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putty-source/keygen/millerrabin.c
Simon Tatham 6520574e58 Side-channel-safe rewrite of the Miller-Rabin test.
Thanks to Mark Wooding for explaining the method of doing this. At
first glance it seemed _obviously_ impossible to run an algorithm that
needs an iteration per factor of 2 in p-1, without a timing leak
giving away the number of factors of 2 in p-1. But it's not, because
you can do the M-R checks interleaved with each step of your whole
modular exponentiation, and they're cheap enough that you can do them
in _every_ step, even the ones where the exponent is too small for M-R
to be interested in yet, and then do bitwise masking to exclude the
spurious results from the final output.
2021-08-27 18:04:49 +01:00

289 lines
10 KiB
C

/*
* millerrabin.c: Miller-Rabin probabilistic primality testing, as
* declared in sshkeygen.h.
*/
#include <assert.h>
#include "ssh.h"
#include "sshkeygen.h"
#include "mpint.h"
#include "mpunsafe.h"
/*
* 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.)
*/
struct MillerRabin {
MontyContext *mc;
mp_int *pm1, *m_pm1;
mp_int *lowbit, *two;
};
MillerRabin *miller_rabin_new(mp_int *p)
{
MillerRabin *mr = snew(MillerRabin);
assert(mp_hs_integer(p, 2));
assert(mp_get_bit(p, 0) == 1);
mr->pm1 = mp_copy(p);
mp_sub_integer_into(mr->pm1, mr->pm1, 1);
/*
* Standard bit-twiddling trick for isolating the lowest set bit
* of a number: x & (-x)
*/
mr->lowbit = mp_new(mp_max_bits(mr->pm1));
mp_sub_into(mr->lowbit, mr->lowbit, mr->pm1);
mp_and_into(mr->lowbit, mr->lowbit, mr->pm1);
mr->two = mp_from_integer(2);
mr->mc = monty_new(p);
mr->m_pm1 = monty_import(mr->mc, mr->pm1);
return mr;
}
void miller_rabin_free(MillerRabin *mr)
{
mp_free(mr->pm1);
mp_free(mr->m_pm1);
mp_free(mr->lowbit);
mp_free(mr->two);
monty_free(mr->mc);
smemclr(mr, sizeof(*mr));
sfree(mr);
}
/*
* The main internal function that implements a single M-R test.
*
* Expects the witness integer to be in Montgomery representation.
* (Since in live use witnesses are invented at random, this imposes
* no extra cost on the callers, and saves effort in here.)
*/
static struct mr_result miller_rabin_test_inner(MillerRabin *mr, mp_int *mw)
{
mp_int *acc = mp_copy(monty_identity(mr->mc));
mp_int *spare = mp_new(mp_max_bits(mr->pm1));
size_t bit = mp_max_bits(mr->pm1);
/*
* The obvious approach to Miller-Rabin would be to start by
* calling monty_pow to raise w to the power q, and then square it
* k times ourselves. But that introduces a timing leak that gives
* away the value of k, i.e., how many factors of 2 there are in
* p-1.
*
* Instead, we don't call monty_pow at all. We do a modular
* exponentiation ourselves to compute w^((p-1)/2), using the
* technique that works from the top bit of the exponent
* downwards. That is, in each iteration we compute
* w^floor(exponent/2^i) for i one less than the previous
* iteration, by squaring the value we previously had and then
* optionally multiplying in w if the next exponent bit is 1.
*
* At the end of that process, once i <= k, the division
* (exponent/2^i) yields an integer, so the values we're computing
* are not just w^(floor of that), but w^(exactly that). In other
* words, the last k intermediate values of this modexp are
* precisely the values M-R wants to check against +1 or -1.
*
* So we interleave those checks with the modexp loop itself, and
* to avoid a timing leak, we check _every_ intermediate result
* against (the Montgomery representations of) both +1 and -1. And
* then we do bitwise masking to arrange that only the sensible
* ones of those checks find their way into our final answer.
*/
unsigned active = 0;
struct mr_result result;
result.passed = result.potential_primitive_root = 0;
while (bit-- > 1) {
/*
* In this iteration, we're computing w^(2e) or w^(2e+1),
* where we have w^e from the previous iteration. So we square
* the value we had already, and then optionally multiply in
* another copy of w depending on the next bit of the exponent.
*/
monty_mul_into(mr->mc, acc, acc, acc);
monty_mul_into(mr->mc, spare, acc, mw);
mp_select_into(acc, acc, spare, mp_get_bit(mr->pm1, bit));
/*
* mr->lowbit is a number with only one bit set, corresponding
* to the lowest set bit in p-1. So when that's the bit of the
* exponent we've just processed, we'll detect it by setting
* first_iter to true. That's our indication that we're now
* generating intermediate results useful to M-R, so we also
* set 'active', which stays set from then on.
*/
unsigned first_iter = mp_get_bit(mr->lowbit, bit);
active |= first_iter;
/*
* Check the intermediate result against both +1 and -1.
*/
unsigned is_plus_1 = mp_cmp_eq(acc, monty_identity(mr->mc));
unsigned is_minus_1 = mp_cmp_eq(acc, mr->m_pm1);
/*
* M-R must report success iff either: the first of the useful
* intermediate results (which is w^q) is 1, or _any_ of them
* (from w^q all the way up to w^((p-1)/2)) is -1.
*
* So we want to pass the test if is_plus_1 is set on the
* first iteration, or if is_minus_1 is set on any iteration.
*/
result.passed |= (first_iter & is_plus_1);
result.passed |= (active & is_minus_1);
/*
* In the final iteration, is_minus_1 is also used to set the
* 'potential primitive root' flag, because we haven't found
* any exponent smaller than p-1 for which w^(that) == 1.
*/
if (bit == 1)
result.potential_primitive_root = is_minus_1;
}
mp_free(acc);
mp_free(spare);
return result;
}
/*
* Wrapper on miller_rabin_test_inner for the convenience of
* testcrypt. Expects the witness integer to be literal, so we
* monty_import it before running the real test.
*/
struct mr_result miller_rabin_test(MillerRabin *mr, mp_int *w)
{
mp_int *mw = monty_import(mr->mc, w);
struct mr_result result = miller_rabin_test_inner(mr, mw);
mp_free(mw);
return result;
}
bool miller_rabin_test_random(MillerRabin *mr)
{
mp_int *mw = mp_random_in_range(mr->two, mr->pm1);
struct mr_result result = miller_rabin_test_inner(mr, mw);
mp_free(mw);
return result.passed;
}
mp_int *miller_rabin_find_potential_primitive_root(MillerRabin *mr)
{
while (true) {
mp_int *mw = mp_unsafe_shrink(mp_random_in_range(mr->two, mr->pm1));
struct mr_result result = miller_rabin_test_inner(mr, mw);
if (result.passed && result.potential_primitive_root) {
mp_int *pr = monty_export(mr->mc, mw);
mp_free(mw);
return pr;
}
mp_free(mw);
if (!result.passed) {
return NULL;
}
}
}
unsigned miller_rabin_checks_needed(unsigned bits)
{
/* Table 4.4 from Handbook of Applied Cryptography */
return (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);
}