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6520574e58
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.
289 lines
10 KiB
C
289 lines
10 KiB
C
/*
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* millerrabin.c: Miller-Rabin probabilistic primality testing, as
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* declared in sshkeygen.h.
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*/
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#include <assert.h>
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#include "ssh.h"
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#include "sshkeygen.h"
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#include "mpint.h"
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#include "mpunsafe.h"
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/*
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* The Miller-Rabin primality test is an extension to the Fermat
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* test. The Fermat test just checks that a^(p-1) == 1 mod p; this
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* is vulnerable to Carmichael numbers. Miller-Rabin considers how
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* that 1 is derived as well.
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*
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* Lemma: if a^2 == 1 (mod p), and p is prime, then either a == 1
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* or a == -1 (mod p).
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*
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* Proof: p divides a^2-1, i.e. p divides (a+1)(a-1). Hence,
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* since p is prime, either p divides (a+1) or p divides (a-1).
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* But this is the same as saying that either a is congruent to
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* -1 mod p or a is congruent to +1 mod p. []
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*
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* Comment: This fails when p is not prime. Consider p=mn, so
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* that mn divides (a+1)(a-1). Now we could have m dividing (a+1)
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* and n dividing (a-1), without the whole of mn dividing either.
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* For example, consider a=10 and p=99. 99 = 9 * 11; 9 divides
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* 10-1 and 11 divides 10+1, so a^2 is congruent to 1 mod p
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* without a having to be congruent to either 1 or -1.
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*
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* So the Miller-Rabin test, as well as considering a^(p-1),
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* considers a^((p-1)/2), a^((p-1)/4), and so on as far as it can
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* go. In other words. we write p-1 as q * 2^k, with k as large as
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* possible (i.e. q must be odd), and we consider the powers
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*
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* a^(q*2^0) a^(q*2^1) ... a^(q*2^(k-1)) a^(q*2^k)
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* i.e. a^((n-1)/2^k) a^((n-1)/2^(k-1)) ... a^((n-1)/2) a^(n-1)
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*
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* If p is to be prime, the last of these must be 1. Therefore, by
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* the above lemma, the one before it must be either 1 or -1. And
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* _if_ it's 1, then the one before that must be either 1 or -1,
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* and so on ... In other words, we expect to see a trailing chain
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* of 1s preceded by a -1. (If we're unlucky, our trailing chain of
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* 1s will be as long as the list so we'll never get to see what
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* lies before it. This doesn't count as a test failure because it
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* hasn't _proved_ that p is not prime.)
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*
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* For example, consider a=2 and p=1729. 1729 is a Carmichael
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* number: although it's not prime, it satisfies a^(p-1) == 1 mod p
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* for any a coprime to it. So the Fermat test wouldn't have a
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* problem with it at all, unless we happened to stumble on an a
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* which had a common factor.
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*
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* So. 1729 - 1 equals 27 * 2^6. So we look at
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*
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* 2^27 mod 1729 == 645
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* 2^108 mod 1729 == 1065
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* 2^216 mod 1729 == 1
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* 2^432 mod 1729 == 1
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* 2^864 mod 1729 == 1
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* 2^1728 mod 1729 == 1
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*
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* We do have a trailing string of 1s, so the Fermat test would
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* have been happy. But this trailing string of 1s is preceded by
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* 1065; whereas if 1729 were prime, we'd expect to see it preceded
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* by -1 (i.e. 1728.). Guards! Seize this impostor.
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*
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* (If we were unlucky, we might have tried a=16 instead of a=2;
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* now 16^27 mod 1729 == 1, so we would have seen a long string of
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* 1s and wouldn't have seen the thing _before_ the 1s. So, just
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* like the Fermat test, for a given p there may well exist values
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* of a which fail to show up its compositeness. So we try several,
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* just like the Fermat test. The difference is that Miller-Rabin
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* is not _in general_ fooled by Carmichael numbers.)
