11.07.2014 Views

Identity-Based Encryption Protocols Using Bilinear Pairing

Identity-Based Encryption Protocols Using Bilinear Pairing

Identity-Based Encryption Protocols Using Bilinear Pairing

SHOW MORE
SHOW LESS

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

Hooray! Your file is uploaded and ready to be published.

Saved successfully!

Ooh no, something went wrong!