680 lines
21 KiB
C
680 lines
21 KiB
C
/*
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* Prime generation using the bignum library and sieving.
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*
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* Copyright (c) 1995 Colin Plumb. All rights reserved.
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* For licensing and other legal details, see the file legal.c.
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*/
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#ifndef HAVE_CONFIG_H
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#define HAVE_CONFIG_H 0
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#endif
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#if HAVE_CONFIG_H
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#include "bnconfig.h"
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#endif
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/*
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* Some compilers complain about #if FOO if FOO isn't defined,
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* so do the ANSI-mandated thing explicitly...
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*/
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#ifndef NO_ASSERT_H
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#define NO_ASSERT_H 0
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#endif
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#if !NO_ASSERT_H
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#include <assert.h>
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#else
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#define assert(x) (void)0
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#endif
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#include <stdarg.h> /* We just can't live without this... */
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#ifndef BNDEBUG
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#define BNDEBUG 1
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#endif
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#if BNDEBUG
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#include <stdio.h>
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#endif
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#include "bn.h"
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#include "lbnmem.h"
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#include "prime.h"
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#include "sieve.h"
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#include "kludge.h"
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/* Size of the shuffle table */
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#define SHUFFLE 256
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/* Size of the sieve area */
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#define SIEVE 32768u/16
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/* Confirmation tests. The first one *must* be 2 */
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static unsigned const confirm[] = {2, 3, 5, 7, 11, 13, 17};
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#define CONFIRMTESTS (sizeof(confirm)/sizeof(*confirm))
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/*
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* Helper function that does the slow primality test.
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* bn is the input bignum; a and e are temporary buffers that are
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* allocated by the caller to save overhead.
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*
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* Returns 0 if prime, >0 if not prime, and -1 on error (out of memory).
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* If not prime, returns the number of modular exponentiations performed.
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* Calls the given progress function with a '*' for each primality test
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* that is passed.
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*
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* The testing consists of strong pseudoprimality tests, to the bases given
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* in the confirm[] array above. (Also called Miller-Rabin, although that's
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* not technically correct if we're using fixed bases.) Some people worry
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* that this might not be enough. Number theorists may wish to generate
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* primality proofs, but for random inputs, this returns non-primes with
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* a probability which is quite negligible, which is good enough.
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*
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* It has been proved (see Carl Pomerance, "On the Distribution of
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* Pseudoprimes", Math. Comp. v.37 (1981) pp. 587-593) that the number of
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* pseudoprimes (composite numbers that pass a Fermat test to the base 2)
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* less than x is bounded by:
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* exp(ln(x)^(5/14)) <= P_2(x) ### CHECK THIS FORMULA - it looks wrong! ###
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* P_2(x) <= x * exp(-1/2 * ln(x) * ln(ln(ln(x))) / ln(ln(x))).
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* Thus, the local density of Pseudoprimes near x is at most
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* exp(-1/2 * ln(x) * ln(ln(ln(x))) / ln(ln(x))), and at least
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* exp(ln(x)^(5/14) - ln(x)). Here are some values of this function
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* for various k-bit numbers x = 2^k:
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* Bits Density <= Bit equivalent Density >= Bit equivalent
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* 128 3.577869e-07 21.414396 4.202213e-37 120.840190
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* 192 4.175629e-10 31.157288 4.936250e-56 183.724558
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* 256 5.804314e-13 40.647940 4.977813e-75 246.829095
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* 384 1.578039e-18 59.136573 3.938861e-113 373.400096
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* 512 5.858255e-24 77.175803 2.563353e-151 500.253110
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* 768 1.489276e-34 112.370944 7.872825e-228 754.422724
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* 1024 6.633188e-45 146.757062 1.882404e-304 1008.953565
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*
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* As you can see, there's quite a bit of slop between these estimates.
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* In fact, the density of pseudoprimes is conjectured to be closer to the
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* square of that upper bound. E.g. the density of pseudoprimes of size
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* 256 is around 3 * 10^-27. The density of primes is very high, from
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* 0.005636 at 256 bits to 0.001409 at 1024 bits, i.e. more than 10^-3.
