Jason Holt writes:

In his paper on Lucre ("2nd defence" against marking): http://anoncvs.aldigital.co.uk/lucre/

Ben Laurie gives this as a (possibly patent-free) blinding technique, where h is the message, and g is the public generator:

r = blind(h) = h^y * g^b (mod p)

To "sign",

s = sign(r) = m^h

To unblind,

(s/g^k^b)^(1/y) (mod p)

(where k is the signer's secret exponent. Of course, nobody but the signer can verify the signature). Unfortunately, this doesn't work with cut and choose where the signer signs the product of unrevealed documents, since the 1/y exponent above would distribute to all the internal terms:

Boy, you've got a lot of faith asking this question on cypherpunks. It's not exactly the intellectual center of the crypto freedom movement these days, you know. The average IQ is rapidly descending into double digits, even not counting Choate. But let's see what we can do for you. First, let's fix your notation. r = blind(h) = h^y * g^b OK s = sign(r) = r^k, not m^h. unblind(s) = (s/g^k^b)^(1/y) = h^k = sign(h). That's what you want to end up with, h^k, as the pseudo-signature on h. Now for a credential system, you apparently want to create a bunch of values which have some structure, and get a signature on a product of them. Using cut and choose, the client will prepare blinded forms of all of the values, then the server will ask for half of the blinding factors to be revealed. This exposes the raw values to be signed and the server can make sure they are in the right form. If so, it then signs the product of the remaining values, which the client unblinds to get back a good signature on the product of the unblinded values. The fundamental problem with this is that the blinding factors have to be different for each of the values. If they are all the same, then when they are revealed for some of the values during cut and choose, that will reveal them for all of them, and so none of them will be effectively blinded any more. But if the blinding factors are all different, we can't unblind since we don't have a unique power 1/y to raise to. That's your problem, right? Here are a couple of possible solutions. First, you could do a cut and choose in which all but one of the blinded values are revealed, and only the remaining (unrevealed) one is signed. This has the problem that it has only a 1/n security factor with n values. That is, the client can just guess which one the server won't ask to check, and if it sent say 100 values, it has a 1/100 chance of getting lucky, which might seem too high. However since credential issuing usually occurs in a non-anonymous context, you can afford to penalize people very heavily if they are caught in this manner. (Cutting the connection and refusing to resume with the previous values has to count as cheating.) Another approach is as follows. Go back to the 50-50 cut and choose with signature on the product. However, use the same y blinding factor for all of the values. Now when the client has to reveal during cut and choose, it keeps the y value secret but reveals all of the h and b values. It then proves in zero knowledge that there exists a y such that the h^y equals the required value. This is a standard ZK proof of knowledge of a discrete logarithm. It is similar to the example Ben's paper gives of how the bank can prove it is raising to the right power. Since you don't have to reveal y, you can use the same y for all of them and successfully perform the unblind operation, getting back the signature on the product of the h's as required. But actually another solution is much simpler, which is to do blinding as just h * g^b, without a y factor. That works fine as long as the bank is known not to be misbehaving. Ben's paper shows how the bank can use a ZK proof to show that it is raising to the same power k every time, basically again that same ZK proof regarding discrete logarithms. If the bank uses such a proof then you can use simpler blinding without a y factor, and you can recover the signature on the product of your h values by dividing by g^k^(sum of b's). So there you go. A little technical for cypherpunks, but unfortunately coderpunks, like the little old lady, has fallen and it can't get up.