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*
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* Put simply, then, the Miller-Rabin test requires us to:
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*
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* 1. write p-1 as q * 2^k, with q odd
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* 2. compute z = (a^q) mod p.
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* 3. report success if z == 1 or z == -1.
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* 4. square z at most k-1 times, and report success if it becomes
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* -1 at any point.
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* 5. report failure otherwise.
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*
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* (We expect z to become -1 after at most k-1 squarings, because
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* if it became -1 after k squarings then a^(p-1) would fail to be
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* 1. And we don't need to investigate what happens after we see a
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* -1, because we _know_ that -1 squared is 1 modulo anything at
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* all, so after we've seen a -1 we can be sure of seeing nothing
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* but 1s.)
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*/
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struct MillerRabin {
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MontyContext *mc;
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mp_int *pm1, *m_pm1;
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mp_int *lowbit, *two;
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};
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MillerRabin *miller_rabin_new(mp_int *p)
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{
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MillerRabin *mr = snew(MillerRabin);
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assert(mp_hs_integer(p, 2));
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assert(mp_get_bit(p, 0) == 1);
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mr->pm1 = mp_copy(p);
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mp_sub_integer_into(mr->pm1, mr->pm1, 1);
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/*
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* Standard bit-twiddling trick for isolating the lowest set bit
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* of a number: x & (-x)
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*/
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mr->lowbit = mp_new(mp_max_bits(mr->pm1));
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mp_sub_into(mr->lowbit, mr->lowbit, mr->pm1);
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mp_and_into(mr->lowbit, mr->lowbit, mr->pm1);
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mr->two = mp_from_integer(2);
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mr->mc = monty_new(p);
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mr->m_pm1 = monty_import(mr->mc, mr->pm1);
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return mr;
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}
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void miller_rabin_free(MillerRabin *mr)
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{
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mp_free(mr->pm1);
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mp_free(mr->m_pm1);
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mp_free(mr->lowbit);
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mp_free(mr->two);
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monty_free(mr->mc);
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smemclr(mr, sizeof(*mr));
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sfree(mr);
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}
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/*
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* The main internal function that implements a single M-R test.
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*
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* Expects the witness integer to be in Montgomery representation.
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* (Since in live use witnesses are invented at random, this imposes
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* no extra cost on the callers, and saves effort in here.)
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*/
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static struct mr_result miller_rabin_test_inner(MillerRabin *mr, mp_int *mw)
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{
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mp_int *acc = mp_copy(monty_identity(mr->mc));
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mp_int *spare = mp_new(mp_max_bits(mr->pm1));
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size_t bit = mp_max_bits(mr->pm1);
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/*
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* The obvious approach to Miller-Rabin would be to start by
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* calling monty_pow to raise w to the power q, and then square it
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* k times ourselves. But that introduces a timing leak that gives
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* away the value of k, i.e., how many factors of 2 there are in
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* p-1.
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*
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* Instead, we don't call monty_pow at all. We do a modular
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* exponentiation ourselves to compute w^((p-1)/2), using the
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* technique that works from the top bit of the exponent
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* downwards. That is, in each iteration we compute
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* w^floor(exponent/2^i) for i one less than the previous
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* iteration, by squaring the value we previously had and then
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* optionally multiplying in w if the next exponent bit is 1.
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*
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* At the end of that process, once i <= k, the division
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* (exponent/2^i) yields an integer, so the values we're computing
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* are not just w^(floor of that), but w^(exactly that). In other
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* words, the last k intermediate values of this modexp are
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* precisely the values M-R wants to check against +1 or -1.
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*
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* So we interleave those checks with the modexp loop itself, and
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* to avoid a timing leak, we check _every_ intermediate result
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* against (the Montgomery representations of) both +1 and -1. And
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* then we do bitwise masking to arrange that only the sensible
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* ones of those checks find their way into our final answer.