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*
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* For those people used to cryptographic levels of security where the
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* 56 bits of DES key space is too small because it's exhaustible with
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* custom hardware searching engines, note that you are not generating
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* 50,000,000 primes per second on each of 56,000 custom hardware chips
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* for several hours. The chances that another Dinosaur Killer asteroid
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* will land today is about 10^-11 or 2^-36, so it would be better to
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* spend your time worrying about *that*. Well, okay, there should be
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* some derating for the chance that astronomers haven't seen it yet,
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* but I think you get the idea. For a good feel about the probability
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* of various events, I have heard that a good book is by E'mile Borel,
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* "Les Probabilite's et la vie". (The 's are accents, not apostrophes.)
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*
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* For more on the subject, try "Finding Four Million Large Random Primes",
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* by Ronald Rivest, in Advancess in Cryptology: Proceedings of Crypto
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* '90. He used a small-divisor test, then a Fermat test to the base 2,
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* and then 8 iterations of a Miller-Rabin test. About 718 million random
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* 256-bit integers were generated, 43,741,404 passed the small divisor
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* test, 4,058,000 passed the Fermat test, and all 4,058,000 passed all
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* 8 iterations of the Miller-Rabin test, proving their primality beyond
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* most reasonable doubts.
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*
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* If the probability of getting a pseudoprime is some small p, then the
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* probability of not getting it in t trials is (1-p)^t. Remember that,
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* for small p, (1-p)^(1/p) ~ 1/e, the base of natural logarithms.
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* (This is more commonly expressed as e = lim_{x\to\infty} (1+1/x)^x.)
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* Thus, (1-p)^t ~ e^(-p*t) = exp(-p*t). So the odds of being able to
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* do this many tests without seeing a pseudoprime if you assume that
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* p = 10^-6 (one in a million) is one in 57.86. If you assume that
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* p = 2*10^-6, it's one in 3347.6. So it's implausible that the density
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* of pseudoprimes is much more than one millionth the density of primes.
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*
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* He also gives a theoretical argument that the chance of finding a
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* 256-bit non-prime which satisfies one Fermat test to the base 2 is
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* less than 10^-22. The small divisor test improves this number, and
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* if the numbers are 512 bits (as needed for a 1024-bit key) the odds
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* of failure shrink to about 10^-44. Thus, he concludes, for practical
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* purposes *one* Fermat test to the base 2 is sufficient.
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*/
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static int
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primeTest(struct BigNum const *bn, struct BigNum *e, struct BigNum *a,
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int (*f)(void *arg, int c), void *arg)
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{
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unsigned i, j;
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unsigned k, l;
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int err;
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#if BNDEBUG /* Debugging */
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/*
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* This is debugging code to test the sieving stage.
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* If the sieving is wrong, it will let past numbers with
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* small divisors. The prime test here will still work, and
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* weed them out, but you'll be doing a lot more slow tests,
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* and presumably excluding from consideration some other numbers
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* which might be prime. This check just verifies that none
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* of the candidates have any small divisors. If this
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* code is enabled and never triggers, you can feel quite
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* confident that the sieving is doing its job.
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*/
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i = bnLSWord(bn);
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if (!(i % 2)) printf("bn div by 2!");
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i = bnModQ(bn, 51051); /* 51051 = 3 * 7 * 11 * 13 * 17 */
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if (!(i % 3)) printf("bn div by 3!");
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if (!(i % 7)) printf("bn div by 7!");
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if (!(i % 11)) printf("bn div by 11!");
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if (!(i % 13)) printf("bn div by 13!");
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if (!(i % 17)) printf("bn div by 17!");
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i = bnModQ(bn, 63365); /* 63365 = 5 * 19 * 23 * 29 */
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if (!(i % 5)) printf("bn div by 5!");
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if (!(i % 19)) printf("bn div by 19!");
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if (!(i % 23)) printf("bn div by 23!");
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if (!(i % 29)) printf("bn div by 29!");
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i = bnModQ(bn, 47027); /* 47027 = 31 * 37 * 41 */
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if (!(i % 31)) printf("bn div by 31!");
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if (!(i % 37)) printf("bn div by 37!");
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if (!(i % 41)) printf("bn div by 41!");
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#endif
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/*
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* Now, check that bn is prime. If it passes to the base 2,
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* it's prime beyond all reasonable doubt, and everything else
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* is just gravy, but it gives people warm fuzzies to do it.
