зеркало из https://github.com/github/putty.git
478 строки
15 KiB
C
478 строки
15 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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/*
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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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static unsigned short primes[6542]; /* # primes < 65536 */
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#define NPRIMES (lenof(primes))
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static void init_primes_array(void)
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{
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if (primes[0])
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return; /* already done */
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bool A[65536];
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for (size_t i = 2; i < lenof(A); i++)
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A[i] = true;
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for (size_t i = 2; i < lenof(A); i++) {
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if (!A[i])
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continue;
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for (size_t j = 2*i; j < lenof(A); j += i)
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A[j] = false;
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}
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size_t pos = 0;
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for (size_t i = 2; i < lenof(A); i++)
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if (A[i])
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primes[pos++] = i;
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assert(pos == NPRIMES);
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}
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static unsigned short mp_mod_short(mp_int *x, unsigned short modulus)
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{
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/*
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* This function lives here rather than in mpint.c partly because
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* this is the only place it's needed, but mostly because it
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* doesn't pay careful attention to constant running time, since
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* as far as I can tell that's a lost cause for key generation
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* anyway.
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*/
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unsigned accumulator = 0;
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for (size_t i = mp_max_bytes(x); i-- > 0 ;) {
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accumulator = 0x100 * accumulator + mp_get_byte(x, i);
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accumulator %= modulus;
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}
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return accumulator;
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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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init_primes_array();
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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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STARTOVER:
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pfn(pfnparam, PROGFN_PROGRESS, phase, ++progress);
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/*
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* Generate a k-bit random number with top and bottom bits set.
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* Alternatively, if `factor' is nonzero, generate a k-bit
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* random number with the top bit set and the bottom bit clear,
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* multiply it by `factor', and add one.
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*/
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mp_int *p = mp_power_2(bits - 1); /* ensure top bit is 1 */
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mp_int *r = mp_random_bits(bits - 1);
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mp_or_into(p, p, r);
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mp_free(r);
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mp_set_bit(p, 0, factor ? 0 : 1); /* set bottom bit appropriately */
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for (size_t i = 0; i < fbsize; i++)
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mp_set_bit(p, bits-fbsize + i, 1 & (firstbits >> i));
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if (factor) {
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mp_int *tmp = p;
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p = mp_mul(tmp, factor);
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mp_free(tmp);
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assert(mp_get_bit(p, 0) == 0);
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mp_set_bit(p, 0, 1);
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}
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/*
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* We need to ensure this random number is coprime to the first
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* few primes, by repeatedly adding either 2 or 2*factor to it
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* until it is. To do this we make a list of (modulus, residue)
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* pairs to avoid, and we also add to that list the extra pair our
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* caller wants to avoid.
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*/
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/* List the moduli */
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unsigned long moduli[NPRIMES + 1];
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for (size_t i = 0; i < NPRIMES; i++)
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moduli[i] = primes[i];
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moduli[NPRIMES] = modulus;
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/* Find the residue of our starting number mod each of them. Also
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* set up the multipliers array which tells us how each one will
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* change when we increment the number (which isn't just 1 if
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* we're incrementing by multiples of factor). */
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unsigned long residues[NPRIMES + 1], multipliers[NPRIMES + 1];
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for (size_t i = 0; i < lenof(moduli); i++) {
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residues[i] = mp_mod_short(p, moduli[i]);
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if (factor)
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multipliers[i] = mp_mod_short(factor, moduli[i]);
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else
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multipliers[i] = 1;
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}
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/* Adjust the last entry so that it avoids a residue other than zero */
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residues[NPRIMES] = (residues[NPRIMES] + modulus - residue) % modulus;
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/*
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* Now loop until no residue in that list is zero, to find a
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* sensible increment. We maintain the increment in an ordinary
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* integer, so if it gets too big, we'll have to give up and go
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* back to making up a fresh random large integer.
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*/
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unsigned delta = 0;
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while (1) {
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for (size_t i = 0; i < lenof(moduli); i++)
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if (!((residues[i] + delta * multipliers[i]) % moduli[i]))
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goto found_a_zero;
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/* If we didn't exit that loop by goto, we've got our candidate. */
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break;
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found_a_zero:
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delta += 2;
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if (delta > 65536) {
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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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/*
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* Having found a plausible increment, actually add it on.
