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Re: [Taler] S, B, B^-1
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From: |
Luis Ressel |
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Subject: |
Re: [Taler] S, B, B^-1 |
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Date: |
Thu, 1 Oct 2015 16:23:34 +0200 |
On Thu, 01 Oct 2015 09:25:15 +0200
Raphael Arias <address@hidden> wrote:
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> Hi,
>
> On 09/30/2015 03:06 PM, address@hidden wrote:
> > On the next to last page:
> >
> > "S_K(.) and B_b(.) commute" i don't think that is true..
> >
> > and unblinding is not the inverse of blinding (at least in RSA).
> >
> > All that is given is: U(S(B(X))) = S(X)
>
> I can agree with at least the last 2 lines of this. I am not sure
> about the first point.
>
> Best regards,
> Raphael
With Chaum's standard RSA blinding scheme, the first claim is true as
well: B_b(S_K(m)) = b^e * m^d mod n /= b * m^d mod n = S_K(B_b(m)).
Regards,
Luis Ressel