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*
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* This starts with verifying Euler's criterion for a base of 2.
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* This is the fastest pseudoprimality test that I know of,
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* saving a modular squaring over a Fermat test, as well as
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* being stronger. 7/8 of the time, it's as strong as a strong
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* pseudoprimality test, too. (The exception being when bn ==
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* 1 mod 8 and 2 is a quartic residue, i.e. bn is of the form
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* a^2 + (8*b)^2.) The precise series of tricks used here is
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* not documented anywhere, so here's an explanation.
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* Euler's criterion states that if p is prime then a^((p-1)/2)
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* is congruent to Jacobi(a,p), modulo p. Jacobi(a,p) is
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* a function which is +1 if a is a square modulo p, and -1 if
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* it is not. For a = 2, this is particularly simple. It's
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* +1 if p == +/-1 (mod 8), and -1 if m == +/-3 (mod 8).
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* If p == 3 mod 4, then all a strong test does is compute
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* 2^((p-1)/2). and see if it's +1 or -1. (Euler's criterion
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* says *which* it should be.) If p == 5 (mod 8), then
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* 2^((p-1)/2) is -1, so the initial step in a strong test,
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* looking at 2^((p-1)/4), is wasted - you're not going to
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* find a +/-1 before then if it *is* prime, and it shouldn't
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* have either of those values if it isn't. So don't bother.
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*
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* The remaining case is p == 1 (mod 8). In this case, we
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* expect 2^((p-1)/2) == 1 (mod p), so we expect that the
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* square root of this, 2^((p-1)/4), will be +/-1 (mod p).
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* Evaluating this saves us a modular squaring 1/4 of the time.
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* If it's -1, a strong pseudoprimality test would call p
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* prime as well. Only if the result is +1, indicating that
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* 2 is not only a quadratic residue, but a quartic one as well,
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* does a strong pseudoprimality test verify more things than
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* this test does. Good enough.
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*
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* We could back that down another step, looking at 2^((p-1)/8)
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* if there was a cheap way to determine if 2 were expected to
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* be a quartic residue or not. Dirichlet proved that 2 is
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* a quartic residue iff p is of the form a^2 + (8*b^2).
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* All primes == 1 (mod 4) can be expressed as a^2 + (2*b)^2,
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* but I see no cheap way to evaluate this condition.
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*/
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if (bnCopy(e, bn) < 0)
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return -1;
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(void)bnSubQ(e, 1);
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l = bnLSWord(e);
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j = 1; /* Where to start in prime array for strong prime tests */
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if (l & 7) {
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bnRShift(e, 1);
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if (bnTwoExpMod(a, e, bn) < 0)
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return -1;
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if ((l & 7) == 6) {
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/* bn == 7 mod 8, expect +1 */
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if (bnBits(a) != 1)
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return 1; /* Not prime */
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k = 1;
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} else {
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/* bn == 3 or 5 mod 8, expect -1 == bn-1 */
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if (bnAddQ(a, 1) < 0)
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return -1;
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if (bnCmp(a, bn) != 0)
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return 1; /* Not prime */
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k = 1;
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if (l & 4) {
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/* bn == 5 mod 8, make odd for strong tests */
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bnRShift(e, 1);
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k = 2;
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}
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}
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} else {
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/* bn == 1 mod 8, expect 2^((bn-1)/4) == +/-1 mod bn */
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bnRShift(e, 2);
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if (bnTwoExpMod(a, e, bn) < 0)
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return -1;
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if (bnBits(a) == 1) {
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j = 0; /* Re-do strong prime test to base 2 */
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} else {
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if (bnAddQ(a, 1) < 0)
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return -1;
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if (bnCmp(a, bn) != 0)
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return 1; /* Not prime */
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}
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k = 2 + bnMakeOdd(e);
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}
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/* It's prime! Now go on to confirmation tests */
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/*
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* Now, e = (bn-1)/2^k is odd. k >= 1, and has a given value
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* with probability 2^-k, so its expected value is 2.
