Identity-Based Encryption Protocols Using Bilinear Pairing
Identity-Based Encryption Protocols Using Bilinear Pairing
Identity-Based Encryption Protocols Using Bilinear Pairing
You also want an ePaper? Increase the reach of your titles
YUMPU automatically turns print PDFs into web optimized ePapers that Google loves.
B declares the public parameters to be<br />
(P, P 1 , P 2 , −→ P 3 , −→ Q 1 , . . . , −→ Q h ),<br />
where −→ P 3 = (P 3,1 , . . . , P 3,h ) and −→ Q i = (Q i,1 , . . . , Q i,ni ). The corresponding master key<br />
αP 2 = Y h+1 + βY 1 is unknown to B. The distribution of the public parameter is as expected<br />
by A.<br />
Phase 1: Suppose A asks for the private key corresponding to an identity v = (v 1 , . . . , v h ′)<br />
for h ′ ≤ h. Note that for any i ≤ η ′ ,<br />
∑n i<br />
V i = P 3,i + v j i Q i,j<br />
j=1<br />
∑n i<br />
= b i,0 P + a i,0 Y h−i+1 + v j i (b i,jP + a i,j Y h−i+1 )<br />
j=1<br />
= F i (v i )Y h−i+1 + J i (v i )P.<br />
Hence, V i is computable from what is known to B.<br />
Recall that A initially committed to sets of identities up to level u before the set-up<br />
phase. If h ′ ≤ u, then there must be a k ≤ h ′ such that F k (v k ) ≠ 0, as otherwise v j ∈ Ij<br />
∗ for<br />
each j ∈ {1, . . . , h ′ } – which the adversary is not allowed by the rules of the Game. In case<br />
h ′ > u, it is possible that F 1 (v 1 ) = · · · = F u (v u ) = 0. Then by construction F u+1 ≠ 0. So, in<br />
either case there is a k such that F k (v k ) ≠ 0. Moreover, k is the first such index in the range<br />
{1, . . . , h ′ }. B picks a random r ∈ Z p and assigns d 0|k = (−J k (v k )/F k (v k ))Y k + βY 1 + rV k and<br />
d 1 = (−1/F k (v k ))Y k + rP. Now,<br />
d 0|k = − J k(v k )<br />
F k (v k ) Y k + βY 1 + α k Y h−k+1 − α k F k(v k )<br />
F k (v k ) Y h−k+1 + rV k<br />
= − J k(v k )<br />
F k (v k ) αk P + αP 2 − α k F k(v k )<br />
F k (v k ) Y h−k+1 + rV k<br />
= αP 2 + ˜rV k<br />
where ˜r = (r −<br />
αk ). Also d F k (v k ) 1 = − 1 Y F k (v k ) k + rP = − αk<br />
F k (v k<br />
P + rP = ˜rP .<br />
)<br />
j ∈ {1, . . . , h ′ } \ {k} we have<br />
For any<br />
˜rV j = (r − αk<br />
F k (v k ) )(F j(v j )Y h−j+1 + J j (v j )P )<br />
= r(F j (v j )Y h−j+1 + J j (v j )P ) − 1<br />
F k (v k ) (F j(v j )Y h+k−j+1 + J j (v j )Y k ).<br />
Recall that, k is the smallest in the range {1, . . . , h ′ }, such that, F k (v k ) ≠ 0. Hence, for<br />
j < k, F j (v j ) = 0 and ˜rV j = rJ j (v j )P − J j(v j )Y k<br />
F k (v k ) . For j > k, Y h+k−j+1 varies between Y 1 to<br />
118