/*
* DH stage 1: invent a number x between 1 and q, and compute e =
* g^x mod p. Return e.
*
* If `nbits' is greater than zero, it is used as an upper limit
* for the number of bits in x. This is safe provided that (a) you
* use twice as many bits in x as the number of bits you expect to
* use in your session key, and (b) the DH group is a safe prime
* (which SSH demands that it must be).
*
* P. C. van Oorschot, M. J. Wiener
* "On Diffie-Hellman Key Agreement with Short Exponents".
* Advances in Cryptology: Proceedings of Eurocrypt '96
* Springer-Verlag, May 1996.
*/
Bignum dh_create_e(void *handle, int nbits)
{
struct dh_ctx *ctx = (struct dh_ctx *)handle;
int i;
int nbytes;
unsigned char *buf;
nbytes = ssh1_bignum_length(ctx->qmask);
buf = snewn(nbytes, unsigned char);
do {
/*
* Create a potential x, by ANDing a string of random bytes
* with qmask.
*/
if (ctx->x)
freebn(ctx->x);
if (nbits == 0 || nbits > bignum_bitcount(ctx->qmask)) {
ssh1_write_bignum(buf, ctx->qmask);
for (i = 2; i < nbytes; i++)
buf[i] &= random_byte();
ssh1_read_bignum(buf, nbytes, &ctx->x); /* can't fail */
} else {
int b, nb;
ctx->x = bn_power_2(nbits);
b = nb = 0;
for (i = 0; i < nbits; i++) {
if (nb == 0) {
nb = 8;
b = random_byte();
}
bignum_set_bit(ctx->x, i, b & 1);
b >>= 1;
nb--;
}
}
} while (bignum_cmp(ctx->x, One) <= 0 || bignum_cmp(ctx->x, ctx->q) >= 0);
sfree(buf);
/*
* Done. Now compute e = g^x mod p.
*/
ctx->e = modpow(ctx->g, ctx->x, ctx->p);
return ctx->e;
}
/*
* This function is a wrapper on modpow(). It has the same effect
* as modpow(), but employs RSA blinding to protect against timing
* attacks.
*/
static Bignum rsa_privkey_op(Bignum input, struct RSAKey *key)
{
Bignum random, random_encrypted, random_inverse;
Bignum input_blinded, ret_blinded;
Bignum ret;
SHA512_State ss;
unsigned char digest512[64];
int digestused = lenof(digest512);
int hashseq = 0;
/*
* Start by inventing a random number chosen uniformly from the
* range 2..modulus-1. (We do this by preparing a random number
* of the right length and retrying if it's greater than the
* modulus, to prevent any potential Bleichenbacher-like
* attacks making use of the uneven distribution within the
* range that would arise from just reducing our number mod n.
* There are timing implications to the potential retries, of
* course, but all they tell you is the modulus, which you
* already knew.)
*
* To preserve determinism and avoid Pageant needing to share
* the random number pool, we actually generate this `random'
* number by hashing stuff with the private key.
*/
while (1) {
int bits, byte, bitsleft, v;
random = copybn(key->modulus);
/*
* Find the topmost set bit. (This function will return its
* index plus one.) Then we'll set all bits from that one
* downwards randomly.
*/
bits = bignum_bitcount(random);
byte = 0;
bitsleft = 0;
while (bits--) {
if (bitsleft <= 0) {
bitsleft = 8;
/*
* Conceptually the following few lines are equivalent to
* byte = random_byte();
*/
if (digestused >= lenof(digest512)) {
unsigned char seqbuf[4];
PUT_32BIT(seqbuf, hashseq);
pSHA512_Init(&ss);
SHA512_Bytes(&ss, "RSA deterministic blinding", 26);
SHA512_Bytes(&ss, seqbuf, sizeof(seqbuf));
sha512_mpint(&ss, key->private_exponent);
pSHA512_Final(&ss, digest512);
hashseq++;
/*
* Now hash that digest plus the signature
* input.
*/
pSHA512_Init(&ss);
SHA512_Bytes(&ss, digest512, sizeof(digest512));
sha512_mpint(&ss, input);
pSHA512_Final(&ss, digest512);
digestused = 0;
}
byte = digest512[digestused++];
}
v = byte & 1;
byte >>= 1;
bitsleft--;
bignum_set_bit(random, bits, v);
}
/*
* Now check that this number is strictly greater than
* zero, and strictly less than modulus.
*/
if (bignum_cmp(random, Zero) <= 0 ||
bignum_cmp(random, key->modulus) >= 0) {
freebn(random);
continue;
} else {
break;
}
}
/*
* RSA blinding relies on the fact that (xy)^d mod n is equal
* to (x^d mod n) * (y^d mod n) mod n. We invent a random pair
* y and y^d; then we multiply x by y, raise to the power d mod
* n as usual, and divide by y^d to recover x^d. Thus an
* attacker can't correlate the timing of the modpow with the
* input, because they don't know anything about the number
* that was input to the actual modpow.
