RSA key generation routines, and the bignum enhancements required to

[u/mdw/putty] / sshrsag.c

/*
 * RSA key generation.
 */

#include "ssh.h"

#define RSA_EXPONENT 37                /* we like this prime */

static void diagbn(char *prefix, Bignum md) {
    int i, nibbles, morenibbles;
    static const char hex[] = "0123456789ABCDEF";

    printf("%s0x", prefix ? prefix : "");

    nibbles = (3 + ssh1_bignum_bitcount(md))/4; if (nibbles<1) nibbles=1;
    morenibbles = 4*md[0] - nibbles;
    for (i=0; i<morenibbles; i++) putchar('-');
    for (i=nibbles; i-- ;)
        putchar(hex[(bignum_byte(md, i/2) >> (4*(i%2))) & 0xF]);

    if (prefix) putchar('\n');
}

int rsa_generate(struct RSAKey *key, struct RSAAux *aux, int bits,
                 progfn_t pfn, void *pfnparam) {
    Bignum pm1, qm1, phi_n;

    /*
     * 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, -1, -0x1D57C4/(bits/2));
    pfn(pfnparam, -2, -0x1D57C4/(bits-bits/2));
    pfn(pfnparam, -3, 5);

    /*
     * We don't generate e; we just use a standard one always.
     */
    key->exponent = bignum_from_short(RSA_EXPONENT);
    diagbn("e = ",key->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.)
     */
    aux->p = primegen(bits/2, RSA_EXPONENT, 1, 1, pfn, pfnparam);
    aux->q = primegen(bits - bits/2, RSA_EXPONENT, 1, 2, pfn, pfnparam);

    /*
     * Ensure p > q, by swapping them if not.
     */
    if (bignum_cmp(aux->p, aux->q) < 0) {
        Bignum t = aux->p;
        aux->p = aux->q;
        aux->q = t;
    }

    /*
     * 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).
     */
    pfn(pfnparam, 3, 1);
    key->modulus = bigmul(aux->p, aux->q);
    pfn(pfnparam, 3, 2);
    pm1 = copybn(aux->p);
    decbn(pm1);
    qm1 = copybn(aux->q);
    decbn(qm1);
    phi_n = bigmul(pm1, qm1);
    pfn(pfnparam, 3, 3);
    freebn(pm1);
    freebn(qm1);
    diagbn("p = ", aux->p);
    diagbn("q = ", aux->q);
    diagbn("e = ", key->exponent);
    diagbn("n = ", key->modulus);
    diagbn("phi(n) = ", phi_n);
    key->private_exponent = modinv(key->exponent, phi_n);
    pfn(pfnparam, 3, 4);
    diagbn("d = ", key->private_exponent);
    aux->iqmp = modinv(aux->q, aux->p);
    pfn(pfnparam, 3, 5);
    diagbn("iqmp = ", aux->iqmp);

    /*
     * Clean up temporary numbers.
     */
    freebn(phi_n);

    return 1;
}