cdot {\varDelta }_{j}^{S}\left(i\right)+q\left(0\right){\varDelta }_{0}^{S}\left(i\right). 此时
\begin{split} {a}_{i,0}=&\left({g}^{q\left(i\right)}{\left({f}_{\mathrm{a}}{f}_{i}\right)}\right)^{{r}_{i}}{\left({h}_{0}\displaystyle\prod\limits _{k=1}^{\left|{b}_{v}\right|}{h}_{k}^{{b}_{v}\left[k\right]}\right)}^{{r}_{i,v}}=\left({g}^{q\left(0\right){\varDelta }_{0}^{S}\left(i\right)}{\left({f}_{\mathrm{a}}{f}_{i}\right)}\right)^{{r}_{i}} \cdot \\& {g}^{\sum\limits _{j\in {\varGamma }' }q\left(j\right){\Delta }_{j}^{S}\left(i\right)}{\left(\displaystyle\prod\limits _{k=1}^{\beta -1}{g}_{\eta -k+1}^{{b}_{{v}_{{t}_{{j}^{*}}}}\left[k\right]-{b}_{v}\left[k\right]}\right)}^{{r}_{i,v}}\left({g}^{{\theta }_{0}+\sum\limits _{k=1}^{\beta }{b}_{v}\left[k\right]{\theta }_{k}} \right)^{{r}_{i,v}} \end{split}. \begin{split} &{\left({g}_{\eta -\beta +1}^{({b}_{{v}_{{t}_{{j}^{*}}}}\left[k\right]-{b}_{v}[k\left]\right){r}_{i,v}}\right)}^{{r}_{i,v}}{\left(\displaystyle\prod\limits _{k=\beta +1}^{l}{g}_{\eta -k+1}^{{b}_{{v}_{{t}_{{j}^{*}}}}\left[k\right]}\right)}^{{r}_{i,{v}}}= \\& {g}^{{\alpha }' {\varDelta }_{0}^{S}\left(i\right)}\left({g}_{\beta }^{\frac{{\varDelta }_{0}^{S}\left(i\right)}{{b}_{v}\left[k\right]-{b}_{{v}_{{t}_{{j}^{*}}}}\left[k\right]}}{g}^{{r}_{i,v}' }\right)^{{\theta }_{0}+\sum\limits _{k=1}^{\beta }{b}_{v}\left[k\right]{\theta }_{k}}{g}^{\sum\limits _{j\in {\varGamma }' }{\rho }_{i}{\varDelta }_{j}^{S}\left(i\right)}\cdot \\& {g}_{\eta -\beta +1}^{{(b}_{{v}_{{t}_{{j}^{*}}}}\left[k\right]-{b}_{v}[k\left]\right){r}_{i,v}' }{\left(\displaystyle\prod\limits _{k=\beta +1}^{l}{g}_{\eta -k+\beta +1}^{{b}_{{v}_{{t}_{{j}^{*}}}}\left[k\right]}\right)}^{\frac{{\varDelta }_{0}^{S}\left(i\right)}{{b}_{v}\left[k\right]-{b}_{{v}_{{t}_{{j}^{*}}}}\left[k\right]}} \cdot \\& {\left({f}_{0}{f}_{i}\right)}^{{r}_{i}}{\left(\displaystyle\prod\limits _{k=\beta +1}^{l}{g}_{\eta -k+\beta +1}^{{b}_{{v}_{{t}_{{j}^{*}}}}\left[k\right]}\right)}^{{r}_{i,v}' }； \\& {a}_{i,1}={g}^{\frac{{a}^{\beta }{\varDelta }_{0}^{S}\left(i\right)}{{b}_{v}\left[k\right]-{b}_{{v}_{{t}_{{j}^{*}}}}\left[k\right]}+{r}_{i,v}' }={g}_{\beta }^{{\varDelta }_{0}^{S}\left(i\right)/({b}_{v}\left[k\right]-{b}_{{v}_{{t}_{{j}^{*}}}}\left[k\right])}{g}^{{r}_{i,v}' }； \\&{a}_{i.k}={\left({g}^{{\theta }_{k}}{g}_{\eta -k+1}^{-1}\right)}^{{a}^{\beta }{\varDelta }_{0}^{S}\left(i\right)/({b}_{v}\left[k\right]-{b}_{{v}_{{t}_{{j}^{*}}}}\left[k\right])+{r}_{i,v}' }=\\&\left({g}_{\beta }^{{\theta }_{k}} {g}_{\eta -k+\beta +1}^{-1}\right)^{{\varDelta }_{0}^{S}\left(i\right)/({b}_{v}\left[k\right]-{b}_{{v}_{{t}_{{js}^{*}}}}\left[k\right])}{\left({g}^{{\theta }_{k}}{g}_{\eta -k+1}^{-1}\right)}^{{r}_{i,v}' } . \end{split} 最后B随机选取 {r}_{i,v}'' \in {\mathbb{Z}}_{p} ，计算 {sk}_{i,v}=\{{a}_{i,0}\displaystyle\prod\limits _{k=\beta +1}^{\left|{b}_{v}\right|}{a}_{i,k}^{{b}_{v}\left[k\right]} \cdot {\left({h}_{0}\displaystyle\prod\limits _{k=1}^{\left|{b}_{v}\right|}{h}_{k}^{{b}_{v}\left[k\right]}\right)}^{{r}_{i,v}'' },{a}_{i,1}{g}^{{r}_{i,v}'' },{a}_{i,\left|{b}_{v}\right|+1}{h}_{\left|{b}_{v}\right|+1}^{{r}_{i,v}'' },{a}_{i,\left|{b}_{v}\right|+2} {h}_{\left|{b}_{v}\right|+2}^{{r}_{i,v}'' }, …, {a}_{i,l}{h}_{l}^{{r}_{i,v}'' }) . 综上所述，B成功模拟 {t}_{y} 时间片段密钥 {S K}_{{t}_{j}} .