*
* The clever bit is that we don't have to do a huge modpow to
* get y and y^d; we will use the number we just invented as
* _y^d_, and use the _public_ exponent to compute (y^d)^e = y
* from it, which is much faster to do.
*/
random_encrypted = modpow(random, key->exponent, key->modulus);
random_inverse = modinv(random, key->modulus);
input_blinded = modmul(input, random_encrypted, key->modulus);
ret_blinded = modpow(input_blinded, key->private_exponent, key->modulus);
ret = modmul(ret_blinded, random_inverse, key->modulus);
freebn(ret_blinded);
freebn(input_blinded);
freebn(random_inverse);
freebn(random_encrypted);
freebn(random);
return ret;
}
Bignum bn_power_2(int n)
{
Bignum ret = newbn(n / BIGNUM_INT_BITS + 1);
bignum_set_bit(ret, n, 1);
return ret;
}
int dsa_generate(struct dss_key *key, int bits, progfn_t pfn,
void *pfnparam)
{
Bignum qm1, power, g, h, tmp;
unsigned pfirst, qfirst;
int progress;
/*
* Set up the phase limits for the progress report. We do this
* by passing minus the phase number.
*
* For prime generation: our initial filter finds things
* coprime to everything below 2^16. Computing the product of
* (p-1)/p for all prime p below 2^16 gives about 20.33; so
* among B-bit integers, one in every 20.33 will get through
* the initial filter to be a candidate prime.
*
* Meanwhile, we are searching for primes in the region of 2^B;
* since pi(x) ~ x/log(x), when x is in the region of 2^B, the
* prime density will be d/dx pi(x) ~ 1/log(B), i.e. about
* 1/0.6931B. So the chance of any given candidate being prime
* is 20.33/0.6931B, which is roughly 29.34 divided by B.
*
* So now we have this probability P, we're looking at an
* exponential distribution with parameter P: we will manage in
* one attempt with probability P, in two with probability
* P(1-P), in three with probability P(1-P)^2, etc. The
* probability that we have still not managed to find a prime
* after N attempts is (1-P)^N.
*
* We therefore inform the progress indicator of the number B
* (29.34/B), so that it knows how much to increment by each
* time. We do this in 16-bit fixed point, so 29.34 becomes
* 0x1D.57C4.
*/
pfn(pfnparam, PROGFN_PHASE_EXTENT, 1, 0x2800);
pfn(pfnparam, PROGFN_EXP_PHASE, 1, -0x1D57C4 / 160);
pfn(pfnparam, PROGFN_PHASE_EXTENT, 2, 0x40 * bits);
pfn(pfnparam, PROGFN_EXP_PHASE, 2, -0x1D57C4 / bits);
/*
* In phase three we are finding an order-q element of the
* multiplicative group of p, by finding an element whose order
* is _divisible_ by q and raising it to the power of (p-1)/q.
* _Most_ elements will have order divisible by q, since for a
* start phi(p) of them will be primitive roots. So
* realistically we don't need to set this much below 1 (64K).
* Still, we'll set it to 1/2 (32K) to be on the safe side.
*/
pfn(pfnparam, PROGFN_PHASE_EXTENT, 3, 0x2000);
pfn(pfnparam, PROGFN_EXP_PHASE, 3, -32768);
/*
* In phase four we are finding an element x between 1 and q-1
* (exclusive), by inventing 160 random bits and hoping they
* come out to a plausible number; so assuming q is uniformly
* distributed between 2^159 and 2^160, the chance of any given
* attempt succeeding is somewhere between 0.5 and 1. Lacking
* the energy to arrange to be able to specify this probability
* _after_ generating q, we'll just set it to 0.75.
*/
pfn(pfnparam, PROGFN_PHASE_EXTENT, 4, 0x2000);
pfn(pfnparam, PROGFN_EXP_PHASE, 4, -49152);
pfn(pfnparam, PROGFN_READY, 0, 0);
invent_firstbits(&pfirst, &qfirst);
/*
* Generate q: a prime of length 160.
*/
key->q = primegen(160, 2, 2, NULL, 1, pfn, pfnparam, qfirst);
/*
* Now generate p: a prime of length `bits', such that p-1 is
* divisible by q.
*/
key->p = primegen(bits-160, 2, 2, key->q, 2, pfn, pfnparam, pfirst);
/*
* Next we need g. Raise 2 to the power (p-1)/q modulo p, and
* if that comes out to one then try 3, then 4 and so on. As
* soon as we hit a non-unit (and non-zero!) one, that'll do
* for g.
*/
power = bigdiv(key->p, key->q); /* this is floor(p/q) == (p-1)/q */
h = bignum_from_long(1);
progress = 0;
while (1) {
pfn(pfnparam, PROGFN_PROGRESS, 3, ++progress);
g = modpow(h, power, key->p);
if (bignum_cmp(g, One) > 0)
break; /* got one */
tmp = h;
h = bignum_add_long(h, 1);
freebn(tmp);
}
key->g = g;
freebn(h);
/*
* Now we're nearly done. All we need now is our private key x,
* which should be a number between 1 and q-1 exclusive, and
* our public key y = g^x mod p.
