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Identity-Based Encryption Protocols Using Bilinear Pairing

Identity-Based Encryption Protocols Using Bilinear Pairing

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Setup: Let P be a generator of G 1 . Choose a random secret x ∈ Z p and set P 1 = xP .<br />

Randomly choose P 2 ; an l 1 × l 2 matrix R where<br />

⎡<br />

⎤<br />

R 1,1 · · · R 1,l2<br />

⎢<br />

⎥<br />

R = ⎣ . . . ⎦<br />

R l1 ,1 · · · R l1 ,l 2<br />

and l 2 many vectors −→ U 1 , . . . , −→ U l2 from G 1 , where each −→ U i = (U i,1 , . . . , U i,n ). The public<br />

parameters are 〈P, P 1 , P 2 , R, −→ U 1 , . . . , −→ U l2 〉, while the master secret is xP 2 .<br />

Key Generation: Given an identity v = (v 1 , . . . , v u ) for any u, this algorithm generates<br />

the private key d v of v as follows.<br />

Let u = k 1 l 2 + k 2 with k 2 ∈ {1, . . . , l 2 }. We represent v by a (possibly incomplete)<br />

(k 1 + 1) × l 2 matrix I where the last row has k 2 elements. Choose (k 1 + 1) many random<br />

elements r 1 , . . . , r k1 , r k2 ∈ Z p and output<br />

(<br />

(<br />

∑k 1 h∑<br />

) (<br />

k2<br />

)<br />

∑<br />

d ID = xP 2 + r i V i,j + R i,j + r k2 V k1 +1,j + R k1 +1,j , r 1 P, . . . , r k1 P, r k2 P,<br />

i=1 j=1<br />

j=1<br />

)<br />

−−−→ −→<br />

r k2 R k1 +1,k 2 +1, . . . , r k2 R k1 +1,h, r k2 U k2 +1, . . . , r k2Ul2 where V i,j = ∑ n<br />

−→<br />

j=1 vj U i,j and r k2Ui denotes that each element of −→ U i is multiplied by the<br />

scalar r k2 .<br />

The private key of v can also be generated given the private key of v |u−1 = v 1 , . . . , v u−1<br />

as required. There are two cases to be considered.<br />

Case 1: Suppose u − 1 = k 1 l 2 + l 2 , then<br />

(<br />

(<br />

k∑<br />

1 +1 h∑<br />

)<br />

)<br />

d v|u−1 = xP 2 + r i V i,j + R i,j , r 1 P, . . . , r k1 P, r k1 +1P<br />

i=1 j=1<br />

= (a 0 , a 1 , . . . , a k1 , a k1 +1) (say)<br />

Choose a random r ∗ ∈ Z p and form d v as<br />

d v = a 0 + r ∗ (V k1 +2,1 + R k1 +2,1), a 1 , . . . , a k1 +1, r ∗ P, r ∗ R k1 +2,2, . . . , r ∗ R k1 +2,h, r ∗ −→<br />

U2 , . . . , r ∗ −→<br />

Ul2<br />

Case 2: Let, u − 1 = k 1 l 2 + k 2 ′ with k 2 ′ < l 2 then,<br />

(<br />

∑k 1 h∑<br />

) ⎛<br />

⎞<br />

k<br />

∑2<br />

′<br />

d v|u−1 = (xP 2 + r i V i,j + R i,j + r k ′ ⎝<br />

2<br />

V k1 ,j + P k1<br />

⎠<br />

+1,j , r 1 P, . . . , r k1 P, r k ′ 2<br />

P,<br />

i=1 j=1<br />

j=1<br />

r k ′ 2<br />

R k1 +1,k 2 ′ +1 , . . . , r k ′ 2<br />

R k1 +1,h, r k ′ −→<br />

2<br />

U k ′<br />

2 +1, . . . , r k ′ −→<br />

2<br />

U l2<br />

= (a 0 , a 1 , . . . , a k1 , a k1 +1, b k ′<br />

2 +1, . . . , b l2 , −→ c k ′<br />

2 +1, . . . , −→ c l2 ) (say)<br />

121

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