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* j = 1 in the usual case when the previous test was as good as
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* a strong prime test, but 1/8 of the time, j = 0 because
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* the strong prime test to the base 2 needs to be re-done.
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*/
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for (i = j; i < CONFIRMTESTS; i++) {
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if (f && (err = f(arg, '*')) < 0)
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return err;
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(void)bnSetQ(a, confirm[i]);
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if (bnExpMod(a, a, e, bn) < 0)
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return -1;
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if (bnBits(a) == 1)
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continue; /* Passed this test */
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l = k;
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for (;;) {
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if (bnAddQ(a, 1) < 0)
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return -1;
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if (bnCmp(a, bn) == 0) /* Was result bn-1? */
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break; /* Prime */
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if (!--l) /* Reached end, not -1? luck? */
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return i+2-j; /* Failed, not prime */
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/* This portion is executed, on average, once. */
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(void)bnSubQ(a, 1); /* Put a back where it was. */
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if (bnSquare(a, a) < 0 || bnMod(a, a, bn) < 0)
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return -1;
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if (bnBits(a) == 1)
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return i+2-j; /* Failed, not prime */
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}
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/* It worked (to the base confirm[i]) */
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}
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/* Yes, we've decided that it's prime. */
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if (f && (err = f(arg, '*')) < 0)
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return err;
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return 0; /* Prime! */
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}
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/*
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* Add x*y to bn, which is usually (but not always) < 65536.
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* Do it in a simple linear manner.
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*/
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static int
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bnAddMult(struct BigNum *bn, unsigned x, unsigned y)
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{
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unsigned long z = (unsigned long)x * y;
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while (z > 65535) {
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if (bnAddQ(bn, 65535) < 0)
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return -1;
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z -= 65535;
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}
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return bnAddQ(bn, (unsigned)z);
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}
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static int
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bnSubMult(struct BigNum *bn, unsigned x, unsigned y)
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{
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unsigned long z = (unsigned long)x * y;
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while (z > 65535) {
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if (bnSubQ(bn, 65535) < 0)
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return -1;
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z -= 65535;
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}
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return bnSubQ(bn, (unsigned)z);
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}
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/*
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* Modifies the bignum to return a nearby (slightly larger) number which
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* is a probable prime. Returns >=0 on success or -1 on failure (out of
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* memory). The return value is the number of unsuccessful modular
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* exponentiations performed. This never gives up searching.
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*
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* All other arguments are optional. They may be NULL. They are:
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*
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* unsigned (*rand)(unsigned limit)
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* For better distributed numbers, supply a non-null pointer to a
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* function which returns a random x, 0 <= x < limit. (It may make it
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* simpler to know that 0 < limit <= SHUFFLE, so you need at most a byte.)
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* The program generates a large window of sieve data and then does
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* pseudoprimality tests on the data. If a rand function is supplied,
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* the candidates which survive sieving are shuffled with a window of
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* size SHUFFLE before testing to increase the uniformity of the prime
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* selection. This isn't perfect, but it reduces the correlation between
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* the size of the prime-free gap before a prime and the probability
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* that that prime will be found by a sequential search.
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*
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* If rand is NULL, sequential search is used. If you want sequential
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* search, note that the search begins with the given number; if you're
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* trying to generate consecutive primes, you must increment the previous
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* one by two before calling this again.
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*
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* int (*f)(void *arg, int c), void *arg
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* The function f argument, if non-NULL, is called with progress indicator
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* characters for printing. A dot (.) is written every time a primality test
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* is failed, a star (*) every time one is passed, and a slash (/) in the
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* (very rare) case that the sieve was emptied without finding a prime
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* and is being refilled. f is also passed the void *arg argument for
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* private context storage. If f returns < 0, the test aborts and returns
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* that value immediately. (bn is set to the last value tested, so you
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* can increment bn and continue.)
