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mirror of https://git.tartarus.org/simon/putty.git synced 2025-01-09 17:38:00 +00:00
putty-source/sshprime.c
Simon Tatham aba52744e4 primegen: fix a small memory leak.
There's always one.
2020-02-23 18:34:18 +00:00

243 lines
8.0 KiB
C

/*
* Prime generation.
*/
#include <assert.h>
#include "ssh.h"
#include "mpint.h"
#include "mpunsafe.h"
#include "sshkeygen.h"
/*
* This prime generation algorithm is pretty much cribbed from
* OpenSSL. The algorithm is:
*
* - invent a B-bit random number and ensure the top and bottom
* bits are set (so it's definitely B-bit, and it's definitely
* odd)
*
* - see if it's coprime to all primes below 2^16; increment it by
* two until it is (this shouldn't take long in general)
*
* - perform the Miller-Rabin primality test enough times to
* ensure the probability of it being composite is 2^-80 or
* less
*
* - go back to square one if any M-R test fails.
*/
/*
* The Miller-Rabin primality test is an extension to the Fermat
* test. The Fermat test just checks that a^(p-1) == 1 mod p; this
* is vulnerable to Carmichael numbers. Miller-Rabin considers how
* that 1 is derived as well.
*
* Lemma: if a^2 == 1 (mod p), and p is prime, then either a == 1
* or a == -1 (mod p).
*
* Proof: p divides a^2-1, i.e. p divides (a+1)(a-1). Hence,
* since p is prime, either p divides (a+1) or p divides (a-1).
* But this is the same as saying that either a is congruent to
* -1 mod p or a is congruent to +1 mod p. []
*
* Comment: This fails when p is not prime. Consider p=mn, so
* that mn divides (a+1)(a-1). Now we could have m dividing (a+1)
* and n dividing (a-1), without the whole of mn dividing either.
* For example, consider a=10 and p=99. 99 = 9 * 11; 9 divides
* 10-1 and 11 divides 10+1, so a^2 is congruent to 1 mod p
* without a having to be congruent to either 1 or -1.
*
* So the Miller-Rabin test, as well as considering a^(p-1),
* considers a^((p-1)/2), a^((p-1)/4), and so on as far as it can
* go. In other words. we write p-1 as q * 2^k, with k as large as
* possible (i.e. q must be odd), and we consider the powers
*
* a^(q*2^0) a^(q*2^1) ... a^(q*2^(k-1)) a^(q*2^k)
* i.e. a^((n-1)/2^k) a^((n-1)/2^(k-1)) ... a^((n-1)/2) a^(n-1)
*
* If p is to be prime, the last of these must be 1. Therefore, by
* the above lemma, the one before it must be either 1 or -1. And
* _if_ it's 1, then the one before that must be either 1 or -1,
* and so on ... In other words, we expect to see a trailing chain
* of 1s preceded by a -1. (If we're unlucky, our trailing chain of
* 1s will be as long as the list so we'll never get to see what
* lies before it. This doesn't count as a test failure because it
* hasn't _proved_ that p is not prime.)
*
* For example, consider a=2 and p=1729. 1729 is a Carmichael
* number: although it's not prime, it satisfies a^(p-1) == 1 mod p
* for any a coprime to it. So the Fermat test wouldn't have a
* problem with it at all, unless we happened to stumble on an a
* which had a common factor.
*
* So. 1729 - 1 equals 27 * 2^6. So we look at
*
* 2^27 mod 1729 == 645
* 2^108 mod 1729 == 1065
* 2^216 mod 1729 == 1
* 2^432 mod 1729 == 1
* 2^864 mod 1729 == 1
* 2^1728 mod 1729 == 1
*
* We do have a trailing string of 1s, so the Fermat test would
* have been happy. But this trailing string of 1s is preceded by
* 1065; whereas if 1729 were prime, we'd expect to see it preceded
* by -1 (i.e. 1728.). Guards! Seize this impostor.
*
* (If we were unlucky, we might have tried a=16 instead of a=2;
* now 16^27 mod 1729 == 1, so we would have seen a long string of
* 1s and wouldn't have seen the thing _before_ the 1s. So, just
* like the Fermat test, for a given p there may well exist values
* of a which fail to show up its compositeness. So we try several,
* just like the Fermat test. The difference is that Miller-Rabin
* is not _in general_ fooled by Carmichael numbers.)
*
* Put simply, then, the Miller-Rabin test requires us to:
*
* 1. write p-1 as q * 2^k, with q odd
* 2. compute z = (a^q) mod p.