*/
qm1 = copybn(key->q);
decbn(qm1);
progress = 0;
while (1) {
int i, v, byte, bitsleft;
Bignum x;
pfn(pfnparam, PROGFN_PROGRESS, 4, ++progress);
x = bn_power_2(159);
byte = 0;
bitsleft = 0;
for (i = 0; i < 160; i++) {
if (bitsleft <= 0)
bitsleft = 8, byte = random_byte();
v = byte & 1;
byte >>= 1;
bitsleft--;
bignum_set_bit(x, i, v);
}
if (bignum_cmp(x, One) <= 0 || bignum_cmp(x, qm1) >= 0) {
freebn(x);
continue;
} else {
key->x = x;
break;
}
}
freebn(qm1);
key->y = modpow(key->g, key->x, key->p);
return 1;
}
/*
* Generate a prime. We arrange to select a prime with the property
* (prime % modulus) != residue (to speed up use in RSA).
*/
Bignum primegen(int bits, int modulus, int residue,
int phase, progfn_t pfn, void *pfnparam) {
int i, k, v, byte, bitsleft, check, checks;
unsigned long delta, moduli[NPRIMES+1], residues[NPRIMES+1];
Bignum p, pm1, q, wqp, wqp2;
int progress = 0;
byte = 0; bitsleft = 0;
STARTOVER:
pfn(pfnparam, phase, ++progress);
/*
* Generate a k-bit random number with top and bottom bits set.
*/
p = newbn((bits+15)/16);
for (i = 0; i < bits; i++) {
if (i == 0 || i == bits-1)
v = 1;
else {
if (bitsleft <= 0)
bitsleft = 8; byte = random_byte();
v = byte & 1;
byte >>= 1;
bitsleft--;
}
bignum_set_bit(p, i, v);
}
/*
* Ensure this random number is coprime to the first few
* primes, by repeatedly adding 2 to it until it is.
*/
for (i = 0; i < NPRIMES; i++) {
moduli[i] = primes[i];
residues[i] = bignum_mod_short(p, primes[i]);
}
moduli[NPRIMES] = modulus;
residues[NPRIMES] = (bignum_mod_short(p, (unsigned short)modulus)
+ modulus - residue);
delta = 0;
while (1) {
for (i = 0; i < (sizeof(moduli) / sizeof(*moduli)); i++)
if (!((residues[i] + delta) % moduli[i]))
break;
if (i < (sizeof(moduli) / sizeof(*moduli))) {/* we broke */
delta += 2;
if (delta < 2) {
freebn(p);
goto STARTOVER;
}
continue;
}
break;
}
q = p;
p = bignum_add_long(q, delta);
freebn(q);
/*
* Now apply the Miller-Rabin primality test a few times. First
* work out how many checks are needed.
*/
checks = 27;
if (bits >= 150) checks = 18;
if (bits >= 200) checks = 15;
if (bits >= 250) checks = 12;
if (bits >= 300) checks = 9;
if (bits >= 350) checks = 8;
if (bits >= 400) checks = 7;
if (bits >= 450) checks = 6;
if (bits >= 550) checks = 5;
if (bits >= 650) checks = 4;
if (bits >= 850) checks = 3;
if (bits >= 1300) checks = 2;
/*
* Next, write p-1 as q*2^k.
*/
for (k = 0; bignum_bit(p, k) == !k; k++); /* find first 1 bit in p-1 */
q = bignum_rshift(p, k);
/* And store p-1 itself, which we'll need. */
pm1 = copybn(p);
decbn(pm1);
/*
* Now, for each check ...
*/
for (check = 0; check < checks; check++) {
Bignum w;
/*
* Invent a random number between 1 and p-1 inclusive.
*/
while (1) {
w = newbn((bits+15)/16);
for (i = 0; i < bits; i++) {
if (bitsleft <= 0)
bitsleft = 8; byte = random_byte();
v = byte & 1;
byte >>= 1;
bitsleft--;
bignum_set_bit(w, i, v);
}
if (bignum_cmp(w, p) >= 0 || bignum_cmp(w, Zero) == 0) {
freebn(w);
continue;
}
break;
}
pfn(pfnparam, phase, ++progress);
/*
* Compute w^q mod p.
*/
wqp = modpow(w, q, p);
freebn(w);
/*
* See if this is 1, or if it is -1, or if it becomes -1
* when squared at most k-1 times.
*/
if (bignum_cmp(wqp, One) == 0 || bignum_cmp(wqp, pm1) == 0) {
freebn(wqp);
continue;
}
for (i = 0; i < k-1; i++) {
wqp2 = modmul(wqp, wqp, p);
freebn(wqp);
wqp = wqp2;
if (bignum_cmp(wqp, pm1) == 0)
break;
}
if (i < k-1) {
freebn(wqp);
continue;
}
/*
* It didn't. Therefore, w is a witness for the
* compositeness of p.
*/
freebn(p);
freebn(pm1);
freebn(q);
goto STARTOVER;
}
/*
* We have a prime!
*/
freebn(q);
freebn(pm1);
return p;
}