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*
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* The "exponent" argument, and following unsigned numbers, are exponents
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* for which an inverse is desired, modulo p. For a d to exist such that
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* (x^e)^d == x (mod p), then d*e == 1 (mod p-1), so gcd(e,p-1) must be 1.
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* The prime returned is constrained to not be congruent to 1 modulo
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* any of the zero-terminated list of 16-bit numbers. Note that this list
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* should contain all the small prime factors of e. (You'll have to test
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* for large prime factors of e elsewhere, but the chances of needing to
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* generate another prime are low.)
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*
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* The list is terminated by a 0, and may be empty.
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*/
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int
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primeGen(struct BigNum *bn, unsigned (*rand)(unsigned),
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int (*f)(void *arg, int c), void *arg, unsigned exponent, ...)
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{
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int retval;
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int modexps = 0;
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unsigned short offsets[SHUFFLE];
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unsigned i, j;
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unsigned p, q, prev;
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struct BigNum a, e;
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#ifdef MSDOS
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unsigned char *sieve;
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#else
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unsigned char sieve[SIEVE];
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#endif
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#ifdef MSDOS
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sieve = lbnMemAlloc(SIEVE);
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if (!sieve)
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return -1;
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#endif
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bnBegin(&a);
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bnBegin(&e);
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#if 0 /* Self-test (not used for production) */
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{
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struct BigNum t;
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static unsigned char const prime1[] = {5};
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static unsigned char const prime2[] = {7};
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static unsigned char const prime3[] = {11};
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static unsigned char const prime4[] = {1, 1}; /* 257 */
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static unsigned char const prime5[] = {0xFF, 0xF1}; /* 65521 */
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static unsigned char const prime6[] = {1, 0, 1}; /* 65537 */
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static unsigned char const prime7[] = {1, 0, 3}; /* 65539 */
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/* A small prime: 1234567891 */
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static unsigned char const prime8[] = {0x49, 0x96, 0x02, 0xD3};
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/* A slightly larger prime: 12345678901234567891 */
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static unsigned char const prime9[] = {
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0xAB, 0x54, 0xA9, 0x8C, 0xEB, 0x1F, 0x0A, 0xD3 };
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/*
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* No, 123456789012345678901234567891 isn't prime; it's just a
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* lucky, easy-to-remember conicidence. (You have to go to