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*/
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if (factor) {
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mp_int *d = mp_from_integer(delta);
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mp_int *df = mp_mul(d, factor);
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mp_add_into(p, p, df);
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mp_free(d);
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mp_free(df);
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} else {
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mp_add_integer_into(p, p, delta);
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}
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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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return p;
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}
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/*
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* Invent a pair of values suitable for use as 'firstbits' in the
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* above function, such that their product is at least 2, and such
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* that their difference is also at least min_separation.
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*
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* This is used for generating both RSA and DSA keys which have
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* exactly the specified number of bits rather than one fewer - if you
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* generate an a-bit and a b-bit number completely at random and
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* multiply them together, you could end up with either an (ab-1)-bit
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* number or an (ab)-bit number. The former happens log(2)*2-1 of the
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* time (about 39%) and, though actually harmless, every time it
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* occurs it has a non-zero probability of sparking a user email along
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* the lines of 'Hey, I asked PuTTYgen for a 2048-bit key and I only
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* got 2047 bits! Bug!'
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*/
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static inline unsigned firstbits_b_min(
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unsigned a, unsigned lo, unsigned hi, unsigned min_separation)
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{
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/* To get a large enough product, b must be at least this much */
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unsigned b_min = (2*lo*lo + a - 1) / a;
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/* Now enforce a<b, optionally with minimum separation */
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if (b_min < a + min_separation)
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b_min = a + min_separation;
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/* And cap at the upper limit */
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if (b_min > hi)
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b_min = hi;
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return b_min;
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}
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void invent_firstbits(unsigned *one, unsigned *two, unsigned min_separation)
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{
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/*
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* We'll pick 12 initial bits (number selected at random) for each
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* prime, not counting the leading 1. So we want to return two
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* values in the range [2^12,2^13) whose product is at least 2^25.
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*
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* Strategy: count up all the viable pairs, then select a random
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* number in that range and use it to pick a pair.
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*
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* To keep things simple, we'll ensure a < b, and randomly swap
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* them at the end.
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*/
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const unsigned lo = 1<<12, hi = 1<<13, minproduct = 2*lo*lo;
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unsigned a, b;
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/*
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* Count up the number of prefixes of b that would be valid for
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* each prefix of a.
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*/
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mp_int *total = mp_new(32);
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for (a = lo; a < hi; a++) {
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unsigned b_min = firstbits_b_min(a, lo, hi, min_separation);
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mp_add_integer_into(total, total, hi - b_min);
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}
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/*
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* Make up a random number in the range [0,2*total).
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*/
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mp_int *mlo = mp_from_integer(0), *mhi = mp_new(32);
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mp_lshift_fixed_into(mhi, total, 1);
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mp_int *randval = mp_random_in_range(mlo, mhi);
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mp_free(mlo);
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mp_free(mhi);
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/*
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* Use the low bit of randval as our swap indicator, leaving the
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* rest of it in the range [0,total).
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*/
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unsigned swap = mp_get_bit(randval, 0);
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mp_rshift_fixed_into(randval, randval, 1);
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/*
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* Now do the same counting loop again to make the actual choice.
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*/
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a = b = 0;
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for (unsigned a_candidate = lo; a_candidate < hi; a_candidate++) {
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unsigned b_min = firstbits_b_min(a_candidate, lo, hi, min_separation);
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unsigned limit = hi - b_min;
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unsigned b_candidate = b_min + mp_get_integer(randval);
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unsigned use_it = 1 ^ mp_hs_integer(randval, limit);
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a ^= (a ^ a_candidate) & -use_it;
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b ^= (b ^ b_candidate) & -use_it;
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mp_sub_integer_into(randval, randval, limit);
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}
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mp_free(randval);
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mp_free(total);
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/*
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* Check everything came out right.
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*/
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assert(lo <= a);
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assert(a < hi);
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assert(lo <= b);
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assert(b < hi);
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assert(a * b >= minproduct);
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assert(b >= a + min_separation);
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/*
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* Last-minute optional swap of a and b.
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*/
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unsigned diff = (a ^ b) & (-swap);
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a ^= diff;
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b ^= diff;
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*one = a;
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*two = b;
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}
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