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freeswitch/libs/libzrtp/third_party/bnlib/germain.c

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/*
* Copyright (c) 1995 Colin Plumb. All rights reserved.
* For licensing and other legal details, see the file legal.c.
*
* Sophie Germain prime generation using the bignum library and sieving.
*/
#ifndef HAVE_CONFIG_H
#define HAVE_CONFIG_H 0
#endif
#if HAVE_CONFIG_H
#include "bnconfig.h"
#endif
/*
* Some compilers complain about #if FOO if FOO isn't defined,
* so do the ANSI-mandated thing explicitly...
*/
#ifndef NO_ASSERT_H
#define NO_ASSERT_H 0
#endif
#if !NO_ASSERT_H
#include <assert.h>
#else
#define assert(x) (void)0
#endif
#define BNDEBUG 1
#ifndef BNDEBUG
#define BNDEBUG 0
#endif
#if BNDEBUG
#include <stdio.h>
#endif
#include "bn.h"
#include "germain.h"
#include "jacobi.h"
#include "lbnmem.h" /* For lbnMemWipe */
#include "sieve.h"
#include "kludge.h"
/* Size of the sieve area (can be up to 65536/8 = 8192) */
#define SIEVE 8192
static unsigned const confirm[] = {2, 3, 5, 7, 11, 13, 17};
#define CONFIRMTESTS (sizeof(confirm)/sizeof(*confirm))
#if BNDEBUG
/*
* For sanity checking the sieve, we check for small divisors of the numbers
* we get back. This takes "rem", a partially reduced form of the prime,
* "div" a divisor to check for, and "order", a parameter of the "order"
* of Sophie Germain primes (0 = normal primes, 1 = Sophie Germain primes,
* 2 = 4*p+3 is also prime, etc.) and does the check. It just complains
* to stdout if the check fails.
*/
static void
germainSanity(unsigned rem, unsigned div, unsigned order)
{
unsigned mul = 1;
rem %= div;
if (!rem)
printf("bn div by %u!\n", div);
while (order--) {
rem += rem+1;
if (rem >= div)
rem -= div;
mul += mul;
if (!rem)
printf("%u*bn+%u div by %u!\n", mul, mul-1, div);
}
}
#endif /* BNDEBUG */
/*
* Helper function that does the slow primality test.
* bn is the input bignum; a, e and bn2 are temporary buffers that are
* allocated by the caller to save overhead. bn2 is filled with
* a copy of 2^order*bn+2^order-1 if bn is found to be prime.
*
* Returns 0 if both bn and bn2 are prime, >0 if not prime, and -1 on
* error (out of memory). If not prime, the return value is the number
* of modular exponentiations performed. Prints a '+' or '-' on the
* given FILE (if any) for each test that is passed by bn, and a '*'
* for each test that is passed by bn2.
*
* The testing consists of strong pseudoprimality tests, to the bases given
* in the confirm[] array above. (Also called Miller-Rabin, although that's
* not technically correct if we're using fixed bases.) Some people worry
* that this might not be enough. Number theorists may wish to generate
* primality proofs, but for random inputs, this returns non-primes with
* a probability which is quite negligible, which is good enough.
*
* It has been proved (see Carl Pomerance, "On the Distribution of
* Pseudoprimes", Math. Comp. v.37 (1981) pp. 587-593) that the number of
* pseudoprimes (composite numbers that pass a Fermat test to the base 2)
* less than x is bounded by:
* exp(ln(x)^(5/14)) <= P_2(x) ### CHECK THIS FORMULA - it looks wrong! ###
* P_2(x) <= x * exp(-1/2 * ln(x) * ln(ln(ln(x))) / ln(ln(x))).
* Thus, the local density of Pseudoprimes near x is at most
* exp(-1/2 * ln(x) * ln(ln(ln(x))) / ln(ln(x))), and at least
* exp(ln(x)^(5/14) - ln(x)). Here are some values of this function
* for various k-bit numbers x = 2^k:
* Bits Density <= Bit equivalent Density >= Bit equivalent
* 128 3.577869e-07 21.414396 4.202213e-37 120.840190
* 192 4.175629e-10 31.157288 4.936250e-56 183.724558
* 256 5.804314e-13 40.647940 4.977813e-75 246.829095
* 384 1.578039e-18 59.136573 3.938861e-113 373.400096
* 512 5.858255e-24 77.175803 2.563353e-151 500.253110
* 768 1.489276e-34 112.370944 7.872825e-228 754.422724
* 1024 6.633188e-45 146.757062 1.882404e-304 1008.953565
*
* As you can see, there's quite a bit of slop between these estimates.
