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Efficient aggregation of encrypted data in Wireless Sensor Network. Author: Einar Mykletun, Gene Tsudik Presented by Yi Cheng Lin. Multi-level network tree (3-ary). Computing statistical data. Computing the Average Computing the Variance. Homomorphic Encryption. M : message space
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Efficient aggregation of encrypted data in Wireless Sensor Network Author: Einar Mykletun, Gene Tsudik Presented by Yi Cheng Lin
Computing statistical data • Computing the Average • Computing the Variance
Homomorphic Encryption M: message space C: ciphertext space M is a group under operation C is a group under operation c1 = Enck1(m1), c2 = Enck2(m2)
problem • One limitation of this proposal is that the identities of the on-responding nodes need to be sent along with the aggregate to the sink • 沒有回應之Node的ID,並排的放在aggregate data之後 Ex: C,ID2,ID5,… Bit數高
Number of bits sent per node for each level in a 3-tree of depth 7
Bit length of the number of no responded nodes • Level 1 • IA(10%) = (10 + 8)/3 + 75 = 81 bit per node • IA(30%) = (10 + 10)/3 + 75 = 82 bit per node Bit length of the number of no responded nodes
Key generation • ID [0, P-1], where P is a large integer • Let i be a randomly generated keystream, where i [0, P-1], sink keep i secret • Compute K = ID – i (mod P) • Let K1 = ID1 – i (mod P) => ID1 = K1 + i (mod P) K2 = ID2 – i (mod P) => ID2 = K2 + i (mod P) For = ID1 + ID2 = (K1+K2) + 2*i (mod P) K1+K2 = (ID1+ID2) – 2*i (mod P)
Additive Homomorphic Encryption Key generation : k
Example sink K = ID – i (mod P) 27, 5 66, 10 c = m + k (mod M) aggregation nodes sensor nodes ID1 ID2 ID3 ID4 No respond i = 40, M = 100, P = 50 In sink : C = 27+66 (mod 100) = 93 ID1 + ID3 = 32 – 15 = 17 => key1 + key3 = 17 – 2*90 (mod 100) = -163 (mod 100) = 37 M = C – (key1+key3) (mod 100) = 93 – 37 (mod100) = 56