# Re: Math deck

```No - you would lose the semantic security. In other words, two encryptions
of the same plaintext would be the same two ciphertexts. With a random
zeta, I can encrypt two equal plaintexts and get two different ciphertexts.
This is particularly important in Wideskies when encrypting the E_i values.```
```
On Thu, Sep 22, 2016 at 5:02 AM, Tim Ellison <t.p.elli...@gmail.com> wrote:

> 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.
>
> 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
> >>>>
> >>>
> >>
> >
>
```