* In fact, the density of pseudoprimes is conjectured to be closer to the
* square of that upper bound. E.g. the density of pseudoprimes of size
* 256 is around 3 * 10^-27. The density of primes is very high, from
* 0.005636 at 256 bits to 0.001409 at 1024 bits, i.e. more than 10^-3.
*
* For those people used to cryptographic levels of security where the
* 56 bits of DES key space is too small because it's exhaustible with
* custom hardware searching engines, note that you are not generating
* 50,000,000 primes per second on each of 56,000 custom hardware chips
* for several hours. The chances that another Dinosaur Killer asteroid
* will land today is about 10^-11 or 2^-36, so it would be better to
* spend your time worrying about *that*. Well, okay, there should be
* some derating for the chance that astronomers haven't seen it yet,
* but I think you get the idea. For a good feel about the probability
* of various events, I have heard that a good book is by E'mile Borel,
* "Les Probabilite's et la vie". (The 's are accents, not apostrophes.)
*
* For more on the subject, try "Finding Four Million Large Random Primes",
* by Ronald Rivest, in Advancess in Cryptology: Proceedings of Crypto
* '90. He used a small-divisor test, then a Fermat test to the base 2,
* and then 8 iterations of a Miller-Rabin test. About 718 million random
* 256-bit integers were generated, 43,741,404 passed the small divisor
* test, 4,058,000 passed the Fermat test, and all 4,058,000 passed all
* 8 iterations of the Miller-Rabin test, proving their primality beyond
* most reasonable doubts.
*
* If the probability of getting a pseudoprime is some small p, then the
* probability of not getting it in t trials is (1-p)^t. Remember that,
* for small p, (1-p)^(1/p) ~ 1/e, the base of natural logarithms.
* (This is more commonly expressed as e = lim_{x\to\infty} (1+1/x)^x.)
* Thus, (1-p)^t ~ e^(-p*t) = exp(-p*t). So the odds of being able to
* do this many tests without seeing a pseudoprime if you assume that
* p = 10^-6 (one in a million) is one in 57.86. If you assume that
* p = 2*10^-6, it's one in 3347.6. So it's implausible that the density
* of pseudoprimes is much more than one millionth the density of primes.
*
* He also gives a theoretical argument that the chance of finding a
* 256-bit non-prime which satisfies one Fermat test to the base 2 is
* less than 10^-22. The small divisor test improves this number, and
* if the numbers are 512 bits (as needed for a 1024-bit key) the odds
* of failure shrink to about 10^-44. Thus, he concludes, for practical
* purposes *one* Fermat test to the base 2 is sufficient.
*/
static int
germainPrimeTest(struct BigNum const *bn, struct BigNum *bn2, struct BigNum *e,
struct BigNum *a, unsigned order, int (*f)(void *arg, int c), void *arg)
{
int err;
unsigned i;
int j;
unsigned k, l, n;
#if BNDEBUG /* Debugging */
/*
* This is debugging code to test the sieving stage.
* If the sieving is wrong, it will let past numbers with
* small divisors. The prime test here will still work, and
* weed them out, but you'll be doing a lot more slow tests,
* and presumably excluding from consideration some other numbers
* which might be prime. This check just verifies that none
* of the candidates have any small divisors. If this
* code is enabled and never triggers, you can feel quite
* confident that the sieving is doing its job.