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* ...4567907 for a prime.)
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*/
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static struct {
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unsigned char const *prime;
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unsigned size;
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} const primelist[] = {
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{ prime1, sizeof(prime1) },
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{ prime2, sizeof(prime2) },
|
|
{ prime3, sizeof(prime3) },
|
|
{ prime4, sizeof(prime4) },
|
|
{ prime5, sizeof(prime5) },
|
|
{ prime6, sizeof(prime6) },
|
|
{ prime7, sizeof(prime7) },
|
|
{ prime8, sizeof(prime8) },
|
|
{ prime9, sizeof(prime9) } };
|
|
|
|
bnBegin(&t);
|
|
|
|
for (i = 0; i < sizeof(primelist)/sizeof(primelist[0]); i++) {
|
|
bnInsertBytes(&t, primelist[i].prime, 0,
|
|
primelist[i].size);
|
|
bnCopy(&e, &t);
|
|
(void)bnSubQ(&e, 1);
|
|
bnTwoExpMod(&a, &e, &t);
|
|
p = bnBits(&a);
|
|
if (p != 1) {
|
|
printf(
|
|
"Bug: Fermat(2) %u-bit output (1 expected)\n", p);
|
|
fputs("Prime = 0x", stdout);
|
|
for (j = 0; j < primelist[i].size; j++)
|
|
printf("%02X", primelist[i].prime[j]);
|
|
putchar('\n');
|
|
}
|
|
bnSetQ(&a, 3);
|
|
bnExpMod(&a, &a, &e, &t);
|
|
p = bnBits(&a);
|
|
if (p != 1) {
|
|
printf(
|
|
"Bug: Fermat(3) %u-bit output (1 expected)\n", p);
|
|
fputs("Prime = 0x", stdout);
|
|
for (j = 0; j < primelist[i].size; j++)
|
|
printf("%02X", primelist[i].prime[j]);
|
|
putchar('\n');
|
|
}
|
|
}
|
|
|
|
bnEnd(&t);
|
|
}
|
|
#endif
|
|
|
|
/* First, make sure that bn is odd. */
|
|
if ((bnLSWord(bn) & 1) == 0)
|
|
(void)bnAddQ(bn, 1);
|
|
|
|
retry:
|
|
/* Then build a sieve starting at bn. */
|
|
sieveBuild(sieve, SIEVE, bn, 2, 0);
|
|
|
|
/* Do the extra exponent sieving */
|
|
if (exponent) {
|
|
va_list ap;
|
|
unsigned t = exponent;
|
|
|
|
va_start(ap, exponent);
|
|
|
|
do {
|
|
/* The exponent had better be odd! */
|
|
assert(t & 1);
|
|
|
|
i = bnModQ(bn, t);
|
|
/* Find 1-i */
|
|
if (i == 0)
|
|
i = 1;
|
|
else if (--i)
|
|
i = t - i;
|
|
|
|
/* Divide by 2, modulo the exponent */
|
|
i = (i & 1) ? i/2 + t/2 + 1 : i/2;
|
|
|
|
/* Remove all following multiples from the sieve. */
|
|
sieveSingle(sieve, SIEVE, i, t);
|
|
|
|
/* Get the next exponent value */
|
|
t = va_arg(ap, unsigned);
|
|
} while (t);
|
|
|
|
va_end(ap);
|
|
}
|
|
|
|
/* Fill up the offsets array with the first SHUFFLE candidates */
|
|
i = p = 0;
|
|
/* Get first prime */
|
|
if (sieve[0] & 1 || (p = sieveSearch(sieve, SIEVE, p)) != 0) {
|
|
offsets[i++] = p;
|
|
p = sieveSearch(sieve, SIEVE, p);
|
|
}
|
|
/*
|
|
* Okay, from this point onwards, p is always the next entry
|
|
* from the sieve, that has not been added to the shuffle table,
|
|
* and is 0 iff the sieve has been exhausted.
|
|
*
|
|
* If we want to shuffle, then fill the shuffle table until the
|
|
* sieve is exhausted or the table is full.
|
|
*/
|
|
if (rand && p) {
|
|
do {
|
|
offsets[i++] = p;
|
|
p = sieveSearch(sieve, SIEVE, p);
|
|
} while (p && i < SHUFFLE);
|
|
}
|
|
|
|
/* Choose a random candidate for experimentation */
|
|
prev = 0;
|
|
while (i) {
|
|
/* Pick a random entry from the shuffle table */
|
|
j = rand ? rand(i) : 0;
|
|
q = offsets[j]; /* The entry to use */
|
|
|
|
/* Replace the entry with some more data, if possible */
|
|
if (p) {
|
|
offsets[j] = p;
|
|
p = sieveSearch(sieve, SIEVE, p);
|
|
} else {
|
|
offsets[j] = offsets[--i];
|
|
offsets[i] = 0;
|
|
}
|
|
|
|
/* Adjust bn to have the right value */
|
|
if ((q > prev ? bnAddMult(bn, q-prev, 2)
|
|
: bnSubMult(bn, prev-q, 2)) < 0)
|
|
goto failed;
|
|
prev = q;
|
|
|
|
/* Now do the Fermat tests */
|
|
retval = primeTest(bn, &e, &a, f, arg);
|
|
if (retval <= 0)
|
|
goto done; /* Success or error */
|
|
modexps += retval;
|
|
if (f && (retval = f(arg, '.')) < 0)
|
|
goto done;
|
|
}
|
|
|
|
/* Ran out of sieve space - increase bn and keep trying. */
|
|
if (bnAddMult(bn, SIEVE*8-prev, 2) < 0)
|
|
goto failed;
|
|
if (f && (retval = f(arg, '/')) < 0)
|
|
goto done;
|
|
goto retry;
|
|
|
|
failed:
|
|
retval = -1;
|
|
done:
|
|
bnEnd(&e);
|
|
bnEnd(&a);
|
|
lbnMemWipe(offsets, sizeof(offsets));
|
|
#ifdef MSDOS
|
|
lbnMemFree(sieve, SIEVE);
|
|
#else
|
|
lbnMemWipe(sieve, sizeof(sieve));
|
|
#endif
|
|
|
|
return retval < 0 ? retval : modexps + CONFIRMTESTS;
|
|
}
|
|
|
|
/*
|
|
* Similar, but searches forward from the given starting value in steps of
|
|
* "step" rather than 1. The step size must be even, and bn must be odd.
