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aba52744e4
There's always one.
243 lines
8.0 KiB
C
243 lines
8.0 KiB
C
/*
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* Prime generation.
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*/
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#include <assert.h>
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#include "ssh.h"
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#include "mpint.h"
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#include "mpunsafe.h"
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#include "sshkeygen.h"
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/*
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* This prime generation algorithm is pretty much cribbed from
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* OpenSSL. The algorithm is:
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*
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* - invent a B-bit random number and ensure the top and bottom
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* bits are set (so it's definitely B-bit, and it's definitely
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* odd)
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*
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* - see if it's coprime to all primes below 2^16; increment it by
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* two until it is (this shouldn't take long in general)
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*
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* - perform the Miller-Rabin primality test enough times to
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* ensure the probability of it being composite is 2^-80 or
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* less
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*
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* - go back to square one if any M-R test fails.
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*/
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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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/*
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* Generate a prime. We can deal with various extra properties of
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* the prime:
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*
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* - to speed up use in RSA, we can arrange to select a prime with
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* the property (prime % modulus) != residue.
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*
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* - for use in DSA, we can arrange to select a prime which is one
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* more than a multiple of a dirty great bignum. In this case
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* `bits' gives the size of the factor by which we _multiply_
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* that bignum, rather than the size of the whole number.
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*
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* - for the basically cosmetic purposes of generating keys of the
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* length actually specified rather than off by one bit, we permit
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* the caller to provide an unsigned integer 'firstbits' which will
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* match the top few bits of the returned prime. (That is, there
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* will exist some n such that (returnvalue >> n) == firstbits.) If
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* 'firstbits' is not needed, specifying it to either 0 or 1 is
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* an adequate no-op.
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*/
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mp_int *primegen(
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int bits, int modulus, int residue, mp_int *factor,
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int phase, progfn_t pfn, void *pfnparam, unsigned firstbits)
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{
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int progress = 0;
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size_t fbsize = 0;
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while (firstbits >> fbsize) /* work out how to align this */
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fbsize++;
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PrimeCandidateSource *pcs = pcs_new(bits, firstbits, fbsize);
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if (factor)
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pcs_require_residue_1(pcs, factor);
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if (modulus)
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pcs_avoid_residue_small(pcs, modulus, residue);
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pcs_ready(pcs);
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STARTOVER:
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pfn(pfnparam, PROGFN_PROGRESS, phase, ++progress);
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mp_int *p = pcs_generate(pcs);
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/*
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* Now apply the Miller-Rabin primality test a few times. First
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* work out how many checks are needed.
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*/
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unsigned checks =
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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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* Next, write p-1 as q*2^k.
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*/
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size_t k;
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for (k = 0; mp_get_bit(p, k) == !k; k++)
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continue; /* find first 1 bit in p-1 */
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mp_int *q = mp_rshift_safe(p, k);
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/*
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* Set up stuff for the Miller-Rabin checks.
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*/
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mp_int *two = mp_from_integer(2);
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mp_int *pm1 = mp_copy(p);
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mp_sub_integer_into(pm1, pm1, 1);
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MontyContext *mc = monty_new(p);
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mp_int *m_pm1 = monty_import(mc, pm1);
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bool known_bad = false;
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/*
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* Now, for each check ...
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*/
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for (unsigned check = 0; check < checks && !known_bad; check++) {
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/*
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* Invent a random number between 1 and p-1.
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*/
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mp_int *w = mp_random_in_range(two, pm1);
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monty_import_into(mc, w, w);
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pfn(pfnparam, PROGFN_PROGRESS, phase, ++progress);
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/*
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* Compute w^q mod p.
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*/
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mp_int *wqp = monty_pow(mc, w, q);
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mp_free(w);
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/*
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* See if this is 1, or if it is -1, or if it becomes -1
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* when squared at most k-1 times.
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*/
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bool passed = false;
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if (mp_cmp_eq(wqp, monty_identity(mc)) || mp_cmp_eq(wqp, m_pm1)) {
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passed = true;
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} else {
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for (size_t i = 0; i < k - 1; i++) {
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monty_mul_into(mc, wqp, wqp, wqp);
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if (mp_cmp_eq(wqp, m_pm1)) {
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passed = true;
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break;
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}
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}
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}
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if (!passed)
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known_bad = true;
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mp_free(wqp);
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}
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mp_free(q);
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mp_free(two);
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mp_free(pm1);
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monty_free(mc);
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mp_free(m_pm1);
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if (known_bad) {
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mp_free(p);
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goto STARTOVER;
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}
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/*
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* We have a prime!
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*/
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pcs_free(pcs);
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return p;
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}
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