*/
i = bnLSWord(bn);
if (!(i % 2)) printf("bn div by 2!");
i = bnModQ(bn, 51051); /* 51051 = 3 * 7 * 11 * 13 * 17 */
germainSanity(i, 3, order);
germainSanity(i, 7, order);
germainSanity(i, 11, order);
germainSanity(i, 13, order);
germainSanity(i, 17, order);
i = bnModQ(bn, 63365); /* 63365 = 5 * 19 * 23 * 29 */
germainSanity(i, 5, order);
germainSanity(i, 19, order);
germainSanity(i, 23, order);
germainSanity(i, 29, order);
i = bnModQ(bn, 47027); /* 47027 = 31 * 37 * 41 */
germainSanity(i, 31, order);
germainSanity(i, 37, order);
germainSanity(i, 41, order);
#endif
/*
* First, check whether bn is prime. This uses a fast primality
* test which usually obviates the need to do one of the
* confirmation tests later. See prime.c for a full explanation.
* We check bn first because it's one bit smaller, saving one
* modular squaring, and because we might be able to save another
* when testing it. (1/4 of the time.) A small speed hack,
* but finding big Sophie Germain primes is *slow*.
*/
if (bnCopy(e, bn) < 0)
return -1;
(void)bnSubQ(e, 1);
l = bnLSWord(e);
j = 1; /* Where to start in prime array for strong prime tests */
if (l & 7) {
bnRShift(e, 1);
if (bnTwoExpMod(a, e, bn) < 0)
return -1;
if ((l & 7) == 6) {
/* bn == 7 mod 8, expect +1 */
if (bnBits(a) != 1)
return 1; /* Not prime */
k = 1;
} else {
/* bn == 3 or 5 mod 8, expect -1 == bn-1 */
if (bnAddQ(a, 1) < 0)
return -1;
if (bnCmp(a, bn) != 0)
return 1; /* Not prime */
k = 1;
if (l & 4) {
/* bn == 5 mod 8, make odd for strong tests */
bnRShift(e, 1);
k = 2;
}
}
} else {
/* bn == 1 mod 8, expect 2^((bn-1)/4) == +/-1 mod bn */
bnRShift(e, 2);
if (bnTwoExpMod(a, e, bn) < 0)
return -1;
if (bnBits(a) == 1) {
j = 0; /* Re-do strong prime test to base 2 */
} else {
if (bnAddQ(a, 1) < 0)
return -1;
if (bnCmp(a, bn) != 0)
return 1; /* Not prime */
}
k = 2 + bnMakeOdd(e);
}
/*
* It's prime! Now check higher-order forms bn2 = 2*bn+1, 4*bn+3,
* etc. Since bn2 == 3 mod 4, a strong pseudoprimality test boils
* down to looking at a^((bn2-1)/2) mod bn and seeing if it's +/-1.
* (+1 if bn2 is == 7 mod 8, -1 if it's == 3)
* Of course, that exponent is just the previous bn2 or bn...
*/
if (bnCopy(bn2, bn) < 0)
return -1;
for (n = 0; n < order; n++) {
/*
* Print a success indicator: the sign of Jacobi(2,bn2),
* which is available to us in l. bn2 = 2*bn + 1. Since bn
* is odd, bn2 must be == 3 mod 4, so the options modulo 8
* are 3 and 7. 3 if l == 1 mod 4, 7 if l == 3 mod 4.
* The sign of the Jacobi symbol is - and + for these cases,
* respectively.
*/
if (f && (err = f(arg, "-+"[(l >> 1) & 1])) < 0)
return err;
/* Exponent is previous bn2 */
if (bnCopy(e, bn2) < 0 || bnLShift(bn2, 1) < 0)
return -1;
(void)bnAddQ(bn2, 1); /* Can't overflow */
if (bnTwoExpMod(a, e, bn2) < 0)
return -1;
if (n | l) { /* Expect + */
if (bnBits(a) != 1)
return 2+n; /* Not prime */
} else {
if (bnAddQ(a, 1) < 0)
return -1;
if (bnCmp(a, bn2) != 0)
return 2+n; /* Not prime */
}
l = bnLSWord(bn2);
}
/* Final success indicator - it's in the bag. */
if (f && (err = f(arg, '*')) < 0)
return err;
/*
* Success! We have found a prime! Now go on to confirmation
* tests... k is an amount by which we know it's safe to shift
* down e. j = 1 unless the test to the base 2 could stand to be
* re-done (it wasn't *quite* a strong test), in which case it's 0.