* 3. report success if z == 1 or z == -1.
* 4. square z at most k-1 times, and report success if it becomes
* -1 at any point.
* 5. report failure otherwise.
*
* (We expect z to become -1 after at most k-1 squarings, because
* if it became -1 after k squarings then a^(p-1) would fail to be
* 1. And we don't need to investigate what happens after we see a
* -1, because we _know_ that -1 squared is 1 modulo anything at
* all, so after we've seen a -1 we can be sure of seeing nothing
* but 1s.)
*/
/*
* Generate a prime. We can deal with various extra properties of
* the prime:
*
* - to speed up use in RSA, we can arrange to select a prime with
* the property (prime % modulus) != residue.
*
* - for use in DSA, we can arrange to select a prime which is one
* more than a multiple of a dirty great bignum. In this case
* `bits' gives the size of the factor by which we _multiply_
* that bignum, rather than the size of the whole number.
*
* - for the basically cosmetic purposes of generating keys of the
* length actually specified rather than off by one bit, we permit
* the caller to provide an unsigned integer 'firstbits' which will
* match the top few bits of the returned prime. (That is, there
* will exist some n such that (returnvalue >> n) == firstbits.) If
* 'firstbits' is not needed, specifying it to either 0 or 1 is
* an adequate no-op.
*/
mp_int *primegen(
int bits, int modulus, int residue, mp_int *factor,
int phase, progfn_t pfn, void *pfnparam, unsigned firstbits)
{
int progress = 0;
size_t fbsize = 0;
while (firstbits >> fbsize) /* work out how to align this */
fbsize++;
PrimeCandidateSource *pcs = pcs_new(bits, firstbits, fbsize);
if (factor)
pcs_require_residue_1(pcs, factor);
if (modulus)
pcs_avoid_residue_small(pcs, modulus, residue);
pcs_ready(pcs);
STARTOVER:
pfn(pfnparam, PROGFN_PROGRESS, phase, ++progress);
mp_int *p = pcs_generate(pcs);
/*
* Now apply the Miller-Rabin primality test a few times. First
* work out how many checks are needed.
*/
unsigned checks =
bits >= 1300 ? 2 : bits >= 850 ? 3 : bits >= 650 ? 4 :
bits >= 550 ? 5 : bits >= 450 ? 6 : bits >= 400 ? 7 :
bits >= 350 ? 8 : bits >= 300 ? 9 : bits >= 250 ? 12 :
bits >= 200 ? 15 : bits >= 150 ? 18 : 27;
/*
* Next, write p-1 as q*2^k.
*/
size_t k;
for (k = 0; mp_get_bit(p, k) == !k; k++)
continue; /* find first 1 bit in p-1 */
mp_int *q = mp_rshift_safe(p, k);
/*
* Set up stuff for the Miller-Rabin checks.
*/
mp_int *two = mp_from_integer(2);
mp_int *pm1 = mp_copy(p);
mp_sub_integer_into(pm1, pm1, 1);
MontyContext *mc = monty_new(p);
mp_int *m_pm1 = monty_import(mc, pm1);
bool known_bad = false;
/*
* Now, for each check ...
*/
for (unsigned check = 0; check < checks && !known_bad; check++) {
/*
* Invent a random number between 1 and p-1.
*/
mp_int *w = mp_random_in_range(two, pm1);
monty_import_into(mc, w, w);
pfn(pfnparam, PROGFN_PROGRESS, phase, ++progress);
/*
* Compute w^q mod p.
*/
mp_int *wqp = monty_pow(mc, w, q);
mp_free(w);
/*
* See if this is 1, or if it is -1, or if it becomes -1
* when squared at most k-1 times.
*/
bool passed = false;
if (mp_cmp_eq(wqp, monty_identity(mc)) || mp_cmp_eq(wqp, m_pm1)) {
passed = true;
} else {
for (size_t i = 0; i < k - 1; i++) {
monty_mul_into(mc, wqp, wqp, wqp);
if (mp_cmp_eq(wqp, m_pm1)) {
passed = true;
break;
}
}
}
if (!passed)
known_bad = true;
mp_free(wqp);
}
mp_free(q);
mp_free(two);
mp_free(pm1);
monty_free(mc);
mp_free(m_pm1);
if (known_bad) {
mp_free(p);
goto STARTOVER;
}
/*
* We have a prime!
*/
pcs_free(pcs);
return p;
}