New answers tagged pairings
i don't know whether it is correct e(c1,d1) = e(g^s, mk.(y^id.h)^r) =e(g^s,mk).e(g^s, (y^id.h)^r) = e(g^s , g1^α ) . e(g^r , (y^id.h)^s) = e(y , g1^α ) . e(g^r , (y^id.h)^s) e(c2,d2) = e((Y^id.h)^s , g^r) = e((Y^id.h)^s , g^r) therefore e(c1,d1)/e(c2,d2) = e(y , g1^α )
Update: My previous answer, although technically true, didn't answer your question. The issue is that the strong-DDH is not hard when using pairing groups, so my answer was merely stating that your problem is at least as hard as an easy problem (duh!) After some thought, I realized your problem is at least as hard as the 2-weak Bilinear Diffie-Hellman ...
There are libraries allowing to do that: PBC: C library JPBC : Java library Hope it helps.
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