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*/
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unsigned active = 0;
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struct mr_result result;
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result.passed = result.potential_primitive_root = 0;
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while (bit-- > 1) {
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/*
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* In this iteration, we're computing w^(2e) or w^(2e+1),
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* where we have w^e from the previous iteration. So we square
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* the value we had already, and then optionally multiply in
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* another copy of w depending on the next bit of the exponent.
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*/
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monty_mul_into(mr->mc, acc, acc, acc);
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monty_mul_into(mr->mc, spare, acc, mw);
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mp_select_into(acc, acc, spare, mp_get_bit(mr->pm1, bit));
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/*
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* mr->lowbit is a number with only one bit set, corresponding
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* to the lowest set bit in p-1. So when that's the bit of the
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* exponent we've just processed, we'll detect it by setting
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* first_iter to true. That's our indication that we're now
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* generating intermediate results useful to M-R, so we also
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* set 'active', which stays set from then on.
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*/
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unsigned first_iter = mp_get_bit(mr->lowbit, bit);
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active |= first_iter;
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/*
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* Check the intermediate result against both +1 and -1.
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*/
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unsigned is_plus_1 = mp_cmp_eq(acc, monty_identity(mr->mc));
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unsigned is_minus_1 = mp_cmp_eq(acc, mr->m_pm1);
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/*
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* M-R must report success iff either: the first of the useful
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* intermediate results (which is w^q) is 1, or _any_ of them
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* (from w^q all the way up to w^((p-1)/2)) is -1.
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*
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* So we want to pass the test if is_plus_1 is set on the
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* first iteration, or if is_minus_1 is set on any iteration.
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*/
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result.passed |= (first_iter & is_plus_1);
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result.passed |= (active & is_minus_1);
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/*
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* In the final iteration, is_minus_1 is also used to set the
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* 'potential primitive root' flag, because we haven't found
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* any exponent smaller than p-1 for which w^(that) == 1.
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*/
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if (bit == 1)
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result.potential_primitive_root = is_minus_1;
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}
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mp_free(acc);
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mp_free(spare);
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return result;
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}
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/*
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* Wrapper on miller_rabin_test_inner for the convenience of
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* testcrypt. Expects the witness integer to be literal, so we
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* monty_import it before running the real test.
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*/
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struct mr_result miller_rabin_test(MillerRabin *mr, mp_int *w)
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{
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mp_int *mw = monty_import(mr->mc, w);
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struct mr_result result = miller_rabin_test_inner(mr, mw);
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mp_free(mw);
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return result;
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}
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bool miller_rabin_test_random(MillerRabin *mr)
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{
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mp_int *mw = mp_random_in_range(mr->two, mr->pm1);
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struct mr_result result = miller_rabin_test_inner(mr, mw);
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mp_free(mw);
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return result.passed;
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}
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mp_int *miller_rabin_find_potential_primitive_root(MillerRabin *mr)
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{
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while (true) {
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mp_int *mw = mp_unsafe_shrink(mp_random_in_range(mr->two, mr->pm1));
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struct mr_result result = miller_rabin_test_inner(mr, mw);
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if (result.passed && result.potential_primitive_root) {
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mp_int *pr = monty_export(mr->mc, mw);
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mp_free(mw);
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return pr;
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}
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mp_free(mw);
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if (!result.passed) {
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return NULL;
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}
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}
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}
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unsigned miller_rabin_checks_needed(unsigned bits)
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{
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/* Table 4.4 from Handbook of Applied Cryptography */
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return (bits >= 1300 ? 2 : bits >= 850 ? 3 : bits >= 650 ? 4 :
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bits >= 550 ? 5 : bits >= 450 ? 6 : bits >= 400 ? 7 :
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bits >= 350 ? 8 : bits >= 300 ? 9 : bits >= 250 ? 12 :
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bits >= 200 ? 15 : bits >= 150 ? 18 : 27);
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}
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