4）密钥更新询问. 为了从当前时间片段 {t}_{j} 获得后续时间片段 {t}_{{j}' } 的密钥 {S K}_{{t}_{{j}' }} ，A向B进行密钥更新询问. B通过原始方案计算更新密钥 {S K}_{{t}_{{j}' }} 并发送给A.
5）签名询问. 给定消息 M=\{{m}_{1},{m}_{2},…, {m}_{{n}_{m}}\} 和签名策略{\varGamma }_{d,S}\left(\cdot \right)，若 J\left(M\right)=0 ，模拟终止；否则，B随机选取 {r}_{\mathrm{a}},s,{z}' ,{r}_{\mathrm{\tau }}\in {\mathbb{Z}}_{p} ，令 z={z}' -\dfrac{{a}^{\eta }}{J\left(M\right)} ，计算
\begin{split} \sigma =&\{{\sigma }_{0},{\sigma }_{1},{\sigma }_{2},{\sigma }_{3},{\sigma }_{4}\}=\Bigg\{{\left({g}_{1}^{J\left(M\right)}{g}^{K\left(M\right)}\right)}^{{z}' }{\left({f}_{\mathrm{\tau }}\prod _{j\in \widehat{{w}_{\mathrm{\tau }}}}{f}_{j}\right)}^{{r}_{\mathrm{\tau }}} \cdot \\& {g}^{{a}' }{\left({f}_{\mathrm{a}}\prod _{j\in \widehat{{w}_{\mathrm{a}}}}{f}_{j}\right)}^{{r}_{a}}{\left({h}_{0}\prod _{k=1}^{l}{h}_{k}^{{b}_{{v}_{{t}_{j}}}\left[k\right]}\right)}^{s}{g}_{\eta }^{-K\left(M\right)/J\left(M\right)},\\& {g}^{s} ,{g}^{{r}_{\mathrm{a}}},{g}^{{r}_{\mathrm{\tau }}} ,{g}^{{z}' }{g}_{\eta }^{-1/J\left(M\right)}\Bigg\}, \end{split} 此时，
\begin{split} {\sigma }_{0}=&{g}^{{a}' }{\left({f}_{\mathrm{a}}\prod _{j\in \widehat{{w}_{\mathrm{a}}}}{f}_{j}\right)}^{{r}_{\mathrm{a}}} {\left({f}_{\mathrm{\tau }}\prod _{j\in \widehat{{w}_{\mathrm{\tau }}}}{f}_{j}\right)}^{{r}_{\mathrm{\tau }}}{\left({g}_{1}^{J\left(M\right)}{g}^{K\left(M\right)}\right)}^{{z}' }{g}_{\eta }^{-\frac{K\left(M\right)}{J\left(M\right)}}\cdot \\&\left(\prod _{k=1}^{l}{h}_{k}^{{b}_{{v}_{{t}_{j}}}\left[k\right]} {h}_{0}\right)^{s}={g}^{\alpha }{\left({f}_{\mathrm{a}}\prod _{j\in \widehat{{w}_{\mathrm{a}}}}{f}_{j}\right)}^{{r}_{\mathrm{a}}}{g}_{\eta }^{-\frac{K\left(M\right)}{J\left(M\right)}}{\left({g}_{1}^{J\left(M\right)}{g}^{K\left(M\right)}\right)}^{{z}' }\cdot \\& {\left({f}_{\mathrm{\tau }}\prod _{j\in \widehat{{w}_{\mathrm{\tau }}}}{f}_{j}\right)}^{{r}_{\mathrm{\tau }}}{\left({h}_{0}\prod _{k=1}^{l}{h}_{k}^{{b}_{{v}_{{t}_{j}}}\left[k\right]}\right)}^{s}{g}^{-{a}^{(\eta +1)}}{g}_{\eta }^{-\frac{K\left(M\right)}{J\left(M\right)}}= \\& {g}^{\alpha }{\left({f}_{\mathrm{a}}\prod _{j\in \widehat{{w}_{\mathrm{a}}}}{f}_{j}\right)}^{{r}_{\mathrm{a}}}{\left({h}_{0}\prod _{k=1}^{l}{h}_{k}^{{b}_{{v}_{{t}_{j}}}\left[k\right]}\right)}^{s}{\left({w}_{0}{\prod _{j=1}^{{n}_{m}}{w}_{j}}^{{m}_{i}}\right)}^{z} \cdot \\& {\left({f}_{\mathrm{\tau }}\prod _{j\in \widehat{{w}_{\mathrm{\tau }}}}{f}_{j}\right)}^{{r}_{\mathrm{\tau }}}；\\& {\sigma }_{4}={g}^{{z}' }{g}_{\eta }^{-1/J\left(M\right)}={g}^{z} . \end{split} 综上所述，模拟签名与原始签名有相同的分布，因此B将签名 \sigma =\left\{{\sigma }_{0},{\sigma }_{1},{\sigma }_{2},{\sigma }_{3},{\sigma }_{4}\right\} 发送给A.
6）伪造. 询问结束后，A输出关于消息{M}^{*}= \{{m}_{1}^{*},{m}_{2}^{*},…,{m}_{{n}_{m}}^{*}\}，满足签名策略 {\varGamma }_{{d}^{*},{S}^{*}}(\cdot ) 和时间片段 {t}_{{j}^{\text{'}*}} 的签名 {\sigma }^{*}=\{{\sigma }_{0}^{*},{\sigma }_{1}^{*},{\sigma }_{2}^{*},{\sigma }_{3}^{*},{\sigma }_{4}^{*},\} . 伪造过程为：选取 {w}_{\mathrm{a}}^{*}\subseteq {S}^{*} 以及 {\varOmega }^{*}\subseteq \varOmega ，令 \widehat{{w}_{\mathrm{a}}^{*}}={w}_{\mathrm{a}}^{*}\cup {\varOmega }^{*} . 要求A没有在 {t}_{{j}^{\text{'}*}} 时间片段并且满足签名策略 {\varGamma }_{{d}^{*},{S}^{*}}(\cdot ) 的条件下对{M}^{*}= \{{m}_{1}^{*},{m}_{2}^{*},…, {m}_{{n}_{m}}^{*}\}进行签名询问. 此时B检查 {t}_{{j}^{\text{'}*}}={t}_{{j}^{*}} 是否成立. 若不成立，则模拟终止；若成立，B计算 J\left({M}^{*}\right) 和 K\left({M}^{*}\right) . 若 J\left({M}^{*}\right)\ne 0 ，模拟终止；否则A输出伪造签名{\sigma }^{*}=\left\{{\sigma }_{0}^{*},\;{\sigma }_{1}^{*},\;{\sigma }_{2}^{*},\;{\sigma }_{3}^{*},\;{\sigma }_{4}^{*}\right\}={\Bigg\{\;{g}^{\alpha }\;\left({f}_{\mathrm{a}}\displaystyle\prod\limits _{j\in \widehat{{w}_{\mathrm{a}}^{*}}}{f}_{j}\right)\;}^{{r}_{a}} \cdot{\big({g}_{1}^{J({M}^{*})} {g}^{K\left({M}^{*}\right)}\big)}^{{z}{'}} {\left( {f}_{\mathrm{\tau }}\displaystyle\prod\limits _{j\in \widehat{{w}_{\mathrm{\tau }}}}{f}_{j}\right) }^{{r}_{\mathrm{\tau }}}{\left({h}_{0}\displaystyle\prod\limits _{k=1}^{l}{h}_{k}^{{b}_{{v}_{{t}_{j}}}\left[k\right]}\right) }^{s}, {g}^{s}, {g}^{{r}_{\mathrm{a}}}, {g}^{{r}_{\mathrm{\tau }}}, {g}^{z}\Bigg\} = \Bigg\{{g}^{{a}{{'}}}{g}_{\eta +1}{g}^{{\delta }_{\mathrm{a}}{r}_{\mathrm{a}}}{g}^{{\delta }_{\mathrm{\tau }}{r}_{\mathrm{\tau }}}{g}^{\left({\theta }_{0}+\sum\limits _{k=1}^{\beta }{b}_{{t}_{{j}^{*}}}\left[k\right]{\theta }_{k}\right)s}{g}^{zK\left({M}^{*}\right)},\;{g}^{s},\;{g}^{{r}_{\mathrm{a}}},\;{g}^{{r}_{\mathrm{\tau }}},{g}^{z}\Bigg\} = \Bigg\{{g}^{{a}{{'}}}{g}_{\eta +1}{\left({\sigma }_{2}^{*}\right)}^{{\delta }_{\mathrm{a}}} {({\sigma }_{3}^{*})}^{{\delta }_{\mathrm{\tau }}} {\left({\sigma }_{1}^{*}\right)}^{{\theta }_{0}+\sum\limits _{k=1}^{\beta }{b}_{{t}_{{j}^{*}}}\left[k\right]{\theta }_{k}}{\left({\sigma }_{4}^{*}\right)}^{K\left({M}^{*}\right)}\cdot,{g}^{s},{g}^{{r}_{\mathrm{a}}},{g}^{{r}_{\mathrm{\tau }}},{g}^{z}\Bigg\}. B通过A提交的伪造签名计算{g}^{{a}_{\eta +1}}={g}_{\eta +1}= \dfrac{{\sigma }_{0}^{*}}{{g}^{{a}{{'}}}{\left({\sigma }_{1}^{*}\right)}^{{\theta }_{0}+\sum\limits _{k=1}^{\beta }{b}_{{t}_{{j}^{*}}}\left[k\right]{\theta }_{k}}{\left({\sigma }_{2}^{*}\right)}^{{\delta }_{\mathrm{a}}}{\left({\sigma }_{3}^{*}\right)}^{{\delta }_{\mathrm{\tau }}}{\left({\sigma }_{4}^{*}\right)}^{K\left({M}^{*}\right)}}. 因此若A能够伪造一个消息的有效签名，那么B就能成功解决 \eta \text{-}\mathrm{D}\mathrm{H}\mathrm{E} 困难问题. 证毕.
5.4 概率分析
为了在前向安全性游戏的交互中不发生终止，需要考虑3个事件：
1） 事件{E}_{1}. 签名询问阶段，满足 J\left(M\right)\ne 0 ，其中 i\in \{\mathrm{1,2},… ,{q}_{\mathrm{s}}\} ；
2） 事件 {E}_{2} . 伪造阶段，满足 J\left({M}^{*}\right)=0 ；
3） 事件 {E}_{3} . 敌手猜测的时间 {t}_{{j}'^{*}} ，满足 {t}_{{j}{'}{^*}}={t}_{{j}^{*}} .
易见，B不发生终止的概率为\mathit{Pr}\left[\overline{abort}\right]= Pr[{\wedge }_{i=1}^{{q}_{\mathrm{s}}}{E}_{1i}\wedge {E}_{2}\wedge {E}_{3}]. 同时，对于所有的i=1,2,… ,{q}_{s}，事件 {E}_{1i} 和事件 {E}_{2} 是相互独立的. 因此，\mathit{Pr}\left[\overline{abort}\right]\ge \mathit{Pr} \left[{\wedge }_{i=1}^{{q}_{\mathrm{s}}}{E}_{1i}\wedge {E}_{2}\right]\mathit{Pr}\left[{E}_{3}\right]=\mathit{Pr}\left[{E}_{3}\right]\mathit{Pr}\left[\left[{\wedge }_{i=1}^{{q}_{\mathrm{s}}}{E}_{1i}\right|{E}_{2}\right] \ \times \mathit{Pr}\left[{E}_{2}\right]\ge Pr \left[{E}_{2}\right](1-\displaystyle\sum _{i=1}^{{q}_{\mathrm{s}}}Pr[\overline{{E}_{1i}}\left|{E}_{2}\right]=\frac{1}{4T{q}_{\mathrm{s}}({n}_{m}+1)}.
综上所述，若存在概率多项式时间敌手以不可忽略概率 \varepsilon 赢得FABSS的前向安全性游戏，那么挑战者就能以{\varepsilon }' \ge \dfrac{\varepsilon }{4T \times {q}_{\mathrm{s}} \times ({n}_{m}+1)}的概率解决\eta \text{-}\mathrm{D}\mathrm{H}\mathrm{E}困难问题假设，其中 T 表示时间片段总数， {q}_{\mathrm{s}} 表示签名询问的次数， {n}_{m} 表示消息的长度.
5.5 不变性
定理2. FABSS方案在 \varepsilon' \text{-} (\eta \text{-} \mathrm{D}\mathrm{H}\mathrm{E}) 困难问题假设下具有 {\varepsilon} \text{-} \mathrm{不}\mathrm{变}\mathrm{性} ，其中存在常数 \psi ，满足 \varepsilon < \psi {\varepsilon}{{'}} .
