static unsigned char *dss_sign(void *key, char *data, int datalen, int *siglen)
{
struct dss_key *dss = (struct dss_key *) key;
Bignum k, gkp, hash, kinv, hxr, r, s;
unsigned char digest[20];
unsigned char *bytes;
int nbytes, i;
SHA_Simple(data, datalen, digest);
k = dss_gen_k("DSA deterministic k generator", dss->q, dss->x,
digest, sizeof(digest));
kinv = modinv(k, dss->q); /* k^-1 mod q */
assert(kinv);
/*
* Now we have k, so just go ahead and compute the signature.
*/
gkp = modpow(dss->g, k, dss->p); /* g^k mod p */
r = bigmod(gkp, dss->q); /* r = (g^k mod p) mod q */
freebn(gkp);
hash = bignum_from_bytes(digest, 20);
hxr = bigmuladd(dss->x, r, hash); /* hash + x*r */
s = modmul(kinv, hxr, dss->q); /* s = k^-1 * (hash + x*r) mod q */
freebn(hxr);
freebn(kinv);
freebn(k);
freebn(hash);
/*
* Signature blob is
*
* string "ssh-dss"
* string two 20-byte numbers r and s, end to end
*
* i.e. 4+7 + 4+40 bytes.
*/
nbytes = 4 + 7 + 4 + 40;
bytes = snewn(nbytes, unsigned char);
PUT_32BIT(bytes, 7);
memcpy(bytes + 4, "ssh-dss", 7);
PUT_32BIT(bytes + 4 + 7, 40);
for (i = 0; i < 20; i++) {
bytes[4 + 7 + 4 + i] = bignum_byte(r, 19 - i);
bytes[4 + 7 + 4 + 20 + i] = bignum_byte(s, 19 - i);
}
freebn(r);
freebn(s);
*siglen = nbytes;
return bytes;
}
static unsigned char *dss_sign(void *key, char *data, int datalen, int *siglen)
{
/*
* The basic DSS signing algorithm is:
*
* - invent a random k between 1 and q-1 (exclusive).
* - Compute r = (g^k mod p) mod q.
* - Compute s = k^-1 * (hash + x*r) mod q.
*
* This has the dangerous properties that:
*
* - if an attacker in possession of the public key _and_ the
* signature (for example, the host you just authenticated
* to) can guess your k, he can reverse the computation of s
* and work out x = r^-1 * (s*k - hash) mod q. That is, he
* can deduce the private half of your key, and masquerade
* as you for as long as the key is still valid.
*
* - since r is a function purely of k and the public key, if
* the attacker only has a _range of possibilities_ for k
* it's easy for him to work through them all and check each
* one against r; he'll never be unsure of whether he's got
* the right one.
*
* - if you ever sign two different hashes with the same k, it
* will be immediately obvious because the two signatures
* will have the same r, and moreover an attacker in
* possession of both signatures (and the public key of
* course) can compute k = (hash1-hash2) * (s1-s2)^-1 mod q,
* and from there deduce x as before.
*
* - the Bleichenbacher attack on DSA makes use of methods of
* generating k which are significantly non-uniformly
* distributed; in particular, generating a 160-bit random
* number and reducing it mod q is right out.
*
* For this reason we must be pretty careful about how we
* generate our k. Since this code runs on Windows, with no
* particularly good system entropy sources, we can't trust our
* RNG itself to produce properly unpredictable data. Hence, we
* use a totally different scheme instead.
*
* What we do is to take a SHA-512 (_big_) hash of the private
* key x, and then feed this into another SHA-512 hash that
* also includes the message hash being signed. That is:
*
* proto_k = SHA512 ( SHA512(x) || SHA160(message) )
*
* This number is 512 bits long, so reducing it mod q won't be
* noticeably non-uniform. So
*
* k = proto_k mod q
*
* This has the interesting property that it's _deterministic_:
* signing the same hash twice with the same key yields the
* same signature.
*
* Despite this determinism, it's still not predictable to an
* attacker, because in order to repeat the SHA-512
* construction that created it, the attacker would have to
* know the private key value x - and by assumption he doesn't,
* because if he knew that he wouldn't be attacking k!
*
* (This trick doesn't, _per se_, protect against reuse of k.
* Reuse of k is left to chance; all it does is prevent
* _excessively high_ chances of reuse of k due to entropy
* problems.)
*
* Thanks to Colin Plumb for the general idea of using x to
* ensure k is hard to guess, and to the Cambridge University
* Computer Security Group for helping to argue out all the
* fine details.
*/
struct dss_key *dss = (struct dss_key *) key;
SHA512_State ss;
unsigned char digest[20], digest512[64];
Bignum proto_k, k, gkp, hash, kinv, hxr, r, s;
unsigned char *bytes;
int nbytes, i;
SHA_Simple(data, datalen, digest);
/*
* Hash some identifying text plus x.
*/
SHA512_Init(&ss);
SHA512_Bytes(&ss, "DSA deterministic k generator", 30);
sha512_mpint(&ss, dss->x);
SHA512_Final(&ss, digest512);
/*
* Now hash that digest plus the message hash.
*/
SHA512_Init(&ss);
SHA512_Bytes(&ss, digest512, sizeof(digest512));
SHA512_Bytes(&ss, digest, sizeof(digest));
SHA512_Final(&ss, digest512);
memset(&ss, 0, sizeof(ss));
/*
* Now convert the result into a bignum, and reduce it mod q.
*/
proto_k = bignum_from_bytes(digest512, 64);
k = bigmod(proto_k, dss->q);
freebn(proto_k);
memset(digest512, 0, sizeof(digest512));
/*
* Now we have k, so just go ahead and compute the signature.
*/
gkp = modpow(dss->g, k, dss->p); /* g^k mod p */
r = bigmod(gkp, dss->q); /* r = (g^k mod p) mod q */
freebn(gkp);
hash = bignum_from_bytes(digest, 20);
kinv = modinv(k, dss->q); /* k^-1 mod q */
hxr = bigmuladd(dss->x, r, hash); /* hash + x*r */
s = modmul(kinv, hxr, dss->q); /* s = k^-1 * (hash + x*r) mod q */
freebn(hxr);
freebn(kinv);
freebn(hash);
/*
* Signature blob is
*
* string "ssh-dss"
* string two 20-byte numbers r and s, end to end
*
* i.e. 4+7 + 4+40 bytes.
*/
nbytes = 4 + 7 + 4 + 40;
bytes = snewn(nbytes, unsigned char);
PUT_32BIT(bytes, 7);
memcpy(bytes + 4, "ssh-dss", 7);
PUT_32BIT(bytes + 4 + 7, 40);
for (i = 0; i < 20; i++) {
bytes[4 + 7 + 4 + i] = bignum_byte(r, 19 - i);
bytes[4 + 7 + 4 + 20 + i] = bignum_byte(s, 19 - i);
}
freebn(r);
freebn(s);
*siglen = nbytes;
return bytes;
}
