1. Efficient aggregation of encrypted data in Wireless Sensor Network Author: Einar Mykletun, Gene Tsudik Presented by Yi Cheng Lin

2. Multi-level network tree (3-ary)

3. Computing statistical data Computing the Average Computing the Variance

4. 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)

5. 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

6. Number of bits sent per node for each level in a 3-tree of depth 7 lev e l num odes A(0%)A(10%)A(30%) IA(10 %) IA(30 %) 137595027008182 29753669507778 3277517236676 4817510717276 5243758510876 672975788576 7218775 76

7. 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

9. Key generation 1. ID [0, P-1], where P is a large integer 2. Let i be a randomly generated keystream, where i [0, P-1], sink keep i secret 3. Compute K = ID – i (mod P) 4. 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)

10. Additive Homomorphic Encryption Key generation : k

11. Example IDKEYmc ID12121527 ID25151732 ID315254166 ID410203555 ID1ID2ID3ID4 sink aggregation nodes sensor nodes No respond K = ID – i (mod P) i = 40, M = 100, P = 50 c = m + k (mod M) 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 27, 566, 10