假设可净化集合 {I}_{N}\subseteq \{\mathrm{1,2},… ,{n}_{m}\} ，净化者已知秘密值集合 S I ，但无法对可净化集合范围之外的数据进行操作. 首先证明引理2.
引理2. 若存在多项式时间的敌手 {A}_{1} 能够对可净化索引集合 {I}_{N} 中的 \kappa 位长度的消息进行操作，其中 0 < \kappa \le {n}_{m} ，并且以 {\varepsilon }_{{A}_{1}} 的优势赢得不变性游戏，那么就存在一个多项式时间敌手A在不可伪造游戏中以 {\varepsilon }_{A}\ge {\varepsilon }_{{A}_{1}} 的优势成功伪造一个长度为 {n}_{m}-\kappa 位消息的有效签名.
证明. 假设 {A}_{1} 在可净化范围内对 \kappa 位长度的消息进行操作，此时A对长度为 {n}_{m}-\kappa 位的消息进行前向安全游戏，在游戏中A模仿挑战者与 {A}_{1} 交互. 在收到 {A}_{1} 提交的相关询问操作后，A通过与前向安全游戏中的挑战者B交互并将结果发送给 {A}_{1} .
1）设置阶段， {A}_{1} 获得可净化索引集合 {I}_{N} ，其中 {I}_{N}\subseteq \left\{\mathrm{1,2},… ,{n}_{m}\right\} . 为简化分析，令{I}_{N}=\left\{{n}_{m}-\kappa +1, {n}_{m}- \kappa +2,… ,{n}_{m}\right\}， \kappa =\left|{I}_{N}\right| . B将公共参数params=\{{G}_{1}, {G}_{2},e,g,{f}_{\mathrm{a}},{f}_{\mathrm{\tau }},{h}_{0},{w}_{0},U,\varOmega ,T,H,{W}_{{n}_{m}-\kappa },Z\}发送给A，其中 {W}_{{n}_{m}-\kappa }=\left\{{w}_{1},{w}_{2},… ,{w}_{{n}_{m}-\kappa }\right\} . A随机选取 {s}_{i}\in {\mathbb{Z}}_{p} ，计算{w}_{i}{{'}}={g}^{{s}_{i}}，其中 i\in \left\{{n}_{m}-\kappa +1,{n}_{m}-\kappa +2,… ,{n}_{m}\right\} . 令W= {W}_{{n}_{m}-\kappa }\cup {W}_{{n}_{m}-\kappa +1}， {W}_{{n}_{m}-\kappa +1}=\left\{{w}_{{n}_{m}-\kappa +1},{w}_{{n}_{m}-\kappa +2},… ,{w}_{{n}_{m}}\right\} . A将公共参数params=\{{G}_{1},{G}_{2},e,g,{f}_{\mathrm{a}},{f}_{\mathrm{\tau }},{h}_{0},{w}_{0},U,\varOmega ,T,H, W,Z\}发送给 {A}_{1} .
在 j=\mathrm{1,2},… ,{q}_{\mathrm{s}} 次的签名询问中，A通过与B的交互来回答 {A}_{1} 的询问. 首先 {A}_{1} 向A询问消息{M}_{j}= \{{m}_{j,1}, {m}_{j,2},… ,{m}_{{j,n}_{m}}\}的签名，A收到询问后向B询问消息\overline{{M}_{j}}= \{{m}_{j,1},{m}_{j,2},…, {m}_{j,{n}_{m}-\kappa }\}的签名. B将签名\mathrm{\sigma }=\{{\sigma }_{j,0}, {\sigma }_{j,1}, {\sigma }_{j,2},{\sigma }_{j,3},{\sigma }_{j,4}\}发送给A，A计算{\mathrm{\sigma }}_{j,0}' ={\sigma }_{j,0}\displaystyle\prod\limits _{i={n}_{m}-\kappa +1}^{{n}_{m}} {{\sigma }_{j,1}}^{{s}_{i}{m}_{j,i}}， {\mathrm{\sigma }}_{j,1}' ={\sigma }_{j,1} ， {\mathrm{\sigma }}_{j,2}' ={\sigma }_{j,2} ， {\mathrm{\sigma }}_{j,3}' ={\sigma }_{j,3} ， {\mathrm{\sigma }}_{j,4}' ={\sigma }_{j,4} . A将签名({\mathrm{\sigma }}_{j,0}' , {\mathrm{\sigma }}_{j,1}' ,{\mathrm{\sigma }}_{j,2}' ,{\mathrm{\sigma }}_{j,3}' ,{\mathrm{\sigma }}_{j,4}' )以及秘密消息\{{{\sigma }_{j,1}}^{{s}_{i}{m}_{j,i}}|i= {n}_{m}-\kappa +1, {n}_{m}- \kappa + 2,… ,{n}_{m}\}发送给 {A}_{1} .
