Chaum's credentials (technical question)
-----BEGIN PGP SIGNED MESSAGE----- In response to the recent discussions about identity, pseudonyms, "is-a-person" credentials, etc., I've been studying Chaum's paper from Auscrypt 90, "Showing Credentials without Identification; Transferring Signatures between Unconditionally Unlinkable Pseudonyms." This is quite a dense and rather cryptic paper which requires careful reading. It doesn't help that the references got left off when the paper was printed. There are also quite a few obvious misprints in some of the printed formulas. I am trying to understand one particular passage, on page 258. Chaum uses the idea of a credential as an RSA signature on a pseudonym, where the pseudonym is a number Px. The RSA modulus has, in this case, two exponents e1 and e2 which mean different things. (Say, e1 means "good credit risk" and e2 means "good driving record".) The corresponding private exponents are d1 and d2. If a person has these two credentials that means that he has the two numbers Px^d1 and Px^d2, from the credentialling organization. These RSA signatures prove that he actually has the characteristics described in the credential. Now, I am having a problem with Chaum's math. This is a little technical but I know we have some people on the list who know some number theory. Here is what Chaum says: "Suppose an organization X were to require that you have each of two credentials, say both that with public exponents e1 and e2. You could send X separatley Px^d1 and Px^d2. It is also possible for you to use the two credentials to form the single credential Px^(d1*d2), which will be called their AND.... To create the AND, you: set g to the multiplicative inverse of d1 modulo d2; set h to the remainder after dividing g*d1-1 by d2; and computing (Px^d1)^g * (Px^d2)^(-h) = Px^(d1*d2)." It would be really nice if this AND credential could be created like this, because it might be applicable to digital cash. Instead of having to go through the complicated spending transaction for each piece of cash, you might be able to combine all the pieces of cash into one, and just spend that. It would be more compact. But Chaum's math doesn't work. First of all, he says "you" should set g to the inverse of d1 modulo d2. But this seems to presume knowledge of d1 and d2. Yet "you" don't know these things; these are the secret exponents of the signing agency. So is Chaum actually talking here about something the signing agency does? It didn't sound that way from the context. If the signing agency wants to compute Px^(d1*d2), given Px^d1 and Px^d2, it can do so easily enough; simply take Px^d1 to the d2 power. You don't need to go through this rigamarole with g and h. So that interpretation doesn't make much sense either. The other possibility I thought of is that he meant that the signing agency would make g and h, as he defined them, public. With g and h then users could combine their credentials as he said. But even that doesn't work; his whole formula doesn't make sense. g is the inverse of d1 mod d2; this means that g*d1 = 1 mod d2, or in other words g*d1 - 1 = k*d2 for some k. That's the definition of the multiplicative inverse. Okay, but then he says h is the remainder when g*d1-1 is divided by d2. But look: g*d1-1 is a MULTIPLE of g2! The remainder will always be zero. So that doesn't make any sense either. So I thought, perhaps he really meant that h should be the quotient rather than the remainder; it would be "k" in the equation I just wrote. Then we'd have g*d1 - 1 = h*d2, which is somewhat encouraging because it resembles his formula. But his formula is (Px^d1)^g * (Px^d2)^(-h), which is Px^(g*d1 - h*d2). Rearranging the equation two lines above, we see that g*d1 - h*d2 = 1. So we end up with just Px, not Px^(d1*d2). So this isn't right, either. In fact, the notion that you can calculate Px^(d1*d2) from Px^d1 and Px^d2 is pretty questionable, since the impossibility of doing this is the basis of Diffie-Hellman key exchange! In short, I haven't found any interpretation of Chaum's math that makes sense. Can anyone shed any light on this? Was this just a mistake in a paper which was, after all, just intended for conference proceedings, not a refereed journal? Thanks - Hal -----BEGIN PGP SIGNATURE----- Version: 2.3a iQCVAgUBLNLlRKgTA69YIUw3AQEmegP9HzQt1vMwuLvVVr2e3LNrL5lPh9jg/cb4 rQkTvh+XVCKlqsI7TJ2pCeAwLygxPMlcw4/3sAeV9K1hWqk0B+bSFU8qWQSmka5+ 2OpJIXt2C+N/qVMKiFzAKMmQf680iVUxdj/TvfV6kZ6hPA5eqHdnHWy45QKEck3B VMNwKRPz2Mo= =TLs8 -----END PGP SIGNATURE-----
Hal Finney writes about a paper of Chaum's, and near the end asks:
In short, I haven't found any interpretation of Chaum's math that makes sense. Can anyone shed any light on this?
I think you're correct, Hal. From everything I can tell, Chaum's confused product and greatest common divisor. First, though, there's a basic fact about the arithmetic of integers which everyone who wants to learn more algebra should know. Z is the set of integers. For every m,n in Z, there is a,b in Z such that a*m + b*n = gcd(m,n). One calculates the gcd by means of the Euclidean algorithm, and the coefficients a and b by an extension of that algorithm. Lots of basic algorithm books contain descriptions.
From an abstract point of view, this is a simple consequence that Z is a principal ideal domain. The ideal (m,n) is composed of the linear span of m and n. Since this ideal is principal, by definition there is a c such that (m,n) = (c). Clearly c is in the linear span of m and n, hence coefficients exist.
Note that reducing mod n gives a*m = gcd(m,n) (mod n). If m and n are relatively prime (means gcd = 1), the a is m inverse mod n. Likewise b is n inverse mod m. This is a standard algorithm for calculating modular inverses. Here's the relevant passage, again:
"Suppose an organization X were to require that you have each of two credentials, say both that with public exponents e1 and e2. You could send X separatley Px^d1 and Px^d2. It is also possible for you to use the two credentials to form the single credential Px^(d1*d2), which will be called their AND.... To create the AND, you: set g to the multiplicative inverse of d1 modulo d2; set h to the remainder after dividing g*d1-1 by d2; and computing (Px^d1)^g * (Px^d2)^(-h) = Px^(d1*d2)."
I believe that Hal is correct when he points out that h is not a remainder (which would be zero, as he points out) but the quotient. I originally misread this as quotient because I recognized the context. First, the multiplicative inverse of d1 (mod d2) exists only if the two are relatively prime. Hal did not quote the whole article, so I don't know if this criterion is stated elsewhere. Let us assume the d's have this property, since the d's can be so chosen. The calculation of d1 (mod d2) is exactly the calculation of the coefficients in the extended Euclidean algorithm above. Consider a*d1 + b*d2 = 1. Reducing mod d2, we have a*d1 = 1 (mod d2). That means that a is d1^(-1) (mod d2). Likewise (a*d1 - 1) / d2 = -b. Chaum's description exactly fits the gcd context.
The other possibility I thought of is that he meant that the signing agency would make g and h, as he defined them, public.
That's my interpretation as well. After the calculation, as Hal observes, you just end up with Px, not Px^(d1*d2). That's totally useless, since you already knew Px. It is certainly possible to create coefficients for combining credentials such that you end up with a product in the exponent. For example, the pairs <d2,0> and <0,d1> both work nicely, with the bad side effect that you've given away a private key. Let's try blinding them. Suppose you have coefficients a and b such that a*d1+b*d2=0; the pair <d2,-d1> works here. Then every such pair of product-combining coefficients can be represented as <d2,0> + r*<a,b>. Since the exponents are mod phi(N), we can suppose that the pair <d2+r*a,r*b> doesn't _directly_ reveal the private keys. But it's unclear to me that this pair of coefficients doesn't reveal d1 and d2. One doesn't know phi(N), but one may not need to. Eric
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