*
* Here, we do the full strong pseudoprimality test. This proves
* that a number is composite, or says that it's probably prime.
*
* For the given base a, find bn-1 = 2^k * e, then find
* x == a^e (mod bn).
* If x == +1 -> strong pseudoprime to base a
* Otherwise, repeat k times:
* If x == -1, -> strong pseudoprime
* x = x^2 (mod bn)
* If x = +1 -> composite
* If we reach the end of the iteration and x is *not* +1, at the
* end, it is composite. But it's also composite if the result
* *is* +1. Which means that the squaring actually only has to
* proceed k-1 times. If x is not -1 by then, it's composite
* no matter what the result of the squaring is.
*
* For the multiples 2*bn+1, 4*bn+3, etc. then k = 1 (and e is
* the previous multiple of bn) so the squaring loop is never
* actually executed at all.
*/
for (i = j; i < CONFIRMTESTS; i++) {
if (bnCopy(e, bn) < 0)
return -1;
bnRShift(e, k);
k += bnMakeOdd(e);
(void)bnSetQ(a, confirm[i]);
if (bnExpMod(a, a, e, bn) < 0)
return -1;
if (bnBits(a) != 1) {
l = k;
for (;;) {
if (bnAddQ(a, 1) < 0)
return -1;
if (bnCmp(a, bn) == 0) /* Was result bn-1? */
break; /* Prime */
if (!--l)
return (1+order)*i+2; /* Fail */
/* This part is executed once, on average. */
(void)bnSubQ(a, 1); /* Restore a */
if (bnSquare(a, a) < 0 || bnMod(a, a, bn) < 0)
return -1;
if (bnBits(a) == 1)
return (1+order)*i+1; /* Fail */
}
}
if (bnCopy(bn2, bn) < 0)
return -1;
/* Only do the following if we're not re-doing base 2 */
if (i) for (n = 0; n < order; n++) {
if (bnCopy(e, bn2) < 0 || bnLShift(bn2, 1) < 0)
return -1;
(void)bnAddQ(bn2, 1);
/* Print success indicator for previous test */
j = bnJacobiQ(confirm[i], bn2);
if (f && (err = f(arg, j < 0 ? '-' : '+')) < 0)
return err;
/* Check that p^e == Jacobi(p,bn2) (mod bn2) */
(void)bnSetQ(a, confirm[i]);
if (bnExpMod(a, a, e, bn2) < 0)
return -1;
/*
* FIXME: Actually, we don't need to compute the
* Jacobi symbol externally... it never happens that
* a = +/-1 but it's the wrong one. So we can just
* look at a and use its sign. Find a proof somewhere.
*/
if (j < 0) {
/* Not a Q.R., should have a = bn2-1 */
if (bnAddQ(a, 1) < 0)
return -1;
if (bnCmp(a, bn2) != 0) /* Was result bn2-1? */
return (1+order)*i+n+2; /* Fail */
} else {
/* Quadratic residue, should have a = 1 */
if (bnBits(a) != 1)
return (1+order)*i+n+2; /* Fail */
}
}
/* Final success indicator for the base confirm[i]. */
if (f && (err = f(arg, '*')) < 0)
return err;
}
return 0; /* Prime! */
}
/*
* Add x*y to bn, which is usually (but not always) < 65536.
* Do it in a simple linear manner.
*/
static int
bnAddMult(struct BigNum *bn, unsigned long x, unsigned y)
{
unsigned long z = (unsigned long)x * y;
while (z > 65535) {
if (bnAddQ(bn, 65535) < 0)
return -1;
z -= 65535;
}
return bnAddQ(bn, (unsigned)z);
}
/*
* Modifies the bignum to return the next Sophie Germain prime >= the
* input value. Sohpie Germain primes are number such that p is
* prime and 2*p+1 is also prime.
*
* This is actually parameterized: it generates primes p such that "order"
* multiples-plus-two are also prime, 2*p+1, 2*(2*p+1)+1 = 4*p+3, etc.
*
* Returns >=0 on success or -1 on failure (out of memory). On success,
* the return value is the number of modular exponentiations performed
* (excluding the final confirmations). This never gives up searching.
