int RSAKey::Generate( int bits )
{
Bignum pm1, qm1, phi_n;
/*
* We don't generate e; we just use a standard one always.
*/
this->exponent = bignum_from_long(RSA_EXPONENT);
/*
* Generate p and q: primes with combined length `bits', not
* congruent to 1 modulo e. (Strictly speaking, we wanted (p-1)
* and e to be coprime, and (q-1) and e to be coprime, but in
* general that's slightly more fiddly to arrange. By choosing
* a prime e, we can simplify the criterion.)
*/
this->p = primegen(bits / 2, RSA_EXPONENT, 1, NULL, 1);
this->q = primegen(bits - bits / 2, RSA_EXPONENT, 1, NULL, 2);
/*
* Ensure p > q, by swapping them if not.
*/
if (bignum_cmp(this->p, this->q) < 0)
swap( p, q );
/*
* Now we have p, q and e. All we need to do now is work out
* the other helpful quantities: n=pq, d=e^-1 mod (p-1)(q-1),
* and (q^-1 mod p).
*/
this->modulus = bigmul(this->p, this->q);
pm1 = copybn(this->p);
decbn(pm1);
qm1 = copybn(this->q);
decbn(qm1);
phi_n = bigmul(pm1, qm1);
freebn(pm1);
freebn(qm1);
this->private_exponent = modinv(this->exponent, phi_n);
this->iqmp = modinv(this->q, this->p);
/*
* Clean up temporary numbers.
*/
freebn(phi_n);
return 1;
}
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;
}
