Scheme
1 Attribute vector $v$ -> $g^v$ Policy vector $x$ -> $rx$, where $r\leftarrow Z_p$ Verify $g^{v\cdot rx}=g^0$ System Model Requester is semi-honest, which means that they honestly perform protocol but may attempt to learn the privacy of other users. Worker may claim the honest CS doesn’t return correct matching tasks. 想要申请自己不满足任务需求的任务,栽赃服务器返回了错误的结果; keyword; collude Server:为了节省算力返回错误的匹配结果;篡改密文 public verifiable outsourced ABE: Worker 拒绝任务后可通过proof验证 Server进行了正确匹配; Verifiability: 对于Worker是否满足Requester制定的访问策略,可将双方的属性向量(任务访问策略向量)秘密共享or使用同态commitment: 对向量$v=(v_0,v_1,v_2)$: Authority授权密钥时声称向量对应位置的commitment。如$AU_0$计算$Com(v_0)=g^{v_0}h^{r_0}$,最后将$Coms=(Com(v_0), Com(v_1), Com(v_2))$发送到区块链存储。 对向量$x(x_0,x_1,x_2)$: 利用Two-party secret sharing,首先Requester对$x$进行两方的加法秘密共享$x_0 = x_0^1+x_0^2$… 验证:$Com(v_0,r_0)^{x_0^1}\cdot Com(v_0,r_0)^{x_0^2}=Com(v_0\cdot x_0,r_0\cdot x_0)$ $\prod \limits_{i=0}^2 Com(v_i x_i)=Com(0)$ 忽略了随机数r ,$Com(v_0,r_0)^{x_0^1}\cdot Com(v_0,r_0)^{x_0^2}=Com(v_0\cdot x_0,r_0\cdot x_0)$ $\prod \limits_{i=0}^2 Com(v_i x_i)=g^{v\cdot x}h^{\sum \limits_{i=0}^n r_i\cdot x_i}=g^0\cdot h^{{\sum \limits_{i=0}^n r_i\cdot x_i}}$ 对于Worker是否满足Requester制定的访问策略,可将双方的属性向量(任务访问策略向量)秘密共享or使用同态commitment: ...