*
* The FILE *f argument, if non-NULL, has progress indicators written
* to it. A dot (.) is written every time a primeality test is failed,
* a plus (+) or minus (-) when the smaller prime of the pair passes a
* test, and a star (*) when the larger one does. Finally, a slash (/)
* is printed when the sieve was emptied without finding a prime and is
* being refilled.
*
* Apologies to structured programmers for all the GOTOs.
*/
int
germainPrimeGen(struct BigNum *bn, unsigned order,
int (*f)(void *arg, int c), void *arg)
{
int retval;
unsigned p, prev;
unsigned inc;
struct BigNum a, e, bn2;
int modexps = 0;
#ifdef MSDOS
unsigned char *sieve;
#else
unsigned char sieve[SIEVE];
#endif
#ifdef MSDOS
sieve = lbnMemAlloc(SIEVE);
if (!sieve)
return -1;
#endif
bnBegin(&a);
bnBegin(&e);
bnBegin(&bn2);
/*
* Obviously, the prime we find must be odd. Further, if 2*p+1
* is also to be prime (order > 0) then p != 1 (mod 3), lest
* 2*p+1 == 3 (mod 3). Added to p != 3 (mod 3), p == 2 (mod 3)
* and p == 5 (mod 6).
* If order > 2 and we care about 4*p+3 and 8*p+7, then similarly
* p == 4 (mod 5), so p == 29 (mod 30).
* So pick the step size for searching based on the order
* and increse bn until it's == -1 (mod inc).
*
* mod 7 doesn't have a unique value for p because 2 -> 5 -> 4 -> 2,
* nor does mod 11, and I don't want to think about things past
* that. The required order would be impractically high, in any case.
*/
inc = order ? ((order > 2) ? 30 : 6) : 2;
if (bnAddQ(bn, inc-1 - bnModQ(bn, inc)) < 0)
goto failed;
for (;;) {
if (sieveBuild(sieve, SIEVE, bn, inc, order) < 0)
goto failed;
p = prev = 0;
if (sieve[0] & 1 || (p = sieveSearch(sieve, SIEVE, p)) != 0) {
do {
/* Adjust bn to have the right value. */
assert(p >= prev);
if (bnAddMult(bn, p-prev, inc) < 0)
goto failed;
prev = p;
/* Okay, do the strong tests. */
retval = germainPrimeTest(bn, &bn2, &e, &a,
order, f, arg);
if (retval <= 0)
goto done;
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 (bnAddMult(bn, (unsigned long)SIEVE*8-prev, inc) < 0)
goto failed;
if (f && (retval = f(arg, '/')) < 0)
goto done;
} /* for (;;) */
failed:
retval = -1;
done:
bnEnd(&bn2);
bnEnd(&e);
bnEnd(&a);
#ifdef MSDOS
lbnMemFree(sieve, SIEVE);
#else
lbnMemWipe(sieve, sizeof(sieve));
#endif
return retval < 0 ? retval : modexps+(order+1)*CONFIRMTESTS;
}
int
germainPrimeGenStrong(struct BigNum *bn, struct BigNum const *step,
unsigned order, int (*f)(void *arg, int c), void *arg)
{
int retval;
unsigned p, prev;
struct BigNum a, e, bn2;
int modexps = 0;
#ifdef MSDOS
unsigned char *sieve;
#else
unsigned char sieve[SIEVE];
#endif
#ifdef MSDOS
sieve = lbnMemAlloc(SIEVE);
if (!sieve)
return -1;
#endif
bnBegin(&a);
bnBegin(&e);
bnBegin(&bn2);
for (;;) {
if (sieveBuildBig(sieve, SIEVE, bn, step, order) < 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;
/* Okay, do the strong tests. */
retval = germainPrimeTest(bn, &bn2, &e, &a,
order, f, arg);
if (retval <= 0)
goto done;
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(&bn2);
bnEnd(&e);
bnEnd(&a);
#ifdef MSDOS
lbnMemFree(sieve, SIEVE);
#else
lbnMemWipe(sieve, sizeof(sieve));
#endif
return retval < 0 ? retval : modexps+(order+1)*CONFIRMTESTS;
}
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