/* * 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 #else #define assert(x) (void)0 #endif #define BNDEBUG 1 #ifndef BNDEBUG #define BNDEBUG 0 #endif #if BNDEBUG #include #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; }