|
|
* Among other possibilities, this can be used to generate "strong"
|
|
* primes, where p-1 has a large prime factor.
|
|
*/
|
|
int
|
|
primeGenStrong(struct BigNum *bn, struct BigNum const *step,
|
|
int (*f)(void *arg, int c), void *arg)
|
|
{
|
|
int retval;
|
|
unsigned p, prev;
|
|
struct BigNum a, e;
|
|
int modexps = 0;
|
|
#ifdef MSDOS
|
|
unsigned char *sieve;
|
|
#else
|
|
unsigned char sieve[SIEVE];
|
|
#endif
|
|
|
|
#ifdef MSDOS
|
|
sieve = lbnMemAlloc(SIEVE);
|
|
if (!sieve)
|
|
return -1;
|
|
#endif
|
|
|
|
/* Step must be even and bn must be odd */
|
|
assert((bnLSWord(step) & 1) == 0);
|
|
assert((bnLSWord(bn) & 1) == 1);
|
|
|
|
bnBegin(&a);
|
|
bnBegin(&e);
|
|
|
|
for (;;) {
|
|
if (sieveBuildBig(sieve, SIEVE, bn, step, 0) < 0)
|
|
goto failed;
|
|
|
|
p = prev = 0;
|
|
if (sieve[0] & 1 || (p = sieveSearch(sieve, SIEVE, p)) != 0) {
|
|
do {
|
|
/*
|
|
* Adjust bn to have the right value,
|
|
* adding (p-prev) * 2*step.
|
|
*/
|
|
assert(p >= prev);
|
|
/* Compute delta into a */
|
|
if (bnMulQ(&a, step, p-prev) < 0)
|
|
goto failed;
|
|
if (bnAdd(bn, &a) < 0)
|
|
goto failed;
|
|
prev = p;
|
|
|
|
retval = primeTest(bn, &e, &a, f, arg);
|
|
if (retval <= 0)
|
|
goto done; /* Success! */
|
|
modexps += retval;
|
|
if (f && (retval = f(arg, '.')) < 0)
|
|
goto done;
|
|
|
|
/* And try again */
|
|
p = sieveSearch(sieve, SIEVE, p);
|
|
} while (p);
|
|
}
|
|
|
|
/* Ran out of sieve space - increase bn and keep trying. */
|
|
#if SIEVE*8 == 65536
|
|
/* Corner case that will never actually happen */
|
|
if (!prev) {
|
|
if (bnAdd(bn, step) < 0)
|
|
goto failed;
|
|
p = 65535;
|
|
} else {
|
|
p = (unsigned)(SIEVE*8 - prev);
|
|
}
|
|
#else
|
|
p = SIEVE*8 - prev;
|
|
#endif
|
|
if (bnMulQ(&a, step, p) < 0 || bnAdd(bn, &a) < 0)
|
|
goto failed;
|
|
if (f && (retval = f(arg, '/')) < 0)
|
|
goto done;
|
|
} /* for (;;) */
|
|
|
|
failed:
|
|
retval = -1;
|
|
|
|
done:
|
|
|
|
bnEnd(&e);
|
|
bnEnd(&a);
|
|
#ifdef MSDOS
|
|
lbnMemFree(sieve, SIEVE);
|
|
#else
|
|
lbnMemWipe(sieve, sizeof(sieve));
|
|
#endif
|
|
return retval < 0 ? retval : modexps + CONFIRMTESTS;
|
|
}
|