2）在伪造阶段，若 {{A}}_{1} 能够成功伪造消息{M}^{*}{'}= \{{m}_{0}^{{*}{{'}}},{m}_{1}^{{*}{{'}}},… ,{m}_{{n}_{m}}^{{*}{{'}}}\}的签名 ({\mathrm{\sigma }}_{j,0}^{*}{'},{\mathrm{\sigma }}_{j,1}^{*}{'},{\mathrm{\sigma }}_{j,2}^{*}{'},{\mathrm{\sigma }}_{j,3}^{*}{'},{\mathrm{\sigma }}_{j,4}^{*}{'}) . A利用该签名进行以下计算. 对于 i=\mathrm{1,2},… ,{q}_{\mathrm{s}} ，\exists i\notin \{{n}_{m}- \kappa +1,{n}_{m}-\kappa +2,… ,{n}_{m}\}，有 {m}_{j,i}\ne {m}_{i}^{*}{'} . 令消息{M}^{*}= \{{m}_{0}^{*},{m}_{1}^{*},…, {m}_{{n}_{m}}^{*}\}，当 i\in \{\mathrm{1,2},… ,{n}_{m}-\kappa \} 时， {m}_{i}^{*}={m}_{i}^{{*}{{'}}} . A计算{\sigma }_{0}^{*}=\dfrac{{\sigma }_{0}^{*}{'}}{\displaystyle\prod\limits _{i={n}_{m}-\kappa +1}^{{n}_{m}}{{\sigma }_{j,1}}^{{s}_{i}{m}_{i}^{*}{'}}}， {\sigma }_{1}^{*}={\sigma }_{1}^{*}{'} ， {\sigma }_{2}^{*}={\sigma }_{2}^{*}{'} ， {\sigma }_{3}^{*}={\sigma }_{3}^{*}{'} ，{\sigma }_{4}^{*}= {\sigma }_{4}^{*}{'} . A将有效签名 {\sigma }^{*}=\{{\sigma }_{0}^{*},{\sigma }_{1}^{*},{\sigma }_{2}^{*},{\sigma }_{3}^{*},{\sigma }_{4}^{*}\} 发送给B. 此时 \forall j\in \{\mathrm{1,2},… ,{q}_{\mathrm{s}}\} ， \exists i\in \{{n}_{m}-\kappa +1,{n}_{m}-\kappa +2,… ,{n}_{m}\} 满足 {m}_{j,i}\ne {m}_{i}^{*}{'} . 可以发现，如果 {A}_{1} 伪造的签名能够通过验证，那么 A 生成的签名也可以通过验证. 因此 A 赢得前向安全性游戏的优势 {\varepsilon}_{A} ，满足{\varepsilon}_{A}\ge {\varepsilon}_{{A}_{1}}，其中{\varepsilon}_{{A}_{1}}表示 {A}_{1} 赢得不变性游戏的优势.
由定理1可得，敌手A赢得前向安全游戏的优势是可忽略的. 因此由引理2可知，敌手 {A}_{1} 赢得不变性游戏的优势也是可忽略的. 证毕.
6. 方案分析
FABSS方案不仅获得细粒度访问控制，缓解了密钥泄露问题，而且具有可净化性，解决了敏感信息泄露问题. 表1给出FABSS方案与文献[21,29,31,33]在匿名性、净化性、前向安全性、透明性以及访问控制方面的优势比较分析. 其中文献[33]给出了支持非单调谓词的高效属性基签名方案，提供签名者的匿名性，同时具有细粒度访问控制，但无法提供前向性和净化性. 文献[21]提出具有前向安全的属性基签名方案，在获得细粒度访问控制的同时缓解了密钥泄露问题，但无法解决敏感信息泄露问题. 文献[29]构造了具有灵活访问结构的属性基可净化签名方案，不仅提供灵活细粒度访问控制，而且还实现了敏感信息隐藏，但无法解决密钥泄露问题. 文献[31]提出可追踪的属性基可净化签名方案，提供净化功能从而实现敏感信息隐藏，同时具有恶意用户追踪功能，避免签名滥用，但无法缓解密钥泄露问题. 本文提出的FABSS方案，不仅具有细粒度访问控制，还具有前向安全性和净化性，而且缓解了密钥泄露问题并保护了敏感数据的隐私.
