Anonymous Hierarchical Identity-Based Encryption ... - Unisinos

where the last equation is predicated on the two following conditions,1+D∑n=1(χ n − 1) = 0 ,1+D∑n=1(χ n − 1) Θ n ¯Θ−1 n = 0 ; (1a,1b)and as required, we find that k 0,(a) = â −ρ 00 and k 0,(b) = ˆb −ρ 00 ; and also, for n = 1, . . . , 1 + D,k n,(a) = â −˜ρnnk n,(b) =ˆb−˜ρnnĝ χn ˜ρ 0 Θ 0/¯Θn βn2 = ân−ρn ĝ −αn χn z 2 ˜ρ 0 Θ 0/¯Θn αn βnĝ χn ˜ρ 0 Θ 0/¯Θn βn2 = â −ρnĝ χn ˜ρ 0 Θ 0/¯Θn αn2 =ˆb−ρnnn ,ĝ −βn χn z 2 ˜ρ 0 Θ 0/¯Θn αn βnĝ χn ˜ρ 0 Θ 0/¯Θn αn −ρn2 = ˆb n .Using analogous calculations, we can derive a similar set of relations, such as, for m = 0, . . . , 1 + D,⎛1+D∏f m,0 = ⎝ ( () L ′ )˜ρ n,m⎞−1 Θ0 ˜ρ−Θn ¯Θ 0,m ∏nĝ ⎠(==n=12ĝ −˜ρ 0,m Θ 021+D∏∏L ′n=0 l=0l=0ŷ I ln,l) P (1+D−1 L ′n=1 (χn,m−1) Θn ¯Θ n ∏ŷ I l ρ n,mn,l,l=0ŷ I l ρ 0,m0,l) P 1+Dn=1 χn,m/(1+D) 1+D L ′where for the last equality to hold we impose that, for all m = 0, . . . , 1 + D,∏ ∏n=1 l=0ŷ I l ρ n,mn,l1+D∑n=1(χ n,m − 1) = 0 ,1+D∑n=1(χ n,m − 1) Θ n ¯Θ−1 n = 0 ; (2a,2b)in addition, we have the required f m,0,(a) = â −ρ 0,m0 and f m,0,(b) = ˆb −ρ 0,m0 ; and furthermore, for alln = 1, . . . , 1 + D,f m,n,(a) = â −˜ρn,mnf m,n,(b) =ˆb−˜ρn,mnĝ χn,m ˜ρ 0,m Θ 0/¯Θn βn2 = ân−ρn,mĝ −αn χn,m z 2 ˜ρ 0,m Θ 0/¯Θn αn βnĝ χn,m ˜ρ 0,m Θ 0/¯Θn βn2 = ân −ρn,m,ĝ χn,m ˜ρ 0,m Θ 0/¯Θn αn2 =ˆb−ρn,mnĝ −βn χn z 2 ˜ρ 0,m Θ 0/¯Θn αn βnĝ χn,m ˜ρ 0,m Θ 0/¯Θn αn −ρn,m2 = ˆb n .As for the remaining components of the key, if, for each l = 1+L ′ , . . . , D, and each m = 0, . . . , 1+D,1+D∑n=1χ n¯θn,l¯Θ n1α n β n= (1 + D) θ 0,lΘ 0,1+D∑n=1χ n,m¯θn,l¯Θ n1α n β n= (1 + D) θ 0,lΘ 0, (3,4)25