On 21/09/16 20:10, Walter Ray-Dulany wrote: > Let me see if I understand you, but, if I do, then I would respond: "Yes." > >> can I simply think of this as a way that one of the encryption values is >> 'selected' during the encrypt process? > > When encrypting, because of our friendly random zeta, we can get many > different ciphertexts from the same message. As you point out, when taking > E(5)*E(7) we can (and in your case, do) get another valid-but-distinct > ciphertext corresponding to the message m=12. Additionally, there is some > choice of zeta that would make E(12) encrypt to 19 instead of 208.

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Yep, that's what I meant. So why does Pirk provoke a different secure random number for each encryption via Pailler#encrypt(BigInteger)? I don't see that it enhances the security, since whatever random value we choose the decryption is based solely upon knowing our p and q. Would it be ok to pick some zeta value when Paillier is initialized, and stick with it? Regards, Tim > On Wed, Sep 21, 2016 at 12:14 PM, Tim Ellison <t.p.elli...@gmail.com> wrote: > >> With apologies for the lazy language...since there can be multiple >> numbers in the encryption space that map back to the same plain text >> number, can I simply think of this as a way that one of the encryption >> values is 'selected' during the encrypt process? >> >> Taking my simple example, if I encrypt E() and decrypt D() the following >> to test the homomorphic properties: >> >> E(5 + 7) mod N = 208 >> D(E(5) * E(7) mod N^2) = D(19) >> >> hmm, but D(19) = D(208) = 12 so we are all good. >> >> Regards, >> Tim (hoping to get to some PIR soon!) >> >> On 21/09/16 13:25, Ellison Anne Williams wrote: >>> Ah, the math-magic of semantic encryption... :) (re: random zeta) >>> >>> We can certainly walk through the proof of the semantic encryption (the >>> random zeta) as it is quite mathematically beautiful, but it will take us >>> even further down the algebraic path. >>> >>> On Wed, Sep 21, 2016 at 8:19 AM, Tim Ellison <t.p.elli...@gmail.com> >> wrote: >>> >>>> On 19/09/16 18:36, Walter Ray-Dulany wrote: >>>> <snip/> >>>>> Let's see what we've got. >>>>> >>>>> ( (16**12)*(7**15) ) mod 225 = 208. >>>>> >>>>> I will leave it as an exercise to check that the decryption of 208 is >> in >>>>> fact 12. >>>> >>>> I like a challenge :-) >>>> >>>> So we got to p=3, q=5, and my encrypted value c=208. >>>> >>>> Following the Wideskies Pallier decryption algorithm, >>>> Step (2): >>>> N = p * q >>>> = 15 >>>> >>>> lambda(N) = lcm(p-1,q-1) >>>> = 4 >>>> >>>> Step (3): >>>> mu = lambda(N) modinverse N >>>> = 4 >>>> >>>> Step (4): >>>> c' = c^lambda(N) mod N^2 >>>> = 208^4 mod 225 >>>> = 46 >>>> >>>> Step(5): >>>> m' = L(c') >>>> = ((c' - 1) / N) mod N >>>> = (45 / 15) mod 15 >>>> = 3 >>>> >>>> Step(6): >>>> m = (m' * mu) mod N >>>> = 12 >>>> >>>> yay! >>>> >>>> The fog is slowly clearing, though I'm totally baffled about how I can >>>> pick a random zeta during encryption, and it plays no part in the >>>> decryption. >>>> >>>> Regards, >>>> Tim >>>> >>> >> >