表 1 方案比较Table 1. Comparison of Schemes7. 性能分析
基于Ubuntu 18.4，在Charm0.5框架下实现了FABSS方案. 利用Charm库中的超奇异椭圆曲线（SS512）测试方案. 实验中群 {G}_{1} 和 {G}_{2} 的阶为 p ， p 为512 b的大素数. 在此参数的计算机上测试主要密码学操作开销，经过1000次测量取平均值后，得到实验中计算双线配对所需时间为1.45 ms，在群 {G}_{1} 和 {G}_{2} 中执行指数运算所需时间分别为1.998 ms和0.2 ms. FABSS与文献[29,31]的通信开销和计算开销比较如表2和表3所示，其中 {|G}_{1}| 表示群 {G}_{1} 中元素的大小， \left|{\widehat{\omega }}_{\mathrm{a}}\right| 表示签名者属性数量， \left|{\widehat{\omega }}_{\mathrm{\tau }}\right| 表示净化者属性数量， l 表示时间二叉树层数. 由表2可知，提出的FABSS方案具有固定的签名长度，减少了通信开销. 由表3可知，提出的方案在验证阶段和净化阶段的指数和配对运算与属性数量无关，降低了计算开销. 实验结果如图3~6所示，由图3和图4可知，随着用户属性数量增加，提出的方案在密钥生成和签名阶段比文献[29,31]需要更大的计算开销，但是密钥生成算法一般只执行1次，所以对方案的性能影响不大；由图5和图6可知，提出的方案在净化以及验证阶段所需的计算时间要小于文献[29,31]，具有较小的计算开销.
表 2 通信开销比较Table 2. Comparison of Communication Cost方案 密钥 签名 净化签名 FABSS [|{\widehat{\omega } }_{\mathrm{a} }|+\eta -1+(l+2)|{\widehat{\omega } }_{\mathrm{a} }|]{|G}_{1}| 5 {|G}_{1}| 5| {G}_{1}| 文献[29] (2+|{\widehat{\omega } }_{\mathrm{a} }|){|G}_{1}| (2{|{\widehat{\omega }}_{\mathrm{a}}|}^{2}+2){|G}_{1}| (2{|{\widehat{\omega }}_{\mathrm{a}}|}^{2}+2){|G}_{1}| 文献[31] (2|{\widehat{\omega } }_{\mathrm{a} }|+2){|G}_{1}|+|{{\mathbb{Z}}}_{p}^{*}| (3|{\widehat{\omega }}_{\mathrm{a}}|+|{\widehat{\omega }}_{\mathrm{\tau }}|+3){|G}_{1}| (3|{\widehat{\omega }}_{\mathrm{a}}|+|{\widehat{\omega }}_{\mathrm{\tau }}|+3){|G}_{1}| 注： {|G}_{1}| 表示群 {G}_{1} 中元素的大小， |{\mathbb{Z}}_{p}^{*}| 表示环 {\mathbb{Z}}_{p}^{*} 中元素的比特大小， |{\widehat{\omega }}_{\mathrm{a}}| 表示签名者属性数量， |{\widehat{\omega }}_{\mathrm{\tau }}| 表示净化者属性数量， l 表示时间二叉树层数， \eta = n + d - 1 ， n 是常数， d 是门限值. 表 3 计算开销比较Table 3. Comparison of Computation Cost方案 密钥生成 签名 验证 净化 FABSS [(4+ \eta +l )| {\widehat{\omega }}_{\mathrm{a}} |]E [(3+l)| {\widehat{\omega }}_{\mathrm{a}} |+| {\widehat{\omega }}_{\mathrm{\tau }} |+13+ {n}_{m} ]E (l+ {n}_{m} )E+5P (8+l+ I+{n}_{m} )E 文献[29] (| {\widehat{\omega }}_{\mathrm{a}} |+1)E (3| {\widehat{\omega }}_{\mathrm{a}} + {n}_{m} +2)E (3+| {\widehat{\omega }}_{\mathrm{a}} )P+(| {\widehat{\omega }}_{\mathrm{a}} |+ {n}_{m} )E (|{\widehat{\omega } }_{\mathrm{a} }|+I+{n}_{m})E 文献[31] (3+3| {\widehat{\omega }}_{\mathrm{a}} |)E (3| {\widehat{\omega }}_{\mathrm{a}} |+2| {\widehat{\omega }}_{\mathrm{\tau }} |+l+4v+6)E (| {\widehat{\omega }}_{\mathrm{a}} |+| {\widehat{\omega }}_{\mathrm{\tau }} |+3)P+lE (| {\widehat{\omega }}_{\mathrm{a}} |+| {\widehat{\omega }}_{\mathrm{\tau }} |+|I|+l+4)E 注： {n}_{m} 表示消息长度， I 表示可净化范围集合，E表示 {G}_{1} 中的指数运算，P表示配对运算， \left|{\widehat{\omega }}_{\mathrm{a}}\right| 表示签名者属性数量， \left|{\widehat{\omega }}_{\mathrm{\tau }}\right| 表示净化者属性数量， l 表示时间二叉树层数， \eta =n+d-1 ， n 是常数， d 是门限值. 8. 结束语
本文形式化了前向安全的属性基可净化签名安全模型. 提出了一种前向安全的高效属性基可净化签名方案，不仅缓解了密钥泄露问题，而且还实现了敏感信息隐藏功能. 基于η-DHE困难问题假设，在标准模型下证明了本文方案的安全性. 通过与现有方案的对比分析可知，提出的方案更适用于电子医疗、电子政务等特殊应用场景中.
作者贡献声明：朱留富提出初步方案、实验设计，以及论文初稿撰写和修改；李继国负责论文思路构建、理论指导、方案分析和论文修改；陆阳和张亦辰负责论文方案分析、论文润色和修改.
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表 1 云计算、边缘计算和边缘智能特点对比
Table 1 Features Comparison of Cloud Computing, Edge Computing , and Edge Intelligence
类别 云计算 边缘计算 边缘智能 架构模型 集中式 分布式 分布式 服务器位置 互联网中 边缘网络中 云—边—端协同网络 目标应用 互联网应用 物联网或移动应用 各种智能应用程序 服务类型 全球信息服务 有限的本地化信息服务 低延时、高可靠的智能服务 设备数量 数百亿 几千万甚至几亿 数百亿甚至上千亿 研究重点 工作流调度、虚拟机管理等 计算卸载、缓存、资源分配等 在边缘侧利用AI实现数据收集、缓存、处理和分析 
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表 2 智能的边缘计算和边缘的智能化特点对比
Table 2 Features Comparison of Intelligent Edge Computing and Edge Intelligence
类别 云/边/端 智能的边缘计算 边缘的智能化 结构层面 云 服务器集群 服务器集群 边缘 基准站、边缘节点 智能化服务 终端 终端设备 智能化应用 内容层面 利用AI技术解决边缘计算相关问题 实现边缘环境中应用的智能化
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表 3 智能的边缘计算相关工作分类
Table 3 Related Work Classification for Intelligent Edge Computing
关键技术 适用场景 问题挑战 优化目标 相应算法 数据来源 计算卸载 车联网 高度动态的车辆拓扑结构. 优化卸载决策和带宽/计算资源分配 深度学习 文献[56] 无人机 无人机与终端用户之间的
计算和信道容量有限.最小化延迟和能耗 深度学习 文献[57] 设备到设备卸载 数据卸载过程效果不稳定. 优化用户体验 强化学习 文献[58] 物联网设备 嵌入式设备处理能力及资源受限； 降低了DNN推理的总延迟 深度学习 文献[59] 设备之间的对抗性竞争；
低延时通信约束.最小化延迟和通信成本 深度强化学习 文献[60] 资源分配 多用户资源约束条件 单个边缘服务器资源受限. 最小化延迟、提高系统实时性 深度学习 文献[61] 移动设备 边缘计算环境复杂多变. 保持MEC架构在不同条件下的稳定性 深度强化学习 文献[46] 车辆边缘计算网络 车辆动态变化. 最大化车辆边缘计算网络的长期效用 深度强化学习 文献[62] 工业物联网 频谱资源有限；
电池容量受限.最大化长期吞吐量 深度学习 文献[63] 无线网络 框架节点之间达成共识的同时
保证系统的性能.最大化系统吞吐量和用户的服务质量 深度强化学习 文献[64] 边缘缓存 边缘计算系统 无线信道的拥塞. 最小化系统成本消耗
系统性能最优深度强化学习 文献[36] 车联网 车辆移动性； 最大化系统效用 深度强化学习 文献[65] 动态网络拓扑；
存储容量和带宽资源有限；最小化系统成本和延迟 深度强化学习 文献[66] 主动缓存的时间变化性； 提高模型性能预测准确率 深度学习 文献[67] 车辆的高移动性. 以最大限度地降低能耗 深度强化学习 文献[68]
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表 4 OpenEI和Edgent的特点对比
Table 4 Features Comparison of OpenEI and Edgent
类别 OpenEI Edgent 可部署的
硬件环境树莓派和集群计算机 树莓派和台式机 适用环境 各种操作系统 静态和动态网络 优化目标 最大化模型准确率 最小化延迟 特点 易于安装、可跨平台使用 超低延时、超高稳定
性及可靠性功能 为边缘提供智能处理和
数据共享功能按需DNN协作推理
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